Timeline for How to calculate 2D standard deviation, with 0 mean, bounded by limits
Current License: CC BY-SA 3.0
19 events
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| Sep 12, 2015 at 18:40 | history | edited | MaxW | CC BY-SA 3.0 |
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| Sep 6, 2015 at 15:46 | history | edited | MaxW | CC BY-SA 3.0 |
added calculated of CDF to prove that Rayleigh funtion gets "right" answeres...
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| Sep 6, 2015 at 15:39 | history | edited | MaxW | CC BY-SA 3.0 |
added calculated of CDF to prove that Rayleigh funtion gets "right" answeres...
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| Jun 21, 2015 at 23:40 | comment | added | MaxW | Here is link to Rice simplification: "Probability, Random Processes, and Statistical Analysis: Applications to Communications, Signal Processing, Queueing Theory and Mathematical Finance" By Hisashi Kobayashi, Brian L. Mark, William Turin, ISBN: 9780521895446, page 170 --- books.google.com/… | |
| Jun 21, 2015 at 23:04 | comment | added | whuber♦ | I don't think it's the Rice distribution. If so, then my answer is entirely wrong--but then it would be difficult to explain away the close agreement with simulations. What we are talking about here is not non-centrality (which is what the Rice distribution models); it's the 2D analog of the distinction between error terms and residuals. | |
| Jun 21, 2015 at 22:02 | comment | added | MaxW | Let me explicitly state that the Rayleigh distribution is the result of converting to polar coordinates about the center point of the balls. The implicit assumption is that the observed center of the balls is the same as the population center which is unobservable. If the sample centers and population centers must be regarded as distinct, then the distribution would be the Rice distribution. | |
| Jun 15, 2015 at 17:12 | history | edited | MaxW | CC BY-SA 3.0 |
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| Jun 15, 2015 at 17:07 | comment | added | MaxW | So the center of mass for the balls is: $$(\mu_x, \mu_y)$$ and the position of any ball i is $$(x_i, y_i)$$ | |
| Jun 15, 2015 at 17:00 | history | edited | MaxW | CC BY-SA 3.0 |
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| Jun 15, 2015 at 16:59 | comment | added | MaxW | I'm going to fix two problems in the analysis above. First I have "use bivariant normal distribution. f(x,v)" which should be f(x,y). Second the equation for r is messed up. It should be the Euclidean distance. | |
| Jun 15, 2015 at 16:42 | comment | added | whuber♦ | No such assumption is made in the question. Your $\mu$ is a distributional parameter which will be unknown to the experimenter, who doesn't even care about it. They are concerned only about distances between the balls after they come to rest and the center of mass of those balls. | |
| Jun 15, 2015 at 15:49 | comment | added | whuber♦ | I think your analysis errs by confounding $\mu$ with the observed center of mass of the balls. | |
| Jun 15, 2015 at 14:32 | comment | added | MaxW | The CDF is setup for one ball of course. From the CDF 39% of the balls will fall within a circle of radius σ, 86% within 2σ, and 99% within 3σ. | |
| Jun 13, 2015 at 19:08 | comment | added | MaxW | The distribution is about the center of mass. | |
| Jun 13, 2015 at 13:40 | comment | added | whuber♦ | Thank you for the clarifications, Max. As a simple check of the correctness of your answer, let's consider one ball instead of $40$. Your answer appears to claim the distribution of the distance between this ball and the center of mass of all balls is a Rayleigh distribution. Unfortunately, in this case that distance is always zero. (The question specifically describes it as "the distance from the center of mass to each ball, which is calculated using simple geometry.") That suggests your answer may be wrong in every case, including for $40$ balls. | |
| Jun 13, 2015 at 5:39 | comment | added | MaxW | filled in the gaps.... | |
| Jun 10, 2015 at 3:46 | history | edited | MaxW | CC BY-SA 3.0 |
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| May 26, 2015 at 13:45 | comment | added | whuber♦ | Thank you for naming the distribution. However, by (1) not differentiating between the distribution's parameter and estimates of that parameter derived from the data, (2) not stating the (strong) assumptions needed about the distribution of the balls, and (3) by being vague, you risk misleading readers. Indeed, it is unclear what the reference of your "this" is: would it be the distribution of locations of the balls? (No.) The distribution of the center of mass? (Yes, but with a scale parameter that differs from the standard deviation of the balls.) Would you like to clarify your answer? | |
| May 26, 2015 at 1:56 | history | answered | MaxW | CC BY-SA 3.0 |