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Not asking for the distribution of $m$th order statistic, for it has been well documented and covered and in fact well-known. What I am rather confused, perhaps basic but failing to see, is the approach using the argument of symmetry in the book, An Introduction to Order Statistics:

Let $f_{m:n}$ be the pdf of the $m$th order statistic. It is given by

$$f_{m:n}(x) =n! f(x) \int\cdots\int \prod_{k=1}^{m-1} f(x_k) \prod_{k=m+1}^n f(x_k) ~\mathrm dx_1\cdots\mathrm dx_{m-1}\mathrm dx_{m+1}\cdots\mathrm dx_n, $$ over the domain $-\infty <x_1<\cdots<x_{m-1}<x<x_{m+1}<\cdots x_n<\infty.$

The authors argued that

the symmetry of $\prod_{k=1}^{m-1} f(x_k) $ with respect to $x_1, \ldots, x_{m-1}$ as well as the symmetry of $\prod_{k=m+1}^n f(x_k)$ with respect to $x_{m+1}, \cdots, x_n$ help us to evaluate the integral as follows: $$\int\cdots\int \prod_{k=1}^{m-1} f(x_k) \prod_{k=m+1}^n f(x_k) ~\mathrm dx_1\cdots\mathrm dx_{m-1}\mathrm dx_{m+1}\cdots\mathrm dx_n=\frac{1}{(m-1)! } \prod_{k=1}^{m-1} \int_{-\infty}^xf(x_k)~\mathrm dx_k\frac{1}{(n-m)!}\prod_{k=m+1}^n \int_x^\infty f(x_k)~\mathrm dx_k$$

and the form of the distribution follows.

I couldn't understand their manipulation of the integral using symmetry: what is actually the argument of symmetry suggested here? Since, they are iid, they are exchangeable but I don't know what actually the authors did. The $x_i$s are ordered in the domain: how could we permute that without not breaking the order?

Finally, how did they calculate the integral: where did the factor of ${(m-1)! }^{-1}$ and ${(n-m)! }^{-1}$ come from?

I would appreciate if someone is able to shed light on this approach.

User1865345
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1 Answers1

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The unrestricted integrals over $(-\infty,x)$ and $(x,\infty)$ allow all $(m-1)!$ possible orders of the first $m-1$ variables and all $(n-m)!$ possible orders of the last $n-m$ variables. Since the variables are exchangeable, the restriction to any given order yields the same contribution as the restriction originally imposed. Thus, all $(m-1)!$ contributions to the first integral are equal, and all $(n-m)!$ contributions to the second integral are equal, so in each case we can use the entire integral and divide by the number of equal contributions.

joriki
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  • +1. That's a nice breakdown of what is happening and I feel I understood your reasoning. But before I accept it, could you tell me the big picture of this approach, if there is any? That is, I want to use such technique in future, so would appreciate if you could generalize the technique, if possible. – User1865345 Mar 21 '24 at 09:22
  • Maybe it will be helpful to link to two other answers where I used the symmetry between all orders of variables: https://math.stackexchange.com/a/53182, https://math.stackexchange.com/a/51406. Often (as in these two examples) it works in the other direction, i.e. the integral is easier to evaluate if you restrict it to one particular order, whereas here the integral is easier if you lift the restriction; but the principle is the same. – joriki Mar 21 '24 at 09:48
  • Thanks, Joriki, would look into those. Appreciate your response. – User1865345 Mar 21 '24 at 09:54