# Correlation of order statistics from Uniform parent

If $(X_1, X_2, \dots, X_n)$ are a random sample from Uniform(0,1) distribution, find the correlation between the order statistics $X$ order $r$ and $X$ order $s$, $r < s$.

I have used the formula for correlation coefficient as:

$$\rho=\frac{E[X_{(r)}X_{(s)}]-E[X_{(r)}]E[X_{(s)}]}{\sqrt{Var(X_{(r)}) Var(X_{(s)})}}$$

I know the order statistic follows a beta distribution and hence found out the mean and variance of $X_{(r)}$ and $X_{(s)}$. The problem is, I'm unable to compute $E[X_{(r)}X_{(s)}]$. I have tried to find the joint expectation by finding the joint probability distribution of the rth and sth order statistics, but when I try to multiply $X_{(r)}$ and $X_{(s)}$ with the joint PDF and integrate it to find the joint expectation, I am unable to integrate it.

• To find $E[X_{(r)}X_{(s)}]$, you will need the joint pdf of $(X_{(r)},X_{(s)})$. Have you derived this? – wolfies May 24 '16 at 17:08
• @wolfies yes, I have. Then I tried to find the E(X(r)X(s)) by integrating the product of the joint distribution and X(r)X(s) dx(r)dx(s) over (0,1). But I couldn't integrate the product – Sweta95 May 24 '16 at 17:13

You appear to understand the steps involved. I am not sure if this is an assignment or exercise, so it may not be appropriate to show workings anyway, but am happy to sketch out the approach ...

Given: $X \sim \text{Uniform}(0,1)$ with pdf $f(x)$:

Then, the joint pdf of the $r^{\text{th}}$ and $s^{\text{th}}$ order statistics is say $g(x_r, x_s)$:

where I am using the OrderStat function from the mathStatica package for Mathematica to automate the calculation.

Given the joint pdf of the $r^{\text{th}}$ and $s^{\text{th}}$ order statistics, you seek their correlation:

All done. This should hopefully help in both forming the appropriate integrals, and checking your working.

• It isn't an assignment, just a practice problem. My approach was exactly similar to yours,sir. Maybe I should check the integration again. Thank you. :) – Sweta95 May 24 '16 at 17:45