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Simple form provided by WHuber: What is the distribution of the diameter of n points in the plane drawn iid from a bivariate Normal distribution? (Diameter is the greatest distance among any pair of the points.)

Original long form: Given a Rayleigh process R($\sigma$) generating cartesian samples $X_i \in \Re^2$ — which is equivalent to a bivariate normal process with 0 correlation and both sigmas = $\sigma$ — what is the distribution of the Extreme Spread of n samples?

Extreme spread $\widehat{ES_n(\sigma)} \equiv \max\limits_{i, j \in n}|X_i - X_j|$

E[$ES_n(\sigma)$], and Pr($ES_n(\sigma)$ > a), seems like it should have a chi(n) distribution, but I am not good enough to derive the exact relationship.

I did find this paper from 1975 which on p.8 suggests as much, but which focuses on empirical solution instead of on the pure math. In contrast, I want to take the parameters of the distribution of X as given and find a formulaic distribution for ES.

I don't know whether order statistics for these distributions have closed forms, but if so perhaps we can express this in terms of the expected value of first and nth order statistics?

Any guidance appreciated!

feetwet
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    Your notation cannot be deciphered by readers unfamiliar with the context. It sounds like the $X_i$ are ordered pairs, whence $|X_i-X_j|$ must be some *two* dimensional distance (Euclidean? $L^p$?), but the use of "spread" suggests it is not. The formula for $EX$ uses $n$ on the left side and $i$ and $j$ on the right but there is no apparent connection among them. "$ES()$" is used in two apparently distinct ways, first with "$\{X\}_n$" as its argument and then immediately with "$(\sigma)$" as its argument (with the subscript $n$ in a different place). Please fix these problems. – whuber Nov 14 '13 at 16:49
  • @whuber -- sorry, just revised it to remove ambiguity. – feetwet Nov 14 '13 at 17:45
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    So, to put it a little more descriptively, would it be correct to say you have $n$ points in the plane drawn iid from a bivariate Normal distribution and you wish to find the distribution of their diameter (which is the greatest distance among any pair of the points)? – whuber Nov 14 '13 at 18:17
  • Yes, precisely (and more elegant)! – feetwet Nov 14 '13 at 18:33
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    Thanks for clarifying the original question feetwet; it helps when useful things from the comments are used to improve the question like that. If I hadn't already upvoted, I would do it for that edit. – Glen_b Nov 14 '13 at 23:04
  • @Glen_b: extra upvote on your behalf (probably would've anyway). – Nick Stauner May 05 '14 at 00:18

1 Answers1

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This statistic is common in shooting such as a rifle. The widest distance between shots is known as the "group size." There is no closed form for the statistic. Thus the paper you referenced by Taylor and Grubbs uses Monte-Carlo techniques to estimate the parameters. In real life for shooting that is really "good enough." Nobody shoots 1,000 groups of 10 shots per group. So the real world sampling in shooting is poor.

MaxW
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    +1 I found plenty of references (in support of this answer) by Googling [rifle "group size" statistic](https://www.google.com/search?q=rifle+"group+size"+statistic). – whuber Jan 04 '14 at 20:15