The joint pdf of three or more order statistics is given by $f_{x_{(1)},x_{(2)},...x_{(n)}}(x_1,x_2,..x_n)=n! f(x_1)f(x_2)..f(x_n) $

Question

The joint pdf of three or more order statistics is given by $f_{x_{(1)},x_{(2)},...x_{(n)}}(x_1,x_2,..x_n)=n! f(x_1)f(x_2)..f(x_n) , \ -\infty<x_1<x_2<...<x_n<\infty $ How can I derive this? I have derived for the bivariate case. Help!

Answer 1

This works under "suitable" conditions (the CDF must be differentiable in all its arguments)

Let $x_1<x_2<\cdots<x_n$ and let $\delta>0$ be small enough so that the intervals $I_k=(x_k-\delta,x_k]$ are disjoint.

Then $$X_{(1)}\in I_1\wedge\cdots\wedge X_{(n)}\in I_n\iff \exists \sigma\in S_n[X_{\sigma(1)}\in I_1\wedge\cdots\wedge X_{\sigma(n)}\in I_n]$$

The probability of the RHS is: $$n!\prod_{k=1}^nP(X\in I_k)=n!\prod_{k=1}^n(F(x_k)-F(x_k-\delta))$$ where $F$ denotes the CDF of the $X_k$.

Further: $$\lim_{\delta\to0}\delta^{-n}P(X_{(1)}\in I_1\wedge\cdots\wedge X_{(k)}\in I_k)=f_{X_{(1)},\dots,X_{(1)}}(x_1,\dots,x_k)$$ and: $$\lim_{\delta\to0}\delta^{-n}n!\prod_{k=1}^nP(X\in I_k)=n!\prod_{k=1}^nf(x_i)$$

Answer 2

This is a standard result available in textbooks covering order statistics.

Here is a proof from Introduction to the Theory of Statistics (3rd edition) by Mood-Graybill-Boes:

(Check the second page)