"DIAMOND Mark R." wrote from [EMAIL PROTECTED]:
>
> I could not find a discussion of the following in any of several text-books
> I looked at in University of West Aust library. I would appreciate some
> guidance, either in the form of a pointer to an appropriate reference, or an
> answer/partial answer/comment or whatever. I did find several web-sites
> which had exam questions on closely related topics, but no relevant
> teaching material! Needless to say, I am not a statistician.
>
> In its most specific form, I would like to know how one determines the
> sampling distribution of the maximum of say, n, random normal deviates. I
> would also like to know something more general. If I take multiple samples
> of n random deviates from some distribution, what will be (or how could I
> determine) the sampling distribution of the k-th ranked deviate. I hope I've
> put this clearly. ... If, say, I take random numbers 6 at a time from some
> distribution, what will be the sampling distribution of the maximum, the
> 2-nd highest, 5-th highest, minimum and such-like.
This is a standard result that *ought* to be in most introductory
_mathematical_ statistics books (but isn't always), under the heading of
"[distribution of] order statistics". (Any introductory _probability_
book that does not contain it should be thrown away as worthless.)
There is a general formula. It is found by first using the cumulative
distribution function to find the probability that n values are less
than or equal to x, which is the same as saying that the nth smallest
value is less than or equal to x. Now differentiate to find the density
function. You should get (after a blackboard full of cancelling terms):
[N!/(n-1)!(N-n)!] F(x)^(n-1) (1-F(x))^(N-n) f(x)
There is also a not-quite-rigorous argument (could be made rigorous
using nonstandard analysis or a squirt of epsilondeltism) that gets
there directly and intuitively. The idea is that for the nth of N values
to be exactly x, we must have:
(n-1) values <= x (probability F(x)^(n-1) )
1 value = x ("probability" f(x) dx, this is where the rigor
slips)
N-n values >= x (probability (1-F(x))^(N-n))
and there are ( N )
(n-1 1 N-n)
ways to choose these values. (The last expression is meant to represent
the trinomial coefficient for N objects split into classes of sizes
n-1, 1, and N-n. It evaluates to N!/(n-1)!(N-n)! )
-Robert Dawson
.
.
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