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# What are the odds of rolling 2 three of a kind using 6 dice in one roll?
Updated: 10/17/2022
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Q: What are the odds of rolling 2 three of a kind using 6 dice in one roll?
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### What is the probability of rolling a 2 on the roll on a die?
If it's a six-sided die (like the kind used in craps and most dice games) the odds are 1 in 6.The probability of rolling a single number on any kind of die* is 1 ÷ (the number of sides).*Except, of course, for weighted, shaved, or otherwise illegal dice.
### What are the odds of rolling 5 of akind with 5 dice?
The odds of rolling five of a kind with five dice is 1 in 1296. The first die is going to roll something, so the odds are 1 in 1. Each of the other dice have a probability of 1 in 6 of matching the first die, so the resultant probability is (1 in 1) times (1 in 6) to the fourth power.
### What is the probability that if you roll the dice you'll get the number 100?
That depends on what kind of dice you are rolling and how many of them you roll. If you roll two 6-sided dice once, the probability of getting the number 100 is exactly zero. You cannot get a 100 on one roll of two 6-sided dice. Other dice and different numbers of them may yield different probabilities.
### What are the odds of rolling four of a kind with six dice in two rolls?
A single roll of six dice comes up with a four of a kind 1 in every 20.736 times. This chance doesn't change on the second throw, only really granting a total of 2 in 20.736.
Yahtzee
### What are the odds of rolling five of any kind with 5 dice?
accumulated binominal distribution 6*f(k;n,p) = 6*f(5;5,1/6)
### If five dice are rolled simultaneously what is the probability of getting 5 of a kind in a single roll?
If we are thinking of getting a '6', here are the odds. Wth one dice, its 1 in 6. So,with two dice its 1 in 216 with three dice its 1 in 7776 with four dice its 1 in 279936 with five dice its a huge 1 in 10077696
### How do you compare and match values of 5 dice for a dice poker game using qbasic?
To compare and match values of 5 dice for a dice poker game using any generalized programming language, consider this possible algorithm... Create an array of five for each throw. Generate the throw using some kind of random number generator. Sort the array. Create another array of six, counting the number of duplicates. Sort that array by count. Compute the score for the throw, considering the ranking appropriate for the game. If you were to match the rules of poker, you would consider the ranking, in increasing value, to be high die, one pair, two pair, three of a kind, straight, full house, four of a kind, and five of a kind. Note that the flush and straight flush are not considered, because dice do not have the concept of suit. Repeat for each throw. Depending on the varient of poker you are trying to implement, add whatever logic is necessary, say to draw and replace one or more dice for draw poker. Add what ever betting and scoring system you want. Note that comparing scores of equal throw, such as two throws getting three of a kind, would require breaking the tie by considering the high die (or set) between the throws.
### What are the odds of rolling 4 of a kind with 5 dice?
It would be 1/6 with 1 dice. To get 5 out of FIVE it would be 1/6 X 1/6 x 1/6 X 1/6 X 1/6 = 1/7776 odds. 4/5 would be 1/216
### What is the probability of getting 4 of a kind if 5 identical dice are tossed simulataneously?
If they are fair dice, the probability is 0.0032If they are fair dice, the probability is 0.0032If they are fair dice, the probability is 0.0032If they are fair dice, the probability is 0.0032
### Is a Yahtzee a full house?
In the game of Yahtzee a full house is worth 25 points. A full house consists of 2 dice with the same number and three dice with the same numbers. | 1,037 | 3,955 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2024-33 | latest | en | 0.931688 |
http://mathhelpforum.com/pre-calculus/103763-comparing-graphs-functions.html | 1,553,653,074,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912207618.95/warc/CC-MAIN-20190327020750-20190327042750-00260.warc.gz | 126,201,498 | 12,670 | # Thread: comparing graphs of functions
1. ## comparing graphs of functions
Compare the functions f(x) = x7 and g(x) = 7x by graphing both functions in several viewing rectangles. (a) Find all points of intersection of the graphs correct to one decimal place.
( 1, 2) (smaller x value)
( 3, 4) (larger x value)
I cant seem to find the points when i put it in my calculator. any help?
2. Originally Posted by kyleu03
Compare the functions f(x) = x7 and g(x) = 7x by graphing both functions in several viewing rectangles. (a) Find all points of intersection of the graphs correct to one decimal place.
( 1, 2) (smaller x value)
( 3, 4) (larger x value)
I cant seem to find the points when i put it in my calculator. any help?
what is the significance of the red 7 ?
is it an exponent in f(x) ?
is
$\displaystyle g(x) = 7x$ or $\displaystyle 7^x$ ?
use a caret (^) symbol to denote exponents ... i.e. x^7 is x to the 7th power.
3. it is just the number, it is red because its changes. idk how to do it
4. Originally Posted by kyleu03
it is just the number, it is red because its changes. idk how to do it
This does not answer the questions raised and will just delay you getting a helpful reply. Re-post the equations. Use the conventional formatting for exponents (that is, powers).
Note: x^7 means $\displaystyle x^7$, 7^x means $\displaystyle 7^x$,
5. Originally Posted by kyleu03
it is just the number, it is red because its changes. idk how to do it
x7 makes no sense. what is f(x) ?
6. ohh im very sorry i didnt see it posted wrong. it is
f(x) = x^7
g(x) = 7^x
7. Originally Posted by kyleu03
ohh im very sorry i didnt see it posted wrong. it is
f(x) = x^7
g(x) = 7^x
one solution should be obvious ...
graph $\displaystyle y = x^7 - 7^x$ and look for the the other zero ... easier to find.
8. "graph and look for the the other zero " does not find the points of intersection for me
9. Originally Posted by kyleu03
"graph and look for the the other zero " does not find the points of intersection for me
An obvious solution is a whole number that lies somewhere between 5 and 8 ....
The other solution solution cannot be found exactly. A decimal approximation lies somewhere between 1 and 2. You need to use your graph (or graphs) to estimate the value correct to 1 decimal place.
As to why you can't find the intersection points when you put it into your calculator, you should check:
1. that you've entered the correct equations.
2. you're using an appropriate window (check that ymin and ymax of the window are OK).
10. I was able to find the smaller, but not larger. any help?
Compare the functions f(x) = x7 and g(x) = 7x by graphing both functions in several viewing rectangles.
(a) Find all points of intersection of the graphs correct to one decimal place.
( 1, 2) (smaller x value)
( 3, 4) (larger x value)
(b) Which function grows more rapidly when x is large? 5
f(x)
g(x)
11. Originally Posted by mr fantastic
An obvious solution is a whole number that lies somewhere between 5 and 8 ....
The other solution solution cannot be found exactly. A decimal approximation lies somewhere between 1 and 2. You need to use your graph (or graphs) to estimate the value correct to 1 decimal place.
[snip]
Originally Posted by kyleu03
I was able to find the smaller, but not larger. any help?
[snip]
There aren't many whole numbers between 5 and 7 that require testing ....
12. are you saying (0,7)
13. Originally Posted by kyleu03
are you saying (0,7)
I am saying that you should check what happens when x = 6 and x = 7.
14. thats wrong blong
15. Originally Posted by kyleu03
thats wrong blong
For crying out loud! $\displaystyle f(x) = x^7$ and $\displaystyle g(x) = 7^x$ have an intersection point at x = 7. The value of each function at x = 7 is 7^7. Given the previous replies, this should not have needed spelling out. | 1,059 | 3,859 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2019-13 | latest | en | 0.918058 |
https://www.esaral.com/q/if-minimum-possible-work-is-done-by-a-refrigerator-in-converting-100-grams-57887 | 1,720,945,478,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514551.8/warc/CC-MAIN-20240714063458-20240714093458-00669.warc.gz | 642,901,826 | 11,570 | # If minimum possible work is done by a refrigerator in converting 100 grams
Question:
If minimum possible work is done by a refrigerator in converting 100 grams of water at $0^{\circ} \mathrm{C}$ to ice, how much heat (in calories) is released to the surroundings at temperature $27^{\circ} \mathrm{C}$ (Latent heat of ice $=80 \mathrm{Cal} /$ gram ) to the nearest integer?
Solution:
(8791)
Given,
Heat absorbed, $Q_{2}=m L=80 \times 100=8000 \mathrm{Cal}$
Temperature of ice, $T_{2}=273 \mathrm{~K}$
Temperature of surrounding,
$T_{1}=273+27=300 \mathrm{~K}$
Efficiency $=\frac{w}{Q_{2}}=\frac{Q_{1}-Q_{2}}{Q_{2}}=\frac{T_{1}-T_{2}}{T_{2}}=\frac{300-273}{273}$
$\Rightarrow \frac{Q_{1}-8000}{8000}=\frac{27}{273} \Rightarrow Q_{1}=8791 \mathrm{Cal}$ | 257 | 763 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2024-30 | latest | en | 0.690896 |
http://mathhelpboards.com/geometry-11/circle-theorems-20206.html?s=937dcc8a83392d4f10f5259d4891b7f7 | 1,490,485,535,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189088.29/warc/CC-MAIN-20170322212949-00632-ip-10-233-31-227.ec2.internal.warc.gz | 236,081,903 | 20,227 | Thanks: 0
1. Hi,
could someone please tell me what theorem I need to be looking at to work out the angle at $\gamma$ please? I've worked out the rest but can't find a theorem for this one.
Thanks
2. Use the vertical angles theorem.
Originally Posted by Euge
Use the vertical angles theorem.
Thanks but that just gives me the angle between the two straight lines. My impression was that $\gamma$ was the part of the angle up to the edge of the circle.
4. That would not make an angle, for an angle is formed by two straight lines. It would only make sense for $gamma$ to be the angle between those two lines.
That would not make an angle, for an angle is formed by two straight lines. It would only make sense for $gamma$ to be the angle between those two lines. | 180 | 769 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2017-13 | longest | en | 0.977289 |
https://zims-en.kiwix.campusafrica.gos.orange.com/wikipedia_en_all_nopic/A/Solar_azimuth_angle | 1,603,632,955,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107889173.38/warc/CC-MAIN-20201025125131-20201025155131-00171.warc.gz | 971,840,438 | 7,726 | # Solar azimuth angle
The solar azimuth angle is the azimuth angle of the Sun's position.[1][2][3] This horizontal coordinate defines the Sun's relative direction along the local horizon, whereas the solar zenith angle (or its complementary angle solar elevation) defines the Sun's apparent altitude.
There are several conventions for the solar azimuth; however, it is traditionally defined as the angle between a line due south and the shadow cast by a vertical rod on Earth. This convention states the angle is positive if the line is east of south and negative if it is west of south.[1][2] For example, due east would be 90° and due west would be -90°. Another convention is the reverse; it also has the origin at due south, but measures angles clockwise, so that due east is now negative and west now positive.[3]
However, despite tradition, the most commonly accepted convention for analyzing solar irradiation, e.g. for solar energy applications, is clockwise from due north, so east is 90°, south is 180°, and west is 270°. This is the definition used by NREL in their solar position calculators[4] and is also the convention used in the formulas presented here. However, Landsat photos and other USGS products, while also defining azimuthal angles relative to due north, take counterclockwise angles as negative.[5]
## Formulas
The following formulas assume the north-clockwise convention. The solar azimuth angle can be calculated to a good approximation with the following formula, however angles should be interpreted with care because the inverse sine, i.e. x = sin−1 y or x = arcsin y, has multiple solutions, only one of which will be correct.
${\displaystyle \sin \phi _{\mathrm {s} }={\frac {-\sin h\cos \delta }{\sin \theta _{\mathrm {s} }}}.}$
The following formulas can also be used to approximate the solar azimuth angle, but these formulas use cosine, so the azimuth angle as shown by a calculator will always be positive, and should be interpreted as the angle between zero and 180 degrees when the hour angle, h, is negative (morning) and the angle between 180 and 360 degrees when the hour angle, h, is positive (afternoon). (These two formulas are equivalent if one assumes the "solar elevation angle" approximation formula).[2][3][4]
{\displaystyle {\begin{aligned}\cos \phi _{\mathrm {s} }&={\frac {\sin \delta \cos \Phi -\cos h\cos \delta \sin \Phi }{\sin \theta _{\mathrm {s} }}}\\[5pt]\cos \phi _{\mathrm {s} }&={\frac {\sin \delta -\cos \theta _{\mathrm {s} }\sin \Phi }{\sin \theta _{\mathrm {s} }\cos \Phi }}.\end{aligned}}}
So practically speaking, the compass azimuth which is the practical value used everywhere (in example in airlines as the so called course) on a compass (where North is 0 degrees, East is 90 degrees, South is 180 degrees and West is 270 degrees) can be calculated as
${\displaystyle {\text{compass }}\phi _{\mathrm {s} }=360-\phi _{\mathrm {s} }.}$
The formulas use the following terminology:
• ${\displaystyle \phi _{\mathrm {s} }}$ is the solar azimuth angle
• ${\displaystyle \theta _{\mathrm {s} }}$ is the solar zenith angle
• ${\displaystyle h}$ is the hour angle, in the local solar time
• ${\displaystyle \delta }$ is the current sun declination
• ${\displaystyle \Phi }$ is the local latitude
## References
1. Sukhatme, S. P. (2008). Solar Energy: Principles of Thermal Collection and Storage (3rd ed.). Tata McGraw-Hill Education. p. 84. ISBN 978-0070260641.
2. Seinfeld, John H.; Pandis, Spyros N. (2006). Atmospheric Chemistry and Physics, from Air Pollution to Climate Change (2nd ed.). Wiley. p. 130. ISBN 978-0-471-72018-8.
3. Duffie, John A.; Beckman, William A. (2013). Solar Engineering of Thermal Processes (4th ed.). Wiley. pp. 13, 15, 20. ISBN 978-0-470-87366-3.
4. Reda, I., Andreas, A. (2004). "Solar Position Algorithm for Solar Radiation Applications". Solar Energy. 76 (5): 577–89. Bibcode:2004SoEn...76..577R. doi:10.1016/j.solener.2003.12.003. ISSN 0038-092X.
5. "Sun Azimuth". Landsat Data Dictionary. USGS.
This article is issued from Wikipedia. The text is licensed under Creative Commons - Attribution - Sharealike. Additional terms may apply for the media files. | 1,088 | 4,167 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 8, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2020-45 | latest | en | 0.924656 |
https://nebusresearch.wordpress.com/tag/nasa/ | 1,643,110,973,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304810.95/warc/CC-MAIN-20220125100035-20220125130035-00348.warc.gz | 456,725,103 | 30,677 | ## Some More Stuff To Read
I’ve actually got enough comics for yet another Reading The Comics post. But rather than overload my Recent Posts display with those I’ll share some pointers to other stuff I think worth looking at.
So remember how the other day I said polynomials were everything? And I tried to give some examples of things you might not expect had polynomials tied to them? Here’s one I forgot. Howard Phillips, of the HowardAt58 blog, wrote recently about discrete signal processing, the struggle to separate real patterns from random noise. It’s a hard problem. If you do very little filtering, then meaningless flutterings can look like growing trends. If you do a lot of filtering, then you miss rare yet significant events and you take a long time to detect changes. Either can be mistakes. The study of a filter’s characteristics … well, you’ll see polynomials. A lot.
For something else to read, and one that doesn’t get into polynomials, here’s a post from Stephen Cavadino of the CavMaths blog, abut the areas of lunes. Lunes are … well, they’re kind of moon-shaped figures. Cavadino particularly writes about the Theorem of Not That Hippocrates. Start with a half circle. Draw a symmetric right triangle inside the circle. Draw half-circles off the two equal legs of that right triangle. The area between the original half-circle and the newly-drawn half circles is … how much? The answer may surprise you.
Cavadino doesn’t get into this, but: it’s possible to make a square that has the same area as these strange crescent shapes using only straightedge and compass. Not That Hippocrates knew this. It’s impossible to make a square with the exact same area as a circle using only straightedge and compass. But these figures, with edges that are defined by circles of just the right relative shapes, they’re fine. Isn’t that wondrous?
And this isn’t mathematics but what the heck. Have you been worried about the Chandler Wobble? Apparently there’s been a bit of a breakthrough in understanding it. Turns out water melting can change the Earth’s rotation enough to be noticed. And to have been noticed since the 1890s.
## Reading the Comics, April 10, 2015: Getting Into The Story Problem Edition
I know it’s been like forever, or four days, since the last time I had a half-dozen or so mathematically themed comic strips to write about, but if Comic Strip Master Command is going to order cartoonists to give me stuff to write about I’m not going to turn them away. Several seemed to me about the struggle to get someone to buy into a story — the thing being asked after in a word problem, perhaps, or about the ways mathematics is worth knowing, or just how the mathematics in a joke’s setup are presented — and how skepticism about these things can turn up. So I’ll declare that the theme of this collection.
Steve Sicula’s Home And Away started a sequence on April 7th about “is math really important?”, with the father trying to argue that it’s so very useful. I’m not sure anyone’s ever really been convinced by the argument that “this is useful, therefore it’s important, therefore it’s interesting”. Lots of things are useful or important while staying fantastically dull to all but a select few souls. I would like to think a better argument for learning mathematics is that it’s beautiful, and astounding, and it allows you to discover new ways of studying the world; it can offer all the joy of any art, even as it has a practical side. Anyway, the sequence goes on for several days, and while I can’t say the arguments get very convincing on any side, they do allow for a little play with the fourth wall that I usually find amusing in comics which don’t do that much.
## Monday, June 4, 1962 – Gemini Mission Begins In Simulation
At the Brooks Air Force Base in Texas two men have begun a simulation of a long-duration Gemini Mission. This program, run by the Air Force School of Aviation Medicine, will have them live for fourteen days in an atmosphere simulating that proposed for the Gemini spacecraft. This will be a 100 percent oxygen atmosphere maintained at five pounds per square inch of pressure.
## Wednesday, April 18, 1962 – Astronaut Applications Open
NASA is accepting applications for additional astronauts and will be doing so through June 1, 1962. The plan is to select between five and ten new astronauts to augment the existing corps of seven. The new astronauts will support Project Mercury operations, and go on to join the Mercury astronauts in piloting the Gemini spacecraft.
Continue reading “Wednesday, April 18, 1962 – Astronaut Applications Open” | 991 | 4,621 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2022-05 | latest | en | 0.918911 |
http://www.dummies.com/how-to/content/how-to-determine-the-dimensions-for-the-least-expe.navId-403863.html | 1,469,952,247,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469258951764.94/warc/CC-MAIN-20160723072911-00110-ip-10-185-27-174.ec2.internal.warc.gz | 429,923,369 | 14,997 | You can use calculus to solve practical problems, such as determining the correct size for a home-improvement project. Here’s an example. A Norman window has the shape of a semicircle above a rectangle. If the straight edges of the frame cost \$20 per linear foot and the circular frame costs \$25 per linear foot and you want a window with an area of 20 square feet, what dimensions will minimize the cost of the frame?
1. Draw a diagram and label the dimensions with variables, as shown in the following figure.
2. 2Express the thing you want to minimize, the cost.
Cost = (length of curved frame) · (cost per linear foot) + (length of straight frame) · (cost per linear foot)
= (ðx)(25) + (2x + 2y)(20)
= 25ðx + 40x + 40y
3. Relate the two variables to each other.
Total Area = Area of Semicircle + Area of Rectangle
4. Solve for y and substitute.
5. Find the domain.
x > 0 is obvious. And when x gets large enough, the entire window of 20 square feet in area will be one big semicircle, so
Thus, x must be less than or equal to 3.57.
6. Find the critical numbers of C(x).
Omit –2.143 because it’s outside the domain. So 2.143 is the only critical number.
7. Evaluate the cost at the critical number and at the endpoints.
So, the least expensive frame for a 20-square-foot window will cost about \$373 and will be 2 times 2.143, or about 4.286 feet or 4’3” wide at the base. Because
the height of the rectangular lower part of the window will be 2.98, or about 3’ tall. The total height will thus be 2.98 plus 2.14, or about 5’1”. | 417 | 1,549 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2016-30 | latest | en | 0.840511 |
http://asphodel.mashvipsmash.xyz/ | 1,611,306,215,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703529179.46/warc/CC-MAIN-20210122082356-20210122112356-00005.warc.gz | 7,311,623 | 6,981 | Texas Hold’em Poker Odds (over 100 Poker Probabilities).
Texas Hold'em Odds. The following Texas Hold em odds table highlights some common probabilities that you may encounter in Hold'em. It is not vital that you learn these probabilities, but it is useful to be aware of the chances of certain situations arising. Holdem Tools Holdem Tools is a web based Texas Holdem odds calculator.
I have reason to believe my opponent has two pair, and I texas holdem how to calculate odds have AA, with four to a flush, my outs are any ace (giving me a set) plus 9 flush cards (giving me a flush), totalling 11 outs.A, K, Q, J, 10, all the same suit.Texas Holdem poker odds chart for after flop outs showing percentages-for and odds-against.Two cards shall be dealt down to each player.
Pot Odds Cheat Sheet. Finally, a favorite method is to use a good cheat sheet. Obviously, carrying a cheat sheet to a brick and mortar casino will practically scream “Shark bait!“ However, if you are playing online, it is a great option. The cheat sheet below shows odds against making your hand with both 2 cards to come as well as 1 card.
I have created an odds calculator and now I want to check if the calculator gives me the correct results. In my program I use Monte Carlo to calculate the odds. The results from the calculator should be compared with the results I am getting by doing the math.
Pot odds in Texas Holdem - YouTube.
Calculating hand odds are your chances of making a hand in Texas Hold'em poker. For example: To calculate your hand odds in a Texas Hold'em game when you hold two hearts and there are two hearts on the flop, your hand odds for making a flush are about 2 to 1.
How To Work Out Hand Probability In Texas Holdem. Ever wondered where some of those odds in the odds charts came from? In this article, I will teach you how to work out the probability of being dealt different types of preflop hands in Texas Holdem. It's all pretty simple and you don't need to be a mathematician to work out the probabilities.
Texas Hold'em Odds Calculator with support to Ranges.
My advice is to not be exact when calculating poker odds. Poker calculations can be taken way too far, and a bit of guesswork never hurt anybody. For example, if I am doing the math on the improvement potential of a hand and the result comes to 23 in 100, it is best to round it to 25 in 100 or 1 in 4, because those are numbers that are easy to deal with.
I need a excel based program that can do the odd calculations for Texas Holdem. It should be able to do from 2 - 10 players preflop odds, after the flop, after the turn, after the river etc. Its pretty simple there are plenty of programs out there that do it but i need it in excel and i need to be able to view the formulas.
Texas Hold'em Tools. Here are a selection of poker tools and stuff to help you improve your game. These tools are aimed mostly toward no limit Texas Hold'em players. If you have any problems or questions about these Texas Hold em tools, please let me know. Hope you enjoy them! Odds charts.
GambleAware offer players and Texas Holdem Odds Calculator Free their families advice and guidance on gambling. They offer information and advice to encourage responsible gambling, both to players and casino operators, and give help tothose whomight have a gambling problem. 50. 500%-Most Lines. | 748 | 3,354 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2021-04 | longest | en | 0.928857 |
http://icpc.njust.edu.cn/Problem/CF/528D/ | 1,603,165,008,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107869785.9/warc/CC-MAIN-20201020021700-20201020051700-00194.warc.gz | 56,681,246 | 10,488 | # Fuzzy Search
Time Limit: 3 seconds
Memory Limit: 256 megabytes
## Description
Leonid works for a small and promising start-up that works on decoding the human genome. His duties include solving complex problems of finding certain patterns in long strings consisting of letters 'A', 'T', 'G' and 'C'.
Let's consider the following scenario. There is a fragment of a human DNA chain, recorded as a string S. To analyze the fragment, you need to find all occurrences of string T in a string S. However, the matter is complicated by the fact that the original chain fragment could contain minor mutations, which, however, complicate the task of finding a fragment. Leonid proposed the following approach to solve this problem.
Let's write down integer k ≥ 0 — the error threshold. We will say that string T occurs in string S on position i (1 ≤ i ≤ |S| - |T| + 1), if after putting string T along with this position, each character of string T corresponds to the some character of the same value in string S at the distance of at most k. More formally, for any j (1 ≤ j ≤ |T|) there must exist such p (1 ≤ p ≤ |S|), that |(i + j - 1) - p| ≤ k and S[p] = T[j].
For example, corresponding to the given definition, string "ACAT" occurs in string "AGCAATTCAT" in positions 2, 3 and 6.
Note that at k = 0 the given definition transforms to a simple definition of the occurrence of a string in a string.
Help Leonid by calculating in how many positions the given string T occurs in the given string S with the given error threshold.
## Input
The first line contains three integers |S|, |T|, k (1 ≤ |T| ≤ |S| ≤ 200 000, 0 ≤ k ≤ 200 000) — the lengths of strings S and T and the error threshold.
The second line contains string S.
The third line contains string T.
Both strings consist only of uppercase letters 'A', 'T', 'G' and 'C'.
## Output
Print a single number — the number of occurrences of T in S with the error threshold k by the given definition.
## Sample Input
Input10 4 1AGCAATTCATACATOutput3
## Sample Output
None
## Hint
If you happen to know about the structure of the human genome a little more than the author of the problem, and you are not impressed with Leonid's original approach, do not take everything described above seriously.
None | 596 | 2,269 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2020-45 | latest | en | 0.828356 |
https://www.usna.edu/Users/oceano/pguth/md_help/so503/lab_1.htm | 1,539,946,527,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512395.23/warc/CC-MAIN-20181019103957-20181019125457-00372.warc.gz | 1,133,502,886 | 2,999 | ## Lab 1: Excel Programming
We will do the same analysis of a data set with three programs:
1. Excel. While not free, Excel in on almost every computer you will use (and there is a free clone in OpenOffice). Thus it is likely to be of use to you in both your professional and personal life (I use Excel to track my budget and estimate taxes. As long as the data set is not too large or complex (our last lab in the course will look at CTD data, which will fail both these criteria), you can do a lot of data analysis in Excel. In addition, by contrasting the steps needed in Excel and Matlab, we will get a better appreciation conceptually for what we must do.
2. Matlab. This is a programming language, with some specialization for matrices, but it works very much like other languages like C/C++, Fortran, Basic, JavaScript, or Python. Matlab is a favorite for science and engineering, but if you understand what you are trying to do, as opposed to just memorizing the code, you could easily translate into another language.
3. A GIS program, which is specialized to deal with this particular kind of data. The program is written in a language similar to Matlab (conceptually Matlab can create compiled programs, but actually taking that step is complicated and beyond what we will do in this course), and allows the use of a GUI to open data sets and manipulate them.
## Graphs
• Graphs should always have labeled axes
• Graphs should only have a legend when there are more than two things on the graph. Otherwise the figure caption suffices.
• Graphs for technical writing should not have a title. The figure caption suffices for technical writing.
• Insure the text on legends and axes is readable. It almost always needs to be make bigger.
• The caption goes below.
## Tables
• You will almost always have to set the number of decimals. By default Excel will put in way too many (which makes it hard to understand the data, and makes a mockery of significant figures), and the decimal points will not line up at they should.
• You need units for each column.
• Need "reader friendly" titles on the columns, not computer geek-speak
• The caption goes above.
# Excel Graph into Word
If you copy a graph from Excel onto the clipboard, and paste it into Word, you will have four choices. You want the one on the far right, to paste it as a picture. Choosing the others gives you an object, which has several disadvantages: The file size can become unreasonably large. Response in Word can become painfully slow. The appearance of the figure could change, and not reflect what you want. If you copy the graph into Paint, and then copy it to Word, you have only one choice, which might look like text but will be a bitmap. While you are in Paint you can make any quick graphical edits (for example combining several graphics into a single one, and putting A, B, C for the figure caption.
last revision 1/11/2017 | 648 | 2,934 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2018-43 | latest | en | 0.933419 |
https://physics.stackexchange.com/questions/382402/how-to-obtain-hamiltonian-formalism-and-phase-space-for-lagrangian-with-second-d?noredirect=1 | 1,628,089,473,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154878.27/warc/CC-MAIN-20210804142918-20210804172918-00417.warc.gz | 460,559,971 | 37,942 | # How to obtain Hamiltonian formalism and phase space for Lagrangian with second-derivatives?
This is a special case of this question of mine, which, I think, might have drawn little attention because it was too general. In this question, I would like to consider a specific case.
Take a classical system given by the action $$S[x] = \int dt \left[ \frac{\alpha}{2} \ddot{x}^2 + \frac{\beta}{2} \dot{x}^2 + \frac{\gamma}{2} x^2 \right].$$
This action is Lagrangian, but it is not what we are usually dealing with in physics, because the Lagrangian contains second time derivatives.
The equations of motion are:
$$\alpha \ddddot{x} - \beta \ddot{x} + \gamma x = 0.$$
This can be solved as usual by employing Fourier transform:
$$x(t) = a_1 e^{i \omega_1 t} + a_1^{*} e^{-i \omega_1 t} + a_2 e^{i \omega_2 t} + a_2^{*} e^{-i \omega_2 t},$$
where $\omega_1^2$ and $\omega_2^2$ are the two roots of $\alpha \omega^4 + \beta \omega^2 + \gamma$, given by
$$\omega_{1,2}^2 = \frac{- \beta \pm \sqrt{\beta^2 - 4 \alpha \gamma}}{2 \alpha} .$$
Now by definition the phase space is the space of solutions of equations of motion. In this case it is parametrized by two complex co-ordinates $a_{1,2}$, meaning that it is 4-dimensional.
I would like to know if there's a natural way to associate the symplectic structure (Poisson bracket) on this phase space. I.e.,
$$\{a_1, a_1^{*}\} = ?$$ $$\{a_1, a_2^{*}\} = ?$$ $$\{a_2, a_1^{*}\} = ?$$ $$\{a_2, a_2^{*}\} = ?$$ $$\{a_1, a_2\} = ?$$ $$\{a_1^{*}, a_2^{*}\} = ?$$ | 523 | 1,513 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2021-31 | latest | en | 0.876107 |
https://socratic.org/questions/how-do-you-change-y-1-4-3-x-1-4-into-slope-intercept-form | 1,722,941,708,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640484318.27/warc/CC-MAIN-20240806095414-20240806125414-00896.warc.gz | 444,129,922 | 6,626 | # How do you change y - -1/4 = -3(x + 1/4) into slope intercept form?
Jul 19, 2018
$y = - 3 x - 1$
#### Explanation:
This is slope-intercept form:
$y - - \frac{1}{4} = - 3 \left(x + \frac{1}{4}\right)$
First, simplify:
$y + \frac{1}{4} = - 3 x - \frac{3}{4}$
Now subtract $\textcolor{b l u e}{\frac{1}{4}}$ from both sides of the equation:
$y + \frac{1}{4} \quad \textcolor{b l u e}{- \quad \frac{1}{4}} = - 3 x - \frac{3}{4} \quad \textcolor{b l u e}{- \quad \frac{1}{4}}$
$y = - 3 x - 1$
This now matches slope-intercept form, $y = m x + b$ where $m = - 3$ and $b = - 1$.
Hope this helps!
Jul 19, 2018
$y = - 3 x - 1$
#### Explanation:
$\text{the equation of a line in "color(blue)"slope-intercept form}$ is.
•color(white)(x)y=mx+b
$\text{where m is the slope and b the y-intercept}$
$\text{distribute and rearrange}$
$y + \frac{1}{4} = - 3 x - \frac{3}{4}$
$\text{subtract "1/4" from both sides}$
$y = - 3 x - \frac{3}{4} - \frac{1}{4}$
$y = - 3 x - 1 \leftarrow \textcolor{red}{\text{in slope-intercept form}}$
Jul 19, 2018
$y = - 3 x - 1$
#### Explanation:
Recall that slope intercept form is given by
$y = m x + b$, with slope $m$ and a $y$-intercept of $b$.
We can start by distributing the $- 3$ on the right to get
$y + \frac{1}{4} = - 3 x - \frac{3}{4}$
Next, let's subtract $\frac{1}{4}$ from both sides to get
$y = - 3 x - 1$
This equation is now in slope-intercept form.
Hope this helps! | 566 | 1,432 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 27, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.90625 | 5 | CC-MAIN-2024-33 | latest | en | 0.749548 |
http://oeis.org/A264436 | 1,568,717,330,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573070.36/warc/CC-MAIN-20190917101137-20190917123137-00480.warc.gz | 147,513,975 | 4,046 | This site is supported by donations to The OEIS Foundation.
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A264436 Triangle read by rows, inverse Bell transform of the complementary Bell numbers (A000587); T(n,k) for n>=0 and 0<=k<=n. 0
1, 0, 1, 0, 1, 1, 0, 3, 3, 1, 0, 14, 15, 6, 1, 0, 89, 100, 45, 10, 1, 0, 716, 834, 405, 105, 15, 1, 0, 6967, 8351, 4284, 1225, 210, 21, 1, 0, 79524, 97596, 52220, 16009, 3080, 378, 28, 1, 0, 1041541, 1303956, 721674, 233268, 48699, 6804, 630, 36, 1 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 0,8 LINKS Peter Luschny, The Bell transform FORMULA Row sums are A029768(n-1) for n>=1. T(n,1) = A007549(n) for n>=1. EXAMPLE Triangle starts: 1, 0, 1, 0, 1, 1, 0, 3, 3, 1, 0, 14, 15, 6, 1, 0, 89, 100, 45, 10, 1, 0, 716, 834, 405, 105, 15, 1, 0, 6967, 8351, 4284, 1225, 210, 21, 1, 0, 79524, 97596, 52220, 16009, 3080, 378, 28, 1 PROG (Sage) # The function bell_transform is defined in A264428. # The function inverse_bell_transform is defined in A264429. def A264436_matrix(dim): uno = [1]*dim complementary_bell_numbers = [sum((-1)^n*b for (n, b) in enumerate (bell_transform(n, uno))) for n in (0..dim)] return inverse_bell_transform(dim, complementary_bell_numbers) A264436_matrix(9) CROSSREFS Cf. A000587, A007549, A029768, A264428, A264429. Sequence in context: A092747 A265608 A184962 * A122850 A132062 A065547 Adjacent sequences: A264433 A264434 A264435 * A264437 A264438 A264439 KEYWORD nonn,tabl AUTHOR Peter Luschny, Dec 01 2015 STATUS approved
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Last modified September 17 06:41 EDT 2019. Contains 327119 sequences. (Running on oeis4.) | 779 | 1,962 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2019-39 | latest | en | 0.431743 |
https://oeis.org/A253015 | 1,721,069,839,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514713.2/warc/CC-MAIN-20240715163240-20240715193240-00028.warc.gz | 390,593,150 | 4,361 | The OEIS is supported by the many generous donors to the OEIS Foundation.
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A253015 Sequence of determinants of matrices based on the digits of nonnegative integers. 1
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, -1, 1, -1, -5, -11, -19, -29, -41, -55, -71, -4, -1, 4, 1, -4, -11, -20, -31, -44, -59, -9, -5, 1, 9, 5, -1, -9, -19, -31, -45, -16, -11, -4, 5, 16, 11, 4, -5, -16 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,3 COMMENTS A given nonnegative integer is transformed into a square matrix whose order equals the quantity of the number's digits. Each element of the main diagonal is a digit of this original number, while other elements are calculated from this diagonal. The determinant of this matrix is the element of the sequence. LINKS Filipi R. de Oliveira, Table of n, a(n) for n = 0..999 FORMULA a(n) = det(B) where B is the n X n matrix with B(i,i) given by the i-th digit of n, B(i,j) = abs(B(i,j-1)-B(i+1,j)) if i < j and B(i,j) = B(i-1,j) + B(i,j+1) if i > j. EXAMPLE For n=124, a(124)=2, as follows: B(1,1) = 1; B(2,2) = 2; B(3,3) = 4; B(1,2) = abs(B(1,1) - B(2,2)) = abs(1-2) = 1; B(2,3) = abs(B(2,2) - B(3,3)) = abs(2-4) = 2; B(1,3) = abs(B(1,2) - B(2,3)) = abs(1-1) = 1; B(2,1) = B(1,1) + B(2,2) = 1 + 2 = 3; B(3,2) = B(2,2) + B(3,3) = 2 + 4 = 6; B(3,1) = B(2,1) + B(3,2) = 3 + 6 = 9. Thus, _______|1 1 1| B(124)=|3 2 2| --> det(B(124)) = a(124) = 2. _______|9 6 4| CROSSREFS See A227876, since the process of matrix construction is this so-called "pyramidalization". Sequence in context: A216587 A174210 A134777 * A257295 A004427 A113230 Adjacent sequences: A253012 A253013 A253014 * A253016 A253017 A253018 KEYWORD sign,base,easy,dumb AUTHOR Filipi R. de Oliveira, Dec 25 2014 STATUS approved
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Last modified July 15 14:56 EDT 2024. Contains 374333 sequences. (Running on oeis4.) | 838 | 2,130 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2024-30 | latest | en | 0.681315 |
https://la.mathworks.com/matlabcentral/cody/problems/42504-data-regularization/solutions/2085253 | 1,611,107,291,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703519843.24/warc/CC-MAIN-20210119232006-20210120022006-00363.warc.gz | 419,727,286 | 17,278 | Cody
# Problem 42504. Data Regularization
Solution 2085253
Submitted on 9 Jan 2020 by Tom Holz
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
filetext = fileread('regular.m'); assert(isempty(strfind(filetext, 'for'))) assert(isempty(strfind(filetext, 'while')))
2 Pass
A = 1; B = 1; assert(isequal(regular(A),B));
3 Pass
A = [2 6 5 3 5 6 3 7]; B = [1 2 3 1 3 2 2 3]; assert(isequal(regular(A),B));
4 Pass
A = [10 2 4 4 2 4 5 6 8 1 6 5 10 3 9 9 9 5 5 5 9 10 3 7 8]; B = [4 1 2 2 2 1 2 4 5 1 2 2 5 1 5 3 3 3 3 3 3 4 1 4 4]; assert(isequal(regular(A),B));
5 Pass
A = randi(100,80,100); B = zeros(size(A)); for iter = 1:size(A,2) [~, ~, B(:, iter)] = unique(A(:,iter)); end assert(isequal(regular(A),B));
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Start Hunting! | 378 | 964 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2021-04 | latest | en | 0.523194 |
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A128409 Triangle read by rows: A000012 * A128408 as infinite lower triangular matrices. 1
1, 0, -1, -1, -1, -1, -1, -1, -1, 0, -2, -1, -1, 0, -1, -1, 0, 0, 0, -1, 1, -2, 0, 0, 0, -1, 1, -1, -2, 0, 0, 0, -1, 1, -1, 0, -2, 0, 0, 0, -1, 1, -1, 0, 0, -1, 1, 0, 0, 0, 1, -1, 0, 0, 1, -2, 1, 0, 0, 0, 1, -1, 0, 0, 1, -1, -2, 1, 0, 0, 0, 1, -1, 0, 0, 1, -1, 0, -3, 1, 0, 0, 0, 1, -1, 0, 0, 1, -1, 0, -1 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 1,11 COMMENTS Left border = the Mertens sequence: A002321: (1, 0, -1, -1, -2, ...). Right border = mu(n), A008683: (1, -1, -1, 0, -1, 1, -1, ...). Row sums = A062563: (1, -1, -3, -3, -5, -1, -3, ...). LINKS Table of n, a(n) for n=1..91. EXAMPLE First few rows of the triangle: 1; 0, -1; -1, -1, -1; -1, -1, -1, 0; -2, -1, -1, 0, -1; -1, 0, 0, 0, -1, 1; -2, 0, 0, 0, -1, 1, -1; ... CROSSREFS Cf. A002321, A008683, A062563, A128407, A128408. Sequence in context: A171400 A271592 A357187 * A133699 A157361 A224444 Adjacent sequences: A128406 A128407 A128408 * A128410 A128411 A128412 KEYWORD tabl,sign,changed AUTHOR Gary W. Adamson, Mar 01 2007 EXTENSIONS Previous a(51) = 0 removed and more terms from Georg Fischer, Jun 08 2023 STATUS approved
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Last modified June 10 06:58 EDT 2023. Contains 363195 sequences. (Running on oeis4.) | 762 | 1,693 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2023-23 | latest | en | 0.511931 |
https://convertoctopus.com/103-feet-to-inches | 1,714,019,802,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712297284704.94/warc/CC-MAIN-20240425032156-20240425062156-00812.warc.gz | 161,569,668 | 7,292 | ## Conversion formula
The conversion factor from feet to inches is 12, which means that 1 foot is equal to 12 inches:
1 ft = 12 in
To convert 103 feet into inches we have to multiply 103 by the conversion factor in order to get the length amount from feet to inches. We can also form a simple proportion to calculate the result:
1 ft → 12 in
103 ft → L(in)
Solve the above proportion to obtain the length L in inches:
L(in) = 103 ft × 12 in
L(in) = 1236 in
The final result is:
103 ft → 1236 in
We conclude that 103 feet is equivalent to 1236 inches:
103 feet = 1236 inches
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 inch is equal to 0.00080906148867314 × 103 feet.
Another way is saying that 103 feet is equal to 1 ÷ 0.00080906148867314 inches.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that one hundred three feet is approximately one thousand two hundred thirty-six inches:
103 ft ≅ 1236 in
An alternative is also that one inch is approximately zero point zero zero one times one hundred three feet.
## Conversion table
### feet to inches chart
For quick reference purposes, below is the conversion table you can use to convert from feet to inches
feet (ft) inches (in)
104 feet 1248 inches
105 feet 1260 inches
106 feet 1272 inches
107 feet 1284 inches
108 feet 1296 inches
109 feet 1308 inches
110 feet 1320 inches
111 feet 1332 inches
112 feet 1344 inches
113 feet 1356 inches | 401 | 1,559 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2024-18 | latest | en | 0.8562 |
https://matheducators.stackexchange.com/questions/tagged/arithmetic?tab=Active | 1,632,149,053,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057039.7/warc/CC-MAIN-20210920131052-20210920161052-00681.warc.gz | 435,021,833 | 38,666 | # Questions tagged [arithmetic]
The tag has no usage guidance.
27 questions
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299 views
### How many zeros do we need to add to get a nonzero value?
A student (kid) of mine asked this question to me. I am not sure what to make of it or how do I answer it. How many zeros do I need to add to get a non-zero value? ...
160 views
### Is there an **official** name for the following "digit reduction" operation? [closed]
In one of my programs I have a function I call reduce(n) which associates to n the recursive sum of ...
7k views
### Is there a virtue to learning how to compute by hand?
I have been professionally tutoring a wide range of students (from elementary school through graduate school) for many years. Most of them are from the United States. I generally focus on helping my ...
193 views
### Sensible amount of repetition 7 year old
We are now in lock-down, so while homeschooling my son I get to se exactly what he does for math. He has been getting a huge amount of repetitive practicing of really simple math, despite being quite ...
212 views
### How do we explain to a little child that a date in 2020 and a date in 2021 are not necessarily a year apart?
I talked with my friend on December 29 2020. Then I talked with him again on January 03, 2021. Q: What was the year when you last talked with your friend? A: 2021. Q: And what was the year the ...
301 views
### Logarithms chronologically before algebra
Do any textbooks or (somewhat?) standard curricula introduce logarithms and their applications in arithmetic without assuming the students know any algebra? (I do not mean just the use of logarithms ...
3k views
### Does this property of subtraction and division have a name?
Addition and multiplication are commutative. Denoting $\circ$ as either such operation, we have $$x \circ y = z \Leftrightarrow y \circ x = z.$$ Subtraction and division have a similar property, where ...
191 views
### Generating system of equations with unique solutions
I have a similar problem addressed in System of Equations Generator. What I need is an automatic way of generating a system of equations with unique solutions, but the equations are not exclusively ...
250 views
### What are other strategies for a 7 year old for addition and subtraction besides counting fingers?
We recently received feedback from our 7 year old daugther's school teacher. One of the things mentioned was that our daughter still counts her fingers when she does addition and subtraction. The ...
6k views
### Is this primarily a "rote computational trick" for multiplication by 9?
I tried uploading a gif, but was unable to do so. What I can do, is share a link to the gif here. (SE software seems to have allowed me to share the link, but not upload it.) What it shows, ...
434 views
### Is there a numerical base that is in any way “better” for simple mathematical calculations than others?
I want to know if there are any numerical bases that are notably well-suited for humans to learn and use at an elementary or grade-school level. I know that different numerical bases (i.e. decimal/...
2k views
### Different ways to multiply decimals
I am an undergraduate secondary math education major. In $2$ weeks I have to give a Number Talk in my math ed class on the problem "$3.9$ times $7.5$". I need to come up with as many different ...
113 views
### Mental/"Paper and Pencil" Arithmetic
Recently, I was watching this video and I began thinking about how much my arithmetic skills have declined in recent years due to over reliance of calculators in upper year (high school) math courses. ...
448 views
### What's the best technique to do math calculations in my head?
I wish to teach myself how to do math calculations–like $99\times 58$ or $2048+1296$ or $506+998$–inside my head. I know there already exist two methods–Abacus and Vedic Maths. I don't know of more, ...
538 views
### How to answer a three-year-old the question "Why is $2+6$ the same as $4+4$"?
I am teaching my daughter, who is currently about $46$ months old, additions. She is very curious and asks a lot of good questions. For example, when I told her that $2+6=8$ and $4+4=8$, she asked me ...
4k views
### Are soroban (Japanese abacus) classes worth doing?
The companies that run these expensive abacus programs for children claim it has all kinds of benefits for their mathematics abilities and speed. Apparently it starts with a child learning the ...
305 views
### How to explain the motivation of parentheses in addition, subtraction and multiplication?
My kid, 5 years old, knows addition, subtraction and multiplication now, of course, in a basic level. Also he understands that parentheses means "whichever inside shall be computed first". When I ...
434 views
### What’s better: number bonds, or addition tables?
I’ve been teaching my kids addition tables (1+3=4, 2+3=5, 3+3=6, etc.) I only just found out about number bonds (1+4=5, 2+3=5, 4+1=5). This seems a better method because it’s mastering all the ...
2k views
### Adding things to bunches of things vs multiplication
"Suppose you bought four boxes of pencils having five pencils in each, how many pencils do you have altogether?" — "Nine." — "How come?" — "Because 4 plus 5 is 9." — "But you cannot add boxes to ...
236 views
### Looking for a video about arithmetic disappearing in a few years
I saw a video 3 or 4 years ago. The video is about the idea that arithmetic will disappear in the future and only will be a sport, like hunting that passed from a need to a sport. In the video there ...
293 views
### Concentrations of Adult Math Phobia
Some professions have more math phobia than others. Few engineers hate math, but many teachers and journalists do. This means that university departments of education and journalism would likely have ...
166 views
### Adding one to numbers bigger than ten
If someone asks you Tell me the next number (add one) after the number one million two hundred thirty-one thousand ninety-nine, do you known if it is a common error that the first number that ...
267 views
### Quantifying arthmetical skill
Question for a research project: What is the standard way of quantifying a student's skill in arithmetic ranging from having to look up numbers on a times-table to computing large sums in their head ...
160 views
### Learning operator priorities by drawing trees
As far as I know (and here I am refering to my own math education), operator priorities of $+$, $-$, $\cdot$, $\div$, power and paranthesis are taught via some simple phrases like "pointy" operators (...
260 views
### Do electronic calculators inhibit mathematical thinking?
I am interested in educators real-world experience or being pointed to any research in this area. I have a student whose arithmetic skills are weak for his/her age. The student counts on his/her ... | 1,580 | 6,899 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2021-39 | latest | en | 0.954039 |
http://aven.amritalearning.com/?sub=100&brch=295&sim=1483&cnt=3417 | 1,638,688,273,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363149.85/warc/CC-MAIN-20211205065810-20211205095810-00447.warc.gz | 8,793,125 | 5,620 | Images Formed By Lens
# Objective
This topic gives an overview of;
• Images Formed by Lenses
• convex lens
• concave lens
• Glass prism
# Images Formed by Lenses
You might have seen a magnifying glass. It is used to read very small print . You might have also used it to observe the body parts of a cockroach or an earthworm. The magnifying glass is actually a type of a lens.Lenses are widely used in spectacles, telescopes and microscopes. Try to add a few more uses of lenses to this list.
Get some lenses. Touch and feel them. Can you find some difference just by touching? Those lenses which feel thicker in the middle than at the edges are convex lenses. Those which feel thinner in the middle than at the edges are concave lenses. Notice that the lenses are transparent and light can pass through them.
## Activity 1
Take a convex lens or magnifying glass. Put it in the path of sunrays. Place a sheet of paper as shown. Adjust the distance between the lens and the paper till you get a bright spot on the paper. Hold the lens and the paper in this position for a few minutes. Does the paper begin to burn?
Now replace the convex lens with a concave lens. Do you see a bright spot on the paper this time, too? Why are you not getting a bright spot this time?
We have seen in the case of mirrors that for different positions of the object the nature and size of the image change. Is it true for lenses also?
## Activity 2
Take a convex lens and fix it on a stand as you did with the concave mirror. Place it on a table. Place a lighted candle at a distance of about 50 cm from the lens. Try to obtain the image of the candle on a paper screen placed on the other side of the lens. You may have to move the screen towards or away from the lens to get a sharp image of the flame. What kind of image did you get? Is it real or virtual?
Now vary the distance of the candle from the lens. Try to obtain the image of the candle flame every time on the paper screen by moving it. Record your observations as you did for the concave mirror.
Did you get in any position of the object an image which was erect and magnified. Could this image be obtained on a screen? Is the image real or virtual? This is how a convex lens is used as a magnifying glass.
In a similar fashion study the images formed by a concave lens. You will find that the image formed by a concave lens is always virtual, erect and smaller in size than the object.
# Sunlight - White Or Coloured ?
Have you ever seen a rainbow in the sky? You might have noticed that it appears usually after the rain when the sun is low in the sky. The rainbow is seen as a large arc in the sky with many colours.
How many colours are present in a rainbow? When observed carefully, there are seven colours in a rainbow, though it may not be easy to distinguish all of them. These are red, orange, yellow, green, blue, indigo and violet.
You might have seen that when you blow soap bubbles, they appear colourful. Similarly, when light is reflected fr om the surface of a Compact Disk (CD), you see many colours.
## Activity 3
Take a glass prism. Allow a narrow beam of sunlight through a small hole in the window of a dark room to fall on one face of the prism. Let the light coming out of the other face of the prism fall on a white sheet of paper or on a white wall. What do you observe? Do you see colours similar to those in a rainbow? This shows that the sunlight consists of seven colours. The sunlight is said to be white light. This means that the white light consists of seven colours. Try to identify these colours and write their names in your notebook.
Can we mix these colours to get white light? Let us try.
## Activity 4
Take a circular cardboard disc of about 10 cm diameter. Divide this disc into seven segments. Paint the seven rainbow colours on these segments. You can also paste, coloured papers on these segments. Make a small hole at the centre of the disc. Fix the disc loosely on the tip of a refill of a ball pen. Ensure that the disc rotates freely . Rotate the disc in the daylight. When the disc is rotated fast, the colours get mixed together and the disc appears to be whitish. Such a disc is popularly known as Newton's disc.
# Summary
• Image formed by a convex mirror is erect, virtual and smaller in size than the object.
• A convex lens can form real and inverted image. When the object is placed very close to the lens, the image formed is virtual, erect and magnified. When used to see objects magnified, the convex lens is called a magnifying glass.
• A concave lens always forms erect, virtual and smaller image than the object.
• White light is composed of seven colours.
Cite this Simulator: | 1,062 | 4,713 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2021-49 | latest | en | 0.936992 |
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# R: Apply Functions
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R: Apply Functions
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### R: Apply Functions
1. 1. Apply functions<br />The Dataminingtools.net Team<br />
2. 2. Apply functions<br />Apply functions are used to execute a function repetitively. "Apply" functions keeps us from having to write loops to perform some operation on every row or every column of a matrix or data frame, or on every element in a list.<br />
3. 3. Apply family<br />sapply()<br />lapply()<br />apply()<br /><ul><li>mapply()
4. 4. tapply()
5. 5. rapply()</li></li></ul><li>Usage<br />Using Loops!<br />> avg <- numeric (8)<br />> avg<br />[1] 0 0 0 0 0 0 0 0<br />> for(i in 1:8)<br />+ avg[i]<-mean(state.x77[,i])<br />> avg[i]<br />[1] 70735.88<br />> avg<br />[1] 4246.4200 4435.8000 1.1700 70.8786 7.3780<br />[6] 53.1080 104.4600 70735.8800<br />
6. 6. Usage<br />Using ‘apply’<br />> apply (state.x77, 2, median)<br />Population Income Illiteracy Life Exp Murder <br /> 2838.500 4519.000 0.950 70.675 6.850 <br /> HS Grad Frost Area <br /> 53.250 114.500 54277.000<br />The 2 means "go by column" -- a 1 would have meant "go by row."<br />
7. 7. Usage<br />We construct a function and pass it to apply. It computes the median and maximum of each column of state.x77.<br />
8. 8. Usage<br /> apply() works on each row, one at a time, to find the smallest number in each row. which() function, returns the indices within a vector for which the vector holds the value TRUE<br />
9. 9. lapplyand sapply<br />The lapply() function works on any list. The "l" in "lapply" stands for "list."<br />The "s" in "sapply" stands for "simplify."<br />
10. 10. tapply<br />tapply() is a very powerful function that lets us break a vector into pieces and apply some function to each of the pieces. It is like sapply(), except that with sapply() the pieces are always elements of a list. With tapply() we get to specify how the breakdown is done. <br />>tapply(barley\$yield, barley\$site, mean) <br />Grand Rapids Duluth University Farm Morris Crookston Waseca <br />24.93167 27.99667 32.66667 35.4 37.42 48.10833<br /> | 814 | 2,520 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2019-26 | latest | en | 0.387345 |
https://www.patriotichackers.com/2024/01/50-force-and-motion-reading.html | 1,726,384,309,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651616.56/warc/CC-MAIN-20240915052902-20240915082902-00830.warc.gz | 849,368,613 | 13,482 | # 50 Force And Motion Reading Comprehension
## Introduction
Understanding force and motion is fundamental to our understanding of the physical world. Whether you're a student studying physics or simply curious about how things work, developing strong reading comprehension skills in this area is crucial. In this article, we will explore various strategies and tips to improve your force and motion reading comprehension.
Before diving into complex concepts, it's important to have a solid foundation. Begin by familiarizing yourself with the basic principles of force and motion, such as Newton's laws and the different types of forces.
### 1.1 Newton's Laws of Motion
Newton's laws of motion are the foundation of classical mechanics. Take the time to understand each law and how they relate to the motion of objects. This will provide a framework for comprehending more advanced topics.
### 1.2 Types of Forces
Learn about the different types of forces, including gravitational, frictional, and electromagnetic forces. Understanding the characteristics and effects of each force will help you interpret and analyze force-related readings more effectively.
## 2. Improve Vocabulary
Developing a strong vocabulary in the field of force and motion will greatly enhance your reading comprehension. Familiarize yourself with key terms and concepts to better understand the material.
### 2.1 Key Terms
Make a list of essential terms related to force and motion, such as acceleration, inertia, and momentum. Look up their definitions and practice using them in context to solidify your understanding.
Read books, articles, and textbooks that discuss force and motion in various contexts. This will expose you to different vocabulary and help you grasp the nuances of how force and motion apply to different scenarios.
## 3. Break Down Complex Texts
When faced with complex texts, it can be overwhelming to try and comprehend everything at once. Break down the text into smaller sections to make the information more digestible.
### 3.1 Identify Main Ideas
Read the introductory paragraph and conclusion to identify the main ideas of the text. This will give you a sense of the overall message and help you connect the dots as you read the rest of the material.
### 3.2 Take Notes
As you read each section, take notes on the main points and any supporting details. This will help you retain the information and make it easier to review later on.
Passively reading through material is often ineffective for comprehension. Practice active reading techniques to engage with the text and deepen your understanding.
Formulate questions as you read, challenging yourself to think critically about the material. This will keep you actively engaged and help you identify any areas of confusion that need further clarification.
### 4.2 Make Connections
Relate the concepts you're reading about to real-life examples or previous knowledge. This will help solidify your understanding and make the material more relatable.
If you're struggling to grasp a particular concept, don't hesitate to seek additional resources. There are numerous online platforms, videos, and tutorials available to supplement your reading comprehension.
### 5.1 Online Videos and Animations
Visualizing the concepts of force and motion can greatly enhance comprehension. Seek out online videos and animations that demonstrate these principles in action.
### 5.2 Tutoring or Study Groups
Consider joining a study group or seeking tutoring if you need additional support. Collaborating with others can provide new perspectives and help clarify any confusion.
## 6. Review and Reflect
After reading a section or completing a practice exercise, take the time to review and reflect on what you've learned.
### 6.1 Summarize the Material
Write a brief summary of the material in your own words. This will not only reinforce your understanding but also allow you to identify any gaps in your knowledge.
### 6.2 Practice Application
Apply the concepts you've learned to solve practice problems or engage in hands-on activities. This will solidify your understanding and build your confidence in applying force and motion principles.
## Conclusion
Improving your force and motion reading comprehension is an ongoing process that requires dedication and practice. By building a strong foundation, expanding your vocabulary, breaking down complex texts, practicing active reading, seeking additional resources, and reviewing and reflecting on what you've learned, you'll be well on your way to mastering force and motion concepts. | 866 | 4,614 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2024-38 | latest | en | 0.894095 |
http://mathhelpforum.com/calculus/213134-limits.html | 1,524,466,070,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945793.18/warc/CC-MAIN-20180423050940-20180423070940-00386.warc.gz | 207,360,164 | 9,502 | # Thread: Limits
1. ## Limits
For (a) is it 2 or undefined?
Is my answer for (b) and (d) correct??
2. ## Re: Limits
Hey asilvester635.
Hint: For (a) consider that a limit has to be the same from both the left and right directions. (Your answers are not correct).
3. ## Re: Limits
I changed (b) to undefined since the the limit from both sides don't equal to each other
I changed (d) into -1
4. ## Re: Limits
The new answers are correct. | 128 | 445 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-17 | latest | en | 0.879884 |
https://www.kodytools.com/units/angularacc/from/gradpns2/to/arcminps2 | 1,713,148,606,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816939.51/warc/CC-MAIN-20240415014252-20240415044252-00713.warc.gz | 775,071,981 | 17,574 | # Gradian/Square Nanosecond to Arcmin/Square Second Converter
1 Gradian/Square Nanosecond = 54000000000000000000 Arcmin/Square Second
## One Gradian/Square Nanosecond is Equal to How Many Arcmin/Square Second?
The answer is one Gradian/Square Nanosecond is equal to 54000000000000000000 Arcmin/Square Second and that means we can also write it as 1 Gradian/Square Nanosecond = 54000000000000000000 Arcmin/Square Second. Feel free to use our online unit conversion calculator to convert the unit from Gradian/Square Nanosecond to Arcmin/Square Second. Just simply enter value 1 in Gradian/Square Nanosecond and see the result in Arcmin/Square Second.
Manually converting Gradian/Square Nanosecond to Arcmin/Square Second can be time-consuming,especially when you don’t have enough knowledge about Angular Acceleration units conversion. Since there is a lot of complexity and some sort of learning curve is involved, most of the users end up using an online Gradian/Square Nanosecond to Arcmin/Square Second converter tool to get the job done as soon as possible.
We have so many online tools available to convert Gradian/Square Nanosecond to Arcmin/Square Second, but not every online tool gives an accurate result and that is why we have created this online Gradian/Square Nanosecond to Arcmin/Square Second converter tool. It is a very simple and easy-to-use tool. Most important thing is that it is beginner-friendly.
## How to Convert Gradian/Square Nanosecond to Arcmin/Square Second (grad/ns2 to arcmin/s2)
By using our Gradian/Square Nanosecond to Arcmin/Square Second conversion tool, you know that one Gradian/Square Nanosecond is equivalent to 54000000000000000000 Arcmin/Square Second. Hence, to convert Gradian/Square Nanosecond to Arcmin/Square Second, we just need to multiply the number by 54000000000000000000. We are going to use very simple Gradian/Square Nanosecond to Arcmin/Square Second conversion formula for that. Pleas see the calculation example given below.
$$\text{1 Gradian/Square Nanosecond} = 1 \times 54000000000000000000 = \text{54000000000000000000 Arcmin/Square Second}$$
## What Unit of Measure is Gradian/Square Nanosecond?
Gradian per square nanosecond is a unit of measurement for angular acceleration. By definition, if an object accelerates at one gradian per square nanosecond, its angular velocity is increasing by one gradian per nanosecond every nanosecond.
## What Unit of Measure is Arcmin/Square Second?
Arcmin per square second is a unit of measurement for angular acceleration. By definition, if an object accelerates at one arcmin per square second, its angular velocity is increasing by one arcmin per second every second.
## What is the Symbol of Arcmin/Square Second?
The symbol of Arcmin/Square Second is arcmin/s2. This means you can also write one Arcmin/Square Second as 1 arcmin/s2.
## How to Use Gradian/Square Nanosecond to Arcmin/Square Second Converter Tool
• As you can see, we have 2 input fields and 2 dropdowns.
• From the first dropdown, select Gradian/Square Nanosecond and in the first input field, enter a value.
• From the second dropdown, select Arcmin/Square Second.
• Instantly, the tool will convert the value from Gradian/Square Nanosecond to Arcmin/Square Second and display the result in the second input field.
## Example of Gradian/Square Nanosecond to Arcmin/Square Second Converter Tool
1
Arcmin/Square Second
54000000000000000000
# Gradian/Square Nanosecond to Other Units Conversion Table
ConversionDescription
1 Gradian/Square Nanosecond = 900000000000000000 Degree/Square Second1 Gradian/Square Nanosecond in Degree/Square Second is equal to 900000000000000000
1 Gradian/Square Nanosecond = 900000000000 Degree/Square Millisecond1 Gradian/Square Nanosecond in Degree/Square Millisecond is equal to 900000000000
1 Gradian/Square Nanosecond = 900000 Degree/Square Microsecond1 Gradian/Square Nanosecond in Degree/Square Microsecond is equal to 900000
1 Gradian/Square Nanosecond = 0.9 Degree/Square Nanosecond1 Gradian/Square Nanosecond in Degree/Square Nanosecond is equal to 0.9
1 Gradian/Square Nanosecond = 3.24e+21 Degree/Square Minute1 Gradian/Square Nanosecond in Degree/Square Minute is equal to 3.24e+21
1 Gradian/Square Nanosecond = 1.1664e+25 Degree/Square Hour1 Gradian/Square Nanosecond in Degree/Square Hour is equal to 1.1664e+25
1 Gradian/Square Nanosecond = 6.718464e+27 Degree/Square Day1 Gradian/Square Nanosecond in Degree/Square Day is equal to 6.718464e+27
1 Gradian/Square Nanosecond = 3.29204736e+29 Degree/Square Week1 Gradian/Square Nanosecond in Degree/Square Week is equal to 3.29204736e+29
1 Gradian/Square Nanosecond = 6.224263236e+30 Degree/Square Month1 Gradian/Square Nanosecond in Degree/Square Month is equal to 6.224263236e+30
1 Gradian/Square Nanosecond = 8.96293905984e+32 Degree/Square Year1 Gradian/Square Nanosecond in Degree/Square Year is equal to 8.96293905984e+32
1 Gradian/Square Nanosecond = 54000000000000000000 Arcmin/Square Second1 Gradian/Square Nanosecond in Arcmin/Square Second is equal to 54000000000000000000
1 Gradian/Square Nanosecond = 54000000000000 Arcmin/Square Millisecond1 Gradian/Square Nanosecond in Arcmin/Square Millisecond is equal to 54000000000000
1 Gradian/Square Nanosecond = 54000000 Arcmin/Square Microsecond1 Gradian/Square Nanosecond in Arcmin/Square Microsecond is equal to 54000000
1 Gradian/Square Nanosecond = 54 Arcmin/Square Nanosecond1 Gradian/Square Nanosecond in Arcmin/Square Nanosecond is equal to 54
1 Gradian/Square Nanosecond = 1.944e+23 Arcmin/Square Minute1 Gradian/Square Nanosecond in Arcmin/Square Minute is equal to 1.944e+23
1 Gradian/Square Nanosecond = 6.9984e+26 Arcmin/Square Hour1 Gradian/Square Nanosecond in Arcmin/Square Hour is equal to 6.9984e+26
1 Gradian/Square Nanosecond = 4.0310784e+29 Arcmin/Square Day1 Gradian/Square Nanosecond in Arcmin/Square Day is equal to 4.0310784e+29
1 Gradian/Square Nanosecond = 1.975228416e+31 Arcmin/Square Week1 Gradian/Square Nanosecond in Arcmin/Square Week is equal to 1.975228416e+31
1 Gradian/Square Nanosecond = 3.7345579416e+32 Arcmin/Square Month1 Gradian/Square Nanosecond in Arcmin/Square Month is equal to 3.7345579416e+32
1 Gradian/Square Nanosecond = 5.377763435904e+34 Arcmin/Square Year1 Gradian/Square Nanosecond in Arcmin/Square Year is equal to 5.377763435904e+34
1 Gradian/Square Nanosecond = 3.24e+21 Arcsec/Square Second1 Gradian/Square Nanosecond in Arcsec/Square Second is equal to 3.24e+21
1 Gradian/Square Nanosecond = 3240000000000000 Arcsec/Square Millisecond1 Gradian/Square Nanosecond in Arcsec/Square Millisecond is equal to 3240000000000000
1 Gradian/Square Nanosecond = 3240000000 Arcsec/Square Microsecond1 Gradian/Square Nanosecond in Arcsec/Square Microsecond is equal to 3240000000
1 Gradian/Square Nanosecond = 3240 Arcsec/Square Nanosecond1 Gradian/Square Nanosecond in Arcsec/Square Nanosecond is equal to 3240
1 Gradian/Square Nanosecond = 1.1664e+25 Arcsec/Square Minute1 Gradian/Square Nanosecond in Arcsec/Square Minute is equal to 1.1664e+25
1 Gradian/Square Nanosecond = 4.19904e+28 Arcsec/Square Hour1 Gradian/Square Nanosecond in Arcsec/Square Hour is equal to 4.19904e+28
1 Gradian/Square Nanosecond = 2.41864704e+31 Arcsec/Square Day1 Gradian/Square Nanosecond in Arcsec/Square Day is equal to 2.41864704e+31
1 Gradian/Square Nanosecond = 1.1851370496e+33 Arcsec/Square Week1 Gradian/Square Nanosecond in Arcsec/Square Week is equal to 1.1851370496e+33
1 Gradian/Square Nanosecond = 2.24073476496e+34 Arcsec/Square Month1 Gradian/Square Nanosecond in Arcsec/Square Month is equal to 2.24073476496e+34
1 Gradian/Square Nanosecond = 3.2266580615424e+36 Arcsec/Square Year1 Gradian/Square Nanosecond in Arcsec/Square Year is equal to 3.2266580615424e+36
1 Gradian/Square Nanosecond = 30000000000000000 Sign/Square Second1 Gradian/Square Nanosecond in Sign/Square Second is equal to 30000000000000000
1 Gradian/Square Nanosecond = 30000000000 Sign/Square Millisecond1 Gradian/Square Nanosecond in Sign/Square Millisecond is equal to 30000000000
1 Gradian/Square Nanosecond = 30000 Sign/Square Microsecond1 Gradian/Square Nanosecond in Sign/Square Microsecond is equal to 30000
1 Gradian/Square Nanosecond = 0.03 Sign/Square Nanosecond1 Gradian/Square Nanosecond in Sign/Square Nanosecond is equal to 0.03
1 Gradian/Square Nanosecond = 108000000000000000000 Sign/Square Minute1 Gradian/Square Nanosecond in Sign/Square Minute is equal to 108000000000000000000
1 Gradian/Square Nanosecond = 3.888e+23 Sign/Square Hour1 Gradian/Square Nanosecond in Sign/Square Hour is equal to 3.888e+23
1 Gradian/Square Nanosecond = 2.239488e+26 Sign/Square Day1 Gradian/Square Nanosecond in Sign/Square Day is equal to 2.239488e+26
1 Gradian/Square Nanosecond = 1.09734912e+28 Sign/Square Week1 Gradian/Square Nanosecond in Sign/Square Week is equal to 1.09734912e+28
1 Gradian/Square Nanosecond = 2.074754412e+29 Sign/Square Month1 Gradian/Square Nanosecond in Sign/Square Month is equal to 2.074754412e+29
1 Gradian/Square Nanosecond = 2.98764635328e+31 Sign/Square Year1 Gradian/Square Nanosecond in Sign/Square Year is equal to 2.98764635328e+31
1 Gradian/Square Nanosecond = 2500000000000000 Turn/Square Second1 Gradian/Square Nanosecond in Turn/Square Second is equal to 2500000000000000
1 Gradian/Square Nanosecond = 2500000000 Turn/Square Millisecond1 Gradian/Square Nanosecond in Turn/Square Millisecond is equal to 2500000000
1 Gradian/Square Nanosecond = 2500 Turn/Square Microsecond1 Gradian/Square Nanosecond in Turn/Square Microsecond is equal to 2500
1 Gradian/Square Nanosecond = 0.0025 Turn/Square Nanosecond1 Gradian/Square Nanosecond in Turn/Square Nanosecond is equal to 0.0025
1 Gradian/Square Nanosecond = 9000000000000000000 Turn/Square Minute1 Gradian/Square Nanosecond in Turn/Square Minute is equal to 9000000000000000000
1 Gradian/Square Nanosecond = 3.24e+22 Turn/Square Hour1 Gradian/Square Nanosecond in Turn/Square Hour is equal to 3.24e+22
1 Gradian/Square Nanosecond = 1.86624e+25 Turn/Square Day1 Gradian/Square Nanosecond in Turn/Square Day is equal to 1.86624e+25
1 Gradian/Square Nanosecond = 9.144576e+26 Turn/Square Week1 Gradian/Square Nanosecond in Turn/Square Week is equal to 9.144576e+26
1 Gradian/Square Nanosecond = 1.72896201e+28 Turn/Square Month1 Gradian/Square Nanosecond in Turn/Square Month is equal to 1.72896201e+28
1 Gradian/Square Nanosecond = 2.4897052944e+30 Turn/Square Year1 Gradian/Square Nanosecond in Turn/Square Year is equal to 2.4897052944e+30
1 Gradian/Square Nanosecond = 2500000000000000 Circle/Square Second1 Gradian/Square Nanosecond in Circle/Square Second is equal to 2500000000000000
1 Gradian/Square Nanosecond = 2500000000 Circle/Square Millisecond1 Gradian/Square Nanosecond in Circle/Square Millisecond is equal to 2500000000
1 Gradian/Square Nanosecond = 2500 Circle/Square Microsecond1 Gradian/Square Nanosecond in Circle/Square Microsecond is equal to 2500
1 Gradian/Square Nanosecond = 0.0025 Circle/Square Nanosecond1 Gradian/Square Nanosecond in Circle/Square Nanosecond is equal to 0.0025
1 Gradian/Square Nanosecond = 9000000000000000000 Circle/Square Minute1 Gradian/Square Nanosecond in Circle/Square Minute is equal to 9000000000000000000
1 Gradian/Square Nanosecond = 3.24e+22 Circle/Square Hour1 Gradian/Square Nanosecond in Circle/Square Hour is equal to 3.24e+22
1 Gradian/Square Nanosecond = 1.86624e+25 Circle/Square Day1 Gradian/Square Nanosecond in Circle/Square Day is equal to 1.86624e+25
1 Gradian/Square Nanosecond = 9.144576e+26 Circle/Square Week1 Gradian/Square Nanosecond in Circle/Square Week is equal to 9.144576e+26
1 Gradian/Square Nanosecond = 1.72896201e+28 Circle/Square Month1 Gradian/Square Nanosecond in Circle/Square Month is equal to 1.72896201e+28
1 Gradian/Square Nanosecond = 2.4897052944e+30 Circle/Square Year1 Gradian/Square Nanosecond in Circle/Square Year is equal to 2.4897052944e+30
1 Gradian/Square Nanosecond = 16000000000000000000 Mil/Square Second1 Gradian/Square Nanosecond in Mil/Square Second is equal to 16000000000000000000
1 Gradian/Square Nanosecond = 16000000000000 Mil/Square Millisecond1 Gradian/Square Nanosecond in Mil/Square Millisecond is equal to 16000000000000
1 Gradian/Square Nanosecond = 16000000 Mil/Square Microsecond1 Gradian/Square Nanosecond in Mil/Square Microsecond is equal to 16000000
1 Gradian/Square Nanosecond = 16 Mil/Square Nanosecond1 Gradian/Square Nanosecond in Mil/Square Nanosecond is equal to 16
1 Gradian/Square Nanosecond = 5.76e+22 Mil/Square Minute1 Gradian/Square Nanosecond in Mil/Square Minute is equal to 5.76e+22
1 Gradian/Square Nanosecond = 2.0736e+26 Mil/Square Hour1 Gradian/Square Nanosecond in Mil/Square Hour is equal to 2.0736e+26
1 Gradian/Square Nanosecond = 1.1943936e+29 Mil/Square Day1 Gradian/Square Nanosecond in Mil/Square Day is equal to 1.1943936e+29
1 Gradian/Square Nanosecond = 5.85252864e+30 Mil/Square Week1 Gradian/Square Nanosecond in Mil/Square Week is equal to 5.85252864e+30
1 Gradian/Square Nanosecond = 1.1065356864e+32 Mil/Square Month1 Gradian/Square Nanosecond in Mil/Square Month is equal to 1.1065356864e+32
1 Gradian/Square Nanosecond = 1.593411388416e+34 Mil/Square Year1 Gradian/Square Nanosecond in Mil/Square Year is equal to 1.593411388416e+34
1 Gradian/Square Nanosecond = 2500000000000000 Revolution/Square Second1 Gradian/Square Nanosecond in Revolution/Square Second is equal to 2500000000000000
1 Gradian/Square Nanosecond = 2500000000 Revolution/Square Millisecond1 Gradian/Square Nanosecond in Revolution/Square Millisecond is equal to 2500000000
1 Gradian/Square Nanosecond = 2500 Revolution/Square Microsecond1 Gradian/Square Nanosecond in Revolution/Square Microsecond is equal to 2500
1 Gradian/Square Nanosecond = 0.0025 Revolution/Square Nanosecond1 Gradian/Square Nanosecond in Revolution/Square Nanosecond is equal to 0.0025
1 Gradian/Square Nanosecond = 9000000000000000000 Revolution/Square Minute1 Gradian/Square Nanosecond in Revolution/Square Minute is equal to 9000000000000000000
1 Gradian/Square Nanosecond = 3.24e+22 Revolution/Square Hour1 Gradian/Square Nanosecond in Revolution/Square Hour is equal to 3.24e+22
1 Gradian/Square Nanosecond = 1.86624e+25 Revolution/Square Day1 Gradian/Square Nanosecond in Revolution/Square Day is equal to 1.86624e+25
1 Gradian/Square Nanosecond = 9.144576e+26 Revolution/Square Week1 Gradian/Square Nanosecond in Revolution/Square Week is equal to 9.144576e+26
1 Gradian/Square Nanosecond = 1.72896201e+28 Revolution/Square Month1 Gradian/Square Nanosecond in Revolution/Square Month is equal to 1.72896201e+28
1 Gradian/Square Nanosecond = 2.4897052944e+30 Revolution/Square Year1 Gradian/Square Nanosecond in Revolution/Square Year is equal to 2.4897052944e+30 | 4,506 | 14,722 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2024-18 | latest | en | 0.898794 |
https://www.question-hub.com/53357/whats-the-value-of-x-3x-5x-5-7-2x-12.html | 1,660,327,896,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571745.28/warc/CC-MAIN-20220812170436-20220812200436-00119.warc.gz | 856,807,714 | 11,101 | Or
# Whats the value of x? 3x - 5(x - 5) = -7 + 2x + 12
Asked 9 months ago
Viewed 943428 times
1
Whats the value of x? 3x - 5(x - 5) = -7 + 2x + 12
### 2 Answer
2
Answer:
x=5 :)
Step-by-step explanation:
##### Shaun Deckow
15.5k 3 10 26
answered 9 months ago
2
Answer:
x = 5
Step-by-step explanation:
Given
3x - 5(x - 5) = - 7 + 2x + 12 ← distribute and simplify left side
3x - 5x + 25 = 2x + 5
- 2x + 25 = 2x + 5 ( subtract 2x from both sides )
- 4x + 25 = 5 ( subtract 25 from both sides )
- 4x = - 20 ( divide both sides by - 4 )
x = 5
##### Jaime Little II
15.5k 3 10 26
answered 9 months ago | 260 | 615 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2022-33 | latest | en | 0.737337 |
http://mathhelpforum.com/calculus/142517-integral-ln-squared.html | 1,481,430,072,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698544097.11/warc/CC-MAIN-20161202170904-00109-ip-10-31-129-80.ec2.internal.warc.gz | 180,130,440 | 11,248 | # Thread: Integral of ln Squared?
1. ## Integral of ln Squared?
$\int\ln^2(x)dx$
2. Looks like a prime candidate for "Parts". It's almost an eyeball problem.
$x\cdot ln^{2}(x) - \int x\cdot 2\cdot ln(x)\cdot \frac{1}{x}\;dx$
3. soo,
$\ln^2(x)-2x\ln(x)+2x+C$
??
4. close, try $x\ln^2(x)-2x\ln(x)+2x+C$
5. So close! Give it another go. This time, don't lose the 'x' on the front.
6. Originally Posted by iPod
soo,
$\ln^2(x)-2x\ln(x)+2x+C$
??
You should get
$x\ln^2(x)-2x\ln(x)+2x+C$
7. woopsy, that was a typo ^^
i meant to right
$x\ln^2(x)-2x\ln(x)+2x+C$
but I have seen what you have done tkhunny, using the 'by parts rule, you have integrated it as if it was
$\int1.\ln^2(x)$
right?
8. Originally Posted by iPod
$\int1.\ln^2(x)$
right?
Yep with $v = \ln^2x$ and $du = 1$
9. Thanks you lot
10. ...except that I don't use that silly American method. The Russians are much more visual and literal. | 337 | 917 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 11, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2016-50 | longest | en | 0.844779 |
https://www.scienceforums.net/profile/147349-noobinmath/content/ | 1,719,331,722,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198866143.18/warc/CC-MAIN-20240625135622-20240625165622-00418.warc.gz | 840,387,945 | 13,796 | # noobinmath
Members
4
1. ## Need description of Prime# distribution in Riemann hypothesis
Primes pattern is simple, just look at the sieve: First you have all numbers starting from 2 until infinity 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 ... now use first number you see (2) and use the number of that number to remove numbers (sieve repeats every 2 numbers) 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 ... 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 ... (1 out of every 2 numbers are removed) now use first number you see (3) and use the number of that number to remove numbers (sieve repeats every 2*3 numbers) 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 ... (all combinations, repeats every 6 numbers) 5 7 11 13 17 19 23 25 29 31 35 37 41 43 47 49 ... (4 out of every 6 numbers are removed) now use first number you see (5) and use the number of that number to remove numbers (sieve repeats every 2*3*5 numbers) 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 ... (apply 5 ) 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 ... (apply 3 ) 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 ... (all combinations, repeats every 30 numbers) 7 11 13 17 19 23 29 31 37 41 43 47 49 ... (22 out of every 30 numbers are removed) now use first number you see (7) and use the number of that number to remove numbers (sieve repeats every 2*3*5*7 numbers) etc.
2. ## Should I Learn C, C++, Go, or Rust?
Go for C if you want to become an electrical engineer, to learn basics or create operating systems. C++ is best for game development. C# is best for enterprise applications. My advice is to don't develop games in unreal engine, gamemaker or godot, instead use a rpg maker when making a game, it is more rewarding and more fun.
3. ## what's a good programming language to learn?
I would say, go to your local job offer site, calculate job offer amount * salary. Than learn the language that has the highest job offers salary combination. It depends on the country you are in. Also try to do some leetcode problems. In my opinion C# is the best language, good pay, many jobs, many documentation. It makes you a valuable asset.
4. ## integer factorisation can't be polynomial for worst case scenario's
Hello I got the following proof, how do I begin to debunk it? Integer factorisation can't be polynomial for worst case scenario's. To prove this theorem, we use the fundamental theorem of arithmetic, which states that every integer greater than 1 can be expressed uniquely as a product of primes. We prove this by contradiction. Assume that there exists an integer n that can be expressed in two different ways as a product of primes, say n = p1 * p2 * ... * pm = q1 * q2 * ... * qn, where the pi and qi are primes and m ≠ n. Without loss of generality, assume that p1 is not equal to any of the qi. Then p1 must divide n, and hence p1 must divide the product q1 * q2 * ... * qn. But since p1 is prime and is not equal to any of the qi, p1 must divide one of the qi, say q1. But now we have p1 divides both n and q1, which contradicts the assumption that the pi and qi are all prime. Therefore, the prime factorization of any integer is unique up to ordering of the factors. Now, suppose we have an algorithm that can find the prime factors of an integer n in polynomial time. We use this algorithm to find the prime factors of a composite integer N, which we can express as N = p * q for two distinct primes p and q. Let D be the set of divisors of N, excluding 1 and N itself. Then all the elements of D are composite numbers and can be expressed as a product of two distinct primes, namely p and q. We can use our polynomial-time algorithm to find the prime factors of each element of D, and then group the results according to the product of the two primes. There will be exactly one group for each element of D. Now, consider the product of the two smallest primes in each group. Since there is exactly one group for each divisor of N, we have accounted for all the factors of N. Therefore, the product of the two smallest primesThe fundamental theorem of arithmetic, also known as the unique factorization theorem, states that every positive integer greater than 1 can be represented uniquely as a product of prime numbers. Specifically, if n is a positive integer greater than 1, then there exist distinct prime numbers p1, p2, ..., pk and positive integers e1, e2, ..., ek such that: n = p1^e1 * p2^e2 * ... * pk^ek Moreover, this representation is unique up to the order of the factors. Now suppose that there exists an efficient algorithm for integer factorization. Let n be a positive integer greater than 1, and let p be the smallest prime factor of n. By the fundamental theorem of arithmetic, we can write: n = p * m where m is a positive integer that is either prime or has a smallest prime factor greater than p. If we could efficiently compute the prime factors of m using our algorithm, then we could also determine the prime factors of n by combining them with the factor p. Moreover, the size of m is strictly smaller than the size of n, so we can repeat this process until we have found all the prime factors of n. However, this algorithm would enable us to factor any positive integer n in polynomial time, which contradicts the well-known fact (proved by Carl Friedrich Gauss) that the number of primes less than n is approximately n/ln(n) for large n. In particular, this implies that the number of prime factors of a positive integer with n bits is roughly n/ln(2), which grows much faster than any polynomial function of n. Therefore, no efficient algorithm for integer factorization can exist, unless P = NP, which is considered unlikely by most experts in computational complexity theory.
× | 1,723 | 6,018 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2024-26 | latest | en | 0.676029 |
https://math.stackexchange.com/questions/2087168/how-to-verify-that-a-basis-for-a-column-space-spans-the-column-space | 1,571,846,122,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987834649.58/warc/CC-MAIN-20191023150047-20191023173547-00141.warc.gz | 579,108,180 | 30,962 | # How to verify that a basis for a column space spans the column space?
I know how to verify that a basis for R^3 spans R^3 -- you just form the equation
(u1, u2, u3) = c1v1 + c2v2 + c3v3 ,
equate corresponding components, and see whether the system has a unique solution.
But for verifying that a set of vectors spans a column space, you're not trying to see whether it spans all of R3 (or R2, or R4, or whatever), correct? You're trying to see whether it spans whatever the column space spans. So you can't just set the linear combination equal to u, right? So then how would you go about verifying that a basis spans a column space?
Suppose you claim that $\{ v_1, v_2, \ldots, v_m \}$ is a basis for the column space.
You just have to verify that $u_1$, $u_2, \ldots, u_n$ can be expressed as linear combinations of $v_1$, $v_2$,$\ldots, v_m$ where $u_i$ are the columns of the matrix.
The column space is formed by linear combinations of $u_1, u_2,\ldots u_n$. If $x$ is in the column space, then $$x = \sum_{i=1}^n a_i u_i$$ for some $a_i$.
if $u_i= \sum_{j=1}^m b_{ij}v_i$,
$$x=\sum_{i=1}^na_i\sum_{j=1}^mb_{ij}v_j=\sum_{j=1}^m\sum_{i=1}^na_ib_{ij}v_j$$ | 387 | 1,165 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2019-43 | latest | en | 0.819276 |
http://talkstats.com/threads/calculating-smallest-detectable-difference-when-mean-is-around-1-and-mse-1.48530/ | 1,670,049,708,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710924.83/warc/CC-MAIN-20221203043643-20221203073643-00340.warc.gz | 43,070,273 | 8,817 | # Calculating Smallest Detectable Difference when mean is around 1 and MSE < 1
#### rodwhiteley
##### New Member
Apologies if this is in the wrong forum, and please move if this is the case.
Just a query that might be intractable.
When calculating Standard Errors of the Measure (SEM's), we take the root of the Mean Square Error of the residual (MSE from the ANOVA table), then multiply by 1.96 and by root 2 to get the Smallest Detectable Change (SDC).
My query is, when the MSE is less than 1, and the mean of the value is around 1, we get some anomalies.
If, say we have a mean of 1, and MSE's of the residual, of say 0.05, now the SDC becomes 0.62, or 62% of the mean, and this looks lousy. If we multiply everything by 100 before we start (say we choose to arbitrarily use Newton-centimetres instead of Newton-metres) then with the same values (mean now of 100, and MSE now of 5) our SDC is 6.2, or only 6% of the mean, and everything is sunshine and roses again.
I am inclined to think that this is a bug in the generation of SDC's, but want to make sure I'm not committing a howler here.
Any thoughts? | 302 | 1,112 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2022-49 | latest | en | 0.915783 |
https://cz.pinterest.com/ideas/centrifugal-force/896878516081/ | 1,718,879,839,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861916.26/warc/CC-MAIN-20240620074431-20240620104431-00559.warc.gz | 170,621,802 | 90,817 | # Centrifugal force
Learn about centrifugal force and how it affects objects in motion. Explore real-world examples and gain a deeper understanding of this fascinating concept.
Microbe Notes
Chris MacAlister
## Centripetal vs Centrifugal Force: Main Differences
### In this guide, you'll find major differences between centripetal vs centrifugal force. Learn calculating the types of forces and examples in daily life.
Hanoch Bar-Nitzan
## Ubud
### Centrifugal Force - Phun Physics - Topics - University of Virginia An object traveling in a circle behaves as if it is experiencing an outward force. This force, known as the centrifugal force, depends on the mass of the object, the speed of rotation, and the distance from the center. The more massive the object, the greater the force; the greater the speed of the object, the greater the force; and the greater the distance from the center, the greater the force.
Desjardin Crump-Gibson
진희 정
Jerry A.
jas brar
Laurie Mayer
Andrew Davis | 224 | 1,001 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-26 | latest | en | 0.847086 |
https://www.maloneymethod.com/tag/counting-skills/page/2/ | 1,720,908,338,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514517.94/warc/CC-MAIN-20240713212202-20240714002202-00538.warc.gz | 775,555,123 | 16,327 | ## Rote Counting Skills from 10 to 20 – the “teens” issue
Rationale: In this exercise we will teach rote counting from 10 to 20 and practice until the fluency level of 150 counts/minute with 2 or fewer errors is reached. Now that the student has met the fluency standard for counting to 10, we can add the next step. Teaching the next ten digits has aContinue reading “Rote Counting Skills from 10 to 20 – the “teens” issue”
## Rote Counting Skills from 1 to 10 (Forward)
Rationale Until a student has well-developed counting skills, all other aspects of arithmetic will be difficult or impossible to do. Fluent counting skills are the basis for most other activities in arithmetic. There are a number of different components that students need to be taught before they attempt to learn more advanced arithmetic operations suchContinue reading “Rote Counting Skills from 1 to 10 (Forward)”
## Teaching Arithmetic Skills â Letâs Get Started!!
In the next number of posts, I am going to provide parents, tutors, teachers, therapists and aides with actual lessons that you can use with children who have non-existent or weak counting skills. I will be posting 2 lessons per week for the foreseeable future. From time to time, I will likely change topics toContinue reading “Teaching Arithmetic Skills â Letâs Get Started!!” | 303 | 1,333 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-30 | latest | en | 0.923404 |
https://community.qlik.com/t5/QlikView-App-Development/Date-set-analysis-problem/m-p/1167665/highlight/true | 1,571,339,981,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986675598.53/warc/CC-MAIN-20191017172920-20191017200420-00305.warc.gz | 435,475,254 | 40,265 | # QlikView App Development
Discussion Board for collaboration related to QlikView App Development.
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## Date set analysis problem
Hi Everyone,
I am trying to make it so that when I select a certain month that the current year's data up until that month is also included in my calculation. For example:
I select February of this year, it is currently June. I just want to see the data from January and February of this year, not January - June. What would be the best way to do this without having to select both January and February.
Any help on this would be greatly appreciated.
Thanks!
Joe
Tags (4)
1 Solution
Accepted Solutions
Honored Contributor
## Re: Date set analysis problem
Hi Joe,
Try this set modifier in your expressions.
{\$<Year = {\$(=max(Year))}, Date = {'<=\$(=max(Date))'}, Month =>}
I'm assuming you have a calendar with a Year field. If you have any other selectable calendar fields like Qtr you should add , 'Qtr =' to the above set so that if you select the 2nd Qtr then select a date in that Qtr your expression will total from the beginning of the year not just from the beginning of the 2nd Qtr..
Cheers
Andrew
4 Replies
Highlighted
MVP
## Re: Date set analysis problem
Do you have date field?
If yes, you can achieve this using date field...
=SUM({<InvoiceYear=, InvoiceMonth=, InvoiceDate = {">=\$(=YearStart(Max(InvoiceDate)))<=\$(=Max(InvoiceDate))"}>}Sales)
Change InvoiceYear, InvoiceMonth and InvoiceDate field names respective to your data model.
Honored Contributor
## Re: Date set analysis problem
Hi Joe,
Try this set modifier in your expressions.
{\$<Year = {\$(=max(Year))}, Date = {'<=\$(=max(Date))'}, Month =>}
I'm assuming you have a calendar with a Year field. If you have any other selectable calendar fields like Qtr you should add , 'Qtr =' to the above set so that if you select the 2nd Qtr then select a date in that Qtr your expression will total from the beginning of the year not just from the beginning of the 2nd Qtr..
Cheers
Andrew
Not applicable
## Re: Date set analysis problem
Thank you! Much appreciated.
Honored Contributor
## Re: Date set analysis problem
You're very welcome Joe! | 529 | 2,207 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2019-43 | latest | en | 0.896 |
https://www.calculatoratoz.com/en/center-of-buoyancy-calculator/Calc-1195 | 1,611,208,782,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703522242.73/warc/CC-MAIN-20210121035242-20210121065242-00613.warc.gz | 704,504,547 | 33,124 | Anirudh Singh
National Institute of Technology (NIT), Jamshedpur
Anirudh Singh has created this Calculator and 100+ more calculators!
## < 11 Other formulas that you can solve using the same Inputs
Impulsive Torque
Impulsive Torque=(Moment of Inertia*(Final Angular Velocity-Angular velocity))/Time Taken to Travel GO
Specific Weight
Specific Weight=Weight of body on which frictional force is applied/Volume GO
Lateral edge length of a Right square pyramid when volume and side length is given
Length of edge=sqrt(Side^2/2+((3*Volume)/Side^2)^2) GO
Slant height of a Right square pyramid when volume and side length are given
Slant Height=sqrt((Side^2/4)+((3*Volume)/Side^2)^2) GO
Angular Momentum
Angular Momentum=Moment of Inertia*Angular Velocity GO
Bottom surface area of a triangular prism when volume and height are given
Bottom Surface Area=Volume/Height GO
Height of a triangular prism when base and volume are given
Height=(2*Volume)/(Base*Length) GO
Top surface area of a triangular prism when volume and height are given
Top Surface Area=Volume/Height GO
Side length of a Right square pyramid when volume and height are given
Side=sqrt((3*Volume)/Height) GO
Height of a right square pyramid when volume and side length are given
Height=(3*Volume)/Side^2 GO
Density
Density=Mass/Volume GO
### Center of Buoyancy Formula
Centre of Buoyancy=Moment of Inertia/(Volume*Centre of gravity)-Metacenter
More formulas
Knudsen Number GO
Kinematic viscosity GO
Pressure Wave Velocity in Fluids GO
Surface tension GO
Bulk Modulus GO
Weight GO
Upthrust Force GO
Viscous Stress GO
Stokes Force GO
Reynolds Number GO
Specific Weight GO
Specific Volume GO
Inertial Force Per Unit Area GO
Body Force Work Rate GO
Heat Loss due to Pipe GO
Dynamic viscosity of fluids GO
Dynamic Viscosity of Gases GO
Viscous Force Per Unit Area GO
Terminal Velocity GO
Poiseuille's Formula GO
Dynamic Viscosity of Liquids GO
Pressure Inside the Liquid Drop GO
Center of Gravity GO
Metacenter GO
Pressure Inside the Soap Bubble GO
Turbulence GO
Height of Capillary Rise GO
Capillarity Through Parallel Plates GO
Capillarity Through an Annular Space GO
Capillarity Through a Circular Tube if inserted in liquid of S1 above a liquid of S2 GO
Cavitation Number GO
Pressure in Excess of Atmospheric Pressure GO
Absolute Pressure at a Height h GO
Normal Stress 1 GO
Normal Stress 2 GO
Differential pressure between two points GO
U-Tube Manometer equation GO
Differential pressure-Differential Manometer GO
Pressure using inclined Manometer GO
Sensitivity of inclined manometer GO
Total Hydrostatic force GO
Center of pressure GO
Buoyancy Force GO
Center of Pressure on Inclined Plane GO
Metacentric Height GO
Metacentric Height when Moment of Inertia is Given GO
Unstable Equilibrium of a Floating Body GO
Experimental determination of Metacentric height GO
Time period of Rolling GO
Rate of Flow GO
Equation of Continuity for Incompressible Fluids GO
Equation of Continuity for Compressible Fluids GO
Vorticity GO
Dynamic Pressure GO
Theoretical Velocity - Pitot Tube GO
Theoretical discharge -Venturimeter GO
Discharge through an Elbow meter GO
Variation of y with x in Free Liquid Jet GO
Time of Flight of Jet GO
Time to Reach Highest Point GO
Maximum Vertical Elevation of a Jet Profile GO
Horizontal Range of the Jet GO
Power Required to Overcome the Frictional Resistance in Laminar Flow GO
Frictional Factor of Laminar flow GO
Head loss due to Laminar Flow GO
Friction velocity GO
Force in direction of jet striking a stationary vertical plate GO
Hydraulic Transmission of Power GO
Efficiency of transmission GO
Bulk Modulus When Velocity Of Pressure Wave Is Given GO
Mass Density When Velocity Of Pressure Wave Is Given GO
Surface Energy When Surface Tension Is Given GO
Surface Area When Surface Tension Is Given GO
Shear Stress When Dynamic Viscosity Of A Fluid Is Given GO
Velocity Of Moving Plates When Dynamic Viscosity Is Given GO
Distance Between Plates When Dynamic Viscosity Of A Fluid Is Given GO
Surface Tension Of Liquid Drop When Change In Pressure Is Given GO
Diameter Of Droplet When Pressure Change Is Given GO
Surface Tension Of Soap Bubble When Pressure Change Is Given GO
The diameter Of Soap Bubble When Pressure Change Is Given GO
Specific Weight Of A Liquid When Absolute Pressure Of That liquid At A height is Given GO
Height Of Liquid When Absolute Pressure Of That Liquid Is Given GO
Specific Weight Of Fluid 1 When Differential Pressure Between Two Points Is Given GO
Specific Weight Of Fluid 2 When Differential Pressure Between Two Points Is Given GO
Height Of Fluid 1 When Differential Pressure Between Two Points Is Given GO
Height Of Fluid 2 When Differential Pressure Between Two Points Is Given GO
Specific Weight of Inclined Manometer Liquid When Pressure at A Point is Given GO
Length of Inclined Manometer When Pressure at a Point is Given GO
Angle of Inclined Manometer When Pressure at a Point is Given GO
Angle of Inclined Manometer When Sensitivity is Given GO
Specific Weight of Liquid When Total Hydrostatic Force is given GO
Depth of Centroid When Total Hydrostatic Force is Given GO
Area of the Surface Wetted When Total Hydrostatic Force is Given GO
Moment of Inertia about Centroid When Center of Pressure is Given GO
Area of Surface Wetted When Center of Pressure is Given GO
## What is Center of Buoyancy?
The center of buoyancy of a floating body is the point about which all the body parts exactly buoy each other—in other words, the effective center of the displaced water. The metacenter remains directly above the center of buoyancy regardless of the tilt of a floating body, such as a ship.
## How to Calculate Center of Buoyancy?
Center of Buoyancy calculator uses Centre of Buoyancy=Moment of Inertia/(Volume*Centre of gravity)-Metacenter to calculate the Centre of Buoyancy, Center of Buoyancy is the point where if you were to take all of the displaced fluid and hold it by that point it would remain perfectly balanced, assuming you could hold a fluid in a fixed shape. Centre of Buoyancy and is denoted by Bcentre symbol.
How to calculate Center of Buoyancy using this online calculator? To use this online calculator for Center of Buoyancy, enter Volume (V), Moment of Inertia (I), Centre of gravity (Gcentre) and Metacenter (Mcenter) and hit the calculate button. Here is how the Center of Buoyancy calculation can be explained with given input values -> -9.998214 = 1.125/(63*10)-10.
### FAQ
What is Center of Buoyancy?
Center of Buoyancy is the point where if you were to take all of the displaced fluid and hold it by that point it would remain perfectly balanced, assuming you could hold a fluid in a fixed shape and is represented as Bcentre=I/(V*Gcentre)-Mcenter or Centre of Buoyancy=Moment of Inertia/(Volume*Centre of gravity)-Metacenter. Volume is the amount of space that a substance or object occupies or that is enclosed within a container, Moment of Inertia is the measure of the resistance of a body to angular acceleration about a given axis, Centre of gravity of the object is the point through which gravitational force is acting and Metacenter , also spelled metacenter, in fluid mechanics, the theoretical point at which an imaginary vertical line passing through the center of buoyancy and center of gravity intersects the imaginary vertical line through a new center of buoyancy created when the body is displaced, or tipped, in the water.
How to calculate Center of Buoyancy?
Center of Buoyancy is the point where if you were to take all of the displaced fluid and hold it by that point it would remain perfectly balanced, assuming you could hold a fluid in a fixed shape is calculated using Centre of Buoyancy=Moment of Inertia/(Volume*Centre of gravity)-Metacenter. To calculate Center of Buoyancy, you need Volume (V), Moment of Inertia (I), Centre of gravity (Gcentre) and Metacenter (Mcenter). With our tool, you need to enter the respective value for Volume, Moment of Inertia, Centre of gravity and Metacenter and hit the calculate button. You can also select the units (if any) for Input(s) and the Output as well.
Let Others Know | 1,888 | 8,112 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2021-04 | latest | en | 0.704617 |
https://aanchalpatial.medium.com/day-25-of-june-leetcode-challenge-3d0e5d2c6252 | 1,718,921,779,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862006.96/warc/CC-MAIN-20240620204310-20240620234310-00447.warc.gz | 66,158,452 | 24,198 | Day 25 of June LeetCode Challenge
Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one.
Example 1:
`Input: [1,3,4,2,2]Output: 2`
Example 2:
`Input: [3,1,3,4,2]Output: 3`
Note:
1. You must not modify the array (assume the array is read only).
2. You must use only constant, O(1) extra space.
3. Your runtime complexity should be less than O(n2).
4. There is only one duplicate number in the array, but it could be repeated more than once.
--
--
We never really grow up, we only learn how to act in public | 185 | 681 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-26 | latest | en | 0.886839 |
https://www.coursehero.com/file/227774/Web-Answers-8/ | 1,513,089,239,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948517181.32/warc/CC-MAIN-20171212134318-20171212154318-00135.warc.gz | 729,609,093 | 328,909 | # Web_Answers_8 - Z C = 1 · Z = X D = 1 Z = 1 E = X = X F =...
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Unit 8 Problem Solutions 8.1 8.2 (a) (contd) Static 1-hazards are: 1101↔1111 and 0100↔0101 8.2 (a) 15 V Z 0 5 10 20 25 30 35 40 t (ns) Y X W A B C D 00 01 11 10 00 01 11 10 1 1 1 1 1 1 1 1 F = A'C'D' + A C + B C'D A B C D 00 01 11 10 00 01 11 10 0 0 0 0 0 0 0 0 F = (A + C') (A'+ C + D ) (B + C + D') Static 0-hazards are: 0001↔0011 and 1000↔1001 8.2 (b) A B C D 00 01 11 10 00 01 11 10 1 1 1 1 1 1 1 1 F t = A'C'D' + A C + B C'D + A'BC' + ABD 8.2 (c) A B C D 00 01 11 10 00 01 11 10 0 0 0 0 0 0 0 0 F t = (A + C') (A' + C + D) (B + C + D') (A' + B + C) (A + B + D') 8.3 (a) 3 F G 1 2 4 5 6 7 t (ns) E C Glitch (static '1' hazard) A B C D 00 01 11 10 00 01 11 10 1 1 1 1 1 1 G = A'C'D + B C + A'BD Modifed circuit (to avoid hazards) 8.3 (b) A = 1; B =
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Unformatted text preview: Z ; C = 1 · Z = X ; D = 1 + Z = 1; E = X ' = X ; F = 1' = 0; G = X · 0 = 0; H = X + 0 = X See FLD Table 8-1, P. 214 8.4 A = B = 0, C = D = 1 So F = AB'D + BC'D' + BCD = But in the fgure, gate 4 outputs F = 1, indicating something is wrong. For the last NAND gate, F = 0 only when all its inputs are 1. But the output o± gate 3 is 0. There±ore, gate 4 is working properly, but gate 3 is connected incorrectly or is mal±unctioning. 8.5...
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Ask a homework question - tutors are online | 728 | 1,504 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2017-51 | latest | en | 0.691468 |
https://reference.wolfram.com/language/ref/Composition.html | 1,686,441,602,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646652.16/warc/CC-MAIN-20230610233020-20230611023020-00575.warc.gz | 553,497,249 | 13,075 | # Composition
Composition[f1,f2,f3,]
represents a composition of the functions f1, f2, f3, .
# Examples
open allclose all
## Basic Examples(2)
Apply a composition of functions to arguments:
Apply a composition using infix notation:
## Scope(6)
Use pure functions inside Composition:
Use Apply with Composition:
Identity inside a composition is simplified:
InverseFunction inside a composition is simplified if possible:
Create a compound operator that looks for symbols and then multiplies them by 2:
Apply that function to a list:
## Applications(2)
Create a composition of a sequence of functions:
Create a sum of numbers to be displayed in held form:
## Properties & Relations(7)
Composition composes on the left:
RightComposition composes on the right:
Pure functions let you set up objects that work like Composition:
Different ways of entering compositions:
Compose a function with itself times using Nest:
Compose a function with itself times using RSolve:
Compose TransformationFunction objects:
Use ComposeSeries to do composition of series expansions:
## Neat Examples(1)
Tabulate square roots of values without using auxiliary variables:
Wolfram Research (1991), Composition, Wolfram Language function, https://reference.wolfram.com/language/ref/Composition.html (updated 2014).
#### Text
Wolfram Research (1991), Composition, Wolfram Language function, https://reference.wolfram.com/language/ref/Composition.html (updated 2014).
#### CMS
Wolfram Language. 1991. "Composition." Wolfram Language & System Documentation Center. Wolfram Research. Last Modified 2014. https://reference.wolfram.com/language/ref/Composition.html.
#### APA
Wolfram Language. (1991). Composition. Wolfram Language & System Documentation Center. Retrieved from https://reference.wolfram.com/language/ref/Composition.html
#### BibTeX
@misc{reference.wolfram_2022_composition, author="Wolfram Research", title="{Composition}", year="2014", howpublished="\url{https://reference.wolfram.com/language/ref/Composition.html}", note=[Accessed: 10-June-2023 ]}
#### BibLaTeX
@online{reference.wolfram_2022_composition, organization={Wolfram Research}, title={Composition}, year={2014}, url={https://reference.wolfram.com/language/ref/Composition.html}, note=[Accessed: 10-June-2023 ]} | 515 | 2,300 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2023-23 | longest | en | 0.652225 |
https://phys.libretexts.org/Courses/University_of_California_Davis/UCD%3A_Physics_7B_-_General_Physics/7%3A_Momentum | 1,721,416,181,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514917.3/warc/CC-MAIN-20240719170235-20240719200235-00746.warc.gz | 401,749,146 | 29,631 | # 7: Momentum
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
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$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$
( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$
$$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$
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$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$
$$\newcommand{\Span}{\mathrm{span}}$$
$$\newcommand{\id}{\mathrm{id}}$$
$$\newcommand{\Span}{\mathrm{span}}$$
$$\newcommand{\kernel}{\mathrm{null}\,}$$
$$\newcommand{\range}{\mathrm{range}\,}$$
$$\newcommand{\RealPart}{\mathrm{Re}}$$
$$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$
$$\newcommand{\Argument}{\mathrm{Arg}}$$
$$\newcommand{\norm}[1]{\| #1 \|}$$
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The conservation of momentum is a fundamental concept of physics along with the conservation of energy and the conservation of mass. Momentum is defined to be the mass of an object multiplied by the velocity of the object. The conservation of momentum states that the amount of momentum remains constant; momentum is neither created nor destroyed, but only changed through the action of forces as described by Newton's laws of motion. Dealing with momentum is more difficult than dealing with mass and energy because momentum is a vector quantity having both a magnitude and a direction. Momentum is conserved in all three physical directions at the same time.
This page titled 7: Momentum is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Dina Zhabinskaya. | 1,914 | 5,157 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2024-30 | latest | en | 0.193171 |
https://www.physicsforums.com/threads/basic-kinematic-to-find-distance.592505/ | 1,508,636,712,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824931.84/warc/CC-MAIN-20171022003552-20171022023552-00215.warc.gz | 972,799,578 | 15,558 | # Basic Kinematic to Find Distance
1. Apr 1, 2012
### 7at1blow
1. The problem statement, all variables and given/known data
A skier reaches a speed of 56 m/s on a 30° slope w/ no friction. What is the minimum distance the skier would have to travel if starting from rest.
2. Relevant equations
Vf2 = Vi2 +2(a)(Δd)
3. The attempt at a solution
I hope no one will scoff too hard, this is my first physics class. I made up a mass of 10kg and calculated the weight in x direction with Wx = W(sin(30))= 98.1N, normal force in x is zero.
Then I found weight in y direction with Wy = W(cos(30))= -85.0N
From there I used ƩFx=M(ax) and came out with ax=4.91 m/s2 and came to a final solution of ≈320m after filling in the kin. equation.
I spent over an hour on this and feel like I may have just made up nonsense and then "math'd" it. Could someone kindly check my work and give me some guidance?
Thanks. | 263 | 908 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2017-43 | longest | en | 0.943925 |
https://effort.academickids.com/encyclopedia/index.php/Regular_space | 1,632,091,246,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056902.22/warc/CC-MAIN-20210919220343-20210920010343-00647.warc.gz | 290,529,663 | 7,521 | # Regular space
In topology and related fields of mathematics, regular spaces and T3 spaces are particularly nice kinds of topological spaces. Both conditions are examples of separation axioms.
Contents
## Definitions
Suppose that X is a topological space.
X is a regular space iff, given any closed set F and any point x that does not belong to F, there are a neighbourhood U of x and a neighbourhood V of F that are disjoint. In fancier terms, this condition says that x and F can be separated by neighbourhoods.
Missing image
Regular_space.png
The point x, represented by a dot to the left of the picture, and the closed set F, represented by a closed disk to the right of the picture, are separated by their neighbourhoods U and V, represented by larger open disks. The dot x has plenty of room to wiggle around the open disk U, and the closed disk F has plenty of room to wiggle around the open disk V, yet U and V do not touch each other.
X is a T3 space if and only if it is both regular and Hausdorff.
Note that some mathematical literature uses different definitions for the terms "regular" and "T3". The definitions that we have given here are the ones usually used today; however, some authors switch the meanings of the two terms, or use both terms synonymously for only one condition. In Wikipedia, we will use the term "regular" freely, but we'll usually say "regular Hausdorff" instead of the less clear "T3". In other literature, you should take care to find out which definitions the author is using. (The phrase "regular Hausdorff", however, is unambiguous.) For more on this issue, see History of the separation axioms.
## Relationships to other separation axioms
A regular space is necessarily also preregular. Since a Hausdorff space is the same as a preregular T0 space, a regular space that is also T0 must be Hausdorff (and thus T3). In fact, a regular Hausdorff space satisfies the slightly stronger condition T. (However, such a space need not be completely Hausdorff.) Thus, the definition of T3 may cite T0, T1, or T instead of T2 (Hausdorffness); all are equivalent in the context of regular spaces.
Speaking more theoretically, the conditions of regularity and T3-ness are related by Kolmogorov quotients. A space is regular iff its Kolmogorov quotient is T3; and, as mentioned, a space is T3 iff it's both regular and T0. Thus a regular space encountered in practice can usually be assumed to be T3, by replacing the space with its Kolmogorov quotient.
There are many results for topological spaces that hold for both regular and Hausdorff spaces. Most of the time, these results hold for all preregular spaces; they were listed for regular and Hausdorff spaces separately because the idea of preregular spaces came later. On the other hand, those results that are truly about regularity generally don't also apply to nonregular Hausdorff spaces.
There are many situations where another condition of topological spaces (such as normality, paracompactness, or local compactness) will imply regularity if some weaker separation axiom, such as preregularity, is satisfied. Such conditions often come in two versions: a regular version and a Hausdorff version. Although Hausdorff spaces aren't generally regular, a Hausdorff space that is also (say) locally compact will be regular, because any Hausdorff space is preregular. Thus from a certain point of view, regularity is not really the issue here, and we could impose a weaker condition instead to get the same result. However, definitions are usually still phrased in terms of regularity, since this condition is more well known than any weaker one.
Most topological spaces studied in mathematical analysis are regular; in fact, they are usually completely regular, which is a stronger condition. Regular spaces should also be contrasted with normal spaces.
## Examples and nonexamples
As described above, any completely regular space is regular, and any T0 space that is not Hausdorff (and hence not preregular) cannot be regular. Most examples of regular and nonregular spaces studied in mathematics may be found in those two articles. On the other hand, spaces that are regular but not completely regular, or preregular but not regular, are usually constructed only to provide counterexamples to conjectures, showing the boundaries of possible theorems. Of course, one can easily find regular spaces that are not T0, and thus not Hausdorff, such as an indiscrete space, but these examples provide more insight on the T0 axiom than on regularity.
Thus, regular spaces are generally not studied because interesting spaces in mathematics are regular without also satisfying some stronger condition. Instead, they are studied to find properties and theorems, such as the ones below, that are actually applied to completely regular spaces, typically in analysis.
## Elementary properties
Suppose that X is a regular space. Then, given any point x and neighbourhood G of x, there is a closed neighbourhood E of x that is a subset of G. In fancier terms, the closed neighbourhoods of x form a local base at x. In fact, this property characterises regular spaces; if the closed neighbourhoods of each point in a topological space form a local base at that point, then the space must be regular.
Taking the interiors of these closed neighbourhoods, we see that the regular open sets form a base for the open sets of the regular space X. This property is actually weaker than regularity; a topological space whose regular open sets form a base is semiregular.
## Extension by continuity
Suppose that A is a set in a topological space X and f is a continuous function from A to some space Y. Suppose that, whenever a net or filter in A converges to a point in X (say x = limn an), then f(an) converges to a point y in Y. Then we would like to be able to extend the domain of definition of f to the closure of A, by letting f(x) = y, and we would like the extension to be continuous as well.
If Y is a regular space, then this is always possible. If Y is regular Hausdorff, then such a continuous extension will not only exist but will be unique. Note that if A is a dense set, then f will be extended to all of X. This is called extension by continuity, since the extension of f is defined (uniquely, in the Hausdorff case) by the requirement that it be continuous.pl:Przestrzeń regularna
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Topic: nlinfit in STATS toolbox
Replies: 1 Last Post: Dec 9, 1996 11:00 AM
Messages: [ Previous | Next ]
John Boccio Posts: 11 Registered: 12/7/04
nlinfit in STATS toolbox
Posted: Dec 9, 1996 9:19 AM
Hi,
Using the m-file below we are trying to fit data to a simple nonlinear
function. See bottom of m-file for function pretrans.m.
The fit is terrible. I have compared the nlinfit result to a result from
Kaleidagraph (see final plot from m-file).
Anyone know what is happening here? Are we using nlinfit wrong?
Thanks,
John Boccio
boccio@swarthmore.edu
% petfit.m
x = [118.9000
119.0000
119.1000
119.2000
119.3000
119.4000
119.6000
119.8000
120.0000
120.2000
120.6000
121.0000
121.4000
121.8000
122.6000
123.4000
124.2000
125.0000];
y = [238.4000
222.9000
208.6000
197.4000
186.3000
176.3000
160.8000
148.8000
136.9000
128.1000
113.3000
102.3000
93.2000
87.5000
76.6000
68.0000
61.3000
55.0000];
beta0=[388,117]';
best=[387.45,117.24]'
[beta,r,j]=nlinfit(x,y,'pretrans',beta0);
plot(x,y,'ro',x,pretrans(beta,x),'y',x,pretrans(best,x),'c')
legend('DATA','NLINFIT','OTHERFIT')
% function yhat=pretrans(beta,x)
% yhat=beta(1)./(x-beta(2));
Date Subject Author
12/9/96 John Boccio
12/9/96 Matt Silvia | 522 | 1,422 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2018-17 | latest | en | 0.540192 |
https://tsfa.co/help-calculate-derivative-94 | 1,680,039,546,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948871.42/warc/CC-MAIN-20230328201715-20230328231715-00077.warc.gz | 650,512,127 | 6,126 | # How to calculate derivative
These can be very helpful when you're stuck on a problem and don't know How to calculate derivative.
x
## Introduction to Derivatives
Step 1: First of all, write the general expression of the first principle method that is helpful in calculating the derivative of the function. d/dz [f (z)] = limh→0 f (z + h) – f (r) / h. Step
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## 3.2: The Derivative as a Function
The derivative formula is defined for a variable 'x' having an exponent 'n'. The exponent 'n' can be an integer or a rational fraction. Hence, the formula to calculate the derivative is: d dx.xn =
## Derivative Calculator • With Steps!
Put in f (x+Δx) and f (x): x2 + 2x Δx + (Δx)2 − x2 Δx. Simplify (x 2 and −x 2 cancel): 2x Δx + (Δx)2 Δx. Simplify more (divide through by Δx): = 2x + Δx. Then, as Δx heads towards 0 we get: = 2x. Result: the derivative of x2 is 2x. In other
Clarify math problem
The math equation is simple, but it's still confusing.
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## Derivative Calculator
The following syntax demonstrates how to use the deriv command to compute the derivative of an expression. For this, we first have to apply the deriv function to a formula as shown below:
## How do users think about us
Actually it's really a smart app, even though u have to pay for the premium, you don't really have to because you can always wait for the ads, and know the steps of ur answer, like let's be honest its free, waiting isn't a big deal for me, so I would highly recommend this app, you'll like have to wait 2 to 5 minutes to get ads, but it's worth it because all the answers are correct.
Juan Bearden
Also after you scan it shows how to solve the problem! I hate my math teacher and now I get to cheat ;). Also, sometimes they aren't able to solve all problems but that's not too often, and actually i think this is the only app i think i will ever use to actually get solutions for my math problems.
Mario Nutting | 678 | 2,680 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2023-14 | latest | en | 0.92233 |
http://www.mapleprimes.com/tags/engineering?page=2 | 1,498,590,045,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128321536.20/warc/CC-MAIN-20170627185115-20170627205115-00635.warc.gz | 577,652,725 | 30,097 | Momentum Linear and Circular with Maple.
Maple 2015
We find recent applications of the components applied to the linear momentum, circular equations applied to engineering. Just simply replace the vector or scalar fields to thereby reasoning and use the right button.
Momento_Lineal_y_Circular.mw
(in spanish)
Atte.
L.AraujoC.
Energy methods with Maple
Maple 2015
Developed and then implemented with open code components. It is very important to note this post is held for students of civil engineering and mechanics. Using advanced mathematical concepts to concepts in engineering.
Metodos_Energeticos_full.mw
(in spanish)
Atte.
L.Araujo.C
How do I solve a system of non-linear equations in...
Hi,
I'm an industrial engineering student who's running into problems with solving a simple system of non-linear equations.
Why do I get this error when issuing the solve command? I'm pretty sure I'm only passing two lists into the function?
Joshua
Mathematical Modeling with Maple Research
Currently calculations: equations, regression analysis, differential equations, etc; to mention a few of them; are developed using traditional methods ie even are proposed and solved by hand and on paper. In teaching our scientists and engineers use the chalkboard as a way to reach students and enable them to solve their calculation. To what extent Maple contributes to research on new mathematical models applied science and engineering ?. Maplesoft appears as a proposal to resolve problems with our traditional proposed intelligent algorithms, development process, embedded components, and not only them but also generates type applications for Apple ipad tablets signature. Based on the computer algebra system Maple Maplesoft gives us the package which works exactly like we were on our work. I will show how mathematics is developed from a purely basic to reach modeling differential equations applied to education and engineering. Also visualizare current techniques for developing applications for mobile devices.
ECI_2015.pdf
Atte.
Lenin Araujo Castillo
Physics Pure
Computer Science
Solutions of Differential Equations with Embedded...
Maple 18
The Embedded Components are containers that currently use industries for modeling complex systems to find viable solutions in real time and thus avoid huge wait times and overload our computer; by this paper should show you how to implement a dynamic worksheet through Embedded Components in Maple; it goes from finding solutions to ordinary differential equations partial; which interact with the researcher using different parameters.
Using graphical programming will find immediate solutions to selected problems in science and engineering criteria of variability and boundary conditions evolving development with buttons on multiple actions.
cimac_2014.pdf
(in spanish)
Solutions_of_Differential_Equations_with_Embedded_Components.mw
Lenin Araujo Castillo
Physics Pure
Computer Science
How to draw isometric figures and orthographic fig...
Good morning.
I am working on engineering drawing project. I request your kind suggestions for
the above cited subject.
With thanks & Regards
M.Anand
Assistant Professor in Mathematics
SR International Institute of Technology,
How do i specify unit output?...
Hey Forums!
Im using maple 16 for some rather simple Engineering calculatioins. One thing about using maple bothers me, and that is the way my result is displayed.
Here is an example
Here is another
Adding "simplify" or "evalf" sometimes does the trick, but id really like...
Create algebraic function using a numeric dsolve s...
Hello,
for my mechanical engineering studies we have to optimize an engine during a workshop. Therefore we have to use a numerical dsolve to solve the equation of motion. I have to get the maximum bearing reaction for a constant rpm. In order to get this I can't think of anything else but use the dsolve solution and use it in maximize.
ysk_2 := t -> -rc*cos(phi(t))-lp*cos(alpha(t))-lkb-y(t);
ysp_2 := t -> -rc*cos(phi(t))-lp2*cos(alpha(t))-y(t);
Entering reference values in tables...
Hi!
I just started using Maple for my engineering study some weeks ago, i have had luck finding help by google and forums so far. But now i ran into the first problem i can't find any answers for.
I want to add some values in a table like you can in excel. The values have to change like everything else in the document when then references changes and
Extension of getdata
Maple
Starting from Maple 15, the useful ?plottools/getdata command is added. It tansforms a Maple plot to a Matrix. Unfortunately, the getdata command deals only with Maple plots. The question arises: "How to get a data from bmp, jpg, tiff, pcx, gif, png and wmf formats?" This is used in medicine and engineering. Such question was asked here
Displaying calculations...
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Gracias | 1,236 | 6,123 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2017-26 | latest | en | 0.933191 |
https://www.univerkov.com/the-middle-line-of-the-trapezoid-is-13-the-area-is-65-find-the-height-of-the-trapezoid/ | 1,642,818,904,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303729.69/warc/CC-MAIN-20220122012907-20220122042907-00274.warc.gz | 1,025,203,946 | 6,268 | # The middle line of the trapezoid is 13. The area is 65. Find the height of the trapezoid.
The area of the trapezoid is equal to the product of the half-sum of the bases and the height.
S = (a + b) / 2 ∙ h.
The midline of a trapezoid is a line segment connecting the midpoints of the sides of the trapezoid. It is parallel to the bases, and its length is equal to half the sum of the bases. Accordingly, the area of the trapezoid is the product of the midline and the height:
S = m ∙ h, where:
S is the area of the trapezoid;
m is the middle line of the trapezoid;
h – height.
Thus, to calculate the height, you need to divide the area of the trapezoid by the length of its midline:
h = S / m;
h = 65/13 = 5 cm.
Answer: The height of the trapezoid is 5 cm.
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# NCERT Solutions for Class 10 Maths Chapter 3: Pair of Linear Equations in Two Variables - Exercise 3.5
Last updated date: 16th Jul 2024
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## NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.5
Exercise 3.5 Class 10 - Linear Equations in Two Variables - is one of the most important exercises of chapter 3 as far as the NCERT syllabus is concerned. This chapter, if you see the textbook, is all about the equation ax + by + c = 0 where a and b both are not equal to zero and a, b and c are real numbers. In NCERT Solutions for Class 10 Maths Chapter 3 exercise 3.5, you get a mixed type of problems based on this equation. To solve these problems, all you have to do is refer to Vedantu’s solution set for 3.5 Maths class 10. It will enable you to get a good grasp of the concept. Eager to get a glimpse of the solved problems in class 10 maths chapter 3 exercise 3.5 solutions? Students can download NCERT Solutions PDF for Ex 3.5 Class 10 Maths at any time. We not only provide Solutions for maths but also NCERT Solutions for Class 10 Science for free only at Vedantu.
Class: NCERT Solutions for Class 10 Subject: Class 10 Maths Chapter Name: Chapter 3 - Pair of Linear Equations in Two Variables Exercise: Exercise - 3.5 Content-Type: Text, Videos, Images and PDF Format Academic Year: 2024-25 Medium: English and Hindi Available Materials: Chapter WiseExercise Wise Other Materials Important QuestionsRevision Notes
NCERT Solutions for Class 10 Maths Chapter 3 Linear Equations In Two Variables Exercise 3.5 will help students to understand the concepts covered in the chapter in an easy manner. The solutions to all questions given in the exercise are provided in the NCERT Solutions Class 10 Maths Exercise 3.5 PDF. Students can download and refer to this PDF in case of doubts regarding the sums.
NCERT Solutions Class 10 Maths Chapter 3- Pair of Linear Equations in Two Variables is an important chapter for the Class 10 Board exams. The Solutions given in the PDF below will help students to evaluate their preparation before the examination. They can solve the exercises after understanding the chapter thoroughly and verify if their solutions are correct.
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## Access NCERT Solutions for Class 10 Maths Chapter 3 – Pair of Linear Equations in Two Variables
Exercise 3.5
1. Which of the following pairs of linear equations has unique solution, no solution or infinitely many solutions? In case there is a unique solution, find it by using cross multiplication method.
(i) $\mathbf{x} - \mathbf{3y} - \mathbf{3}=\mathbf{0};\text{ }\mathbf{3x}- \mathbf{9y} - \mathbf{2}=\mathbf{0}$
Ans: Calculating $\frac{{{a}_{1}}}{{{a}_{2}}},\frac{{{b}_{1}}}{{{b}_{2}}}$ and $\frac{{{c}_{1}}}{{{c}_{2}}}$,
$\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{1}{3}$
$\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{1}{3}$
$\frac{{{c}_{1}}}{{{c}_{2}}}=\frac{3}{2}$
Hence, $\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}\ne \frac{{{c}_{1}}}{{{c}_{2}}}$.
Therefore there are no solution for the given pair of linear equation as the given lines are parallel to each other and do not intersect.
(ii) $\mathbf{2x}+\mathbf{y}=\mathbf{5};\text{ }\mathbf{3x}+\mathbf{2y}=\mathbf{8}$
Ans: Calculating $\frac{{{a}_{1}}}{{{a}_{2}}},\frac{{{b}_{1}}}{{{b}_{2}}}$ and $\frac{{{c}_{1}}}{{{c}_{2}}}$,
$\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{2}{3}$
$\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{1}{2}$
$\frac{{{c}_{1}}}{{{c}_{2}}}=\frac{-5}{-8}$
Hence, $\frac{{{a}_{1}}}{{{a}_{2}}}\ne \frac{{{b}_{1}}}{{{b}_{2}}}$.
Therefore the given pair of linear equations have a unique solution as the given lines intersect each other at a unique point.
Using cross-multiplication method,
$\frac{x}{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}=\frac{y}{{{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}}}=\frac{1}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}$
$\frac{x}{-8-\left( -10 \right)}=\frac{y}{15+16}=\frac{1}{4-3}$
$\frac{x}{2}=\frac{y}{1}=1$
$x=2,y=1$
Therefore, $x=2$ and $y=1$.
(iii) $\mathbf{3x} - \mathbf{5y}=\mathbf{20};\text{ }\mathbf{6x} - \mathbf{10y}=\mathbf{40}$
Ans: Calculating $\frac{{{a}_{1}}}{{{a}_{2}}},\frac{{{b}_{1}}}{{{b}_{2}}}$ and $\frac{{{c}_{1}}}{{{c}_{2}}}$,
$\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{1}{2}$
$\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{1}{2}$
$\frac{{{c}_{1}}}{{{c}_{2}}}=\frac{1}{2}$
Hence, $\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{{{c}_{1}}}{{{c}_{2}}}$.
Therefore there are infinite number of solution for the given pair of linear equation as the given lines are overlapping each other.
(iv) $\mathbf{x} - \mathbf{3y} - \mathbf{7}=\mathbf{0};\text{ }\mathbf{3x} - \mathbf{3y} - \mathbf{15}=\mathbf{0}$
Ans: Calculating $\frac{{{a}_{1}}}{{{a}_{2}}},\frac{{{b}_{1}}}{{{b}_{2}}}$ and $\frac{{{c}_{1}}}{{{c}_{2}}}$,
$\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{1}{3}$
$\frac{{{b}_{1}}}{{{b}_{2}}}=1$
$\frac{{{c}_{1}}}{{{c}_{2}}}=\frac{-7}{-15}$
Hence, $\frac{{{a}_{1}}}{{{a}_{2}}}\ne \frac{{{b}_{1}}}{{{b}_{2}}}$.
Therefore the given pair of linear equations have a unique solution as the given lines intersect each other at a unique point.
Using cross-multiplication method,
$\frac{x}{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}=\frac{y}{{{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}}}=\frac{1}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}$
$\frac{x}{45-\left( 21 \right)}=\frac{y}{-21-\left( -15 \right)}=\frac{1}{-3+\left( 9 \right)}$
$\frac{x}{24}=\frac{y}{-6}=\frac{1}{6}$
$x=4,y=-1$
Therefore, $x=4$ and $y=-1$.
2. (i) For which values of $a$ and $b$ does the following pair of linear equations have an infinite number of solutions?
$\mathbf{2x}+\mathbf{3y}=\mathbf{7};\text{ }\left( \mathbf{a}\text{ } - \text{ }\mathbf{b} \right)\mathbf{x}+\left( \mathbf{a}\text{ }+\text{ }\mathbf{b} \right)\mathbf{y}=\mathbf{3a}+\mathbf{b} - \mathbf{2}$
Ans: Calculating $\frac{{{a}_{1}}}{{{a}_{2}}},\frac{{{b}_{1}}}{{{b}_{2}}}$ and $\frac{{{c}_{1}}}{{{c}_{2}}}$,
$\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{2}{a-b}$
$\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{3}{a+b}$
$\frac{{{c}_{1}}}{{{c}_{2}}}=\frac{-7}{-\left( 3a+b-2 \right)}$
Condition for infinitely many solutions,
$\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{{{c}_{1}}}{{{c}_{2}}}$
$\frac{2}{a-b}=\frac{7}{\left( 3a+b-2 \right)}$
$6a+2b-4=7a-7b$
$a-9b=-4$ …… (i)
$\frac{2}{a-b}=\frac{3}{a+b}$
$2a+2b=3a-3b$
$a-5b=0$ …… (ii)
Subtracting equation (i) from equation (ii), we get
$4b=4$
$b=1$ …… (iii)
Substituting (iii) in equation (ii), we get
$a-5=0$
$a=5$
Therefore, the given equations will have infinite number of solutions for $a=5$ and $b=1$.
(ii) For which value of $\mathbf{k}$ will the following pair of linear equations have no solution?$\mathbf{3x}+\mathbf{y}=\mathbf{1};\text{ }\left( \mathbf{2k}\text{ } - \text{ }\mathbf{1} \right)\mathbf{x}+\left( \mathbf{k}\text{ } - \text{ }\mathbf{1} \right)\mathbf{y}=\mathbf{2k}+\mathbf{1}$.
Ans: Calculating $\frac{{{a}_{1}}}{{{a}_{2}}},\frac{{{b}_{1}}}{{{b}_{2}}}$ and $\frac{{{c}_{1}}}{{{c}_{2}}}$,
$\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{3}{2k-1}$
$\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{1}{k-1}$
$\frac{{{c}_{1}}}{{{c}_{2}}}=\frac{1}{2k+1}$
Condition for no solution,
$\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}\ne \frac{{{c}_{1}}}{{{c}_{2}}}$
$\frac{3}{2k-1}=\frac{1}{k-1}\ne \frac{1}{2k+1}$
$\frac{3}{2k-1}=\frac{1}{k-1}$
$3k-3=2k-1$
$k=2$
Therefore, the given equations will not have any solutions for $k=2$.
3. Solve the following pair of linear equations by the substitution and cross multiplication methods:
$\mathbf{8x}+\mathbf{5y}=\mathbf{9};\text{ }\mathbf{3x}+\mathbf{2y}=\mathbf{4}$
Ans :
Using substitution method,
$8x+5y=9$ …… (i)
$3x+2y=4$ …… (ii)
From equation (ii) we get
$x=\frac{4-2y}{3}$ …… (iii)
Substituting (iii) in equation (i), we get
$8\left( \frac{4-2y}{3} \right)+5y=9$
$32-16y+15y=27$
$y=5$ …… (iv)
Substituting (iv) in equation (ii), we obtain
$3x+10=4$
$x=-2$
Therefore, $x=-2$ and $y=5$.
Using cross multiplication method:
$8x+5y=9$
$3x+2y=4$
$\frac{x}{-20-\left( -18 \right)}=\frac{y}{-27-\left( -32 \right)}=\frac{1}{16-15}$
$\frac{x}{-2}=\frac{y}{5}=1$
$x=-2,y=5$
Therefore, $x=-2$ and $y=5$.
4. Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method:
(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student $\mathbf{A}$ takes food for $\mathbf{20}$ days she has to pay $\mathbf{Rs}\text{ }\mathbf{1000}$ as hostel charges whereas a student $\mathbf{B}$, who takes food for $\mathbf{26}$ days, pays $\mathbf{Rs}\text{ }\mathbf{1180}$ as hostel charges. Find the fixed charges and the cost of food per day.
Ans: Assuming that the fixed charge is $Rs\text{ }x$ and the charge for food per day is $Rs\text{ y}$.
Writing the algebraic representation using the information given in the question:
$x+20y=1000$ …… (i)
$x+26y=1180$ …… (ii)
Subtracting equation (i) from equation (ii), we obtain
$6y=180$
$y=30$ …… (iii)
Substituting (iii) in equation (i), we obtain
$x+20\left( 30 \right)=1000$
$x=400$ …… (iv)
Therefore, $x=400$ and $y=30$.
Hence, fixed charges are $Rs\text{ 400}$ and charges per day are $Rs\text{ 30}$.
(ii) A fraction becomes $\frac{1}{3}$ when $\mathbf{1}$ is subtracted from the numerator and it becomes $\frac{1}{4}$ when $\mathbf{8}$ is added to its denominator. Find the fraction.
Ans: Assuming that the fraction is $\frac{x}{y}$.
Writing the algebraic representation using the information given in the question:
$\frac{x-1}{y}=\frac{1}{3}$
$3x-y=3$ …… (i)
$\frac{x}{y+8}=\frac{1}{4}$
$4x-y=8$ …… (ii)
Subtracting equation (i) from equation (ii), we obtain
$x=5$ …… (iii)
Substituting (iii) in equation (i), we obtain
$15-y=3$
$y=12$ …… (iv)
Therefore, $x=5$ and $y=12$.
Hence, the fraction is $\frac{5}{12}$.
(iii) Yash scored $\mathbf{40}$ marks in a test, getting $\mathbf{3}$ marks for each right answer and losing $\mathbf{1}$ mark for each wrong answer. Had $\mathbf{4}$ marks been awarded for each correct answer and $\mathbf{2}$ marks been deducted for each incorrect answer, then Yash would have scored $\mathbf{50}$ marks. How many questions were there in the test?
Ans: Say the number of right answers be $\text{ }x$ and the number of wrong answers be $\text{ y}$.
Writing the algebraic representation using the information given in the question:
$3x-y=40$ …… (i)
$2x-y=25$ …… (ii)
Subtracting equation (ii) from equation (i), we obtain
$x=15$ …… (iii)
Substituting (iii) in equation (ii), we obtain
$y=5$ …… (iv)
Therefore, $x=15$ and $y=5$.
Hence, number of right answers = $\text{15}$ and number of wrong answers = $\text{ 5}$.
Therefore, the total number of questions = $\text{20}$
(iv) Places A and B are $\mathbf{100}\text{ }\mathbf{km}$ apart on a highway. One car starts from $\mathbf{A}$ and another from $\mathbf{B}$ at the same time. If the cars travel in the same direction at different speeds, they meet in $\mathbf{5}$ hours. If they travel towards each other, they meet in $\mathbf{1}$ hour. What are the speeds of the two cars?
Ans: Assuming that the speed of first car is $u\text{ }km/h$ and the speed of second car is $\text{v }km/h$.
Relative speed when both cars are travelling in same direction $\text{=}\left( u-v \right)\text{ }km/h$
Relative speed when both cars are travelling in opposite direction $\text{=}\left( u+v \right)\text{ }km/h$
Writing the algebraic representation using the information given in the question:
$5\left( u-v \right)=100$
$u-v=20$ …… (i)
$1\left( u+v \right)=100$ …… (ii)
Adding equation (i) and (ii), we get
$2u=120$
$u=60$ …… (iii)
Substituting (iii) in equation (ii), we obtain
$v=40$
Hence, speed of one car is $\text{60 }km/h$ and speed of the other car is $\text{40 }km/h$.
(v) The area of a rectangle gets reduced by $\mathbf{9}$ square units, if its length is reduced by $\mathbf{5}$ units and breadth is increased by $\mathbf{3}$ units. If we increase the length by $\mathbf{3}$ units and the breadth by $\mathbf{2}$ units, the area increases by $\mathbf{67}$ square units. Find the dimensions of the rectangle.
Ans: Assuming length of the rectangle be $x$ and the breadth be $\text{y}$.
Writing the algebraic representation using the information given in the question:
$\left( x-5 \right)\left( y+3 \right)=xy-9$
$3x-5y-6=0$ …… (i)
$\left( x+3 \right)\left( y+2 \right)=xy+67$ …… (ii)
Using cross multiplication method:
$\frac{x}{305-\left( -18 \right)}=\frac{y}{-12-\left( -183 \right)}=\frac{1}{9-\left( -10 \right)}$
$\frac{x}{323}=\frac{y}{171}=\frac{1}{19}$
$x=17,y=9$
Hence, the rectangle has length of 17 units and breadth of 9.
## Class 10 Chapter 3 Exercise 3.5 Solution: A Brief Overview
Exercise 3.5 class 10 Maths consists of various questions related to Linear Equations in Two Variables. In this section, you will get familiar with the few questions related to the above-mentioned topic. Rigorous practising of class 10 Maths ex. 3.5 will definitely help you to understand the entire chapter. This will also increase your problem-solving ability along with coping up with the latest CBSE question pattern. Ex. 3.5 class 10 Maths mainly deals with the following types of questions:
Question 1: Which of the following pairs of linear equations consists of a unique solution, without solution or infinitely numerous solutions? Derive the solution in case of a unique solution applying a cross multiplication method.
i) x – 3y – 3= 0 and 3x – 9y - 2 = 0 ii) 2x+ y = 5 and 3x + 2y = 8 iii) x – 3y – 7= 0 and 3x – 3y -15 = 0 iv) 3x – 5y = 20 and 6x – 10y = 40.
Solution:
i) Given, x -3y -3 = 0 and 3x -9y – 2= 0.
a1/a2 = 1/3, b1/b2 = -3/-9 = 1/3, c1/c2 = -3/-2 = 3/2
(a1/a2) = (b1/b2) ≠ (c1/c2)
There is no solution for the above equation as the said pair of lines are parallel to each other and there will not be any intersection between them.
Here are a few more solved problems from exercise 3.5 Maths class 10.
ii) Given, 2x + y = 5 and 3x + 2y = 8
a1/a2 = 2/3, b1/b2 = 1/2 and c1/c2 = -5/-8
(a1/a2) ≠ (b1/b2)
These equations should have a unique solution by applying cross multiplication technique as there is an intersection between them at a unique point.
x/(b1c2 – c1b2) = y/(c1a2 – c2a1) = 1/ (a1b2 – a2b1)
x/(-8-(-10)) = y/(15+16) = 1/(4-3)
x/2 = y/1 = 1
Hence, x = 2 and y = 1
iii) Given, x - 3y - 7 = 0 and 3x - 3y - 15 = 0
(a1/a2) = 1/3
(b1/b2) = -3/-3 = 1
(c1/c2) = -7/-15
(a1/a2) ≠ (b1/b2
There will definitely be a unique solution as the set of lines are intersecting each other at a unique point.
Applying cross multiplication method we get,
x/(45-21) = y /(-21+ 15) = 1/(-3 + 9)
x/24 = y/-6 = 1/6
x/24 = 1/6 and y/-6 = 1/6
Hence, x = 4 and y = 1
iv) Given, 3x - 5y = 20 and 6x – 10y = 40
(a1/a2) = 3/6 = 1/2
(b1/b2) = -5/-10 = 1/2
(c1/c2) = 20/40 = 1/2
(a1/a2) = (b1/b2) = (c1/c2)
There will be an infinite number of solutions as the given pair of lines is overlapping each other.
### What is a Cross Multiplication Method To Solve Linear Equations in One Variable?
In Maths, a cross multiplication method that solves linear equations in two variables is the easiest method and gives the accurate value of the variables.
Let us consider, we have a1x1 + b1y1 + c1 = 0 and a2x + b2x + cz = 0 are the two equations which have to be solved. By using cross multiplication, we will get the values x and y such as:
$X= \frac{b_1c_2-b_2c_1}{b_2a_1-b_1a_2}$
$Y= \frac{c_1a_2-c_2a_1}{b_2a_1-b_1a_2}$
Where,
b2a1- ba2 is not equal to 0.
### Why Should You Choose Vedantu?
If you are looking for the solutions of Ex. 3.5 class 10 Maths Vedantu will be the best option for you. There are various reasons in support of the above statement.
• Vedantu provides the exercise 3.5 class 10 Maths NCERT solutions which are entirely exam-oriented and based on the latest CBSE guidelines.
• The class 10 Maths ch 3 ex. 3.5 is designed by the pool of experienced teachers of Vedantu.
• You can easily access the ex. 3.5 class 10 Maths solutions from the official website of Vedantu.
• Class 10 Maths exercise 3.5 solutions are given in PDF format and are completely free of cost.
• In the new era of digital learning, Vedantu will assist you till the fag end of the exam schedule.
• The questions of Maths class 10 chapter 3 exercise 3.5 are architected from moderate to difficult problems so that students can get familiar with the exam pattern gradually.
Not only NCERT class 10 maths chapter 3 exercise 3.5, but all the study materials of Vedantu are going to be equally helpful for you in your journey towards the CBSE board exam.
## FAQs on NCERT Solutions for Class 10 Maths Chapter 3: Pair of Linear Equations in Two Variables - Exercise 3.5
1. What is the Procedure of Accessing Class 10 Chapter 3 Exercise 3.5 Solutions?
If you want to access the class 10th maths chapter 3 exercise 3.5 you have to follow the following steps:
1. Go to the official site of Vedantu and hit the search bar.
2. Type the words NCERT solutions class 10 maths chapter 3 exercise 3.5.
3. You will be redirected to the page containing NCERT solutions for class 10 maths chapter 3.5.
4. You can either practise Maths class 10 ex 3.5 online or download it for future reference.
5. Solve the ex 3.5 class 10 maths on your own first and then, refer to the solutions.
Ex 3.5 class 10 maths solutions are not the only study materials. You get all the chapter-wise solutions available on Vedantu’s website.
2. How Helpful are Maths PDFs like Class 10 Maths Ex 3.5 Solutions?
Class 10 Maths ch 3 ex 3.5 solutions are just one of the many study materials that are considered as the Bible of CBSE Mathematics. The questions covered in class 10 Maths Chapter 3 ex 3.5 and are completely designed as per the latest CBSE guidelines. If you practise the NCERT class 10 Maths 3.5 rigorously you will definitely score high marks in this portion of CBSE Maths paper.
3. What are the different processes of solving a pair of linear equations in two variables?
Any pair of linear equations with two variables can be solved by the following processes:
Graphical process: A graph of the pair of linear equations with two variables can be presented with two lines on the graphical plane.
Algebraic processes: There are three methods to find the solution of any given pair of linear equations:
• Substitution
• Elimination
• Cross-multiplication
4. What is the underlying concept of Chapter 3 of Class 10 Maths?
Students learn about linear equations with two variables in Chapter 3 of Class 10 Maths. Any equation may be written as ax + by + c = 0.
When a, b and c have to be real numbers, with a and b being non zero, it is referred to as a linear equation with two variables, i.e, x and y. The solution to these kinds of equations is two values, one to replace x and another to replace y. Students will learn that all solutions to the equation are points on the representing line.
5. What are the topics and sub-topics included in Chapter 3 of Class 10 Maths?
Chapter 3 of Class 10 Maths is important to understand future concepts easily. The following are the topics and sub-topics of this Chapter:
• Pair of Linear Equations with Two Variables
• Solving a Pair of Linear Equations with Graphical method
• Substitution
• Elimination
• Cross-Multiplication
• Reducing equations to a Pair of Linear Equations with Two Variables
6. What are the most important formulas that come in Chapter 3 of Class 10 Maths?
If
a1 + b1y + c1 = 0
a2 + b2y + c2 = 0,
the kind of roots or the solutions can be determined by:
If a1/a2 ≠ b1/b2
• Unique solution
• Intersecting lines
If a1/a2 ≠ b1/b2 ≠ c1/c2
• No solution
• Parallel lines
If a1/a2 = b1/b2 = c1/c2
• Infinite solutions
• Coincident lines
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StatsCasts are narrated screen video recordings of explanations of statistical concepts. They are produced by Swinburne University of Technology, the University of the Sunshine Coast, and the University of Southern Queensland. They are part of an ongoing collaborative research project to develop high quality resources and investigate the effectiveness of StatsCasts to support statistics learning. They are targeted at prerequisite to first year level, in a range of subjects such as: Engineering, Sciences, Health Sciences and Business.
# StatsCasts Swinburne University of Technology
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StatsCasts are narrated screen video recordings of explanations of statistical concepts. They are produced by Swinburne University of Technology, the University of the Sunshine Coast, and the University of Southern Queensland. They are part of an ongoing collaborative research project to develop high quality resources and investigate the effectiveness of StatsCasts to support statistics learning. They are targeted at prerequisite to first year level, in a range of subjects such as: Engineering, Sciences, Health Sciences and Business.
• video
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In this video we look at an example of conducting a Chi-square test. We set up the null and alternative hypotheses, determine the significance level, calculate expected frequencies and the Chi-square statistic and use statistical tables to determine the critical value of the Chi-square statistic. We then use this information to make a decision about H0 and to write a conclusion.
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In this video we look at how to use statistical tables to calculate probabilities in a Poisson distribution. This includes an example of using the table for the probability density function to determine the probability the random variable is equal to particular value in a case where the average number of events per interval needs to be adjusted to match the units specified in the question and an example of using the table for the cumulative distribution function to determine the probability the random variable takes a value between two specified numbers.
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## Calculation of probabilities in a Poisson distribution, using tables - Part 1 (StatsCasts)
In this video we look at how to use statistical tables to calculate probabilities in a Poisson distribution. This includes an example of using the table for the probability density function to determine the probability the random variable is equal to a particular value and an example of using the table for the cumulative distribution function to determine the probability the random variable is less than a certain value and an example determining the probability it is greater than or equal to a certain value.
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• video
Calculation of probabilities in a Binomial distribution, using tables (StatsCasts)
## Calculation of probabilities in a Binomial distribution, using tables (StatsCasts)
In this video we look at how to use statistical tables to calculate probabilities in a Binomial distribution. This includes an example of using the table for the probability density function to determine the probability the random variable takes a particular value and an example of using the table for the cumulative distribution function to determine the probability the random variable is less than or equal to a certain value and an example determining the probability it is greater than or equal to a certain value.
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Deciding if a distribution is Binomial or Poisson (StatsCasts)
## Deciding if a distribution is Binomial or Poisson (StatsCasts)
In this video we look at how to decide for a given scenario (worded problem) if the distribution described is a Binomial distribution or Poisson distribution and whether its probability distribution function or its cumulative distribution function is required to calculate a specified probability.
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## Background to test statistic and p value in z test (StatsCasts)
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# Statistically Speaking - QP's and PP's
Topic closed. 1161 replies. Last post 6 years ago by Todd.
Page 35 of 78
Kentucky
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Posted: July 26, 2010, 9:36 pm - IP Logged
I don't want to start another brouhaha but here's what I've gotten out of this:
PP's are capable of outperforming QP's by having all the numbers in play (even though as shown, that edge still does not guarantee it).
But having that many lines in play is going to cost a fortune with regular play (a lot more than most people want to spend, anyway).
So while in theory and while playing on paper it can have it's benefits, it would seem impractical from a sound financial prospective to go that route.
If PP players are spending \$46 per drawing or even boiling it down to \$20 per drawing to win \$7 or \$10 or whatever and I can't beat them with \$2 per drawing worth of QP's it really doesn't bother me.
And maybe that's what the difference is: The QP players are only buying a couple bucks worth (or \$5 or \$10) and on that basis alone the PP players can't compete because at that level they can't get enough numbers in play.
Which would then return them to the Wheelhouse of Lady Luck.
Disclaimer: I am not an advocate or proponent of either method. I play both ways.
"So while in theory and while playing on paper it can have it's benefits, it would seem impractical from a sound financial prospective to go that route "
The 70% to 80% QP purchases and winners is a mathematical probability, Ridge. A QP player has to buy on average 75 QPs before they can expect to get all 46 bonus balls. A PP player would have to purchase at least 12 PP before they can have all 56 numbers. It's reasonable to assume the average player purchases on average \$10 or less so we won't see very many QP players buying 75 QPs. It's more likely to see a systems player buying 12 PPs, but for every \$12 spent on PPs, \$36 will be spent on QPs.
"PP's are capable of outperforming QP's by having all the numbers in play (even though as shown, that edge still does not guarantee it)."
The 46 combo wheel is a mathematical probability too, but there is no guarantee it can out perform Tigg's \$1 QP. The one thing both QP and PP players have in common is both groups are trying to win millions so while getting back a few bucks helps, it's not exactly why they are playing.
light on my feet
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Posted: July 26, 2010, 9:37 pm - IP Logged
07-14-21-36-54 + 07
1 Quick Pick for Powerball
Generated by Lottery Post's exclusive Quick Picks Generator
http://www.lotterypost.com/quickpicks
for all you excuse makers, that's 46 QUICK PICKS from the LP generator.
now what OTHER excuse(s) will you come up with ???
"i am .........."meant to"
P.S., that RJoH is a stand up guy. thanks, vision
until further notice, it's france everyday
Way back up in them dadgum hills, son!
United States
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14903 Posts
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Posted: July 26, 2010, 9:57 pm - IP Logged
07-14-21-36-54 + 07
1 Quick Pick for Powerball
Generated by Lottery Post's exclusive Quick Picks Generator
http://www.lotterypost.com/quickpicks
for all you excuse makers, that's 46 QUICK PICKS from the LP generator.
now what OTHER excuse(s) will you come up with ???
Dude, lol, do you know how many of your detractors are going to go out and play that same line just to make sure you can't claim the jackpot alone, just in case?
I know that kinda stuff happens cuz tiggs does it all the time to everybody. He's probably at the store getting the ticket right now.
No wait, it's almost 10:00, he's probably strapped down for the night already.
"The only thing necessary for evil to triumph is for good men to do nothing"
--Edmund Burke
light on my feet
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May 20, 2002
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Posted: July 26, 2010, 11:10 pm - IP Logged
Dude, lol, do you know how many of your detractors are going to go out and play that same line just to make sure you can't claim the jackpot alone, just in case?
I know that kinda stuff happens cuz tiggs does it all the time to everybody. He's probably at the store getting the ticket right now.
No wait, it's almost 10:00, he's probably strapped down for the night already.
even iffin i wanted to, i couldn't but a ticket for powerball, because they only sell mega da millions in this nutty state.
what's completely hysterical, is the "others" kept saying i had to purchase the tickets in order to get "46 combo's" of QP's, otherwise it would technically be a "PP".
i only wish i knew what their first thought was when "they" saw my 46 QP combo accomplishment.
i suspect a change of underwear for each of them. probably twice for stack.
ha ha.
anyway, "they" are putting their collective excuse super colliders together thru PM's as we speak.
who knew that one guy, armed with only a high school edumacation, would draw a demarcation factual line in the LP sand.
tiggs, go get em bud. take them random numbers, and ride off into LP history.
you have my blessing.
btw, now that these people have witnessed my "talents" at keepings things honest, since i know peoples behaviors so well, i expect less and less "conversation" from "them".
why? because "they" know i am one that's going to make sure they can prove.
the fact that even less people will be lining up, will serve to prove i was right.
see how fast i busted out those 46 combo's? it came to me at work, and when i got home, i busted those out toot suite.
that's how people act when they "have something".
they don't talk about it, they throw it up ASAP
"i am .........."meant to"
P.S., that RJoH is a stand up guy. thanks, vision
until further notice, it's france everyday
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Posted: July 26, 2010, 11:36 pm - IP Logged
What I would like to see is dollar-for-dollar tests. Put it up, or shut it up. Post the picks PICS and let’s see how you do. Give 20 events to rule the point. Then let's look at the results.
DD
light on my feet
United States
Member #356
May 20, 2002
2744 Posts
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Posted: July 27, 2010, 1:29 am - IP Logged
What I would like to see is dollar-for-dollar tests. Put it up, or shut it up. Post the picks PICS and let’s see how you do. Give 20 events to rule the point. Then let's look at the results.
DD
not a problem
you can ask stack if he will do it with you. lol
(stack is the one that keeps insisting it will take \$\$\$\$\$\$\$ in order to "make it work")
i do like that "put up, or shut" up thingy you mentioned. (it always interested me that people who say that, always seem to push people ahead of them in the line......."hey, you go do it")
despite bravado ad hominem continuum run amokitis you should you should you should you should, only one person ever has.
it took me probably 12.3 secs to gather all my QP's up.
all 46 of them
excuses are like systems. everyone supposedly has one
"i am .........."meant to"
P.S., that RJoH is a stand up guy. thanks, vision
until further notice, it's france everyday
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Posted: July 27, 2010, 1:30 am - IP Logged
Number 10,000.
Yahoo!
DD
mid-Ohio
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Posted: July 27, 2010, 9:04 am - IP Logged
What I would like to see is dollar-for-dollar tests. Put it up, or shut it up. Post the picks PICS and let’s see how you do. Give 20 events to rule the point. Then let's look at the results.
DD
As a premium member you have access to the complete history of every lottery in the country, that's more than enough resources to run any kind of test you want as many times as you want.
Some members are already posting as many or more PB and MM combinations on the prediction board every drawing and their results are available for every one to see.
* you don't need to buy more tickets, just buy a winning ticket *
Harbinger
D.C./MD.
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July 30, 2006
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Posted: July 27, 2010, 9:39 am - IP Logged
not a problem
you can ask stack if he will do it with you. lol
(stack is the one that keeps insisting it will take \$\$\$\$\$\$\$ in order to "make it work")
i do like that "put up, or shut" up thingy you mentioned. (it always interested me that people who say that, always seem to push people ahead of them in the line......."hey, you go do it")
despite bravado ad hominem continuum run amokitis you should you should you should you should, only one person ever has.
it took me probably 12.3 secs to gather all my QP's up.
all 46 of them
excuses are like systems. everyone supposedly has one
We tried something similar a couple years back, I am not fond of the DC lotttery because it is grade F RNG.
http://www.lotterypost.com/topic/160905
As you can see only one member put up. This isn't the only time either.
New Member
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Posted: July 27, 2010, 9:49 am - IP Logged
What I would like to see is dollar-for-dollar tests. Put it up, or shut it up. Post the picks PICS and let’s see how you do. Give 20 events to rule the point. Then let's look at the results.
DD
Areyou people implying that the QPs aren't completely random? Why shouldvision have to put up money? A random number is a random number. Itshouldn't matter which machine picked it. There should be no need tobuy that many QPs to test this.
Harbinger
D.C./MD.
United States
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5583 Posts
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Posted: July 27, 2010, 9:58 am - IP Logged
Areyou people implying that the QPs aren't completely random? Why shouldvision have to put up money? A random number is a random number. Itshouldn't matter which machine picked it. There should be no need tobuy that many QPs to test this.
Welcome to Lottery Post! I like your member name, right off the periodic table. Si silicon atomic #14 Bi bismuth atomic #83, todays lucky P4 1483! Am I close? Sunset Intl. Bible Institute? Urdu?
Kentucky
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Posted: July 27, 2010, 12:19 pm - IP Logged
What I would like to see is dollar-for-dollar tests. Put it up, or shut it up. Post the picks PICS and let’s see how you do. Give 20 events to rule the point. Then let's look at the results.
DD
If the test was based on coverage and we tested RJ's 20 PPs over the next 10,000 drawings it would show those 20 combos having 5 wbs in every drawing because RJ used all the wbs. We already know Dude's 20 RNG picks failed to match all 5 wbs in the first test. You can't hit the jackpot without matching all 5 wbs.
Since the topic is about 70% to 80% of purchased QPs winning 70% to 80% of all the prizes, the only logical way to test PPs vs QPs would to purchase the QPs. There is no logical reason to test a wheel using all 56 wbs and all 46 bonus for coverage and because Dude's 20 combo RNG included the same bonus ball multiple times, we can conclude 46 purchased QPs would do the same.
As for testing performance if no tickets are purchased, the results are useless and I've tested purchasing 20 QPs many times and the only reason I got anything back was because I got lucky. The odds against us are terrible, but we keep trying because someone is going to get very lucky and it might be one of us.
Patrick mentioned Maddog's Challenges are the real tests and I agree, but I don't play them because if I happened to match all 5 numbers plus the bonus ball and not buy the tickets, there isn't a building tall enough in the world for me to jump off. And for that very reason, I'm not going to waste my time playing meaningless games with Visiondude or anybody else.
light on my feet
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Posted: July 27, 2010, 1:51 pm - IP Logged
If the test was based on coverage and we tested RJ's 20 PPs over the next 10,000 drawings it would show those 20 combos having 5 wbs in every drawing because RJ used all the wbs. We already know Dude's 20 RNG picks failed to match all 5 wbs in the first test. You can't hit the jackpot without matching all 5 wbs.
Since the topic is about 70% to 80% of purchased QPs winning 70% to 80% of all the prizes, the only logical way to test PPs vs QPs would to purchase the QPs. There is no logical reason to test a wheel using all 56 wbs and all 46 bonus for coverage and because Dude's 20 combo RNG included the same bonus ball multiple times, we can conclude 46 purchased QPs would do the same.
As for testing performance if no tickets are purchased, the results are useless and I've tested purchasing 20 QPs many times and the only reason I got anything back was because I got lucky. The odds against us are terrible, but we keep trying because someone is going to get very lucky and it might be one of us.
Patrick mentioned Maddog's Challenges are the real tests and I agree, but I don't play them because if I happened to match all 5 numbers plus the bonus ball and not buy the tickets, there isn't a building tall enough in the world for me to jump off. And for that very reason, I'm not going to waste my time playing meaningless games with Visiondude or anybody else.
"There is no logical reason to test a wheel using all 56 wbs and all 46 bonus for coverage
what? first the guy says for several posts that the only way is to do it that way. now, because he doesn't want to do it, he has verbage interruptus majorus
why do people put themselves thru integrity schizophrenia needlessly?
Patrick mentioned Maddog's Challenges are the real tests and I agree
i bet you do, stack. "hey, you go do it, because i refuse". lol
As for testing performance if no tickets are purchased, the results are useless
that's a complete crock of an excuse, and now everyone knows it, stack.
the "picks" have no way of knowing if there is a pile of cash behind them or not.
yours obviously wouldn't, because you don't even have any. LOL
"And for that very reason, I'm not going to waste my time playing meaningless games with Visiondude or anybody else".
you never "were" going to do anything anyway, other than self aggrandizement postulation nation.
so other than reading your own "hyperBOIL", if i were you, i wouldn't either.
it's really sad, that you are afraid of the big bad LP random generator. lol
this is a test of the emergency LP broadcast station. this is only a test
"i am .........."meant to"
P.S., that RJoH is a stand up guy. thanks, vision
until further notice, it's france everyday
mid-Ohio
United States
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March 24, 2001
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Posted: July 27, 2010, 2:06 pm - IP Logged
If the test was based on coverage and we tested RJ's 20 PPs over the next 10,000 drawings it would show those 20 combos having 5 wbs in every drawing because RJ used all the wbs. We already know Dude's 20 RNG picks failed to match all 5 wbs in the first test. You can't hit the jackpot without matching all 5 wbs.
Since the topic is about 70% to 80% of purchased QPs winning 70% to 80% of all the prizes, the only logical way to test PPs vs QPs would to purchase the QPs. There is no logical reason to test a wheel using all 56 wbs and all 46 bonus for coverage and because Dude's 20 combo RNG included the same bonus ball multiple times, we can conclude 46 purchased QPs would do the same.
As for testing performance if no tickets are purchased, the results are useless and I've tested purchasing 20 QPs many times and the only reason I got anything back was because I got lucky. The odds against us are terrible, but we keep trying because someone is going to get very lucky and it might be one of us.
Patrick mentioned Maddog's Challenges are the real tests and I agree, but I don't play them because if I happened to match all 5 numbers plus the bonus ball and not buy the tickets, there isn't a building tall enough in the world for me to jump off. And for that very reason, I'm not going to waste my time playing meaningless games with Visiondude or anybody else.
"it would show those 20 combos having 5 wbs in every drawing because RJ used all the wbs"
Correction: I made an effort to cover all 59 WBs but could only cover 54 of them with the parameters I used.
The distribution of the WBs would based on the last 40 drawings because that were the number of previous drawings that covered all 59 WB. I had 20 lines because I had 20 MB. I used 20 MB because they covered the most popular MB repeat positions and the the top most popular MBs. That's how my system/strategy works.
For Wednesday drawing the distribution of the WBs will be based on the last 41 drawings because that's the number of previous drawings that covers all 59 WB. I will play 18 lines because I have 18 MB. I have 18 MB because they cover the most popular MB repeat positions and the the top most popular MBs.
Each time I play I pick a unique group of combinations based on the most recent history of a game so comparing my picks to the next 100 drawings or more wouldn't be a fair evaluation of my system/strategy since I do a new workup each time.
Even though Visiondude is only running a simulation of playing a game where he spends less than 20 seconds coming up with his QP combinations using the LP QP generator, the system players will have to spend the same amount of time picking combinations as if they were really playing which may be a lot of time they consider wasted if they aren't really playing as you point out. Since I am actually playing or posting on the prediction board those combinations, it isn't that much extra time for me.
* you don't need to buy more tickets, just buy a winning ticket *
light on my feet
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May 20, 2002
2744 Posts
Offline
Posted: July 27, 2010, 2:07 pm - IP Logged
We tried something similar a couple years back, I am not fond of the DC lotttery because it is grade F RNG.
http://www.lotterypost.com/topic/160905
As you can see only one member put up. This isn't the only time either.
thanks jarasan.
now, after reading about the LP generator, i now know why stack is shirking his responsibility.
the LP generator is the king of "generators" technology wise (evidently), and he most likely knows it. ha ha
a "cryptographically strong random numbers" generator.
man, that sounds like the obi wan kanobi of random generators.
plus, obviously "the force" is with me, so there's that. lol
shoot, i would even let stack steer me to the minor leagues of random generators, so i could give him another "edge"
one thing i don't want on my conscience, is that guy jumping off a tall building because of me.
.......it's just a game
"i am .........."meant to"
P.S., that RJoH is a stand up guy. thanks, vision
until further notice, it's france everyday
Page 35 of 78 | 5,162 | 19,090 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2016-50 | latest | en | 0.962216 |
https://ru-facts.com/how-do-you-round-to-even-numbers/ | 1,709,587,952,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476532.70/warc/CC-MAIN-20240304200958-20240304230958-00249.warc.gz | 507,697,079 | 12,113 | How do you round to even numbers?
How do you round to even numbers?
The round-to-even method works like this:
1. If the difference between the number and the nearest integer is less than 0.5, round to the nearest integer. This familiar rule is used by many rounding methods.
2. If the difference between the number and the nearest integer is exactly 0.5, look at the integer part of the number.
How do you round a number in C++?
round() in C++ round is used to round off the given digit which can be in float or double. It returns the nearest integral value to provided parameter in round function, with halfway cases rounded away from zero. Instead of round(), std::round() can also be used .
What is the round function in C++?
The round() function in C++ returns the integral value that is nearest to the argument, with halfway cases rounded away from zero. It is defined in the cmath header file.
What is half even rounding?
In the case of the round-half-even algorithm (which is often referred to as Banker’s Rounding because it is commonly used in financial calculations), half-way values are rounded toward the nearest even number. Thus, 3.5 will round up to 4 and 4.5 will round down to 4.
What is the even odd rounding rule?
The intention of the odd-even rule is to average the effects of rounding off. Odds and evens rule. EVEN: If the last retained digit is even, its value is not changed, the 5 and any zeros that follow are dropped. ODD: if the last digit is odd, its value is increased by one.
What is fixed in C++?
The fixed() method of stream manipulators in C++ is used to set the floatfield format flag for the specified str stream. This flag sets the floatfield to fixed. It means that the floating-point values will be written in fixed point notations.
How do you write Setprecision in C++?
Let’s see the simple example to demonstrate the use of setprecision:
1. #include // std::cout, std::fixed.
2. #include // std::setprecision.
3. using namespace std;
4. int main () {
5. double f =3.14159;
6. cout << setprecision(5) << f << ‘\n’;
7. cout << setprecision(9) << f << ‘\n’;
8. cout << fixed;
Does C++ round up or down?
C++ always truncates, aka rounds down. If you want it to round to the nearest intager, add 0.5 or 0.5f before casting.
How do you round down a double in C++?
Round a Double to an Int in C++
1. Use the round() Function for Double to Int Rounding.
2. Use the lround() Function to Convert Double to the Nearest Integer.
3. Use the trunc() Function for Double to Int Rounding.
What is the round off of 6?
Rules for Rounding Here’s the general rule for rounding: If the number you are rounding is followed by 5, 6, 7, 8, or 9, round the number up.
Begin typing your search term above and press enter to search. Press ESC to cancel. | 690 | 2,791 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2024-10 | latest | en | 0.892245 |
https://itslinuxfoss.com/use-and-operator-python-if/ | 1,726,229,679,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651513.89/warc/CC-MAIN-20240913101949-20240913131949-00551.warc.gz | 294,016,848 | 32,275 | In Python, different operations are performed on variables and values using logical operators such as “And”, “Or”, and “Not”, etc. The values of logical operators in Python are Boolean (True or False). If the condition is satisfied, the value returns will be “True”, and if the condition does not satisfy, the value returns will be “False”.
In this write-up, we will provide a comprehensive overview of the “And” operator in Python “If” with multiple examples. The below-listed contents are explained in this post:
So let’s begin!
## What is Logical And Operator in Python?
The Logical “AND” operator is applied on a minimum of two operands. These operands refer to a condition (which returns a True or False value). Let’s see how the “AND” operator works in Python:
The following conclusion can be withdrawn from the above table:
• If both values return “True” then the output of the “AND” operator will also be “True”.
• If any one value is “Not True / False” then the Logical “AND” operator output will be “False”.
In Python, the Logical “And” operator can combine with an “if” statement to replace multiple lines of code with a single line expression. For example, the nested if statement containing two conditions can easily be replaced with one if condition with a “Logical And Operator”.
Let’s demonstrate the usage of “AND” with Python “If”.
## Example 1: Comparing Two Values Using AND in Python If
In the example code below, the “And” operator is used along with the “if” statement to check two number values.
Code
```#Logical And Operator with Python If
Number_1 = 50
Number_2 = -12
if Number_1==50 and Number_2<0:
print('Number_1 is equal to',Number_1,'and',Number_2,'is smaller than zero.')
```
In the above example:
• Two integer variables named “number_1” and “number_2” are used in the program..
• The “And” operator is used with the “if” to execute the condition. If both left and right side conditions are “True” then the print statement will run and show the results.
• If one of the conditions is “False” then the code will not run..
Note: In the above code example, we are defining the if statement only to show the concept of And Operator with if statement. We may use an “else statement” to show the result when the condition of the “if statement” is not satisfied.
Output
The above output shows that number 1 is equal to “50” and number 2 is smaller than “0”.
## Example 2: Logical And Operator with Python if-else
In the example code below, the “AND” operator is used along with the “If-else” statement to operate on three number values.
Code
```# Logical And Operator with Python If-Else
Number_1 = 52
Number_2 = 23
Number_3 = -92
if Number_1 > 0 and Number_2 > 0:
print("Numbers are greater than 0")
if Number_1 > 0 and Number_2 > 0 and Number_3 > 0:
print("Numbers are greater than 0")
else:
print("one number is smaller than 0")
```
In the above code:
• Three variables named “number_1”, “number_2” and “number_3” are initialized.
• Two “if statements” with “AND” Operator is used in the above code.
• The first condition states that both “number_1” and “number_2” are greater than zero, and the second “if” condition with the “AND” operator states that all three numbers are greater than zero.
• The “else” statement prints the result when both conditions are unsatisfied.
Output
The above output shows that two numbers are greater than “Zero” and the third number is smaller than “Zero”.
That’s all from this Python guide!
## Conclusion
In Python, the “AND” operator along with the “if” statement is used to initialize/compare more than one condition in a single line of program code. The “AND” operator will give us a “True” value when both conditions become “True”. If any of the operand values are “False”, then the “AND” operator returns the “False” value. The “if-else” statement or multiple if statement can also be used with the “AND” Operator according to the requirement of program logic. In this article, “Python IF” along with “AND” Operator is discussed in detail with numerous examples. | 978 | 4,063 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-38 | latest | en | 0.848811 |
https://socratic.org/questions/58c8851311ef6b14cc847843 | 1,601,155,254,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400245109.69/warc/CC-MAIN-20200926200523-20200926230523-00470.warc.gz | 639,906,644 | 6,281 | #### Explanation:
We can express the ratio as an equation in $\textcolor{b l u e}{\text{fractional form}}$
$\Rightarrow \frac{9}{3060} = \frac{8}{x}$
where x represents the amount her son received.
$\textcolor{b l u e}{\text{cross-multiply}}$
$\Rightarrow 9 x = 8 \times 3060$
$\Rightarrow x = \frac{8 \times 3060}{9}$
$\textcolor{w h i t e}{\Rightarrow x} = 2720$
Total given to children =3060+2720=$5780 Mar 16, 2017 $5780
#### Explanation:
There are several ways of approaching this.
The monies are split into 8 parts for the son and 9 parts for the daughter. This makes a whole of:
$8 + 9 = 17 \text{ parts}$
Let the whole sum be $x$
Then we have for the daughter -> 9/17x=$3060 Multiply both sides by $\frac{17}{9}$[17/9xx9/17]xx x=17/9xx$3060
1xx x =\$5780 | 261 | 777 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 15, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2020-40 | latest | en | 0.814117 |
https://www.emathhelp.net/sacks-us-to-register-tons/ | 1,620,715,531,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991904.6/warc/CC-MAIN-20210511060441-20210511090441-00305.warc.gz | 777,722,530 | 4,048 | # Sacks (US) to register tons
This free conversion calculator will convert sacks (US) to register tons, i.e. sack (US) to register ton. Correct conversion between different measurement scales.
Convert
The formula is $$V_{\text{register ton}} = \frac{107521}{2880000} V_{\text{sack (US)}}$$$, where $$V_{\text{sack (US)}} = 15$$$.
Therefore, $$V_{\text{register ton}} = \frac{107521}{192000}$$$. Answer: $$15 \text{sack (US)} = \frac{107521}{192000} \text{register ton}\approx 0.560005208333333 \text{register ton}.$$$ | 162 | 521 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2021-21 | latest | en | 0.58246 |
http://ufdc.ufl.edu/UFE0045709/00001 | 1,513,012,897,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948513784.2/warc/CC-MAIN-20171211164220-20171211184220-00774.warc.gz | 288,576,155 | 76,654 |
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# Bayesian Methods for Modeling Dependence Structures in Longitudinal Data
## Material Information
Title:
Bayesian Methods for Modeling Dependence Structures in Longitudinal Data
Physical Description:
1 online resource (142 p.)
Language:
english
Creator:
Publisher:
University of Florida
Place of Publication:
Gainesville, Fla.
Publication Date:
## Thesis/Dissertation Information
Degree:
Doctorate ( Ph.D.)
Degree Grantor:
University of Florida
Degree Disciplines:
Statistics
Committee Chair:
Daniels, Michael Joseph
Committee Members:
Doss, John
Khare, Kshitij
Manini, Todd M
## Subjects
Subjects / Keywords:
bayesian -- correlation -- covariance -- longitudinal -- sparsity
Statistics -- Dissertations, Academic -- UF
Genre:
Statistics thesis, Ph.D.
bibliography ( marcgt )
theses ( marcgt )
government publication (state, provincial, terriorial, dependent) ( marcgt )
born-digital ( sobekcm )
Electronic Thesis or Dissertation
## Notes
Abstract:
In the modeling of longitudinal data from several groups, appropriate handling of the dependence structure is of central importance. In this dissertation we consider two important situations where estimation of the dependence structure is particularly important. We develop Bayesian prior distributions for a correlation matrix that favors parsimonious models. We additionally introduce two new estimation methods for the situation where data are composed of multiple groups each of which requires its own covariance matrix. Modeling a correlation matrix can be a difficult statistical task due to both the positive definite and the unit diagonal constraints. Because the number of parameters increases quadratically in the dimension, it is often useful to consider a sparse parameterization. In Chapter 2, we introduce prior distributions on the set of correlation matrices through the partial autocorrelations (PACs), each of which vary independently over -1,1. The first of the two proposed priors shrinks each of the PACs toward zero with increasingly aggressive shrinkage in lag. The second prior (a selection prior) is a mixture of a zero point mass and a continuous component for each PAC, allowing for a sparse representation. The structure implied under our priors is readily interpretable because each zero PAC implies a conditional independence relationship in the distribution of the data. For ordered data selection priors on the PACs provide a computationally attractive alternative to selection on the elements of the correlation matrix or its inverse. These priors allow for data-dependent shrinkage/selection under an intuitive parameterization in an unconstrained setting. The proposed priors are compared to standard methods through a simulation study and a multivariate probit data example. In Chapter 3, we focus on the challenge of estimating multiple covariance matrices. Standard methods include specifying a single covariance matrix for all groups or independently estimating the covariance matrix for each group without regard to the others, but when these model assumptions are incorrect, these techniques can lead to biased mean effects or loss of efficiency, respectively. Thus, it is desirable to develop methods to simultaneously estimate the covariance matrix for each group that will borrow strength across groups in a way that is ultimately informed by the data. In addition, for several groups with covariance matrices of even medium dimension, it is difficult to manually select a single best parametric model among the huge number of possibilities given by incorporating structural zeros and/or commonality of individual parameters across groups. In this chapter we develop a family of nonparametric priors using the matrix stick-breaking process of Dunson et al. (2008) that seeks to accomplish this task by parameterizing the covariance matrices in terms of the parameters of their modified Cholesky decomposition (Pourahmadi, 1999). We establish some theoretic properties of these priors, examine their effectiveness via a simulation study, and illustrate the priors using data from a longitudinal study of a depression treatment. Chapter 4 proposes a second method to handle the situation of simultaneous covariance estimation. We introduce a covariance partition prior which provides a partition of the groups at each measurement time. Groups in a common set of the partition share dependence parameters for the distribution of the current measurement given the preceding ones, and the sequence of partitions is modeled as Markov chain to encourage them to vary smoothly across measurement times. This approach additionally encourages a lower-dimensional structure of the covariance matrices by using a sparse Cholesky structure. We demonstrate the performance of our model through a simulation study and analysis of the depression study data.
General Note:
In the series University of Florida Digital Collections.
General Note:
Includes vita.
Bibliography:
Includes bibliographical references.
Source of Description:
Description based on online resource; title from PDF title page.
Source of Description:
This bibliographic record is available under the Creative Commons CC0 public domain dedication. The University of Florida Libraries, as creator of this bibliographic record, has waived all rights to it worldwide under copyright law, including all related and neighboring rights, to the extent allowed by law.
Statement of Responsibility:
Thesis:
Thesis (Ph.D.)--University of Florida, 2013.
Local:
Electronic Access:
RESTRICTED TO UF STUDENTS, STAFF, FACULTY, AND ON-CAMPUS USE UNTIL 2015-08-31
## Record Information
Source Institution:
UFRGP
Rights Management:
Applicable rights reserved.
Classification:
lcc - LD1780 2013
System ID:
UFE0045709:00001
## Material Information
Title:
Bayesian Methods for Modeling Dependence Structures in Longitudinal Data
Physical Description:
1 online resource (142 p.)
Language:
english
Creator:
Publisher:
University of Florida
Place of Publication:
Gainesville, Fla.
Publication Date:
## Thesis/Dissertation Information
Degree:
Doctorate ( Ph.D.)
Degree Grantor:
University of Florida
Degree Disciplines:
Statistics
Committee Chair:
Daniels, Michael Joseph
Committee Members:
Doss, John
Khare, Kshitij
Manini, Todd M
## Subjects
Subjects / Keywords:
bayesian -- correlation -- covariance -- longitudinal -- sparsity
Statistics -- Dissertations, Academic -- UF
Genre:
Statistics thesis, Ph.D.
bibliography ( marcgt )
theses ( marcgt )
government publication (state, provincial, terriorial, dependent) ( marcgt )
born-digital ( sobekcm )
Electronic Thesis or Dissertation
## Notes
Abstract:
In the modeling of longitudinal data from several groups, appropriate handling of the dependence structure is of central importance. In this dissertation we consider two important situations where estimation of the dependence structure is particularly important. We develop Bayesian prior distributions for a correlation matrix that favors parsimonious models. We additionally introduce two new estimation methods for the situation where data are composed of multiple groups each of which requires its own covariance matrix. Modeling a correlation matrix can be a difficult statistical task due to both the positive definite and the unit diagonal constraints. Because the number of parameters increases quadratically in the dimension, it is often useful to consider a sparse parameterization. In Chapter 2, we introduce prior distributions on the set of correlation matrices through the partial autocorrelations (PACs), each of which vary independently over -1,1. The first of the two proposed priors shrinks each of the PACs toward zero with increasingly aggressive shrinkage in lag. The second prior (a selection prior) is a mixture of a zero point mass and a continuous component for each PAC, allowing for a sparse representation. The structure implied under our priors is readily interpretable because each zero PAC implies a conditional independence relationship in the distribution of the data. For ordered data selection priors on the PACs provide a computationally attractive alternative to selection on the elements of the correlation matrix or its inverse. These priors allow for data-dependent shrinkage/selection under an intuitive parameterization in an unconstrained setting. The proposed priors are compared to standard methods through a simulation study and a multivariate probit data example. In Chapter 3, we focus on the challenge of estimating multiple covariance matrices. Standard methods include specifying a single covariance matrix for all groups or independently estimating the covariance matrix for each group without regard to the others, but when these model assumptions are incorrect, these techniques can lead to biased mean effects or loss of efficiency, respectively. Thus, it is desirable to develop methods to simultaneously estimate the covariance matrix for each group that will borrow strength across groups in a way that is ultimately informed by the data. In addition, for several groups with covariance matrices of even medium dimension, it is difficult to manually select a single best parametric model among the huge number of possibilities given by incorporating structural zeros and/or commonality of individual parameters across groups. In this chapter we develop a family of nonparametric priors using the matrix stick-breaking process of Dunson et al. (2008) that seeks to accomplish this task by parameterizing the covariance matrices in terms of the parameters of their modified Cholesky decomposition (Pourahmadi, 1999). We establish some theoretic properties of these priors, examine their effectiveness via a simulation study, and illustrate the priors using data from a longitudinal study of a depression treatment. Chapter 4 proposes a second method to handle the situation of simultaneous covariance estimation. We introduce a covariance partition prior which provides a partition of the groups at each measurement time. Groups in a common set of the partition share dependence parameters for the distribution of the current measurement given the preceding ones, and the sequence of partitions is modeled as Markov chain to encourage them to vary smoothly across measurement times. This approach additionally encourages a lower-dimensional structure of the covariance matrices by using a sparse Cholesky structure. We demonstrate the performance of our model through a simulation study and analysis of the depression study data.
General Note:
In the series University of Florida Digital Collections.
General Note:
Includes vita.
Bibliography:
Includes bibliographical references.
Source of Description:
Description based on online resource; title from PDF title page.
Source of Description:
This bibliographic record is available under the Creative Commons CC0 public domain dedication. The University of Florida Libraries, as creator of this bibliographic record, has waived all rights to it worldwide under copyright law, including all related and neighboring rights, to the extent allowed by law.
Statement of Responsibility:
Thesis:
Thesis (Ph.D.)--University of Florida, 2013.
Local:
Electronic Access:
RESTRICTED TO UF STUDENTS, STAFF, FACULTY, AND ON-CAMPUS USE UNTIL 2015-08-31
## Record Information
Source Institution:
UFRGP
Rights Management:
Applicable rights reserved.
Classification:
lcc - LD1780 2013
System ID:
UFE0045709:00001
Full Text
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Idedicatethisdissertationtomyfamilyandfriends,whoseconsistentsupporthasbeeninvaluabletomeonthisjourney. 3
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TABLEOFCONTENTS page ACKNOWLEDGMENTS .................................. 4 LISTOFTABLES ...................................... 8 LISTOFFIGURES ..................................... 9 ABSTRACT ......................................... 10 CHAPTER 1INTRODUCTION ................................... 13 1.1LiteratureReviewforCorrelationEstimationandPriorDistributions .... 15 1.2LiteratureReviewforCovarianceEstimationandPriorDistributions .... 17 1.3LiteratureReviewforSimultaneousCovarianceEstimationandPriorDistributions ................................... 20 2SPARSEPRIORDISTRIBUTIONSFORCORRELATIONMATRICESTHROUGHTHEPARTIALAUTOCORRELATIONS ....................... 23 2.1BayesianCorrelationEstimation ....................... 23 2.2PartialAutocorrelations ............................ 25 2.3PartialAutocorrelationShrinkagePriors ................... 28 2.3.1SpecicationoftheShrinkagePrior .................. 28 2.3.2SamplingundertheShrinkagePrior ................. 29 2.4PartialAutocorrelationSelectionPriors .................... 30 2.4.1SpecicationoftheSelectionPrior .................. 30 2.4.2NormalizingConstantforPriorsonR ................. 31 2.4.3SamplingundertheSelectionPrior .................. 32 2.5Simulations ................................... 33 2.6DataAnalysis .................................. 40 2.7Discussion ................................... 44 3ANONPARAMETRICPRIORFORSIMULTANEOUSCOVARIANCEESTIMATION ..................................... 47 3.1SimultaneousCovarianceEstimation ..................... 47 3.2TheMatrixStick-BreakingProcess ...................... 50 3.3CovarianceGroupingPriors .......................... 51 3.3.1Lag-BlockGroupingPriorfor .................... 51 3.3.2Correlated-LognormalGroupingPriorfor)]TJ ET 0 G 0 g BT /F1 11.955 Tf 319.22 -583.28 Td[(.............. 53 3.4TheoreticalProperties ............................. 54 3.4.1GeneralizedAutoregressiveParameterProperties ......... 54 3.4.2InnovationVarianceProperties .................... 57 3.5ComputationalConsiderations ........................ 58 5
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3.6RiskSimulation ................................. 59 3.7DataExample .................................. 63 3.8Discussion ................................... 67 4COVARIANCEPARTITIONPRIORS:ABAYESIANAPPROACHTOSIMULTANEOUSCOVARIANCEESTIMATION .................. 68 4.1SimultaneousCovarianceEstimationandaDrawbackoftheCovarianceGroupingPriors ................................. 68 4.2CovariancePartitionPrior ........................... 70 4.2.1PriorontheSequenceofPartitions .................. 70 4.2.2PriorontheCholeskyParameters .................. 74 4.3SamplingAlgorithm .............................. 76 4.4SimulationStudy ................................ 80 4.5DepressionDataExample ........................... 83 4.6Discussion ................................... 87 5CONCLUSIONSANDFUTUREWORK ...................... 89 APPENDIX ACALCULATINGTHEDICSTATISTICFORCTQDATA .............. 91 BADDITIONALCOVARIANCEGROUPINGPRIORSANDTHEIRPROPERTIES 96 B.1SparsityGroupingPriorfor ......................... 96 B.2Non-SparseGroupingPriorfor ....................... 97 B.3InvGammaGroupingPriorfor)]TJ ET 0 G 0 g BT /F1 11.955 Tf 226.21 -367.61 Td[(........................ 98 B.4FurtherGroupingPriorExtensions ...................... 99 CDERIVATIONOFCOVARIANCEGROUPINGPRIORPROPERTIES ...... 101 C.1ProofsforGeneralizedAutoregressiveParameterProperties ....... 101 C.2ProofsforInnovationVarianceProperties .................. 107 DDETAILSOFMCMCALGORITHMFORCOVARIANCEGROUPINGPRIORS 109 D.1Preliminaries .................................. 109 D.2SamplingStepsforLag-BlockGroupingPrior ................ 109 D.3SamplingStepsforCorrelated-LognormalGroupingPrior ......... 114 D.4SamplingStepsforSparsityGroupingPrior ................. 117 D.5SamplingStepsforNon-SparseGroupingPrior ............... 118 D.6SamplingStepsforInvGammaGroupingPrior ............... 118 D.7FinalCommentsaboutComputationalAlgorithm .............. 119 EADDITIONALRISKSIMULATIONSFORCOVARIANCEGROUPINGPRIORS 121 E.1AdditionalDetailsforRiskSimulationofSection3.6 ............ 121 E.2RiskSimulation2 ................................ 123 6
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E.3RiskSimulation3 ................................ 125 E.4RiskSimulation4 ................................ 126 E.5ExtendedAnalysisofDepressionStudyData ................ 128 FMODELPARAMETERSFORRISKSIMULATIONOFCHAPTER4 ....... 131 REFERENCES ....................................... 133 BIOGRAPHICALSKETCH ................................ 139 7
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LISTOFTABLES Table page 2-1RiskestimatesforsimulationstudywithdimensionJ=6. ............ 37 2-2SpecicationofD0. ................................. 38 2-3RiskestimatesforsimulationstudywithdimensionJ=10. ........... 39 2-4ModelcomparisonstatisticsfortheCTQdata. .................. 42 3-1Riskestimatesfromsimulationstudy. ....................... 62 3-2Modeltstatisticsandtreatmenteffectsforthedepressiondata. ........ 64 4-1Riskestimatesfromsimulationstudy. ....................... 82 4-2Modelselectionstatisticsfordepressionstudy. .................. 85 E-1ParametervaluesforrisksimulationofSection 3.6 ................ 122 E-2ProbabilityYitismissingbygroupmforrisksimulationofSection 3.6 ..... 122 E-3ExpandedriskestimatesforSection 3.6 simulationstudy. ............ 123 E-4Riskestimatesforsimulation2. ........................... 124 E-5Riskestimatesforsimulation3. ........................... 126 E-6Parametervaluesforsimulation4. ......................... 127 E-7Riskestimatesforsimulation4. ........................... 127 E-8Expandedmodeltstatisticsandtreatmenteffectsforthedepressiondata. .. 129 8
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LISTOFFIGURES Figure page 2-1BoxplotsoftheobservedlossfromsimulationstudywithJ=6. ........ 36 2-2BoxplotsoftheobservedlossfromsimulationstudywithJ=10. ........ 39 3-1Posteriorprobabilitiesofmatchingfortheinnovationvariances. ......... 65 3-2Posteriorprobabilitiesofmatchingforthegeneralizedautoregressiveparameters. ...................................... 66 4-1MarginalprobabilitiesforthreechoicesofqwithM=8groups. ......... 74 4-2Boxplotsoflossfromsimulationstudy. ....................... 82 4-3Posteriorprobabilitythatm1andm2areinthesamesetofthepartitionPt. .. 86 9
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usingasparseCholeskystructure.Wedemonstratetheperformanceofourmodelthroughasimulationstudyandanalysisofthedepressionstudydata. 12
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CHAPTER1INTRODUCTIONWhenworkingwithlongitudinal(ortime-ordered)data,specifyingand/ormodelingofthedependencestructureisofprimeimportance.Inmostsituationsthemethodsoflinearandgeneralizedlinearmodelscanbeadaptedtodescribethemeanstructure.Aslongitudinaldataconsistsofmultiplemeasurementswithinanexperimentalunit,methodstohandlethedependenceacrossthemeasurementsbecomesnecessary.Here,wedistinguishlongitudinaldataanalysisfromtherelatedeldofmultivariateanalysis(seee.g., Anderson 1984 ).Wedenelongitudinaldatatobeamultivariateresponseforanexperimentalunit(patientorcase)wheretheresponsesaremeasurementsofthesameoutcomeatdifferenttimepoints,e.g.,whetherapatientsmokesduringagivenweekoverthecourseof8weeksornumberofdepressionsymptomsinaweekover16weeks.Dataofthistypehaveaclearorderingintime,whereasgeneralmultivariatedatamaynot.Oneshouldnotethatthemethodswedevisehereandmostinthelongitudinaldataliteraturemakeuseofassumptionsandintuitionthatareonlyreasonablewhenthemeasurementsfollowaxedorderingandmaynotbeappropriateinmoregeneralmultivariatesituations.Itisnotuncommonforstatisticianstomistakenlyassumethatestimationofthecovarianceparametersperformsasecondaryroletothemeanestimation.Infact,theyshouldgenerallybetreatedjointly.Insituationswithcompletedata(i.e.,allpatientsareobservedatallobservationtimes)andmultivariatenormality,themeanandcovarianceparametersareorthogonalinthesenseof Cox&Reid ( 1987 ),andtheestimatesofthemeanparameterswillbeconsistentundermisspecicationofthedependencestructure.However,whenoneanalyzesreal-worldlongitudinaldatasuchasthatfromaclinicaltrial,thedatawillusuallyexhibitsomeamountofmissingness.Ifthereismissingnessinthedata,thereisnolongerorthogonalitybetweenmeananddependenceparameters,evenatthetruevalueofthecovariancematrix( Little&Rubin 2002 ).Hence,forthe 13
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posteriordistributionofthemeanparameterstobeconsistent,thedependencestructuremustbecorrectlyspecied.So,eveninthemissingatrandomcase(MAR; Daniels&Hogan 2008 ),wheremissingnessdependsonlyontheobservedvaluesnottheunobserveddata,itisnolongerappropriatetotreatthecovariancestructureasanuisanceparameter.Inthiscasebiasedmeanestimatescanresultifwedonotusethecorrectmodelforthedependence( Daniels&Hogan 2008 ,Section6.2).Tofurthermotivatethenecessityofmethodologyforimprovedcovarianceestimation,notethateveninthecompletedatacasewherethemeananddependenceareasymptoticallyindependentundernormality,efciencygainsmaybepossibleforsmallormoderatelysizeddatasets.Throughfoursimulationexampleswithrelativelysmallsamplesizes, Crippsetal. ( 2005 )demonstratedimprovementsinestimatingregressioncoefcients,ttedvalues,andthepredictivedensityusingthe Wongetal. ( 2003 )covarianceselectionprioroveramoredispersedcovariancepriorchoice.InthisdissertationwewilldevelopBayesianmethodstomodellongitudinaldependencestructures.Thesedependonthespecicationofanappropriatepriordistributionforthecovariancematrixoritsparameters.AkeyconsiderationindevelopingthesepriorsistheabilitytoincorporatethemaspartofaMarkovchainMonteCarlo(MCMC)schemetoobtainposteriorinference.Wealsowantthepriorstobestructuredsuchthattheyarecenteredatintuitivepriorbeliefsspecictolongitudinaldata,suchasdecreasingdependenceacrosstimeorpositive(ratherthannegative)correlation.Additionally,wedesirepriorsthatpromotesparse(orlower-dimensional)parameterizationsofthedependencestructure.Throughoutweconsidertwoimportantsituationsinlongitudinaldata.FirstinChapter 2 ,welookatsituationswherethedependencestructureisconstrainedtobecorrelationmatrixforidentiability.Thiscomplicationisencounteredinmultivariateprobitmodels( Chib&Greenberg 1998 ),Gaussiancopularegression( Pittetal. 2006 ),certainlatentvariablemodels( Daniels&Normand 2006 ),amongothers.Theother 14
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problemweconsiderinChapters 3 and 4 isthatofjointlyestimatingmultiplecovariancematrices.Oftenlongitudinaldatamaybeviewedascomposedofseveralgroupseachofwhichmayneeditsowncovariancestructure.Itisdesirabletodevelopmethodologytojointlyestimatethesecovariancematricesallowingsharingofinformation.Wenowreviewtheimportantcontributionstotheliteratureforeachofthisproblems. 1.1LiteratureReviewforCorrelationEstimationandPriorDistributionsFirst,weexploretheproblemofmodelingacorrelationmatrix.Consideramean-zero,J-dimensionalrandomvectorYwithcorrelationmatrixR.TherearetwomainconstraintsontheJJmatrixR:thediagonalelementsmustallequaloneandRispositive-denite.LetRJdenotethesetofallrealmatricesthatsatisfytheserequirements.Positive-denitenessgenerallyprovidesthegreatestdifcultyinanalysis,asthesetofvaluesforaparticularelementijthatsatisfythepositive-deniteconstraintdependsonthechoiceoftheremainingelementsofR.Additionally,becausethenumberofparametersinRisquadraticinthedimensionJ,methodstondaparsimoniousorlower-dimensionalstructurecanbebenecial.Oneoftheearliestattemptstondlower-dimensionalstructureforthemoregeneralproblemofestimatingacovariancematrixistheideaofcovarianceselection( Dempster 1972 ).Bysettingsomeoftheoff-diagonalelementsoftheconcentrationmatrix=)]TJ /F9 7.97 Tf 6.59 0 Td[(1tozero,amoreparsimoniouschoiceforthecovariancematrixofYisachieved.Azerointhe(i,j)-thpositionofimplieszerocorrelation(andfurther,independenceundermultivariatenormality)betweenYiandYj,conditionalontheremainingcomponentsofY.Thisproperty,alongwithitsrelationtographicalmodeltheory( Lauritzen 1996 ),hasledtotheuseofcovarianceselectionasastandardpartofanalysisinmultivariateproblems(forinstance, Rothmanetal. 2008 ; Wongetal. 2003 ; Yuan&Lin 2007 ).However,onemustbecautiouswhenusingsuchselectionmethodsasnotallproducepositivedeniteestimators.Forinstance,thresholdingthesamplecorrelationmatrixwillnotnecessarilybepositivedenite( Bickel&Levina 2008 ). 15
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Eveninsituationswherethefocusisonageneralcovariancematrix,modelspecicationmaydependonthecorrelationstructurethroughtheso-calledseparationstrategy( Barnardetal. 2000 ).Theseparationstrategyinvolvesreparameterizingby=SRS,withSadiagonalmatrixcontainingthemarginalstandarddeviationsofYandRthecorrelationmatrix.Separationcanalsobeperformedontheconcentrationmatrix,=TCTsothatTisdiagonalandC2RJ.ThediagonalelementsofTgivethepartialstandarddeviations,whiletheelementscijofCarethe(full)partialcorrelations.ThecovarianceselectionproblemforisequivalenttochoosingelementsofthepartialcorrelationmatrixCtobenull.SeveralauthorshaveconstructedpriorstoefcientlyestimatebyallowingCtobeasparsematrix( Carteretal. 2011 ; Wongetal. 2003 ).Inmanycasesthefullpartialcorrelationmatrixmaynotbeconvenienttouse.Whenthecovariancematrixisxedtoacorrelationmatrixsuchasforthemultivariateprobitmodel,theelementsoftheconcentrationmatrixTandCareconstrainedtomaintainaunitdiagonalfor.Thisiseasytoseesince=RandCeachhaveJ(J)]TJ /F4 11.955 Tf 11.85 0 Td[(1)=2parametersbutTaddsanadditionalJparameters.Additionally,interpretationofparametersinthefullpartialcorrelationmatrixcanbechallenging,particularlyforlongitudinalsettingsasthepartialcorrelationsaredenedconditionalonfuturevalues.Forexample,c12givesthecorrelationbetweenY1andY2conditionalonthefuturemeasurementsY3,...,YJ.AfurtherissuewithBayesianmethodsthatpromotesparsityinCiscalculatingthevolumeofthespaceofcorrelationmatriceswithaxedzeropattern;seeSection 2.4.2 fordetails.PreviousBayesiansolutionsareconcernedwithchoicesofanappropriatepriordistributionp(R)onRJ.CommonlyusedpriorsincludeplacingequalweightonallelementsofRJ( Barnardetal. 2000 )andJeffrey'spriorp(R)/jRj)]TJ /F9 7.97 Tf 6.59 0 Td[((J+1)=2.InthesecasesthesamplingstepsforRcansometimesbenetfromparameterexpansiontechniques( Liu 2001 ; Liu&Daniels 2006 ; Zhangetal. 2006 ). Liechtyetal. ( 2004 )alsodevelopacorrelationmatrixpriorbyspecifyingeachelementijofRasan 16
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independentnormalsubjecttoR2RJ. Pittetal. ( 2006 )extendthecovarianceselectionprior( Wongetal. 2003 )tothecorrelationmatrixcasebyxingtheelementsofTtobeconstrainedbyCsothatTisthediagonalmatrixsuchthatR=(TCT))]TJ /F9 7.97 Tf 6.58 0 Td[(1hasunitdiagonal.Thedifcultyofjointlydealingwiththepositive-deniteandunitdiagonalconstraintsofacorrelationmatrixhasledsomeresearcherstoconsiderpriorsforRbasedonthepartialautocorrelations(PACs).ThePACbetweenYiandYj(i
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isonesuchmethod( Carteretal. 2011 ; Pittetal. 2006 ; Wongetal. 2003 ).Othernon-Bayesiantechniquesthatencouragesparseversionsofthedependencestructureincludebandingthesamplecovarianceorconcentrationmatrix( Bickel&Levina 2008 ),usingalasso-typepenaltyontheelementsof( Mazumder&Hastie 2012 ; Meinhausen&Buhlmann 2006 )ortheelementstheCholeskydecompositionof( Rothmanetal. 2008 ),andbandingtheCholeskydecompositionof( Rothmanetal. 2010 ).Again,onemustbecautiouswhenusingselectionmethodsasnotallproducepositivedeniteestimators.Aspreviouslymentioned,theideasfromcovarianceselectionareconnectedtothoseofgraphicalmethods.Bayesianmethodsforgraphicalmodelsgenerallyconsistofxingthezerostructureof(or)torepresentaparticulargraphG.Apriorfor()isthenconstructedtorangeoverthespaceofconcentration(covariance)matriceswiththeappropriatezerostructure.Therearemanysuchpriorswithvaryingdegreesofexibility( Dawid&Lauritzen 1993 ; Khare&Rajaratnam 2011 ; Letac&Massam 2007 ; Rajaratnametal. 2008 ). Giudici&Green ( 1999 )proposeahierarchicalpriorwherethegraphGisassumedrandomoverspaceofdecomposablegraphs.Mostgraphicalmethodsdonotincorporatetheorderingoftheresponses,andconsequently,Y1andY2areaslikelytobeuncorrelatedasareY1andYJ.Thisisundesirableforlongitudinaldata,andsowedonotconsiderfurtherthegraphicalmodelmethodology.Bayesianfactormodelsprovidesanotheroptiontodeveloplower-dimensionspecicationsof.Themostcommonofthesemodelsfactorthecovariancematrixas=0+D,whereDisdiagonal(sometimesoftheform2I)andisJkwithk
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exhibitidentiabilityproblemswhenusedaspartofaMCMCanalysis( Lopes&West 2004 ).AparameterizationbasedontheCholeskydecompositionofhasbeenproposedby Pourahmadi ( 1999 2000 ).Inthisparameterizationthecovariancematrixdependsontwosetsofparameters:,thegeneralizedautoregressiveparameters(GARPs),and)]TJ /F1 11.955 Tf 6.94 0 Td[(,thesetofinnovationvariances(IVs),suchthat(,)]TJ /F4 11.955 Tf 6.94 0 Td[())]TJ /F9 7.97 Tf 6.58 0 Td[(1=T()D()]TJ /F4 11.955 Tf 6.95 0 Td[()T()0=2666666641)]TJ /F11 11.955 Tf 9.3 0 Td[(12)]TJ /F11 11.955 Tf 9.3 0 Td[(131)]TJ /F11 11.955 Tf 9.3 0 Td[(231...3777777752666666641 11 2...1 J3777777752666666641)]TJ /F11 11.955 Tf 9.3 0 Td[(121)]TJ /F11 11.955 Tf 9.3 0 Td[(13)]TJ /F11 11.955 Tf 9.3 0 Td[(231............377777775.TheT()matrixisupper-triangularwithonesonitsdiagonal.NotethatthereareJparametersforeach)]TJ /F4 11.955 Tf 11.46 0 Td[(=(1,...,J)andK=J(J)]TJ /F4 11.955 Tf 12.44 0 Td[(1)=2parametersassociatedwitheach=(12,...,J)]TJ /F9 7.97 Tf 6.59 0 Td[(1,J).OneofthekeymotivationsbehindtheuseofthemodiedCholeskydecompositionisthattheonlyconstraintonand)]TJ /F1 11.955 Tf 10.27 0 Td[(neededtoguarantee(,)]TJ /F4 11.955 Tf 6.94 0 Td[()ispositivedeniteisthatj>0forallj=1,...,J.AdditionallytheGARPsandIVsareinterpretedasparametersfromsequentialregressions;i.e.,EfYjjy1,...,yj)]TJ /F9 7.97 Tf 6.59 0 Td[(1g=1jy1++j)]TJ /F9 7.97 Tf 6.59 0 Td[(1,jyj)]TJ /F9 7.97 Tf 6.59 0 Td[(1andVarfYjjy1,...,yj)]TJ /F9 7.97 Tf 6.59 0 Td[(1g=j(assumingwithoutlossofgeneralityYhasmeanzero).Again,thedependenceontheorderingofthecomponentsofYisclear.TheinterpretabilityoftheGARPsreliesonanassumedorderoftheJcomponentsofY,whichisnaturalinthecaseoflongitudinalmeasurements.Priorsforcanthenbeenformedbyspecifyingpriorsonand)]TJ /F1 11.955 Tf 6.94 0 Td[(. Daniels&Pourahmadi ( 2002 )developpriorsthatexploitconjugacyfortheGARPsandIVs. Smith&Kohn ( 2002 )formparsimoniouspriorsthatstochasticallysetelementsoftozero.Notethatjk=0(j
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Y1,...,Yj)]TJ /F9 7.97 Tf 6.58 0 Td[(1,Yj+1,...,Yk)]TJ /F9 7.97 Tf 6.58 0 Td[(1,sothatthesparsityisinterpretableasanindependencerelationship.OtherBayesianmethodsforcovariancepriorsincludespecifyingthepriorintermsofthematrixlogarithmofsothatitisunconstrained( Leonard&Hsu 1992 ),priorsbasedonthespectraldecomposition( Daniels&Kass 1999 ),aatprioronortheJeffrey'spriorp()/jj)]TJ /F9 7.97 Tf 6.59 0 Td[((J+1)=2,andthereferenceprior( Yang&Berger 1994 ). 1.3LiteratureReviewforSimultaneousCovarianceEstimationandPriorDistributionsInSection 1.2 wedescribedtechniquestomodelasinglecovariancematrix.Arelatedchallengeissimultaneouslyestimatingmultiplecovariancematrices.Frequentlyinlongitudinalproblemsdataiscomposedofseveralgroups,suchasdifferingtreatmentsinaclinicaltrial.Inmanycases,particularlyifonedoesnothavemanyobservationspergroup,oneassumesthatthecovariance(orcorrelation)structureisconstantacrossallgroups.However,thisassumption,ifitfailstohold,canhaveadramaticeffectontheinferenceofmeaneffects,evenleadingtobiasifdataareincomplete( Daniels&Hogan 2008 ).Conversely,ifonespecieseachofthecovariancematriceswithoutregardtotheothergroups,thiscanleadtoalossofinformation.Soitisimportanttondmethodsthatcanndamiddlegroundbetweenthesetwoextremesbysharinginformationaboutthedependenceacrossgroups.Letmdenotethecovariancematrixforgroupm(m=1,...,M)and=f1,...,Mgbethecollectionofcovariancematrices.Manyauthorshavedevelopedfrequentistestimatorsforthiscollectionbyinducingcommonalityamongsomefeatureofthem. Boik ( 2002 2003 )proposedmodelstoinducestructurebyimposingcommonalityonsome(orall)oftheprincipalcomponentsofthecovarianceorcorrelationmatrix.Othershaveusedthevariance-correlationdecompositionforestimationbyimposingstructuressuchasproportionalityofallmorcommonalityamongthecorrelationmatrices( Manly&Rayner 1987 ). Pourahmadietal. ( 2007 ) 20
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developedestimationandtestingproceduresforequalityamongtheGARPsandsubsetsoftheGARPs.Inaclusteringcontext McNicholas&Murphy ( 2010 )advocateasimilarcovarianceestimationprocedurethatincludesbandingtheT()matrices. Daniels ( 2006 )consideredaBayesianperspectivebyintroducingpriorsfortheGARPsandIVs,aswellastheprincipalcomponentsofthecovariancematrices,thatinducepoolingacrossgroups.Unfortunately,itiscomputationallychallengingtoselectamongallthepossiblemodelswithintheseclasses. Hoff ( 2009 )proposedahierarchicaleigenmodelforthatpoolstheeigenvectorsofeachgrouptowardacommonstructure. Guoetal. ( 2011 )consideranautomatedapproachusingthelassotoestimatesparsegraphicalmodelsbyselectingsetsofedgescommontoallgroups,aswellasgroup-specicedges.Inthelongitudinaldatasettingwewishtondmorecovariancestructurethanjustcommonzerosacrossallgroups.Wewanttoconsidermodelsthatallowsubsetsofthemodelparameterstobeequalacross(asubsetofthe)groupsatnon-zerovalues;theGuoetal.estimatorsdonotaccommodatethisgoal. Danaheretal. ( 2012 )extendthisworktoallowtheelementsofmtobeequaltoasinglenon-zerovalue,zero,ordistinctacrossgroups.Butthisstilldoesnotallowforastructurecontainingsubsetsofgroups.WeadditionallynotethatitisnotclearhowonecouldeasilyadapteitherpenaltytermintoaBayesianprioronthesetcovariancematricesforoursetting.Othertechniqueshavebeenproposedthatmodelthecovariancematrixasafunctionofoneofmorecontinuouscovariates.AnumberofmodelsofthisavorhavebeendevelopedthatarespeciedthroughregressionsontheGARPsand/orIVs( Daniels 2006 ; Pourahmadi 1999 2000 ; Pourahmadi&Daniels 2002 ). Chiuetal. ( 1996 )devisesuchamodelbyregressingontolog(m).OtherregressionframeworksincludemodelsonthePACsandmarginalvariances( Wang&Daniels 2013a ),regressingwithinafactormodel( Fox&Dunson 2011 ),andmodelbasedonam=B+xmx0m0factorization( Hoff&Niu 2012 ).Additionalmethodstreat 21
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thecovariancematricesasrealizationsofastochasticvolatilityprocess( Lopesetal. 2011 ; Philipov&Glickman 2006a b ).However,covarianceregressionmodelsareoftenplaguedbythedifcultyofinterpretingtheregressionparameters.ToformourpriorsonwewillmakeuseofthemodiedCholeskyparameterizationbecauseoftheunrestrictednessoftheparameters,theinterpretabilityforlongitudinaldata,andthecomputationaladvantagesviaconjugacy( Daniels&Pourahmadi 2002 ).OurgoalistodeveloppriorsforthesetofGARPsandIVsinsuchawaythatweborrowstrengthacrosstheMgroups.Additionally,wewanttoshareinformationacross)]TJ /F7 7.97 Tf 6.94 -1.79 Td[(mandmvalues(m=(m,)]TJ /F7 7.97 Tf 6.94 -1.79 Td[(m),m=1,...,M),particularlythoseGARPsofacommonlag.AnotherconsiderationforpriordevelopmentistoencouragesparsityoftheelementsofT(m),thatis,containingfewnon-zeroelements.BecauseeachGARPrepresentsaconditionaldependency,settingm;jktozeroestablishesaconditionalindependencerelationshipbetweenapairofcomponentsofY.Itisnecessarytoconsiderpriorsthatallowthedatatoinformthebalancebetweenthesetwogoals:poolingacrossgroupsandintroducingsparsity.Aboveall,weseektoaccomplishthisinanautomated,stochasticfashion.Weproposetwosolutionstothisproblem.InChapter 3 wedevelopanonparametricpriorbasedonthematrixstick-breakingprocess( Dunsonetal. 2008 ).ThispriorallowsforclusteringoftheGARPssimultaneouslyacrossgroupsmandij'sofacommonlagj)]TJ /F5 11.955 Tf 13.23 0 Td[(i,whileallowingsomeGARPstobeidenticallyzero(implyingaconditionalindependence).ThesecondsolutiontothesimultaneousestimationproblempresentedinChapter 4 considersclusteringbasedonthet-thcolumnofT(m)andm;t,whichdenescollectionsofthegroups1,...,MthathavethesamedependenceparametersforthedistributionofYtgivenY1,...,Yt)]TJ /F9 7.97 Tf 6.58 0 Td[(1.TheequalityrelationshipsinthiscovariancepartitionpriorarespecieddirectlythroughaMarkovchainonthesequenceofpartitionsof1,...,M. 22
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CHAPTER2SPARSEPRIORDISTRIBUTIONSFORCORRELATIONMATRICESTHROUGHTHEPARTIALAUTOCORRELATIONS 2.1BayesianCorrelationEstimationDeterminingthestructureofanunknownJJcovariancematrixisalongstandingstatisticalchallenge.Akeydifcultyindealingwiththecovariancematrixisthepositivedenitenessconstraint.Thisisbecausethesetofvaluesforaparticularelementijthatyieldapositivedenitedependsonthechoiceoftheremainingelementsof.Additionally,becausethenumberofparametersinisquadraticinthedimensionJ,methodstondaparsimonious(lower-dimensional)structurecanbebenecial.Oneoftheearliestattemptsinthisdirectionistheideaofcovarianceselection( Dempster 1972 ).Bysettingsomeoftheoff-diagonalelementsoftheconcentrationmatrix=)]TJ /F9 7.97 Tf 6.59 0 Td[(1tozero,amoreparsimoniouschoiceforthecovariancematrixoftherandomvectorYisachieved.Azerointhe(i,j)-thpositionofimplieszerocorrelation(andfurther,independenceundermultivariatenormality)betweenYiandYj,conditionalontheremainingcomponentsofY.Thisproperty,alongwithitsrelationtographicalmodeltheory(e.g., Lauritzen 1996 ),hasledtotheuseofcovarianceselectionasastandardpartofanalysisinmultivariateproblems( Rothmanetal. 2008 ; Wongetal. 2003 ; Yuan&Lin 2007 ).However,oneshouldbecautiouswhenusingsuchselectionmethodsasnotallproducepositivedeniteestimators.Forinstance,thresholdingthesamplecovariance(concentration)matrixwillnotgenerallybepositivedenite,andadjustmentsareneeded( Bickel&Levina 2008 ).Modelspecicationformaydependonacorrelationstructurethroughtheso-calledseparationstrategy( Barnardetal. 2000 ).Theseparationstrategyinvolvesreparameterizingby=SRS,withSadiagonalmatrixcontainingthemarginalstandarddeviationsofYandRthecorrelationmatrix.LetRJdenotethesetofvalidcorrelationmatrices,thatis,thecollectionofJJpositivedenitematriceswithunit 24
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diagonal.Separationcanalsobeperformedontheconcentrationmatrix,=TCTsothatTisdiagonalandC2RJ.ThediagonalelementsofTgivethepartialstandarddeviations,whiletheelementscijofCarethe(full)partialcorrelations.ThecovarianceselectionproblemisequivalenttochoosingelementsofthepartialcorrelationmatrixCtobenull.SeveralauthorshaveconstructedpriorstoestimatebyallowingCtobeasparsematrix( Carteretal. 2011 ; Wongetal. 2003 ).Inmanycasesthefullpartialcorrelationmatrixmaynotbeconvenienttouse.Incaseswherethecovariancematrixisxedtobeacorrelationmatrixsuchasthemultivariateprobitcase,theelementsoftheconcentrationmatrixTandCareconstrainedtomaintainaunitdiagonalfor( Pittetal. 2006 ).Additionally,interpretationofparametersinthepartialcorrelationmatrixcanbechallenging,particularlyforlongitudinalsettingsasthepartialcorrelationsaredenedconditionalonfuturevalues.Forexample,c12givesthecorrelationbetweenY1andY2conditionalonthefuturemeasurementsY3,...,YJ.AnadditionalissuewithBayesianmethodsthatpromotesparsityinCiscalculatingthevolumeofthespaceofcorrelationmatriceswithaxedzeropattern;seeSection 2.4.2 fordetails.InadditiontotheroleRplaysintheseparationstrategy,insomedatamodelsthecovariancematrixisconstrainedtobeacorrelationmatrixforidentiability.Thisisthecaseforthemultivariateprobitmodel( Chib&Greenberg 1998 ),Gaussiancopularegression( Pittetal. 2006 ),certainlatentvariablesmodels(e.g. Daniels&Normand 2006 ),amongothers.Thus,itisnecessarytomakeuseofmethodsspecicforestimatingand/ormodelingacorrelationmatrix.WeconsiderthisproblemofcorrelationmatrixestimationinaBayesiancontextwhereweareconcernedwithchoicesofanappropriatepriordistributionp(R)onRJ.CommonlyusedpriorsincludeauniformprioroverRJ( Barnardetal. 2000 )andJeffrey'spriorp(R)/jRj)]TJ /F9 7.97 Tf 6.59 0 Td[((J+1)=2.InthesecasesthesamplingstepsforRcansometimesbenetfromparameterexpansiontechniques( Liu 2001 ; Liu&Daniels 25
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2006 ; Zhangetal. 2006 ). Liechtyetal. ( 2004 )developacorrelationmatrixpriorbyspecifyingeachelementijofRasanindependentnormalsubjecttoR2RJ. Pittetal. ( 2006 )extendthecovarianceselectionprior( Wongetal. 2003 )tothecorrelationmatrixcasebyxingtheelementsofTtobeconstrainedbyCsothatTisthediagonalmatrixsuchthatR=(TCT))]TJ /F9 7.97 Tf 6.59 0 Td[(1hasunitdiagonal.ThedifcultyofjointlydealingwiththepositivedeniteandunitdiagonalconstraintsofacorrelationmatrixhasledsomeresearcherstoconsiderpriorsforRbasedonthepartialautocorrelations(PACs)insettingswherethedataareordered.PACssuggestapracticalalternativebyavoidingthecomplicationofthepositivedeniteconstraint,whileprovidingeasilyinterpretableparameters( Joe 2006 ). Kurowicka&Cooke ( 2003 2006 )framethePACideaintermsofavinegraphicalmodel. Daniels&Pourahmadi ( 2009 )constructaexibleprioronRthroughindependentshiftedBetapriorsonthePACs. Wang&Daniels ( 2013a )constructunderlyingregressionsforthePACs,aswellasatriangularpriorwhichshiftsthepriorweighttoamoreintuitivechoiceinthecaseoflongitudinaldata.InsteadofsettingpartialcorrelationsfromCtozerotoincorporatesparsity,ourgoalistoencourageparsimonythroughthePACs.AsthePACsareunconstrained,selectiondoesnotleadtothecomputationalissuesassociatedwithndingthenormalizingconstantforasparseC.WeintroduceandcomparepriorsforbothselectionandshrinkageofthePACsthatextendspreviousworkonsensibledefaultchoices( Daniels&Pourahmadi 2009 ).Thelayoutofthischapterisasfollows.Inthenextsectionwewillreviewtherelevantdetailsofthepartialautocorrelationparameterization.Section 2.3 proposesapriorforRinducedbyshrinkagepriorsonthePACs.Section 2.4 introducestheselectionpriorforthePACs.SimulationresultsshowingtheperformanceofthepriorsappearinSection 2.5 .InSection 2.6 theproposedPACpriorsareappliedtoadatasetfromasmokingcessationclinicaltrial.Section 2.7 concludesthechapterwithabriefdiscussion. 26
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2.2PartialAutocorrelationsForageneralrandomvectorY=(Y1,...,YJ)0thepartialautocorrelationbetweenYiandYj(i
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forj)]TJ /F5 11.955 Tf 12.43 0 Td[(i>1.AstherelationshipbetweenRandisone-to-one,theJacobianforthetransformationfromRtocanbecomputedeasily.ThedeterminantoftheJacobianisgivenby jJ()j=Yi
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whichhasacontributionfromijof(1)]TJ /F11 11.955 Tf 11.89 0 Td[(2ij)[J)]TJ /F9 7.97 Tf 6.58 0 Td[(1)]TJ /F9 7.97 Tf 6.58 0 Td[((j)]TJ /F7 7.97 Tf 6.58 0 Td[(i)]=2.NotethatpfR()istheproductofindependentSBeta(ij,ij)distributionsforeachij,whereij=ij=1+[J)]TJ /F4 11.955 Tf 9.77 0 Td[(1)]TJ /F4 11.955 Tf 9.77 0 Td[((j)]TJ /F5 11.955 Tf 9.78 0 Td[(i)]=2.Thisprovidesanunconstrainedrepresentationoftheat-Rprior.Inlongitudinal/ordereddatacontexts,weexpectthePACstobenegligibleforelementsthathavelargelags.Weexploitthisconceptviatwotypesofpriors.First,weintroducepriorsthatshrinkPACstowardzerowiththeaggressivenessoftheshrinkagedependingonthelag.Next,wepropose,inthespiritof Wongetal. ( 2003 ),aselectionpriorthatwillstochasticallychoosePACstobesettozero. 2.3PartialAutocorrelationShrinkagePriors 2.3.1SpecicationoftheShrinkagePriorUsingthePACframework,weformpriorsthatwillshrinkthePACijtowardzero.Ithaslongbeenknownthatshrinkageestimatorscanproducegreatlyimprovedestimation( James&Stein 1961 ).Aspreviouslynoted,ij=0impliesthatYiandYjareuncorrelatedgiventheinterveningvariables(Yi+1,...,Yj)]TJ /F9 7.97 Tf 6.58 0 Td[(1).InthecasewhereYhasamultivariatenormaldistribution,thisimpliesindependencebetweenYiandYj,given(Yi+1,...,Yj)]TJ /F9 7.97 Tf 6.59 0 Td[(1).Weanticipatethatvariablesfartherapartintime(andconditionalonmoreintermediatevariables)aremorelikelytobeuncorrelated,sowewillmoreaggressivelyshrinkijforlargervaluesofthelagj)]TJ /F5 11.955 Tf 11.95 0 Td[(i.WeleteachijSBeta(ij,ij)independently.Aswewishtoshrinktowardzero,wewantEfijg=0,sowexij=ij.ItiseasilyshownthatVarfijg=4ijij (ij+ij)2(ij+ij+1),whichwedenotebyij.WerecovertheSBetashapeparametersbyij=ij=()]TJ /F9 7.97 Tf 6.59 0 Td[(1ij)]TJ /F4 11.955 Tf 12.35 0 Td[(1)=2.Hence,thedistributionofijisdeterminedbyitsvarianceij.RatherthanspecifyingtheseJ(J)]TJ /F4 11.955 Tf 11.96 0 Td[(1)=2differentvariances,weparameterizethemthrough Varfijg=ij=0jj)]TJ /F5 11.955 Tf 11.95 0 Td[(ij)]TJ /F12 7.97 Tf 6.59 0 Td[(,(2) 29
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where02(0,1)and>0.Clearly,ijisdecreasinginlagsothathigherlagtermswillgenerallybeclosertozero.Weletthepositiveparameterdeterminetheratethatijdecreasesinlag.TofullyspecifytheBayesianset-up,wemustintroducepriordistributionsonthetwoparameters,0and.Tospecifythesehyperpriors,weuseauniform(orpossiblyamoregeneralbeta)for0andagammadistributionfor.Werequire>0,soij=0jj)]TJ /F5 11.955 Tf 12.87 0 Td[(ij)]TJ /F12 7.97 Tf 6.59 0 Td[(remainsandecreasingfunctionoflag.InthesimulationsanddataanalysisofSections 2.5 and 2.6 ,weuseGamma(5,5),sothathasapriormeanof1andpriorvarianceof1=5.Weuseamoderatelyinformativepriortokeepfromdominatingtheroleof0inij=0jj)]TJ /F5 11.955 Tf 12.8 0 Td[(ij)]TJ /F12 7.97 Tf 6.59 0 Td[(.Alargevalueofwillforceallijoflaggreaterthanonetobeapproximatelyzero,regardlessofthevalueof0. 2.3.2SamplingundertheShrinkagePriorTheutilityofourpriordependsonourabilitytoincorporateitintoaMarkovchainMonteCarlo(MCMC)scheme.ForsimplicityweassumethatthedataconsistsofY1,...,YN,whereeachYiisaJ-dimensionalnormalvectorwithmeanzeroandcovarianceR,whichisacorrelationmatrixsoastomimicthecomputationsforthemultivariateprobitcase.LetL(jY)denotethelikelihoodfunctionforthedata,parameterizedbythePACs,.TheMCMCchainweproposeinvolvessequentiallyupdatingeachoftheJ(J)]TJ /F4 11.955 Tf 12.04 0 Td[(1)=2PACs,followedbyupdatingthehyperparametersdetermermingthevarianceoftheSBetadistributions.Tosampleaparticularij,wemustdrawthenewvaluefromthedistributionproportionaltoL(ij,()]TJ /F7 7.97 Tf 6.59 0 Td[(ij)jY)pij(ij),wherepij(ij)istheSBeta(ij,ij)densityand()]TJ /F7 7.97 Tf 6.58 0 Td[(ij)representsthesetofPACsexceptij.Duetothesubtleroleofijinthelikelihoodpiece,thereisnosimpleconjugatesamplingstep.InordertosamplefromL(ij,()]TJ /F7 7.97 Tf 6.58 0 Td[(ij)jY)pij(ij),weintroduceanauxiliaryvariableUij( Damienetal. 1999 ; Neal 30
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2003 ),andnotethatwecanrewritetheconditionaldistributionas L(ij,()]TJ /F7 7.97 Tf 6.59 0 Td[(ij)jY)pij(ij)=Z10Ifuij
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( 2013a )by=2,=1;alternatively,independenthyperpriorsfor,couldbespecied.Thevalueofijgivestheprobabilitythatijwillbenon-zero,i.e.willbedrawnfromthecontinuouscomponentinthemixturedistribution.Hence,wehavetheprobabilitythatYiandYjareuncorrelated,giventheintercedingvariables,is1)]TJ /F11 11.955 Tf 13.21 0 Td[(ij.Asthevaluesofthe'sdecrease,theselectionpriorplacesmoreweightonthepoint-mass0componentofthedistribution( 2 ),yieldingmoresparsechoicesfor.AswithourparameterizationsofthevarianceijinSection 2.3.1 ,wemakeastructuralchoiceoftheformofijsothatthisprobabilitydependsonthelag-value.Welet ij=0jj)]TJ /F5 11.955 Tf 11.96 0 Td[(ij)]TJ /F12 7.97 Tf 6.58 0 Td[(,(2)similartoourchoiceofijintheshrinkageprior.Thischoice( 2 )speciesthecontinuouscomponentprobabilitytobeanpolynomialfunctionofthelag.Becauseijisdecreasingasthelagj)]TJ /F5 11.955 Tf 12.85 0 Td[(iincreases,pr(ij=0)increases.Conceptually,thismeansthatweanticipatethatvariablesfartherapartintime(andconditionalonmoreintermediatevariables)aremorelikelytobeuncorrelated.Aswiththeshrinkageprior,wechoosehyperpriorsof0Unif(0,1)andGamma(5,5). 2.4.2NormalizingConstantforPriorsonROneofthekeyimprovementsofourselectionprioroverothersparsepriorsforRisthesimplicityofthenormalizingconstant,asmentionedintheintroduction.PreviouscovariancepriorswithasparseC( Carteretal. 2011 ; Pittetal. 2006 ; Wongetal. 2003 )placeaatprioronthenon-zerocomponentscijforagivenpatternofzeros.However,theneedednormalizingconstantrequiresndingthevolumeofthesubspaceofRJcorrespondingtothepatternofzerosinC.Thisturnsouttobeaquitedifculttaskandprovidesmuchofthechallengeintheworkofthethreepreviouslycitedpapers. 32
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WeareabletoavoidthisissuebyspecifyingourselectionpriorintermsoftheunrestrictedPACparameterization.Asthevalueofanyoftheij'sdoesnoteffectthesupportoftheremainingPACs,thevolumeof[)]TJ /F4 11.955 Tf 9.3 0 Td[(1,1]J(J)]TJ /F9 7.97 Tf 6.59 0 Td[(1)=2correspondingtoanycongurationofwithJ0(J(J)]TJ /F4 11.955 Tf 12.78 0 Td[(1)=2)non-zeroelementsis2J0,thevolumeofaJ0-dimensionalhypercube.Becausethisconstantdoesnotdependonwhichelementsarenon-zero,weneednotexplicitlydealwithitintheMCMCalgorithmtobeintroducedinthenextsubsection.Further,weareabletheexploitstructureintheorderofthePACsinselection(i.e.higherlagtermsaremorelikelytobenull),whereasin Pittetal. ( 2006 ),theprobabilitythatcijiszeroischosentominimizetheeffortrequiredtondthenormalizingconstant.AnadditionalbenetofperformingselectiononthepartialautocorrelationasopposedtothepartialcorrelationsCisthatthezeropatternsholdundermarginalizationsofthebeginningand/orendingtimepoints.Forinstance,ifwemarginalizeouttheJthtimepoint,thecorrespondingmatrixofPACsistheoriginalafterremovingthelastrowandcolumn.However,anyzeroelementsinCwillnotbepreservedbecausecorr(Y1,Y2jY3,...,YJ)=0doesnotgenerallyimplythatcorr(Y1,Y2jY3,...,YJ)]TJ /F9 7.97 Tf 6.59 0 Td[(1)=0. 2.4.3SamplingundertheSelectionPriorSamplingwiththeselectionpriorproceedssimilarlytotheshrinkagepriorschemewiththemaindifferencebeingtheintroductionofthepointmassin( 2 ).AsbeforewesequentiallyupdateeachofthePACs,bydrawingthenewvaluefromthedistributionproportionaltoL(ij,()]TJ /F7 7.97 Tf 6.59 0 Td[(ij)jY)pij(ij),wherepij(ij)givesthedensitycorrespondingthepriordistributionin( 2 )(withrespecttotheappropriatemixturedominatingmeasure).Wecannotusetheslicesamplingstepaccordingto( 2 )butmustwritethedistributionas L(ij,()]TJ /F7 7.97 Tf 6.59 0 Td[(ij)jY)pij(ij)=Z10Ifuij
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Fortheselectionprior,wesampleUijuniformlyovertheintervalfromzerotoL(ij,()]TJ /F7 7.97 Tf 6.58 0 Td[(ij)jY),usingthecurrentvalueofij,andthendrawijfrompij(),restrictedtotheslicesetP=f:uij
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Thesamplingdistributionsof0anddependononlythroughthesetofindicatorvariablesij.Aswiththevarianceparametersoftheshrinkagepriors,weincorporateapairofslicesamplingstepstoupdatethehyperparameters. 2.5SimulationsTobetterunderstandthebehaviorofourproposedpriors,weconductedasimulationstudytoassessthe(frequentist)riskoftheirposteriorestimators.WeconsiderfourchoicesADforthetruecovariancematrixinthecaseofsix-dimensional(J=6)data.RAwillhaveanautoregressive(AR)structurewithAij=0.7jj)]TJ /F7 7.97 Tf 6.58 0 Td[(ij.ThecorrespondingAhasvaluesof0.7forthelag-1termsandzerofortheothers,asparseparameterization.ForthesecondcorrelationmatrixRBwechoosetheidentitymatrixsothatallofPACsarezerointhiscase.TheChasastructurethatdecaystozero.Forthelag-1termsCi,i+1=0.7,andfortheremainingterms,Cij=0.4j)]TJ /F7 7.97 Tf 6.59 0 Td[(i)]TJ /F9 7.97 Tf 6.59 0 Td[(1,j)]TJ /F5 11.955 Tf 11.96 0 Td[(i>1.NeitherCnorRChavezeroelements,butCijdecreasequicklyinlagj)]TJ /F5 11.955 Tf 12.04 0 Td[(i.Finally,weconsideracorrelationmatrixthatcomesfromasparseD,D=266666641.9.30000.901.8.4.100.800.801.6.200.620.670.601.8.30.580.630.580.801.70.460.500.450.690.70137777775,wheretheupper-triangularelementscorrespondtoDandthelower-triangularelementsdepictthemarginalcorrelationsfromRD.NotethatwhileDissomewhatsparse,RDhasonlynon-zeroelements.ForeachofthesefourchoicesofthetruedependencestructureandforsamplesizesofN=20,50,and200,wesimulate50datasets.Foreachdatasetaposteriorsamplefor(andhence,R)isobtainedbyrunninganMCMCchainfor5000iterations,afteraburn-inof1000.Weuseeverytenthiterationforinference,givingasampleof500valuesforeachdataset.Weconsidertheperformanceofboththeselectionandshrinkagepriorson.Fortheselectionprior,weperformanalyseswithSBeta(1,1) 35
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(i.e.,Unif()]TJ /F4 11.955 Tf 9.3 0 Td[(1,1))andSBeta(2,1)(triangularprior)forthecontinuouscomponentofthemixturedistributions( 2 ).Inboththeselectionandshrinkagepriors,thehyperpriorsare0Unif(0,1)andGamma(5,5).Theestimatorsfromtheshrinkageandselectionpriorsarecomparedwiththeestimatorsresultingfromtheat-R,at-PAC,andtriangularpriors.Finally,weconsideranaiveshrinkagepriorwhereisxedatzeroin( 2 ).Here,allPACsareequallyshrunkwithvarianceij=0independentoflag.Weconsidertwolossfunctionsincomparingtheperformanceofthesixpriorchoices:L1(^R,R)=tr(^RR)]TJ /F9 7.97 Tf 6.59 0 Td[(1))]TJ /F4 11.955 Tf 12.96 0 Td[(logj^RR)]TJ /F9 7.97 Tf 6.59 0 Td[(1j)]TJ /F5 11.955 Tf 19.93 0 Td[(pandL2(^,)=Pi
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Figure2-1. BoxplotsoftheobservedlossusingL1(^R1,R)fortheJ=6cases.Thepriordistributionscomparedare(1)shrinkage,(2)selection(2,1),(3)selection(1,1),(4)at-R,(5)at-,(6)triangular,and(7)naiveshrinkage. 37
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Table2-1. RiskestimatesforsimulationstudywithdimensionJ=6.Correlationmatrices:Aautoregressivestructure;Bindependence;Cnon-zerodecaying;Dsparse.Lossfunctions:L1(^R,R)=tr(^RR)]TJ /F9 7.97 Tf 6.58 0 Td[(1))]TJ /F4 11.955 Tf 11.96 0 Td[(logj^RR)]TJ /F9 7.97 Tf 6.59 0 Td[(1j)]TJ /F5 11.955 Tf 17.93 0 Td[(p;L2(^,)=Pi
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Table2-2. 1010PACmatrixD0shownabovethediagonalanditsrespectivecorrelationmatrixRD0shownbelowthediagonal.D0=26666666666666641.9.300000000.901.8.4.1000000.800.801.6.2000000.620.670.601.8.300000.580.630.580.801.700000.460.500.450.690.701.8.4.100.370.400.360.550.560.801.6.200.310.340.300.460.470.670.601.8.30.290.320.290.430.440.630.580.801.70.230.250.230.340.350.500.450.690.7013777777777777775 estimatedriskforat-Risvisiblyworsethantheothers.RecallthatCijisdecreasinginlagbutisnotequaltozero.Infact,thesmallestelementC16=(0.4)4=0.0256whichmaynotbecloseenoughtozerotobeeffectivelyzeroedout,explainingwhytheselectionpriorsarelesseffectiveforCthanintheotherscenarios.WhenweconsiderestimatingthesparsecorrelationmatrixD,theshrinkageandselectionpriorsoutperformthefourotherpriors.FromTable 2-1 weseethatforlossfunction1andtheN=20samplesizetheestimatedriskdecreasesby45(25),45(24)and39(16)percentfortheestimatesfromtheshrinkage,selection(2,1),andselection(1,1)priorsovertheat-R(at-)priors.Thisisquiteasubstantialdropforthesmallsamplesize.Fortheothersamplesizeswestillobservedacleardecreaseovertheatpriors.ForN=50thereisadropof32(20),26(14),and22(9)percentforthesparsepriorsovertheatpriors,andwithN=200adecreaseof13(9),10(7),and7(4)percent.ToinvestigatehowourpriorsbehaveasJincreases,werepeattheanalysisusingthenon-sparsedecayingRCandasparseRD0withthedimensionofthematrixincreasedtoJ=10.Again,Ci,i+1=0.7forthelag-1termsandCij=0.4j)]TJ /F7 7.97 Tf 6.59 0 Td[(i)]TJ /F9 7.97 Tf 6.59 0 Td[(1forallj)]TJ /F5 11.955 Tf 12.56 0 Td[(i>1,andweexpandthepreviousRDtothe1010RD0showninTable 2-2 .AsbeforetheabovediagonalelementsarefromD0andthebelowdiagonalelements 39
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Table2-3. RiskestimatesforsimulationstudywithdimensionJ=10.Correlationmatrices:Cnon-zerodecaying;D0sparse.Lossfunctions:L1(^R,R)=tr(^RR)]TJ /F9 7.97 Tf 6.59 0 Td[(1))]TJ /F4 11.955 Tf 11.96 0 Td[(logj^RR)]TJ /F9 7.97 Tf 6.58 0 Td[(1j)]TJ /F5 11.955 Tf 17.93 0 Td[(p;L2(^,)=Pi
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missingnessduetostudydropout.Asinpreviousanalysesofthisdata( Daniels&Hogan 2008 ),weassumethismissingnessisignorable.Forpatienti=1,...,N(N=281),wedenotethevectorofquitstatusesbyQi=(Qi1,...,QiJ)0.Weonlyconsidertheresponsesafterpatientsareaskedtoquit,weeks5through12(J=8).HereQit=1indicatesasuccess(notsmoking)forpatientiattimet(1tJ,correspondingtoweekt+4),Qit=)]TJ /F4 11.955 Tf 9.3 0 Td[(1forafailure(smokingduringtheweek),andQit=0iftheobservationismissing.Followingtheusualconventionsofthemultivariateprobitregressionmodel( Chib&Greenberg 1998 ),weletYibetheJ-dimensionalvectoroflatentvariablescorrespondingtoQi.Thus,Qit=1impliesthatYit0,andQit=)]TJ /F4 11.955 Tf 9.3 0 Td[(1givesYit<0.WhenQit=0,thesignofYitrepresentsthe(unobserved)quitstatusfortheweek.WeassumethelatentvariablesfollowamultivariatenormaldistributionYiNJ(i,R)fori=1,...,N,wherei=Xi,XiisaJqmatrixofcovariatesandaq-vectorofregressioncoefcients.AsthescaleofYisunidentied,thecovariancematrixofYisconstrainedtobeacorrelationmatrixR.WeconsidertwochoicesofXi:`time-varying'whichspeciesadifferentitforeachtimewithineachtreatmentgroup(q=2J)and`time-constant'whichgivesthesamevalueofitacrossalltimeswithintreatmentgroup(q=2).Withthetime-constantandtime-varyingchoicesofthemeanstructure,weconsiderthefollowingpriorsforR:shrinkage,selection,at-R,at-,triangular,naiveshrinkage,andanautoregressive(AR)prior.TheARpriorassumesanAR(1)structureforR,thatis,ij=jj)]TJ /F7 7.97 Tf 6.59 0 Td[(ijandi,i+1=andij=0ifjj)]TJ /F5 11.955 Tf 11.95 0 Td[(ij>1.WeassumeaUnif()]TJ /F4 11.955 Tf 9.3 0 Td[(1,1)distributionfor.Asintherisksimulation,weconsidertheselectionpriorwithbothSBeta(1,1)andwithSBeta(2,1)forthecontinuouscomponent.Theremainingpriordistributionstobespeciedare0Unif(0,1),Gamma(5,5),andthepriorontheregressioncoefcientsisat. 42
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ToanalyzethedatawerunanMCMCchainfor12,000iterationsafteraburn-inof3000,retainingeverytenthobservation.Convergencewasassessedthroughgraphicaldiagnosticsanddeemedadequate.TherearethreesetsofparameterstosampleintheMCMCchain:theregressioncoefcients,thecorrelationmatrix,andthelatentvariables.TheconditionalforgivenYandRismultivariatenormal.SamplingthecorrelationmatrixevolvesasdiscussedinSections 2.3.2 and 2.4.3 usingtheresidualsYi)]TJ /F16 11.955 Tf 12.05 0 Td[(i.ThelatentvariablesYi,whichareconstrainedbyQi,aresampledaccordingtothestrategyof Liuetal. ( 2009 ,Proposition1).Tocomparethespecicationbasedonourpriorchoices,wemakeuseofthedevianceinformationcriterion(DIC; Spiegelhalteretal. 2002 ).TheDICstatisticcanbeviewedsimilarlytotheBayesianorAkaikeinformationcriterion,butDICdoesnotrequiretheusertocountthenumberofmodelparameters.ThisiskeyforBayesianmodelsthatutilizeshrinkageand/orsparsitypriorsasitisnotclearwhetherorhowoneshouldcountaparameterthathasbeensettoorshrunktowardzero.Tothatend,let Dev=)]TJ /F4 11.955 Tf 9.3 0 Td[(2loglik(^,^RjQ)=Xi)]TJ /F4 11.955 Tf 9.3 0 Td[(2loglik(^,^RjQi)(2)bethedevianceortwicethenegativelog-likelihoodwiththeparameters^and^R.Here^istheposteriormean,andforthecorrelationestimate^R,weusetherstoftheestimatorsweconsideredinSection 2.5 ,^R=SEfR)]TJ /F9 7.97 Tf 6.59 0 Td[(1g)]TJ /F9 7.97 Tf 6.58 0 Td[(1SwithS=[diag(EfR)]TJ /F9 7.97 Tf 6.59 0 Td[(1g)]1=2.ThecomplexityofthemodelismeasuredbythetermpD,sometimescalledtheeffectivenumberofparameters.ThispDiscalculatedas pD=Ef)]TJ /F4 11.955 Tf 15.27 0 Td[(2loglik(,RjQ)g)]TJ /F1 11.955 Tf 20.59 0 Td[(Dev,(2)wheretheexpectationisovertheposteriordistributionoftheparameters(,R).TheDICmodelcomparisonstatisticisDIC=Dev+2pD,thesumoftermsmeasuringmodeltandcomplexity.SmallervaluesofDICarepreferred. 43
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Table2-4. ModelcomparisonstatisticsfortheCTQdata. MeanStructureCorrelationPriorDevpDDIC Time-constantShrinkage1031141060Time-constantSelection(2,1)1042121066Time-constantSelection(1,1)1044121068Time-constantTriangular1029201068Time-constantat-1029201069Time-constantNaiveshrinkage1033201074Time-constantAR107131078Time-constantat-R1043211086Time-varyingShrinkage1022251071Time-varyingTriangular1017301077Time-varyingSelection(2,1)1033221077Time-varyingSelection(1,1)1036221080Time-varyingat-1019301080Time-varyingNaiveshrinkage1023311085Time-varyingAR1068131093Time-varyingat-R1034311097 As Wang&Daniels ( 2011 )pointout,DICshouldbecalculatedusingtheobserveddata,whichinthiscaseisthequitstatusresponsesQinotthelatentvariablesYi.Hencethelog-likelihoodforQiatparameters(,R)isequalto loglik(,RjQi)=logZ(,1)JIfQityt08tg(yjXi,R)dy,(2)where(j,)istheJ-dimensionalmultivariatenormaldensitywithmeanandcovariancematrix.Theintegralin( 2 )isnottractablebutcanbeestimatedusingimportancesampling( Robert&Casella 2004 ,Section3.3).SeeAppendix A fordetailsaboutestimatingtheDIC.Themodelt(Dev),complexity(pD),andcomparison(DIC)statisticsareinTable 2-4 ;DICstatisticswereestimatedwithastandarderrorofapproximately0.5.WeseethatthemodelsthatuseameanstructurethatdependsonlyontreatmentandnottimettendtohavelowerDICvalues.Thetime-varyingmodelsarepenalizedinthepDtermforhavingtoestimatetheadditional14regressioncoefcients.Ofthe 44
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correlationpriorstheat-RandARpriorsperformmuchworsethantheshrinkage,selection,triangular,andat-PACpriorswiththesamemeanstructure.Additionally,theselectionpriorthatusesthetriangularformforSBeta(=2,=1)tendtohaveasmallerDICthantheSBeta(1,1)priors.FromTable 2-4 wedeterminethepriorchoicethatbestbalancesmodeltwithparsimonyisclearlythemodelwithtime-constantmeanstructureandtheshrinkageprioronthecorrelationmatrixprior.Usingthisbestttingmodel,theposteriormeanofis()]TJ /F4 11.955 Tf 9.3 0 Td[(0.504,)]TJ /F4 11.955 Tf 9.3 0 Td[(0.295)implyingthatthemarginalprobability(95%credibleinterval)ofnotsmokingduringagivenstudyweekis()]TJ /F4 11.955 Tf 9.29 0 Td[(0.504)=0.307(0.24,0.37)forthecontrolgroupand()]TJ /F4 11.955 Tf 9.3 0 Td[(0.295)=0.384(0.32,0.45)fortheexercisegroup,where()isthedistributionfunctionofthestandardnormaldistribution.Thetestofthehypothesisthatthecontroltreatmentisaseffectiveastheexercisetreatment(i.e.,H0:12)hasaposteriorprobabilityof0.06,providingsomeevidencetotheclaimthatexerciseimprovescessationresults.Wenowexamineinmoredetailtheeffecttheshrinkagepriorhasonmodelingthecorrelationmatrix.Theposteriormeans(credibleinterval)oftheshrinkageparametersare^0=0.406(0.25,0.60)and^=2.44(1.6,3.4).Withavalueofgreaterthan1,thevarianceofijisdecayingtozerofairlyrapidly.Theposteriormeanofis^=2666666666641.000.700.120.020.050.000.00-0.010.711.000.830.160.090.020.010.000.640.841.000.810.120.100.060.020.560.740.821.000.780.240.090.030.510.640.690.791.000.810.370.040.480.610.660.740.831.000.880.210.480.610.670.740.830.891.000.780.400.520.570.630.700.770.801.00377777777775,withthelowerdiagonalvaluesgivingtheelementsof^R.WeseethatthePACsarefarfromzeroinonlythersttwolagsandtheremaining'sareclosetozero.Thisisbecausethesepartialautocorrelationshavebeenshrunkalmosttozeroinmostiterations. 45
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2.7DiscussionInthispaperwehaveintroducedtwonewpriorsforcorrelationmatrices,ashrinkagepriorandaselectionprior.ThesepriorschooseasparseparameterizationofthecorrelationmatrixthroughthesetofPACs.Intheselectioncontext,bystochasticallyselectingtheelementsoftozeroout,ourmodelndsinterpretableindependencerelationshipsfornormaldataandavoidstheneedforcomplexmodelselectionofthedependencestructure.Akeyimprovementoftheselectionprioroverexistingmethodsforsparsecorrelationmatricesisthatourapproachavoidsthecomplexnormalizingconstantsseeninpreviouswork.Additionally,insettingswithtime-ordereddata,thepartialautocorrelationsaremoreinterpretablethanthefullpartialcorrelations,astheydonotinvolveconditioningonfuturevalues.Whiletheexampleswehaveconsideredhereinvolvesituationswherethecovariancematrixwasconstrained(asinthedataexample)orknown(asinthesimulations)tobeacorrelationmatrix,theextensiontoarbitraryissimple.Returningtotheseparationstrategy=SRS( Barnardetal. 2000 ),apriorforcanbeformedbyplacingindependentpriorsonSandR,i.e.p()=p(R)p(S).Usingoneoftheproposedpriorsforp(R),sensiblechoicesofp(S)includeanindependentinversegammaforeachofthejjoraatprioronfS=diag(11,...,JJ):jj>0g.ThisleadstoaprioronwithsparsePACs.ThesimulationsanddatawehaveconsideredheredealwithYoflowormoderatedimension.WeprovideafewcommentsregardingthescalabilityofourapproachfordatawithlargerJ.AswebelievethatPACsoflargerlagplayaprogressivelysmallerroleindescribingthe(temporal)dependence,itmaybereasonabletospecifyamaximumallowablelagfornon-zeroPACs.Thatis,wechoosesomeksuchthatij=0forallj)]TJ /F5 11.955 Tf 12.09 0 Td[(i>kandsampleij(j)]TJ /F5 11.955 Tf 12.09 0 Td[(ik)fromeitherourshrinkageorselectionprior.Bandingthematrixisrelatedtotheideaofbandingthecovariancematrix( Bickel&Levina 2008 ),concentrationmatrix( Rothmanetal. 2008 ),ortheCholeskydecomposition 46
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of)]TJ /F9 7.97 Tf 6.59 0 Td[(1( Rothmanetal. 2010 ).Bandinghasalsobeenstudiedby Wang&Daniels ( 2013b ).Inadditiontoreducingthenumberofparametersthatmustbesampled,othermatrixcomputationswillbefasterbyusingpropertiesofbandedmatrices.Relatedtothis,modicationstotheshrinkagepriormaybeneededforlargerdimensionJ.Recallthatthevarianceofijisij=0jj)]TJ /F5 11.955 Tf 12.48 0 Td[(ij)]TJ /F12 7.97 Tf 6.58 0 Td[(.Forlargelags,thiscanbeveryclosetozeroleadingtonumericalinstability;recalltheparametersoftheSBetadistributionareinverselyrelatedtoijthroughij=ij=()]TJ /F9 7.97 Tf 6.59 0 Td[(1ij)]TJ /F4 11.955 Tf 12.32 0 Td[(1)=2.Replacing( 2 )withij=0minfjj)]TJ /F5 11.955 Tf 12.36 0 Td[(ij,kg)]TJ /F12 7.97 Tf 6.59 0 Td[(orij=0+1jj)]TJ /F5 11.955 Tf 12.36 0 Td[(ij)]TJ /F12 7.97 Tf 6.59 0 Td[(toboundthevariancesawayfromzeroorbandingaftertherstklagsprovidetwopossibilitiestoavoidsuchnumericalissues.Further,wehaveparametrizedthevariancecomponentandtheselectionprobabilityinsimilarwaysinourtwosparsepriors.Thequantityisoftheform0jj)]TJ /F5 11.955 Tf 12.12 0 Td[(ij)]TJ /F12 7.97 Tf 6.59 0 Td[(forbothijin( 2 )andijin( 2 ),butotherparameterizationsarepossible.Wehaveconsideredsomesimulations(notincluded)allowingthevariance/selectionprobabilitytobeuniqueforlag,i.e.ij=jj)]TJ /F7 7.97 Tf 6.59 0 Td[(ij.ApriorneedstobespeciedforeachoftheseJ)]TJ /F4 11.955 Tf 12.47 0 Td[(1's,ideallydecreasinginlag.Alternatively,onecoulduse0=jj)]TJ /F5 11.955 Tf 12.88 0 Td[(ij,whichcanbeviewedasaspecialcasewheretheprioronisdegenerateat1.Inourexperienceresultswerenotverysensitivetothechoiceoftheparameterization,andposteriorestimatesofandRweresimilar.Inaddition,wehavefocusedourdiscussiononthecorrelationestimationprobleminthecontextofanalysiswithmultivariatenormaldata.WenotethatthesepriorsareadditionallyapplicableinthecontextofestimatingaconstrainedscalematrixforthemultivariateStudentt-distribution.ConsidertherandomvariableYtJ(,R,).Thatis,YfollowsaJ-dimensionalt-distributionwithlocation(mean)vector,scalematrixR(constrainedtobeacorrelationmatrix),anddegreesoffreedom(eitherxedorrandom).Usingthegamma-mixture-of-normalstechnique( Albert&Chib 1993 ),werewritethedistributionofYtobeYjNJ(,)]TJ /F9 7.97 Tf 6.58 0 Td[(1R)andGamma(=2,=2). 47
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SamplingforRaspartofanMCMCchainfollowsasinSections 2.3.2 and 2.4.3 usingY?=p (Y)]TJ /F16 11.955 Tf 12.19 0 Td[()asthedata.However,oneshouldnotethatazeroPACijimpliesthatYiandYjareuncorrelatedgivenYi+1,...,Yj)]TJ /F9 7.97 Tf 6.58 0 Td[(1,butthisisnotequivalenttoconditionalindependenceasinthenormalcase. 48
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CHAPTER3ANONPARAMETRICPRIORFORSIMULTANEOUSCOVARIANCEESTIMATION 3.1SimultaneousCovarianceEstimationWhenworkingwithlongitudinaldata,specifyingthemodelforthedependencestructureisamajorconsideration.Oftenthedataarecomposedofseveralgroups,suchasdifferingtreatmentsinaclinicaltrial.Inmanycases,particularlyifonedoesnothavemanyobservationspergroup,oneassumesthatthecovarianceorcorrelationstructureisconstantacrossallgroups.However,thisassumption,ifitfailstohold,canhaveadramaticeffectontheinferenceformeaneffects,evensometimesleadingtobias.Conversely,ifonespecieseachofthecovariancematriceswithoutregardtotheothergroups,thiscanleadtoalossofinformation.Dealingwiththesecompetingmodelsforthecovariancestructureisaconcerninmanystatisticalapplications,suchasclassicationandmodel-basedclustering.Therefore,itisdesirabletodevelopmethodstosimultaneouslyestimatethesetofcovariancematricesthatwillborrowinformationacrossgroupsinacoherent,automatedmannerallowingforstructuralzeros,commonalityacrosssubsetsofthegroups,andappropriateequalityofparameterswithinagroup.Whenthedataarefullyobservedundermultivariatenormality,themeanandcovarianceparametersareorthogonalinthesenseof Cox&Reid ( 1987 ),andthemeanparameterswillbeconsistentundermisspecicationofthecovariancestructure.However,ifthereismissingness,asisoftenthecaseforlongitudinaldata,thereisnolongerorthogonality,evenatthetruevalueofthecovariancematrix( Little&Rubin 2002 ).Hence,fortheposteriordistributionofthemeanparameterstobeconsistent,thedependencestructuremustbecorrectlyspecied,anditisnotappropriatetotreatthecovariancematrixasanuisanceparameter( Daniels&Hogan 2008 ,Section6.2).Further, Crippsetal. ( 2005 )demonstrateefciencygainsfortheregressionparametersforfullyobserveddatabyusingparsimoniousmodelsforthecovariancematrix. 49
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AssumethatwehaveMgroupsofnormallydistributedlongitudinaldatawithnmresponsesofdimensionp,Ymiforthemthgroup.Weassumewithoutlossofgeneralitythatthemeanvectorforeachgroupiszero.ThedistributionoftheYmiisYmijmNp)]TJ /F4 11.955 Tf 5.48 -9.68 Td[(0,m(i=1,...,nm;m=1,...,M),withthecovariancematrixm=(m,)]TJ /F7 7.97 Tf 12.25 -1.79 Td[(m)parameterizedbythegeneralizedautoregressiveparameters,mandinnovationvariances,)]TJ /F7 7.97 Tf 6.78 -1.8 Td[(m,asdescribedby Pourahmadi ( 1999 2000 ).Forbrevity,wesometimesrefertothegeneralizedautoregressiveparametersastheautoregressiveparameters.WealsorefertothisasthemodiedCholeskyparameterization,sincetheparametersarederivedbyperformingaCholeskydecompositiononm,(m,)]TJ /F7 7.97 Tf 12.25 -1.79 Td[(m))]TJ /F9 7.97 Tf 6.59 0 Td[(1=T(m)D()]TJ /F7 7.97 Tf 11.66 -1.79 Td[(m)T(m)>.Here,)]TJ /F7 7.97 Tf 6.77 -1.79 Td[(m=(m1,...,mp),andD()]TJ /F7 7.97 Tf 11.65 -1.8 Td[(m)isappdiagonalmatrixwith(j,j)-element(mj))]TJ /F9 7.97 Tf 6.58 0 Td[(1.TheT(m)matrixisupper-triangularwithonesonthemaindiagonal,andtheabove-diagonalelementsaregivenbythenegativesofm.Theelementsofm=(m1,...,mJ)areindexedbyj=1,...,J(J=p(p)]TJ /F4 11.955 Tf 12.08 0 Td[(1)=2)correspondingtolocation(j1,j2)inT(m)(1
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toestimatesparsegraphicalmodelsbyselectingsetsofedgescommontoallgroups,aswellasgroup-specicedges.However,theirmethodshareslittleinformationaboutnon-zeroparametersacrossthegroups. Pourahmadietal. ( 2007 )developedestimationandtestingproceduresforequalityamongsubsetsofthemj's.Inaclusteringcontext,modelsassumingT(m)and/orD()]TJ /F7 7.97 Tf 11.66 -1.79 Td[(m)tobeeitherconstantordistinctacrossallgroupsweredevelopedby McNicholas&Murphy ( 2010 ). Daniels ( 2006 )consideredaBayesianperspectivebyintroducingpriorsfortheparametersoftheCholeskydecomposition,aswellastheprincipalcomponentsofthecovariancematrices,thatinducepoolingacrossgroups.Unfortunately,itiscomputationallychallengingtoselectamongallthepossiblemodelswithintheseclasses. Hoff ( 2009 )alsoconsidersamodelthatshrinkstowardacommoneigenvectorstructure,allowingtheextentofthepoolingtovaryacrosseachprincipleaxis.Othermethodshavebeenproposedthatmodelthecovariancematrixasaregressionfunctionofacontinuouscovariate( Chiuetal. 1996 ; Daniels 2006 ; Fox&Dunson 2011 ; Hoff&Niu 2012 ).However,covarianceregressionmodelsareoftenplaguedbythedifcultyofinterpretingtheregressionparameters.InthischapterwefocussolelyonthemodiedCholeskyparameterizationbecauseoftheunrestrictednessoftheparameters,theinterpretabilityforlongitudinaldata,andthecomputationaladvantagesviaconjugacy( Daniels&Pourahmadi 2002 ).Ourgoalistodevelopapriorforthesets=f1,...,Mgand)-406(=f)]TJ /F9 7.97 Tf 6.78 -1.79 Td[(1,...,)]TJ /F7 7.97 Tf 30.19 -1.79 Td[(MginsuchawaythatweborrowstrengthacrosstheMgroups.Additionally,wewanttoshareinformationacross)]TJ /F7 7.97 Tf 6.78 -1.79 Td[(mandmvalues,particularlythoseautoregressiveparametersofacommonlag.AnotherconsiderationforpriordevelopmentistoencouragesparsityoftheelementsofT(m).Becauseeachmjrepresentsaconditionaldependency,settingmjtozeroestablishesaconditionalindependencerelationshipbetweenapairofcomponentsofY.Itisnecessarytoconsiderpriorsthatallowthedatatoinformthebalancebetweenthesetwogoals:poolingacrossgroupsandintroducingsparsity.Aboveall,weseektoaccomplishthisinanautomated,stochasticfashion.Toform 51
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XjH=1forallj,sothatthestick-breakingweightssumtoone,guaranteeingFmjisavaliddistribution.Thematrixstick-breakingprocessisthendenedusingtheabovespecicationasH!1,andtheauthorsrefertotheniteHcaseasthetruncationapproximationtothematrixstick-breakingprocess.Wecanconsidertheadequacyofthisapproximationusingamethodsimilartothatemployedby Ishwaran&James ( 2001 ). Dunsonetal. ( 2008 )showthatforasetfmjhgdrawnfromthefullprocess, E 1Xh=Hmjh!=1)]TJ /F4 11.955 Tf 48.06 8.09 Td[(1 (1+)(1+)H)]TJ /F9 7.97 Tf 6.58 0 Td[(1.(3)WemaychoosethenumberofclustersHsuchthatthisexpectedapproximationerror( 3 )isarbitrarilysmall,sotheeffectoftheapproximationisnegligible.BecausetheprobabilitymeasuresFmjandFm0jfortwogroupsmandm0sharethesamesetofatomsfj1,...,jHg,thereisapositiveprobabilitythatmjwillequalm0j.Thisoccurswhenmjandm0jaredrawnfromthesamecluster,thatis,ifmj=m0j=jhforsomehin1,...,H.Theprobabilityofthisoccurringisaknownfunctionofthestick-breakingparametersand. 3.3CovarianceGroupingPriors 3.3.1Lag-BlockGroupingPriorforWenowproposepriorstouseforsimultaneouscovarianceestimationbasedonthematrixstick-breakingprocess.Thesepriorsarereferredtoasgroupingpriorsbecausetheyinducegroupingamongthevaluesofthevariousparameters.Tothisend,weindependentlyplacepriorsonand)]TJ /F1 11.955 Tf 10.1 0 Td[(withtheprioroninducedbythemappingm=(m,)]TJ /F7 7.97 Tf 12.26 -1.8 Td[(m).Becauseand)]TJ /F1 11.955 Tf 10.1 0 Td[(areorthogonalparameters( Pourahmadi 2007 ),itissensibletochooseindependentpriors. 53
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Thepriorfor,referredtoasthelag-blockgroupingprior,isdenedasfollows. mjFmj()=HXh=1mjhq(j)h()(m=1,...,M;j=1,...,J), (3) qhq0+(1)]TJ /F11 11.955 Tf 11.95 0 Td[(q)N(0,2)(q=1,...,p)]TJ /F4 11.955 Tf 11.96 0 Td[(1;h=1,...,H), (3) mjh=UmhXjhYl
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Weformtheprobabilitiesfmjhgasin Dunsonetal. ( 2008 ).Theandstick-breakingparametersservethesameroleasandbefore.Wesubscriptthemwithtodistinguishthestick-breakingparametersfortheprioronfromtheparameterstobedenedfortheprioron)]TJ /F1 11.955 Tf 6.77 0 Td[(.TheUmhandXjhparametersfollowthesameinterpretationasinthematrixstick-breakingprocess,butwhilewesharecandidatesacrossautoregressiveparametersofthesamelag,eachparameterhasitsownvaluesXjh.Akeydistinctionbetweenourpriorandtheoriginalprocessof Dunsonetal. ( 2008 )istheuseofthesamesetofcandidatevaluesfordifferentparameters.Thishasimportantconsequencesforthetheoreticalpropertiesofourpriors.Inparticular,formjandmj0withj6=j0andq(j)=q(j0),i.e.differentautoregressiveparametersforacommongroupandlag,theirdistributionsFmjandFmj0arepositivelycorrelated,whereasundertheoriginalspecicationtheywouldbeuncorrelated.Thisimplicationisquiteattractiveforlongitudinaldataasitfollowscommonintuition.Forexample,itmaybereasonabletoconsiderthattheregressioneffectofYtontoYt)]TJ /F9 7.97 Tf 6.59 0 Td[(1tobethesamefordifferentvaluesoft.WediscussthesepropertiesfurtherinSection 3.4 3.3.2Correlated-LognormalGroupingPriorfor)]TJ /F1 11.955 Tf -251.13 -24.53 Td[(Wenowdenethepriorfortheinnovationvariances)]TJ /F1 11.955 Tf 10.1 0 Td[(asfollows. mjGmj()=HXh=1mjhjh()(m=1,...,M;j=1,...,p), (3) jh=exp(!jh)(j=1,...,p;h=1,...,H),!h=(!1h,...,!ph)TNpf 1p,R()g(h=1,...,H), (3) mjh=WmhZjhYl
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Wedrawtheinnovationvariancemjfromthestick-breakingmeasureGmj,wherethecandidateatomsaredrawnbyexponentiatingamultivariatenormalvariable!h.Theprobabilitymjhofeachoftheatomsisformedusingthestick-breakingmethodontheproductofWandZ.Thesebetarandomvariablesdependontheparametersand.Thecandidatesjharedrawninacorrelatedfashionunliketheoriginalmatrixstick-breakingprocessandmarginallyfollowalognormaldistribution,providingthenameofthisprior.Weintroducetheintermediatevariable!hin( 3 ),whichisap-dimensionalnormallydistributedrandomvectorwithmeanvector 1pandcovariancematrixR().Here, andarescalarquantities,>0,andR()isthecorrelationmatrixcorrespondingtoanautoregressivefunctionoforder1.The(i,j)componentofR()isji)]TJ /F7 7.97 Tf 6.58 0 Td[(jj.Thischoiceismotivatedbythefactthatonesometimesconsiderstheinnovationvariancesasrealizedvaluesofsomeunknownsmoothfunctionoftime.Similartothelag-blockpriorwewillobtaintheatomsjhfortherandommeasureGmjinadependentway,whileleavingtheconstructionoftheprobabilityweightsmjhunchanged.Inthespecialcasewhere=0,thecomponentsofthe!hvectorareindependent.Consequently,theinnovationvariancecandidatesjharedistributedaccordingtothelognormal( ,)distribution,andthisspecialcasefollowsthematrixstick-breakingprocessframework.Inadditiontothegroupingpriorsthatwehavedenedhere,thereareotherpossibilitiestoformsimilarpriorsonthesetf1,...,Mgusingthematrixstick-breakingprocessframework.WeexploresomeoftheseinAppendix B 3.4TheoreticalProperties 3.4.1GeneralizedAutoregressiveParameterPropertiesWenowexploresomeofthetheoreticalpropertiesoftheproposedgroupingpriorsinthecasewhereH,H!1.Recallthatthematrixstick-breakingprocessisformally 56
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denedtobethelimitingdistributionasthenumberofclustersapproachesinnity,andthenitenumberofclusterscase,whilenecessaryforimplementation,isviewedasanapproximation.Ourgroupingpriorsfollowinthesameway.Thefollowingproperties,( 3 )( 3 ),arederivedfortheselimitingdistributions,andweensurethatthenumberofclustersischosenlargeenoughthatthesepropertiesmaybeconsideredtoholdapproximately.TheinitialpropertiesmirrorPropositions1,2,and4of Dunsonetal. ( 2008 ).Partialderivationsof( 3 )( 3 )areprovidedinAppendix C .First,weconsiderthebehavioroffromthelag-blockgroupingprior.Forthefollowingcalculations,weassumethattheq'sandallhyperparametersarexed.Additionally,foreaseofnotation,weignorethesubscriptonq,,whenitisclearfromcontext,andlet()denotetheprobabilitymeasurefortheNormal(0,2)distribution.Dene()=0()+(1)]TJ /F11 11.955 Tf 12.31 0 Td[()(),theprobabilitymeasureforthemixturedistributionoftheqh's.ForallsetsAintheBoreleldofthereallineB(R), EfFmj(A)g=(A),VarfFmj(A)g=2 (2+)(2+))]TJ /F9 7.97 Tf 6.58 0 Td[(2(A)f1)]TJ /F4 11.955 Tf 11.96 0 Td[((A)g. (3) Thisunbiasednesspropertyshowsthatitisappropriatetorefertothe0-normalmixtureasthebasedistributionfor.TheformofthevarianceshowsthatandcontroltheextenttowhichtherandommeasureFmjdiffersfromthebasedistribution.Aseitherorapproachinnity,thedistributionofmjcollapsestotheparametricbase;smallvaluesofandallowforamoreexibleprior.Fortwodifferentgroupsm6=m0, corrfFmj(A),Fm0j(A)g=+=2++1 2+++1. (3) BecausethiscorrelationbetweenamountofmassthedistributionfunctionsassigntothesetAdoesnotdependonthechoiceofA,itmaybeusedasasimpleunivariatemeasureofthedegreetowhichinformationissharedacrossgroups.Simplealgebra 57
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showsthat1=2corrfFmj(A),Fm0j(A)g1.Inparticular,corrfFmj(A),Fm0j(A)gapproaches1=2asaseitherorapproachinnityandapproaches1as!0.Forgroupsm6=m0,theprobabilityofmatchingforthejthautoregressiveparameteris pr(mj=m0j)=2+1)]TJ /F12 7.97 Tf 6.58 0 Td[(2 (1+)(2+))]TJ /F9 7.97 Tf 6.59 0 Td[(1. (3) Thepresenceofthezeropointmassincausesourpropertiestodifferfromthosederivedin Dunsonetal. ( 2008 ).Aseitherorapproachinnity,thisprobabilityapproaches2,theprobabilitythatbothmjandm0jarezeroifdrawnindependentlyfromtheparametricbasedistribution.Therighthandsideof( 3 )isincreasingin,aslargervaluesofindicatethatbothtermsaremorelikelytobezerowhetherornottheycomefromthesamecluster.Additionally,( 3 )increaseswheneitheranddecreases,coincidingwiththeincreasein( 3 ).Consideringtwodifferentautoregressiveparametersj6=j0ofthesamelagq=q(j)=q(j0)fromthesamegroupm,wehavethat corrfFmj(A),Fmj0(A)g=+=2++1 2+++1,pr(mj=mj0)=2q+1)]TJ /F12 7.97 Tf 6.59 0 Td[(2q (2+)(1+))]TJ /F9 7.97 Tf 6.59 0 Td[(1. (3) Thegroupingpriorhasimposedacorrelationstructureonthedistributionfunctionsofthe'sofacommonlag,allowingustoborrowstrengthintheestimationofthedependenceparametersfromthesamelag.Thiscorrelationisthesameas( 3 )fortheearlierm6=m0casewiththeroleofandinreverse.Likewise,theprobabilityofmatchingacrossparametersofcommonlag( 3 )isalsoequivalenttotheprobabilityofmatchingacrossgroupforcommonparameter( 3 )withandexchanged.Thisisakeydistinctionfromtheprocessof Dunsonetal. ( 2008 )wherethecorrelationandmatchingprobabilitieswouldbezero. 58
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Fordifferentgroupsm6=m0anddifferentautoregressiveparametersj6=j0ofthesamelag, corrfFmj(A),Fm0j0(A)g==2+++1 2+2+2+1,pr(mj=m0j0)=2q+1)]TJ /F12 7.97 Tf 6.59 0 Td[(2q 2(1+)(1+))]TJ /F9 7.97 Tf 6.59 0 Td[(1. (3) Somealgebrashowsthatthiscorrelationislessthanboth( 3 )and( 3 ).Likewise,pr(mj=m0j0)issmallerthan( 3 )and( 3 ).Thatis,thecorrelationsofthedistributionfunctionsandtheprobabilityofmatchingacrossbothgroupandautoregressiveparameterarestrictlysmallerthanthecorrelationandmatchingprobabilityacrossjustone.If>,thencorrfFmj(A),Fm0j(A)g
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3.4.2InnovationVariancePropertiesWenowexplorethebehavioroftheinnovationvariancesandtheirdistributionsGmj.LetR+denotethepositiverealline,logAbethesetflogx:x2AgforanyA2B(R+),and()theprobabilityfunctionfortheN( ,)distribution,assumingthehyperparameters ,arexed.Properties( 3 )( 3 )holdasintheautoregressiveparametercasewith(A)replaced(logA)andsetto0.Forinnovationvariancesofthesamegroupanddifferenttimesj6=j0, corrfGmj(A),Gmj0(A)g=+=2++1 2+++1corrn!j1(logA),!j01(logA)o, (3) andforbothdifferentgroupsm6=m0anddifferenttimesj6=j0, corrfGmj(A),Gm0j0(A)g==2+++1 2+2+2+1corrn!j1(logA),!j01(logA)o. (3) ThecorrelationofthesedistributionsnowdependsonthechoiceofBorelsetA.However,theyaretheproductsofatermthatdependssolelyonthestick-breakingparametersandandatermthatdependsonlyonAandthedistributionof(!j1,!j01)N2f 12,R()g,whereR()isthe22correlationmatrixwithoff-diagonalelementsjj)]TJ /F7 7.97 Tf 6.58 0 Td[(j0j.Thehighercorrelationsforneighboringtermsimpliesasmoothingofthevariancesasafunctionofjfor>0.Weobservethattheleadingtermgivesthesamecorrelationstructureasin( 3 ).Additionally,withthechoiceof=0,thetermdependingonAiszero,andthedistributionsareuncorrelatedasin Dunsonetal. ( 2008 ).Forj6=j0and1m,m0M,pr(mj=m0j0)=0,thatis,thereisnomatchingoftheinnovationvariancesacrosstimepoints.Thisisaconsequenceofthefactthattwopointsdrawnfromacorrelatednormaldistributionwithjj<1willbeequalwithprobabilityzero. 60
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3.5ComputationalConsiderationsRecallthatequation( 3 )provideduswiththeexpectedapproximationerrorwhichweemploytochoosethenumberofclustersnecessaryforthematrixstick-breakingprocesstruncation.Thisformulacontinuestoholdfortheproposedgroupingpriors,sincethestick-breakingweightsareformedusingthesameframework.Hence,ifthevaluesof,foreitherforthelag-blockorcorrelated-lognormalpriorareassumedknown,thenwechoosethenumberofclustersHsuchthat( 3 )islessthansomethreshold,suchas0.01.Aswegenerallydonothaveanyknowledgeorpriorbeliefaboutthesestick-breakingparameters,itwilloftenbeinappropriatetoprespecifyvalues,sowefollowthesuggestionof Dunsonetal. ( 2008 )andspecifyindependentGamma(1,1)priorsforand.AnalysesusingGamma(10,10)andGamma(0.1,0.1)indicatelittleeffectofthispriorchoiceontheestimatesof.TochoosethevalueofHwhenusingapriorforthestick-breakingparameters,werunapreliminaryMarkovchainforapproximately10%ofthedesiredchainlengthandusetheposteriormeanstotestwhether( 3 )isbelowourthreshold.Ifso,wexthisvalueofHforremainingcomputation.Oneofthenicepropertiesofthematrixstick-breakingprocessisthatintroducingappropriatelatentvariablesleadstoacomputationalalgorithmthatgenerallysamplesfromwell-knownconjugatedistributions( Dunsonetal. 2008 ).Becauseanormalpriorforthegeneralizedautoregressiveparametersprovidesconjugacy,thesamplingforisfromarelativelyeasytosamplezero-normalmixture.Withthelognormaldistribution,conjugacyfortheinnovationvariancesisnotobtainedsinceinversegammaistheconjugatedistributionfor,butwecansampleefcientlybyincorporatingaslicesamplingstep( Neal 2003 ).Weappropriatelymodifythealgorithmof Dunsonetal. ( 2008 )forposteriorsamplingfromourgroupingpriorsanddiscussfurthercomputationalchallengesinAppendix D 61
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Oneissueinthesamplingalgorithmisthebehaviorofsamplingthecorrelation.Inourexperiencewhenconsideringpriorswithrandom,didnotseemtobewellinformedbythedata.Hence,weopttotreatasatuningparameter.Werecommendspecifyingadefaultvaluesuchas=0.75,possiblytryingafewotherchoicesandselectingthevaluebasedonsomemodelselectioncriterion.AsshowninthedepressiondatastudyinSection 3.7 ,thethreechoicesof=0.5,0.75,and0.9leadtosimilarmodeltsasmeasuredbythedeviance.Basedonoursimulationstudies,itappearsthatthecorrelated-lognormalpriorisfairlyrobusttothechoiceof. 3.6RiskSimulationWenowexaminetheoperatingcharacteristicsoftheproposedgroupingpriorsviaarisksimulationdesignedtomimictheanalysisofatypicallongitudinaldatascenario.Weincorporateanon-zeromean,andthesimulateddatawillsufferfromignorabledropout.ThereareM=8groupseachwithnm=50measurementsofdimensionp=6.LetDidenotethetimet=2,...,p+1ofdropoutforsubjecti,whereDi=p+1indicatesasubjectwhocompletesthestudy.Dropoutisinducedaccordingtothemodel logitfpr(Di=t+1jDi>t,yit,m)g=0t+1tyit+2m(t=1,...,p)]TJ /F4 11.955 Tf 11.95 0 Td[(1). (3) Thismissingdatamechanismismissingatrandombecausethedropouttimedependsonlyonobservedvalues.Themean,covariance,anddropoutparametersforthesimulation,aswellastheprobabilitiesofmissingnessattimetforeachgroup,areprovidedinAppendix E .Thechoicesofand)]TJ /F1 11.955 Tf 10.1 0 Td[(donothaveanyequalitiesacrossgroupsbutsomewithinlag.Howeverwiththesmallsamplesizes,itwillgenerallystillbeadvantageoustoshareinformationacrosstheeightgroups.Also,thereisamoderateamountofsparsityin,asistypicalforordereddata.Allgroupshaveameanofzeroattime1,andthemeanfunctionsincreaseatdifferingratestothenaltimet=6.Thedropoutratesvaryacrossgroupswithmostgroupslosing35to50%oftheirsubjectsbyt=6.Groups3and 62
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Underthisdatamodel,wegenerate50datasetsandrunourMarkovchainMonteCarloalgorithmoneachdatasetwitheachpriorfor50,000iterationskeepingeverytenthiteration,usingaburn-inof10,000.Weplacethefollowingpriorsonthehyperparameterswhenappearinginthepriorspecication:q,independentUnif(0,1);,,,,1,and2,independentGamma(1,1);2InvGamma(0.1,0.1);InvGamma(0.1,0.1); N(0,c2),withc2=1000.Wexthevalueoftobe0.75.Weassumeaatprioronthegroup-specicmeanvectorsm.Tohandleincompletedata,weusedataaugmentationtosamplethemissingdatavaluesfromnormaldistributionsconditionalontheobserveddata.WemeasuretheperformanceofourproposedpriorsbyestimatingtheriskassociatedwiththeBayesestimatorsundertwocommonlossfunctions( Yang&Berger 1994 ),L1(m,^m1)=tr()]TJ /F9 7.97 Tf 6.59 0 Td[(1m^m1))]TJ /F4 11.955 Tf 12.86 0 Td[(logj)]TJ /F9 7.97 Tf 6.59 0 Td[(1m^m1j)]TJ /F5 11.955 Tf 19.75 0 Td[(pandL2(m,^m2)=trf()]TJ /F9 7.97 Tf 6.58 0 Td[(1m^m2)]TJ /F5 11.955 Tf 10.83 0 Td[(I)2g.Sincetheselossesaredenedintermsofasinglecovariancematrix,weconsiderthelossforestimatingthesetofcovariancematricestobetheaverageacrossgroupsofthelossesfromtheindividualcovariancematrices.Tostudytheabilitytorecoverthemeanfunctionwithdifferingpriorson,weusetheposteriormeanofmandthelossfunctionL(^m,m)=(^m)]TJ /F11 11.955 Tf 12.14 0 Td[(m)>)]TJ /F9 7.97 Tf 6.59 0 Td[(1m(^m)]TJ /F11 11.955 Tf 12.14 0 Td[(m),takinganaverageacrossgroupsstandardizedbythetruecovariancematrixm.TheestimatedrisksassociatedwithestimatingthecovariancematricesforlossfunctionsL1andL2areshowninTable 3-1 .Thegroupingpriorshowsariskimprovementof30and25%overthenaiveBayes1priorand52and41%overthegroup-specicatprior.WhilethenaiveBayesprior1accommodatessparsity,itdoesnotpromoteanyequalityrelationshipsintheautoregressiveparametersacrossgroupsasinthetrueparametervalues,unlikethegroupingprior.ThenalcolumnofTable 3-1 displaystheriskinmeanestimationshowingaclearimprovementunderthegroupingprior.Thelag-block/correlated-lognormalpriorproducesarisk14%smallerthanthenaiveBayes1priorand29%smallerthanthe 64
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Table3-1. EstimatedrisksforeachchoiceofcovariancepriorfromthesimulationinSection 3.6 .TheestimatedriskiscalculatedastheaveragelossusinglossfunctionsL1(m,^m1)=tr()]TJ /F9 7.97 Tf 6.58 0 Td[(1m^m1))]TJ /F4 11.955 Tf 11.95 0 Td[(logj)]TJ /F9 7.97 Tf 6.58 0 Td[(1m^m1j)]TJ /F5 11.955 Tf 17.93 0 Td[(p,L2(m,^m2)=trf()]TJ /F9 7.97 Tf 6.58 0 Td[(1m^m2)]TJ /F5 11.955 Tf 11.95 0 Td[(I)2g,andL(^m,m)=(^m)]TJ /F11 11.955 Tf 11.95 0 Td[(m)>)]TJ /F9 7.97 Tf 6.58 0 Td[(1m(^m)]TJ /F11 11.955 Tf 11.96 0 Td[(m). PriorEstimatedRiskL1L2L CovarianceGroupingPrior0.4250.7420.175NaiveBayes10.6050.9870.203NaiveBayes20.6301.0100.210Group-specicat*0.8921.2550.248Common-at8.10584.3390.925 *Thegroup-specicatpriorisonlyover49datasetsbecausetheMarkovchainfailedtoconvergeforonedataset. group-specicestimator.Theriskassociatedwiththecommon-priorisalmostvetimesthatassociatedwiththegroupingpriors.Thiscorrespondswithourobservationsintheintroductionthatbyconsideringamoreaccuratestructureonthedependence,weestimatethemeanfunctionmoreefcientlyandwithlessbias.Additionalrisksimulationsusingthegroupingpriorundersimplerdatamodelswithfullyobserveddatahavebeenperformed,someofwhichareincludedinAppendix E .Thesecontinuedtoshowthatourprioroutperformsthenaivecompetitorsundermanydifferenttypesofcovariancematrixspecicationssuchassituationswithnosparsityanddissimilarcovariancematricesacrossgroups,andunderincreasingnm,M,andp.Inparticular,webelievethatasthenumberofgroupsMandthedimensionofthecovariancematrixpincreases,thegroupingestimatorsforwilloutperformthenaiveBayesestimatorsandthemarginbywhichtheydosoincreases.Thechoiceoftheatpriorscontinuedtoperformpoorlycomparedtothegroupingandnaivechoices. 3.7DataExampleWenowdemonstratetheuseofthegroupingpriorsinthettingofalongitudinaldatasetfromadepressionstudy.Thedata,originallypresentedby Thaseetal. ( 1997 ) 65
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Figure3-1. Theposteriorprobabilitiesofmatchingfortheinnovationvariancesattimes1,9,and15.Thesizeoftheboxesareproportionaltopr(mj=m0jjyobs). Wealsoconsiderhowthechoiceofcovarianceprioreffectsmeanestimation.Weshowthetreatmenteffect,calculatedasthedifferenceinmeanvaluebetweenbaselineandweek16,and95%credibleintervalsforthersttwogroupsinTable 3-2 .Ingroupm=2weseethattherearecleardifferencesforthiseffectacrossthedifferentpriors.Thetreatmenteffectunderthegroupingpriorsisestimatedtobearound9.5points,butitis10.2forthecommon-atand6.9forthegroup-specicatprior.Evenbetweenthetwonaivepriors,theestimatedtreatmenteffectsdifferby0.6.Forgroup1,aswellasgroups3and4,wedonotobservemuchdifferenceinthemeaneffect,exceptforsomedeviationwiththecommon-prior,althoughthecondenceintervalismorenarrowforthegroupingpriorsthantheatversions.Thesetwogroupsdemonstratethebiasandefciencyissuesrelevanttocovariancematrixestimationwithmissingdataasdiscussedintheintroduction.Wenotethatthedifferencesdonotrisetothelevelofstatisticalsignicancehere,buttheyarelargeenoughtobeofpracticalimportance.Figures 3-1 and 3-2 showthegroupingnatureoftheproposedpriors.Figure 3-1 showstheposteriorprobabilitiesofpr(mj=m0j)foreachm,m0combinationattimesj=1,9,15.Thesetimeswerechosenasrepresentativeoftheoverallpatternsinthedata.Forj=1andmostoftheundisplayedtimes,thereissubstantialmatchingforthegroups1and2,thelowinitialseveritygroup,aswellasforgroups3and4,thehigh 68
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Figure3-2. Theposteriorprobabilitiesofmatchingforthegeneralizedautoregressiveparameters.Panel(a)containsthematchingfortherstfourlag-1terms,and(b)displaystherstfourlag-4terms.Thesizeofthegrayboxesareproportionaltopr(mj=m0j0jyobs).Theblackboxesoverlayingthediagonalareproportionaltotheposteriorofpr(mj=0).Theaxesindicategroupm(topline)andautoregressiveparameterjoflag-q(bottomline). initialseveritygroups,withlessmatchingacrossthepairs.Thevariancesatj=9and15showastrongerpropensitytomatchacrossallgroups.Figure 3-2 givestheposteriorprobabilitiesofmatchingforthelag-1andlag-4autoregressiveparameters.Weshowonlytherstfourofeachduetospacelimitations.Theblackboxesthatoverlaythey=xdiagonalareproportionaltotheposteriorofpr(mj=0).Weseethatthelag-1termsarerarelysettozero,whilethelag-4termsforgroups2and3arefrequentlyzeroedout.Duetothenatureofthegroupingprior,thereisapositiveprobabilityofequalityacrossmj'sofacommonlag.InFigure 3-2 (a)wenotethepairwiseprobabilityofequalityisveryhighforallcombinationsoftherstautoregressiveparameterofallgroups,i.e.theregressionoftime2ontotime1,andtheotherlag-1,group2terms.Onewouldbeunlikelytolearnofthis 69
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relationshiportoconsideramodelwithequalityacrossall,orevenalargesubset,oftheseparametersusingotherapproaches.ConsideringFigure 3-2 (b),therearelargermatchingprobabilitiesforthelag-4parameters,muchofwhichisduetomatchingwithbothparameterssettozero.However,thematchingisnotalwaysduetoequalityatzero,ascanbeseenfromthelargeprobabilitiesofmatchingacrossthegroup1's.Thereissimilarbehaviorforthegroup4generalizedautoregressiveparameters. 3.8DiscussionWehavedevelopedaprioronthesetofMcovariancematricesthatsimultaneouslyexploitssparsityandmatchingofdependenceparametersacrossgroups.Themodelspacecontainingallcombinationswhereeachautoregressiveparameter/varianceisconstantacrossallpossiblesubsetsofthegroupshasBp(p+1)=2Mmodels,whereBMistheMthBellnumber( Stanley 1997 ,p.33).Infact,thegroupingpriorsconsideraspacethatisevenlargersinceweallowmatchingacrossautoregressiveparametersofacommonlag.Withthismanymodelswehavelittlehopeofndingthemostappropriateone.Ourgroupingpriorsavoidthisproblembystochasticallyconsideringthepossibilityofeachofthesemodelsinasingleanalysisandaccountingforuncertaintyappropriately.ItisourbeliefthatrunningaMarkovchainwithoneofthesegroupingpriorsisanecessaryalternativetotheunreasonabletimeandenergyrequiredtotandcomparethisextremelylargeclassofmodels. 70
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CHAPTER4COVARIANCEPARTITIONPRIORS:ABAYESIANAPPROACHTOSIMULTANEOUSCOVARIANCEESTIMATION 4.1SimultaneousCovarianceEstimationandaDrawbackoftheCovarianceGroupingPriorsWhenmodelinglongitudinaldata,estimationofthecovariancematricesisofprimeimportance.Inmanycases,thedatamayconsistofmultiplegroupsdenedbydifferencesintreatmentsand/orbaselinecovariates.Animportantconsiderationfortheanalystiswhether,andhow,thedependencestructurevariesacrossthesegroups.Often,oneperformsinferenceundertheassumptionthatthecovariancestructuresareequalacrossthesegroups,butifthisassumptionisincorrect,theresultinginferencesmaybeinvalidated,evenforinferencesonthemeanmodelifthereismissingness( Daniels&Hogan 2008 ).Conversely,modelingthedependencestructuresindependentlywithoutregardtotheothergroupscanleadtoinefciencyifthegroupsamplesizesaresmallorthedimensionislarge.Ourgoalistodevelopmethodologythatwillallowforthesharingofinformationacrossgroupstoimproveestimationefciency.ConsiderMgroupscontainingnmobservations,Ymi(i=1,...,nm;m=1,...,M),ofT-dimensional,multivariatenormaldata.Withoutlossofgenerality,weassumeYmihasmeanzero;otherwise,weletYmirepresenttheresidualaftercentering.Eachgroupmhasitsowncovariancematrixm,andwelet=f1,...,Mgdenotethecollectionofcovariancematrices.WeparameterizeeachmthroughthemodiedCholeskydecomposition( Pourahmadi 1999 2000 ),so)]TJ /F9 7.97 Tf 6.59 0 Td[(1m=TmDmT>mwithDmadiagonalmatrixwithpositiveentriesandTmanupper-triangularmatrixwithunitdiagonals.TheparametersofthisCholeskyparameterizationareinterpretablebyconsideringthesequentialdistributionsofYmi, f(Ymi1,...,YmiT)=f(Ymi1)f(Ymi2jYmi1)f(YmiTjYmi1,...,Ymi,T)]TJ /F9 7.97 Tf 6.59 0 Td[(1).(4) 71
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UndermultivariatenormalityeachoftheseTsequentialdistributions,f(YmitjYmi1,...,Ymi,t)]TJ /F9 7.97 Tf 6.59 0 Td[(1),isanormaldistributionwithmeanPt)]TJ /F9 7.97 Tf 6.59 0 Td[(1j=1m;jtYmijandvariancemt.Letmt=(m;1t,...,m;t)]TJ /F9 7.97 Tf 6.59 0 Td[(1,t)>bethevectorofregressioncoefcientsfromthesequentialregressionforthetthresponse.FromtheCholeskydecomposition,theunconstrainedelementsofthetthcolumnofTmare)]TJ /F16 11.955 Tf 9.3 0 Td[(mt,andthe(t,t)elementofDmis1=mt.Hence,thesesequentialdistributionsfullyanduniquelydeterminem.Further,weneednotworryaboutthepositivedeniteconstraintonmaslongasmt>0forallt,andthenormalandinversegammadistributionsprovideconditionallyconjugatepriorsformtandmt,respectively( Daniels&Pourahmadi 2002 ).Wedenotethem;jt'sasthegeneralizedautoregressiveparameters(GARPs)andthemt'sastheinnovationvariances(IVs).Therehasbeenasubstantialamountofresearchonthesimultaneouscovarianceestimationproblem,butrelativelylittlefocusonthelongitudinaldatascenario.Methodshavebeensuggestedbyimposingcommonalitiesonaparticularfeatureofthem's:equalityacrosssubsetsoftheprinciplecomponentsof( Boik 2002 )oritscorrelationmatrix( Boik 2003 );equalityacrosscorrelationmatricesorthevolumesof( Manly&Rayner 1987 );equalityacrossallTmand/orallDm( McNicholas&Murphy 2010 );equalityamongarbitrarysubsetsofTmandDm( Pourahmadietal. 2007 ).Thelaterisinthespiritofourmethod,buttherethesubsetsmustbeprovidedbytheuser.Thispresentsasignicantcomputationalchallengebecausetherearemuchtoomanypossiblesubsetstondthebestcongurationwithoutanautomatedmethod. Daniels ( 2006 )and Hoff ( 2009 )proposeshrinkagepriorsontheCholeskytermsandtheeigenvectors,respectively. Guoetal. ( 2011 )and Danaheretal. ( 2012 )proposepenalizedlikelihoodmodelsthatinduceasparsitystructurecommonacrossgroups.The Danaheretal. ( 2012 )techniquealsoallowsequalityacrossgroupsatasinglenon-zerovalue.Thisisunsatisfactoryinourcase,asweexpectthattheremaybesubsetsofthegroupsthatshouldshareparametervaluesatdistinctnon-zerovalues; Guoetal. ( 2011 ) 72
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providesnoinformationacrossgroupsbeyondacommongraphicalstructure.Otherauthorshavemodeledthecovariancematricesthroughregressionsofacontinuouscovariate( Chiuetal. 1996 ; Daniels 2006 ; Fox&Dunson 2011 ; Hoff&Niu 2012 ),buttheregressionparametersinthesemodelsoftenlackinterpretation.InChapter 3 (seealso Gaskins&Daniels 2013 ),weproposedanonparametricpriorfortheCholeskytermsbasedonthematrixstick-breakingprocess( Dunsonetal. 2008 ).Thismethodologyproposesasparsepriorthatallowsformatchingofthem;jtGARPs(m=1,...,M;j=1,...,t)]TJ /F4 11.955 Tf 11.64 0 Td[(1;t=2,...,T)acrossallmandallj,twithinaxedvalueoft)]TJ /F5 11.955 Tf 11.98 0 Td[(j.Thevaluet)]TJ /F5 11.955 Tf 11.98 0 Td[(jiscalledthelagandrepresentsthetimedistancebetweenthetheresponseYtandtheregressorYj.TheIVsmtmayalsomatchacrossgroups.Thispriorconsidersanenormousmodelspaceandrequiresmanylatentvariablesforanefcientsamplingscheme.Inthischapterwesimplifythesetofmodelsunderconsiderationwhilestillconsideringarichsetofsparsemodelswithcommonstructuresacrossgroups.ThisleadstofastercomputationsthanthemethodofChapter 3 ,andhence,willbetteraccommodatedatawithlargerdimensionT.InSection 4.2 weintroduceourapproachbasedonacovariancepartitionprior.Toshareinformationacrossgroups,thispriorspeciespartitionsofthegroupssuchthatthosegroupscontainedinthesamesetofthepartitionwillhavecommonvaluesforsomesubsetofthedependenceparameters.Wedeneapartitioncorrespondingtoeachofthesequentialdistributionsin( 4 )andconsiderthesequenceofpartitionstobehaveasaMarkovchain.Section 4.3 containsdetailsofthecomputationalalgorithmneededtogenerateaposteriorsamplefortheproposedprior.PerformanceofthecovariancepartitionpriorisstudiedinSections 4.4 and 4.5 throughasimulationstudyandtheanalysisofdatafromadepressionstudy.AbriefdiscussioninSection 4.6 concludesthechapter. 73
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4.2CovariancePartitionPrior 4.2.1PriorontheSequenceofPartitionsFirst,itisnecessarytodenesomenotation.LetM=f1,...,Mgdenotethecollectionofallgroups.Weareinterestedinpartitioningthegroupsintosetsthatshareasimilardependencestructure.LetPdenoteapartitionofMandthecollectionofallpossibleP.ThecardinalityofisBM,theMthBellnumber( Stanley 1997 ,p.33).BMisequaltothesumoftheStirlingnumbersofthesecondkind.AnypartitionPcanbewrittenasthecollectionofitssetsP=fS1,...,Sdg,whereeachSiisnon-empty,disthedegree(thenumberofdistinctsets)ofP,Si\Sj=;forall1i
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clusteringofaDirichletprocess(Si)=(ni!),whereistheconcentrationparameterandnithedegreeofsetSi.However,ourfocusisonajointprior(P1,...,PT)forTtime-orderedpartitions.Wedesireapriorthatwillencouragesimilarstructuresacrosst.Ifgroupsm1andm2areinthesamesetinPt,andthereforesharethesameGARPsandIVforresponset,itshouldbemorelikelythattheyareinthesamesetofPt+1.Tothatend,weconsiderthesequenceofpartitionsfP1,...,PTgtobeaMarkovprocessonthestatespace.BytheMarkovpropertyandtime-invariance,weletthedistributionofPtgivenallpreviouspartitionsdependsonlyonthemostrecentpartition,thatis,pr(PtjP1,...,Pt)]TJ /F9 7.97 Tf 6.59 0 Td[(1)=pr(PtjPt)]TJ /F9 7.97 Tf 6.58 0 Td[(1)=pr(P2jP1)forallt.Hence,thefullconditionaldistributionofPtgivenalltheotherpartitionsdependsonlyontheadjacentpartitionsPt)]TJ /F9 7.97 Tf 6.59 0 Td[(1andPt+1.Tospecifythetransitionprobabilitypr(P2jP1),weemployacommonlyusedmetricondenedbyd(P1,P2)=2jP1\P2j)-303(jP1j)-302(jP2j,wherejPjgivesthedegreeofapartitionPandtheintersectionpartitionisP1\P2=fS:S6=;;S=S1\S2forsomesetsS12P1andS22P2g( Day 1981 ).Notethatfortwogroupsm16=m2tobeinthesamesetofthepartitionP1\P2,theymusthavebeeninthesamesetinbothP1andP2.Thedistanced(P1,P2)isinterpretedastheminimumnumberofmergesandsplitsofthesetsofP1neededtoobtainP2.Othermetricsforcanbefoundin Arabie&Boorman ( 1973 )or Denud&Guenoche ( 2006 ).Usingthed(,)metric,wedenetheclosenessbetweenthetwopartitionsbycq(P1,P2)=[1+fd(P1,P2)gq])]TJ /F9 7.97 Tf 6.59 0 Td[(1,whereqisanon-negativeconstantdeterminingtherelativestrengthofthedistance.Noteforallniteq,thatcq(P,P)=1forallP,cq(P1,P2)2(0,1)forP16=P2,andcq(,)isdecreasingind(,).Deneaq(P)=B)]TJ /F9 7.97 Tf 6.58 0 Td[(1MPP0cq(P,P0)tobetheattractivenessofthepartitionP,givenbytheaverageclosenessofPover.ThetransitionprobabilityfromPt)]TJ /F9 7.97 Tf 6.59 0 Td[(1toPt(t>1) 75
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isproportionaltotheclosenessofthepartitionsandisgivenby pr(PtjPt)]TJ /F9 7.97 Tf 6.59 0 Td[(1)=cq(Pt)]TJ /F9 7.97 Tf 6.59 0 Td[(1,Pt) aq(Pt)]TJ /F9 7.97 Tf 6.58 0 Td[(1)BM.(4)ItiseasytoverifythatthestationaryprobabilityofPisproportionaltoitsattractivenessaq(P).Hence,wechoosethedistributionfortheinitialpartitionP1tobethesetofstationaryprobabilitiespr(P1)=aq(P1)=(AqBM),whereAq=B)]TJ /F9 7.97 Tf 6.58 0 Td[(1MPPaq(P)istheaverageattractiveness.BecausewearestartingaMarkovchainatitsstationarydistribution,themarginalprobabilityforpartitionPtis pr(Pt)=aq(Pt) AqBM,(Pt2,t=1,...,T).(4)Combining( 4 )and( 4 ),thedistributionoftheentirepartitionprocessisgivenby (P1,...,PT)=B)]TJ /F7 7.97 Tf 6.59 0 Td[(TMaq(P1) AqTYt=2cq(Pt)]TJ /F9 7.97 Tf 6.59 0 Td[(1,Pt) aq(Pt)]TJ /F9 7.97 Tf 6.59 0 Td[(1).(4)Ourpartitionpriorclearlydependsonthevalueofq.Whenq=0,c0(P1,P2)=1=2forallP1,P2undertheusualconvention00=1.Hence,aq(P)=1=2forallP,andpr(PtjPt)]TJ /F9 7.97 Tf 6.59 0 Td[(1)=pr(Pt)=1=BM.Thepreviouspartitionprovidesnoinformationaboutthenewpartition,andallpartitionsareequallylikely.Wecallthiscasetheindependent-uniformsprior,becauseeachpartitionisindependentoftheothersandfollowsauniformdistributionover.Because0qisdiscontinuousatq=0,cq(P1,P2)!0.5+0.5I(P1=P2)asqapproacheszero.Inthislimit,pr(PtjPt)]TJ /F9 7.97 Tf 6.58 0 Td[(1)!f1+I(Pt=Pt)]TJ /F9 7.97 Tf 6.58 0 Td[(1)g=(BM+1).Theprobabilityofmovingtoanydifferentpartitionis1=(BM+1)withprobabilityofremainingatthesamepartitionistwiceaslikely.SinceBMislarge,thereislittlepracticaldifferencebetweentheresultsatq=0andq!0.Conversely,asqgetslarger,cq(P1,P2)0ifd(P1,P2)>1.Hence,movesfromPt)]TJ /F9 7.97 Tf 6.59 0 Td[(1toPtthatrequiremorethanonemergeorsplitareincreasinglyunlikely.Forlargeqthisplacesahighlyrestrictivestructureonthepartitionprocess.Hence,weonlyallow 76
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Figure4-1. MarginalprobabilitiesforthreechoicesofqwithM=8groups.TheseprobabilitiesarescaledbyBMandcanbeviewedastheratioofmarginalprobabilitiesunderqandundertheindependent-uniforms(q=0)case,i.e.pr(Pjq)=pr(Pjq=0).Partitionsareorderedbyincreasingdegreeonthex-axiswithnoinformativeorderingbetweenpartitionsofcommondegree;partitionsofodddegreeareinblackandthoseofevendegreeareingray. valuesofqbetween0and10sothatpr(PtjPt)]TJ /F9 7.97 Tf 6.59 0 Td[(1)isboundedawayfromzeroforallqandallt,Pt)]TJ /F9 7.97 Tf 6.58 0 Td[(1,Pt.WenotethatXPpr(Pjq=10))]TJ /F4 11.955 Tf 11.95 0 Td[(pr(Pjq=1).001,andsothereislittledifferencebetweenthemarginalprobabilitiesforqgreaterthan10.Tobetterseetheeffectthechoiceofqhasonthemarginalprobabilitiespr(P)of( 4 ),weplotthemarginalprobabilitiesforthreechoicesofqforM=8groupsinFigure 4-1 .TheseprobabilitiesarescaledbyBM=4140andcanbeviewedastheratioofmarginalprobabilitiesundertheparticularqtotheprobabilityundertheindependent-uniformsq=0case,i.e.pr(Pjq)=pr(Pjq=0).ForeachqthehighestprobabilitypartitionisPpool=fMg,thepartitionthatpoolsalldataintoasinglegroup.Forsmallerqallmarginalprobabilitiesarerelativelyclose,butasqincreases,thedisparityacrossPincreases.Forq=5,maxP,P0fpr(P)=pr(P0)g=11.4,butsincethesupportislarge,thisdifferenceisnotsogreatastoleaveanypartitionswithnegligible 77
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support.Thisdisparitydoesnotincreasesubstantiallyafterq=5aseachcq(P1,P2)isapproximatelyeither1,1=2,or0forq>5.Asitisnotclearwhatanoptimumvalueofqwouldbe,weletqbearandomvariabletobesampledaspartoftheanalysis.BecauseoursamplingschemewillrequirethevaluesofAqandaq(P1),...,aq(PT)foreachq,wespecifyanitesupporttominimizethecomputationalcomplexity.Tothatend,wechoosethesequenceQfrom0to10withstepsof0.025(jQj=401)tobethesupportofq.ThepriorforqisuniformoverthesetQ. 4.2.2PriorontheCholeskyParametersGivenaparticularpartitionPt,groupsinacommonsetwillsharecommonvaluesforthedependenceparametersassociatedwiththesequentialdistributionf(YmitjYmi1,...,Ymi,t)]TJ /F9 7.97 Tf 6.59 0 Td[(1)from( 4 ).Theseparametersare(mt,mt),wheremtisemptywhent=1.Tothatend,foreachsetSit2Pt=fS1t,...,Sdttg,weassociatetheparameters(?it,?it),sothat(?it,?it)=(mt,mt)forallm2Sit.As(mt,mt)aredeterminedbyf(?it,?it)gdti=1andPt,wenowspecifytheprioron(?it,?it)conditionalonthepartitionPt.Inadditiontothematchingacrossgroupsthatisinducedbyourpartitions,wedevelopourpriortoinducesparsityintheTmmatrices.Undermultivariatenormalitym;jt=0impliesYmijandYmitareindependentgiventheYmik'swithk
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Thepriorfor(?it,?it)conditionalonthepartitionPtisasfollows. pr(it=kjPt)=kt=expf)]TJ /F11 11.955 Tf 15.28 0 Td[(1I(k=0))]TJ /F11 11.955 Tf 11.95 0 Td[(2kg Pt)]TJ /F9 7.97 Tf 6.59 0 Td[(1l=0expf)]TJ /F11 11.955 Tf 15.27 0 Td[(1I(l=0))]TJ /F11 11.955 Tf 11.95 0 Td[(2lg(k=0,...,t)]TJ /F4 11.955 Tf 11.96 0 Td[(1), (4) ?itjPtInvGamma(1,2), (4) ~itjit,?it,PtNit(0it,?it2Iit). (4) itdeterminesonhowmanyofthepreviousmeasurementstheresponseYmitisdependent(form2Sit).Welet~itgivetheitnon-zeroGARPs,andso?it=(0>t)]TJ /F9 7.97 Tf 6.59 0 Td[(1)]TJ /F12 7.97 Tf 6.59 0 Td[(it,~>it)>givesthefullvectorofGARPscorrespondingtothenegativesoftheunconstrainedelementsofthetthcolumnofTm.Forconditionalconjugacyweuseanormalpriorfor~itandtheinversegammadistributionfortheinnovationvariances.Whent=1therearenoGARPs,andonlythepriorin( 4 )isneeded.ForafullyBayesiananalysis,weneedhyperpriorsforthehyperparameters2,1,2,1,and2.Wespecify2InvGamma(0.1,0.1)andindependentGamma(1,1)priorsfortheinnovationvarianceparameters1,2.Todenethehyperpriorsfor1and2,werstexploretheinterpretationoftheseparameters.Algebrashowsthat2=logpr(it=k) pr(it=k+1)(k=1,...,t)]TJ /F4 11.955 Tf 11.96 0 Td[(1;t=3,...,T),providinganinterpretationof2asthelogoddsthatYmitdependingononefewerpastresponse.Toencouragesparsityandsmallervaluesofit,wechooseanexponentialdistributionwithratelog(2)asthepriordistributionfor2.Thisguaranteesthatthislogoddsisstrictlypositivewithapriormeanoflog(2).Further,1)]TJ /F11 11.955 Tf 11.95 0 Td[(2k=logpr(it=k) pr(it=0)(k=1,...,t)]TJ /F4 11.955 Tf 11.96 0 Td[(1;t=2,...,T).Wechooseapriorof1j2Unif(2,(T)]TJ /F4 11.955 Tf 12.12 0 Td[(1)2).Theleftendpointrepresentsthecasethatpr(it=1)=pr(it=0),i.e.thattheresponsedependsononlythemostrecentmeasurementisaslikelyastheresponsebeingindependentofallpastmeasurements,andtherightendpointisequivalenttopr(iT=T)]TJ /F4 11.955 Tf 12.16 0 Td[(1)=pr(iT=0),thenalresponse 79
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dependingonthefullhistoryisaslikelyasindependencefromthehistory.Further,foranyk=1,...,T)]TJ /F4 11.955 Tf 12.45 0 Td[(1,2k2[2,(T)]TJ /F4 11.955 Tf 12.45 0 Td[(1)2]soitisequallylikelythat1=2kforanyk.Thatis,itisaslikelythatYmitdependsonthekpreviousresponsesasthecasethatYmitisindependentofitshistory,foreachchoiceofk.Simulationshavedemonstratedrobustnesstothesehyperpriorchoices. 4.3SamplingAlgorithmPosteriorinferenceusingthecovariancepartitionpriorswillrelyonaposteriorsamplegeneratedfromaMarkovchainMonteCarlo(MCMC)algorithm.InthissectionwedescribetheGibbssamplernecessarytoobtainaposteriorsample.LetH=(q,2,1,2,1,2)denotethesetofmodelhyperparameters,andCt=f(mt,mt)gMm=1denotethesetofCholeskyparametersforthetthsequentialdistributionsofallMgroups.Further,C()]TJ /F7 7.97 Tf 6.59 0 Td[(t)representsthesetofCholeskyparametersC1,...,CTexcludingCt;P()]TJ /F7 7.97 Tf 6.59 0 Td[(t)isdenedsimilarly.ThesamplingalgorithmconsistsofstepsoftheformPt,CtjP()]TJ /F7 7.97 Tf 6.59 0 Td[(t),C()]TJ /F7 7.97 Tf 6.59 0 Td[(t),H,DthatjointlyupdatethepartitionandCholeskyparametersassociatedwiththetthsequentialdistributiongiventhedataD.WeperformthisstepintwopartsbyfactoringPt,CtjP()]TJ /F7 7.97 Tf 6.59 0 Td[(t),C()]TJ /F7 7.97 Tf 6.58 0 Td[(t),H,D=PtjP()]TJ /F7 7.97 Tf 6.59 0 Td[(t),C()]TJ /F7 7.97 Tf 6.59 0 Td[(t),H,DCtjPt,P()]TJ /F7 7.97 Tf 6.59 0 Td[(t),C()]TJ /F7 7.97 Tf 6.58 0 Td[(t),H,D.Finally,weupdatethehyperparametersHgivenP1,...,PT,C1,...,CT,D.Beforebeginningthesamplingscheme,therearesomecomputationsthatshouldbeperformedandstoredrst.ToupdatethevalueofqrequiresknowledgeofAqandaq()foreachPt.Wechoosetocomputethefullsetofattractivenessesforeachq2QandallP2rst.WhenweupdateqintheMCMCscheme,welookuptheneededvalues.First,wedescribethePtjP()]TJ /F7 7.97 Tf 6.58 0 Td[(t),C()]TJ /F7 7.97 Tf 6.59 0 Td[(t),H,DstepthatsamplesthepartitionPTgiventheremainingpartitions,marginalizedovertheGARPsandIVsCt.ItcanbeshownthatPtjP()]TJ /F7 7.97 Tf 6.59 0 Td[(t),C()]TJ /F7 7.97 Tf 6.59 0 Td[(t),H,Ddependsonlyontheotherpartitionsandq, 80
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PtjP()]TJ /F7 7.97 Tf 6.59 0 Td[(t),C()]TJ /F7 7.97 Tf 6.59 0 Td[(t),H,D=PtjP()]TJ /F7 7.97 Tf 6.59 0 Td[(t),q,D;wesamplefromthisdistributionwithareversiblejumpMCMCstep( Green 1995 ).LetPtdenotethecurrentvalueofthepartition.WeproposeacandidatepartitionP?t=fS?1t,...,S?d?ttgasfollows.UniformlyselectmfromM,andletS02Ptdenotethesetofthepartitionthatcontainsgroupm.ThechoiceofthecandidateP?tdependsontheformofS0. IfS0=M,thenP?t=fMnfmg,fmgg.Thatit,thecandidatepartitionsplitsgroupmfromM. IfS0=fmg,thenuniformlysampleS00fromtheremainingsetsinPt.ThecandidateP?tisfS0[S00g[(PtnfS0,S00g).Thatis,wemergethesingletonsetS0withthesetS00. IfS0isneitherasingletonorM,thenweuniformlysampleS00from(PtnfS0g)[f;g.ThecandidatepartitionP?tisgivenbyfS0nfmgg[fS00[fmgg[(PtnfS0,S00g).Thatis,wemovethegroupmfromsetS0tothe(possiblyempty)setS00.Itcanbeshownthatthecorrespondingratiooftransitionprobabilitiesispr(Pt!P?t) pr(P?t!Pt)=dt)]TJ /F5 11.955 Tf 11.96 0 Td[(I(fmg2Pt) d?t)]TJ /F5 11.955 Tf 11.95 0 Td[(I(fmg2P?t).Byconstruction,PtandP?tdifferonlyinthelocationofgroupm.BecausewearedealingwithpartitionsonM,arelativelysmallspace,weareabletotraversethehighprobabilityregionsofbymakingthesesinglestepmoves.Forproblemswherethemodelspaceislarger,itmaybenecessarytousesplit-mergemoves(e.g. Kimetal. 2006 ; Richardson&Green 1997 )oramixtureofsinglestepandsplit-mergemoves(asin Boothetal. 2008 ).However,theposteriorsamplesfromoursimulationsanddataanalysisdoesnotindicatethatthisisneededinoursetting.ComputingtheprobabilityofacceptingthemovetoP?trequiresthelikelihoodimpliedbyPtandP?tconditionalontheotherpartitions,lik(PtjP()]TJ /F7 7.97 Tf 6.59 0 Td[(t),data)/pr(PtjPt)]TJ /F9 7.97 Tf 6.59 0 Td[(1)pr(Pt+1jPt)dtYi=1marg(Ymi,m2Sit), 81
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wheremarg(Ymi,m2Sit)isthelikelihoodofthedatafromgroupsinsetSitaftermarginalizingouttheGARPsandIVs, marg(Ymi,m2Sit)=t)]TJ /F9 7.97 Tf 6.59 0 Td[(1X=0ZRZR+(Ym2SitnmYi=1f(YmitjYmi1,...,Ymi,t)]TJ /F9 7.97 Tf 6.58 0 Td[(1;,))(~j,)()td~d=t)]TJ /F9 7.97 Tf 6.59 0 Td[(1X=0t(2))]TJ /F7 7.97 Tf 6.59 0 Td[(n=2)]TJ /F12 7.97 Tf 6.58 0 Td[(j)]TJ /F9 7.97 Tf 6.59 0 Td[(2I+X>Xj)]TJ /F9 7.97 Tf 6.59 0 Td[(1=212\(1))]TJ /F9 7.97 Tf 6.58 0 Td[(1\(1+n=2)2+1 2y>nIn)]TJ /F3 11.955 Tf 11.96 0 Td[(X)]TJ /F11 11.955 Tf 5.48 -9.68 Td[()]TJ /F9 7.97 Tf 6.58 0 Td[(2I+X>X)]TJ /F9 7.97 Tf 6.58 0 Td[(1X>oy)]TJ /F9 7.97 Tf 6.59 0 Td[((1+n=2), (4) wheren=Pm2Sitnm,yisthen-vectorcontainingYmit(i=1,...,nm,m2Sit),andXisthenmatrixwithrow(Ymi,t)]TJ /F12 7.97 Tf 6.59 0 Td[(,...,Ymi,t)]TJ /F9 7.97 Tf 6.59 0 Td[(1)>.For=0,theX0matrixisempty,andthesummandin( 4 )reducesto0t(2))]TJ /F7 7.97 Tf 6.58 0 Td[(n=212\(1))]TJ /F9 7.97 Tf 6.58 0 Td[(1\(1+n=2)2+1 2y>y)]TJ /F9 7.97 Tf 6.58 0 Td[((1+n=2).ThereversiblejumpMCMCstepacceptsthemovefromPttoP?tifUmin(1,pr(P?tjPt)]TJ /F9 7.97 Tf 6.58 0 Td[(1)pr(Pt+1jP?t)Qd?ti=1marg(Ymi,m2S?it) pr(PtjPt)]TJ /F9 7.97 Tf 6.58 0 Td[(1)pr(Pt+1jPt)Qdti=1marg(Ymi,m2Sit)pr(Pt!P?t) pr(P?t!Pt))forUUnif(0,1).Thevaluespr(P?tjPt)]TJ /F9 7.97 Tf 6.59 0 Td[(1),pr(Pt+1jP?t),pr(PtjPt)]TJ /F9 7.97 Tf 6.59 0 Td[(1),pr(Pt+1jPt)aregivenin( 4 )(forthecurrentvalueofq).FortheCtjPt,P()]TJ /F7 7.97 Tf 6.58 0 Td[(t),C()]TJ /F7 7.97 Tf 6.59 0 Td[(t),H,Dstep,wecanexpressthisdistributionasQdti=1[(?it,?it,it)jPt,H,D]andsampleeach(?it,?it,it)exactlyfromitsconditionaldistribution.Thesamplingdistributionsare pr(it=kjSit,H,D)/kt)]TJ /F7 7.97 Tf 6.59 0 Td[(kj)]TJ /F9 7.97 Tf 6.58 0 Td[(2Ik+X>kXkj)]TJ /F9 7.97 Tf 6.58 0 Td[(1=22+1 2y>nIn)]TJ /F3 11.955 Tf 11.95 0 Td[(Xk)]TJ /F11 11.955 Tf 5.48 -9.69 Td[()]TJ /F9 7.97 Tf 6.58 0 Td[(2Ik+X>kXk)]TJ /F9 7.97 Tf 6.59 0 Td[(1X>koy)]TJ /F9 7.97 Tf 6.59 0 Td[((1+n=2), (4) ?itjSit,it,H,DInvGamma1+n 2,2+1 2y>nIn)]TJ /F3 11.955 Tf 11.95 0 Td[(X)]TJ /F11 11.955 Tf 5.48 -9.68 Td[()]TJ /F9 7.97 Tf 6.59 0 Td[(2I+X>X)]TJ /F9 7.97 Tf 6.59 0 Td[(1X>oy,~itjSit,it,?it,H,DNit)]TJ /F11 11.955 Tf 5.48 -9.68 Td[()]TJ /F9 7.97 Tf 6.58 0 Td[(2I+X>X)]TJ /F9 7.97 Tf 6.58 0 Td[(1X>y,?it)]TJ /F11 11.955 Tf 5.48 -9.68 Td[()]TJ /F9 7.97 Tf 6.59 0 Td[(2I+X>X)]TJ /F9 7.97 Tf 6.59 0 Td[(1. 82
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Wethenset?it=(0>t)]TJ /F9 7.97 Tf 6.59 0 Td[(1)]TJ /F12 7.97 Tf 6.59 0 Td[(it,~>it)>and(mt,mt)=(?it,?it)form2Sit.Both( 4 )and( 4 )requirethecalculationoftterms.FordatawithlargeT,thismayadverselyeffectcomputationalspeed.Insuchcaseswerecommendmodifying( 4 )topr(it=k)/expf)]TJ /F11 11.955 Tf 15.28 0 Td[(1I(k=0))]TJ /F11 11.955 Tf 12.75 0 Td[(2kgI(kk0),wherek0isaxedvaluerepresentingthemaximumnumberofpreviousmeasurementsonwhichYmitcandepend.Hence,( 4 )and( 4 )willonlyrequirethecalculationofminft,k0+1gterms.ThisisrelatedtotheideaofbandingtheCholeskymatrix( Rothmanetal. 2010 ; Wu&Pourahmadi 2003 ),whichxesallm;jt,t)]TJ /F5 11.955 Tf 12.33 0 Td[(j>k0,tobezeroandallm;jt,t)]TJ /F5 11.955 Tf 12.33 0 Td[(jk0,tobenon-zero.TheadditionoftheI(kk0)termtopr(it=k)inourmodelprovidesamoreexiblestructurethanbandingasitallowssomem;jt=0,t)]TJ /F5 11.955 Tf 11.96 0 Td[(jk0.WenowupdatethehyperparametersH.For2and2,conjugacygivestheconditionalsof2InvGamma 0.1+TXt=2dtXi=1it,0.1+TXt=2ditXi=1?it>?it!,2Gamma 1+1TXt=1dt,1+TXt=1dtXi=1(?it))]TJ /F9 7.97 Tf 6.59 0 Td[(1!.Theconditionaldistributionof1isavailableinclosedform, (1j)/\(1))]TJ /F27 7.97 Tf 8 5.98 Td[(PTt=1dtPTt=11dt2exp()]TJ /F11 11.955 Tf 9.3 0 Td[(1 1+TXt=1dtXi=1log(?it)!),whichcanbesampledbyslicesampling( Neal 2003 ).Theconditionalfor(1,2)doesnothaveastandardform,butdependssolelyonthesetofit's.Letting(1,2)denotethehyperprior,weupdate(1,2)byslicesamplingfromtheconditionaldistribution (1,2j)/(1,2)TYt=2dtYi=1expf)]TJ /F11 11.955 Tf 15.28 0 Td[(1I(it=0))]TJ /F11 11.955 Tf 11.95 0 Td[(2itg Pt)]TJ /F9 7.97 Tf 6.58 0 Td[(1l=0expf)]TJ /F11 11.955 Tf 15.27 0 Td[(1I(l=0))]TJ /F11 11.955 Tf 11.96 0 Td[(2lg.SamplingqrequiresaMetropolis-Hastingsstep.WecreateaproposalsetaroundthecurrentvalueqbyQ?=fq+0.025n:n=1,...,50g,andthenwedrawq? 83
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uniformlyfromQ?.Weacceptthemovetoq?ifUmin(1,I(q?2Q)Aq Aq?aq?(P1) aq(P1)TYt=2cq?(Pt)]TJ /F9 7.97 Tf 6.59 0 Td[(1,Pt) cq(Pt)]TJ /F9 7.97 Tf 6.59 0 Td[(1,Pt)aq(Pt)]TJ /F9 7.97 Tf 6.59 0 Td[(1) aq?(Pt)]TJ /F9 7.97 Tf 6.59 0 Td[(1))forUUnif(0,1).Otherwise,weretainq. 4.4SimulationStudyWeexaminetheperformanceofourpriorsusingasimulationstudy.WesimulatedataconsistingofM=8groupswithsamplesizesofn1==n5=60,n6==n8=30.Becausethereislessdataforthenalthreegroups,weexpectthatleveraginginformationacrossgroupswillaidintheestimationespeciallyforthesesmallergroups.ObservationsaremeasuredatT=10timepoints.ThetruepartitionschosenareP1=fMg,P2==P5=ff1,2,3,4g,f5,6,7,8gg,P6==P8=ff1,2,3,4g,f5,6g,f7,8gg,andP9=P10=ff1,2g,f3,4g,f5,6g,f7,8gg.Giventhepartitions,thetruevaluesofTmandDmarespeciedtorepresentalongitudinaldatascenariowithserialcorrelation.Thetruecovariancematriceshaveasparsestructurebyspecifyingittobebetween1and3,mostoften2.ExactdetailsareprovidedinAppendix F .Underthisdatadistribution,wegenerate50datasetsandrunanMCMCchainforeachusingthealgorithmofSection 4.3 .Afteraburn-inof3000iterations,thechainrunsfor9000iterations,andweretaineverytenthforposteriorinference.Tomeasuretheperformanceoftheresultingestimators,weusethelossfunctions Lall(,^)=Lall(f1,...,Mg,f^1,...,^Mg)=MXm=1nm NL(m,^m),L(m,^m)=tr()]TJ /F9 7.97 Tf 6.59 0 Td[(1m^m))]TJ /F4 11.955 Tf 11.96 0 Td[(logj)]TJ /F9 7.97 Tf 6.59 0 Td[(1m^mj)]TJ /F5 11.955 Tf 17.93 0 Td[(T,whereN=Pmnm,and^m=Epost(TmDmT>m))]TJ /F9 7.97 Tf 6.59 0 Td[(1istheBayesestimatorforgroupm( Yang&Berger 1994 ).L(m,^m)isthestandardlog-likelihoodlossfunctionforasinglecovarianceestimator,andLall(,^)measuresthelosstoestimatingthecollectionofcovariancematricesbytakingaweightedaveragewithweightsproportionaltogroup 84
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Table4-1. Riskestimatesfromsimulationstudy. LabelinPartitionPriorCholeskyPriorRiskunderRiskunderFig. 4-2 Lall(,^)L(8,^8) 1cov.partitionpriorsparsity0.3040.3552indep.-uniformspriorsparsity0.3470.4503indep.-uniformspriornon-sparse0.4630.5354cov.partitionpriornon-sparse0.4660.5125Pt=Pindsparsity0.5150.6916Pt=Ppoolsparsity0.7691.1587Pt=Pindnon-sparse0.7801.0598Pt=Ppoolnon-sparse0.8201.187 size.Theestimatedriskofthecovariancepartitionpriormethodistheaveragelossoverthe50datasets.Wecomparethecovariancepartitionpriorswithseveralcommonlyusedmethods.LetPind=ff1g,,fMggrepresentthepartitionwhereeachgroupisinasingletonset,correspondingtothecasewherenoinformationissharedacrossgroups,andrecallPpool=fMgpoolsthedataintoasinglegroup.Weconsideratotalofeightpriorstructures:thedistributiononthepartitionsisgivenby( 4 )withthepriorforquniformonQ,calledthe`covariancepartitionprior';xingq=0,the`independent-uniformsprior';PtisxedasPindforallt;PtisxedasPpoolforallt.Witheachofthesefourstructuresonthepartitions,weconsidertwopriorsontheCholeskyparameters:sparsityisinducedaccordingto( 4 )( 4 )andanon-sparsechoicebyxingit=t)]TJ /F4 11.955 Tf 12.16 0 Td[(1.Table 4-1 andFigure 4-2 containtheestimatedriskofestimatingthecollectionofcovariancematricesandtheriskforestimatingthenalgroupundereachofthesepriorspecications.Itisclearthatestimatesbasedonpartitionpriorshavelowerrisk.NeitherofthePpoolpriorsworkwellbecausetheyproduceinconsistentestimators.Further,thePindpriorwithoutsparsityhastoomanyparameterstoestimateefciently;addingsparsityalleviatesthissomewhat.Thecovariancepartitionpriorwithsparsityshowsthebestperformance;theriskofthePindnon-sparsechoiceis2.5timeslarger.Theriskof 85
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Table4-2. Modelselectionstatisticsfordepressionstudy. PartitionPriorCholeskyPriorDevpDDIC cov.partitionpriornon-sparse39,46471140,887cov.partitionpriorsparsity39,51471540,945Pt=Ppoolnon-sparse40,25834540,949Pt=Ppoolsparsity40,28535941,004indep.-uniformspriornon-sparse39,34883641,020indep.-uniformspriorsparsity39,43583141,096Pt=Pindnon-sparse39,051116441,381Pt=Pindsparsity39,183115941,501 missingvaluesgiventheobservedmeasurementsandcurrentparametervalues.Eachgroupisassumedtohaveaunique,unstructuredmeanvector,whichisgivenaatprior.AnMCMCanalysisisrunforeachoftheeightpriorspecicationsfromtherisksimulation.Again,asinglechainisrunfor9000iterationsafteraburn-inof3000,andthevaluesfromeverytenthiterationareretainedforposteriorinference.Withtheincreaseddimensionofthisdata(T=17),thesparsecovariancepartitionpriorcansamplearound4.5iterationsinthetimenecessaryforthecovariancegroupingpriors( Gaskins&Daniels 2013 )methodtosampleone.Thepartitionpriorwiththenon-sparseCholeskyparameterizationsamplesabout4timesfasterthanthesparseversion.Todeterminethemodelthatbesttsthedata,weusethedevianceinformationcriterion(DIC; Spiegelhalteretal. 2002 ).TheDICiscalculatedasDIC=Dev+2pD,whereDevisthedevianceevaluatedattheposteriorexpectationoftheparametervaluesandpDisamodelcomplexityterm.ThispDiscomputedasthedifferencebetweentheexpecteddevianceandDev,andthetermisofteninterpretedastheeffectivenumberofmodelparameters.Weusetheobserveddatalikelihoodtocomputethedeviances( Wang&Daniels 2011 ).ModelswithsmallerDICareconsideredtobetterbalancethemodeltandcomplexity.Table 4-2 containsthemodelcomparisonstatistics. 88
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Thecovariancepartitionpriorwiththenon-sparseCholeskystructureprovidestheoverallbestmodelt.ThePpoolmodelsperformsecondbestbyutilizingfewerparametersthantheothermodelsbutatacostofapoorermodeltDev.Theindependent-uniformspriorshowssimilarmodelttothecovariancepartitionpriorbutusesabout120moreparameters,asthedegreeofthepartitionsP1,...,PTtendstobelargerwiththeindependent-uniformspriorthanunderthepriorwithrandomq.ThePindperformsworstduetothelargenumberofparametersrequired.Wenotethatforthisparticulardatasetourspecializedsparsitystructuredoesnotprovideanimprovementinmodelt.ThisisclearlyseensincetheDICarelargerthanthenon-sparsepriorforeachofthefourpartitionpriors.Additionally,wehavecomparablevaluesofpDunderbothCholeskychoices,indicatingthatformostiterationstheTmmatricesarefullorclosetofull.Furtherinspectionoftheparameterestimatesindicatethatrestrictingsparsitytotheformf(YmitjYmi1,...,Ymi,t)]TJ /F9 7.97 Tf 6.58 0 Td[(1)=f(YmitjYmi,t)]TJ /F7 7.97 Tf 6.59 0 Td[(k,...,Ymi,t)]TJ /F9 7.97 Tf 6.58 0 Td[(1)isnotwellsupportedbythisdata.ItispossibletoexploitsparsityinthecolumnsofTm(aswasshowninSection 3.7 ),buttheappropriatesparsestructurerequireszerosspreadthroughoutthecolumnsinsteadofjustintherstt)]TJ /F5 11.955 Tf 9.69 0 Td[(k)]TJ /F4 11.955 Tf 9.68 0 Td[(1positions.Weconcludethediscussionofthedepressiondatabyexaminingthestructurethatthecovariancepartitionpriorinducesforthisdataunderthebestttingmodel,thecovariancepartitionpriorwiththenon-sparseCholeskystructure.Themeanofq,theparameterdeterminingthesmoothnessofthestochasticprocessofpartitions,is8.09witha95%credibleintervalof(5.53,9.95).Figure 4-3 showstheclusteringpropertiesofourprioratbaselinethroughweek11;theremaining,undisplayedweeksbehavesimilarlytoweeks8.Eachpaneldepictsthepairwiseprobabilitythatm1andm2aregroupedtogetherattimet(weekt)]TJ /F4 11.955 Tf 12.61 0 Td[(1),thatis,theprobabilitythatthereexistsasetS2Ptwithfm1,m2gS.Wenotethatatbaselineandthenexttwoweeksgroupsaresplitbyinitialseveritywithalowseveritysetf1,2,5,6gandahighseverityset 89
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Figure4-3. Theposteriorprobabilitythatm1andm2areinthesamesetofthepartitionPtforbaselinethroughweek11(t=1,...,12). f3,4,7,8g.Fromweeks2through7thereislessstabilityinthepartitions,butfromweek8on,therearetwostrongclusters:onecontainsgroups4(drug,highseverity,male),5(nodrug,lowseverity,female),and6(drug,lowseverity,female)andtheothercontainsgroups2(drug,lowseverity,male),7(nodrug,highseverity,female),and8(drug,highseverity,female).Thepartitionlocationsofgroups1and3(nodrug,low/highseverity,male)arelessstableastvaries.Whileourmodelmakesuseofthepartitionsprimarilyasatoolforsharinginformationacrossgroups,thematchingstructurethatourmodelcapturesmaybe 90
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CHAPTER5CONCLUSIONSANDFUTUREWORKInthisdissertationwehavedevelopedthreenovelmethodstoestimatethedependencestructuresinlongitudinaldata.InChapter 2 weintroducedBayesianpriordistributionsforthecorrelationmatrixRthatinducessparsitythroughthepartialautocorrelations( Daniels&Pourahmadi 2009 ).ByallowingshrinkageorselectiononthePACs,weimproveestimationbyworkinginalower-dimensionspace.Inthenexttwochapters,weconsideredtheproblemofsimultaneouscovarianceestimation.Chapter 3 introducesanon-parametricmodelthroughthematrixstick-breakingprocess( Dunsonetal. 2008 )thatallowsequalityofparametersacrossgroupsandacrossGARPsofacommonlag.InChapter 4 weproposeamethodrelyingonpartitioningthegrouplabelsateachtimepoint,sothatthesequentialdistributionscoincideforgroupsinthesamesetofthepartition.BothpriordistributionsareformedintermsofthemodiedCholeskyparameterization( Pourahmadi 1999 )andfavorsparsechoicesoftheT()matrix.Intermsoffutureresearchdirections,thePACparameterizationofRisquiteintuitivebuthasyettoattractalotofresearchforlongitudinaldata.Inparticular,algorithmsformoreefcientcomputationsoftheposteriorareneeded,especiallyforcorrelationmatricesoflargerdimension.Also,modelsthatseekstructureinthePACsbeyondsparsitycanbedevelopedsuchasshrinkingorclusteringPACswithinlag.Additionalmodelsthatimposesparsitythroughthemarginalorpartialcorrelations,beyond Pittetal. ( 2006 ),couldbedeveloped.Regardingthesimultaneouscovarianceestimationproblem,thereareanumberofextensionsandfuturedirectionstoexplore.Possibleextensionsofthegroupingandcovariancepartitionmodelsincludesituationswherethedimensionofmchangesacrossgroupsorifthetimebetweenmeasurementsdiffersacrossgroups.Oftenlongitudinaldataarenotmeasuredatxedtimes,andsooneshouldconsiderwhat 92
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kindofmethodologyisneededtoallowsharingofinformationacrossgroupsforthesesituations.InbothChapters 3 and 4 weconsiderthegroupsasexchangeable,butdependingoncontext,theremaybeaninformativeorderingtothegroupssuchasincreasinglevelsoftreatmentthatshouldbeexploited.RatherthanrelyingongroupingsoftheGARPandIVparametersthatareequal,modelscouldbedevelopedthatrelyontargetedshrinkage.Further,theideaofusingajointdistributiononpartitions(P1,...,PT)ispotentiallyapplicableinavarietyofcontextsoutsideofcovarianceestimationwhensimultaneousclusteringisrequired. 93
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APPENDIXACALCULATINGTHEDICSTATISTICFORCTQDATAWedescribeinthisappendixthedetailsinvolvedinapproximatingtheDICtermusedformodelcomparisonoftheCTQIdata,andinparticular,theestimationoftheintegralin( 2 )introducedinSection 2.6 .First,weintroducenotation.Let=(,R)bethesetofparameters,^=(^,^R)thesetoftheposteriorestimates,andgthevalueofattheg-thiterationoftheMarkovchain(g=1,...,G).ThefunctionIi(Y)=IfQitYt08tgindicateswhetherYisasetoflatentvariableswhosesignsagreewithQi.Denepi()tobetheprobabilityofobservingQiundertheparameters.Asin( 2 ),thisispi()=pi(,R)=Z(,1)JIi(y)(yj)dy,where(j)isthemultivariatenormaldensitywithmeanXiandcovariancematrixRwhen=(,R).Hence,loglik(jQi)=logpi().Aspreviouslynoted,thisintegralinintractable.Usingthedenitionsofthedevianceandcomplexityparameter(equations( 2 )and( 2 ))andthenewnotation,wecanwriteDICasthesumofthecontributionsDICiforeachpatient,DIC=XiDICi=Xih2logpi(^))]TJ /F4 11.955 Tf 11.95 0 Td[(4Eflogpi()gi.AsobservationsQiareindependent,itsufcestoconsidertheperpatientcontributionDICi.Notethattheexpectationinthenaltermiswithrespecttotheposteriordistributionoftheparametersandwillbeestimatedbyitsaverageoverthevaluesfromtheposteriorsample1,...,G.Additionally,wewillhavetoapproximatepi()withsomeestimate^pi().So,dDICi=2log^pi(^))]TJ /F4 11.955 Tf 11.95 0 Td[(4G)]TJ /F9 7.97 Tf 6.59 0 Td[(1GXg=1log^pi(g). 94
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NotethatcalculationofDICwillrequire(G+1)estimatesoftheintegralpi()foreachi=1,...,N.Toevaluatethisintegralweuseimportancesampling( Robert&Casella 2004 ,Section3.3).Wetakeasoursamplingdensityt(j),themultivariatet-distributionwith5degreesoffreedom,locationparameterXi^,andscalematrixk^Rforsomeconstantk>1.Wedene=(^,k^R)tobethesetofparametersforthesamplingdistribution.Wechoosethet-distributionsothatt(j)willhaveheaviertailsthan(jg)forg=1,...,Gand^.Thisalsomotivatesthechoicetouseascalematrixthatisaninatedversionof^R.Notethatwecanwritepi()aspi()=Z(,1)JIi(y)(yj)dy=Z(,1)JIi(z)(zj) t(zj)t(zj)dz.Toestimatethis,drawZ1,...,ZHiindependentlyfromt(j),and^pi()=H)]TJ /F9 7.97 Tf 6.59 0 Td[(1iHiXh=1Ii(Zh)(Zhj) t(Zhj)isanunbiasedandconsistentestimatorofpi().EvaluationofdDICiinvolves,foreachg=1,...,G,simulatingadatasetZ=fZhgHih=1andcalculating^pi(g),followedbydrawinganalZtoestimate^pi(^).DrawingG+1independentdatasetsturnsouttobecomputationallyslow.Instead,wewilldrawasinglesampleZtousetocalculateall^pi(1),...,^pi(G),^pi(^).Itisclearthattheseestimatesremainunbiasedandconsistent.WhatremainsistoconsiderwhateffectthiswillhaveonthevariabilityoftheindividualcontributionstotheDIC,dDICi.FirstwederivethevarianceofdDICiinthesituationwherewedrawanewdatasetZforeach^pi(g).Inthiscase, VarfdDICig=4Varflog^pi(^)g+16G)]TJ /F9 7.97 Tf 6.58 0 Td[(2XgVarflog^pi(g)g,(A) 95
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wheretheexpectation(inthevariance)iswithrespecttothesamplingdistributionofZ.Foranygor^,thisvarianceis Varflog^pi()gpi())]TJ /F9 7.97 Tf 6.59 0 Td[(2Varf^pi()g=pi())]TJ /F9 7.97 Tf 6.58 0 Td[(2H)]TJ /F9 7.97 Tf 6.58 0 Td[(1iVarIi(Z1)(Z1j) t(Z1j)=pi())]TJ /F9 7.97 Tf 6.59 0 Td[(2H)]TJ /F9 7.97 Tf 6.59 0 Td[(1iEIi(Z1)(Z1j)2 t(Z1j)2)]TJ /F5 11.955 Tf 11.95 0 Td[(pi()2,wheretheapproximationintherstlineisduetothedeltamethod.Thisquantitycanbeconsistentlyestimatedby dVarflog^pi()g=^pi())]TJ /F9 7.97 Tf 6.59 0 Td[(2H)]TJ /F9 7.97 Tf 6.59 0 Td[(1i"H)]TJ /F9 7.97 Tf 6.59 0 Td[(1iHiXh=1Ii(Zh)(Zhj)2 t(Zhj)2)]TJ /F4 11.955 Tf 12.1 0 Td[(^pi()2#.(A)TocalculatethevarianceofdDICiunderoursamplingschemewithasinglesampleZ,note VarfdDICig=4Varflog^pi(^)g+16G)]TJ /F9 7.97 Tf 6.59 0 Td[(2XgVarflog^pi(g)g+16G)]TJ /F9 7.97 Tf 6.59 0 Td[(2XgXg06=gCovflog^pi(g),log^pi(g0)g (A) )]TJ /F4 11.955 Tf 9.29 0 Td[(16G)]TJ /F9 7.97 Tf 6.59 0 Td[(1XgCovflog^pi(g),log^pi(^)g.Thequantitiesonthesecondandthirdlinesof( A )representtheadditionaltermsduetosharingthedatasetZacrosscalculationsof^pi().DeneCOVitobethesumofthesecovarianceterms.WemaywriteCOViasCOVi=16G)]TJ /F9 7.97 Tf 6.59 0 Td[(2XgXg06=gCovflog^pi(g),log^pi(g0)g)]TJ /F5 11.955 Tf 32.23 8.09 Td[(G G)]TJ /F4 11.955 Tf 11.96 0 Td[(1Covflog^pi(g),log^pi(^)g.FromthedeltamethodwehaveCovflog^pi(g),log^pi()gCov^pi(g) pi(g),^pi() pi(), 96
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andso COVi16G)]TJ /F9 7.97 Tf 6.59 0 Td[(2XgXg06=g"Cov^pi(g) pi(g),^pi(g0) pi(g0))]TJ /F5 11.955 Tf 23.6 8.09 Td[(G G)]TJ /F4 11.955 Tf 11.96 0 Td[(1Cov(^pi(g) pi(g),^pi(^) pi(^))#=16G)]TJ /F9 7.97 Tf 6.59 0 Td[(2XgXg06=gCov(^pi(g) pi(g),^pi(g0) pi(g0))]TJ /F5 11.955 Tf 23.59 8.09 Td[(G G)]TJ /F4 11.955 Tf 11.95 0 Td[(1^pi(^) pi(^))=16G)]TJ /F9 7.97 Tf 6.59 0 Td[(2H)]TJ /F9 7.97 Tf 6.59 0 Td[(1iXgXg06=gCovIi(Z1)(Z1jg) pi(g)t(Z1j),Ii(Z1) (Z1jg0) pi(g0)t(Z1j))]TJ /F5 11.955 Tf 23.59 8.09 Td[(G G)]TJ /F4 11.955 Tf 11.96 0 Td[(1(Z1j^) pi(^)t(Z1j)!)=16G)]TJ /F9 7.97 Tf 6.59 0 Td[(2H)]TJ /F9 7.97 Tf 6.59 0 Td[(1iXgXg06=gG G)]TJ /F4 11.955 Tf 11.96 0 Td[(1 (A) +E(Ii(Z1)(Z1jg) pi(g)t(Z1j) (Z1jg0) pi(g0)t(Z1j))]TJ /F5 11.955 Tf 23.59 8.09 Td[(G G)]TJ /F4 11.955 Tf 11.95 0 Td[(1(Z1j^) pi(^)t(Z1j)!)#.Aslongasthisquantity( A )issmall(relativetotheindependencevarianceestimator( A )),wemaysavecomputationaltimebyonlysamplingonedatasetZwithoutsacricingprecision.InthesituationoftheanalysisoftheCTQdata,estimationofCOViwitharepresentativesampleofobservationsshowedCOVitobesmallrelativetotheindependencevarianceestimator( A ).Infact,thistermisoftennegative,indicatingthatusingthecommondatasetZmayimproveestimationforsomeobservationsi.Intherepresentativesampleweconsidered,theadditionoftheCOVitermtendedtoleadtochangesinthestandarderrorofdDICirangingfromadecreaseof5%toanincreaseof25%.AsitiscomputationallyinfeasibletocomputeCOViforallobservations(anestedloopoverg0insidealoopoverg),weestimatethestandarderroroftheDICestimateusingtheindependenceestimatorobtainedfrom( A )and( A ).TheDev,pD,andDICestimatesinTable 2-4 arecomputedinthisway.TheimportancesamplingsizeHiischosenbyrstdrawing200,000valuesofZht(j),wherevariancescalingfactorkis1.52.ThischoiceofkismadewithconsiderationtothedimensionJofZ(increasingJshouldcorrespondtoincreasingk),howfarfromthe 97
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origintendstobe,howlikelyIi(Z)istobeone,amongotherconsiderations;ultimately,trial-and-errorwithsmallchoicesofHiledusconcludethatk=1.52worksreasonablywell.Havingdrawn200,000valuesofZh,ifPhIi(Zh)2000(i.e.,atleast2000oftheZh'shavesignsappropriateforalatentvariableofQi),thenHi=200,000andthissetfZhg200,000h=1istheimportancesampleZ.Ifnot,wecontinuetodrawadditionalsetsof200,000Zh'stoappendtothedatasetuntilPhIi(Zh)2000.Thisimpliesthatwehavelargersamplesforthosepatientsiwithsmallvaluesofpi().ThishelpstocontrolthevarianceofDICisincethetermisprecededbypi())]TJ /F9 7.97 Tf 6.59 0 Td[(2(seeequation( A )).Withthisscheme,weestimatethestandarderrorsforourDICestimatesinTable 2-4 tobearound0.5. 98
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APPENDIXBADDITIONALCOVARIANCEGROUPINGPRIORSANDTHEIRPROPERTIES B.1SparsityGroupingPriorforWeintroducethreeadditionalgroupingpriorspecicationstotheoneintroducedinSection 3.3 ,twofortheautoregressiveparametersandonefortheinnovationvariances.Thesethreepriorsfollowcloselytothematrixstick-breakingprocess( Dunsonetal. 2008 ).Herewerelyonfewerlongitudinalassumptionsintheconstructionofthesemodels,andsotheymaybeviewedasamiddlegroundbetweenthelag-blockandcorrelated-lognormalpriorsandthenaiveBayespriorsusedinSections 3.6 and 3.7 .Thesparsitygroupingpriorisdenedbyreplacingequations( 3 )and( 3 )fromthelag-blockgropingpriorwith mjFmj()=HXh=1mjhjh()(m=1,...,M;j=1,...,J),jhq(j)0+(1)]TJ /F11 11.955 Tf 11.95 0 Td[(q(j))N(0,2)(j=1,...,J;h=1,...,H). (B) Thekeydistinctionisthatin( B )wesamplecandidatesjhforeachautoregressiveparameterj.Previouslywehadasetofcandidatesqhforallautoregressiveparameterswithinlag.Becausethecandidatesaredistinctacrosslags,thispriorwillnolongerallowclusteringofautoregressiveparametersacrosslags,exceptforcommonzerovalues.Asbefore,weuseazero-normalmixtureforthebasedistributionsofthecandidatestoencouragesparsity.Thissparsitygroupingpriorforisdenedsimilarlytothematrixstick-breakingprocess,withthekeydifferencebeingthat Dunsonetal. ( 2008 )specifythatthebasedistributionofthejh'sbenonatomic.Thisisnotthecasewithourpriorsinceweuseadistributionforthecandidatesthatcontainapointmassatzero.Thisdoesnotleadtoaproblem,butitdoesaltersomeofthetheoreticproperties.Whensamplingfromthesparsitygroupingprior,properties( 3 )( 3 )ofthelag-blockpriorcontinuetohold.Thismeansthatthedistribution'scorrelation, 99
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corrfFmj(A),Fm0j(A)g,andclusteringproperties,pr(mj=m0j),acrossgroupswithinthesameautoregressiveparametercontinuetohold.However,properties( 3 )and( 3 )whichconsiderbehavioracrossautoregressiveparametersj6=j0withinthesamelagchange.ItisnowthecasethatcorrfFmj(A),Fmj0(A)g=corrfFmj(A),Fm0j0(A)g=0.Additionally,pr(mj=mj0)=pr(mj=m0j0)=2q,theprobabilitythatbothareindependentlysettozero.Hence,thispriorprovidesaless-richgroupingstructurethanthelag-blockprior.Recallthatinthespecialcase!0and!1,eachofthemodelparametersissampledindependentlyfromthecandidatebasedistribution.Thatis,mjisdistributedasequation( 3 ),whichisthenaiveBayesprior1thatweusedfortherisksimulationanddataanalysis.Hence,thesparsitygroupingpriorsubsumesthisnaivemodelasaspecialcase. B.2Non-SparseGroupingPriorforThenon-sparsegroupingpriorisaslightsimplicationofthesparsitygroupingthatnolongerencouragessparsity.Toformthisprior,weusethesparsitygroupingpriorandreplaceequation( B )withjhN(0,2).Havingremovedthezeropointmassfromthe-level,mjisalmostsurelynon-zero.Thus,wenolongerallowforconditionalindependencerelationshipsorgainsparsityintheT()matrix,butbyxingthemeanatzeroforthedistributionofthecandidates,westillencourageshrinkagetowardindependence.Thispriorfollowsexactlytheframeworkof Dunsonetal. ( 2008 ).Weadditionallypointoutthatthenon-sparsegroupingpriorisequivalenttothesparsitygroupingpriorifwexq=0forallq.Hence,therespectivepropertiesforthispriorareobtainedbysubstituting=0,andconsequently,()=(),intothepropertiesofthesparsitygroupingprior.Becausethenon-sparsepriorisamatrixstick-breakingprocesspriorwithanon-atomicbasedistribution,thepropertiesmay 100
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alternativelybetakenfromPropositions1,2,and4in Dunsonetal. ( 2008 ).Additionally,thenaiveBayes2prioristhespecialcaseofthisnon-sparsegroupingpriorwhen!0and!1. B.3InvGammaGroupingPriorfor)]TJ /F1 11.955 Tf -308.03 -24.53 Td[(Weconstructanadditionalpriorfortheinnovationvariancesdifferingfromthecorrelated-lognormalpriorintwokeyways.First,wedonotstrivetotreattheinnovationvariancesascomingfromasmoothfunction.Inthepreviouspriorthevariancecandidatesjhweresampledfromacorrelated-lognormaldistribution,butwenowtreatthecandidatesindependentlyacrossj.Secondly,becausewechooseindependentcandidates,wechoosetheconjugateinversegammaforthebasedistributioninsteadofthelognormal.TheInvGammagroupingpriorisdenedbyreplacinglines( 3 )and( 3 )with mjGmj()=HXh=1mjhjh()(m=1,...,M;j=1,...,p),jhInvGamma(1,2),(j=1,...,p;h=1,...,H). (B) Wenolongerrequireany!hin( 3 )becausewedonotdesireintoinduceacorrelationamongthe's.Aswiththenon-sparsepriorthisisaspecialcaseofthematrixstick-breakingprocess.Wecomparethispriorwiththecorrelated-lognormalpriorinthespecialcasewhere=0.Recallthatif=0,thenthecandidatesjhareindependentvariablesfromthelognormal( ,)distribution,andthecorrelated-lognormalpriorgivesamatrixstick-breakingprocess.Betweenthesetwopriors,theInvGammagroupingpriorandthecorrelated-lognormalgroupingpriorwith=0,werecommendusingtheInvGammabecauseoftheconjugacythatisobtained.Thebenetofusingthenormal(equivalently,lognormal)distributionfor!()isthatinducingthecorrelationinsideaclusterisstraightforward,againthatoutweighsthelossofconjugacy.Forsituationswhereitisreasonabletoconsidertheinnovationvariancesasfollowingasmoothprogress,such 101
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asmostlongitudinaldatamodels,werecommendthecorrelated-lognormalpriorwithanon-zerochoiceof.Toobtainthetheoreticalpropertiesofthisprior,letI\()denotetheprobabilityfunctionoftheInvGamma(1,2)distribution,withxedvaluesforthehyperparameters.BecausetheInvGammagroupingpriorisamatrixstick-breakingprocess,themostrelevantpropertiesfollowimmediatelyfrom Dunsonetal. ( 2008 ).ForA2B(R+)theunbiasedandvariancepropertiesaregivenbyEfGmj(A)g=I\(A),VarfGmj(A)g=2 (2+)(2+))]TJ /F4 11.955 Tf 11.96 0 Td[(2I\(A)f1)]TJ /F1 11.955 Tf 11.95 0 Td[(I\(A)g.Thebehavioracrossgroupsm6=m0withcommontimejasin( 3 )and( 3 )continuetoholdwith=0.ThekeydifferenceisthatcorrfGmj(A),Gmj0(A)g=corrfGmj(A),Gm0j0(A)g=0,contrastedwith( 3 )and( 3 ).Thatis,thedistributionsfortimesjandj0areuncorrelated.Itremainstruethatpr(mj=mj0)=0.As!0and!1,thispriorconvergestothenaivepriorusedfortheinnovationvariancesinSections 3.6 and 3.7 ofthearticle. B.4FurtherGroupingPriorExtensionsInadditiontothegroupingpriorspreviouslydenedinChapter 3 andhere,thereareanumberofothernaturalextensionsandpossiblevariationsthatonecouldconstruct.Forinstance,onecouldallowfordifferingvaluesof2,thevarianceoftheautoregressiveparametercandidates,thatdependonthelagvalueoftheassociatedautoregressiveparameter.Thismightbebenecialinasituationwherepislargeandonebelievesthatthe'saftertherstfewlagsvarymoretightlyaroundzero.Additionally,wecanremovethesparsityfromthelag-blockgroupingpriorbydeletingthepointmassatzerofromthedistributionofthe's.Asanotherchoice,insteadofspecifyingtheand)]TJ /F1 11.955 Tf 10.1 0 Td[(asseparateblockswithdifferentvaluesofthestick-breakingparametersand,onecoulddrawbothsetsofparameterswiththesamevaluesofand.Insteadofspecifyingthatthecandidate'sarezeroaccording 102
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totheprobability,anotherextensionistomodifythegroupingpriorbyintroducingazero-thclusterwherej0=0forallj.Theselectionofmjwouldthenfollowbypr(mj=0)=pr(mj=j0)=q(j)andforh=1,...,H,pr(mj=jh)=(1)]TJ /F11 11.955 Tf 12.28 0 Td[(q(j))mjh.Thepropertieswehaveconsideredareeasilyobtainedandcomparedtothoseobtainedinthesparsitygroupingcase.Whiletheseorothersmaybemorenaturalincertaincontexts,webelievethatthelag-blockandcorrelated-lognormalgroupingpriorsarethemostapplicablechoicesforgenerallongitudinaldata. 103
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APPENDIXCDERIVATIONOFCOVARIANCEGROUPINGPRIORPROPERTIES C.1ProofsforGeneralizedAutoregressiveParameterPropertiesWeprovidepartialproofstothepropertiesofthelag-blockgroupingpriornotedinSection 3.4 ofthearticle.Theproofsforproperties( 3 )and( 3 )canbefoundintheappendixto Dunsonetal. ( 2008 ).Duetothenon-atomicnatureofthebasedistribution( 3 ),wearenotabletodirectlyapplytheirconclusiontoourpriortoprove( 3 ).Detailsfollow.pr(mj=m0j)=pr(mj=m0j6=0)+pr(mj=m0j=0)=E(Xhmjhm0jhjh(Rn0))+E(Xhmjhjh(0)Xim0jiji(0))=E(Xhmjhm0jhjh(Rn0))+E(Xhmjhm0jhjh(0))+2E(Xh1Xi=h+1mjhm0jijh(0)ji(0))=E(Xhmjhm0jhjh(R))+2E(Xh1Xi=h+1mjhm0jijh(0)ji(0))=E(Xhmjhm0jh)+22E(Xh1Xi=h+1mjhm0ji)=(I)+22(II),whereexpressions(I)and(II)arecalculatedbelowandRdenotestherealline. 104
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(I)=E(XhUmhUm0hX2jhYl
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Using(I)and(II),wehavepr(mj=m0j)=(I)+22(II)=2+1)]TJ /F11 11.955 Tf 11.96 0 Td[(2 (1+)(2+))]TJ /F4 11.955 Tf 11.95 0 Td[(1.Toestablishthecorrelationpropertyin( 3 ),letq=q(j)=q(j0).FirstnoteEfFmj(A)Fmj0(A)g=E(Xhmjhmj0hqh(A))+2E(Xh1Xi=h+1mjhmj0iqh(A)q0i(A))=(A)E(Xhmjhmj0h)+2(A)2E(Xh1Xi=h+1mjhmj0i)=(A)(III)+2(A)2(IV),whereformulas(III)and(IV)aredenedbelow.(III)=E(XhU2mhXjhXj0hYl
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(IV)=E"Xh1Xi=h+1UmhXjh(1)]TJ /F5 11.955 Tf 11.96 0 Td[(UmhXj0h)(Yl
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Toprovethematchingprobabilityof( 3 ),notepr(mj=mj0)=pr(mj=mj06=0)+pr(mj=mj0=0)=E(Xhmjhmj0hqh(Rn0))+E(Xhmjhqh(0)Ximj0iqi(0))=E(Xhmjhmj0hqh(R))+2E(Xh1Xi=h+1mjhmj0iqh(0)qi(0))=(III)+22(IV)=2q+1)]TJ /F11 11.955 Tf 11.95 0 Td[(2q (2+)(1+))]TJ /F4 11.955 Tf 11.95 0 Td[(1.Toobtain( 3 ),wehaveEfFmj(A)Fm0j0(A)g=E(Xhmjhm0j0hqh(A))+2E(Xh1Xi=h+1mjhm0j0iqh(A)qi(A))=(A)E(Xhmjhm0j0h)+2(A)2E(Xh1Xi=h+1mjhm0j0i)=(A)(V)+2(A)2(VI),where(V)=E(XhUmhUm0hXjhXj0hYl
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and(VI)=E"Xh1Xi=h+1UmhXjh(1)]TJ /F5 11.955 Tf 11.96 0 Td[(Um0hXj0h)(Yl
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Toseethatthedistributionsareuncorrelated,assumem=m0.ThenEfFmj(A)Fmj0(A)g=E(Xhmjhmj0hqh(A)q0h(A))+2E(Xh1Xi=h+1mjhmj0iqh(A)q0i(A))=(A)2(III)+2(A)2(IV)=(A)2.Whenm6=m0,theproofproceedssimilarlybutrequiresexpressions(V)and(VI)insteadof(III)and(IV). C.2ProofsforInnovationVariancePropertiesToestablishthepropertiesforthecorrelated-lognormalprior,notethatforaxedcommonvalueofand,thedistributionsofUmhandWmh,aswellasXjhandZjh,arethesame.Hence,thesetfmjhgwillbedistributedthesameasthesetfmjhg,andwemayusetheexpressions(I)(VI)toobtainexpectationsoftheinnovationvariancestick-breakingweights.Toprove( 3 ),noticeEfGmj(A)Gmj0(A)g=E(Xhmjhmj0hjh(A)j0h(A))+2E(Xh1Xi=h+1mjhmj0ijh(A)j0i(A))=Enj1(A)j01(A)oE(Xhmjhmj0h)+2fEj1(A)gfEj01(A)gE(Xh1Xi=h+1mjhmj0i)=Enj1(A)j01(A)o(III)+2fEj1(A)gfEj01(A)g(IV)=1 (2+)(1+))]TJ /F4 11.955 Tf 11.95 0 Td[(1hEn!j1(logA)!j01(logA)ofE!j1(logA)gfE!j01(logA)gi+fE!j1(logA)gfE!j01(logA)g=1 (2+)(1+))]TJ /F4 11.955 Tf 11.95 0 Td[(1covn!j1(logA),!j01(logA)o+(logA)2. 110
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Applyingvarf!j1(logA)g=(logA)f1)]TJ /F4 11.955 Tf 12.36 0 Td[((logA)gandastheformulasforEfGmj(A)gandvarfGmj(A)ggivesthenalresult.Theproofof( 3 )followsthesameasabove,exceptoneusesexpressions(V)and(VI)inplaceof(III)and(IV).Finally,pr(mj=m0j0)=0followsfromtheobservationthat!jh6=!j0h0almostsurelyasaconsequenceofthemultivariatenormaldistributionwithanon-degeneratecorrelation.Proofsofthepropertiesofthesparsity,non-sparse,andInvGammagroupingpriorsdenedinAppendix B areexcluded.Thesecanbederivedfollowingstepssimilartothoseaboveorintheappendixto Dunsonetal. ( 2008 ). 111
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APPENDIXDDETAILSOFMCMCALGORITHMFORCOVARIANCEGROUPINGPRIORS D.1PreliminariesAsmentionedinSection 3.5 ,weintroduceseverallatentvariablestofacilitatesamplingfromthedistributionsFmj()andGmj()inequations( 3 )and( 3 ),followingthealgorithmof Dunsonetal. ( 2008 ).Tothisend,considerthefollowingfoursetsofbinarydummyvariables:umjhBern(Umh)(m=1,...,M;j=1,...,J;h=1,...,H),xmjhBern(Xjh)(m=1,...,M;j=1,...,J;h=1,...,H),wmjhBern(Wmh)(m=1,...,M;j=1,...,p;h=1,...,H),zmjhBern(Zjh)(m=1,...,M;j=1,...,p;h=1,...,H).NowdeneRmj=minfh:1=umjh=xmjhgandAmj=minfh:1=wmjh=zmjhg.Byconstruction,pr(Rmj=h)=mjhandpr(Amj=h)=mjh.WeletRmjdesignatewhichjhoutoftheHcandidateswechooseasmj,andlikewise,Amjgivesthejhtoselectasmj.Hence,isdeterminedbyfRmjgandfjhgand)]TJ /F1 11.955 Tf 10.1 0 Td[(byfAmjgandfjhg.Thus,aftersamplingthevaluesoffRmjg,fjhg,fAmjg,andfjhg,thevaluesofand)]TJ /F1 11.955 Tf 10.1 0 Td[(aredetermined.NowwecalculatetheconditionaldistributionsthatwewillneedforourGibbssamplerforeachofthegroupingpriors.Notationally,wedenotetheconditionaldistributionforarandomvariable,sayC,conditionalontheremainingrandomvariablesbyCj)]TJ /F1 11.955 Tf 15.94 0 Td[(. D.2SamplingStepsforLag-BlockGroupingPriorSamplingfromthelag-blockgroupingpriorproceedsinfoursteps:theautoregressiveparametersthoughtheparametercandidatesandthedummyvariablesR,u,x;thecandidateprobabilitiesmjhbysamplingtheUmhandXjh;thestick-breakingparameters 112
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,;thehyperparametersofthebasedistribution( 3 ).Therststepproceedsintwosubsteps.Algorithmdetailsfollow.Step1a:Parametercandidates.Itisimportanttorecallthedenitionoftheautoregressiveparametersasconditionalregressioncoefcients.Forinstance,therstparameterm1istheregressioncoefcientforymi1ontoymi2withinnovationvariancem1.Likewise,m2andm3arethecoefcientsofymi1andymi2formodelingymi3withvariancem2.Weletxmijdenotethecomponentofymithatcorrespondstothejthautoregressiveparameterregressor,e.g.xmij=ymi1forj=1,2andxmi3=ymi2.Similarly,weletmjdenotetherelevantinnovationvariance.Thatis,m1=m1,andforj=2and3,mj=m2.Finally,wedeneemijtobetheresidualfortheregressionequation,excludingthecontributionofxmij.Thatis,forj=1,emi1=ymi2,forj=2,emi2=ymi3)]TJ /F11 11.955 Tf 13.09 0 Td[(m3ymi2,andforj=3,emi3=ymi3)]TJ /F11 11.955 Tf 13.09 0 Td[(m2ymi1.The-variablesaredenedinthenaturalwaysothatemijN(mjxmij,mj)foreachj.Havingestablishedthenecessarynotation,weseethatthecontributiontothedatalikelihoodofmjisproportionaltoexp()]TJ /F4 11.955 Tf 18.65 8.09 Td[(1 2mjnmXi=1)]TJ /F5 11.955 Tf 5.48 -9.68 Td[(emij)]TJ /F11 11.955 Tf 11.96 0 Td[(mjxmij2).However,wedonotdrawthemj'sbutqh.Forh=1,...,Handxedq=1,...,p)]TJ /F4 11.955 Tf 9.59 0 Td[(1,letPqhdenotesthesetof(m,j)suchthatq(j)=qandRmj=h,whichisthesetofgroupandautoregressiveparameterpairsthatcontributetothelikelihoodofqh.Thus,thecontributionofqhis exp8<:X(m,j)2PqhnmXi=1)]TJ /F4 11.955 Tf 9.3 0 Td[(1 2mj(emij)]TJ /F11 11.955 Tf 11.95 0 Td[(qhxmij)29=;. (D) 113
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Thissummationover(m,j)2Pqhmeansthatweareonlyincludingthe(m,j)pairssuchthatqh=mj.Fromthisobservation,wehavethattheconditionaldistributionofqhis (qhj)]TJ /F4 11.955 Tf 12.62 0 Td[()/exp8<:X(m,j)2PqhnmXi=1)]TJ /F4 11.955 Tf 9.3 0 Td[(1 2mj(emi)]TJ /F11 11.955 Tf 11.96 0 Td[(jhxmi)29=;q0(qh)+(1)]TJ /F11 11.955 Tf 11.96 0 Td[(q)(22))]TJ /F14 5.978 Tf 7.79 3.26 Td[(1 2exp)]TJ /F11 11.955 Tf 12.37 9.1 Td[(2qh 22/q0(qh)+(1)]TJ /F11 11.955 Tf 11.95 0 Td[(q) exp()2 22N(,2), (D) where =2X(m,j)2PqhnmXi=1emijxmij mj,2=8<:1 2+X(m,j)2PqhnmXi=1(xmij)2 mj9=;)]TJ /F9 7.97 Tf 6.59 0 Td[(1.(D)Thus,tosamplefromthisconditional( D ),wesetqhtozerowithprobabilityq q+(1)]TJ /F11 11.955 Tf 11.96 0 Td[(q) expn()2 22o,andotherwise,drawqhfromN(,2).IfPqhisempty,thentheconditionalisq0+(1)]TJ /F11 11.955 Tf 11.96 0 Td[(q)N(0,2),theoriginalpriorforqhgivenby( 3 ).Step1b:Dummyvariables.Havingupdatedthecandidatesvaluesfqhgh,wenowsamplethevariablesthatdeterminewhichcandidatewechoosefortheautoregressiveparameter.First,form=1,...,Mandjsuchthatq=q(j),wesampleRmjaftermarginalizingoutfumjh,xmjhgHh=1.Theconditionalprobabilitydistributionisgivenby pr(Rmj=hj)-222(nfumjh,xmjhgh)/mjhexp()]TJ /F4 11.955 Tf 18.65 8.09 Td[(1 2mjnmXi=1(emij)]TJ /F11 11.955 Tf 11.96 0 Td[(q(j)hxmij)2).(D)Hence,wedrawRmjfromthemultinomialdistributionwithprobabilitiesfrom( D ),normalizedtosumtoone.GiventhevalueofRmj,wedrawthesetfumjh,xmjhghtorequirethatRmjistherstoccasionwherebothumjhandxmjhareone.Forh>Rmj,drawumjhBern(Umh)andxmjhBern(Xjh),andwhenh=Rmj,1=umjh=xmjh.Forh
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thenwejointlydrawumjhandxmjhinaccordancetothefollowingprobabilities:pr(umjh=0,xmjh=0)=(1)]TJ /F5 11.955 Tf 11.96 0 Td[(Umh)(1)]TJ /F5 11.955 Tf 11.96 0 Td[(Xjh)=(1)]TJ /F5 11.955 Tf 11.95 0 Td[(UmhXjh),pr(umjh=1,xmjh=0)=Umh(1)]TJ /F5 11.955 Tf 11.95 0 Td[(Xjh)=(1)]TJ /F5 11.955 Tf 11.96 0 Td[(UmhXjh),pr(umjh=0,xmjh=1)=(1)]TJ /F5 11.955 Tf 11.96 0 Td[(Umh)Xjh=(1)]TJ /F5 11.955 Tf 11.96 0 Td[(UmhXjh).Steps1aand1bshouldbeperformconsecutivelywiththesamevalueofq.Step2:Candidateprobabilities.ToupdatetheprobabilitiesmjhwemustsamplenewvaluesforthecomponentsUmhandXjh.Tothatend,giventhevaluesoftheumjh'sandtheothervariables,theconditionalforUmhforh
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and.Thentheconditionalforisj)-277(Gamma0@M(H)]TJ /F4 11.955 Tf 11.96 0 Td[(1)+1,1)]TJ /F7 7.97 Tf 16.95 14.95 Td[(MXm=1H)]TJ /F9 7.97 Tf 6.58 0 Td[(1Xh=1log(1)]TJ /F5 11.955 Tf 11.95 0 Td[(Umh)1A.Likewise,j)-278(Gamma0@J(H)]TJ /F4 11.955 Tf 11.96 0 Td[(1)+1,1)]TJ /F7 7.97 Tf 18.25 14.94 Td[(JXj=1H)]TJ /F9 7.97 Tf 6.58 0 Td[(1Xh=1log(1)]TJ /F5 11.955 Tf 11.95 0 Td[(Xjh)1A.WecanchooseadifferentGamma(a,b)priorinsteadofGamma(1,1),andwewillmaintainthegamma-gammaconjugacy.Step4:Basedistributionhyperparameters.Thebasedistributionforqhin( 3 )dependsontwosetsofhyperparameters:thevarianceofthecontinuouspiece2andthesparsityprobabilitiesq.WechoosetheInvGamma(a,b)familyofdistributionsfortheprioron2,sothatwewillhaveconjugacy.Thisyieldsthefollowingconditionaldistribution,2j)-278(InvGamma a+1 2Xq,hf1)]TJ /F11 11.955 Tf 11.95 0 Td[(0(qh)g,b+1 2Xq,h2qh!.Onemustnowspecifythevaluesofa,b.WerecommendInvGamma(0.1,0.1),sothatourpriorapproximatesthecommonly-usedimproperprior(2)/)]TJ /F9 7.97 Tf 6.59 0 Td[(2.ByplacingaBeta(q,q)prioronq,theconditionalforqisqj)-278(Beta0@q+HXh=10(qh),q+HXh=1f1)]TJ /F11 11.955 Tf 11.95 0 Td[(0(qh)g1A.Itisnecessarytospecifythevaluesofqandq.Werecommendusingq=q=1forallq,whichgivesaUnif(0,1)priorforeachq.Alternatively,onecouldchoosethevaluesofqandqtomoreaggressivelyshrinkqforlowerlagstowardzeroandqforhigherlagstowardone. 116
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D.3SamplingStepsforCorrelated-LognormalGroupingPriorThesamplingfortheinnovationvariancepriorfollowssimilarlytothepreviousprior.However,wenowlongerhavetoperformthecandidatevalueanddummyvariablesamplingstepconcurrentlyasforthelag-blockprior.Step1:Parametercandidates.Let~emijbetheresidualobtainedfromthedifferenceofymijandthepreviouscomponentsofymimultipliedbytheappropriateautoregressiveparameters.Forinstance,whenj=1,~emi1=ymi1,andforj=2,~emi2=ymi2)]TJ /F11 11.955 Tf -421.13 -23.91 Td[(m1ymi1,andsoon.Notethatthisisadifferentdenitionofthese~e-residualsfromthee-residualsusedintheautoregressiveparametersamplingsteps.Foreachvalueofj,thisyields~emijN(0,mj).Thecontributiontothelikelihoodofjhfromthedataisproportionalto )]TJ /F14 5.978 Tf 7.79 3.26 Td[(1 2Pmnmh(Amj)jhexp()]TJ /F4 11.955 Tf 16.99 8.09 Td[(1 2jhXmnmXi=1(~emij)2h(Amj)).(D)Insteadofconsideringtheconditionalforjh,weinsteadchoosetolookintermsof!jh=logjh.Foreachsamplingset,wepartition!hinto(!hA,!hB)sothat!hAcontainsthecollectionof!jhsuchthatAmj=hforatleastonem.Thisdivides!hintothe!hB,whichcanbedrawneasilythroughaconjugatedistribution,andthe!hA,whichrequireamoreadvancedsamplingmethod.Tosample!hBgiventheremainingvariables,weletadenotethelengthof!hAandb=p)]TJ /F5 11.955 Tf 10.17 0 Td[(adenotethelengthof!hB.DeneRAAtobethesubmatrixofR()correspondingtotheelementsof!hA,RBBcorrespondingtotheelementsof!hB,andRBAcontaintheelementsoftherowsof!hBandcolumnsof!hA.Then,usingstandardmultivariatenormalresults,!hBj!hA,)-277(Nb)]TJ /F11 11.955 Tf 5.48 -9.68 Td[( 1b+RBAR)]TJ /F9 7.97 Tf 6.59 0 Td[(1AA(!hA)]TJ /F11 11.955 Tf 11.95 0 Td[( 1a),(RBB)]TJ /F5 11.955 Tf 11.96 0 Td[(RBAR)]TJ /F9 7.97 Tf 6.59 0 Td[(1AAR0BA).Jointlydrawingthevector!hBleadstobettermixingthandrawingeachcomponentseparately. 117
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Tosample!hA,wecyclethroughthecomponents!hof!hAfor=1,...,a.Werecognizethatthecontributiontotheconditionalof!hfromthepriorisexp)]TJ /F4 11.955 Tf 17.39 8.09 Td[(1 2(!h)]TJ /F11 11.955 Tf 11.96 0 Td[( )2,where = +R,()]TJ /F12 7.97 Tf 6.59 0 Td[()R)]TJ /F9 7.97 Tf 6.59 0 Td[(1()]TJ /F12 7.97 Tf 6.59 0 Td[(),()]TJ /F12 7.97 Tf 6.59 0 Td[()(!h()]TJ /F12 7.97 Tf 6.58 0 Td[())]TJ /F11 11.955 Tf 11.96 0 Td[( 1p)]TJ /F9 7.97 Tf 6.59 0 Td[(1),=1)]TJ /F5 11.955 Tf 11.95 0 Td[(R,()]TJ /F12 7.97 Tf 6.59 0 Td[()R)]TJ /F9 7.97 Tf 6.58 0 Td[(1()]TJ /F12 7.97 Tf 6.59 0 Td[(),()]TJ /F12 7.97 Tf 6.58 0 Td[()R0,()]TJ /F12 7.97 Tf 6.58 0 Td[(),!h()]TJ /F12 7.97 Tf 6.59 0 Td[()isthe!hvectorafterremoving!h,R()]TJ /F12 7.97 Tf 6.58 0 Td[(),()]TJ /F12 7.97 Tf 6.59 0 Td[()istheR()matrixformedbyremovingtherowandcolumncorrespondingto,andR,()]TJ /F12 7.97 Tf 6.59 0 Td[()isthevectordenedbytakingtherowofR()andremovingthecomponent.Wecanequivalentlyviewthisash=exp(!h)lognormal( ,),andcalculatetheconditionaldistributionintermsofh.Thisgives(hjh()]TJ /F12 7.97 Tf 6.59 0 Td[(),)]TJ /F4 11.955 Tf 9.3 0 Td[()/)]TJ /F14 5.978 Tf 7.78 3.26 Td[(1 2Pmnmh(Amj))]TJ /F9 7.97 Tf 6.59 0 Td[(1hexp()]TJ /F4 11.955 Tf 18.63 8.09 Td[(1 2hXmnmXi=1(~emij)2h(Amj))]TJ /F4 11.955 Tf 20.05 8.09 Td[(1 2(logh)]TJ /F11 11.955 Tf 11.95 0 Td[( )2).Samplingfromthisdistributionrequiresanapproximatesamplingstep.Werecommendslicesampling( Neal 2003 ),althoughanalternativesamplingstrategycouldbeused.Step2:Dummyvariables.TodrawAmjwewillproceedsimilarlytothepreviousStep1bbylookingattheconditionalmarginallyoverfwmjh,zmjhgh. pr(Amj=hjnfwmjh,zmjhgh)/mjh)]TJ /F14 5.978 Tf 7.78 3.26 Td[(1 2nmjhexp )]TJ /F4 11.955 Tf 17 8.09 Td[(1 2jhnmXi=1~e2mi!. (D) Hence,wedrawAmjfromthemultinomialdistributionwithprobabilitiesfrom( D ),normalizedtosumtoone.Asbefore,wesimulatethesetswmjhandzmjhconditionalonAmjbeingtherstoccasionwherebothwmjhandzmjhareone.Forh>Amj,drawwmjhBern(Wmh)andzmjhBern(Zjh),andwhenh=Amj,1=wmjh=zmjh.For 118
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h
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Inthesimulationanddataexample,weusea=b=0.1,c2=1000.AsmentionedinSection 3.5 ,ithasbeenourexperiencethatsamplingleadstoinstability,andwegenerallyrecommendxingit. D.4SamplingStepsforSparsityGroupingPriorThesparsitygroupingpriorintroducedinAppendix B.1 followsasimilarstructuretothelag-blockpriorexceptusesanewsetsofautoregressiveparametercandidatesforeachj.Consequently,thealgorithmrequiredforsamplingfromthispriorwillbequitesimilar.WedenethenecessarystepsiftheydifferfromtheprocedureintroducedinAppendix D.2 .Step1a:Parametercandidates.Usingthesamenotationasbefore,wenotethatthecontributionfromthedataaboutjhis exp8<:Xm:Rmj=h)]TJ /F4 11.955 Tf 9.3 0 Td[(1 2mjnmXi=1(emij)]TJ /F11 11.955 Tf 11.96 0 Td[(jhxmij)29=;,whichwecompareto( D ).Incorporatingthezero-normalmixtureprior,theposteriorconditionaldistributionwillalsobeazero-normalmixture.Wesetjhtozerowithprobabilityq(j) q(j)+(1)]TJ /F11 11.955 Tf 11.96 0 Td[(q(j)) expn()2 22o,andsamplefromaN(,2)otherwise.Themeanandvarianceforthecontinuouscomponentaregivenby =2Xm:Rmj=hnmXi=1emijxmij mj,2=8<:1 2+Xm:Rmj=hnmXi=1(xmij)2 mj9=;)]TJ /F9 7.97 Tf 6.58 0 Td[(1.(D)Step1b:Dummyvariables.ThestepisthesameasStep1bforthelag-blockpriorexceptthat( D )becomespr(Rmj=hjnfumjh,xmjhgh)/mjhexp()]TJ /F4 11.955 Tf 18.65 8.09 Td[(1 2mjnmXi=1(emij)]TJ /F11 11.955 Tf 11.96 0 Td[(jhxmij)2).NowwecycleSteps1aand1boverjinsteadofq. 120
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Step4:Basedistributionhyperparameters.Thesamplingdistributionsforthehyperparametersaregivenby2j)-278(InvGamma a+1 2Xj,hf1)]TJ /F11 11.955 Tf 11.95 0 Td[(0(jh)g,b+1 2Xj,h2jh!andqj)-278(Beta0@q+Xj:q(j)=qHXh=10(jh),q+Xj:q(j)=qHXh=1f1)]TJ /F11 11.955 Tf 11.95 0 Td[(0(jh)g1A,forpriorsdistributionsof2InvGamma(a,b)andqBeta(q,q).Notethatthesummationsinthedistributionforqsumovertheautoregressivetermsthatcorrespondtolag-q. D.5SamplingStepsforNon-SparseGroupingPriorSamplingunderthispriorproceedsasunderthelag-blockandsparsitypriors.Wedescribethestepsthatdifferfromlag-blockprior.Step1a:Parametercandidates.Becausethispriordoesnotimposesparsity,theconditionaldistributionforjhisN(,2)whereand2aregivenbyequation( D ).Step4:Basedistributionhyperparameters.Theonlyhyperparameterfortheautoregressivecandidatesis2.WithapriordistributionofInvGamma(a,b),thesamplingdistributionis2j)-278(InvGamma a+1 2JH,b+1 2Xj,h2jh!. D.6SamplingStepsforInvGammaGroupingPriorThesamplingalgorithmproceedsasforthecorrelated-lognormalprior.WedescribeonlythosestepsthatrequiremodicationfromthealgorithminAppendix D.3 .Step1:Parametercandidates.Usingthelikelihoodcontributionofjhfrom( D )andapriordistributionofInvGamma(1,2),wehavethattheconditionalsampling 121
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distributionisjhj)-278(InvGamma0@1+1 2Xm:Amj=hnm,2+1 2Xm:Amj=hnmXi=1~e2mij1A.Step5:Basedistributionhyperparameters.Thehyperparametersassociatedwiththisinnovationvariancegroupingare1and2in( B ).WeplaceindependentGamma(1,1)priorsoneach.Theconditionalfor2is2j)-277(Gamma 1pH+1,1+Xj,h)]TJ /F9 7.97 Tf 6.59 0 Td[(1jh!.Theconditionalfor1is(1j)]TJ /F4 11.955 Tf 15.94 0 Td[()/\(1))]TJ /F7 7.97 Tf 6.59 0 Td[(pH)]TJ /F12 7.97 Tf 6.59 0 Td[(1pH2exp")]TJ /F11 11.955 Tf 9.29 0 Td[(1(1+Xj,hlog(jh))#,butthisisnotastandarddistributiontosample.So,itbecomesnecessarytoimplementanalternativesamplingmethod,andwechoosetointroduceaMetropolis-in-Gibbssteptoapproximatelysimulatefromthisconditional.Drawthecandidatevalue1toreplacethecurrentvalue1fromtheN(1,)distribution,andacceptthemoveto1withprobabilityminf1,~g,where~=I(1>0)"exp(log\(1) \(1)+1 pH(1)]TJ /F11 11.955 Tf 11.96 0 Td[(1) 1+Xj,hlog(jh)+pHlog(2)!)#pH.Itisnecessarytoprespecifyacandidatevariancesuchthattheacceptancerateis20to40%( Gelmanetal. 1996 ). D.7FinalCommentsaboutComputationalAlgorithmWenallynotethatonecanviewourgroupingpriorsinahierarchicalfashionwithmultiplelevels.Asisoftenthecaseinhierarchicalmodels,theremaybelittleinformationabouttheparametersinthelowestlevels.Wehaveoftenfoundthistobethecaseforthegroupingpriorsresultinginpoormixingforsomeofthelowerlevelmodelparameters.Whilethevaluesoftheautoregressiveparametersand 122
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APPENDIXEADDITIONALRISKSIMULATIONSFORCOVARIANCEGROUPINGPRIORS E.1AdditionalDetailsforRiskSimulationofSection3.6Mean,covariance,anddropoutmodel( 3 )parametersaredisplayedinTable E-1 .Recallthatthedropoutmodel( 3 )isgivenbylogitfpr(Di=t+1jDi>t,yit,m)g=0t+1tyit+2m.ThedropoutprobabilitiesthatthismodelinducesaregiveninTable E-2 .RecallthattheT(m)upper-triangularmatrixisdenedthroughtheJ-dimensionalvectormbyT(m)=2666666666641)]TJ /F11 11.955 Tf 9.3 0 Td[(m1)]TJ /F11 11.955 Tf 9.3 0 Td[(m2)]TJ /F11 11.955 Tf 9.3 0 Td[(m41)]TJ /F11 11.955 Tf 9.3 0 Td[(m3)]TJ /F11 11.955 Tf 9.3 0 Td[(m51)]TJ /F11 11.955 Tf 9.3 0 Td[(m61...377777777775.RecallthatinSection 3.6 ,weanalyzedtheriskunder5differentpriorchoices:thetwonaivepriors,thetwoatpriors,andthegroupingpriorformedbythelag-blockandthecorrelated-lognormalgroupingpriors.WeextendtherisksimulationsfromTable 3-1 byaddingfouradditionalgroupingpriorsbasedonthenewpriorsintroducedinAppendix B .Thesenewpriorsarecomposedofmixinganautoregressiveparametergroupingpriorwithaninnovationvariancegroupingprior.Thesefouradditionalpriorsarelag-block/InvGamma,sparsity/correlated-lognormal,sparsity/InvGamma,andnon-sparse/InvGamma.SpecicationsforhyperpriorsandothersimulationchoicesarethesameasSection 3.6 .Table E-3 containsthecovarianceandmeanriskforthisextendedsimulation.Allveofthegroupingpriorsbeatthenaiveandatpriors.Thelag-block/correlated-lognormalgroupingprioristhemosteffectiveofourgroupingpriorforthisanalysis.Theabilitytoborrowstrengthacrossgroupsimprovestheestimationsuchthateventhe 124
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TableE-1. ParametervaluesforrisksimulationofSection 3.6 .1=(0,1.9,5.2,9.9,16.0,23.5)2=(0,1.8,4.8,9.0,14.4,21.0)3=(0,1.8,5.6,11.4,19.2,29.0)4=(0,2.0,5.0,9.0,14.0,20.0)5=(0,2.0,5.2,9.6,15.2,22.0)6=(0,3.0,6.0,9.0,12.0,15.0)7=(0,1.8,4.8,9.0,14.4,21.0)8=(0,2.8,7.2,13.2,20.8,30.0)1=(0.7,0.2,0.7,0,0.2,0.7,0,0,0.2,0.7,0,0,0,0.2,0.7)2=(0.6,0.1,0.6,0.1,0.1,0.6,-0.1,0.1,0.1,0.6,-0.1,-0.1,0.1,0.1,0.6)3=(0.4,0.3,0.4,-0.2,0.3,0.4,0,-0.2,0.3,0.4,-0.2,0,-0.2,0.3,0.4)4=(0.3,0,0.3,-0.1,0,0.3,0,-0.1,0,0.3,0,0,-0.1,0,0.3)5=(1,-0.5,1,0.2,-0.5,1,0,0.2,-0.5,1,0,0,0.2,-0.5,1)6=(0.8,-0.4,0.8,0.3,-0.4,0.8,0,0.3,-0.4,0.8,0,0,0.3,-0.4,0.8)7=(0.9,-0.2,1,-0.2,-0.2,1,-0.2,-0.2,-0.2,1,-0.2,-0.2,-0.2,-0.2,1)8=(-0.9,0.1,-0.9,0,0.1,-1,0.2,0,0.1,-0.8,-0.2,0.2,0,0.1,-0.8))]TJ /F9 7.97 Tf 6.18 -1.64 Td[(1=(1,1,1,1,1,1))]TJ /F9 7.97 Tf 6.18 -1.64 Td[(2=(1.5,1.5,1.5,1.5,1.5,1.5))]TJ /F9 7.97 Tf 6.18 -1.64 Td[(3=(3.4,3.1,2.8,2.5,2.2,1.8))]TJ /F9 7.97 Tf 6.18 -1.64 Td[(4=(3,3,2,2,2,1))]TJ /F9 7.97 Tf 6.18 -1.64 Td[(5=(3.5,3.2,2.9,3.5,3.2,2.9))]TJ /F9 7.97 Tf 6.18 -1.64 Td[(6=(5,3.7,3,3,2,2))]TJ /F9 7.97 Tf 6.18 -1.64 Td[(7=(2,1.8,1.6,1.4,1.2,1))]TJ /F9 7.97 Tf 6.18 -1.64 Td[(8=(3.3,3,2.7,2.4,2.2,1.9)0=(-2.5,-3.5,-9,-13,-20)1=(0.4,0.5,0.8,1.0,1.2)2=(0,0.2,-2,0,0,0,0.1,-4) TableE-2. ProbabilityYitismissingbygroupmforrisksimulationofSection 3.6 mt=2t=3t=4t=5t=6 10.0800.1560.1670.2390.49820.1000.1880.1990.2410.35630.0140.0290.0340.1120.67040.0900.1800.1900.2250.30350.0930.1990.2240.3230.49660.1000.2510.2820.3330.34870.0940.1830.1970.2510.37480.0020.0070.0140.1720.726 125
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TableE-3. EstimatedrisksforeachchoiceofcovariancepriorfromtheextensionoftheSection 3.6 simulation.TheestimatedriskiscalculatedastheaveragelossusinglossfunctionsL1(m,^m1)=tr()]TJ /F9 7.97 Tf 6.59 0 Td[(1m^m1))]TJ /F4 11.955 Tf 11.96 0 Td[(logj)]TJ /F9 7.97 Tf 6.59 0 Td[(1m^m1j)]TJ /F5 11.955 Tf 17.94 0 Td[(p,L2(m,^m2)=trf()]TJ /F9 7.97 Tf 6.59 0 Td[(1m^m2)]TJ /F5 11.955 Tf 11.96 0 Td[(I)2g,andL(^m,m)=(^m)]TJ /F11 11.955 Tf 11.95 0 Td[(m)>)]TJ /F9 7.97 Tf 6.59 0 Td[(1m(^m)]TJ /F11 11.955 Tf 11.96 0 Td[(m). CovariancePriorEstimatedRiskPrioronPrioron)]TJ /F5 11.955 Tf 60.69 0 Td[(L1L2L Lag-blockCorr-lognormal0.4250.7420.175Lag-blockInvGamma0.4370.7590.174SparsityCorr-lognormal0.5530.9150.196SparsityInvGamma0.5650.9340.196Non-sparseInvGamma0.5510.9120.200NaiveBayes10.6050.9870.203NaiveBayes20.6301.0100.210Group-specicat*0.8921.2550.248Common-at8.10584.3390.925 *Thegroup-specicatpriorisonlyover49datasetsbecausetheMarkovchainfailedtoconvergeforonedataset. non-sparsegroupingprior,whichdoesnotallowthecorrectindependencerelationships,beatstherstnaiveprior,whichcorrectlyincorporatesthepotentialindependence.Thelag-block/correlated-lognormalpriorcontinuestobeattheremainderofthegroupingpriors,withariskimprovementof30and25%overthenaive1priorand52and41%overthegroup-specicatprior.Intermsoftheriskassociatedwithmeanestimation,wenotethatallvegroupingpriorsalsodominatethenaiveandatpriors.Thetwolag-blockpriorsbeattherstnaivepriorby14%.Thegroupingpriorswithsparseandnon-sparsepriorsforonlydoslightlybetterthanthenaivechoicesbutarestillclearlysuperiortheatpriors. E.2RiskSimulation2Wenowdescribethreesomewhatsimplerrisksimulationstofurtherdemonstratethatourgroupingpriorsperformwellinavarietyofsituations.ConsiderM=5groupsandp=4four-dimensionalnormallydistributedmean-zerorandomvariables.Theve 126
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TableE-4. Riskestimatesforsimulation2. CovariancePriorEstimatedRiskPrioronPrioron)]TJ /F5 11.955 Tf 55.7 0 Td[(L1L2 Lag-blockCorr-lognormal0.2470.429Lag-blockInvGamma0.2570.448SparsityCorr-lognormal0.2700.462SparsityInvGamma0.2810.480NaiveBayes10.2910.488Non-sparseInvGamma0.2920.493NaiveBayes20.3220.530Group-specicat0.4630.700Common-at1.5606.623 covariancematricesaredenedbythefollowingspecicationoftheautoregressiveandinnovationvarianceparameters:1=(0.7,0.2,0.7,0,0.2,0.7),)]TJ /F9 7.97 Tf 6.77 -1.79 Td[(1=(1,1,1,1),2=(0.7,0,0.3,0,0,0.7),)]TJ /F9 7.97 Tf 6.77 -1.8 Td[(2=(2,2,2,2),3=(0.3,0,0.3,0,0,0.3),)]TJ /F9 7.97 Tf 6.77 -1.79 Td[(3=(2,2,1,1),4=(0.7,0.2,0.7,0.1,0.2,0.7),)]TJ /F9 7.97 Tf 6.77 -1.79 Td[(4=(5,5,5,5),5=(0.7,0,0.7,0,0,0.3),)]TJ /F9 7.97 Tf 6.77 -1.79 Td[(5=(1,1,2,2).Weusesamplesizesofn1=...=n4=30,n5=15.Forthisspecicationmanyoftheparametersacrossgroupsareequal,andmanyofthehigherlagautoregressivetermsarezero.Additionally,withthesmallersamplesizeforthefthgroup,thegroupingpriorsshouldimproveestimationof5bysharinginformationacrosssimilargroups.Withgroupsofdifferentsizes,wemeasurethelosstoestimatingthecollectiontobeweightedaverageoftheindividuallosses,Lk(m,^mk)(k=1,2),withweightsproportionaltothesamplesizenm.Usingthesameset-upastheprevioussimulationproducestheriskestimatesgiveninTable E-4 .Thepriorcomposedofthelag-blockstructureontheautoregressiveandthecorrelated-lognormalspecicationforthevarianceshasthebestriskestimatesofthecollection.Comparingthelag-block/InvGammaandsparsity/correlated-lognormalpriorstothesparsity/InvGammagroupingprior,themodicationoneithertheautoregressive 127
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ortheinnovationvariancesproducesimprovedriskperformance.Thelag-block/correlated-lognormalanalysisproducesriskestimatesthatare15%and12%lowerthanthenaive1prior.Itisnaturaltocomparethesparsenaivepriortothesparsity/InvGammabecausetherstisalimitingcaseofthelatter.Likewise,wecomparenaive2priorwithgrouping/InvGamma.Forbothlossfunctions,thesparsity/InvGammabeatsnaive1andgrouping/InvGammabeatsnaive2,indicatingtheborrowingofinformationacrossgroupsinducedbythegroupingpriorsimprovestheestimation.Wealsoseethatthesparsitypriorperformsbetterthanthenon-sparsegroupingprior,butthisistobeexpectedsinceweknowthatthereareautoregressiveparametersthatareequaltozerointhetruemodel.Comparatively,theestimatorsfromtheatpriorsperformverypoorly;therisksforthegroupingpriorsare37%smallerthanthegroup-specicestimatorforL1and30%forL2. E.3RiskSimulation3Weperformanotherrisksimulationsimilartothepreviouswithagainvegroupsandp=4.Wexthemeantozeroandtheparametersofthetruecovariancematricesaregivenby1=(1,0.5,1,0.5,0.5,1),)]TJ /F9 7.97 Tf 6.77 -1.8 Td[(1=(2,2,2,2),2=(1,0.5,1,0.5,0.5,1),)]TJ /F9 7.97 Tf 6.77 -1.79 Td[(2=(4,4,4,4),3=(1,-0.5,1,-0.5,-0.5,1),)]TJ /F9 7.97 Tf 6.77 -1.79 Td[(3=(2,2,2,2),4=(1,-0.5,1,-0.5,-0.5,1),)]TJ /F9 7.97 Tf 6.77 -1.79 Td[(4=(4,4,4,4),5=(2,-1.0,2,-0.5,-1.0,1),)]TJ /F9 7.97 Tf 6.77 -1.79 Td[(5=(2,2,1,1).Againweusethesamplesizesn1=...=n4=30,n5=15.Thereshouldbealargeamountofclusteringinthiscase,sincethereisagreatdealofcommonalityamongautoregressiveparametersandinnovationvariancesfordifferentsamples.Thesecovariancematricesalsodonothaveanyconditionalindependencerelationshipstoexploitsinceeachofthe'sarenonzero.RiskestimatesareshowninTable E-5 .Asinthepreviousrisksimulation,thelag-block/correlated-lognormalpriorproducesthebestriskat15%and20%lowerthan 128
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TableE-6. Parametervaluesforsimulation4.1=(0.7,0.2,0.7,0,0.2,0.7,0,0,0.2,0.7,0,0,0,0.2,0.7)2=(0.7,0.2,0.7,0.1,0.2,0.7,0,0.1,0.2,0.7,0,0,0.1,0.2,0.7)3=(0.3,0,0.3,0,0,0.3,0,0,0,0.3,0,0,0,0,0.3)4=(0.3,0,0.3,-0.1,0,0.3,0,-0.1,0,0.3,0,0,-0.1,0,0.3)5=(1,-0.5,1,0,-0.5,1,0,0,-0.5,1,0,0,0,-0.5,1)6=(1,-0.5,1,0.3,-0.5,1,0,0.3,-0.5,1,0,0,0.3,-0.5,1)7=(1,-0.2,1,-0.2,-0.2,1,-0.2,-0.2,-0.2,1,-0.2,-0.2,-0.2,-0.2,1)8=(1,-0.2,1,-0.2,-0.2,1,-0.2,-0.2,-0.2,1,-0.2,-0.2,-0.2,-0.2,1))]TJ /F9 7.97 Tf 6.78 -1.79 Td[(1=(1,1,1,1,1,1))]TJ /F9 7.97 Tf 6.78 -1.79 Td[(2=(1,1,1,1,1,1))]TJ /F9 7.97 Tf 6.78 -1.8 Td[(3=(3.4,3.1,2.8,2.5,2.2,1.8))]TJ /F9 7.97 Tf 6.78 -1.79 Td[(4=(3,3,2,2,2,1))]TJ /F9 7.97 Tf 6.78 -1.8 Td[(5=(5,3,3,4,4,4))]TJ /F9 7.97 Tf 6.78 -1.79 Td[(6=(5,5,3,3,2,2))]TJ /F9 7.97 Tf 6.78 -1.79 Td[(7=(2,1.8,1.6,1.4,1.2,1))]TJ /F9 7.97 Tf 6.78 -1.79 Td[(8=(3.4,3.1,2.8,2.5,2.2,1.8) TableE-7. Riskestimatesforsimulation4. CovariancePriorEstimatedRiskPrioronPrioron)]TJ /F5 11.955 Tf 55.7 0 Td[(L1L2 Lag-blockCorr-lognormal0.4680.781Lag-blockInvGamma0.4920.816SparsityCorr-lognormal0.5560.904SparsityInvGamma0.5830.939Non-sparseInvGamma0.6020.963NaiveBayes10.6641.013NaiveBayes20.7611.121Group-specicat1.3001.584Common-at3.03614.149 asintherisksimulationofthearticle.Here,weallowforM=8groupsandconsider66covariancematrices,denedbythedependenceparametersinTable E-6 .Weassumeameanofzeroandfullyobservealldata.Thischoiceforincorporatescommonalitybothwithinlagandacrossgroups,aswellaspossessingmanyconditionalindependencerelationshipsamongthehigherlagterms.Wechooseasamplesizeofthirtyfortherstvegroupsandfteenforthenalthreegroups,andthirtyclustersforthegroupingpriors.TheestimatedriskassociatedwithestimatingthecovariancematricesforeachofthetwolossfunctionsisshowninTable E-7 130
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TableE-8. Modeltstatisticsandtreatmenteffectsforthersttwogroupsforthedepressiondatausingeachofthepriors. CovariancePriorModelFitTreatmentEffectPrioronPrioron)]TJ /F5 11.955 Tf 79.22 0 Td[(DevpDDICGroup1Group2 Lag-blockCorr-logN(=0.90)39,00634239,6909.23(7.03,11.48)9.51(6.85,12.19)Lag-blockInvGamma38,99935039,6989.22(6.98,11.45)9.39(6.73,12.13)Lag-blockCorr-logN(=0.75)39,00634739,7009.22(6.99,11.42)9.56(6.85,12.27)Lag-blockCorr-logN(=0.50)39,00334939,7009.24(7.00,11.53)9.41(6.74,12.14)SparsityCorr-logN(=0.90)38,88746439,8169.25(7.02,11.49)8.78(6.33,11.24)SparsityCorr-logN(=0.75)38,88746639,8199.23(6.96,11.42)8.82(6.39,11.27)SparsityCorr-logN(=0.50)38,88347239,8279.25(7.01,11.51)8.68(6.23,11.20)SparsityInvGamma38,88447539,8349.25(7.02,11.53)8.64(6.16,11.15)NaiveBayes138,87548139,8379.25(7.11,11.46)8.56(6.16,10.99)Non-sparseInvGamma38,81852939,8769.29(7.01,11.59)8.81(6.21,11.52)NaiveBayes238,76556339,8909.24(7.02,11.49)7.99(5.59,10.46)Common-at39,90722040,3479.44(6.21,12.53)10.17(7.02,13.24)Group-specicat39,178102141,2199.20(6.44,12.08)6.93(4.22,9.77) 132
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APPENDIXFMODELPARAMETERSFORRISKSIMULATIONOFCHAPTER4WedescribethecovariancematricesusedtocreatedatausedintherisksimulationofSection 4.4 .Foreachgroupm,weprovidetheTmCholeskymatrixwhereinsteadofdenoting1onthediagonalweincludetheIVmtinbold.Emptyupper-triangularelementsarezero.RecallthatthetruepartitionstructureisP1=fMg,P2==P5=ff1,2,3,4g,f5,6,7,8gg,P6==P8=ff1,2,3,4g,f5,6g,f7,8gg,andP9=P10=ff1,2g,f3,4g,f5,6g,f7,8gg.GroupsareinthesamesetofthepartitionifmtandthetthcolumnoftheTmmatrixareequal.Groupsm=1,2:26666666666666641.0)]TJ /F4 11.955 Tf 9.3 0 Td[(.7)]TJ /F4 11.955 Tf 9.29 0 Td[(.21.0)]TJ /F4 11.955 Tf 9.29 0 Td[(.7)]TJ /F4 11.955 Tf 9.3 0 Td[(.21.0)]TJ /F4 11.955 Tf 9.3 0 Td[(.7)]TJ /F4 11.955 Tf 9.3 0 Td[(.21.0)]TJ /F4 11.955 Tf 9.3 0 Td[(.7)]TJ /F4 11.955 Tf 9.3 0 Td[(.21.0)]TJ /F4 11.955 Tf 9.3 0 Td[(.7)]TJ /F4 11.955 Tf 9.3 0 Td[(.21.0)]TJ /F4 11.955 Tf 9.3 0 Td[(.7)]TJ /F4 11.955 Tf 9.3 0 Td[(.21.0)]TJ /F4 11.955 Tf 9.3 0 Td[(.7)]TJ /F4 11.955 Tf 9.3 0 Td[(.21.0)]TJ /F4 11.955 Tf 9.3 0 Td[(.7)]TJ /F4 11.955 Tf 9.3 0 Td[(.21.0)]TJ /F4 11.955 Tf 9.3 0 Td[(.71.03777777777777775Groupsm=3,4:26666666666666641.0)]TJ /F4 11.955 Tf 9.3 0 Td[(.7)]TJ /F4 11.955 Tf 9.29 0 Td[(.21.0)]TJ /F4 11.955 Tf 9.29 0 Td[(.7)]TJ /F4 11.955 Tf 9.3 0 Td[(.21.0)]TJ /F4 11.955 Tf 9.3 0 Td[(.7)]TJ /F4 11.955 Tf 9.3 0 Td[(.21.0)]TJ /F4 11.955 Tf 9.3 0 Td[(.7)]TJ /F4 11.955 Tf 9.3 0 Td[(.31.0)]TJ /F4 11.955 Tf 9.3 0 Td[(.8)]TJ /F4 11.955 Tf 9.3 0 Td[(.30.7)]TJ /F4 11.955 Tf 9.3 0 Td[(.8)]TJ /F4 11.955 Tf 9.3 0 Td[(.30.7)]TJ /F4 11.955 Tf 9.3 0 Td[(.8)]TJ /F4 11.955 Tf 9.3 0 Td[(.30.7)]TJ /F4 11.955 Tf 9.3 0 Td[(.8)]TJ /F4 11.955 Tf 9.3 0 Td[(.30.7)]TJ /F4 11.955 Tf 9.3 0 Td[(.80.73777777777777775 134
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Groupsm=5,6:26666666666666641.0)]TJ /F4 11.955 Tf 9.3 0 Td[(.5)]TJ /F4 11.955 Tf 9.29 0 Td[(.21.0)]TJ /F4 11.955 Tf 9.29 0 Td[(.5)]TJ /F4 11.955 Tf 9.3 0 Td[(.20.8)]TJ /F4 11.955 Tf 9.3 0 Td[(.5)]TJ /F4 11.955 Tf 9.3 0 Td[(.20.8)]TJ /F4 11.955 Tf 9.3 0 Td[(.5)]TJ /F4 11.955 Tf 9.3 0 Td[(.20.8)]TJ /F4 11.955 Tf 9.3 0 Td[(.5)]TJ /F4 11.955 Tf 9.3 0 Td[(.20.8)]TJ /F4 11.955 Tf 9.3 0 Td[(.5)]TJ /F4 11.955 Tf 9.3 0 Td[(.20.8)]TJ /F4 11.955 Tf 9.3 0 Td[(.5)]TJ /F4 11.955 Tf 9.3 0 Td[(.20.8)]TJ /F4 11.955 Tf 9.3 0 Td[(.5)]TJ /F4 11.955 Tf 9.3 0 Td[(.20.8)]TJ /F4 11.955 Tf 9.3 0 Td[(.50.83777777777777775Groupsm=7,8:26666666666666641.0)]TJ /F4 11.955 Tf 9.3 0 Td[(.5)]TJ /F4 11.955 Tf 9.29 0 Td[(.21.0)]TJ /F4 11.955 Tf 9.29 0 Td[(.5)]TJ /F4 11.955 Tf 9.3 0 Td[(.20.8)]TJ /F4 11.955 Tf 9.3 0 Td[(.5)]TJ /F4 11.955 Tf 9.3 0 Td[(.20.8)]TJ /F4 11.955 Tf 9.3 0 Td[(.5)]TJ /F4 11.955 Tf 9.3 0 Td[(.20.8)]TJ /F4 11.955 Tf 9.3 0 Td[(.5)]TJ /F4 11.955 Tf 9.3 0 Td[(.20.8)]TJ /F4 11.955 Tf 9.3 0 Td[(.5)]TJ /F4 11.955 Tf 9.3 0 Td[(.2.20.8)]TJ /F4 11.955 Tf 9.3 0 Td[(.5)]TJ /F4 11.955 Tf 9.3 0 Td[(.2.20.8)]TJ /F4 11.955 Tf 9.3 0 Td[(.7)]TJ /F4 11.955 Tf 9.3 0 Td[(.21.2)]TJ /F4 11.955 Tf 9.3 0 Td[(.71.23777777777777775 135
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PAGE 140 | 61,729 | 173,973 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2017-51 | latest | en | 0.732739 |
http://www.chegg.com/homework-help/questions-and-answers/sum-vector-magnitude-145-m-angled405-counterclockwise-x-direction-vector-magnitude-150-m-a-q640405 | 1,475,105,051,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738661768.10/warc/CC-MAIN-20160924173741-00014-ip-10-143-35-109.ec2.internal.warc.gz | 392,367,984 | 13,724 | In the sum + = , vector has a magnitude of 14.5 m and is angled40.5°counterclockwise from the +x direction, and vector has a magnitude of 15.0 m and is angled 28.0° counterclockwisefrom the -x direction. What are the magnitude and theangle (relative to +x) of ?
________m at 2______° counterclockwise from the+x direction | 95 | 321 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2016-40 | latest | en | 0.921175 |
http://skywired.net/blog/tag/design-2/ | 1,721,733,820,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518029.87/warc/CC-MAIN-20240723102757-20240723132757-00363.warc.gz | 24,325,186 | 25,856 | ## Worst case design and the BITX 40 mic amp
The BITX 40’s microphone amp has a flaw that may shorten the lifespan of microphones by overstressing their components. Here’s what I found and how I fixed it so far.
Here’s the BITX 40 mic amp schematic:
The mic bias is supplied by the TX line through 4.9K of resistance. TX is the DC power supply, 12-15V.
Most electret microphones have a maximum voltage of 10 V, though I’ve seen a few datasheets with a 9 V limit. The data sheets also say that the mics draw 0.5 mA maximum. No minimum is given.
Now, applying Ohm’s law,
$(0.5 \mathrm{mA}) * (4.9 \mathrm{k\Omega}) = 2.45 \mathrm{V}$
Subtracting that from 12 V gives us 9.55 V, so one might think that everything is great.
Not really.
From the books of Bob Pease I learned the importance of worst case design. In worst case design, one looks at the datasheet minimum and maximum specifications and designs the circuit to work under the worst combination of specifications. Ignore the datasheet typical ratings, because those describe the best case. No one wants a circuit that only works in the best possible circumstances.
The BITX 40 docs specify a maximum power supply of 15 V, so that is the worst-case TX. The worst case microphone current is the minimum current, but I haven’t seen a mic datasheet that lists a minimum. Therefore, I have to make a guess. I need the guess to be on the safe side without being ridiculous. Knowing that the current is set by a JFET and knowing about both the variability of JFETs and how data sheet maximums are chosen, I could guess that the typical current is maybe half of the maximum and the minimum half of that, so 0.125 mA. If I want to make a truly robust circuit I could assume a minimum current of zero. I know a JFET can’t create current out of nothing, so the lowest possible current it draws is zero.
Let’s look at what voltages that creates:
$(15 V) - (0.125 \mathrm{mA})(4.9 \mathrm{k\Omega}) = 14.388$
Uh-oh. That’s well above the 10 V maximum for the mic.
The zero current case is easier to work out. With zero current through the resistor, the mic sees the full 15 V.
A simple fix is to put a resistor in parallel with MIC1, making a voltage divider. I use a computer headset with my BITX, and I know that computer mics are typically fed with 5 V through a 2.2 kΩ resistor. Knowing that, I picked a 4.3 kΩ resistor across MIC1 to form a voltage divider equivalent to 2.3 kΩ fed by 7 V. That’s close enough, and 7 V is well below the maximum for most electret elements.
Unfortunately, the mic now drives an impedance about 30% lower than before, reducing its output. The RF drive, R136, should be adjusted to compensate.
One could fix the issue a different way, of course. A Zener diode in series with R121 could drop the voltage just as effectively without lowering the amplifier’s input impedance. An LM780x regulator could do the job, too, albeit with more components. I happened to have 1% resistors within reach and Zener diodes on a different floor of the house. I chose the resistor.
### Is this a real problem?
No one else seems to have noticed this issue, so it’s worth asking whether it matters. One answer is that it is worthwhile, and pleasing, to do things right. There is a place for bodging a circuit together that works as a one-off, but I enjoy the n-dimensional puzzle of getting the details right.
The other answer is that it might be killing microphones, but not often enough that anyone has noticed. The BITX 20 mailing list, home to BITX 40 discussion, has seen a few comments from hams whose microphone capsules worked for a while, then failed. Some have been able to trace the problems to bad solder joints or physical damage, but I have to wonder if any of the remaining unsolved cases were caused by overvoltage.
Manufacturer maximum specifications are based on an assumption about the device’s expected lifetime. Operating beyond that specification can be expected to shorten the device’s lifetime. Sometimes that shortened lifetime is dramatic, with a flash of light, a puff of smoke, or an outpouring of heat. Other times it is subtle and takes longer. Perhaps mic elements will fail faster than usual. Perhaps their average lifespan will be 2 years instead of 20.
The difference might not be enough to notice, but we can fix it anyway. Worst case design will save the day.
## How Delta-Sigma Works, part 3: The controls-system perspective
This post is part of a series on delta-sigma techniques: data converters, modulators, and more. A complete list of posts in the series are in the How Delta-Sigma Works tutorial page.
In the first installment of How Delta-Sigma Works, I presented the basic first-order delta-sigma converter loop. Now it is time to begin digging a little deeper and look at how the loop works. To do this, we will need the mathematical tools of closed-loop control systems. Even without the mathematics, thinking about a delta-sigma modulator as a closed-loop controller can bring insight into how it works.
In engineering, closed-loop control is often used to keep a system working at a setpoint despite environmental disturbances or variations in the system itself. A furnace thermostat is a simple closed-loop controller, turning on the heat when the temperature drops below a setpoint and turning it back off when the temperature is above. Another example is vehicle cruise control. Unlike a thermostat, which usually has a binary output (on/off), cruise control adjusts the engine fuel control to maintain a roughly constant speed. In my car, it does this by physically moving the gas pedal. The environmental variations cruise control can encounter include hills, the quality of the fuel, and headwinds or tailwinds. The closed-loop controller adjusts the gas pedal as needed to maintain constant speed despite these effects.
A simple closed-loop control system can be drawn as in the figure below. The reference input, r, is the command input to the controller. In the thermostat example, r is the setpoint temperature, while for cruise control, it is the set speed. The output, y, is the controlled value. This is not necessarily in the same units as the reference input r. For example, in the cruise control, y might be a direct measurement of speed, or it might be something else related, such as a cumulative count of revolutions of the vehicle’s wheels.
In between the input r and the output y is the control loop. At the bottom of the loop, in its feedback path, a block h processes the output y into a form that is subtracted from the input r to create an error signal e. This error signal is an indication of how far the controller is from the desired operating point. The goal of a controller design is to keep e near, if not at, 0. The feedback block h can represent many different things. In the example of a cruise control system with the output y in units of distance, h might calculate speed by computing the derivative of y. In other systems, h might simply scale the value to convert its units. If y and r are already in the same units, such as in a thermostat example, h can pass through y unchanged ($h = y$).
Now that the error signal has been calculated, it is processed by the controller gc, the output of which goes to the “plant” being controlled. (To a controls engineer, anything being controlled, whether a car, a heating system, or a giant factory, is a plant.) The plant is represented by the box gp. The processing in the controller gc is easy to grasp. The controller calculates some function of its input in order to find the command it should give the plant.
What happens in this plant, gp, may be a little harder to imagine. The function gp is a mathematical model of the physics of the actual plant. It may be calculated from basic principles (a physicist’s delight!), or it may be an empirical model derived from the inputs and outputs of the actual plant. In any event, a reasonable guess at the function gp is necessary before one can design a closed-loop controller.
A delta-sigma modulator also has a closed loop, which suggests that perhaps insight can be gained by comparing it to a controller. The first-order delta-sigma analog-to-digital converter from the first installment, is shown again below.
The resemblance to a closed-loop controller is clear when one groups the blocks as in the next figure. The subtractor has the same function in both diagrams, comparing the input to the output. The integrator functions as the controller, gc, and the analog-to-digital convertor and its register are the plant, gp, being controlled. The digital-to-analog converter is the feedback path, h.
This grouping gives immediate insight into how the delta-sigma modulator does its magic: It is a linear controller for an analog-to-digital converter, which is the plant. The controller is always trying to drive that plant’s output as close as possible to the setpoint, and does so by adding up (integrating) the error signal. Also, since the integrator is adding up the history of the error signal, it can be seen that although the output at any given moment may not equal the input, the long-term average will be equal. That integrator will try to keep the long-term average of the error, e, equal to 0.
There are many controllers that can control a given plant. PID (proportional-integral-derivative) controllers are simple and very popular, while more sophisticated controllers can be designed using other techniques. If a delta-sigma modulator is a control loop, it is reasonable to ask if controllers other than an single integrator will result in a better-performing modulator. In fact, other control functions can be used in delta-sigma modulators and can give lower noise or more desirable characteristics in the frequency domain.
Finally, one should not get too carried away with a linear control model. Analog-to-digital and digital-to-analog converters are inherently non-linear, while the mathematics of control theory primarily deals with linear systems. Assuming linear behavior will get us a long way towards understanding delta-sigma techniques, but it is important not to take the analogy too far.
Next in this series: Noise shaping, the frequency-domain secret behind delta-sigma data converters.
## AK5388 audio ADC breakout board design
A few weeks ago, I picked the AKM Semiconductor AK5388 as the analog-to-digital converter for my receiver design. This high-performance audio DAC has a 24-bit output with up to 123 dB signal-to-noise ratio (A-weighted). I hope that the narrower bandwidths of a communications receiver will beat that. The idea is to do the automatic gain control (AGC) in digital, so that the receiver will not need an analog AGC loop. I hope that a preamp or attenuator at the front end will be enough to make up for the limited dynamic range of the AGC.
Now, I am aware that getting performance like that requires great care in the details of the design and layout, but if this wasn’t a challenge, it wouldn’t be nearly as much fun! The AK5388 datasheet does give me some concern, because it does not make much mention of the techniques needed for a high-performance DAC. Admittedly, datasheets from Japanese manufacturers are often thinner on details than those from the leading US suppliers, and this one is much more complete than some. On the other hand, it may be that the datasheet is glossing over any coddling this chip will need.
There is one way to find out, and that is to start breadboarding. This is a surface-mount part, so I have designed a breakout board very much like the board I built for the Actel Microsemi A3PN250 FPGA. The schematic is below.
The schematic largely follows the “System Design” section of the datasheet and the example of the AKD5388-A evaluation kit (PDF). It’s not strictly a breakout because I included on-board analog and digital power supplies. An ADC like this needs quiet, local power, so I picked some reasonably quiet supplies and put them on-board. I chose 7800-family regulators from TI because they have reasonable noise specifications. I almost went with other choices that were explicitly designed for low noise, but they were quite a bit more expensive for not much less noise. The AKD5388-A kit uses an uA78M05 and claims to be able to obtain the full dynamic range of this ADC, so there is little reason to spend more at this point.
I’m using the analog power supply as the voltage reference. I considered a dedicated reference, which would be more expensive but perhaps lower noise. Here, too, I followed the lead of the evaluation kit.
The PCB layout should be straightforward. I like to build by skywiring components over a groundplane, so this breakout board will be designed to sit directly on a groundplane, with all components and connections on the top side. All external wiring points will be pads on the PCB, ready for soldering, just as with the A3PN250 FPGA board.
I’m excited about trying out this part. What do you think? Comments welcome!
## Choosing a high-performance audio ADC
As I discussed two weeks ago, it is time to refocus my efforts on the DSP-based ham radio project that started this blog. Let’s take a look at the architecture I had in mind originally:
This is a popular topology for radios that put their intermediate frequency at or near 0 Hz. It is also very similar to the most popular ham software-defined radio topology. In fact, those ham SDRs simply substitute a PC sound card for the ADC and the PC’s CPU for the FPGA. Here, though, I don’t want to bring a PC into the picture yet. Radios that require a PC don’t feel like “real radios” to me. I want to end up with a self-contained box, though I won’t mind if it optionally integrates with a PC.
For simplicity, I want to implement the receiver’s automatic gain control (AGC) functional digitally, after the ADCs. This means that with a well-designed front end, the ADC dynamic range will become the radio’s dynamic range.
With that in mind, I went looking for the best dynamic range audio ADC I could find, or at least the best one I could afford. I have identified seven manufacturers of 24-bit audio ADCs. Here is the best that they have:
Part SNR, A-weighted THD+N Package Price
TI PCM4222 123 dB -108 dB TQFP-48 $29.99 D* TI PCM4220 123 dB -108 dB TQFP-48$19.10 M
AKM AK5388 123 dB (Note 1) -110 dB LQFP-44 $10.93 D AKM AK5394A 123 dB -110 dB SOP-28$22.00 D
Cirrus CS5381 120 dB -110 dB SOIC-24, TSSOP-24 $32.22 D/M Wolfson WM8786 111 dB -102 dB SSOP-20$3.48 M
ADI AD1974 105 dB -96 dB LQFP-48 $10.03 D NXP UDA1361 100 dB -88 dB SSOP-16$1.37 M
Notes:
Figures are typical values as shown in the manufacturer’s data sheet. All are for two channels active and 24-bit PCM output. Prices are quantity 1 from the cheaper of Digi-Key or Mouser.
1. The AK5388 is a four-channel ADC with 120 dB SNR. The 123 dB SNR requires the use of “mono mode”, which connects each stereo pair in parallel with a single input, effectively creating a two-channel ADC with 3 dB better SNR.
* Non-stock item, but a limited number of units are currently in stock.
The AK5388 looks like the price/performance winner, at $10.93 for 123 dB SNR and the best THD+N. One catch is figuring out the mono mode, which didn’t have crystal-clear documentation in the data sheet. On the other hand, even in four-channel mode it has 120 dB SNR, which still puts it at an excellent price/performance point. Lurking in the SNR specification is A-weighting, which is a specification used for audio that isn’t much seen in the measurement world. The idea behind A-weighting is to reflect the human ear’s varying perception of noise at different frequencies by doing a weighted sum of the noise in the SNR measurement. Thus, frequencies where noise is more audible count worse than frequencies where it isn’t. A-weighting is not quite the right measurement for a near-zero IF ADC, first because a communications receiver’s bandwidth is quite a bit smaller than the 20 kHz used for the A-weighted measurement, and second because the communications signal being digitized is not necessarily at its final audible frequency yet. I used A-weighting for the comparison because all of the ADCs were specified that way. Some of them also had an SNR in 20 kHz bandwidth specification, which does not weight the noise spectrum. For those ADCs, the 20 kHz SNR was 3 dB lower than the A-weighted SNR. , I decided to compare the A-weighted SNR to put the ADCs on equal ground, even though the 20 kHz bandwidth SNR is closer to what I would like to know for a communications receiver. It’s worth mentioning that I looked at precision ADCs as well. I couldn’t find any 24-bit precision ADCs from Analog Devices or Linear Technologies that had a high-enough sample rate to be usable. In contrast, TI offers the single-channel ADS1281 and ADS1282, which offer a stunning 130 dB SNR (unweighted) at 250 samples per second. These might be reasonable for a Morse code receiver with a 100 Hz passband. When used for sideband, these ADCs would have to be operated in their 4000 SPS mode, at which their SNR drops to 118 dB. When one considers that filtering the output of the audio ADCs down to 100 Hz would provide an extra 6+ dB of SNR (because less bandwidth means less noise power), the ADS1281/1282 no longer has such an advantage. Worse, a pair of ADS1281 (two are needed because they are single-channel) will set you back$108 at Digi-Key.
In the end, I’ve found a combination of excellent performance and a good price. The AK5388 it is. The next step is to build or buy a board for it so I can start experimenting.
## QST performance measurements side-by-side in a table
Hans Remeeus PA1HR has compiled QST‘s performance measurements for dozens of recent rigs into a single, easy-to-read table. It’s perfect for anyone shopping for a new rig. It’s also great for numerically inclined homebrewers. I have been wishing for something like this so I have an idea of what kind of goals to set for a design. Have a look!
(via Hans Brakob K0HB and the Twin Cities DX Association e-mail list)
## Humbled by a sine wave oscillator
The old saw goes, “oscillators don’t, amplifiers do”, but that’s an analog saying. It doesn’t apply to the digital world. After all, digital is easy, while analog is hard, right? (Be careful not to answer too quickly!)
For the last few weeks, I’ve spent my tinkering time on adding sine wave sidetone to the Verilog iambic keyer. As you may recall, sidetone is audio feedback for the keying, in other words, audible Morse code. My thinking was that while this would appear to be a side trip on the way to FPGA-DSP radio nirvana, it would actually be valuable. My vision for the radio will require sine wave synthesis in a couple of places. Also, to have sine wave sidetone, I will need a digital-to-analog converter (DAC), so why not put theory into practice and build a delta-sigma DAC?
I looked around a bit to find a good way to synthesize a sine wave. I didn’t want to do a simple ROM lookup table, because the radio will need memories for storing filter data and coefficients. Instead, I settled on an algorithm that uses two integrators and a multiplier. For some ratios of clock to output frequencies, the multiplication by $-m$ can be reduced to a simple right-shift, which is always appealing to a hardware designer because it can be implemented without any gates. The negation can then be absorbed into the first adder, costing little more than a set of inverters and a carry input.
Then I found this elegant topology in Delta-Sigma Modulators: Modelling, Design, and Applications, by Bourdopoulos et al:
In this design, the multiplier has been replaced by a delta-sigma modulator and two constants. This is gorgeous! First of all, the multiplier is gone, replaced by a few adders and a mux. Secondly, the delta-sigma bitstream is a built-in output for a DAC. There are a couple of wrinkles. First of all, there needs to be a scaling by $2^{-b}$ before and $2^b$ after the delta-sigma modulator. Otherwise, the modulator will see signals outside of its stable range. (Stability is a topic for a future tutorial in the How Delta-Sigma Works series.) Second, I’m not sure about putting the delay of the delta-sigma converter into the loop. It seems to me that the two integrators should be non-delaying. Third, are the two integrators sufficient filtering to remove the delta-sigma’s high-frequency noise, or is there a risk of high-frequency feedback making the loop unstable?
It sounds great, but when I coded it up in Verilog I couldn’t get it to be stable. I fiddled with constants and bit widths for a while and couldn’t get it going. I also tried two non-delaying integrators in the loop. That didn’t work, either, and I couldn’t figure out why. Rather than do any stability analysis or model it in Octave, I decided to cut my losses and go to a more direct approach.
I took the original sine wave oscillator topology, which I easily got working, and tacked on a delta-sigma modulator to the output. I decided to increase the clock rate in order to get a better signal-to-noise ratio (SNR) out of the DAC. The oscillator blew up – unstable again! After some more fiddling, I realized I was going to have to run with 24 or more bits of precision if I stuck to the design in the first figure. To reduce the precision I needed to carry, I instead split the scaling between both integrators:
I fiddled with fewer bits, but with the clock-to-output frequency ratio I was working with (1 MHz clock, 800 Hz output), I needed all 16 bits to see a pretty sine wave in the ModelSim waveform viewer.
It looked great in the waveform viewer, so I burned it onto the FPGA and… noise. There is a brief tone buried in the noise, then it turns to all noise, and a little later, to a high-pitched whine without noise. The sine wave buried in the noise might mean I need better filtering between the FPGA and the audio amp (I didn’t use any, relying solely on the audio amp’s rolloff), but the all-noise and whining stages tell me that the oscillator or modulator is unstable. I probably didn’t run the testbench long enough in simulation to see it.
Meanwhile, at work, I’ve been having a great time introducing Test-Driven Development (TDD) into our firmware development process. The process was tedious at first, but now that I’ve been doing it long enough to see the results, I love it! My code is better organized, better tested, and more solid than ever before.
That, combined with my frustration with the sine wave project, led me to think about applying Agile software engineering techniques to Verilog development. I did some Google searches to see if anyone else is applying TDD or full Agile to hardware. They are! (agilesoc.com: Why Agile is a good fit for FPGA and ASIC development)
I learned some technical things while working on the sidetone. More importantly, though, I was reminded that methodical, well-tested development is not only better than seat-of-the-pants hacking, it is faster as well. Hobby projects benefit from discipline just as much as professional ones. I am going to set aside the sidetone project for a while and look at some other things. Meanwhile I am going to think a bit about how to apply TDD and some other Agile concepts to my humble basement tinkering.
## How Delta-Sigma Works, part 2: The Anti-Aliasing Advantage
This post is part of a series on delta-sigma techniques: analog-to-digital and digital-to-analog converters, modulators, and more. A complete list of posts in the series are in the How Delta-Sigma Works tutorial page.
Today, let’s take another look at delta-sigma conversion. The first part of this series showed how a one-bit, first-order delta-sigma modulator creates a bitstream, the average value of which equals the input voltage. It turns out that how we find that average value makes a big difference in the performance of a delta-sigma analog-to-digital converter. In fact, if done right, not only does it improve performance, but it greatly simplifies the analog circuitry preceding the A-to-D. Let’s take a look at why this works!
## Aliasing Explained
One phenomenon that happens with any analog-to-digital conversion is known as aliasing. A-to-D is inherently a sampling process, in which an analog signal that is continuous in time is converted to a digital signal that exists in discrete chunks, or samples. The rate at which those samples are taken is known as the sampling rate. The figure below shows three analog sine waves, in red, being sampled at the times marked with the blue vertical lines. The top sine wave has a frequency of 1 Hz, and it is being sampled at 7 Hz. (The seventh sample is not obvious, because it is zero and the left- hand or the right-hand edge of the graph. Which edge you choose is unimportant.)
The second and third plots in the figure show aliasing in action. When any signal with a frequency above one-half the sampling rate is sampled, the signal is <i>aliased</i> down to a frequency between 0 Hz and one-half the sampling rate. The second line shows a 6 Hz sine wave in red, which is being sampled at 7 samples per second (sps). The blue lines show where the samples fall. Because 6 Hz is above 3.5 sps (one half of 7 sps), the sine wave will be aliased. As you can see, the blue samples are identical to those you see in the top trace, except that they have the opposite polarity. The dark blue trace connects them and shows that a 6 Hz sine wave is indistinguishable from a 1 Hz sine wave when both are sampled at 7 sps. That is aliasing in action.
The same process happens for any input frequency above one-half the sampling rate. The third plot in the figure, for example, shows an 8 Hz sine wave. Again, it is sampled at 7 Hz. This time the samples are identical to those in the top line. The sampled waveform (in dark blue once again) is indistinguishable from the sampled version of our original 1 Hz sine wave.
Aliasing is so important that one-half the sampling rate has become known as the “Nyquist frequency” for a system. You may see people refer to signals as being “above Nyquist” or “below Nyquist”. Aliasing happens in a repeated pattern as the frequency rises, and each repetiion of that pattern is called a “Nyquist zone”. All of this is in memory of Harry Nyquist (1889-1976), who with Claude Shannon discovered much of the mathematics behind sampling.
Because of aliasing, analog-to-digital converters are usually preceded by an anti-aliasing filter. This filter removes the frequency content outside of the desired Nyquist zone, so that noise and interfering signals do not alias into the passband of the ADC. Most often, a low-pass filter is used, so that the selected Nyquist zone runs from DC (0 Hz) to 1/2 the sampling frequency. There are some exotic RF applications in which a bandpass filter selects frequencies in a higher Nyquist zone, which the ADC then aliases down, but these are uncommon.
## How Oversampling Makes Anti-Aliasing Easier
In order to get good accuracy, delta-sigma converters need to run at a much higher frequency than the input signal. The first part of this series introduced an idea for an ADC built from a delta-sigma modulator followed by a digital counter:
For this to work, the delta-sigma modulator has to <i>oversample</i>. In order to have the counter reflect the input signal accurately, there have to be many 1 and 0 bits in the modulator’s output for the counter to count. In other words, each sample from the counter’s output has to reflect many samples in the modulator. This is called oversampling.
The nice thing about oversampling is that the Nyquist frequency goes up with the sample rate, even when oversampling! The classic example of this happens in CD audio systems. CDs carry audio signals of up to 20 kHz, with a 44.1 ksps sample rate and a Nyquist frequency of 22.05 kHz. Furthermore, CD audio has a dynamic range of about 97 dB. In order to avoid aliasing signals back into the 0 Hz – 20 kHz audio band, the antialiasing filter at the input of a CD audio ADC needs to have a rolloff frequency (-3 dB) at 20 kHz, and should be down to -97 dB by 24.1 kHz. This is an impractical filter to design and build.
However, if an oversampled delta-sigma ADC is used, then the Nyquist frequency goes up to one-half the oversampled sample rate. A typical choice might be 64x oversampling, in which case the ADC will sample at 2.8224 MHz. Then the Nyquist frequency is 1.4112 MHz. The anti-aliasing filter still needs to have its -3 dB rolloff at 20 kHz, but it does not need to be -97 dB down until 2.82 MHz. That is a much easier filter to design. In fact, a 3-pole filter, easily and cheaply implemented with an op amp, is sufficient.
## Moving Anti-Aliasing to the Digital Domain
To go from 2.8 Msps to 44.1 ksps requires another round of sampling, this time in the digital domain. Remember the counter? The process of reading its count, then resetting it for another round of counting, is a form of sampling, and aliasing can result. The figure below shows an example. In this case, an 8 Hz sine wave is being sampled at 70 sps by an oversampling ADC. Then, that digital signal is being downsampled, by taking every tenth sample, to 7 sps. The result is that the 8 Hz input is aliased to 1 Hz, just as if it was sampled at 7 sps in the first place.
Just as in the analog domain, there are times when the aliasing resulting from downsampling can be useful, but often it is not. To prevent it, we need a low-pass filter, but this time the filter can be digital. In the CD-audio example, the filter needs to have the same rolloff characteristics as the challenging analog filter (-3 dB at 20 kHz, and -97 dB at 24.1 kHz). Doing that in a digital filter, though, is much easier than in analog. Digital arithmetic can produce a filter of arbitrarily good performance, without the precision components or careful tuning adjustments that might be required in analog. All it takes is throwing enough logic gates at the problem, and thanks to Moore’s Law, logic gates are cheap.
Since low-pass filters have an averaging effect, the filter will turn the bitstream of 1’s and 0’s into a series of multi-bit samples. The counter becomes unnecessary. Instead, it is enough to keep one sample from the low-pass filter’s output every so often, discarding the rest. The ADC now looks like this. (A simple anti-aliasing filter before the input is needed, but not shown.)
The figure below shows the principle. A 1 Hz sine wave is oversampled at 70 Hz in the top graph, then 9 out of every 10 samples are discarded, leaving only the highlighted ones. Those samples are plotted in the bottom graph. The result is identical to the sine wave samples in the first graph at the top of this article.
Technology similar to this, with some additional improvements to reduce the amount of math needed, makes it possible to put CD-quality ADCs in every desktop and laptop computer. Instead of an expensive analog filter, cheap digital gates on an IC provide most of the filtering, reducing the cost of the ADC to only a dollar or two.
## Wrapping Up
In this article, we have come full circle to find out how delta-sigma techniques can make analog design easier. I’ve shown how aliasing happens and explained the need for anti-aliasing filters. Then we looked at the oversampling inherent to delta-sigma modulators, and how that permits a simpler analog antialiasing filter, at long as a digital low-pass filter is included after the modulator. This is just one of the reasons why delta-sigma principles are very cool. Coming up in this series: Simulating a delta-sigma modulator and an introduction to noise shaping.
## Reference
Bourdopoulos, George I., Aristodemos Pnevmatikakis, Vassilis Anastassopoulos, and Theodore Deliyannis. Delta-Sigma Modulators: Modeling, Design and Applications. London: Imperial College Press, 2003 | 7,536 | 32,049 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 6, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2024-30 | latest | en | 0.86961 |
http://support.sas.com/rnd/app/stat/examples/SurveyVariance/new_example/ | 1,550,474,483,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247484689.3/warc/CC-MAIN-20190218053920-20190218075920-00577.warc.gz | 260,668,203 | 21,869 | # SAS/STAT Software Examples
## Estimating the Variance of a Variable in a Finite Population
Contents | SAS Program | PDF
## Overview
The finite population variance of a variable provides a measure of the amount of variation in the corresponding attribute of the study population’s members, thus helping to describe the distribution of a study variable. Whether you are studying a population’s income distribution in a socioeconomic study, rainfall distribution in a meteorological study, or scholastic aptitude test (SAT) scores of high school seniors, a small population variance is indicative of uniformity in the population while a large variance is indicative of a more diverse population. Another use for the population variance is to determine sample size. For example, the U.S. Environmental Protection Agency uses estimated population variances from pilot studies such as the Environmental Monitoring and Assessment Program–Surface Waters Northeast Lakes Pilot study to assist in planning future sampling strategies (Courbois and Urquhart; 2004).
Suppose you have data that were sampled according to some complex survey design. The SURVEYMEANS procedure enables you to estimate finite population totals, means, and ratios in addition to the design-based variances of the estimated quantities, but it does not directly estimate the finite population variance of a variable. However, because a variance can be expressed mathematically as a total, you can easily estimate the finite population variance of a variable by using PROC SURVEYMEANS plus a little SAS programming.
Whenever you estimate a population parameter such as a mean or a variance, you should also report the precision of the estimate. The most commonly reported measure of precision is the variance (or its square root, the standard error). The survey analysis procedures in SAS/STAT software currently provide three different variance estimation methods for complex survey designs: the Taylor series linearization method, the delete-one jackknife method, and the balanced repeated replication (BRR) method. This example demonstrates how to use all three methods to estimate the variance .
Because the finite population parameter of interest in this example is the variance of a variable, the measure of precision of the estimate is the variance of a variance. Therefore, as you consider the example, it is important to keep in mind the distinction between the two different meanings of the word variance. In one context, a variance is estimated in order to describe the distribution of a variable. A variance used in this context is denoted and its estimator is denoted . In the other context, a variance is estimated in order to describe the sampling distribution of an estimator. A variance used in this context is denoted and its estimator is denoted .
## Analysis
Suppose you want to estimate the variance of a variable from a finite population using data that were sampled according to some complex survey design. The finite population variance of is
(1)
where is the total number of elements in the population, is the th observation of the variable , and is the population mean of . A sample-based estimator of is
(2)
where is an estimator of the population total , is an estimator of the population mean, is the number of elements in the sample, and is the probability that element is observed in the sample.
To estimate , you first estimate both and with PROC SURVEYMEANS. Next, you generate a variable (call it ) such that each observation is equal to
(3)
Now you use PROC SURVEYMEANS to estimate the total of . The estimated weighted total of is equal to . However, the variance of the weighted total of that is computed by PROC SURVEYMEANS, regardless of which VARMETHOD= option you select, is not equal to , the variance of the estimate . Computing requires some additional SAS programming.
## Using the Taylor Series Linearization Method to Estimate
To estimate by using the Taylor series linearization method, construct a variable , such that
(4)
where is computed as in equation (2). Use PROC SURVEYMEANS to estimate the total (and the variance of the total) of . The total that is computed by PROC SURVEYMEANS is of no interest, but the variance of the total is equal to , the variance of the estimate (Särndal, Swensson, and Wretman 1992 , chap. 5.5).
The following steps summarize how you estimate , the finite population variance of a variable , and , the variance of the finite population variance estimator (using the Taylor series linearization method):
1. Use PROC SURVEYMEANS to estimate the sample mean of the variable , and save the estimated mean. PROC SURVEYMEANS also computes the sum of the sampling weights, which is the value of in the analysis. Save that value also; it is used in the construction of .
2. Using the sample mean from step 1, construct the variable as in equation (3).
3. Use PROC SURVEYMEANS to estimate the weighted total of the variable . Save the estimated total, which is the estimate of the population variance ().
4. Using the sample mean from step 1 and the estimate of obtained in step 3, construct the variable as in equation (4).
5. Use PROC SURVEYMEANS to estimate the weighted total of the variable . The estimated variance of this total obtained from PROC SURVEYMEANS is an estimator of the variance of .
## Example
### Ice Cream Study Data Set
This example uses the IceCreamStudy data set from the example "Stratified Cluster Sample Design" in the chapter "The SURVEYMEANS Procedure" of the SAS/STAT User's Guide.
The study population is a junior high school with a total of 4,000 students in grades 7, 8, and 9. In the original example, researchers want to know how much these students spend weekly for ice cream, on the average, and what percentage of students spend at least \$10 weekly for ice cream. This example measures the variability of the students’ expenditures by estimating , the variance of the variable that contains the students’ expenditures.
Suppose that every student belongs to a study group and that study groups are formed within each grade level. Each study group contains between two and four students. Table 1 shows the total number of study groups and the total number of students for each grade.
Table 1 Study Groups and Students by Grade
Number of Study Groups
Number of Students
7
608
1,824
8
252
1,025
9
403
1,151
It is quicker and more convenient to collect data from students in the same study group than to collect data from students individually. Therefore, this study uses a stratified clustered sample design. The primary sampling units are study groups. The list of all study groups in the school is stratified by grade level. From each grade level, a sample of study groups is randomly selected, and all students in each selected study group are interviewed. The sample consists of eight study groups from the 7th grade, three groups from the 8th grade, and five groups from the 9th grade.
The SAS data set IceCreamStudy saves the responses of the selected students:
```data IceCreamStudy;
input Grade StudyGroup Spending Weight @@;
datalines;
7 34 7 76.0 7 34 7 76.0 7 412 4 76.0 9 27 14 80.6
7 34 2 76.0 9 230 15 80.6 9 27 15 80.6 7 501 2 76.0
9 230 8 80.6 9 230 7 80.6 7 501 3 76.0 8 59 20 84.0
7 403 4 76.0 7 403 11 76.0 8 59 13 84.0 8 59 17 84.0
8 143 12 84.0 8 143 16 84.0 8 59 18 84.0 9 235 9 80.6
8 143 10 84.0 9 312 8 80.6 9 235 6 80.6 9 235 11 80.6
9 312 10 80.6 7 321 6 76.0 8 156 19 84.0 8 156 14 84.0
7 321 3 76.0 7 321 12 76.0 7 489 2 76.0 7 489 9 76.0
7 78 1 76.0 7 78 10 76.0 7 489 2 76.0 7 156 1 76.0
7 78 6 76.0 7 412 6 76.0 7 156 2 76.0 9 301 8 80.6
;
```
Table 2 identifies the variables contained in the data set IceCreamStudy.
Table 2 Variables in IceCreamStudy Data Set
Variable
Description
StudyGroup
Student’s study group (PSU)
Spending
Student’s expenditure per week for ice cream, in dollars
Weight
Sampling weights
The SAS data set StudyGroup is created to provide PROC SURVEYMEANS with the sample design information shown in Table 1. The variable Grade identifies the strata, and the variable _TOTAL_ contains the total number of study groups in each stratum.
```data StudyGroups;
datalines;
7 608
8 252
9 403
;
```
### Step 1: Compute and
Use PROC SURVEYMEANS to obtain an estimate of the sample mean. Specify the MEAN and STACKING options in the PROC SURVEYMEANS statement. The STACKING option causes the procedure to create an output data set with a single observation. This table structure makes it easy in later steps to identify the saved estimates and to assign their values to macro variables. The WEIGHT statement specifies that the variable Weight contains the sampling weights. The STRATA statement specifies that the variable Grade identifies strata membership. The CLUSTER statement specifies that the variable StudyGroup identifies cluster (or PSU) membership. The ODS OUTPUT statement requests output data sets for the statistics and data summary tables, to be named Statistics and Summary, respectively. The sample mean is stored in the data set Statistics. The data set Summary contains the sum of the sampling weights, the number of strata, and the number of clusters. The sum of the sampling weights is needed to compute ; the number of strata and the number of clusters are used later when computing confidence limits for .
```proc surveymeans data=IceCreamStudy mean stacking ;
weight Weight;
cluster StudyGroup;
var Spending;
ods output Statistics = Statistics
Summary = Summary;
run;
```
The following DATA step saves the sample mean of the variable Spending in a macro variable named Spending_Mean:
```data _null_;
set Statistics;
call symput("Spending_Mean",Spending_Mean);
run;
```
The next DATA step saves the sum of the sampling weights in a macro variable named N, the number of strata in a macro variable named H, and the number of clusters in a macro variable named C:
```data Summary;
set Summary;
if Label1="Sum of Weights" then call symput("N",cValue1);
if Label1="Number of Strata" then call symput("H",cValue1);
if Label1="Number of Clusters" then call symput("C",cValue1);
run;
```
### Step 2: Construct the Variable
Construct the variable in a DATA step by using the macro variables Spending_Mean and N:
```data Working;
set IceCreamStudy;
z=(1/(&N-1))*(Spending-&Spending_Mean)**2;
run;
```
### Step 3: Estimate the Total of
Use PROC SURVEYMEANS to estimate the weighted total of the variable . Specify the SUM and STACKING options in the PROC SURVEYMEANS statement. The ODS OUTPUT statement saves the statistics table to a data set named Result.
```proc surveymeans data = Working sum stacking;
weight Weight;
var z;
ods output Statistics = Result(keep=z_Sum);
run;
```
The following DATA step retrieves the estimated total of z and stores it in a macro variable named Variance. The total of z is equal to .
```data _null_;
set Result;
call symput("Variance",z_Sum);
run;
```
### Step 4: Construct the Variable
Construct the variable by using the macro variables Spending_Mean, N, and Variance:
```data Taylor;
set IceCreamStudy;
u=(1/(&N-1))*((Spending-&Spending_Mean)**2 - &Variance);
run;
```
### Step 5: Estimate the Total of
Use PROC SURVEYMEANS to estimate the total of the variable . Specify the SUM, VARSUM, TOTAL=, and STACKING options in the PROC SURVEYMEANS statement. The VARSUM option computes the variance of the total. In this step, the computation of interest is the variance of the estimated total rather than the total itself. Therefore, the sampling design must be appropriately represented in the SURVEYMEANS procedure. The TOTAL= option is specified to enable the procedure to apply a finite population correction in the variance computation. The STRATA statement specifies that the strata be identified by the variable Grade, and the CLUSTER statement specifies that cluster membership be identified by the variable StudyGroup. The ODS OUTPUT statement saves the statistics table in a data set named TaylorResult.
```proc surveymeans data = Taylor sum varsum stacking total=StudyGroups;
cluster StudyGroup;
weight Weight;
var u;
ods output Statistics = TaylorResult;
run;
```
The following DATA step creates the variable Estimate in the TaylorResult data set and assigns it the value of , which is stored in the macro variable Variance. The confidence limits are computed, and the TaylorResult data set is prepared for printing.
Note: The confidence limits are computed in this example by using a distribution with degrees of freedom. This results in a confidence interval that is symmetric about the estimated parameter. Confidence intervals constructed in this manner have good coverage properties, however negative lower confidence limits are possible. There are alternative methods for computing confidence intervals that will exclude the possibility of negative lower confidence limits. For example, if the study variable is approximately normally distributed, confidence limits can be computed using a chi-square distribution. Another possibility is to to use the distribution with the lower confidence limit computed as . In the simple case that is presented in this example, the latter method is acceptable. However, there are situations where it is not. Whatever method you choose, it is important that the confidence intervals be constructed in a manner that is consistent with any assumptions you make about the underlying data and the parameter estimation method.
```%let df=%eval(&C - &H);
data TaylorResult;
set TaylorResult(rename=(u_VarSum=Variance
u_StdDev=StdErr));
Estimate=&Variance;
LowerCL= Estimate + StdErr*TINV(.025,&df);
UpperCL= Estimate + StdErr*TINV(.975,&df);
label Estimate=Population Variance Estimate
Variance=Variance of Estimate
StdErr=Standard Error of Estimate
LowerCL=Lower Confidence Limit
UpperCL=Upper Confidence Limit;
Variable='Spending';
run;
```
Use PROC PRINT to print the contents of the data set TaylorResult:
```title 'Parameter Estimates';
proc print data=TaylorResult label noobs;
var Variable Estimate Variance StdErr LowerCL UpperCL;
run;
title ;
```
Output 1 displays the results. The estimate of the population variance of the variable Spending is 28.46. The variance of the estimate is 27.87. The standard error of the estimate is 5.28, and the estimated lower and upper 95% confidence limits are 17.05 and 39.86, respectively.
Output 1 Estimate of Population Variance
Parameter Estimates
Variable Population Variance
Estimate
Variance of Estimate Standard Error of
Estimate
Lower Confidence
Limit
Upper Confidence
Limit
Spending 28.4604 27.869473 5.279155 17.0555 39.8653
## Using the Delete-One Jackknife Method to Estimate
The delete-one jackknife resampling method of variance estimation deletes one primary sampling unit (PSU) at a time from the full sample to create replicates, where is the total number of PSUs. In each replicate, the sample weights of the remaining PSUs are modified by the jackknife coefficient . The modified weights are called replicate weights.
If is the estimate of obtained by using only the data and the replicate weights from the th replicate, the jackknife variance estimate is
(5)
with degrees of freedom, where is the jackknife coefficient for the th replicate, is the number of replicates, and is the number of strata (or when there is no stratification). See the section "Jackknife Method" in the chapter "The SURVEYMEANS Procedure" of the SAS/STAT User's Guide for more details.
Recall that when you construct , you use estimates of and that are computed by using the full sample. However, the jackknife variance estimator requires that the be computed from the th replicate. Thus, the jackknife estimate of the variance of the total of is not equal to the jackknife estimate of the variance of .
The following steps summarize how you estimate , the finite population variance of a variable , and , the variance of the finite population variance estimator (using the delete-one jackknife method):
1. Use PROC SURVEYMEANS to estimate the sample mean and the sum of the weights for the full sample. Save both estimates as they are used in the construction of .
2. Construct as in equation (3), using the full-sample estimates of and obtained in step 1.
3. Use PROC SURVEYMEANS to estimate the weighted total of the variable . Save the estimated total, which is the full-sample estimate of the population variance (). When you estimate the total, specify the VARMETHOD=JACKKNIFE option and the OUTWEIGHTS= and OUTJKCOEFS= method-options in the PROC SURVEYMEANS statement. Both the OUTWEIGHTS= and OUTJKCOEFS= data sets are used in later steps.
4. For each replicate, use PROC SURVEYMEANS to compute the sample mean and the sum of the weights by using only the data and replicate weights for the th replicate. Save the estimates for later use.
5. For each replicate, using the estimates for and that were obtained in step 4, construct the variable such that
(6)
6. Use PROC SURVEYMEANS to estimate the weighted total of by replicate, and save the estimates for later use. The estimated weighted total of is equal to for the th replicate.
7. Construct a variable (call it ) by using the estimates from step 6, the jackknife coefficients, and the full-sample estimate from step 3 such that
8. Use PROC SURVEYMEANS to estimate the unweighted total of the variable from step 7. The estimated unweighted total of is , the delete-one jackknife estimate of the variance of .
## Example
This example uses the same IceCreamStudy data set that was described in the section Ice Cream Study Data Set and reproduces the steps described in the section Using the Delete-One Jackknife Method to Estimate . Steps 1 and 2 are identical to the first two steps in the previous example but are repeated here for completeness.
### Step 1: Compute and for the Full Sample
Use PROC SURVEYMEANS to obtain an estimate of the sample mean. Specify the MEAN and STACKING options in the PROC SURVEYMEANS statement. The WEIGHT statement specifies that the variable Weight contain the sampling weights. The STRATA statement specifies that the variable Grade identifies strata membership. The CLUSTER statement specifies that the variable StudyGroup identifies cluster (or PSU) membership. The ODS OUTPUT statement creates output data sets for the statistics and data summary tables, to be named Statistics and Summary, respectively. The sample mean is stored in the data set Statistics. The data set Summary contains the sum of the sampling weights and the number of strata. The sum of the sampling weights is needed to compute ; the number of strata is used later when computing confidence limits for .
```proc surveymeans data=IceCreamStudy mean stacking ;
weight Weight;
cluster StudyGroup;
var Spending;
ods output Statistics = Statistics
Summary = Summary;
run;
```
The following DATA step saves the sample mean of the variable Spending in a macro variable named Spending_Mean:
```data _null_;
set Statistics;
call symput("Spending_Mean",Spending_Mean);
run;
```
The next DATA step saves the sum of the sampling weights in a macro variable named N and the number of strata in a macro variable named H:
```data Summary;
set Summary;
if Label1="Sum of Weights" then call symput("N",cValue1);
if Label1="Number of Strata" then call symput("H",cValue1);
run;
```
### Step 2: Construct the Variable by Using the Full-Sample Estimates of and
Construct the variable in a DATA step by using the macro variables Spending_Mean and N:
```data Working;
set IceCreamStudy;
z=(1/(&N-1))*(Spending-&Spending_Mean)**2;
run;
```
### Step 3: Estimate the Total of for the Full Sample
Use PROC SURVEYMEANS to estimate the weighted total of the variable . Specify the SUM and STACKING options in the PROC SURVEYMEANS statement. Also specify the VARMETHOD=JACKKNIFE option with the OUTJKCOEFS= and OUTWEIGHTS= method-options. The OUTJKCOEFS= method-option saves the jackknife coefficients in a SAS data set named Jkcoefs. The OUTWEIGHTS= method-option saves the replicate weights in a SAS data set named Jkweights.
In this step you must fully specify the sampling design so that the jackknife coefficients and replicate weights are computed correctly. The STRATA statement specifies that the strata be identified by the variable Grade. The CLUSTER statement specifies that the PSUs be identified by the variable StudyGroup. The WEIGHT statement specifies that the full-sample sampling weights be contained in the variable Weight. The ODS OUTPUT statement saves the statistics table to a data set named Result and the variance estimation table to a data set named VarianceEstimation.
```proc surveymeans data=Working sum stacking
varmethod=JACKKNIFE(outjkcoefs=Jkcoefs
outweights=Jkweights);
cluster StudyGroup;
weight Weight;
var z;
ods output Statistics = Result
VarianceEstimation=VarianceEstimation;
run;
```
```data _null_;
set Result;
call symput("Variance",z_Sum);
run;
```
You can see from the "Variance Estimation" table in Output 2 that there are 16 replicates.
Output 2 Estimate of Population Variance
The SURVEYMEANS Procedure
Data Summary
Number of Strata 3
Number of Clusters 16
Number of Observations 40
Sum of Weights 3162.6
Variance Estimation
Method Jackknife
Number of Replicates 16
The next DATA step retrieves the number of replicates and stores the value in a macro variable named R:
```data _null_;
set VarianceEstimation;
where label1="Number of Replicates";
call symput("R",cvalue1);
run;
%let R=%eval(&R);
```
The data set Jkcoefs has 16 observations, one for each replicate. The th observation contains the jackknife coefficient for the th replicate. The data set Jkweights contains the original variables from the IceCreamStudy data set and 16 new variables named RepWgt_1 through RepWgt_16; there are observations.
### Step 4: Compute and for Replicate Samples
Before computing and , use the following DATA step to convert the data set Jkweights from wide form to long form; doing so enables you to use BY-group processing with PROC SURVEYMEANS.
```data Long(drop= RepWt_1 - RepWt_&R Z);
set Jkweights;
array num (*) RepWt_1 - RepWt_&R;
do replicate=1 to dim(num);
Jkweight=num(replicate);
output;
end;
run;
```
The data set Long has observations. There are 16 copies of the original variables from the IceCreamStudy data set stacked on top of each other, and each copy is identified by the variable Replicate. Instead of the 16 replicate weight variables, RepWgt_1 through RepWgt_16, there is now one variable, Jkweight, which is constructed by stacking the variables RepWgt_1 through RepWgt_16 on top of each other. Thus, the first 40 observations contain a copy of the original variablesand the contents of RepWgt_1, and the variable Replicate has a value of 1. The second 40 observations contain a copy of the original variables and the contents of RepWgt_2, and the variable Replicate has a value of 2. The remaining observations are constructed and identified similarly.
Next, sort the data set Long by Replicate:
```proc sort data=Long out=Long;
by Replicate;
run;
```
Use PROC SURVEYMEANS to estimate the mean of Spending by Replicate. Doing so produces the estimates of and for each replicate. The WEIGHT statement specifies that the sampling weights be contained in the variable Jkweight. The ODS OUTPUT statement saves the sample means () in a SAS data set named JKMeans and saves the sum of the replicate weights () in a data set named JKN. By default, the means are stored in a variable named Mean and the sum of the replicate weights are stored in a variable named N.
```proc surveymeans data=Long mean;
weight Jkweight;
var Spending;
by Replicate;
ods output Statistics = JKMeans(keep=Replicate Mean)
Summary = JKN;
run;
```
### Step 5: Construct the Variable for the Replicate Samples
Before you can construct the variable for the replicate samples, you must merge the data sets JKMeans and JKN with Long, by Replicate:
```proc sort data=JKMeans out=JKMeans;
by Replicate;
run;
```
```data JKN(keep=N replicate );
set JKN(rename=(nvalue1=N));
where Label1="Sum of Weights";
run;
```
```proc sort data=JKN out=JKN;
by Replicate;
run;
```
```data Long;
merge Long JKN JKMeans;
by Replicate;
run;
```
Now construct the variable using the merged data set:
```data Long;
set Long;
z=(1/(N-1))*(Spending-Mean)**2;
run;
```
### Step 6: Estimate the Total of for Replicate Samples
Use PROC SURVEYMEANS to estimate the total of the variable by Replicate. The WEIGHT statement specifies that the sampling weights be contained in the variable Jkweight. You do not need to specify the STRATA and CLUSTER statements. The ODS OUTPUT statement saves the estimated totals in the variable JKEstimate in a SAS data set named Statistics. The estimated totals are the estimates for each replicate.
```proc surveymeans data=Long sum stacking;
weight Jkweight;
var z;
by Replicate;
ods output Statistics=Statistics(drop=Z_StdDEV rename=(Z_Sum=JKEstimate));
run;
```
### Step 7: Construct the Variable
Before you can construct the variable , you must sort and merge, by Replicate, the data sets Statistics and Jkcoefs:
```proc sort data=Statistics out=Statistics;
by Replicate;
run;
```
```proc sort data=Jkcoefs out=Jkcoefs;
by Replicate;
run;
```
```data Statistics;
merge Statistics Jkcoefs;
by Replicate;
run;
```
The data set Statistics now contains the jackknife coefficients in the variable JKcoefficients and the estimates in the variable JKEstimate. Construct the variable by using these variables and the full-sample estimate that is saved in the macro variable Variance:
```data Statistics;
set Statistics;
u=JKcoefficient*(JKEstimate-&Variance)**2;
run;
```
### Step 8: Estimate the Total of
Use PROC SURVEYMEANS to compute the unweighted total of . Specify the SUM option in the PROC SURVEYMEANS statement. The ODS OUTPUT statement saves the total in a variable named Variance in a SAS data set named JKResult.
```proc surveymeans data=Statistics sum;
var u;
ods output Statistics=JKResult(rename=(sum=Variance));
run;
```
The following DATA step computes the standard error of the estimate and the upper and lower 95% confidence limits. In this example, the confidence limits are computed using a distribution with degrees of freedom. The variable Estimate is generated and assigned the estimated value of , which is stored in the macro variable Variance. Labels are created for the existing variables. A new variable Variable is generated, and its value is specified to be the name of the variable that is being analyzed (Spending).
```%let df=%eval(&R-&H);
data JKResult;
set JKResult;
StdErr=sqrt(Variance);
Estimate=&Variance;
LowerCL= Estimate + StdErr*TINV(.025,&df);
UpperCL= Estimate + StdErr*TINV(.975,&df);
label Estimate=Population Variance Estimate
Variance=Variance of Estimate
StdErr=Standard Error of Estimate
LowerCL=Lower Confidence Limit
UpperCL=Upper Confidence Limit;
Variable='Spending';
run;
```
Use the PRINT procedure to print the contents of the data set JKResult:
```title 'Parameter Estimates';
proc print data=JKResult label noobs;
var Variable Estimate Variance StdErr LowerCL UpperCL;
run;
title ;
```
Output 3 displays the results. The estimate of the population variance for the variable Spending is 28.46. The variance of the estimate is 30.27, and the standard error of the estimate is 5.50. The estimated lower and upper 95% confidence limits are 16.57 and 40.35, respectively.
Output 3 Estimate of Population Variance
Parameter Estimates
Variable Population Variance
Estimate
Variance of Estimate Standard Error
of Estimate
Lower Confidence
Limit
Upper Confidence
Limit
Spending 28.4604 30.267500 5.50159 16.5750 40.3459
## Using the BRR Method to Estimate
The BRR method requires that the full sample be drawn by using a stratified sample design with two PSUs per stratum. If is the total number of strata, the total number of replicates is the smallest multiple of four that is greater than . Each replicate is obtained by deleting one PSU per stratum according to the corresponding Hadamard matrix and adjusting the original weights for the remaining PSUs. The new weights are called replicate weights.
If is the estimate of obtained by using only the data and the replicate weights from the th replicate, the BRR variance estimate is
(7)
with degrees of freedom. See the section "Balanced Repeated Replication (BRR) Method" in the chapter "The SURVEYMEANS Procedure" of the SAS/STAT User's Guide for more details.
Recall that when you construct , you use estimates of and that are computed by using the full sample. However, the BRR variance estimator requires that the be computed from the th replicate. Thus, the BRR estimate of the variance of the total of is not equal to the BRR estimate of the variance of .
The following steps summarize how you estimate , the finite population variance of a variable , and , the variance of the finite population variance estimator (using the BRR method):
1. Use PROC SURVEYMEANS to estimate the sample mean and the sum of the weights for the full sample. Save both estimates for later use: they are used in the construction of .
2. Construct as in equation (3) by using the full-sample estimates of and obtained in step 1.
3. Use PROC SURVEYMEANS to estimate the weighted total of the variable , and save the estimated total. This total is the full-sample estimate of the population variance (). When you estimate the total, specify the VARMETHOD=BRR option and the OUTWEIGHTS= method-option in the PROC SURVEYMEANS statement. The OUTWEIGHTS= SAS data set is used in later steps. Also save the number of strata and the number of replicates for later use.
4. For each replicate, use PROC SURVEYMEANS to estimate the sample mean and the sum of the weights by using only the data and replicate weights for the th replicate. Save the estimates for later use.
5. For each replicate, using the estimates for and that were obtained in step 4, construct the variable such that
(8)
6. Use PROC SURVEYMEANS to estimate the weighted total of by replicate, and save the estimates for later use. The estimated weighted total of is equal to for the th replicate.
7. Construct a variable (call it ) by using the estimates from step 6, the number of replicates , and the full-sample estimate from step 3 such that
8. Use PROC SURVEYMEANS to estimate the unweighted total of the variable from step 7. The estimated unweighted total of is , the BRR estimate of the variance of .
## Example
### The MUNIsurvey Data Set
This example uses the MUNIsurvey data set from the section "Variance Estimation Using Replication Methods" in the chapter "The SURVEYMEANS Procedure" of the SAS/STAT User's Guide. The data are not shown here, but a SAS program that generates the data is included in the sample SAS code that you can download for this example.
In the original example, the San Francisco Municipal Railway (MUNI) conducted a survey to estimate the average waiting time for MUNI subway system’s passengers. This example estimates the variance of the passengers’ waiting time.
The study uses a stratified cluster sample design. Each MUNI subway line is a stratum. The subway lines included in the study are 'J-Church,' 'K-Ingleside,' 'L-Taraval,' 'M-Ocean View,' 'N-Judah,' and the street car 'F-Market & Wharves.' The MUNI vehicles in service for these lines during a day are the primary sampling units. Within each stratum, two vehicles (PSUs) are randomly selected. Then the waiting times of passengers for a selected MUNI vehicle are collected.
The collected data are saved in the SAS data set MUNIsurvey. Table 3 identifies the variables contained in the data set.
Table 3 Variables in MUNIsurvey Data Set
Variable
Description
Line
The MUNI line that a passenger is riding (strata)
Vehicle
The vehicle that a passenger is boarding (PSU)
Waittime
The time (in minutes) that a passenger waited
Weight
Sampling weights
### Step 1: Compute and for the Full Sample
Use PROC SURVEYMEANS to obtain estimates of the sample mean () and the sum of the sampling weights () for the full sample. Specify the MEAN and STACKING options in the PROC SURVEYMEANS statement. The WEIGHT statement specifies that the variable Weight contain the sampling weights. The STRATA statement specifies that the variable Line identify stratum membership. The CLUSTER statement specifies that the variable Vehicle identify PSU or cluster membership. The ODS OUTPUT statement produces output data sets for the statistics and data summary tables, to be named Statistics and Summary, respectively. The sample mean is stored in the data set Statistics, and the sum of the sampling weights is stored in the data set Summary.
```proc surveymeans data=MUNIsurvey mean stacking ;
weight Weight;
strata Line;
cluster Vehicle;
var Waittime;
ods output Statistics = Statistics
Summary = Summary;
run;
```
The following DATA step saves the sample mean () of the variable Waittime in a macro variable named Waittime_Mean:
```data _null_;
set Statistics;
call symput("Waittime_Mean",Waittime_Mean);
run;
```
The next DATA step saves the sum of the sampling weights in a macro variable named N and the number of strata in a macro variable named H:
```data Summary;
set Summary;
if Label1="Sum of Weights" then call symput("N",cValue1);
if Label1="Number of Strata" then call symput("H",cValue1);
run;
```
### Step 2: Construct the Variable by Using the Full-Sample Estimates of and
Construct the variable in a DATA step by using the macro variables Waittime_Mean and N:
```data Working;
set MUNIsurvey;
Z=(1/(&N-1))*(Waittime-&Waittime_Mean)**2;
run;
```
### Step 3: Estimate the Total of for the Full Sample
Use PROC SURVEYMEANS to estimate the total of the variable . Specify the SUM and STACKING options in the PROC SURVEYMEANS statement. Also specify the VARMETHOD=BRR(OUTWEIGHTS=) option. The OUTWEIGHTS= method-option saves the replicate weights in a SAS data set named BRRweights.
In this step you must fully specify the sampling design so that the replicate weights are computed correctly. The STRATA statement specifies that the strata be identified by the variable Line. The CLUSTER statement specifies that the PSUs be identified by the variable Vehicle. The WEIGHT statement specifies that the full-sample sampling weights be contained in the variable Weight. The ODS OUTPUT statement saves the statistics table to a data set named Estimate and the variance estimation table to a data set named VarianceEstimation.
```proc surveymeans data=Working sum stacking
varmethod=brr(outweights=BRRweights);
strata Line;
cluster Vehicle;
weight Weight;
var z;
ods output Statistics = Estimate
VarianceEstimation=VarianceEstimation;
run;
```
Output 4 Estimate of Population Variance
The SURVEYMEANS Procedure
Data Summary
Number of Strata 6
Number of Clusters 12
Number of Observations 1937
Sum of Weights 143040
Variance Estimation
Method BRR
Number of Replicates 8
You can see from Output 4 that there are eight replicates and 1,937 observations.
The data set BRRweights contains the original variables from the Munisurvey data set and eight new variables named RepWgt_1 through RepWgt_8.
The following DATA step retrieves the estimated total of the variable and stores it in a macro variable named Variance. The total of the variable is equal to .
```data _null_;
set Estimate;
call symput("Variance",Z_Sum);
run;
```
The next DATA step retrieves the number of replicates and stores the value in a macro variable named R: The number of replicates is used later to construct the variable .
```data _null_;
set VarianceEstimation;
where label1="Number of Replicates";
call symput("R",cvalue1);
run;
%let R=%eval(&R);
```
### Step 4: Compute and for Replicate Samples
Before computing and , use the following DATA step to convert the data set BRRweights from wide form to long form; doing so enables you to use BY-group processing with PROC SURVEYMEANS.
```data Long(drop= RepWt_1 - RepWt_&R Z);
set BRRweights;
array num (*) RepWt_1 - RepWt_&R;
do replicate=1 to dim(num);
BRRweight=num(replicate);
output;
end;
run;
```
The data set Long has observations. There are eight copies of the original variables from the Munisurvey data set stacked on top of each other, and each copy is identified by the variable Replicate. Instead of the eight replicate weight variables, RepWgt_1 through RepWgt_8, there is now one variable, BRRweight, which is constructed by stacking the variables RepWgt_1 through RepWgt_8 on top of each other. Thus, the first 1,937 observations contain a copy of the original variables and the contents of RepWgt_1, and the variable Replicate has a value of 1. The second 1,937 observations contain a copy of the original variables and the contents of RepWgt_2, and the variable Replicate has a value of 2. The remaining observations are constructed and identified similarly.
Next, sort the data set Long by Replicate:
```proc sort data=Long out=Long;
by Replicate;
run;
```
Use PROC SURVEYMEANS to estimate the mean of Waittime by Replicate. Doing so produces the estimates of and for each replicate. The WEIGHT statement specifies that the sampling weights be contained in the variable BRRweight. The ODS OUTPUT statement saves the sample means in a SAS data set named BRRMeans and the sum of the replicate weights in a data set named BRRN.
```proc surveymeans data=Long mean;
weight BRRweight;
var Waittime;
by Replicate;
ods output Statistics = BRRMeans(keep=Replicate Mean)
Summary = BRRN;
run;
```
### Step 5: Construct the Variable
Before you can construct the variable , you must merge the data sets BRRMeans and BRRN with Long by Replicate:
```proc sort data=BRRMeans out=BRRMeans;
by Replicate;
run;
```
```data BRRN(keep=N replicate );
set BRRN(rename=(nvalue1=N));
where Label1="Sum of Weights";
run;
```
```proc sort data=BRRN out=BRRN;
by Replicate;
run;
```
```data Long;
merge Long BRRN BRRMeans;
by Replicate;
run;
```
Now construct the variable using the merged data set:
```data Long;
set Long;
z=(1/(N-1))*(Waittime-Mean)**2;
run;
```
### Step 6: Estimate the Total of for the Replicate Samples
Use PROC SURVEYMEANS to estimate the total of the variable by Replicate. The WEIGHT statement specifies that the variable BRRweight contain the sampling weights. You do not need to specify the STRATA and CLUSTER statements. The ODS OUTPUT statement saves the estimated totals in the variable BRREstimate in a SAS data set named Statistics. The estimated totals are the estimates for each replicate.
```proc surveymeans data=Long sum stacking;
weight BRRweight;
var z;
by Replicate;
ods output Statistics=Statistics(drop=Z_StdDEV
rename=(Z_Sum=BRREstimate));
run;
```
### Step 7: Construct the Variable
```data Statistics;
set Statistics;
u=(1/&R)*(BRREstimate-&Variance)**2;
run;
```
### Step 8: Estimate the Total of
Use PROC SURVEYMEANS to compute the unweighted total of . Specify the SUM option in the PROC SURVEYMEANS statement. The ODS OUTPUT statement saves the total in a variable named Variance in a SAS data set named BRRResult.
```proc surveymeans data=Statistics sum;
var u;
ods output Statistics=BRRResult(rename=(sum=Variance));
run;
```
The following DATA step computes the standard error of the estimate and the upper and lower 95% confidence limits. The confidence limits for this example are computed by using a distribution with H=6 degrees of freedom. The variable Estimate is generated and assigned the estimated value of , which is stored in the macro variable Variance. The data set is also prepared for printing.
```data BRRResult;
set BRRResult;
StdErr=sqrt(Variance);
Estimate=&Variance;
LowerCL= Estimate + StdErr*TINV(.025,&H);
UpperCL= Estimate + StdErr*TINV(.975,&H);
Variable='Waittime';
label Estimate=Population Variance Estimate
Variance=Variance of Estimate
StdErr=Standard Error of Estimate
LowerCL=Lower Confidence Limit
UpperCL=Upper Confidence Limit;
run;
```
Use the PRINT procedure to print the contents of the data set BRRResult:
```title 'Parameter Estimates';
proc print data=BRRResult label noobs;
var Variable Estimate Variance StdErr LowerCL UpperCL;
run;
title ;
```
Output 5 displays the results. The estimate of the population variance for the variable Waittime is 18.02. The variance of the estimate is 2.17, and the standard error of the estimate is 1.47. The estimated lower and upper 95% confidence limits are 14.41 and 21.63, respectively.
Output 5 Estimate of Population Variance
Parameter Estimates
Variable Population Variance
Estimate
Variance of Estimate Standard Error
of Estimate
Lower Confidence
Limit
Upper Confidence
Limit
Waittime 18.0196 2.172780 1.47404 14.4128 21.6264
## Appendix: Estimating the Finite Population Standard Deviation and Computing by Using the Delta Method
After you have an estimate of the finite population variance of a variable and a design-based estimator of the variance , you can estimate the finite population standard deviation of the variable and a design-based estimator of its variance by means of a simple transformation. Specifically, an estimator of the finite population standard deviation is
and, by application of the so-called delta method, an estimator of the variance of is
where is the derivative of with respect to evaluated at . Substituting the sample-based estimators and for and , respectively, yields
and
## Example
Consider the BRR example provided in the section Using the BRR Method to Estimate . The estimation results are stored in the data set BRRResult. To compute the finite population standard deviation, its variance, and confidence limits, perform the transformations in the following DATA step. Note that the order of the first two assignment statements is critical.
```data BRRStdDev;
set BRRResult;
Variance=(1/(4*Estimate))*Variance;
Estimate=sqrt(Estimate);
StdErr=sqrt(Variance);
LowerCL= Estimate + StdErr*TINV(.025,&H);
UpperCL= Estimate + StdErr*TINV(.975,&H);
label Estimate=Population Standard Deviation Estimate;
run;
```
Use the PRINT procedure to print the contents of the data set BRRStdDev:
```title 'Parameter Estimates';
proc print data=BRRStdDev label noobs;
var Variable Estimate Variance StdErr LowerCL UpperCL;
run;
title ;
```
Output 6 displays the results. The estimate of the population standard deviation for the variable Waittime is 4.24. The variance of the estimate is 0.03, and the standard error of the estimate is 0.17. The estimated lower and upper 95% confidence limits are 3.82 and 4.67, respectively.
Output 6 Estimate of Population Standard Deviation
Parameter Estimates
Variable Population Standard
Deviation Estimate
Variance of Estimate Standard Error
of Estimate
Lower Confidence
Limit
Upper Confidence
Limit
Waittime 4.24495 0.030145 0.17362 3.82011 4.66979
## References
Courbois, J.-Y. P. and Urquhart, N. S. (2004), “Comparison of Survey Estimates of the Finite Population Variance,” Journal of Agricultural, Biological, and Environmental Statistics, 9(2), 236–251.
Särndal, C. E., Swensson, B., and Wretman, J. (1992), Model Assisted Survey Sampling, New York: Springer-Verlag. | 10,245 | 43,308 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2019-09 | latest | en | 0.871246 |
https://www.cram.com/essay/Quartic-Polynomial-Analysis/F3JUACYLG6E45 | 1,709,626,846,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707948223038.94/warc/CC-MAIN-20240305060427-20240305090427-00064.warc.gz | 712,931,481 | 13,436 | Quartic Function: What To Do With The Sports Car
Superior Essays
In 1540, a man by the name of Lodovico Ferrari, please be aware that I don’t think his name has anything to do with the sports car, was an Italian mathematician known for discovering the solutions to quartic functions. A quartic function is a function of the form ax^4 + bx^3 +cx^2 +dx+e, where a is a nonzero, which is defined by a polynomial raised to the fourth degree, called quartic polynomial. We will probably go more in depth about these quartic polynomials soon in class. My quartic polynomial was 3x^4 -7x^3 -3x^2 +17x+10, and in this project, I was asked to analyze this polynomial.
Finding the end behavior was one of the first steps of this analysis. The end behavior refers to where the tail ends of the graph are pointing on both the
…show more content…
The zeros are where the graph crosses the X axis. These zeros are also known as the solutions or the roots. The types of zeros that functions can have are rational zeros, irrational zeros, or complex zeros. Rational zeros are just the basic integers. Irrational zeros have radicals that can’t be removed. Complex zeros have imaginary numbers. Out of these 3 types of zeros, a quartic function can only have 4 rational/irrational zeros, 2 rational/irrational zeros and 2 complex zeros, or 4 complex zeros. I was able to find how many positive and negative zeros I had using the Descartes Rule of Signs. Using the original equation, I counted from left to right, the amount of sign changes, and that gave me number of positive real zeros. I discovered 2 positive and 2 negative real zeros using this method. Now to find the negative number, the exact same process is repeated, only this time, the –X value is substituted in. The next step would ultimately be to use Interval Value Theorem to find what intervals the roots are in. However, I don’t have any sign changes in my table, which is what you would look for to determine the intervals, so I pretty much don’t have any intervals. I think that it may have to do with the fact that it is a parabola, and its specific end behavior. One tail end is going to negative infinity, so all the values are negative and the other tail end is going to positive infinity, so the values are positive. If the graph were to keep going there would never be any sign changes, so there can’t be any intervals. Since there aren’t any intervals, we can move onto the next step of using the Rational Zero Theorem. In order to use this theorem, you would take the P values which are the factors of the constant in the function, and take the q values which are the factors of the coefficient. Then, take each P value and divide it by q, and these will become the possible zeros. These possible zeros
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## » Pre University Math Help
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Math Discussion | 1,150 | 2,931 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2018-22 | latest | en | 0.664262 |
https://wisdomessays.com/centre-of-gravity-question/ | 1,719,237,720,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865383.8/warc/CC-MAIN-20240624115542-20240624145542-00573.warc.gz | 529,959,171 | 11,700 | # CENTRE OF GRAVITY QUESTION
CENTRE OF GRAVITY QUESTION
just see if you can handle it , you can assume all dimensions.
I need help in finding the Center of gravity of 3-D object.
I need to solve in 2 different ways. One of them has to be by summing the moments.
I need a 3-D Free body diagram that explains the CG location/determination.
The Pallet has 2 lifting points indicated by doted circle as shown in the pic.
Please let me know if can do and how much?
Determine the CG by summing the moments in terms of variable. then plug in any reasonable dimensions and weight. weight could be 100 lb.
Please have a FBD that represent the given drawing. Note there are 2 lifting points (the doted circles) | 170 | 708 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-26 | latest | en | 0.945583 |
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Lately I’ve been spending way too much time playing 2048 by Gabriel Cirulli. It’s a ridiculously simple (to play; not to win) puzzle game: you use the arrow keys to slide numbered tiles around on a 4 by 4 grid; tiles with the same number merge when they collide to form one tile with double the number; a 2 or 4 tile drops randomly on each turn. You “win” if you make a 2048 tile, but you can then go on to try to increase your score after that.
Given the random element there’s no guaranteed winning strategy. But some general principles tend to work for me:
I try to keep my highest numbers in the bottom row, with the highest one in the lower right corner. The point is that, say, a 512 can’t merge with anything but another 512; if there’s a 512 next to a 2, the 2 gets “starved” because you can’t “feed” it from that direction until you’ve built it up to 512 (or you move the 512 and 2 apart). On the other hand, it’s great to have a 256 next to a 512 because once you “feed” the 256 and turn it into a 512, you can then merge the two 512s right away.
Keeping the high numbers in the bottom row means I almost never use the up arrow. Lift the bottom row and you’re likely to get a 2 or 4 placed under it! Where it’ll starve, taking up space; meanwhile the high number it’s shoved up is starving low numbers around it.
I also try not to use the left arrow unless the bottom row is filled, or there are no legal right arrow or down arrow moves. Again, if you move the bottom row left, you run the risk of getting a 2 or 4 shoved in on the lower right next to your biggest number. It’s not likely, unless there are few vacant spaces in the grid, but it can happen.
If the bottom row is full (and there are no pairs to merge in it) then the left arrow is safe, of course.
So most of my action happens on the lower left. I build up a number on the left side (first or second position from the left) of the second row I can merge with the number below it. Then merge. Repeat until there’s a mergeable pair in the bottom row. Arrow right to merge them… and now things are dangerous, the bottom row’s not full and I try to avoid left arrows until a tile appears in the left column and I can drop it to the bottom. Occasionally I run out of right and down moves in that situation and have to hold my breath, hit the left arrow, and hope the new tile doesn’t show up in the bottom row. If it does it’s hard to recover the game, once you’ve built up the highest number to 256 or so. If it doesn’t, of course I always move right on the next move.
Of course it’s good to have numbers in a powers-of-two sequence orthogonally connected. Then when you get two of the lowest number next to one another you can merge them, then merge that with the next, merge that with the next, and so on, eating up the whole sequence at once.
Other than that, I pretty much play by learned instinct based on previous games. Lately if I can avoid stupid moves (like an ill-conceived up, or left with the bottom row unfilled) I find I can get to 2048 a fair fraction of the time. Maybe more than half.
An easier version of the game is here: http://siva1456.github.io/2048/ (thanks, Louis Nick) (Actually it’s easier to win, but about as hard to play once you keep going.) Another version, very doge, such hard, is here: http://doge2048.com/
I’m going to brag by putting up an image of my best game. Feel free to avoid looking at it if it’ll just irritate you. Feel free not to rub my face in it if you’ve done better.
Update: Of course my best has gotten better this past week, and the other day I astonished myself with this result:
I’ve now retired from 2048. No really. | 912 | 3,666 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2019-04 | longest | en | 0.909561 |
xajenarelax.agronumericus.com | 1,548,151,630,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583835626.56/warc/CC-MAIN-20190122095409-20190122121409-00020.warc.gz | 1,008,785,462 | 6,145 | Write a system of equations to represent the situation jersey
The maneuver forced the Brystol to take evasive action, a move to place more of the mass of the Earth directly behind it in relation to the incoming Jovan forces.
Well, actually the report says megawatts so obviously I made a mistake somewhere. During the Cold Warproxy wars were standard operating procedure. But does Man have any "right" to spread through the universe?
Asymmetric warfare is a conflict between two populations of drastically different levels of military capability or size. Notes in the text emphasize unusual or special procedures or conditions.
For example, when using the HERO Safe Separation Distance Calculator, it is important to use average power and not peak power as the input parameter for calculating a safe separation distance.
It likewise may not exist among developed post-industrial societies, its place taken by forms of nastiness we today can only vaguely guess at. Solution A straight line can be constructed through the three transient points to yield a slope of 1. Moore listened to the communication between Luna Authority and the Supreme Commander of the Jovan fleet broadcast on an open channel.
Many thanks are also due to my students and colleagues at the University of Maryland, in the US and around the world, and in particular in Sweden and Germany, for their collaboration, constructive criticism and influence through the years.
The foregoing situation should, ideally, be worked out as trade rather than war, but the parties involved will have to get into intelligible conversation first.
If the invaders want real estate, it is counterproductive to dust the planet with enough radioactive material to render it uninhabitable for the next ten thousand years.
This is equal to 8.
At least during the initial portion of galactic enculturation, interstellar lebensraum cannot be ruled out. The Jovians might reasonably resent such devices plowing through their equivalents of orchards, gardens, and flocks, not to mention their families and themselves. It was about as stupid a conflict as you're likely to find, during which the real principals licked their lips and chuckled while well-meaning idealists wrecked their own societies in pursuit of unobtainable goals by improper means.
But efforts to determine what causes seas to rise are marred by poor data and disagreements about methodology. The methods of storing this energy are dictated by the degrees of freedom of the particle itself energy modes. The high-strangeness category carries another serious implication.
Any shift from territorial states to other forms of power-political relations would certainly render war-as-we-know-it a doubtful proposition — we have contemporary evidence of the problems that very formidable militaries face when they don't know who the hell the enemy is.
This volume provides technical guidance to assist commanding officers in carrying out their responsibilities for safety from a radio frequency hazards standpoint. Still, in the excitement of battle, the Lunan forces panicked in the face of the onslaught of the carrier Saratoga barreling in on a full-frontal attack, decelerating engines of over fifty million tons of thrust burning like full-fledged supernovas.
They also sent what appears to have been a small tugboat to deal with Vaduz. Without them and my loving wife and family, I would not be here today.
Proxy War A related concept is " Proxy War. The idea is "Let's you and him fight.
Not a simple nothingness; rather, Ishrail explained, an unfathomable interplay of forces, fields and planes. Since a gas fills any container in which it is placed, volume is an extensive property. But Jon not only knew the real reason for the challenge, he had anticipated it. Since gamma rays are rays, not particles, they have that pesky exponential attenuation with shielding.
They merely informed their vassals that they had become the property of whatever Vilk it happened to be, and levied tribute accordingly.
Luna Nation chose Earth as a backdrop to their defensive position, forcing the Jovan forces to attack at a shallow angle across the interference provided by the face of a full-sized planet. If each one were the equivalent of a missile submarine or even a Boeingeven a very low batting average might give a Jovian victory.
There, I worked with a student of Bryson, Walter Denham. Make covert contact or send your agents into a couple of the most powerful nations and encourage them to attack each other.
Classical mechanics assumes that time has a constant rate of passage that is independent of the state of motion of an observeror indeed of anything external. We chose DA as an analytic tool because of its ubiquitous past successes in solving complex problems of physics, engineering, mathematical biology, and biophysics [ 16 - 21 ].
The test consists of a series of flow rates. The basic deliverability test method that uses all stabilized data is the flow-after-flow test. What then is the real cause of sea-level rise of 1 to 2 millimeters a year?
The Navy categorizes all ordnance in terms of the relative immunity.Writing Equations Write an equation to represent the situation. 2. FAMILY Katie is twice as old as her sister Mara. The sum of their ages is Write a one-variable equation to represent the situation. 3. GEOMETRY The formula F + V = E + 2 shows the relationship between.
– Algebra 1 Midterm Review 1. The expression 13x + 5 represents the number of marbles you have after purchasing 13 c. What does the slope of the graph above represent in this situation? a. Write a system of equations that can be used to.
The Bellman Award is given for distinguished career contributions to the theory or application of automatic control. It is the highest recognition of professional achievement for US control systems engineers and scientists.
The Real Number System (N-RN) Represent and solve equations and inequalities graphically (10, 11, 12) Interpreting Functions (F-IF) Understand the concept of a function and use function notation (1, 2, 3) Common Core State Standards – Mathematics Content Focus by Course.
Pdf Pass Homework Practice and Problem-Solving Practice Workbook i_0iv_G4_TP_indd i0i_0iv_G4_TP_indd i 44/7/08 AM/7/08 AM. equations Represent and What are extraneous solutions?
What is the difference between solving equations and inequalities? How do you solve quadratic equations? What is a solution to a system of equations? Why is the October-June (30 days) Cooperative learning, open ended CURRICULUM GUIDE DISCIPLINE: Algebra 1 CP GRADE LEVEL:
Write a system of equations to represent the situation jersey
Rated 3/5 based on 76 review | 1,371 | 6,712 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2019-04 | longest | en | 0.948663 |
https://quantumcomputing.stackexchange.com/questions/9027/how-to-create-a-q-operation-to-generate-a-random-number-from-1-to-max/9069 | 1,713,325,676,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817128.7/warc/CC-MAIN-20240417013540-20240417043540-00889.warc.gz | 460,315,828 | 40,773 | # How to create a q# operation to generate a random number from 1 to max?
I have the following problem: I want to create a q# operation for generating a random integer from 1 to max and return the generated number. What algorithm do I need? What does the q# code look like? I am very new to quantum computing so sorry if this is a simple question!
• Do you need to implement a quantum operation for doing this, or will a classical one do? Q# has a built-in operation RandomInt which does exactly that: docs.microsoft.com/qsharp/api/qsharp/… Dec 2, 2019 at 17:16
With Q# you can generate random numbers in two ways:
1. Using a classical pseudorandom number generator, which is exactly the same that a classical language like Python does when you use the library random. As Mariia Mykhailova says in the comments, Q# has a built-in operation RandomInt that does exactly this: RandomInt
2. Using a quantum operation that uses measurements on qubits to extract random bits to build a random number. Unfortunately, right now Q# is not yet compatible with any real quantum hardware, so to implement this method you will need to use a classical simulator of a quantum computer that ultimately will implement a classical pseudorandom function like in (1).
However, since you're new in quantum computing and you want to learn, choosing (2), even if at the end is just a convoluted version of (1), makes sense for pedagogical purposes. So let's see how to implement (2) using Q# and Python.
• a) First we need to create a Q# file (Operation.qs) to obtain random bits from measuring qubits. To see how you have a nice tutorial here. Your Q# code should look like this:
namespace Quantum { open Microsoft.Quantum.Intrinsic; operation QuantumRandomNumberGenerator() : Result { using(q = Qubit()) { H(q); let r = M(q); Reset(q); return r; } } }
• b) Secondly we need to create a Python file (in the same folder than the .qs file) that is going to build an integer by calling the quantum operation defined above several times. The code is the following (comments inside the code):
Text code:
import qsharp
from Quantum import QuantumRandomNumberGenerator
bit_string = []
max = 50
for i in range(0, max.bit_length()):
bit_string.append(QuantumRandomNumberGenerator.simulate())
number = int("".join(str(x) for x in bit_string), 2)
print(number)
Now if you execute this Python code in the terminal (supposing you have all the dependencies correctly installed) the following message should appear in the command promt:
Preparing Q# environment...
33
My random number was 33. | 592 | 2,571 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2024-18 | latest | en | 0.883434 |
https://electronics.stackexchange.com/questions/605047/how-to-find-bldc-motor-power-and-torque-rating-with-torque-constants-and-other-g | 1,713,709,127,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817780.88/warc/CC-MAIN-20240421132819-20240421162819-00739.warc.gz | 199,972,191 | 38,250 | # How to find BLDC motor power and torque rating with torque constants and other given parameters
I am currently working on simulating a solar pv fed bldc motor in simulink software. I have the datasheet values of bldc motor from a iet journal as Number of poles, P = 6; rated speed, Nrated = 3000 rpm; stator resistance, Rs = 0.41 Ω; stator inductance, Ls = 2.13 mH; voltage constant, Ke = 78 VL-L/krpm.
But in Simulink I took the PMSM motor with trapezoidal back emf but I have option only to input the a) pole pairs b) Rs & Ls (phase to phase values), c) Voltage constant or torque constant or flux linkage as given in matlab documentation. Kindly check the link. I have no option to enter the rated power and speed of the bldc motor in simulation model.
1) In this case, if I enter the parameters that can be inputted to the model, then how we can find that motor with that given voltage constant (say 78 VL-L/krpm), Rs, Ls, pole pairs is a machine with rated speed of 3000 rpm?
2) Also, how to find the rated torque for this machine if the frequency of the input voltage to bldc motor from inverter is changed periodically based on gate pulses from hysteresis current controller?
I don't know much about the bldc motor & simulink so please go easy on me. If there is any specific relation that connects these terms kindly let me know.
• From the voltage constant and the speed, you can work out the rated voltage. Or from the constant and the voltage, you can work out the speed. From Power = V * I = Torque * Speed you can find a very simple relationship between the voltage constant and the torque constant : from the latter and the current you can derive the torque ( or from that and the torque, you can derive the current).
– user16324
Jan 20, 2022 at 14:52 | 447 | 1,773 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2024-18 | latest | en | 0.888468 |
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## rebecca1233 2 years ago The hypotenuse of a 30°- 60°- 90° right triangle measures 22 cm. What are the lengths of the longer leg and shorter leg? @tcarroll010
• This Question is Closed
1. rebecca1233
@tcarroll010
2. cwest50813
short leg =11 longer leg= 11$\sqrt{3}$
3. rebecca1233
huh
4. tcarroll010
The shorter leg of a 3-60-90 degree right triangle is half the hypotenuse, so the shorter leg is 11. Let "x" be the length of the longer leg, then: x^2 + 11^2 = 22^2 x^2 = 22^2 - 11^2
5. cwest50813
you multiply the 11 and the square root of three together to get longer leg
6. rebecca1233
x^2 = 22^2 - 11^2 gives u 19.05?
7. tcarroll010
yes.
8. rebecca1233
so thats the longer leg?
9. tcarroll010
yes.
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Privacy Policy | 298 | 890 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2015-40 | longest | en | 0.784638 |
https://livingscented.com/how-many-cups-of-chocolate-chips-in-a-bag/ | 1,660,768,709,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573104.24/warc/CC-MAIN-20220817183340-20220817213340-00432.warc.gz | 355,052,903 | 16,612 | # How Many Cups Of Chocolate Chips In A Bag
Assuming you are talking about a standard sized bag of chocolate chips, there are generally 2 cups in a bag. This is a rough estimate, as the size of a cup can vary and the amount of chips in a bag can also vary slightly. However, 2 cups is a good estimate of how many chocolate chips are in a bag.
Chocolate chips are a baking staple that can be used in a variety of recipes. Whether you are making cookies, brownies, or even pancakes, chocolate chips are a great addition. If you are looking for a quick and easy recipe that uses chocolate chips, check out this recipe for chocolate chip cookies.
## Math 1530 Section 7.2 HW #23
Assuming you are talking about a standard sized bag of chocolate chips, there are approximately 2 cups in a bag. This is a rough estimate, as the amount can vary slightly depending on the brand and size of the bag.
## How many cups of chocolate chips in a 12 oz bag
A 12 oz bag of chocolate chips typically contains 2 cups of chocolate chips.
## How many cups of chocolate chips in a 10oz bag
A 10oz bag of chocolate chips typically contains 2 cups of chocolate chips.
Read Also: Why Is Bacon So Salty
## How many chocolate chips in a bag
How many chocolate chips in a bag? This is a question that I get asked a lot! The answer really depends on the size of the bag and the size of the chocolate chips.
A bag of mini chocolate chips will have more chocolate chips than a bag of regular size chocolate chips. So, if you are looking for a quick answer, it really depends on the size of the bag and the size of the chocolate chips.
## How many chocolate chips are in a bag of nestle
A bag of Nestle chocolate chips contains 12 ounces, or 340 grams. There are approximately 2,000 chips in a bag.
## 1 cup of chocolate chips in grams
A cup of chocolate chips weighs approximately 120 grams. This varies depending on the size and density of the chips, but is generally around this amount.
## How many cups of chocolate chips are in a bag
A bag of chocolate chips typically contains between 2 and 4 cups. This varies depending on the brand and size of the bag.
## However, a good rule of thumb is that there are approximately 2 cups of chocolate chips in a standard size bag
Chocolate chips are a key ingredient in many desserts, from cookies to cakes to brownies. But how many chocolate chips are in a standard size bag? A good rule of thumb is that there are approximately 2 cups of chocolate chips in a standard size bag.
This can vary slightly depending on the brand of chocolate chips, but 2 cups is a good general estimate. Keep in mind that when a recipe calls for a specific amount of chocolate chips, like 1 cup or 2 cups, it is referring to the standard size bag. So if a recipe calls for 1 cup of chocolate chips, you would need to use half of a standard size bag.
Read Also: Is The Ninja Dishwasher Safe
Now that you know how many chocolate chips are in a standard size bag, you can be sure to use the right amount when baking your favorite desserts!
## Conclusion
In a standard bag of chocolate chips, there are 2 cups of chocolate chips.
John Davis
John Davis is the founder of this site, Livings Cented. In his professional life, he’s a real-estate businessman. Besides that, he’s a hobbyist blogger and research writer. John loves to research the things he deals with in his everyday life and share his findings with people. He created Livings Cented to assist people who want to organize their home with all the modern furniture, electronics, home security, etc. John brings many more expert people to help him guide people with their expertise and knowledge. | 817 | 3,675 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2022-33 | latest | en | 0.944817 |
https://mpra.ub.uni-muenchen.de/9446/ | 1,708,575,852,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473690.28/warc/CC-MAIN-20240222030017-20240222060017-00517.warc.gz | 431,669,865 | 25,653 | # Duration Models: Specification, Identification, and Multiple Durations
Van den Berg, Gerard J. (2000): Duration Models: Specification, Identification, and Multiple Durations.
Since the early 1980s, the econometric analysis of duration variables has become widespread. This chapter provides an overview of duration analysis, with an emphasis on the specification and identification of duration models, and with special attention to models for multiple durations. Most of the chapter deals with so-called reduced-form duration models, notably the popular Mixed Proportional Hazard (MPH) model and its multivariate extensions. The MPH model is often used to describe the relation between the empirical exit rate and background variables'' in a concise way. However, since the applications usually interpret the results in terms of some economic-theoretical model, we examine to what extent the deep structural parameters of some important theoretical models can be related to reduced-form parameters. We subsequently examine the specification and identification of the MPH model in great detail, we provide intuition on what drives identification, and we infer to what extent biases may occur because of misspecifications. This examination is carried out separately for the case of single-spell data and the case of multi-spell data. We also compare different functional forms for the unobserved heterogeneity distribution. Next, we examine models for multiple durations. In the applied econometric literature on the estimation of multiple-duration models, the range of different models is actually not very large. Typically, the models allow for dependence between the duration variables by way of their unobserved determinants, with each single duration following its own MPH model. In addition to this, the model may allow for an interesting causal'' effect of one duration on the other, as motivated by an underlying economic theory. For all these models we examine the conditions for identification. Some of these are intimately linked to particular estimation strategies. The multiple-duration model where the marginal duration distributions each satisfy an MPH specification, and the durations can only be dependent by way of their unobserved determinants, is called the Multivariate Mixed Proportional Hazard (MMPH) model. For this model, we address the issue of the dimensionality of the heterogeneity distribution and we compare the flexibility of different parametric heterogeneity distributions. On a number of occasions, we incorporate recent insights from the biostatistical literature on duration analysis, and we contrast points of view in this literature to those in the econometric literature. Finally, throughout the chapter, we discuss the importance of the possible collection of additional data. | 510 | 2,818 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-10 | latest | en | 0.924764 |
http://mathoverflow.net/revisions/71190/list | 1,369,414,483,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368704818711/warc/CC-MAIN-20130516114658-00015-ip-10-60-113-184.ec2.internal.warc.gz | 167,185,035 | 6,571 | 5 deleted 270 characters in body
Denote by $\Psi : k^n\to k^n$ the map of affine spaces corresponding to $\Phi$, and without loss of generality assume $\Psi(0) = 0$. Putting $M = (x_1,\ldots,x_n)$ and $N = (y_1,\ldots,y_n)$, this means that $\Phi^{-1}(N) = M$, so $\Phi(M) = N$ since $\Phi$ is surjective. We then get an induced map $\Phi_\ast:M/M^2 \Phi_a:M^a/M^{a+1} \to N/N^2$N^{a}/N^{a+1}$for any$a\geq 1$. Here both$M/M^2$M^a/M^{a+1}$ and $N/N^2$ N^{a}/N^{a+1}$are$n$-dimensional$k$-vector spaces of the same dimension, and$\Phi_\ast$\Phi_a$ is clearly thus an isomorphism since it is clearly surjective. But now if $\Phi(f) = 0$ for some $f$, then $\Phi(f) \in N$ and hence $f\in M$. We then have Choose $\Phi_\ast(f) = 0$, a contradiction. Thus a$so that$\Phi$is injective. Geometrically, saying f$ lies in $\Phi$ is surjective means M^a$but not in$\Psi$is M^{a+1}$ (such an immersion, and so its differential $a$ clearly exists: it is an isomorphism. But if the degree of the lowest degree homogeneous piece of $\Phi$ fails to be injective, f$). We then$\Psi$maps$k^n$onto a lower dimensional subvariety of have$k^n$, \Phi_a(f) = 0$ and so its differential isn't an isomorphism.
I feel obliged to add that we are somehow still using "dimension theory" here$f\notin M^{a+1}$, in the sense contradicting that the dimension of affine space is encoded in the dimension of its tangent space. However, knowledge of chains of primes etc. $\Phi_a$ is clearly overkillan isomorphism.
4 edited body
Denote by $\Psi : k^n\to k^n$ the map of affine spaces corresponding to $\Phi$, and without loss of generality assume $\Psi(0) = 0$. Putting $M = (x_1,\ldots,x_n)$ and $N = (y_1,\ldots,y_n)$, this means that $\Phi^{-1}(N) = M$, so $\Phi(M) = N$ since $\Phi$ is surjective. We then get an induced map $\Phi_\ast:M/M^2 \to N/N^2$. Here both $M/M^2$ and $N/N^2$ are $n$-dimensional $k$-vector spaces, and $\Phi_\ast$ is clearly an isomorphism. But now if $\Phi(f) = 0$ for some $f$, then $\Phi(f) \in N$ and hence $f\in M$. We then have $\Phi^\ast(f) \Phi_\ast(f) = 0$, a contradiction. Thus $\Phi$ is injective.
Geometrically, saying $\Phi$ is surjective means $\Psi$ is an immersion, and so its differential is an isomorphism. But if $\Phi$ fails to be injective, then $\Psi$ maps $k^n$ onto a lower dimensional subvariety of $k^n$, and so its differential isn't an isomorphism.
I feel obliged to add that we are somehow still using "dimension theory" here, in the sense that the dimension of affine space is encoded in the dimension of its tangent space. However, knowledge of chains of primes etc. is clearly overkill.
3 added 246 characters in body
Denote by $\Psi : k^n\to k^n$ the map of affine spaces corresponding to $\Phi$, and without loss of generality assume $\Psi(0) = 0$. Putting $M = (x_1,\ldots,x_n)$ and $N = (y_1,\ldots,y_n)$, this means that $\Phi^{-1}(N) = M$, so $\Phi(M) = N$ since $\Phi$ is surjective. We then get an induced map $\Phi_\ast:M/M^2 \to N/N^2$. Here both $M/M^2$ and $N/N^2$ are $n$-dimensional $k$-vector spaces, and $\Phi_\ast$ is clearly an isomorphism. But now if $\Phi(f) = 0$ for some $f$, then $\Phi(f) \in N$ and hence $f\in M$. We then have $\Phi^\ast(f) = 0$, a contradiction. Thus $\Phi$ is injective.
Geometrically, saying $\Phi$ is surjective means $\Psi$ is an immersion, and so its differential is an isomorphism. But if $\Phi$ fails to be injective, then $\Psi$ maps $k^n$ onto a lower dimensional subvariety of $k^n$, and so its differential isn't an isomorphism.
I feel obliged to add that we are somehow still using "dimension theory" here, in the sense that the dimension of affine space is encoded in the dimension of its tangent space. However, knowledge of chains of primes etc. is clearly overkill.
2 added 273 characters in body; edited body
1 | 1,188 | 3,811 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2013-20 | latest | en | 0.823567 |
https://juegosdelogica.com/index.php/en/mathematical-puzzles/23-solution-riddles/233-solution-puzzle-battle | 1,591,235,328,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347436828.65/warc/CC-MAIN-20200604001115-20200604031115-00325.warc.gz | 398,587,589 | 10,948 | # Puzzles, situation puzzles and recreational mathematics.
## Neuron juice.
The important thing is not to stop questioning. (Albert Einstein.)
#### Riddle.
21.- In an extraordinary battle, at least 70 % of the combatants lost an eye ; 75% one ear , at least 80 % lost a hand and a leg 85 %.
How many , at least they lost the four bodies ?
#### Solution
At least 45% lost the eye and ear.
At least 65% lost his hand and leg.
At least 10% lost all four bodies . | 123 | 468 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2020-24 | latest | en | 0.885851 |
https://www.geeksforgeeks.org/python-group-list-elements-based-on-frequency/?ref=ml_lbp | 1,702,329,150,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679518883.99/warc/CC-MAIN-20231211210408-20231212000408-00179.warc.gz | 845,537,619 | 52,074 | # Python | Group list elements based on frequency
Given a list of elements, write a Python program to group list elements and their respective frequency within a tuple.
Examples:
```Input : [1, 3, 4, 4, 1, 5, 3, 1]
Output : [(1, 3), (3, 2), (4, 2), (5, 1)]
Input : ['x', 'a', 'x', 'y', 'a', 'x']
Output : [('x', 3), ('a', 2), ('y', 1)]```
Method #1: List comprehension We can use list comprehension to form tuples of each element and the count of its occurrence and store it in ‘res’, but that will contain the duplicate first element. Thus, to remove the duplicate first element, we use OrderedDict(res).items().
## Python3
`# Python3 program to Grouping list ` `# elements based on frequency` `from` `collections ``import` `OrderedDict ` `def` `group_list(lst):` ` ` ` ``res ``=` `[(el, lst.count(el)) ``for` `el ``in` `lst]` ` ``return` `list``(OrderedDict(res).items())` ` ` `# Driver code` `lst ``=` `[``1``, ``3``, ``4``, ``4``, ``1``, ``5``, ``3``, ``1``]` `print``(group_list(lst))`
Output:
`[(1, 3), (3, 2), (4, 2), (5, 1)]`
Time Complexity: O(n^2) as we are using two for loops to count the frequency of each element in the list and store it in a new list.
Auxiliary Space: O(n) as we are using an additional list to store the frequency of each element in the original list.
Method #2: Using collections.Counter() collections.Counter() provides two direct methods keys() and values() that provides the elements and its occurrences. At last, zip them together using Python zip() method.
## Python3
`# Python3 program to Grouping list ` `# elements based on frequency` `from` `collections ``import` `Counter` `def` `group_list(lst):` ` ` ` ``return` `list``(``zip``(Counter(lst).keys(), Counter(lst).values()))` ` ` `# Driver code` `lst ``=` `[``1``, ``3``, ``4``, ``4``, ``1``, ``5``, ``3``, ``1``]` `print``(group_list(lst))`
Output:
`[(1, 3), (3, 2), (4, 2), (5, 1)]`
The time complexity of the group_list() function is O(n), where n is the length of the input list.
The space complexity of the group_list() function is O(k), where k is the number of unique elements in the input list.
Method #3: Using itertools.groupby()
This method uses the itertools.groupby() function to group the elements of the list by their value, and then uses a list comprehension to create a list of tuples containing each element and the length of its group (i.e. its frequency).
## Python3
`from` `itertools ``import` `groupby` `def` `group_list(lst):` ` ``return` `[(el, ``len``(``list``(group))) ``for` `el, group ``in` `groupby(``sorted``(lst))]` `lst ``=` `[``1``, ``3``, ``4``, ``4``, ``1``, ``5``, ``3``, ``1``]` `print``(group_list(lst))` `#This code is contributed by Edula Vinay Kumar Reddy`
Output
`[(1, 3), (3, 2), (4, 2), (5, 1)]`
Time complexity: O(nlogn)
Auxiliary Space: O(n)
Method #4: Using dictionary
We can also solve the problem by using a dictionary to keep track of the frequency of each element. We can iterate through the list and for each element, we check if it’s already in the dictionary. If it is, we increment its value by 1, otherwise, we add it to the dictionary with a value of 1. Finally, we can create a list of tuples from the dictionary’s items.
Step-by-step approach:
• Initialize an empty dictionary called freq_dict to keep track of the frequency of each element in the list.
• Iterate through the elements in the list
• For each element, check if it’s already in freq_dict.
• If it is, increment its value by 1.
• If it’s not, add it to the dictionary with a value of 1.
• Create a list of tuples from the dictionary’s items using the items() method.
• Sort the list of tuples in descending order based on the frequency of the elements.
• Return the sorted list of tuples.
Below is the implementation of the above approach:
## Python3
`def` `group_list(lst):` ` ``freq_dict ``=` `{}` ` ``for` `el ``in` `lst:` ` ``if` `el ``in` `freq_dict:` ` ``freq_dict[el] ``+``=` `1` ` ``else``:` ` ``freq_dict[el] ``=` `1` ` ``res ``=` `list``(freq_dict.items())` ` ``res.sort(key``=``lambda` `x: x[``1``], reverse``=``True``)` ` ``return` `res` `lst ``=` `[``1``, ``3``, ``4``, ``4``, ``1``, ``5``, ``3``, ``1``]` `print``(group_list(lst))`
Output
`[(1, 3), (3, 2), (4, 2), (5, 1)]`
Time complexity: O(n*logn) due to the sorting step. The dictionary operations take O(1) time.
Auxiliary space: O(n) to store the dictionary.
Method #5: Using set() and count()
Steps:
1. Define a function called “group_list” that takes a list “lst” as input.
2. Create an empty list called “result” to store the grouped elements.
3. Create a set of unique elements in the input list “lst” using the “set” function.
4. Use a for loop to iterate over each element in the set.
5. Use the “count” method to count the number of occurrences of the current element in the input list “lst”.
6. Append a tuple of the current element and its frequency to the “result” list using the “append” method.
7. Once all the elements in the set have been processed, return the “result” list.
8. Define the input list “lst”.
9. Call the “group_list” function with the “lst” argument and store the result in “grouped_list”.
10. Print the “grouped_list” using the “print” statement.
## Python3
`def` `group_list(lst):` ` ``result ``=` `[]` ` ``unique_elements ``=` `set``(lst)` ` ``for` `ele ``in` `unique_elements:` ` ``frequency ``=` `lst.count(ele)` ` ``result.append((ele, frequency))` ` ``return` `result` `# Driver code` `lst ``=` `[``1``, ``3``, ``4``, ``4``, ``1``, ``5``, ``3``, ``1``]` `grouped_list ``=` `group_list(lst)` `print``(grouped_list)`
Output
`[(1, 3), (3, 2), (4, 2), (5, 1)]`
Time complexity: O(n^2) where n is the length of the input list “lst” as we need to count the occurrences of each unique element in the list.
Auxiliary space: O(n) as we need to store the frequencies of all unique elements in the input list in the result list.
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Next | 1,868 | 6,420 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2023-50 | latest | en | 0.654457 |
https://www.askmattrab.com/notes/1056-elementary-properties-of-a-group | 1,660,478,730,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572033.91/warc/CC-MAIN-20220814113403-20220814143403-00134.warc.gz | 599,732,283 | 8,754 | # Elementary Properties of a Group
Elementary Properties of a Group
Theorem I: The identity element in a group (G,౦) is unique.
Proof:
Let e and e’ be two identity elements in a group (G,౦), if possible.
Then,
When e is the identity element,
e ౦ e’ = e’ = e’ ౦ e
When e’ is the identity element,
e ౦ e’ = e = e’ ౦ e
∵ e ౦ e’ = e’ ౦ e
∴ e = e’
Hence, an identity element is unique.
Theorem II:
Each element in a group (G,౦) has a unique inverse.
Proof:
Let ‘a’ be an element of group (G,౦) with an identity element ‘e’.
Let b and c be two inverses of ‘a’, if possible.
Then,
a ౦ b = e = b ౦ a
a ౦ c = e = c ౦ a
Now,
c = c ౦ e [ Existence of identity element, c = c ౦ e ]
= c ౦ (a ౦ b) [ Inverse Law, a ౦ b = e ]
= (c ౦a) ౦ b [ Associative Law ]
= e ౦ b [ Inverse Law, c ౦ a = e ]
∴ c = b
Hence, this shows that each element in a group has a unique inverse.
Theorem III:
Cancellation Law:
If a, b, c are the elements of a group (G,౦) and if a ౦ b = a ౦ c, then b=c
Also if b౦a = c౦a, them b=c
Proof:
For a ∈ G, there exists an inverse element.
As a ∈ (G,౦), a has an inverse element a-1 such that,
a ౦ a-1 = a-1 ౦ a = e [here, e is the identity element]
We have,
a ౦ b = a ౦ c
Operating both sides by a-1 on the left, we get,
a-1 ౦ (a ౦ b) = a-1 ౦ (a ౦ c)
or, (a-1 ౦ a) ౦ b = (a-1 ౦ a) ౦ c [ Associative Law ]
or, e ౦ b = e ౦ c [ Inverse Law, a-1 ౦ a = e ]
∴ b = c [ Identity Law, e * b = b and e * c = c ]
Similarly, we have,
b ౦ a = c ౦ a
Operating both sides by a-1 on the right, we get,
(b ౦ a) ౦ a-1 = (c ౦ a) ౦ a-1
or, b ౦ (a ౦ a=1) = c ౦ (a ౦ a-1) [ Associative Law ]
or, b ౦ e = c ౦ e [ Inverse Law, a ౦ a-1 = e ]
∴ b = c [ Identity Law, b ౦ e = b and c ౦ e = c ]
Hence, proved.
Theorem IV:
If a, b ∈ (G, ౦), then
(i) (a ౦ b)-1 = b-1 ౦ a-1
(ii) (a-1)-1 = a
Proof:
(i) To prove: (a ౦ b)-1 = b-1 ౦ a-1
If the inverse of (a ౦ b) is equal to (b-1 ౦ a-1), then (a ౦ b) ౦ (b-1 ౦ a-1) must be equal to the identity element [ ∵ a ౦ a-1 = e ]. In this case, the identity element is e.
So, to prove: (a ౦ b) ౦ (b-1 ౦ a-1) = e
Taking, (a ౦ b) ౦ (b-1 ౦ a-1)
= a ౦ (b ౦ b-1) ౦ a-1 [ Associative Law ]
= a ౦ e ౦ a-1 [ Inverse Law, b ౦ b-1 = e ]
= (a ౦ e) ౦ a-1
= a ౦ a-1 [ Identity Law, a ౦ e = a ]
=e
∴ (a ౦ b) ౦ (b-1 ౦ a-1) = e
∴ a ౦ b is the inverse of b-1 ౦ a-1
i.e. (a ౦ b)-1 = b-1 ౦ a-1
(ii) To prove: (a-1)-1 = a
We know that, a ౦ a-1 = e [Inverse Law, since a-1 is the inverse of a]
or, a-1 ౦ a = e
Operating both sides by (a-1)-1 on the right, we get,
or, (a-1)-1 a-1 ౦ a = (a-1)-1 e
or, {(a-1)-1 a-1} ౦ a = (a-1)-1 e
or, e ౦ a = (a-1)-1 e [ Inverse Law, (a-1)-1 a-1 = e ]
∴ a = (a-1)-1 [ Identity Law, e ౦ a = a and (a-1)-1 e = (a-1)-1 ]
NEXT METHOD:
To prove: (a-1)-1 = a
We know that, a ౦ a-1 = e = a-1 ౦ a [Inverse Law, since a-1 is the inverse of a]
Now, RHS = a
= a ౦ e [ Identity Law, a = a ౦ e ]
= a ౦ a-1 ౦ (a-1)-1 [ Inverse Law, a-1 ౦ (a-1)-1 = e ]
= {a ౦ a-1} ౦ (a-1)-1
= e ౦ (a-1)-1 [ Inverse Law, a ౦ a-1 = e ]
= (a-1)-1 [ Identity Law, e ౦ (a-1)-1 = (a-1)-1 ]
= LHS
Hence, proved.
Theorem V:
If a and b are the elements of a group (G,౦), then
a ౦ x = b and x ౦ a = b
have unique solutions in (G,౦)
Proof:
We have, a ౦ x = b
Operating both sides by a-1 on the left, we get,
(a-1 ౦ a) ౦ x = a-1 ౦ b
or, e ౦ x = a-1 ౦ b [ Inverse Law, a-1 ౦ a = e ]
∴ x = a-1 ౦ b [ Identity Law, e ౦ x = x ]
This is the required solution.
To show uniqueness,
Let x1 and x2 be two solutions of a ౦ x = b, then
a ౦ x1 = b
a ౦ x2 = b
From results, a ౦ x1 = a ౦ x2
Operating both sides by a-1 on the left, we get,
a-1 ౦ (a ౦ x1) = a-1 ౦ (a ౦ x2
or, (a-1 ౦ a) ౦ x1 = (a-1 ౦ a) ౦ x2 [ Associative Law ]
or, e ౦ x1 = e ౦ x2 [ Inverse Law, a-1 ౦ a = e ]
∴ x1 = x2 [ Identity Law, x1 ౦ e = x1 and x2 ౦ e = x2 ]
NEXT METHOD:
For x ౦ a = b
Operating both sides by a-1 on the right, we get,
x ౦ (a ౦ a-1) = b ౦ a-1
or, x ౦ e = b ౦ a-1 [ Inverse Law, a ౦ a-1 = e ]
∴ x = b ౦ a-1 [ Identity Law, x ౦ e = x ]
This is the required solution.
To show uniqueness,
Let x1 and x2 be two solutions of x ౦ a = b, then
x1 ౦ a = b
x2 ౦ a = b
From results, x1 ౦ a = x2 ౦ a
Operating both sides by a-1 on the right, we get,
(x1 ౦ a) ౦ a-1 = (x2 ౦ a) ౦ a-1
or, x1 ౦ (a ౦ a=1) = x2 ౦ (a ౦ a-1) [ Associative Law ]
or, x1 ౦ e = x2 ౦ e [ Inverse Law, a ౦ a-1 = e ]
∴ x1 = x2 [ Identity Law, x1 ౦ e = x1 and x2 ౦ e = x2 ]
| 2,270 | 5,180 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2022-33 | latest | en | 0.580702 |
https://tr.tradingview.com/script/ufrYbFVK/ | 1,618,069,119,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038057142.4/warc/CC-MAIN-20210410134715-20210410164715-00336.warc.gz | 669,716,820 | 96,478 | 471 görüntülenme
This indicator basicly usind ADX ( Average Directional Index )
ADX can show us how trend is strong
ADX below 20: the market is currently not trending
ADX crosses above 20: signifies that a new trend is emerging. Traders may start placing sell or buy orders in the direction of the price movement.
ADX between 20 and 40: When the ADX is growing between 20 and 40 it is considered as a confirmation of an emerging trend. Traders should use this opportunity to buy or short sell in the trend's direction.
ADX above 40: the trend is very strong.
ADX crosses 50: the trend is extremely strong.
ADX crosses 70: a very rare occasion, which is called a “Power Trend.”
If we use ADX with DI+ and DI+ indactor can tell us to buy.
How can we calculate this all?
Directional Movement (DI) is defined as the largest part of the current period’s price range that lies outside the previous period’s price range. For each period calculate:
+DI = positive or plus DI = High - Previous High
-DI = negative or minus DI = Previous Low - Low
The smaller of the two values is reset to zero, i.e., if +DI > -DI , then -DI = 0. On an inside bar (a lower high and higher low), both +DI and -DI are negative values, so both get reset to zero as there was no directional movement for that period.
The True Range ( TR ) is calculated for each period, where:
TR = Max of ( High - Low ), ( High -PreviousClose ), ( PreviousClose - Low )
The +DI , -DI and TR are each accumulated and smoothed using a custom smoothing method proposed by Wilder. For an n period smoothing, 1/n of each period’s value is added to the total each period, similar to an exponential smoothing:
+DIt = (+DIt-1 - (+DIt-1 / n)) + (+DIt)
-DIt = (-DIt-1 - (-DIt-1 / n)) + (-DIt)
TRt = (TRt-1 - (TRt-1 / n)) + ( TRt )
Compute the positive/negative Directional Indexes, +DI and -DI , as a percentage of the True Range:
+DI = ( +DI / TR ) * 100
-DI = ( -DI / TR ) * 100
Compute the Directional Difference as the absolute value of the differences: DIdiff = | (( +DI ) - ( -DI )) |
Sum the directional indicator values: DIsum = (( +DI ) + ( -DI )) .
Calculate the Directional Movement index: DX = ( DIdiff / DIsum ) * 100 . The DX is always between 0 and 100.
Finally, apply Wilder’s smoothing technique to produce the final ADX value:
ADXt = ( ( ADXt-1 * ( n - 1) ) + DXt ) / n
When indicator tell us to buy?
If when DI+ crosses DI- and ADX is bigger than DI- indicator tell us to buy.
Açık kaynak kodlu komut dosyası
Gerçek TradingView ruhuyla, bu komut dosyasının yazarı açık kaynak olarak yayınladı, böylece yatırımcılar bunu anlayabilir ve doğrulayabilir. Yazara tebrikler! Ücretsiz olarak kullanabilirsiniz, ancak bu kodun bir yayında yeniden kullanımı Kullanım Koşulları tarafından belirlenmektedir. Bir grafikte kullanmak için favorilerinize ekleyebilirsiniz.
Bu komut dosyasını bir grafikte kullanmak ister misiniz?
## Yorumlar
Profil Profil Ayarları Hesap ve Ödemeler Tavsiye edilen arkadaşlar Koinler Destek Kayıtlarım Destek Merkezi Özel Mesajlar Sohbet Çıkış | 830 | 3,057 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2021-17 | longest | en | 0.915936 |
https://www.enotes.com/homework-help/evaluate-function-indicated-value-x-functionvaluef-327235 | 1,484,930,343,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280835.60/warc/CC-MAIN-20170116095120-00108-ip-10-171-10-70.ec2.internal.warc.gz | 905,744,491 | 12,296 | Evaluate the function at the indicated value of x FunctionValue f(x) = `log_8(x)` x = 2 f(2)=______Can someone help me solve this?
hala718 | High School Teacher | (Level 1) Educator Emeritus
Posted on
`f(x)= log_8 x `
`==gt x= 2`
We need to find f(2).
Then, we will plug in 2 into the function.
`==gt f(2)= log_8 2 `
Now we will use the change base property to simplify.
`==gt f(2)= log_8 2 = (log_2 2)/(log_2 8) = 1/3 `
`==gt f(2)= 1/3` | 174 | 458 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2017-04 | longest | en | 0.74531 |
https://allresearchers.com/critical-value/ | 1,716,080,251,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057631.79/warc/CC-MAIN-20240519005014-20240519035014-00146.warc.gz | 78,704,307 | 40,669 | Critical Value
1. A goodness-of-fit test involves three different categories.
a. How many degrees of freedom are associated with this test?
b. What is the critical value for χ2 at α=.05?
c. What is the critical value at α=.01?
2. 6. If a test is reliable, each subject will tend to get the same score each time he or she takes the test. Therefore, the correlation between two administrations of the test should be high. The reliability of the verbal GRE was tested using five subjects, as shown in the table below:
verbal GRE (1) verbal GRE (2)
540 570
510 520
580 600
550 530
520 520
a. What is the value of the correlation coefficient? What is the value of SP (sum of the products – numerator of the correlation coefficient)? What is the value of SSx and SSy ?
b. Is the correlation significantly different from zero? what is the critical r value?
c. calculate the regression line predicting the second exam from the first exam. What is the value of the slope what is the value of the Y intercept
d. Does the slope provide support for a significant linear relationship between exam 1 and exam 2?
SS regression =
df regression=
MS regression =
SS residual =
df residual=
F=?(1pt)
SS total=
What is the critical F value? | 289 | 1,222 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2024-22 | latest | en | 0.900118 |
https://web2.0calc.com/questions/squaring-a-square-root | 1,656,521,705,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103640328.37/warc/CC-MAIN-20220629150145-20220629180145-00300.warc.gz | 637,934,166 | 5,724 | +0
# Squaring a square root?
+1
169
4
+61
Okay, I stumbled upon a question that eventually led to this, can someone please help me out? Thank you.
$$\sqrt (160-64\sqrt3)$$
Nov 27, 2021
#1
+2
What do you want to do with this expression?
Nov 27, 2021
#2
0
if you want to square
sqrt ( 160 - 64 sqrt 3) you will get 160 - 64 sqrt 3
Guest Nov 27, 2021
#3
+117466
+2
I also ask, what do you want to do with it?
$$\sqrt{160-64\sqrt3}\\ =4\sqrt{10-4\sqrt3}$$
Nov 28, 2021
#4
+61
+1
:< It was part of a problem, but I guess I did the first part wrong. Oof.
Ooflord Dec 2, 2021 | 234 | 596 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2022-27 | latest | en | 0.915614 |
https://discuss.codechef.com/t/candivide-editorial/105594 | 1,686,398,099,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224657169.98/warc/CC-MAIN-20230610095459-20230610125459-00492.warc.gz | 246,461,975 | 4,259 | # CANDIVIDE - Editorial
Authors: krypto_ray
Testers: iceknight1093, tabr
Editorialist: iceknight1093
TBD
None
# PROBLEM:
Chef has three friends and N candies. Can he distribute all the candies equally to them?
# EXPLANATION:
Since each candy should be distributed, and each friend should receive an equal number, the total number of candies should be divisible by 3.
So, the answer is “Yes” if N is divisible by 3, and “No” otherwise.
In most programming languages, this can be checked using the statement if (N%3 == 0)
# TIME COMPLEXITY
\mathcal{O}(1) per test case.
# CODE:
Editorialist's code (Python)
for _ in range(int(input())):
n = int(input())
print('Yes' if n%3 == 0 else 'No') | 197 | 698 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2023-23 | latest | en | 0.807835 |
https://stage.edx.org/course/applications-of-linear-algebra-part-2 | 1,627,547,815,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153854.42/warc/CC-MAIN-20210729074313-20210729104313-00325.warc.gz | 529,168,870 | 85,509 | # Applications of Linear Algebra Part 2
Explore applications of linear algebra in the field of data mining by learning fundamentals of search engines, clustering movies into genres and of computer graphics by posterizing an image.
This course is archived
Estimated 4 weeks
12–18 hours per week
Instructor-paced
Instructor-led on a course schedule
Free
Our world is in a data deluge with ever increasing sizes of datasets. Linear algebra is a tool to manage and analyze such data.
This course is part 2 of a 2-part course, with this part extending smoothly from the first. Note, however, that part 1, is not a prerequisite for part 2. In this part of the course, we'll develop the linear algebra more fully than part 1. This class has a focus on data mining with some applications of computer graphics. We'll discuss, in further depth than part 1, sports ranking and ways to rate teams from thousands of games. We’ll apply the methods to March Madness. We'll also learn methods behind web search, utilized by such companies as Google. We'll also learn to cluster data to find similar groups and also how to compress images to lower the amount of storage used to store them. The tools that we learn can be applied to applications of your interest. For instance, clustering data to find similar movies can be applied to find similar songs or friends. So, come to this course ready to investigate your own ideas.
Courses offered via edX.org are not eligible for academic credit from Davidson College. A passing score in a DavidsonX course(s) will only be eligible for a verified certificate generated by edX.org.
### At a glance
• Language: English
• Video Transcript: English
# What you'll learn
Skip What you'll learn
• How to solve least-square systems, about eigenvectors of matrix, how to Markov Chains, and the matrix decomposition called the singular value decomposition.
• To apply the least-squares method to finding a presidential look-alike
• To use an eigenvector to cluster a dataset into groups or downsample an image
• To use Markov Chains to analyze a board game
• How Markov Chains were proposed by Google as part of their search engine process
• Applications of the singular value decomposition in image compression and data mining.
• Explore applications with Matlab codes provided with the course. | 486 | 2,321 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2021-31 | latest | en | 0.933803 |
http://www.mathynomial.com/problem/1725 | 1,529,355,132,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267861163.5/warc/CC-MAIN-20180618203134-20180618223134-00036.warc.gz | 477,518,156 | 3,389 | # Problem #1725
1725 Externally tangent circles with centers at points A and B have radii of lengths 5 and 3, respectively. A line externally tangent to both circles intersects ray AB at point C. What is BC? $\textbf{(A)}\ 4\qquad\textbf{(B)}\ 4.8\qquad\textbf{(C)}\ 10.2\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 14.4$ This problem is copyrighted by the American Mathematics Competitions.
Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.
• Reduce fractions to lowest terms and enter in the form 7/9.
• Numbers involving pi should be written as 7pi or 7pi/3 as appropriate.
• Square roots should be written as sqrt(3), 5sqrt(5), sqrt(3)/2, or 7sqrt(2)/3 as appropriate.
• Exponents should be entered in the form 10^10.
• If the problem is multiple choice, enter the appropriate (capital) letter.
• Enter points with parentheses, like so: (4,5)
• Complex numbers should be entered in rectangular form unless otherwise specified, like so: 3+4i. If there is no real component, enter only the imaginary component (i.e. 2i, NOT 0+2i). | 316 | 1,078 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2018-26 | latest | en | 0.845653 |
http://www.victoriaweather.ca/monthlymeans_data.php?id=55&month=5 | 1,579,846,750,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250615407.46/warc/CC-MAIN-20200124040939-20200124065939-00439.warc.gz | 288,088,919 | 6,184 | This message is an acknowledgement of the long list of station problems. Work to repair sites with problems will begin in the second half of January. Thanks for your patience.
# Station Monthly Statistics -- Savory Elementary School
## May at Savory Elementary School
Station name Month
May at Savory Elementary School
Year Minimum
°C
Mean
(Std. Dev.) °C
Maximum
°C
Count
20189.914.91 (3.3)20.531
20199.514.68 (3.4)20.231
20169.114.30 (3.5)20.131
20158.914.20 (3.6)20.431
20149.414.17 (3.2)19.431
20178.513.24 (3.0)18.031
20138.613.02 (3.1)18.131
20088.512.76 (2.8)17.431
20067.512.69 (3.4)18.331
20097.112.47 (3.5)18.131
20076.611.97 (3.5)17.731
20126.411.64 (3.4)16.931
20107.211.49 (2.8)16.231
20116.811.10 (2.7)15.731
Means8.113.018.4
## How are Monthly Means Calculated?
The table shows values for the selected month from each year in the database. Values are given in degrees Celsius. Averages shown here are calculated from daily averages for the month. The value in parentheses following the monthly mean value is the standard deviation of the daily averages used to compute the mean. The minimum (maximum) is the average of the minimum (maximum) temperature observed on each day of the selected month. Table rows are shown by default in descending order of Mean value. Click on the column heads to change the ordering. In some cases the months may be incomplete. Months that have fewer than 24 daily values are shown in red. This may include the present month.
The last row in the table gives the averages of the columns. Click through the links in the column headers to sort by the different quanitities.
The values are plotted so that the minimum and maximum are marked at the end of the whiskers, the mean value is shown with a horizontal tick across each box, and the boxes show the range of plus and minus one standard deviation from the mean value. The average temperature is shown with a dashed horizontal line and the average extremes are drawn with dotted lines. | 547 | 1,988 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2020-05 | latest | en | 0.849469 |
www.healthsomeness.com | 1,723,206,061,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640763425.51/warc/CC-MAIN-20240809110814-20240809140814-00295.warc.gz | 634,990,718 | 25,519 | # Weight loss calorie calculator
The calculator below will provide you with a fairly accurate estimate of the number of calories you should be consuming in order to lose weight.
Gender Age Weight Switch to lbs Activity Level
This calculator will never show a number below 1000 calories per day. Please speak to a qualified health professional before attempting to eat less than that.
The calculator is based on the Mifflin-St Jeor equation, an algorithm that is currently thought to be one of the most accurate for calculating calorie requirements, according to studies such as this and this. The formula is as follows:
• Males: (10 x w) + (6.25 x h) – (5 x a) + 5
• Females: (10 x w) + (6.25 x h) – (5 x a) – 161
where w = weight in kg, h = height in cm and a = age in years.
As you can see, gender, age, height and weight all influence calorie intake. Much of this has to do with body size and body composition. Men in general tend to have larger and more muscular bodies than women, so need to consume a few extra calories. Additionally, as you get older your metabolism naturally slows down, so you require fewer calories.
The Mifflin-St Jeor equation provides an indication of the average number of calories an individual would burn whilst at rest. Of course, most people aren’t resting the entire day, so their activity levels need to be factored in as well. This is done by multiplying the calories required at rest by a number between 1.2 and 1.9. The former would be for someone who leads a very inactive lifestyle whereas the latter would be for someone who is extremely active. This new result tells you how many calories are needed for weight maintenance.
In order to lose weight, you need to consume fewer calories than you burn. When you do so, your body is forced to make use of its fat stores as a source of energy; as a result, you end up burning fat and losing weight. The calculator above takes a very simplistic approach to calculate the number of calories needed for weight loss. It simply multiplies the calories needed for weight maintenance by 0.8 (ordinary pace) or 0.6 (fast pace).
The calculator does not provide a number less than 1000 calories. This is because consuming too few calories can actually slow down weight loss efforts by affecting one’s metabolism. If you wish to consume fewer than 1000 calories over long periods of time, you should speak to a qualified medical professional so that he or she can provide assistance along the way.
The speed at which you lose weight is influenced by many factors, but by adhering to the calorie recommendations showed using the calculator above, you can expect to lose between 1 and 2 pounds per week.
## What foods have calories?
All foods have calories. This is because they contain one or more of the three macro-nutrients: protein, carbohydrate and fat. The quantity of each macro-nutrient in various foods does vary significantly. For example, a medium sized apple (182 g) has 0.5 g of protein, 25 g of carbs and 0.3 g of fat; an average size avocado (200 g) has 4 g of protein, 17 g of carbs and 29.5 g of fat.
Fat provides you with 9 calories per gram, whereas protein and carbs provide you with 4. Because of this, the apple has 94 calories and the avocado has 322.
It is also important to note that water has 0 calories, so foods that contain large quantities of it (like most vegetables and fruits) tend to contain far fewer calories than low-water foods. This can clearly be seen if you take a look at the nutritional profile of watercress (which is made up of 95% water) – it only has 4 calories per cup!
Because fat contains more calories than protein or carbs, a natural inclination would be to eliminate all fat rich foods from the diet. However, whilst reducing fat intake to fit within a certain caloric intake may be a good idea, it should not be avoided entirely. This is because fat rich foods such as avocados, oily fish, nuts and seeds are very nutritious and can actually come in handy when trying to lose weight, if eaten in moderation.
Instead of just cutting out fat, follow the tips mentioned in the next section to easily reduce your calorie intake.
## How to naturally reduce calorie intake
Eating more of certain foods and less of others can have a dramatic effect on your overall calorie consumption, and therefore, your weight loss efforts.
Using the calculator above you may have calculated that you need to eat 1700 calories per day. However if you eat 1700 calories worth of junk food, chances are you will be left feeling hungry and unsatisfied. Instead, you want to eat nutritious and filling food.
### Eat enough protein
Out of the three macro-nutrients, protein is arguably the most beneficial for weight loss. This is because it helps on both ends of the calories in – calories out equation: numerous studies have shown that protein consumption can boost metabolism whilst reducing hunger.
For example in this study, participants increased protein intake from 15% to 30% of total calories. Doing so resulted in 400+ fewer calories being consumed, which is a significant amount. Another study involving obese men showed that those who followed a high protein diet had less of a late night desire to snack and fewer obsessive thoughts about food.
It is well-known that not all calories are created equal. For example, a can of cola and two large boiled eggs (which are high in protein) have approximately the same number of calories. However, eating the eggs will leave you feeling much more satisfied than the cola will.
Studies seem to suggest that a protein intake of 30% of total calories is ideal, so this is a number you may want to aim for. There are plenty of protein rich foods you can eat, including legumes, fish, unprocessed meat, nuts, seeds and grains.
### Drink fewer sugary beverages, replacing them with water
One way that calories sneak into your body is via the beverages that you drink. Sugar sweetened beverages such as soft drinks and fruit juices tend to be high in calories, but as mentioned in the previous section, don’t have as large of an effect on your appetite as solid foods do.
When you eat solid food, the act of chewing and the bulk that the food adds to your stomach sends signals to your brain, telling it that you should feel less hungry. Unfortunately this doesn’t happen as efficiently when you consume liquid calories.
Studies have found that sugar consumption has been strongly linked with weight gain, and much of the sugar in these beverages is in the form of fructose, which when consumed in excessive amounts overburdens the liver and leads to fat retention.
For this reason, it is a good idea to minimise your consumption of sugary beverages. In place of them, drink water. Because it has 0 calories, you can drink larger quantities of it, without having to worry about going overboard on calories. In fact, studies have shown that having a glass or two of water before a meal can help reduce the amount of food that you eat.
### Eat real food
By this we mean eating foods that are as close to their natural state as possible. These are the ones that are loaded with vitamins, minerals, antioxidants and plenty of other beneficial compounds. Unprocessed foods also tend to contain large quantities of fiber, which like protein, can help to satisfy the appetite. The foods shown here are a good representation of the kinds that you should be eating.
When trying to lose weight, vegetables should be a large part of your diet. They are bursting with nutrition, are low in calories and are very affordable, making them the most weight loss friendly foods out there. As a general rule of thumb, half of your plate at each meal should consist of a variety of vegetables. Luckily, there is no shortage of vegetables to choose from.
The reason why vegetables are key to weight loss is because you can eat so much of them. For example, an average cheese burger has 300 calories, but a cup of broccoli only has 31 calories. This means you would have to eat 10 cups of broccoli to make up for the calories in the burger. Guess which of the two would take up more space in your stomach?
The foods that you want to minimise consumption of or avoid entirely are those that have been highly processed. Examples include things like cake, candy, doughnuts and potato chips. These foods contain large quantities of unhealthy fat and added sugar, but provide very little nutritional value, making them a source of empty calories. | 1,802 | 8,507 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2024-33 | latest | en | 0.959419 |
http://scienceworld.scholastic.com/news/2012/12/force | 1,441,338,227,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440645335509.77/warc/CC-MAIN-20150827031535-00057-ip-10-171-96-226.ec2.internal.warc.gz | 206,654,250 | 13,783 | The force acting on an object can be defined as the object's mass—the amount of matter in it—times its acceleration.
United Launch Alliance / NASA
Force
From Grolier's New Book of Knowledge
A torsion balance can be used to measure the tiny gravitational force between small objects.
Grolier
When two balloons are given the same electrical charge, they will push one another apart.
Grolier
A force is a push or a pull. Forces make things move, stop, and change direction. They also can make objects accelerate. In fact, the force acting on an object can be defined as the object's mass—the amount of matter in it—times its acceleration. This is expressed mathematically as F = m × a.
One force we experience every moment of our lives is gravity. Like any force, gravity can make an object accelerate. You could prove this by dropping a ball from different heights and measuring how long it takes the ball to reach the ground. For example, you could drop a ball from a height of 4 feet (1.2 meters). And then you could do it again from 8 feet (2.4 meters). You might think that the ball would take twice as long to reach the ground the second time. But you would be wrong. When dropped from 8 feet (2.4 meters), the ball would actually take less than twice as long to reach the ground. This proves that the ball has accelerated under the force of gravity.
MEASURING FORCE
Since force may be defined as mass times acceleration, you can find the force acting on an object if you know its mass and acceleration. However, more than one force acts on an object at a time. One of the most common forces besides gravity is friction. Friction is due to electrical forces. It opposes the motion of an object being pushed or pulled. Part of the force applied to an object to make it move must be used to overcome friction. As a result, the force that makes the mass accelerate is less than the force applied.
The unit used to measure force is called the newton (N). The unit's name honors Isaac Newton, an English scientist. He was the first person to explain motion satisfactorily. A newton is the force needed to make a 1 kilogram (2.2 pound) mass accelerate at a rate of 1 meter (3.3 feet) per second per second. This means that the speed increases by 1 meter (3.3 feet) per second for every second the force is applied.
One way to measure forces is with a spring balance. In this device, the spring's stretch is proportional to the force acting on it. If the force doubles, the stretch also doubles. With a spring balance, it is easy to measure the friction opposing an object's motion. Simply pull the object so that it moves at a constant speed. Since the object is not accelerating, the net force (the sum of all the forces acting on the body) is zero. The applied force is just enough to overcome friction.
Forces can also be measured with balances. These devices consist of pans suspended from opposite ends of a bar with a fulcrum (pivot point) at the center. With these devices, an unknown force acting on one pan can be balanced by a known force applied to the opposite pan. Very small forces can be measured with a type of balance known as a torsion balance. This device is a long beam that is suspended from a fine wire. The angle through which the beam turns and the time period of its swing depend on the force. In 1798, British physicist Henry Cavendish used a torsion balance to measure the gravitational force between known masses. Two small lead balls at each end of the balance were attracted to larger ones placed near the smaller balls. With this balance, Cavendish measured forces as small as one-billionth of a newton.
Another way to measure forces is with crystals known as piezoelectric crystals. When a force is applied to such a crystal, a small measurable voltage develops across the crystal.
HOW FORCES REACT
If you hang a 1 kilogram (2.2 pound) mass from a calibrated spring balance, the balance reads 9.8 N. This reading indicates that the Earth pulls on a 1-kilogram mass with a force of 9.8 N. If we release the mass, it will accelerate toward the Earth at a rate of 9.8 meters (32.2 feet) per second per second. But when an object hangs from a spring balance, it is motionless. Does this mean that there is no force acting on it? No, it simply means that there is no unbalanced, or unopposed, force. As the object hangs from the spring balance, there is gravitational force pulling the object downward. But this force is balanced by the opposing upward pull of the spring. These two opposing forces add up to zero. Thus there is no net force on the object.
FORCE: A VECTOR QUANTITY
Things that can be added or subtracted using ordinary arithmetic are said to be scalar quantities. If you walk 2 miles (3.2 kilometers) and then walk another 2 miles, the total distance you have walked is 4 miles (6.4 kilometers). Distance is a scalar quantity.
However, suppose you walk 2 miles east and then walk 2 miles north. How far are you from your starting point? The answer is not 4 miles. To find your position, you must take into account direction as well as distance. In this case your final position is 2.8 miles (4.5 kilometers) northeast of where you started. The drawingshows how you can use scaled arrows that give both direction and distance to find what is called your displacement. A quantity, such as displacement, that involves both a magnitude (size) and a direction is called a vector quantity.
A force is a vector quantity. It can be represented by an arrow (vector). The arrow's length is scaled to represent the strength of the force. And the head of the arrow gives its direction. If a weight is suspended from a spring, the forces can be represented by arrows of equal length. One arrow points up, and one points down. Vector quantities are added by placing the head of one arrow on the tail of the other. In the case of two equal vectors that point in opposite directions, the sum is zero.
If many forces act on an object, you can find the sum of the forces. You do so by adding all the arrows that represent these forces. If an object is not moving, the sum of the forces (and the arrows used to represent them) must be zero.
FOUR BASIC FORCES
There appear to be four basic forces in the universe. The one most people are familiar with is the force of gravity. The three other basic forces are the electromagnetic force, the strong nuclear force, and the weak nuclear force.
The force of gravity pulls objects toward the Earth. It also pulls all the planets toward the sun. But it is the weakest of the four basic forces. When Isaac Newton first proposed his Law of Universal Gravitation, many scientists were deeply troubled.
According to Newton's view, the force of gravity acts across empty space. There is no need for contact between bodies that attract one another. How does the sun pull on the Earth? Newton never answered that question. He assumed that gravitational forces can act at a distance.
Eventually German physicist Albert Einstein developed his General Theory of Relativity. Einstein eliminated the need for gravitational force as an explanation for why the planets circle the sun. Instead, he proposed that their motion arises from the natural geometrical properties of space.
The electromagnetic force is more than 4 million, quintillion, quintillion times stronger than the force of gravity. It is the force that holds together the atoms and molecules that make up matter. It is also the force that resists attempts to push atoms and molecules closer together or pull them apart. That is, it resists compression and extension. Consequently, it is the force that explains most of the muscular pushes and pulls that we exert on objects in our daily lives.
You can see the effects of electrical and magnetic forces quite easily. Use thread to suspend two balloons side by side. Give both balloons like charges by rubbing them with a paper towel or cloth. Because both balloons carry like charges, they will push each other apart. If they were oppositely charged, they would attract one another. Magnets behave in a similar way. The north pole of one magnet will pull on the south pole of another magnet and vice versa. However, like poles—two north or two south poles—will push one another apart.
At one time scientists thought that electrical and magnetic forces were different and unrelated. Now we know that magnetic forces exist because of the motion of electric charges, or currents. In a bar magnet, for example, the flow of charges in the atoms that make up the magnet are aligned. Because of this alignment, a strong magnetic field is produced.
The strong nuclear force is 137 times stronger than the electromagnetic force. Normally, the electrical repulsion between the positive charges that make up the nucleus would cause them to fly apart. But the strong force holds nuclei together. Although it is powerful, it acts only over very short distances. These distances are comparable to the size of nuclear particles.
The fourth force is the weak nuclear force. It was discovered during studies of beta radiation, which is the emission of electrons from some atomic nuclei. Like the strong force, it acts over a very short range. But it is only about one-thousandth as strong.
According to the quantum theory, what we refer to as forces are caused by the emission (giving off) and absorption of particles called field quanta. The electron in the hydrogen atom is attracted to the proton by an exchange of photons. Photons are the field quanta for what we call the electromagnetic force.
Similarly, the other three forces are the result of the interaction of their field quanta. Gravitons account for gravity. Gluons account for the strong nuclear force. And bosons account for the weak nuclear force. Physicists familiar with quantum theory prefer to speak of interactions rather than forces. But most people will continue to refer to the pushes and pulls in their lives as forces.
By Robert Gardner | 2,118 | 9,989 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2015-35 | longest | en | 0.953961 |
https://www.askiitians.com/forums/Algebra/sale-plants-4-saplings-in-a-row-in-her-garden-t_278438.htm | 1,621,187,323,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991178.59/warc/CC-MAIN-20210516171301-20210516201301-00176.warc.gz | 657,033,770 | 35,432 | #### Thank you for registering.
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# Sale plants 4 saplings, in a row, in her garden. The distance between two adjacent saplings is ¾ m. Find the distance between the first and the last sapling.
Harshit Singh
one month ago
Dear Student
From the question, it is given that,
The distance between two adjacent saplings = 3⁄4 m
Number of saplings planted by Saili in a row = 4
Then, number of gap in saplings = 3⁄4 × 4 =3
∴The distance between the first and the last saplings = 3 × 3⁄4
= (9/4) m
= 2 whole 1⁄4 m
Hence, the distance between the first and the last saplings is 2 whole 1⁄4 m.
Thanks | 281 | 910 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2021-21 | latest | en | 0.881992 |
https://fr.mathworks.com/matlabcentral/cody/groups/77/problems/44819 | 1,569,202,302,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514575844.94/warc/CC-MAIN-20190923002147-20190923024147-00092.warc.gz | 484,562,295 | 19,223 | Cody
# Problem 44819. Relative pose in 2D: problem 1
We consider a world reference frame denoted by {0} which has its x-axis pointing east and its y-axis pointing north. There is a robot with an attached body-fixed coordinate frame {B} whose origin is in the centre of the robot, and whose x-axis points in the robot's forward direction.
With respect to the world frame origin , the robot's centre is at a distance of 123m in the x-direction and -74.6m in the y-direction. The car's heading angle is exactly SSW.
Write a 3x3 matrix homogeneous transformation matrix that expresses the pose of the robot frame {B} with respect to the world frame {O}.*
### Solution Stats
38.71% Correct | 61.29% Incorrect
Last solution submitted on Sep 18, 2019 | 187 | 749 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2019-39 | latest | en | 0.920858 |
http://www.jiskha.com/math/trigonometry/?page=12 | 1,369,136,554,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368699924051/warc/CC-MAIN-20130516102524-00013-ip-10-60-113-184.ec2.internal.warc.gz | 554,924,600 | 12,866 | Tuesday
May 21, 2013
# Homework Help: Math: Trigonometry
Trig/Math
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Monday, January 3, 2011 at 8:36pm
trig
Third time is the charm? I'll try again. Could someone show me how, (- sin (x/2) /( 2 sin (x/2) + cos (x/2)) is an alternate representation for, 1 / ( 4 tan (x/2) + 2 ) TIA Carol This doesn't require the solving of any equations. For example, ( same concept for what I ...
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I asked below but I guess it was not understood what I was asking. I'll try again. Could someone show me how, (- sin (x/2) /( 2 sin (x/2) + cos (x/2)) is an alternate representation for, 1 / ( 4 tan (x/2) + 2 ) TIA Carol I know the half-angle formulas are used but I cannot...
Monday, January 3, 2011 at 5:30pm
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Monday, October 25, 2010 at 4:15am
trig
suppose sin(x)=4/5 and cos(y)=-12/13, and x and y are in the second quadrant. determine the value of each expression: sin(x+y), cos(x-y), cos2x, sin2y.
Sunday, October 24, 2010 at 11:23pm
trig
prove this (1+cosx)/(1-cosx)=(1+secx)/(secx-1)
Sunday, October 24, 2010 at 5:53pm
trig
suppose sin(x)=5/6, and x is in the first quadrant. Determine cos(x), sin(2x), and cos(2x).
Sunday, October 24, 2010 at 2:50pm
trig
Suppose that sin x = 1/5 and cos y = 2/3, and x and y are each angles in Quadrant 1. Determine sin(x+y).
Friday, October 22, 2010 at 8:48pm
College Algebra
I REALLY don't understand the reason/basis/use of logarithms. I have listened to my teacher, who never is clear on much of anything (it would help if he spoke better English), and a more advanced student, who couldn't explain them to me. The problem I am working on is...
Thursday, October 21, 2010 at 10:54am
trig
how do you establish the identity of cos(3pi/2+0)=sin0
Thursday, October 21, 2010 at 10:44am
trig
If sinθ>0 and tangθ<0, explain how to find the Quadrant in which θ lies.
Thursday, October 21, 2010 at 2:18am
Math 12
trig question 2sin^2x+sinx-6=0 how to solve for x when the restriction is x is greater than or equal to zero but less than 2pi
Thursday, October 21, 2010 at 12:55am
pre calc trig check my work please
sin x + cos x -------------- = ? sin x sin x cos x ----- + ----- = sin x sin x cos x/sin x = cot x this is what i got, the problem is we have a match the expression to the equation work sheet and this is not one of the answers. need to figure out what im doing wrong so i can ...
Wednesday, October 20, 2010 at 11:32pm
trig
if 8 tan Q=15 then sinA-cosA=?
Wednesday, October 20, 2010 at 11:22am
Trig
A force of 323 newtons makes an angle of 54 degrees and 20 minutes with a second force. The resultant of the two forces makes an angle of 39 degrees forty minutes with the first force. Find the magnitude of the second force. I have the answer, 814.3022, but I need to know how ...
Tuesday, October 19, 2010 at 11:30pm
trig
how do i find six trig functions for cos(pi over 2-x)=3/5, cos x=4/5?
Tuesday, October 19, 2010 at 11:28pm
Trig
Tuesday, October 19, 2010 at 12:16am
Trig
Solve for 0≤x≤2π ---> (2√2)cos²(x)-(√2-2)cos(x)-1=0 much thanks .
Monday, October 18, 2010 at 11:46pm
Trig
Find the general solution for the equation 4tanx+cotx=5?????? and the steps please? THANYOO!
Monday, October 18, 2010 at 11:07pm
Trig
If b=5 , the angle C=50 and the angle A=50 find the other angle B and the remaining sides and . Give your answer to at least 3 decimal places.
Monday, October 18, 2010 at 7:58pm
trig
f(x)= 2sec(x-3.14/4)-1
Monday, October 18, 2010 at 2:52pm
trig
find all values of degrees in the set (o,2 pi) satisfying cos of degree over 3 equal to square root of 3 over two and -cosdegree + 2sindegres=-2
Sunday, October 17, 2010 at 7:43pm
trig
prove that cos 52 + cos 68 + cos172 = 0
Sunday, October 17, 2010 at 2:13am
trig
how to do the calculation while taking 'A' as 15 in the half angle formula used for calculating the value of sin 7.5 ?
Sunday, October 17, 2010 at 2:05am
trig
a hedgehog wishes to cross a road without being run over. he observes the angle of elevation of a lamp post on the other side of the road to be 27 degrees from the edge of the road and 15 degrees from a point 10m back from the raod. How wide is the road??
Saturday, October 16, 2010 at 4:00pm
trig
given cosine (-2/5) and pie<t<(3pie/2) find sine (2x-1)
Friday, October 15, 2010 at 12:54pm
trig
wha tis the exact value of sin 9pi/4
Thursday, October 14, 2010 at 2:39pm
College Algebra and Trig
a rancher has 12,000 feet of fence to enclose 5 adjacent pens. determine the outer rectangular dimensions that result the maximum enclosed area.
Thursday, October 14, 2010 at 1:05pm
Calculus....
can you show me how to maniplate this using Trig subtitution... integral dx/4+x^2 x is bounded on [-2, 2]
Tuesday, October 12, 2010 at 11:24am
trig
Find the graph y = 2sin x + cos(2.5x) over the interval
Tuesday, October 12, 2010 at 1:53am
trig
What is the graph of y = 3 sin (3x + 180)?
Tuesday, October 5, 2010 at 9:54am
Trig
Find the value of the six trigonometric functions. t = 7(pi)/4 Can you show me step by step so I can see how this is solved? Also, do I need to graph this first in order to understand or begin the problem? I am trying to self teach - but having problems.
Tuesday, October 5, 2010 at 12:12am
Trig
Find the reference number and the terminal point determined by t. a. t = - 7(pi)/6
Tuesday, October 5, 2010 at 12:09am
college trig
I need to find the domain of the following function: y=arcsin((x-3)/(2x+1)) My math is a bit rusty after holidays...This is what I did: (I'll use <> for the sign of 'not equal') -> 2x+1<>0 <=> 2x<>-1 <=> x <> -1/2 But i'm ...
Monday, October 4, 2010 at 5:08pm
Maths - trig
Solve 3cos2x - 7cosx = 0, when 0<=x<=360 And also, find the exact values for x when 0<=x<=360 if 3tan^2x=1 Thankyou!
Sunday, October 3, 2010 at 5:37pm
trig
how do you solve: find cos2x if sinx is equal to 1/5
Saturday, October 2, 2010 at 11:57am
trig
How would I solve the following equation for x? 2 sin^2(x) + 3 tanx secx = 2 I've tried the problems from different approaches, but couldn't come up with a solution. Could you please provide your thought process. It would be greatly appreciated. Thanks!
Friday, October 1, 2010 at 12:38pm
trig
25q to the second power -81=0
Tuesday, September 28, 2010 at 6:40pm
trig
The angle of elevation from the top of a house to a jet flying 2 miles above the house is x radians. If d represents the horizontal distance, in miles, of the jet from the house, express d in terms of a trigonmetric function of x.
Sunday, September 26, 2010 at 6:16pm
trig
which ser of lenths will form a triangle with greater area? 6ft,8ft, 10ft or 6ft,8ft,12ft...
Saturday, September 25, 2010 at 9:33pm
trig
4=log[7]2401
Saturday, September 25, 2010 at 11:38am
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http://www.vias.org/crowhurstba/crowhurst_basic_audio_vol1_031.html | 1,701,166,387,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679099281.67/warc/CC-MAIN-20231128083443-20231128113443-00037.warc.gz | 92,427,117 | 3,189 | Basic Audio is a free introductory textbook to the basics of audio physics and electronics. See the editorial for more information....
# The Dynamic (Moving-Coil) Microphone
Author: N.H. Crowhurst
The principle of dynamic microphone operation
If a wire connected to a meter that will indicate when current flows is moved about near to a magnet, the meter will show current fluctuations. When the wire is moved, the meter deflects. Holding the wire still produces no current. The direction of current indicated on the meter depends on the direction in which the wire is moved. This is an ideal basis for converting movement into electrical current. Because movement is the essential feature for conversion, a microphone using this principle is called a dynamic microphone. The problem in making a microphone of this type is that a large movement is needed to produce even a small current, while the movement of the air particles due to sound waves is small.
Increasing the effectiveness of the magnet
This problem is overcome, to some extent, by increasing the intensity of the magnetic field. A North and South pole are brought close together, and the wire moves in the narrow space between them. To increase its effectiveness, many turns of wire move in the same gap. It is convenient to make the gap circular, because this simplifies construction of the coil, gets the poles close together, and gives the coil free space in which to move.
Converting movement into current is only part of the job. We must first move the coil by means of the sound waves, which requires a diaphragm. To move freely, the diaphragm must be light - as little heavier than air as possible. Because the coil is also attached to the diaphragm, it, also, must be as light as possible, or it would load the diaphragm down. Hence, a small coil must be used.
Cross section of a simple moving-coil microphone
The use of a small coil requires a very intense magnetic field to get the best results. To accomplish this, the gap is made very small. To prevent the coil rubbing against the magnet poles, a centering "spider" or suspension is used, which allows free movement in the direction of vibration, while preventing the coil from moving against the pole faces.
Last Update: 2010-11-03 | 467 | 2,272 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2023-50 | latest | en | 0.935334 |
https://jeopardylabs.com/play/multiplication-review-754 | 1,670,627,482,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711552.8/warc/CC-MAIN-20221209213503-20221210003503-00753.warc.gz | 360,235,387 | 9,383 | Basic Multiplication
Equal Groups
Word Problems
Random Review!
100
2x0=?
0
100
Write the multiplication sentence
3x4=12
100
2 groups of 4
4+4=8
100
Sarah's classroom has 4 tables of 4 kids. How many students are there altogether? Write the multiplication sentence and answer.
4x4=16
100
322+305
627
200
10x4=?
40
200
Write the multiplication sentence
4x5=20
200
3 groups of 5
5+5+5=15
200
There are 3 tilt-a-whirl cars at the fair. Each car can fit 2 people. How many people can fit on the ride at once? Write the multiplication sentence and answer.
3x2=6
200
862-241
621
300
3x2=?
6
300
Write the multiplication sentence
5x3=15
300
4 groups of 6
6+6+6+6=24
300
Mrs. Cross bought 5 pizzas for her class. Each pizza is cut into 7 slices. How many pieces are there altogether? Write the multiplication sentence and answer
5x7=35
300
Round 784 to the nearest ten
780
400
5x8=?
40
400
Write the multiplication sentence
3x6=18
400
7 groups of 3
3+3+3+3+3+3+3=21
400
At the farm there are 6 trees, each with 7 apples on each tree. How many apples are there total? Write the multiplication sentence and answer
6x7=42
400
900-735
50
500
9x9=?
81
500
Write the multiplication sentence
2x8=16
500
5 groups of 6
6+6+6+6+6=30
500
Each student has 8 piles of M&Ms. There are 4 M&Ms in each pile. How many M&Ms does each student have? Write the multiplication sentence and the answer.
8x4=32
500
Draw a picture and solve 10 divided by 5
8-0
Click to zoom | 484 | 1,507 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2022-49 | latest | en | 0.87742 |
https://www.nickzom.org/blog/tag/screw-rotation-speed/ | 1,590,737,017,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347402457.55/warc/CC-MAIN-20200529054758-20200529084758-00135.warc.gz | 827,679,078 | 27,200 | ## How to Calculate and Solve for Pressure Gradient | Polymer & Textile
The image above represents pressure gradient.
To compute for pressure gradient, four essential parameters are needed and these parameters are value (α), viscosity (μ), screw rotation speed (N) and proportionality constant that depends on screw geometry (B).
The formula for calculating pressure gradient:
ΔP = αμN / B
Where;
ΔP = Pressure Gradient
α = Value
μ = Viscosity
N = Screw Rotation Speed
B = Proportionality Constant that Depends on Screw Geometry
Let’s solve an example;
Find the pressure gradient when the value is 2, viscosity is 9, screw rotation speed is 20 and proportionality constant that depends on screw geometry is 24.
This implies that;
α = Value = 2
μ = Viscosity = 9
N = Screw Rotation Speed = 20
B = Proportionality Constant that Depends on Screw Geometry = 24
ΔP = αμN / B
ΔP = (2)(9)(20) / 24
ΔP = 360 / 24
ΔP = 15
Therefore, the pressure gradient is 15.
Calculating the Value when the Pressure Gradient, Viscosity, Screw Rotation Speed and Proportionality Constant that Depends on Screw Geometry is Given.
α = ΔP x B / μN
Where;
α = Value
ΔP = Pressure Gradient
μ = Viscosity
N = Screw Rotation Speed
B = Proportionality Constant that Depends on Screw Geometry
Let’s solve an example;
Find the value when the pressure gradient is 20, viscosity is 5, screw rotation speed is 11 and the proportionality constant that depends on screw geometry is 7.
This implies that;
ΔP = Pressure Gradient = 20
μ = Viscosity = 5
N = Screw Rotation Speed = 11
B = Proportionality Constant that Depends on Screw Geometry = 7
α = ΔP x B / μN
α = 20 x 7 / (5)(11)
α = 140 / 55
α = 2.54
Therefore, the value is 2.54. | 468 | 1,713 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2020-24 | latest | en | 0.78585 |
https://math.codidact.com/posts/279044 | 1,624,418,020,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488528979.69/warc/CC-MAIN-20210623011557-20210623041557-00383.warc.gz | 356,553,856 | 10,442 | Q&A
# Why does “unless” mean “if not”?
+2
−1
Harry Gensler. Introduction to Logic (2017 3 ed). p 169.
“Unless” is also equivalent to “if not”; so we also could use “(∼B ⊃ D) (“If you don’t breathe, then you’ll die”).”
Nicholas JJ Smith, Logic: The Laws of Truth (2012). p 115.
The statement “P unless Q” means that if Q is not true, P is true—so we translate it as $¬ , Q→P$.
Using solely the original meaning of "unless" below, please expound why? How does definition 1 below ≡ if not? I know that definition 1 is obsolete, but I'm interested in the etymology. OED Third Edition, June 2017. Screenshot.
A. adv. Only in conjunctional phrases followed by than or that.
1. Forming a conjunctional phrase introducing a case in which an exception to a preceding negative statement (expressed or implied) will or may exist: (not) on a less or lower condition, requirement, etc., than (what is specified). Obsolete.
Why does this post require moderator attention?
Why should this post be closed?
#### 1 comment
-1. "Using solely the original meaning of 'unless' below" is a restrictive condition you've added in for no apparent reason. msh210 7 months ago
+2
−0
The quotes from the books are exactly the kind of ridiculously naive and over-simplified linguistics in introductions to logic that I criticize in this blog article. Particularly for everyday natural language expressions, you will find no shortage of ways where this "translation" provides only a clumsy interpretation, is inadequate, is outright wrong, or is just nonsensical. This also has nothing to do with how mathematicians/logicians use formal logic.
Even then, your question is strange. There's no reason to expect every definition of "unless" to be even clumsily equivalent to this "translation". There's also no reason to expect the etymology of "unless" or this obsolete form have anything at all to do with this logical interpretation. This is almost certainly not the form of "unless" the authors of these books were intending, especially as the given example doesn't use "than" or "that".
Either you're asking how the natural language meaning of "if not" relates to this obsolete form of "unless", in which case this is not a mathematics question; or you're unreasonably asking how this "translation" corresponds to a meaning of "unless" that is clearly not the meaning the authors intended. It would be like asking why "not" is "translated" to logical negation but under the condition that "not" is defined as "nothing" because "not" (allegedly) derives from "nought" which means (among other definitions) "nothing".
Why does this post require moderator attention?
+2
−0
According to the definition you quote, “unless” gives an exception to a preceding negative statement. An exception to a statement is a condition in which the statement does not apply. Therefore the statement does apply only if the condition is not fulfilled (because otherwise the exception applies, and therefore the statement does not). Therefore, as far as logic is concerned, “unless” translates into “if not”.
In describing an exception, the “unless” phrase also suggests that the condition is generally not fulfilled, but that part is irrelevant from a logical point of view (the logic only cares about the truth/falsehood of statements, not about the probability that they will be found to be true).
Why does this post require moderator attention? | 778 | 3,417 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2021-25 | latest | en | 0.932211 |
https://crypto.stackexchange.com/questions/76208/not-using-rsa-in-ot1-2-oblivious-transfer/76209 | 1,585,641,529,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370500331.13/warc/CC-MAIN-20200331053639-20200331083639-00107.warc.gz | 364,614,737 | 31,327 | # Not Using RSA In OT1/2 Oblivious Transfer?
I was reading Oblivious Transfer (Wikipedia) and the 1-2 oblivious transfer. I was confused why RSA was needed. Why wouldn't this work?:
A:m1, m0
A:x1, x2 -> B
B:k, b
B:Xb+k -> v -> A
A:Xb+k-X1, Xb+k-X2 -> k1, k2
A:m0+k0, m1+k1 -> m'1, m'2 -> B
B:m'b-k -> mb
• @JohnHao: it works because (assuming $b=1$) the Bob receives $m'_1 = m_1 + (k^e + x_0 - x_1)^d$, but $k^e + x_0 - x_1$ is essentially a random number, and so Bob cannot recover the e'th root of it (this is the RSA problem). – poncho Dec 5 '19 at 14:42 | 224 | 566 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2020-16 | latest | en | 0.861783 |
http://thetopsites.net/article/54180865.shtml | 1,601,374,279,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401641638.83/warc/CC-MAIN-20200929091913-20200929121913-00543.warc.gz | 126,122,350 | 6,176 | ## Javascript Filter integers
javascript parseint
javascript filter object
javascript array.filter multiple arguments
javascript filter array multiple values
javascript integer
isfloat javascript
javascript filter array of objects by another array of objects
javascript check if string is integer
I'm learning javascript on FreecodeCamp. And, the solution of my problem doesn't help me more. We have to filter integers in an array which is saved in const variable.
```const realNumberArray = [4, 5.6, -9.8, 3.14, 42, 6, 8.34, -2];
const squareList = (arr) => {
"use strict";
const squaredIntegers = arr.filter( (num) => num > 0 && num % parseInt(num) === 0 ).map( (num) => Math.pow(num, 2) );
return squaredIntegers;
};
const squaredIntegers = squareList(realNumberArray);
console.log(squaredIntegers);
```
and I cannot understand really this part `num % parseInt(num) === 0 ).map( (num) => Math.pow(num, 2) );`
Why using parseint() method and Math.pow. Someone could explain me why?
thanks.
`parseInt(num)` gives an integer part of num, for example, `parseInt(3.14) === 3 //true`. Using `num % parseInt(num)` basically gives a difference between the number and its integer part. If it isn't `0`, the number is thrown out. `Math.pow(num)` gives a squared number, which is returned to the new array. Though, `num * num` is faster in that regard, not having to include a module and to call an object property.
Other than that, the code is very crammed in the solution, and I would suggest to break it down to improve readability. Seems like the style in which it is written adds to the confusion.
What are integers in JavaScript?, only means that the numbers don't have a decimal fraction, and 32-bit means that they are within a certain range. Filter Definition & Syntax. The filter() method returns a new array created from all elements that pass a certain test preformed on an original array. Here’s what the syntax looks like: let newArr = oldArr.filter(callback); newArr — the new array that is returned; oldArr — the array to run the filter function on
```const squaredIntegers = arr.filter( (num) => num > 0 && num % parseInt(num) === 0 ).map( (num) => Math.pow(num, 2) );
return squaredIntegers;
```
Here inside the `filter`, it is checked if `num` is positive `(num>0)` and `num` is an `integer`. For checking for integer. `num % parseInt(num)` parseInt changes the num to an `integer`, and the `modulus` of a number with itself is 0 so the condition `num % parseInt(num)==0`. `Math.pow(num,2)` is used to square the num.
filter() Array Method in JavaScript ← Alligator.io, () checks it against the condition. This is useful for accessing properties, in the case of objects. Definition and Usage. The filter() method creates an array filled with all array elements that pass a test (provided as a function). Note: filter() does not execute the function for array elements without values.
Many people have explained this in a very good way. You can also do this part
```const squaredIntegers = arr.filter( (num) => num > 0 && num % parseInt(num) === 0 ).map( (num) => Math.pow(num, 2) );
```
Like this, and it should work without problems. Hopefully, this helps.
```const squaredIntegers = arr.filter( (num) => num > 0 && num % 1 === 0 ).map( (num) => num * num );
```
Number.isInteger(), How do you check if value is a number JavaScript? Summary: in this tutorial, you will learn how to use the JavaScript Array filter() method to filter elements in an array.. Introduction to JavaScript array filter() method. One of the most common tasks when working with an array is to create a new array that contains a subset of elements of the original array.
```const squaredIntegers = arr.filter(
(num) => {
if (Number.isInteger(num) && num >0){
return num;
}}
).map((num) => num *num); // Math.pow(num, 2) can also be used here
return squaredIntegers;
};
```
This could be an easier solution for your problem, although all the above solutions work fine as well. `num*num` could also be replaced with `Math.pow(num, 2)`.
Checking whether a value is an integer in JavaScript, Integers lead an odd life in JavaScript. In the ECMAScript specification, they only exist conceptually: All numbers are always floating point and array - the original array the filter was called on Having these arguments in mind, we can also write the syntax as: let newArray = array.filter(callback(element, index, array)); or let newArray = array.filter((element, index, array) => { //filter 'em elements })
JavaScript Array filter() Method, Well organized and easy to understand Web building tutorials with lots of examples of how to use HTML, CSS, JavaScript, SQL, PHP, Python, Bootstrap, Java In Javascript we have map, filter and reduce, all functions that given an initial list (array of things), transform it into something else, while keeping that same original list intact. Map
JavaScript Number isInteger() Method, This method returns true if the value is of the type Number, and an integer (a number without decimals). Otherwise it returns false. Browser Support. Method. filter () calls a provided callback function once for each element in an array, and constructs a new array of all the values for which callback returns a value that coerces to true. callback is invoked only for indexes of the array which have assigned values; it is not invoked for indexes which have been deleted or which have never been assigned values.
How to filter out only Numbers in an Array using JavaScript, An array in JavaScript can hold different types of value in a single variable. The value can be a string or a number. Sometimes you might have to separate the If you’re starting in JavaScript, maybe you haven’t heard of .map(), .reduce(), and .filter().For me, it took a while as I had to support Internet Explorer 8 until a couple years ago. | 1,363 | 5,844 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2020-40 | latest | en | 0.829235 |
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Comment by Wei_Dai2 on Fairness vs. Goodness · 2009-02-23T16:25:45.000Z · LW · GW
This is fascinating. JW plays C in the last round, even though AA just played D in the next-to-last round. What explains that? Maybe JW's belief in his own heroic story is strong enough to make him sacrifice his self-interest?
Theoretically, of course, utility functions are invariant up to affine transformation, so a utility's absolute sign is not meaningful. But this is not always a good metaphor for real life.
So you're suggesting that real life has some additional structure which is not representable in ordinary game theory formalism? Can you think of an extension to game theory which can represent it? (Mathematically, not just metaphorically.)
Comment by Wei_Dai2 on Epilogue: Atonement (8/8) · 2009-02-07T22:43:00.000Z · LW · GW
Nicholas, our own universe may have an infinite volume, and it's only the speed of light that limits the size of the observable universe. Given that infinite universes are not considered implausible, and starlines are not considered implausible (at least as a fictional device), I find it surprising that you consider starlines that randomly connect a region of size 2^(10^20) to be implausible.
Starlines have to have an average distance of something, right? Why not 2^(10^20)?
Comment by Wei_Dai2 on Epilogue: Atonement (8/8) · 2009-02-07T02:55:35.000Z · LW · GW
Nicholas, suppose Eliezer's fictional universe contains a total of 2^(10^20) star systems, and each starline connects two randomly selected star systems. With a 20 hour doubling speed, the Superhappies, starting with one ship, can explore 2^(t36524/20) random star systems after t years. Let's say the humans are expanding at the same pace. How long will it take, before humans and Superhappies will meet again?
According to the birthday paradox, they will likely meet after each having explored about sqrt(2^(10^20)) = 2^(510^19) star systems, which will take 510^19/(365*24/20) or approximately 10^17 years to accomplish. That should be enough time to get over our attachment to "bodily pain, embarrassment, and romantic troubles", I imagine.
Comment by Wei_Dai2 on Epilogue: Atonement (8/8) · 2009-02-06T19:54:54.000Z · LW · GW
But the tech in the story massively favors the defense, to the point that a defender who is already prepared to fracture his starline network if attacked is almost impossible to conquer (you’d need to advance faster than the defender can send warnings of your attack while maintaining perfect control over every system you’ve captured). So an armed society would have a good chance of being able to cut itself off from even massively superior aliens, while pacifists are vulnerable to surprise attacks from even fairly inferior ones.
I agree, and that's why in my ending humans conquer the Babyeaters only after we develop a defense against the supernova weapon. The fact that the humans can see the defensive potential of this weapon, but the Babyeaters and the Superhappies can't, is a big flaw in the story. The humans sacrificed billions in order to allow the Superhappies to conquer the Babyeaters, but that makes sense only if the Babyeaters can't figure out the same defense that the humans used. Why not?
Also, the Superhappies' approach to negotiation made no game theoretic sense. What they did was, offer a deal to the other side. If they don't accept, impose the deal on them anyway by force. If they do accept, trust that they will carry out the deal without try to cheat. Given these incentives, why would anyone facing a Superhappy in negotiation not accept and then cheat? I don't see any plausible way in which this morality/negotiation strategy could have become a common one in Superhappy society.
Lastly, I note that the Epilogue of the original ending could be named Atonement as well. After being modified by the Superhappies (like how the Confessor was "rescued"?), the humans would now be atoning for having forced their children suffer pain. What does this symmetry tell us, if anything?
Comment by Wei_Dai2 on True Ending: Sacrificial Fire (7/8) · 2009-02-05T22:48:00.000Z · LW · GW
So, what about the fact that all of humanity now knows about the supernova weapon? How is it going to survive the next few months?
Comment by Wei_Dai2 on Three Worlds Decide (5/8) · 2009-02-04T10:25:00.000Z · LW · GW
In case it wasn't clear, the premise of my ending is that the Ship's Confessor really was a violent thief and drug dealer from the 21th century, but his "rescue" was only partially successful. He became more rational, but only pretended to accept what became the dominant human morality of this future, patiently biding time his whole life for an opportunity like this.
Comment by Wei_Dai2 on Three Worlds Decide (5/8) · 2009-02-04T10:03:00.000Z · LW · GW
The Ship's Confessor uses the distraction to anesthetizes everyone except the pilot. He needs the pilot to take command of the starship and to pilot it. The ship stays to observe which star the Superhappy ship came from, then takes off for the nearest Babyeater world. They let the Babyeaters know what happened, and tell them to supernova the star that Superhappies came from at all costs.
When everyone wakes up, the Ship's Confessor convinces the entire crew to erase their memory of the true Alderson's Coupling Constant, ostensibly for the good of humanity. He pretends to do so himself, but doesn't. After the ship returns to human space, he uses his accumulated salary to build a series of hidden doomsday devices around every human colony, and becomes the dictator of humanity through blackmail. Everyone is forced to adopt an utility function of his choosing as their own. With every resource of humanity devoted to the subject, scientists under his direction eventually discover a defense against the supernova weapon, and soon after that, the Babyeaters are conquered, enslaved, and farmed for their crystal brains. Those brains, when extracted and networked in large arrays, turn out to be the cheapest and most efficient computing substrate in the universe. These advantages provide humanity with such a strong competitive edge, that it never again faces an alien that is its match, at least militarily.
Before the universe ends in a big crunch, the Confessed (humanity's eventual name) goes on to colonize more than (10^9)^(10^9) star systems, and to meet and conquer almost as many alien species, but the Superhappy people are never seen again. Their fate becomes one of the most traded futures in the known universe, but those bets will have to remain forever unsettled.
Comment by Wei_Dai2 on The Allais Paradox · 2009-02-01T22:52:00.000Z · LW · GW
Eliezer, I see from this example that the Axiom of Independence is related to the notion of dynamic consistency. But, the logical implication goes only one way. That is, the Axiom of Independence implies dynamic consistency, but not vice versa. If we were to replace the Axiom of Independence with some sort of Axiom of Dynamic Consistency, we would no longer be able to derive expected utility theory. (Similarly with dutch book/money pump arguments, there are many ways to avoid them besides being an expected utility maximizer.)
I'm afraid that the Axiom of Independence cannot really be justified as a basic principle of rationality. Von Neumann and Morgenstern probably came up with it because it was mathematically necessary to derive Expected Utility Theory, then they and others tried to justify it afterward because Expected Utility turned out to be such an elegant and useful idea. Has anyone seen Independence proposed as a principle of rationality prior to the invention of Expected Utility Theory?
Comment by Wei_Dai2 on Value is Fragile · 2009-02-01T03:06:13.000Z · LW · GW
To expand on my categorization of values a bit more, it seems clear to me that at least some human value do not deserved to be forever etched into the utility function of a singleton. Those caused by idiosyncratic environmental characteristics like taste for salt and sugar, for example. To me, these are simply accidents of history, and I wouldn't hesitate (too much) to modify them away in myself, perhaps to be replaced by more interesting and exotic tastes.
What about reproduction? It's a value that my genes programmed into me for their own purposes, so why should I be obligated to stick with it forever?
Or consider boredom. Eventually I may become so powerful that I can easily find the globally optimal course of action for any set of goals I might have, and notice that the optimal course of action often involves repetition of some kind. Why should I retain my desire not to do the same thing over and over again, which was programmed into me by evolution back when minds had a tendency to get stuck in local optimums?
And once I finally came to that realization, I felt less ashamed of values that seemed 'provincial' - but that's another matter.
Eliezer, I wonder if this actually has more to do with your current belief that rationality equals expected utility maximization. For an expected utility maximizer, there is no distinction between 'provincial' and 'universal' values, and certainly no reason to ever feel ashamed of one's values. One just optimizes according to whatever values one happens to have. But as I argued before, human beings are not expected utility maximizers, and I don't see why we should try to emulate them, especially this aspect.
Comment by Wei_Dai2 on Value is Fragile · 2009-02-01T01:33:01.000Z · LW · GW
Tim and Tyrrell, do you know the axiomatic derivation of expected utility theory? If you haven't read http://cepa.newschool.edu/het/essays/uncert/vnmaxioms.htm or something equivalent, please read it first.
Yes, if you change the spaces of states and choices, maybe you can encode every possible agent as an utility function, not just those satisfying certain axioms of "rationality" (which I put in quotes because I don't necessarily agree with them), but that would be to miss the entire point of expected utility theory, which is that it is supposed to be a theory of rationality, and is supposed to rule out irrational preferences. That means using state and choice spaces where those axiomatic constraints have real world meaning.
Comment by Wei_Dai2 on Value is Fragile · 2009-01-31T01:24:37.000Z · LW · GW
A utility function is like a program in a Turing-complete language. If the behaviour can be computed at all, it can be computed by a utility function.
Tim, I've seen you state this before, but it's simply wrong. A utility function is not like a Turing-complete language. It imposes rather strong constraints on possible behavior.
Consider a program which when given the choices (A,B) outputs A. If you reset it and give it choices (B,C) it outputs B. If you reset it again and give it choices (C,A) it outputs C. The behavior of this program cannot be reproduced by a utility function.
Here's another example: When given (A,B) a program outputs "indifferent". When given (equal chance of A or B, A, B) it outputs "equal chance of A or B". This is also not allowed by EU maximization.
Comment by Wei_Dai2 on The Baby-Eating Aliens (1/8) · 2009-01-30T21:03:16.000Z · LW · GW
If the aliens' wetware (er, crystalware) is so efficient that their children are already sentient when they are still tiny relative to adults, why don't the adults have bigger brains and be much more intelligent than humans? Given that they also place high values on science and rationality, had invented agriculture long before humans did, and haven't fought any destructive wars recently, it makes no sense that they have a lower level of technology than humans at this point.
Other than that, I think the story is not implausible. The basic lesson here is the same as in Robin's upload scenarios: when sentience is really cheap, no one will be valued (much) just for being sentient. If we want people to be valued just for being sentient, either the wetware/crystalware/hardware can't be too efficient, or we need to impose some kind of artificial scarcity on sentience.
Comment by Wei_Dai2 on Value is Fragile · 2009-01-30T19:08:15.000Z · LW · GW
Maybe we don't mean the same thing by boredom?
I'm using Eliezer's definition: a desire not to do the same thing over and over again. For a creature with roughly human-level brain power, doing the same thing over and over again likely means it's stuck in a local optimum of some sort.
Genome equivalents which don't generate terminally valued individual identity in the minds they descrive should outperform those that do.
I don't understand this. Please elaborate.
Why not just direct expected utility? Pain and pleasure are easy to find but don't work nearly as well.
I suppose you mean why not value external referents directly instead of indirectly through pain and pleasure. As long as wireheading isn't possible, I don't see why the latter wouldn't work just as well as the former in many cases. Also, the ability to directly value external referents depends on a complex cognitive structure to assess external states, which may be more vulnerable in some situations to external manipulation (i.e. unfriendly persuasion or parasitic memes) than hard-wired pain and pleasure, although the reverse is probably true in other situations. It seems likely that evolution would come up with both.
Define sexual. Most sexual creatures are too simple to value the first two. Most plausible posthumans aren't sexual in a traditional sense.
I mean reproduction where more than one party contributes genetic material and/or parental resources. Even simple sexual creatures probably have some notion of beauty and/or status to help attract/select mates, but for the simplest perhaps "instinct" would be a better word than "value".
- likely values for intelligent creatures with sexual reproduction (music, art, literature, humor)
Disagree.
These all help signal fitness and attract mates. Certainly not all intelligent creatures with sexual reproduction will value exactly music, art, literature, and humor, but it seems likely they will have values that perform the equivalent functions.
Comment by Wei_Dai2 on Value is Fragile · 2009-01-30T09:04:53.000Z · LW · GW
We can sort the values evolution gave us into the following categories (not necessarily exhaustive). Note that only the first category of values is likely to be preserved without special effort, if Eliezer is right and our future is dominated by singleton FOOM scenarios. But many other values are likely to survive naturally in alternative futures.
• likely values for all intelligent beings and optimization processes (power, resources)
• likely values for creatures with roughly human-level brain power (boredom, knowledge)
• likely values for all creatures under evolutionary competition (reproduction, survival, family/clan/tribe)
• likely values for creatures under evolutionary competition who cannot copy their minds (individual identity, fear of personal death)
• likely values for creatures under evolutionary competition who cannot wirehead (pain, pleasure)
• likely values for creatures with sexual reproduction (beauty, status, sex)
• likely values for intelligent creatures with sexual reproduction (music, art, literature, humor)
• likely values for intelligent creatures who cannot directly prove their beliefs (honesty, reputation, piety)
• values caused by idiosyncratic environmental characteristics (salt, sugar)
• values caused by random genetic/memetic drift and co-evolution (Mozart, Britney Spears, female breasts, devotion to specific religions)
The above probably isn't controversial, rather the disagreement is mainly on the following:
• the probabilities of various future scenarios
• which values, if any, can be preserved using approaches such as FAI
• which values, if any, we should try to preserve
I agree with Roko that Eliezer has made his case in an impressive fashion, but it seems that many of us are still not convinced on these three key points.
Take the last one. I agree with those who say that human values do not form a consistent and coherent whole. Another way of saying this is that human beings are not expected utility maximizers, not as individuals and certainly not as societies. Nor do most of us desire to become expected utility maximizers. Even amongst the readership of this blog, where one might logically expect to find the world's largest collection of EU-maximizer wannabes, few have expressed this desire. But there is no principled way to derive an utility function from something that is not an expected utility maximizer!
Is there any justification for trying to create an expected utility maximizer that will forever have power over everyone else, whose utility function is derived using a more or less arbitrary method from the incoherent values of those who happen to live in the present? That is, besides the argument that it is the only feasible alternative to a null future. Many of us are not convinced of this, neither the "only" nor the "feasible".
Comment by Wei_Dai2 on Eutopia is Scary · 2009-01-12T20:35:43.000Z · LW · GW
Why make science a secret, instead of inventing new worlds with new science for people to explore? Have you heard of "Theorycraft"? It's science applied to the World of Warcraft, and for some, Theorycraft is as much fun as the game it's based on.
Is there something special about the science of base-level reality that makes it especially fun to explore and discover? I think the answer is yes, but only if it hasn't already been explored and then covered up again and made into a game. It's the difference between a natural and an artificial challenge.
Comment by Wei_Dai2 on Complex Novelty · 2008-12-20T04:28:50.000Z · LW · GW
One day we'll discover the means to quickly communicate insights from one individual to another, say by directly copying and integrating the relevant neural circuitry. Then, in order for an insight to be Fun, it will have to be novel to transhumanity, not just the person learning or discovering it. Learning something the fast efficient way will not be Fun because there's not true effort. Pretending that the new way doesn't exist, and learning the old-fashioned way, will not be Fun because there's not true victory.
I'm not sure there are enough natural problems in the universe to supply the whole of transhumanity with an adequate quantity of potential insights. "Natural" meaning not invented for the sole purpose of providing an artificial challenge. Personally, I can't see how solving the n-th random irrelevant mathematical problem is any better than lathing the n-th table leg.
Comment by Wei_Dai2 on Not Taking Over the World · 2008-12-16T21:29:11.000Z · LW · GW
Anna Salamon wrote: Is it any safer to think ourselves about how to extend our adaptation-executer preferences than to program an AI to figure out what conclusions we would come to, if we did think a long time?
First, I don't know that "think about how to extend our adaptation-executer preferences" is the right thing to do. It's not clear why we should extend our adaptation-executer preferences, especially given the difficulties involved. I'd backtrack to "think about what we should want".
Putting that aside, the reason that I prefer we do it ourselves is that we don't know how to get an AI to do something like this, except through opaque methods that can't be understood or debugged. I imagine the programmer telling the AI "Stop, I think that's a bug." and the AI responding with "How would you know?"
g wrote: Wei Dai, singleton-to-competition is perfectly possible, if the singleton decides it would like company.
In that case the singleton might invent a game called "Competition", with rules decided by itself. Anti-prediction says that it's pretty unlikely those rules would happen to coincide with the rules of base-level reality, so base-level reality would still be controlled by the singleton.
Comment by Wei_Dai2 on Not Taking Over the World · 2008-12-16T09:24:06.000Z · LW · GW
Robin wrote: Having to have an answer now when it seems an likely problem is very expensive.
(I think you meant to write "unlikely" here instead of "likely".)
Robin, what is your probability that eventually humanity will evolve into a singleton (i.e., not necessarily through Eliezer's FOOM scenario)? It seems to me that competition is likely to be unstable, whereas a singleton by definition is. Competition can evolve into a singleton, but not vice versa. Given that negentropy increases as mass squared, most competitors have to remain in the center, and the possibility of a singleton emerging there can't ever be completely closed off. BTW, a singleton might emerge from voluntary mergers, not just one competitor "winning" and "taking over".
Another reason to try to answer now, instead of later, is that coming up with a good answer would persuade more people to work towards a singleton, so it's not just a matter of planning for a contingency.
Comment by Wei_Dai2 on Not Taking Over the World · 2008-12-16T08:18:33.000Z · LW · GW
An expected utility maximizer would know exactly what to do with unlimited power. Why do we have to think so hard about it? The obvious answer is that we are adaptation executioners, not utility maximizers, and we don't have an adaptation for dealing with unlimited power. We could try to extrapolate an utility function from our adaptations, but given that those adaptations deal only with a limited set of circumstances, we'll end up with an infinite set of possible utility functions for each person. What to do?
James D. Miller: But about 100 people die every minute!
Peter Norvig: Refusing to act is like refusing to allow time to pass.
What about acting to stop time? Preserve Earth at 0 kelvin. Gather all matter/energy/negentropy in the rest of the universe into secure storage. Then you have as much time as you want to think.
Comment by Wei_Dai2 on What I Think, If Not Why · 2008-12-14T20:45:00.000Z · LW · GW
Maybe we don't need to preserve all of the incompressible idiosyncrasies in human morality. Considering that individuals in the post-Singularity world will have many orders of magnitude more power than they do today, what really matter are the values that best scale with power. Anything that scales logarithmically for example will be lost in the noise compared to values that scale linearly. Even if we can't understand all of human morality, maybe we will be able to understand the most important parts.
Just throwing away parts of one's utility function seems bad. That can't be optimal right? Well, as Peter de Blanc pointed out, it can be if there is no feasible alternative that improves expected utility under the original function. We should be willing to lose our unimportant values to avoid or reduce even a small probability of losing the most important ones. With CEV, we're supposed to implement it with the help of an AI that's not already Friendly, and if we don't get it exactly right on the first try, we can't preserve even our most important values. Given that we don't know how to safely get an unFriendly AI to do anything, much less something this complicated, the probability of failure seems quite large.
Comment by Wei_Dai2 on What I Think, If Not Why · 2008-12-14T06:58:00.000Z · LW · GW
(Eliezer, why do you keep using "intelligence" to mean "optimization" even after agreeing with me that intelligence includes other things that we don't yet understand?)
Morality does not compress
You can't mean that morality literally does not compress (i.e. is truly random). Obviously there are plenty of compressible regularities in human morality. So perhaps what you mean is that it's too hard or impossible to compress it into a small enough description that humans can understand. But, we also have no evidence that effective universal optimization in the presence of real-world computational constraints (as opposed to idealized optimization with unlimited computing power) can be compressed into a small enough description that humans can understand.
Comment by Wei_Dai2 on What I Think, If Not Why · 2008-12-13T03:44:00.000Z · LW · GW
Eliezer, you write as if there is no alternative to this plan, as if your hand is forced. But that's exactly what some people believe about neural networks. What about first understanding human morality and moral growth, enough so that we (not an AI) can deduce and fully describe someone's morality (from his brain scan, or behavior, or words) and predict his potential moral growth in various circumstances, and maybe enough to correct any flaws that we see either in the moral content or in the growth process, and finally program the seed AI's morality and moral growth based on that understanding once we're convinced it's sufficiently good? Your logic of (paraphrasing) "this information exists only in someone's brain so I must let the AI grab it directly without attempting to understand it myself" simply makes no sense. First the conclusion doesn't follow from the premise, and second if you let the AI grab and extrapolate the information without understanding it yourself, there is no way you can predict a positive outcome.
In case people think I'm some kind of moralist for harping on this so much, I think there are several other aspects of intelligence that are not captured by the notion of "optimization". I gave some examples here. We need to understand all aspects of intelligence, not just the first facet for which we have a good theory, before we can try to build a truly Friendly AI.
Comment by Wei_Dai2 on What I Think, If Not Why · 2008-12-12T19:24:00.000Z · LW · GW
Eliezer, as far as I can tell, "reflective equilibrium" just means "the AI/simulated non-sentient being can't think of any more changes that it wants to make" so the real question is what counts as a change that it wants to make? Your answer seems to be whatever is decided by "a human library of non-introspectively-accessible circuits". Well the space of possible circuits is huge, and "non-introspectively-accessible" certainly doesn't narrow it down much. And (assuming that "a human library of circuits" = "a library of human circuits") what is a "human circuit"? A neural circuit copied from a human being? Isn't that exactly what you argued against in "Artificial Mysterious Intelligence"?
(It occurs to me that perhaps you're describing your understanding of how human beings do moral growth and not how you plan for an AI/simulated non-sentient being to do it. But if so, that understanding seems to be similar in usefulness to "human beings use neural networks to decide how to satisfy their desires.")
Eliezer wrote: I don't think I'm finished with this effort, but are you unsatisfied with any of the steps I've taken so far? Where?
The design space for "moral growth" is just as big as the design space for "optimization" and the size of the target you have to hit in order to have a good outcome is probably just as small. More than any dissatisfaction with the specific steps you've taken, I don't understand why you don't seem to (judging from your public writings) view the former problem to be as serious and difficult as the latter one, if not more so, because there is less previous research and existing insights that you can draw from. Where are the equivalents of Bayes, von Neumann-Morgenstern, and Pearl, for example?
Comment by Wei_Dai2 on What I Think, If Not Why · 2008-12-12T07:47:00.000Z · LW · GW
Isn't CEV just a form of Artificial Mysterious Intelligence? Eliezer's conversation with the anonymous AIfolk seems to make perfect sense if we search and replace "neural network" with "CEV" and "intelligence" with "moral growth/value change".
How can the same person that objected to "Well, intelligence is much too difficult for us to understand, so we need to find some way to build AI without understanding how it works." by saying "Look, even if you could do that, you wouldn't be able to predict any kind of positive outcome from it. For all you knew, the AI would go out and slaughter orphans." be asking us to to place our trust in the mysterious moral growth of nonsentient but purportedly human-like simulations?
Comment by Wei_Dai2 on Artificial Mysterious Intelligence · 2008-12-09T06:59:39.000Z · LW · GW
Eliezer, MacKay's math isn't very difficult. I think it will take you at most a couple of hours to go through how he derived his equations, understand what they mean, and verify that they are correct. (If I knew you were going to put this off for a year, I'd mentioned that during the original discussion.) After doing that, the idea that sexual reproduction speeds up evolution by gathering multiple bad mutations together to be disposed of at once will become pretty obvious in retrospect.
Jeff, I agree with what you are saying, but you're using the phrase "sexual selection" incorrectly, which might cause confusion to others. I think what you mean is "natural selection in a species with sexual reproduction". "Sexual selection" actually means "struggle between the individuals of one sex, generally the males, for the possession of the other sex".
Comment by Wei_Dai2 on The Weighted Majority Algorithm · 2008-11-15T01:57:00.000Z · LW · GW
Even if P=BPP, that just means that giving up randomness causes "only" a polynomial slowdown instead of an exponential one, and in practice we'll still need to use pseudorandom generators to simulate randomness.
It seems clear to me that noise (in the sense of randomized algorithms) does have power, but perhaps we need to develop better intuitions as to why that is the case.
Comment by Wei_Dai2 on The Weighted Majority Algorithm · 2008-11-14T22:41:00.000Z · LW · GW
To generalize Peter's example, a typical deterministic algorithm has low Kolmogorov complexity, and therefore its worst-case input also has low Kolmogorov complexity and therefore a non-negligible probability under complexity-based priors. The only possible solutions to this problem I can see are:
2. redesign the deterministic algorithm so that it has no worst-case input
3. do a cost-benefit analysis to show that the cost of doing either 1 or 2 is not justified by the expected utility of avoiding the worst-case performance of the original algorithm, then continue to use the original deterministic algorithm
The main argument in favor of 1 is that its cost is typically very low, so why bother with 2 or 3? I think Eliezer's counterargument is that 1 only works if we assume that in addition to the input string, the algorithm has access to a truly random string with a uniform distribution, but in reality we only have access to one input, i.e., sensory input from the environment, and the so called random bits are just bits from the environment that seem to be random.
My counter-counterargument is to consider randomization as a form of division of labor. We use one very complex and sophisticated algorithm to put a lower bound on the Kolmogorov complexity of a source of randomness in the environment, then after that, this source of randomness can be used by many other simpler algorithms to let them cheaply and dramatically reduce the probability of hitting a worst-case input.
Or to put it another way, before randomization, the environment does not need to be a malicious superintelligence for our algorithms to hit worst-case inputs. After randomization, it does.
Comment by Wei_Dai2 on Complexity and Intelligence · 2008-11-05T06:59:00.000Z · LW · GW
Rolf, I was implicitly assuming that even knowing BB(k), it still takes O(k) bits to learn BB(k+1). But if this assumption is incorrect, then I need to change the setup of my prediction game so that the input sequence consists of the unary encodings of BB(1), BB(2), BB(4), BB(8), â¦, instead. This shouldnât affect my overall point, I think.
Comment by Wei_Dai2 on Complexity and Intelligence · 2008-11-05T06:23:00.000Z · LW · GW
After further thought, I need to retract my last comment. Consider P(next symbol is 0|H) again, and suppose you've seen 100 0's so far, so essentially you're trying to predict BB(101). The human mathematician knows that any non-zero number he writes down for this probability would be way too big, unless he resorts to non-constructive notation like 1/BB(101). If you force him to answer "over and over, what their probability of the next symbol being 0 is" and don't allow him to use notation like 1/BB(101) then he'd be forced to write down an inconsistent probability distribution. But in fact the distribution he has in mind is not computable, and that explains how he can beat Solomonoff Induction.
Comment by Wei_Dai2 on Complexity and Intelligence · 2008-11-05T03:18:10.000Z · LW · GW
Good question, Eliezer. If the human mathematician is computable, why isn't it already incorporated into Solomonoff Induction? It seems to me that the human mathematician does not behave like a Bayesian. Let H be the hypothesis that the input sequence is the unary encodings of Busy Beaver numbers. The mathematician will try to estimate, as best as he can, P(next symbol is 0|H). But when the next symbol turns out to be 1, he doesn't do a Bayesian update and decrease P(H), but instead says "Ok, so I was wrong. The next Busy Beaver number is bigger than I expected."
I'm not sure I understand what you wrote after "to be fair". If you think a Solomonoff inductor can duplicate the above behavior with an alternative setup, can you elaborate how?
Comment by Wei_Dai2 on Complexity and Intelligence · 2008-11-05T00:33:14.000Z · LW · GW
A halting oracle is usually said to output 1s or 0s, not proofs or halting times, right?
It's easy to use such an oracle to produce proofs and halting times. The following assumes that the oracle outputs 1 if the input TM halts, and 0 otherwise.
For proofs: Write a program p which on inputs x and i, enumerates all proofs. If it finds a proof for x, and the i-th bit of that proof is 1, then it halts, otherwise it loops forever. Now query the oracle with (p,x,0), (p,x,1), ..., and you get a proof for x if it has a proof.
Halting times: Write a program p which on inputs x and i, runs x for i steps. If x halts before i steps, then it halts, otherwise it loops forever. Now query the oracle with (p,x,2), (p,x,4), (p,x,8), ..., until you get an output of "1" and then use binary search to get the exact halting time.
I don't recall if I've mentioned this before, but Solomonoff induction in the mixture form makes no mention of the truth of its models. It just says that any computable probability distribution is in the mixture somewhere, so you can do as well as any computable form of cognitive uncertainty up to a constant.
Eliezer, if what you say is true, then it shouldn't be possible for anyone, using just a Turing machine, to beat Solomonoff Induction at a pure prediction game (by more than a constant), even if the input sequence is uncomputable. But here is a counterexample. Suppose the input sequence consists of the unary encodings of Busy Beaver numbers BB(1), BB(2), BB(3), …, that is, BB(1) number of 1s followed by a zero, then BB(2) number of 1s followed by a 0, and so on. Let’s ask the predictor, after seeing n input symbols, what is the probability that it will eventually see a 0 again, and call this p(n). With Solomonoff Induction, p(n) will approach arbitrarily close to 0 as you increase n. A human mathematician on the other hand will recognize that the input sequence may not be computable and won’t let p(n) fall below some non-zero bound.
Comment by Wei_Dai2 on Complexity and Intelligence · 2008-11-04T06:51:20.000Z · LW · GW
Nick wrote: Good point, but when the box says "doesn't halt", how do I know it's correct?
A halting-problem oracle can be used for all kinds of things besides just checking whether an individual Turing machine will halt or not. For example you can use it to answer various mathematical questions and produce proofs of the answers, and then verify the proofs yourself. You should be able to obtain enough proofs to convince yourself that the black box is not just giving random answers or just being slightly smarter than you are.
If P!=NP, you should be able to convince yourself that the black box has at least exponentially more computational power than you do. So if you are an AI with say the computational resources of a solar system, you should be able to verify that the black box either contains exotic physics or has access to more resources than the rest of the universe put together.
Eliezer wrote: So once again I say: it is really hard to improve your math abilities with eyes open in a way that you couldn't theoretically do with eyes closed.
It seems to me that an AI should/can never completely rule out the possibility that the universe contains physics that is mathematically more powerful than what it has already incorporated into itself, so it should always keep its eyes open. Even if it has absorbed the entire universe into itself, it might still be living inside a simulation, right?
Comment by Wei_Dai2 on Complexity and Intelligence · 2008-11-04T04:15:29.000Z · LW · GW
In fact, it's just bloody hard to fundamentally increase your ability to solve math problems in a way that "no closed system can do" just by opening the system. So far as I can tell, it basically requires that the environment be magic and that you be born with faith in this fact.
Eliezer, you're making an important error here. I don’t think it affects the main argument you're making in this article (that considerations of "complexity" doesn't rule out self-improving minds), but this error may have grave consequences elsewhere. The error is that while the environment does have to be magic, you don't need to have faith in this, just not have faith that it's impossible.
Suppose you get a hold of a black box that seems to act as a halting-problem oracle. You’ve thrown thousands of problems at it, and have never seen in incorrect or inconsistent answer. What are the possibilities here? Either (A) the environment really is magic (i.e. there is uncomputable physics that enables implementation of actual halting-problem oracles), or (B) the box is just giving random answers that happen to be correct by chance, or (C) you’re part of a simulation where the box is giving all possible combinations of answers and you happen to be in the part of the simulation where the box is giving correct answers. As long as your prior probability for (A) is not zero, as you do more and more tests and keep getting correct answers, it’s posterior probability will eventually dominate (B) and (C).
Why is this so important? Well in standard Solomonoff Induction, the prior for (A) is zero, and if we program that into an AI, it won’t do the right thing in this situation. This may have a large effect on expected utility (of us, people living today), because while the likelihood of us living in an uncomputable universe with halting-problem oracles is low, the utility we gain from correctly recognizing and exploiting such a universe could be huge.
Comment by Wei_Dai2 on Fundamental Doubts · 2008-07-16T00:52:23.000Z · LW · GW
Some problems are hard to solve, and hard even to define clearly. It's possible that "qualia" is not referring to anything meaningful, but unless you are able to explain why it feels meaningful to someone, but isn't really, I don't think you should demand that they stop using it.
Having said that, here's my attempt at an operational definition of qualia.
Comment by Wei_Dai2 on Living in Many Worlds · 2008-06-12T11:55:00.000Z · LW · GW
Eliezer, suppose the nature of the catastrophe is such that everyone on the planet dies instantaneously and painlessly. Why should such deaths bother you, given that identical people are still living in adjacent branches? If avoiding death is simply a terminal value for you, then I don't see why encouraging births shouldn't be a similar terminal value.
I agree that the worlds in which we survive may not be pleasant, but average utilitarianism implies that we should try to minimize such unpleasant worlds that survive, rather than the existential risk per se, which is still strongly counterintuitive.
I don't know what you are referring to by "hard to make numbers add up on anthropics without Death events". If you wrote about that somewhere else, I've missed it.
A separate practical problem I see with the combination of MWI and consequentialism is that due to branching, the measure of worlds a person is responsible for is always rapidly and continuously decreasing, so that for example I'm now responsible for a much smaller portion of the multiverse than I was just yesterday or even a few seconds ago. In theory this doesnât matter because the costs and benefits of every choice I face are reduced by the same factor, so the relative rankings are preserved. But in practice this seems pretty demotivational, since the subjective mental cost of making an effort appears to stay the same, while the objective benefits of such effort decreases rapidly. Eliezer, I'm curious how you've dealt with this problem.
Comment by Wei_Dai2 on Living in Many Worlds · 2008-06-11T18:49:00.000Z · LW · GW
Put me down as a long time many-worlder who doesn't see how it makes average utilitarianism more attractive.
It seems to me that MWI poses challenges for both average utilitarianism and sum utilitarianism. For sum utilitarianism, why bother to bring more potential people into existence in this branch, if those people are living in many other branches already?
But I wonder if Eliezer has considered that MWI plus average utilitarianism seems to imply that we don't need to worry about certain types of existential risk. If some fraction of the future worlds that we're responsible for gets wiped out, that wouldn't lower the average utility, unless for some reason the fraction that gets wiped out would otherwise have had an average utility that's higher than the average of the surviving branches. Assuming that's not the case, the conclusion follows that we don't need to worry about these risks, which seems pretty counterintuitive.
Comment by Wei_Dai2 on Argument Screens Off Authority · 2007-12-20T11:00:47.000Z · LW · GW
Eliezer, what is your view of the relationship between Bayesian Networks and Solomonoff Induction? You've talked about both of these concepts on this blog, but I'm having trouble understanding how they fit together. A Google search for both of these terms together yields only one meaningful hit, which happens to be a mailing list post by you. But it doesn't really touch on my question.
On the face of it, both Bayesian Networks and Solomonoff Induction are "Bayesian", but they seem to be incompatible with each other. In the Bayesian Networks approach, conditional probabilities are primary, and the full probability distribution function is more of a mathematical formalism that stays in the background. Solomonoff Induction on the other hand starts with a fully specified (even if uncomputable) prior distribution and derives any conditional probabilities from it as needed. Do you have any idea how to reconcile these two approaches?
Comment by Wei_Dai2 on Not for the Sake of Happiness (Alone) · 2007-11-23T21:37:45.000Z · LW · GW
Toby, how do you get around the problem that the greatest sum of happiness across all lifes probably involves turning everyone into wireheads and putting them in vats? Or in an even more extreme scenario, turning the universe into computers that all do nothing but repeatedly runs a program that simulates a person in an ultimate state of happiness. Assuming that we have access to limited resources, these methods seem to maximize happiness for a given amount of resources.
I'm sure you agree that this is not something we do want. Do you think that it is something we should want, or that the greatest sum of happiness across all lifes can be achieved in some other way?
Comment by Wei_Dai2 on Not for the Sake of Happiness (Alone) · 2007-11-23T01:29:28.000Z · LW · GW
I agree with Eliezer here. Not all values can be reduced to desire for happiness. For some of us, the desire not to be wireheaded or drugged into happiness is at least as strong as the desire for happiness. This shouldn't be a surprise since there were and still are pyschoactive substances in our environment of evolutionary adaptation.
I think we also have a more general mechanism of aversion towards triviality, where any terminal value that becomes "too easy" loses its value (psychologically, not just over evolutionary time). I'm guessing this is probably because many of our terminal values (art, science, etc.) exist because they helped our ancestors attract mates by signaling genetic superiority. But you can't demonstrate genetic superiority by doing something easy.
Toby, I read your comment several times, but still can't figure out what distinction you are trying to draw between the two senses of value. Can you give an example or thought experiment, where valuing happiness in one sense would lead you to do one thing, and valuing it in the other sense would lead you to do something else?
Michael, do you have a more specific reference to something Parfit has written?
Comment by Wei_Dai2 on Natural Selection's Speed Limit and Complexity Bound · 2007-11-08T22:08:00.000Z · LW · GW
Eliezer, I just noticed that you've updated the main post again. The paper by Worden that you link to makes the mistake of assuming no crossing or even chromosomal assortment, as you can see from the following quotes. It's not surprising that sex doesn't help under those assumptions.
(being quote)
Next consider what happens to one of the haploid genotypes j in one generation. Through random mating, it gets paired with another haploid genotype k, with probability q; then the pair have a probability of surviving sigmajk.
...
(b) Crossing: Similarly, in a realistic model of crossing, we can show that it always decreases the diploid genotype information Jµ. This is not quite the same as proving that crossing always decreases Iµ, but is a powerful plausibility argument that it does so. In that case, crossing will not violate the limit.
(end quote)
As for not observing species gaining thousands of bits per generation, that might be due to the rarity of beneficial mutations. A dog not apparently having greater morphological or biochemical complexity than a dinosaur can also be explained in many other ways.
If you have the time, I think it would be useful to make another post on this topic, since most people who read the original article will probably not see the detailed discussions in the comments or even notice the Addendum. You really should cite MacKay. His paper does provide a theoretical explanation for what happens in the simulations, if you look at the equations and how they are derived.
Comment by Wei_Dai2 on Natural Selection's Speed Limit and Complexity Bound · 2007-11-07T18:55:00.000Z · LW · GW
Eliezer, MacKay actually does predict what we have observed in the simulations. Specifically equation 28 predicts it if you let δf=f instead of δf=f-1/2. You need to make that change to the equation because in your simulation with Beneficial=0, mutations only flips 1 bits to 0, whereas in MacKay's model mutations also flip 0 bits to 1 with equal probability.
Comment by Wei_Dai2 on Natural Selection's Speed Limit and Complexity Bound · 2007-11-07T11:43:00.000Z · LW · GW
I only ran 300 generations, but I just redid them with 5000 generations (which took a few hours), and the results aren't much different. See plots at http://www.weidai.com/fitness/plot3.png and http://www.weidai.com/fitness/plot4.png.
I also reran the simulations with Number=100. Fitness is lower at all values of Mutation (by about 1/3), but it's still linear in 1/Mutation^2, not 1/Mutation. The relationship between Fitness and Number is not clear to me at this point. As Eliezer said, the combinatorial argument I gave isn't really relevant.
Also, with Number=1000, Genome=1000, Mutate=0.005, Fitness stabilizes at around 947. So, extrapolating from this, when 1/Mutate^2 is much larger than Genome, which is the case for human beings, almost the entire genome can be maintained against mutations. It doesn't look like this line of inquiry gives us a reason to believe that most of human DNA is really junk.
Elizer, I've noticed an apparent paradox in information theory that may or may not be related to your "disconnect between variance going as the square root of a randomized genome, and the obvious argument that eliminating half the population is only going to get you 1 bit of mathematical information." It may be of interest in either case, so I'll state it here.
Suppose Alice is taking a test consisting of M true/false questions. She has no clue how to answer them, so it seems that the best she can do is guess randomly and get an expected score of M/2. Fortunately her friend Bob has broken into the teacher's office and stolen the answer key, but unfortunately he can't send her more than 1 bit of information before the test ends. What can they do, assuming they planned ahead of time?
The naive answer would be to have Bob send Alice the answer to one of the questions, which raises the expected score to 1+(M-1)/2.
A better solution is to have Bob tell Alice whether "true" answers outnumber "false" answers. If the answers are uniformly distributed, the variance of the number of "true" answers is M/4, which means Alice can get an expected score of M/2+sqrt(M)/2 if she answers all "true" or all "false" according to what Bob tells her. So here's the paradox: how did Alice get sqrt(M)/2 more answers correct when Bob only sent her 1 bit of information?
(What if the teacher knew this might happen and made the number of "true" answers exactly equal to the number of "false" answers? Alice and Bob should have established a common random bit string R of length M ahead of time. Then Bob can send Alice a bit indicating whether she should answer according to R or the complement of R, with the same expected outcome.)
Comment by Wei_Dai2 on Natural Selection's Speed Limit and Complexity Bound · 2007-11-07T02:37:00.000Z · LW · GW
Eliezer, the simulation is a great idea. I've used it to test the following hypothesis: given sufficiently large population and genome size, the number of useful bits that a sexual species can maintain against a mutation probability (per base) of m is O(1/m^2). The competing hypothesis is the one given in your opening post, namely that it's O(1/m).
To do this, I set Number=1000, Genome=1000, Beneficial=0, and let Mutation range from 0.03 to 0.14 in steps of 0.01. Then I plotted Fitness (which in the program equals the number of useful bits in the genome) against both 1/Mutation and 1/Mutation^2. I think the results [1] are pretty clear: when 1/Mutation^2 is small compared to Number and Genome, Fitness is linear in 1/Mutation^2, not 1/Mutation.
Oh, I rewrote the simulation code in C++ to make it run faster. It's available at http://www.weidai.com/fitness/adhoc.cpp.
[1] http://www.weidai.com/fitness/fitness.htm, first plot is Fitness vs 1/Mutation, second is Fitness vs 1/Mutation^2.
Comment by Wei_Dai2 on Natural Selection's Speed Limit and Complexity Bound · 2007-11-06T21:47:00.000Z · LW · GW
Tom McCabe, having 100 chromosomes with no recombination lets you maintain the genome against a mutation pressure of 10 bits per generation, not 100. (See my earlier comment.) But that's still much better than 1 bit per generation, which is what you get with no sex.
Comment by Wei_Dai2 on Natural Selection's Speed Limit and Complexity Bound · 2007-11-06T21:27:00.000Z · LW · GW
Eliezer, you are mistaken that eliminating half of the population gives only 1 bit of mathematical information. If you have a population of N individuals, there are C(N,N/2) = N!/((N/2)!*(N/2)!) different ways to eliminate half of them. See http://en.wikipedia.org/wiki/Combination. Therefore it takes log_2(C(N,N/2)) (which is O(N)) bits to specify how to eliminate half of the population.
So, it appears that with sexual recombination, the maximum number of bits a species can gain per generation is min(O(G^0.5), O(N)).
Oh, sending 10 bits each with a 60% probability of being correct actually lets you send a total of just 0.29 bits of information. Each bit in this communications channel gives the receiver only 0.029 bits of information, using a formula from http://en.wikipedia.org/wiki/Binary_symmetric_channel. But, the amount of information actually received still grows linearly with the number of bits put into the channel.
Comment by Wei_Dai2 on Natural Selection's Speed Limit and Complexity Bound · 2007-11-06T11:02:00.000Z · LW · GW
I think MacKay's "If variation is created by recombination, the population can gain O(G^0.5) bits per generation." is correct. Here's my way of thinking about it. Suppose we take two random bit strings of length G, each with G/2-G^0.5 zeros and G/2+G^0.5 ones, randomly mix them twice, then throw away the result has fewer ones. What is the expected number of ones in the surviving mixed string? It's G/2+G^0.5+O(G^0.5).
Or here's another way to think about it. Parent A has 100 good (i.e., above average fitness) genes and 100 bad genes. Same with parent B. They reproduce sexually and have 4 children, two with 110 good genes and 90 bad genes, the other two (who do not survive to further reproduce) with 90 good genes and 110 bad genes. Now in one generation they've managed to eliminate 10 bad genes instead of just 1.
This seems to imply that the human genome may have much more than 25 MB of information.
Comment by Wei_Dai2 on Torture vs. Dust Specks · 2007-11-01T01:38:00.000Z · LW · GW
I have argued in previous comments that the utility of a person should be discounted by his or her measure, which may be based on algorithmic complexity. If this "torture vs specks" dilemma is to have the same force under this assumption, we'd have to reword it a bit:
Would you prefer that the measure of people horribly tortured for fifty years increases by x/3^^^3, or that the measure of people who get dust specks in their eyes increases by x?
I argue that no one, not even a superintelligence, can actually face such a choice. Because x is at most 1, x/3^^^3 is at most 1/3^^^3. But how can you increase the measure of something by more than 0 but no more than 1/3^^^3? You might, perhaps, generate a random number between 0 and 3^^^3 and do something only if that random number is 0. But algorithmic information theory says that for any program (even a superintelligence), there are pseudorandom sequences that it cannot distinguish from truly random sequences, and the prior probability that your random number generator is generating such a pseudorandom sequence is much higher than 1/3^^^3. Therefore the probability of that "random" number being 0 (or being any other number that you can think of) is actually much larger than 1/3^^^3.
Therefore, if someone tells you "measure of ... increases by x/3^^^3", in your mind you've got to be thinking "... increases by y" for some y much larger than 1/3^^^3. I think my theories explains both those who answer SPECKS and those who say no answer is possible.
Comment by Wei_Dai2 on Pascal's Mugging: Tiny Probabilities of Vast Utilities · 2007-10-24T05:15:00.000Z · LW · GW
Regarding the comments about exploding brains, it's a wonder to me that we are able to think about these issues and not lose our sanity. How is it that a brain evolved for hunting/gathering/socializing is able to consider these problems at all? Not only that, but we seem to have some useful intuitions about these problems. Where on Earth did they come from?
Nick> Does your proposal require that one accepts the SIA?
Yes, but using a complexity-based measure as the anthropic probability measure implies that the SIA's effect is limited. For example, consider two universes, the first with 1 observer, and the second with 2. If all of the observers have the same complexity you'd assign a higher prior probability (i.e., 2/3) to being in the second universe. But if the second universe has an infinite number of observers, the sum of their measures can't exceed the measure of the universe as a whole, so the "presumptuous philosopher" problem is not too bad.
Nick> If I understand your suggestion correctly, you propose that the same anthropic probability measure should also be used as a measure of moral importance.
Yes, in fact I think there are good arguments for this. If you have an anthropic probability measure, you can argue that it should be used as the measure of moral importance, since everyone would prefer that was the case from behind the veil of ignorance. On the other hand, if you have a measure of moral importance, you can argue that for decisions not involving externalities, the global best case can be obtained if people use that measure as the anthropic probability measure and just consider their self interests.
BTW, when using both anthropic reasoning and moral discounting, it's easy to accidentally apply the same measure twice. For example, suppose the two universes both have 1 observer each, but the observer in the second universe has twice the measure of the one in the first universe. If you're asked to guess which universe you're in with some payoff if you guess right, you don't want to think "There's 2/3 probability that I'm in the second universe, and the payoff is twice as important if I guess 'second', so the expected utility of guessing 'second' is 4 times as much as the EU of guessing 'first'."
I think that to avoid this kind of confusion and other anthropic reasoning paradoxes (see http://groups.google.com/group/everything-list/browse_frm/thread/dd21cbec7063215b/), it's best to consider all decisions and choices from a multiversal objective-deterministic point of view. That is, when you make a decision between choices A and B, you should think "would I prefer if everyone in my position (i.e., having the same perceptions and memories as me) in the entire multiverse chose A or B?" and ignore the temptation to ask "which universe am I likely to be in?".
But that may not work unless you believe in a Tegmarkian multiverse. If you don't, you may have to use both anthropic reasoning and moral discounting, being very careful not to double-count.
Comment by Wei_Dai2 on Pascal's Mugging: Tiny Probabilities of Vast Utilities · 2007-10-23T01:11:00.000Z · LW · GW
Eliezer, what if the mugger (Matrix-claimant) also says that he is the only person who has that kind of power, and he knows there is just one copy of you in the whole universe? Is the probability of that being true less than 1/3^^^^3?
Comment by Wei_Dai2 on Pascal's Mugging: Tiny Probabilities of Vast Utilities · 2007-10-21T21:29:00.000Z · LW · GW
Eliezer, I think Robin's guess about mangled worlds is interesting, but irrelevant to this problem. I'd guess that for you, P(mangled worlds is correct) is much smaller than P(it's right that I care about people in proportion to the weight of the branches they are in). So Robin's idea can't explain why you think that is the right thing to do.
Nick, your paper doesn't seem to mention the possibility of discounting people by their algorithmic complexity. Is that an option you considered? | 13,577 | 59,486 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2021-25 | latest | en | 0.928151 |
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#### Resources tagged with Combinations similar to Cross-country Race:
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Broad Topics > Decision Mathematics and Combinatorics > Combinations
### Cross-country Race
##### Stage: 3 Challenge Level:
Eight children enter the autumn cross-country race at school. How many possible ways could they come in at first, second and third places?
### Colour Building
##### Stage: 3 Challenge Level:
Using only the red and white rods, how many different ways are there to make up the other colours of rod?
### Pattern of Islands
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In how many distinct ways can six islands be joined by bridges so that each island can be reached from every other island...
### Clocked
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Is it possible to rearrange the numbers 1,2......12 around a clock face in such a way that every two numbers in adjacent positions differ by any of 3, 4 or 5 hours?
### Dicing with Numbers
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In how many ways can you arrange three dice side by side on a surface so that the sum of the numbers on each of the four faces (top, bottom, front and back) is equal?
### Painting Cubes
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Imagine you have six different colours of paint. You paint a cube using a different colour for each of the six faces. How many different cubes can be painted using the same set of six colours?
### Sam Again
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Here is a collection of puzzles about Sam's shop sent in by club members. Perhaps you can make up more puzzles, find formulas or find general methods.
##### Stage: 2, 3 and 4 Challenge Level:
Three dice are placed in a row. Find a way to turn each one so that the three numbers on top of the dice total the same as the three numbers on the front of the dice. Can you find all the ways to do. . . .
### 1 Step 2 Step
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Liam's house has a staircase with 12 steps. He can go down the steps one at a time or two at time. In how many different ways can Liam go down the 12 steps?
### Flight of the Flibbins
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Blue Flibbins are so jealous of their red partners that they will not leave them on their own with any other bue Flibbin. What is the quickest way of getting the five pairs of Flibbins safely to. . . .
### Brailler
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The machine I use to produce Braille messages is faulty and one of the pins that makes a raised dot is not working. I typed a short message in Braille. Can you work out what it really says?
### Small Change
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In how many ways can a pound (value 100 pence) be changed into some combination of 1, 2, 5, 10, 20 and 50 pence coins?
### Flagging
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How many tricolour flags are possible with 5 available colours such that two adjacent stripes must NOT be the same colour. What about 256 colours?
### Take Three from Five
##### Stage: 4 Challenge Level:
Caroline and James pick sets of five numbers. Charlie chooses three of them that add together to make a multiple of three. Can they stop him?
### Man Food
##### Stage: 2 and 3 Challenge Level:
Sam displays cans in 3 triangular stacks. With the same number he could make one large triangular stack or stack them all in a square based pyramid. How many cans are there how were they arranged?
### The Secret World of Codes and Code Breaking
##### Stage: 2, 3 and 4
When you think of spies and secret agents, you probably wouldn’t think of mathematics. Some of the most famous code breakers in history have been mathematicians.
### The Olympic Torch Tour
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Imagine you had to plan the tour for the Olympic Torch. Is there an efficient way of choosing the shortest possible route?
### Elevenses
##### Stage: 3 Challenge Level:
How many pairs of numbers can you find that add up to a multiple of 11? Do you notice anything interesting about your results?
##### Stage: 3 Challenge Level:
In a league of 5 football teams which play in a round robin tournament show that it is possible for all five teams to be league leaders.
### Teddy Town
##### Stage: 1, 2 and 3 Challenge Level:
There are nine teddies in Teddy Town - three red, three blue and three yellow. There are also nine houses, three of each colour. Can you put them on the map of Teddy Town according to the rules?
### Lesser Digits
##### Stage: 3 Challenge Level:
How many positive integers less than or equal to 4000 can be written down without using the digits 7, 8 or 9?
### Even Up
##### Stage: 3 Challenge Level:
Consider all of the five digit numbers which we can form using only the digits 2, 4, 6 and 8. If these numbers are arranged in ascending order, what is the 512th number?
### Factoring a Million
##### Stage: 4 Challenge Level:
In how many ways can the number 1 000 000 be expressed as the product of three positive integers?
##### Stage: 3 Challenge Level:
Is it possible to use all 28 dominoes arranging them in squares of four? What patterns can you see in the solution(s)?
### Last Biscuit
##### Stage: 3 and 4 Challenge Level:
A game that demands a logical approach using systematic working to deduce a winning strategy
### Coins
##### Stage: 3 Challenge Level:
A man has 5 coins in his pocket. Given the clues, can you work out what the coins are?
### More Children and Plants
##### Stage: 2 and 3 Challenge Level:
This challenge extends the Plants investigation so now four or more children are involved.
### Scratch Cards
##### Stage: 4 Challenge Level:
To win on a scratch card you have to uncover three numbers that add up to more than fifteen. What is the probability of winning a prize?
### Tea Cups
##### Stage: 2 and 3 Challenge Level:
Place the 16 different combinations of cup/saucer in this 4 by 4 arrangement so that no row or column contains more than one cup or saucer of the same colour.
### Crossing the Bridge
##### Stage: 3 Challenge Level:
Four friends must cross a bridge. How can they all cross it in just 17 minutes?
### Six Times Five
##### Stage: 3 Challenge Level:
How many six digit numbers are there which DO NOT contain a 5?
### More Plant Spaces
##### Stage: 2 and 3 Challenge Level:
This challenging activity involves finding different ways to distribute fifteen items among four sets, when the sets must include three, four, five and six items.
### Chances Are
##### Stage: 4 Challenge Level:
Which of these games would you play to give yourself the best possible chance of winning a prize?
### Coin Tossing Games
##### Stage: 4 Challenge Level:
You and I play a game involving successive throws of a fair coin. Suppose I pick HH and you pick TH. The coin is thrown repeatedly until we see either two heads in a row (I win) or a tail followed by. . . .
### Nines and Tens
##### Stage: 3 Challenge Level:
Explain why it is that when you throw two dice you are more likely to get a score of 9 than of 10. What about the case of 3 dice? Is a score of 9 more likely then a score of 10 with 3 dice?
### Trominoes
##### Stage: 3 and 4 Challenge Level:
Can all but one square of an 8 by 8 Chessboard be covered by Trominoes? | 1,708 | 7,268 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2017-43 | latest | en | 0.905455 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/3610/2/a/e/ | 1,708,888,999,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474641.34/warc/CC-MAIN-20240225171204-20240225201204-00797.warc.gz | 860,825,697 | 58,180 | # Properties
Label 3610.2.a.e Level $3610$ Weight $2$ Character orbit 3610.a Self dual yes Analytic conductor $28.826$ Analytic rank $0$ Dimension $1$ CM no Inner twists $1$
# Related objects
Show commands: Magma / PariGP / SageMath
## Newspace parameters
comment: Compute space of new eigenforms
[N,k,chi] = [3610,2,Mod(1,3610)]
mf = mfinit([N,k,chi],0)
lf = mfeigenbasis(mf)
from sage.modular.dirichlet import DirichletCharacter
H = DirichletGroup(3610, base_ring=CyclotomicField(2))
chi = DirichletCharacter(H, H._module([0, 0]))
N = Newforms(chi, 2, names="a")
//Please install CHIMP (https://github.com/edgarcosta/CHIMP) if you want to run this code
chi := DirichletCharacter("3610.1");
S:= CuspForms(chi, 2);
N := Newforms(S);
Level: $$N$$ $$=$$ $$3610 = 2 \cdot 5 \cdot 19^{2}$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 3610.a (trivial)
## Newform invariants
comment: select newform
sage: f = N[0] # Warning: the index may be different
gp: f = lf[1] \\ Warning: the index may be different
Self dual: yes Analytic conductor: $$28.8259951297$$ Analytic rank: $$0$$ Dimension: $$1$$ Coefficient field: $$\mathbb{Q}$$ Coefficient ring: $$\mathbb{Z}$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 190) Fricke sign: $$-1$$ Sato-Tate group: $\mathrm{SU}(2)$
## $q$-expansion
comment: q-expansion
sage: f.q_expansion() # note that sage often uses an isomorphic number field
gp: mfcoefs(f, 20)
$$f(q)$$ $$=$$ $$q - q^{2} + 3 q^{3} + q^{4} - q^{5} - 3 q^{6} - 5 q^{7} - q^{8} + 6 q^{9}+O(q^{10})$$ q - q^2 + 3 * q^3 + q^4 - q^5 - 3 * q^6 - 5 * q^7 - q^8 + 6 * q^9 $$q - q^{2} + 3 q^{3} + q^{4} - q^{5} - 3 q^{6} - 5 q^{7} - q^{8} + 6 q^{9} + q^{10} - 4 q^{11} + 3 q^{12} + q^{13} + 5 q^{14} - 3 q^{15} + q^{16} - 3 q^{17} - 6 q^{18} - q^{20} - 15 q^{21} + 4 q^{22} + 7 q^{23} - 3 q^{24} + q^{25} - q^{26} + 9 q^{27} - 5 q^{28} + 3 q^{29} + 3 q^{30} + 2 q^{31} - q^{32} - 12 q^{33} + 3 q^{34} + 5 q^{35} + 6 q^{36} + 2 q^{37} + 3 q^{39} + q^{40} + 6 q^{41} + 15 q^{42} + 6 q^{43} - 4 q^{44} - 6 q^{45} - 7 q^{46} + 3 q^{48} + 18 q^{49} - q^{50} - 9 q^{51} + q^{52} + 13 q^{53} - 9 q^{54} + 4 q^{55} + 5 q^{56} - 3 q^{58} + 9 q^{59} - 3 q^{60} - 12 q^{61} - 2 q^{62} - 30 q^{63} + q^{64} - q^{65} + 12 q^{66} + 3 q^{67} - 3 q^{68} + 21 q^{69} - 5 q^{70} - 6 q^{72} + 11 q^{73} - 2 q^{74} + 3 q^{75} + 20 q^{77} - 3 q^{78} + 2 q^{79} - q^{80} + 9 q^{81} - 6 q^{82} - 10 q^{83} - 15 q^{84} + 3 q^{85} - 6 q^{86} + 9 q^{87} + 4 q^{88} - 2 q^{89} + 6 q^{90} - 5 q^{91} + 7 q^{92} + 6 q^{93} - 3 q^{96} + 2 q^{97} - 18 q^{98} - 24 q^{99}+O(q^{100})$$ q - q^2 + 3 * q^3 + q^4 - q^5 - 3 * q^6 - 5 * q^7 - q^8 + 6 * q^9 + q^10 - 4 * q^11 + 3 * q^12 + q^13 + 5 * q^14 - 3 * q^15 + q^16 - 3 * q^17 - 6 * q^18 - q^20 - 15 * q^21 + 4 * q^22 + 7 * q^23 - 3 * q^24 + q^25 - q^26 + 9 * q^27 - 5 * q^28 + 3 * q^29 + 3 * q^30 + 2 * q^31 - q^32 - 12 * q^33 + 3 * q^34 + 5 * q^35 + 6 * q^36 + 2 * q^37 + 3 * q^39 + q^40 + 6 * q^41 + 15 * q^42 + 6 * q^43 - 4 * q^44 - 6 * q^45 - 7 * q^46 + 3 * q^48 + 18 * q^49 - q^50 - 9 * q^51 + q^52 + 13 * q^53 - 9 * q^54 + 4 * q^55 + 5 * q^56 - 3 * q^58 + 9 * q^59 - 3 * q^60 - 12 * q^61 - 2 * q^62 - 30 * q^63 + q^64 - q^65 + 12 * q^66 + 3 * q^67 - 3 * q^68 + 21 * q^69 - 5 * q^70 - 6 * q^72 + 11 * q^73 - 2 * q^74 + 3 * q^75 + 20 * q^77 - 3 * q^78 + 2 * q^79 - q^80 + 9 * q^81 - 6 * q^82 - 10 * q^83 - 15 * q^84 + 3 * q^85 - 6 * q^86 + 9 * q^87 + 4 * q^88 - 2 * q^89 + 6 * q^90 - 5 * q^91 + 7 * q^92 + 6 * q^93 - 3 * q^96 + 2 * q^97 - 18 * q^98 - 24 * q^99
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
comment: embeddings in the coefficient field
gp: mfembed(f)
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
1.1
0
−1.00000 3.00000 1.00000 −1.00000 −3.00000 −5.00000 −1.00000 6.00000 1.00000
$$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Atkin-Lehner signs
$$p$$ Sign
$$2$$ $$1$$
$$5$$ $$1$$
$$19$$ $$-1$$
## Inner twists
This newform does not admit any (nontrivial) inner twists.
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 3610.2.a.e 1
19.b odd 2 1 190.2.a.b 1
57.d even 2 1 1710.2.a.g 1
76.d even 2 1 1520.2.a.j 1
95.d odd 2 1 950.2.a.c 1
95.g even 4 2 950.2.b.a 2
133.c even 2 1 9310.2.a.u 1
152.b even 2 1 6080.2.a.b 1
152.g odd 2 1 6080.2.a.x 1
285.b even 2 1 8550.2.a.bm 1
380.d even 2 1 7600.2.a.a 1
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
190.2.a.b 1 19.b odd 2 1
950.2.a.c 1 95.d odd 2 1
950.2.b.a 2 95.g even 4 2
1520.2.a.j 1 76.d even 2 1
1710.2.a.g 1 57.d even 2 1
3610.2.a.e 1 1.a even 1 1 trivial
6080.2.a.b 1 152.b even 2 1
6080.2.a.x 1 152.g odd 2 1
7600.2.a.a 1 380.d even 2 1
8550.2.a.bm 1 285.b even 2 1
9310.2.a.u 1 133.c even 2 1
## Hecke kernels
This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{2}^{\mathrm{new}}(\Gamma_0(3610))$$:
$$T_{3} - 3$$ T3 - 3 $$T_{7} + 5$$ T7 + 5 $$T_{13} - 1$$ T13 - 1
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$T + 1$$
$3$ $$T - 3$$
$5$ $$T + 1$$
$7$ $$T + 5$$
$11$ $$T + 4$$
$13$ $$T - 1$$
$17$ $$T + 3$$
$19$ $$T$$
$23$ $$T - 7$$
$29$ $$T - 3$$
$31$ $$T - 2$$
$37$ $$T - 2$$
$41$ $$T - 6$$
$43$ $$T - 6$$
$47$ $$T$$
$53$ $$T - 13$$
$59$ $$T - 9$$
$61$ $$T + 12$$
$67$ $$T - 3$$
$71$ $$T$$
$73$ $$T - 11$$
$79$ $$T - 2$$
$83$ $$T + 10$$
$89$ $$T + 2$$
$97$ $$T - 2$$ | 2,816 | 5,705 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-10 | latest | en | 0.370862 |
http://www.personal.psu.edu/mgz/Finance405/Free%20Cash%20Flow.htm | 1,580,087,802,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251694071.63/warc/CC-MAIN-20200126230255-20200127020255-00192.warc.gz | 271,946,829 | 2,876 | Free Cash Flow
What is free cash flow?
It is the annual amount of cash that a firm has available for all of its investors after covering all expenses including new investment.
How can I calculate free cash flow?
FCF = EBIT*(1-tax rate) + Depreciation – New Investment
Notes:
1. EBIT*(1-tax rate) is the cash flow from the firm’s operations assuming no debt financing. We have been calling it NOPLAT. The taxes included in EBIT*(1-tax rate) are a little bit too high for firms that have debt and utilize the interest tax deduction. We use EBIT*(1-tax rate) because we use the after-tax cost of debt in our WACC calculation.
2. New Investment has two components:
a) The increase in net working capital
b) The increase in property, plant, and equipment
In class, we defined free cash flow as EBIT*(1-tax rate)+Depreciation – Capital Expenditures. This definition fails to deduct from cash flows the expenditures required to finance increases in working capital. While capital expenditures can be very difficult to estimate, working capital can be assumed to be a constant fraction of sales or EBIT and increases in working capital can be measured accordingly.
What do I do with free cash flow?
The above explanation makes it easy to calculate free cash flow for previous years. You want to project free cash flow into the future and take its present value using WACC.
How do I project future free cash flows?
Maybe the easiest thing to do is estimate a growth rate for EBIT*(1-tax rate). Most analyst would probably say this is the same as the growth rate for sales. You could either base your estimate of future growth on past growth rates or on management estimates, or some method of your own choosing. A catch might be in estimating the growth rate for new investment. If Target is in a “growth phase”, it is building lots of new stores meaning capital expenditures are substantial and expected to remain that way until a sufficient number of new stores have been opened. You need to assume some future levels for this key variable. Probably, percentage increases in working capital are the same as percentage increases in sales.
Sales growth: Same store sales versus overall sales growth
Target and similar firms can make their sales grow either by growing same-store sales or by increasing the number of stores or both. Notice that growing sales by increasing the number of stores tends to be the more expensive way to grow sales because of the capital expenditures needed to construct a new store. Suppose we have two similar firms with equal sales for this year. They also are projected to have the same rate of growth in future sales. However, one of these firms achieves its sales growth by building new stores while the other achieves its sales growth by growing same-store sales. Can you see that one firm has larger cash flows than the other? | 607 | 2,890 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2020-05 | latest | en | 0.91617 |
https://www.jiskha.com/display.cgi?id=1354487248 | 1,503,217,621,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886106358.80/warc/CC-MAIN-20170820073631-20170820093631-00235.warc.gz | 910,128,769 | 4,218 | physics
posted by .
1.
A football player weighing 75 kg running at 2 m/s toward west tackles a 70 kg player running at 1.5 m/s in the opposite direction. What is the final velocity of the players if they both fall together?
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More Similar Questions | 699 | 2,705 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2017-34 | latest | en | 0.967546 |
https://www.youphysics.education/ideal-gas/ | 1,680,122,800,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949025.18/warc/CC-MAIN-20230329182643-20230329212643-00724.warc.gz | 1,185,595,384 | 6,856 | # Equation of state of an ideal gas - Isotherms of an ideal gas
Real substances show a varied and often complex behavior which is difficult to translate into a simple equation. For example, they may exist in different states or undergo phase changes; depending on the type of substance, their density varies with temperature and can increase or decrease with it.
However, for certain simplified situations it is possible to find equations of state that relate the thermodynamic variables used to describe the state of a substance. The simplest case is that of a gas.
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When a gas has low density (its molecules are very far from each other) and the interactions between its molecules are weak, we can explain its behavior using the concept of ideal gas.
An ideal gas is a theoretical model of gas whose equation of state is derived under the following hypotheses:
• The interactions between the molecules are negligible and they only undergo elastic collisions between them.
• The volume of the gas molecules is negligible.
The equation of state obtained by applying the above assumptions describes reasonably well the behavior of real gases for pressure and temperature conditions far from those of a change of state. That is, a pressure low enough so that the molecules are distant from each other and can therefore be considered as point particles and temperatures high enough so that the gas will not change phase to a liquid.
The equation of state of an ideal gas is given by:
where n is the number of moles and R is the ideal gas constant, whose value in SI units is:
We can graph the equation of state of an ideal gas in a PV diagram. To do so, we isolate p from the equation of state:
In this equation, we can assign different values to the temperature (in kelvin), which gives us an equation of two variables, p and V for each temperature value, for example: | 390 | 1,998 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2023-14 | latest | en | 0.948238 |
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## #26 2013-04-28 17:44:35
gAr
Star Member
Offline
### Re: Marbles without replacement (probability)
For k colors with n marbles each, I find it to be:
May require some modification when number of marbles are different.
Last edited by gAr (2013-04-28 20:53:23)
"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?
"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."
## #27 2013-04-28 19:03:43
bobbym
Offline
### Re: Marbles without replacement (probability)
Hi gAr;
That is a nice one, it will come in handy.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #28 2013-04-28 20:04:07
gAr
Star Member
Offline
### Re: Marbles without replacement (probability)
Hi,
Got a recursion to work!
Here's a sage code:
#### Code:
```A = 10
B = 10
C = 10
D = 10
def f(a,b,c,d):
if (A-a == 2 or B-b == 2 or C-c == 2 or D-d == 2):
return (A+B+C+D-(a+b+c+d))
return a/Integer(a+b+c+d)*f(a-1,b,c,d) + b/Integer(a+b+c+d)*f(a,b-1,c,d) + c/Integer(a+b+c+d)*f(a,b,c-1,d) + d/Integer(a+b+c+d)*f(a,b,c,d-1)
print f(A,B,C,D)```
It returns : 30008/9139
"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?
"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."
## #29 2013-04-28 20:09:23
bobbym
Offline
### Re: Marbles without replacement (probability)
Hi gAr;
Yeparoo! Very good!
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #30 2013-04-28 20:15:09
gAr
Star Member
Offline
### Re: Marbles without replacement (probability)
Yeah, that looks much neater than the formula!
I think this will help in many problems involving probability..
"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?
"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."
## #31 2013-04-28 20:16:56
bobbym
Offline
### Re: Marbles without replacement (probability)
Hi;
Have you tried the formula in post#26 on this problem?
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #32 2013-04-28 20:20:40
gAr
Star Member
Offline
### Re: Marbles without replacement (probability)
Yes, and we must multiply that by 2, I forgot to include that.
"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?
"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."
## #33 2013-04-28 20:23:32
bobbym
Offline
### Re: Marbles without replacement (probability)
I am not getting the correct answer then, when I multiply by 2 I get:
I am using n = 10 and k = 4.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #34 2013-04-28 20:31:56
gAr
Star Member
Offline
### Re: Marbles without replacement (probability)
Perhaps I made a mistake when translating my program to the formula? Or you took out n from the sum ?
Here's the code I used:
#### Code:
```def ans(n,k):
summ = 0
for i in range(1,k+1):
pr = 1
for j in range(1,i):
pr *= (k-j)*n/Integer(4*n-j)
summ += (i)*(i+1)/2 * pr * (n-1)/Integer(4*n-i)
return 2*summ```
ans(10,4) returns 30008/9139
"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?
"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."
## #35 2013-04-28 20:37:47
bobbym
Offline
### Re: Marbles without replacement (probability)
Hi;
Why the Integer() command?
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #36 2013-04-28 20:40:39
gAr
Star Member
Offline
### Re: Marbles without replacement (probability)
It causes problem like flooring if it's not written..
In python, we need to convert it to float.
"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?
"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."
## #37 2013-04-28 20:44:31
bobbym
Offline
### Re: Marbles without replacement (probability)
Hi;
There is a typo in the formula. In post #26 there is (k-i), in post #34, (k-j).
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Pearl2013
Novice
Offline
## #39 2013-04-28 20:52:45
gAr
Star Member
Offline
### Re: Marbles without replacement (probability)
Hi bobbym,
Ah yes, sorry!
"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?
"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."
## #40 2013-04-28 21:00:16
bobbym
Offline
### Re: Marbles without replacement (probability)
No problem! Thank you for the formula.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #41 2013-04-28 21:06:38
gAr
Star Member
Offline
### Re: Marbles without replacement (probability)
You're welcome!
I think we can't write a formula like that when the number of marbles are different for different colors, since it involves permutation?
Anyway, I'll stop here, spent almost a day on this one!
"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?
"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."
## #42 2013-04-28 21:09:58
bobbym
Offline
### Re: Marbles without replacement (probability)
Hi;
I will play with it a little in a bit. Right now there is a little bit of an administrative matter to watch for.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #43 2013-04-28 21:19:00
gAr
Star Member
Offline
### Re: Marbles without replacement (probability)
Okay..
"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?
"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."
## #44 2013-04-28 21:23:47
bobbym
Offline
### Re: Marbles without replacement (probability)
Hi;
Looks like the worst is over. Now about that formula...
If we hold k = 4 and vary n then we come up with this:
for n = 2, 3, 4, ...
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #45 2013-04-28 21:42:48
gAr
Star Member
Offline
### Re: Marbles without replacement (probability)
That's a good one!
"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?
"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."
## #46 2013-04-28 21:44:54
bobbym
Offline
### Re: Marbles without replacement (probability)
Oh yes and quite useless unless you have exactly 4 colors.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #47 2013-04-28 21:56:48
gAr
Star Member
Offline
### Re: Marbles without replacement (probability)
Anyway, we can easily extend the recursion if not the formula for any number of variables and experiment with the terms.
"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?
"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."
## #48 2013-04-28 22:03:25
bobbym
Offline
### Re: Marbles without replacement (probability)
There are a lot of possibilities here.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #49 2013-04-28 22:05:37
gAr
Star Member
Offline
### Re: Marbles without replacement (probability)
Yes.
"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?
"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."
## #50 2013-04-28 22:30:18
bobbym
Offline
### Re: Marbles without replacement (probability)
Hi;
I am going to get a little sleep. See you later.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof. | 2,946 | 10,608 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2014-15 | longest | en | 0.872187 |
https://www.physicsforums.com/threads/about-the-concept-of-work-done-high-school-student.748902/ | 1,596,698,837,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439736883.40/warc/CC-MAIN-20200806061804-20200806091804-00402.warc.gz | 796,970,627 | 15,765 | # About the concept of work done (High school student)
## Main Question or Discussion Point
Hi I am kind of Confused about an example in work done:
It says that a man applies 250 N parallel to the slope to push a box up a slope for 3 m, Now I know that the formula for work done if Force by Perpendicular distance which is 250 x 3 = 750 (which is the correct answer in my book). The thing is, isn't the object "Gaining" Gravitational potential energy ie change in its energy ie transfer of energy and therefor we should add that energy to the 750??
This is where I am confused.
Can any one help please :)
Edit: Or is the Force x Distance formula includes all changes in energy together?
Related Classical Physics News on Phys.org
berkeman
Mentor
Hi I am kind of Confused about an example in work done:
It says that a man applies 250 N parallel to the slope to push a box up a slope for 3 m, Now I know that the formula for work done if Force by Perpendicular distance which is 250 x 3 = 750 (which is the correct answer in my book). The thing is, isn't the object "Gaining" Gravitational potential energy ie change in its energy ie transfer of energy and therefor we should add that energy to the 750??
This is where I am confused.
Can any one help please :)
Edit: Or is the Force x Distance formula includes all changes in energy together?
Welcome to the PF.
Since you are explicitly given the force needed to push the box up the ramp, that includes both the force needed to push against the friction of the box on the ramp, and the force needed to lift the box up the delta-height that it gains. Does that make more sense?
Welcome to the PF.
Since you are explicitly given the force needed to push the box up the ramp, that includes both the force needed to push against the friction of the box on the ramp, and the force needed to lift the box up the delta-height that it gains. Does that make more sense?
So what you mean is that if we get the component of each force and see the distance moved by each to get the total energy transfered it would add up to the Force x distance? so this formula should really be
Resultant force by displacement right?
Doc Al
Mentor
So what you mean is that if we get the component of each force and see the distance moved by each to get the total energy transfered it would add up to the Force x distance? so this formula should really be
Resultant force by displacement right?
It depends on what you want to compute.
If you want the work done by the man, then that is simply the force he exerts times the displacement. But to determine the final energy of the box, you'll need to consider all the forces acting, which includes gravity and friction. | 613 | 2,698 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2020-34 | latest | en | 0.942686 |
https://vintage-kitchen.com/food/faq-is-milk-denser-than-water/ | 1,653,033,238,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662531762.30/warc/CC-MAIN-20220520061824-20220520091824-00050.warc.gz | 690,483,858 | 43,601 | # FAQ: Is milk denser than water?
A gallon is a measure of volume, and density is directly proportional to the mass of a fixed volume. Milk consists of about 87% water and contains other substances that: are heavier than water, excluding fat. A gallon of milk is heavier than a gallon of water.
## What is the densest water or milk?
Answer: Milk is denser than water. Simply put, fat molecules and proteins somehow cling to water molecules to form a uniform mixture, increasing the density of milk, which is heavier than water.
## What is the densest water or milk Why?
milk is denser than water.
## Is milk more liquid than water?
For many people, milk is just a white liquid that looks and feels like water, but feels a little “thicker.” Milk can even consist of 90% water, while the rest is a mixture of proteins, fats and lactose (milk sugar).
## Does milk float in water?
Most of the milk is water and the rest is mostly fat. Both ice cream and fat are lighter than water, so frozen milk would float.
## Which milk is the densest?
Cream or milk fat has a lighter density than water and will float on the surface of non-homogenized milk. If you skim the surface, some of the fat remains, the denser parts, and the milk is denser. This explains why skim milk is denser.
## Which liquid has the highest density?
So the liquid with the highest density is mercury. It is the only metal that exists in liquid form at room temperature. Other metals are hard and rigid and exist in solid phase. It has a density of about 13,546 grams per cubic centimeter.
## What is the density of milk or water?
Literature shows that the density of non-fat solids in milk at 10°C is about 1.6 times that of water, while the density of milk fat is 0.93 times that of milk. ‘water.
## Is milk denser than coffee?
Coffee is slightly less dense than milk, so it tends to float on top. However, if it is poured quickly enough, it will penetrate deeper into the milk and the two liquids will form a turbulent, chaotic mixture.
## Is blood thicker than milk?
We in the West are used to saying that “blood is thicker than water”; but the Arabs have the idea that blood is thicker than milk than mother’s milk. Among them, two children fed from the same breast are called “milk brothers” or “milk brothers”; and the bond between them is very strong.
## Is honey denser than water?
Honey is also a liquid; with few dense particles, so it is denser than water.
## How to make milk thicker?
One of the easiest ways to thicken milk is to boil it on the stove. As it heats up, the liquid parts of the milk begin to evaporate. Don’t forget to stir all the time! If you want to make condensed milk, add sugar before heating it.
## Is cream heavier than milk?
We think cream is “heavier” than milk because we think mentally about what the fat content can do to us, but it’s actually lighter than milk, which is why cream will come up. The cream, which is lighter than the milk, will rise to the top where it can be skimmed off by hand.
## Is the milk good?
Cow’s milk is a good source of protein and calcium, as well as nutrients such as vitamin B12 and iodine. It also contains magnesium, which is important for bone development and muscle function, and whey and casein, which play a role in lowering blood pressure. | 761 | 3,309 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-21 | latest | en | 0.955739 |
https://www.coursehero.com/file/14070117/Final-Math-Problems/ | 1,542,172,444,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039741628.8/warc/CC-MAIN-20181114041344-20181114063344-00520.warc.gz | 852,644,074 | 99,294 | Final Math Problems
# Final Math Problems - 1 1 SFL 260 Section 1,2,5 Dr E...
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1 1 SFL 260- Section 1,2,5 Dr. E. Jeffrey Hill Final Exam Example Math Problems Winter 2016 85. You receive a \$8,400 gift from your long lost uncle and invest it in a savings account at Bank of American Fork which pays 1.8% annual interest. After one year how much interest will you have earned? 86. You and your spouse have brand new twin baby girls, and that makes three children. You realize that soon the time will come to become a real Mormon family and purchase a mini-van. If the estimated price of a two-year old, low-mileage Honda Odyssey mini-van in four years will be \$18,000, how much money must you invest at the beginning of each month over the next four years if you can earn 3.75% annual interest with monthly compounding? 87. Dr. Hill is kind of weird with numbers. He likes the number 9. It’s his lucky number. So he invests \$9,999.99 as a lump sum into a savings account which has monthly compounding. The interest rate is 9.99 percent per year. After 9 years how much interest will he have earned?
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2 88. You are applying for a loan and are required to submit a balance sheet with your net worth. You own a 2013 Lexus ES350 that you bought last month for \$22,995. The Kelly Bluebook value for this car is \$25,995. You owe \$19,150 on the car loan for the Lexus. You pay off your Visa credit card every month and have not paid any credit card interest this year. The current Visa credit card balance is \$5,522 and the next statement is due in 15 days. You have a student loan balance of \$2,500. You presently have \$925 in your checking account and \$1,140 in your savings account. You own 200 shares of WalMart stock that you purchased for \$95.50 per share. It is now selling for \$188.42 per share. You own computers and other electronics that you purchased for \$11,000 but could probably sell today on E-bay for \$1,200. Your gross income is \$110,000 per year. What is
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https://plainmath.net/93084/find-a-formula-for-the-exponential-funct | 1,670,311,367,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711074.68/warc/CC-MAIN-20221206060908-20221206090908-00332.warc.gz | 478,655,134 | 12,184 | # Find a formula for the exponential function passing through the points (-2, 1) and (3, 32).
Find a formula for the exponential function passing through the points (-2, 1) and (3, 32).
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domino671v
Let the exponential function be
$y=a{b}^{x}$
At
At
Now,
$\frac{a{b}^{3}}{a{b}^{-2}}=\frac{32}{1}\phantom{\rule{0ex}{0ex}}⇒{b}^{5}=32={2}^{5}\phantom{\rule{0ex}{0ex}}⇒b=2\phantom{\rule{0ex}{0ex}}\therefore a=\frac{32}{{b}^{3}}=\frac{32}{8}=4$
Hence the exponential function is
$y={4.2}^{x}$ | 246 | 697 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 28, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2022-49 | latest | en | 0.777786 |
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sta108hw2_2011
# sta108hw2_2011 - Measures of association(chap 2.9 some...
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Topics to be covered this week (Due on Wednesday, July 6) Statistics 108 Summer Session I Monday, June 27 Inference in regression (chaps 2.1-2.3), confidence interval for the mean response and prediction interval (chaps 2.4-2.6). Tuesday, June 28 Analysis of variance for regression (chap 2.7), general linear test (chap 2.8), measures of association (chap 2.9). Wednesday, June 29
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Unformatted text preview: Measures of association (chap 2.9), some considerations (chap 2.10), Diagnostics (chaps 3.1-3.5, 3.8), Transformations (chap 3.9). Homework 2 (due on Tuesday, June 5) Problems 2.16, 2.17, 2.18, 2.19, 2.26, 3.6 (skip parts (d) and (e)), 3.17 (in part (b), you may calculate R 2 instead of SSE and then choose the transformation that maximizes R 2 )....
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# Problem 1272. The almost-birthday problem.
• What is Size?
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## calculating ball charge in a ball mill
Calculate ball charge ball mill Henan Mining. ball mill charge calculation gh ineffi ciency This raises the point that in some cases the Bond EF4 oversize factor calculation is not suf cient to compensate fi for the milling inef ciencies that mill charge calculation residentialpaintersHow To Calculate Charge In Ball.
## Mill Steel Charge Volume Calculation
We can calculate the steel charge volume of a ball or rod mill and express it as the % of the volume within the liners that is filled with grinding media. While the mill is stopped, the charge volume can be gotten by measuring the diameter inside the liners and the distance from the top of the charge to the top of the mill. The % loading or change volume can then be read off the graph below or ...
## calculating steel ball charge in ball mill
Exploring ball size distribution in coal grinding mills - ResearchGate. ABSTRACT Tube mills use steel balls as grinding media. ... new balls periodically to maintain a steady balanced ball charge in the mill. ..... of the calculation of the size distribution of the equilibrium mixture of balls in a ball mill is developed.
## Ball charges calculators - thecementgrindingoffice.com
- Ball top size (bond formula): calculation of the top size grinding media (balls or cylpebs):-Modification of the Ball Charge: This calculator analyses the granulometry of the material inside the mill and proposes a modification of the ball charge in order to improve the mill efficiency:
## Calculate and Select Ball Mill Ball Size for Optimum Grinding
In Grinding, selecting (calculate) the correct or optimum ball size that allows for the best and optimum/ideal or target grind size to be achieved by your ball mill is an important thing for a Mineral Processing Engineer AKA Metallurgist to do. Often, the ball used in ball mills is oversize “just in case”. Well, this safety factor can cost you much in recovery and/or mill liner wear and ...
## calculating a sag mill critical speed
calculation of ball mill charge volume. critical speed calculation ball mill 4229. Ball mill critical speed Henan Mining Machinery Co., Ltd. ball mill critical speed black powder Ball mill Wikipedia A ball mill, a type of grinder, is a cylindrical device used in grinding or mixing materials like ores, chemicals, ceramic raw materials and paintsBall mills rotate around a horizontal axis ...
## calculating load on a ball mill - welkominhetlandvancuijk.nl
Ball Mill Circulating Load. Calculating a grinding circuit’s circulating loads based on Screen Analysis of its slurries. Compared to %Solids or Density based Circulating load equations, a more precise method of determining grinding circuit tonnages uses the screen size distributions of …
## how to calculate charge in ball mill - airsoftschwyz.ch
how to calculate mill charge level. How To Calculate Mill Charge Level How to Calculate Charge Volume in Ball or Rod MillMining and However as seen in Figure 4 the power curve becomes very flat in the range above 45 As a result mills are seldom run with charge levels greater than 45 How to. how to compute cement mill ball charge
## A Method to Determine the Ball Filling, in Miduk Copper ...
was to locate the variation of ball filling in the mill. Ball Charge Program Abrasion In this section, ball abrasion was calculated via manufacture`s ball charge program. At the time of this research, mill ball charged, feed rate, and average moisture were 7 tons (ball size was 100 …
## How To Calculate Grinding Media In Cement Mill
How To Calculate Grinding Media In Cement Mill. Calculation of grinding media of cement mill. batch grinding have been crucial to a better understanding of the variables that affect figure 36 examples of movement of the media inside a ball mill simulated using indicate cement grinding is responsible for 185 billion kwh of energy the calculation of power in ball mills and the energy efficiency ...
## calculating a sag mill critical speed
calculation of ball mill charge volume. critical speed calculation ball mill 4229. Ball mill critical speed Henan Mining Machinery Co., Ltd. ball mill critical speed black powder Ball mill Wikipedia A ball mill, a type of grinder, is a cylindrical device used in grinding or mixing materials like ores, chemicals, ceramic raw materials and paintsBall mills rotate around a horizontal axis ...
## How To Calculate Charge In Ball Mill
How To Calculate Charge In Ball Mill. We can calculate the steel charge volume of a ball or rod mill and express it as the of the volume within the liners that is filled with grinding media while the mill is stopped the charge volume can be gotten by measuring the diameter inside the liners and the distance from the top of the charge to the top of the mill
## calculating load on a ball mill - welkominhetlandvancuijk.nl
Ball Mill Circulating Load. Calculating a grinding circuit’s circulating loads based on Screen Analysis of its slurries. Compared to %Solids or Density based Circulating load equations, a more precise method of determining grinding circuit tonnages uses the screen size distributions of …
## How To Calculate Grinding Media In Cement Mill
How To Calculate Grinding Media In Cement Mill. Calculation of grinding media of cement mill. batch grinding have been crucial to a better understanding of the variables that affect figure 36 examples of movement of the media inside a ball mill simulated using indicate cement grinding is responsible for 185 billion kwh of energy the calculation of power in ball mills and the energy efficiency ...
## how to calculate charge in ball mill - airsoftschwyz.ch
how to calculate mill charge level. How To Calculate Mill Charge Level How to Calculate Charge Volume in Ball or Rod MillMining and However as seen in Figure 4 the power curve becomes very flat in the range above 45 As a result mills are seldom run with charge levels greater than 45 How to. how to compute cement mill ball charge
## how to calculate p ball mill - eetcafeflores.nl
How To Calculate For Ball Mill Media Charge. How To Calculate For Ball Mill Media Charge A&C Machinery is professional mineral processing equipment manufacturer in the world, not our equipment has the excellent quality, but also our product service is very thorough.
## Ball Mill Design/Power Calculation - LinkedIn
12-12-2016· Ball Mill Power Calculation Example A wet grinding ball mill in closed circuit is to be fed 100 TPH of a material with a work index of 15 and a size …
## Ball Mill Initial Charge Calculation - treppen-linke.de
Ball Mill Initial Charge Calculation. Cement mill grinding media calculation vegaholdings.Determination of media charge grinding, in other words, the grinding mill must produce an optimum mesh of grind, regardless of the grinding media utilized all tumbling mills effect breakage, 5 to calculate the top ball size once this top, ferrosilicon, limestone, taconite and cement.
## calculates the grinding charge of a ball mill | Mining ...
To calculate grinding media charge for continuous type ball mill, M = 0.000676 x D2 x L Example … how to calculate cement grinding mill balls charge – kefid Mining FLshanghai ball mill for cement grinding. cement mill is a corrugated lining designed to obtain maximum power …
## A Method to Determine the Ball Filling, in Miduk Copper ...
was to locate the variation of ball filling in the mill. Ball Charge Program Abrasion In this section, ball abrasion was calculated via manufacture`s ball charge program. At the time of this research, mill ball charged, feed rate, and average moisture were 7 tons (ball size was 100 …
## calculating load on a ball mill - welkominhetlandvancuijk.nl
Ball Mill Circulating Load. Calculating a grinding circuit’s circulating loads based on Screen Analysis of its slurries. Compared to %Solids or Density based Circulating load equations, a more precise method of determining grinding circuit tonnages uses the screen size distributions of …
## how to calculate p ball mill - eetcafeflores.nl
How To Calculate For Ball Mill Media Charge. How To Calculate For Ball Mill Media Charge A&C Machinery is professional mineral processing equipment manufacturer in the world, not our equipment has the excellent quality, but also our product service is very thorough.
## Ball Mill Design/Power Calculation - LinkedIn
12-12-2016· Ball Mill Power Calculation Example A wet grinding ball mill in closed circuit is to be fed 100 TPH of a material with a work index of 15 and a size …
## how to calculate charge in ball mill - airsoftschwyz.ch
how to calculate mill charge level. How To Calculate Mill Charge Level How to Calculate Charge Volume in Ball or Rod MillMining and However as seen in Figure 4 the power curve becomes very flat in the range above 45 As a result mills are seldom run with charge levels greater than 45 How to. how to compute cement mill ball charge
## how to calculate grinding rate of ball mill
how to calculate grinding rate of ball mill HFC Refrigerants (55) HST Hydraulic Cone CrusherHST series hydraulic cone crusher is combined with technology such as machinery, hydraulic pressure, electricity, automation, intelligent control, etc. , representing the most advanced crusher technology in the world.
## grinding ball charge in tube mill
Pdf grinding media balls charge calculation in ball mill. Cement grinding Vertical roller mills versus ball mills- how to calculate grinding media in a ball mill in cement industry,invention in France which involved a tube mill with a charge of steel balls or flint cement industry the ball mill was really an epoch-making breakthrough as for products, enabling a high grinding efficiency and stable.
## Ball Mill Initial Charge Calculation - treppen-linke.de
Ball Mill Initial Charge Calculation. Cement mill grinding media calculation vegaholdings.Determination of media charge grinding, in other words, the grinding mill must produce an optimum mesh of grind, regardless of the grinding media utilized all tumbling mills effect breakage, 5 to calculate the top ball size once this top, ferrosilicon, limestone, taconite and cement.
## A Method to Determine the Ball Filling, in Miduk Copper ...
was to locate the variation of ball filling in the mill. Ball Charge Program Abrasion In this section, ball abrasion was calculated via manufacture`s ball charge program. At the time of this research, mill ball charged, feed rate, and average moisture were 7 tons (ball size was 100 …
## calculates the grinding charge of a ball mill | Mining ...
To calculate grinding media charge for continuous type ball mill, M = 0.000676 x D2 x L Example … how to calculate cement grinding mill balls charge – kefid Mining FLshanghai ball mill for cement grinding. cement mill is a corrugated lining designed to obtain maximum power …
## ball mill ball charge sizes - praxis-terrasol.ch
Calculate and Select Ball Mill Ball Size for Optimum . 2020-9-8 In Grinding, selecting (calculate) the correct or optimum ball size that allows for the best and optimum/ideal or target grind size to be achieved by your ball mill is an important thing for a Mineral Processing Engineer AKA Metallurgist to do. | 2,341 | 11,427 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2021-17 | latest | en | 0.840632 |
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