Compute the correlation coefficient $r(X_{(1)},X_{(3)})$ .

Question

I got this problem where:

The random variables $X_1, X_2,$ and $X_3$ are independent and $Exp(1)-$ distributed.
Compute the correlation coefficient $r(X_{(1)},X_{(3)})$ .

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I know through research that: $\mathrm E(X_{(k)})=\sum\limits_{i=n-k+1}^n\frac1i,\qquad \mbox{Var}(X_{(k)})=\sum\limits_{i=n-k+1}^n\frac1{i^2}.$

I could find everything except, $E(X_{(1)}X_{(3)})$.

I got: $E(X_{(1)})=\frac {1}{3},\qquad E(X_{(3)})=1+\frac {1}{2}+\frac {1}{3}$

$Var(X_{(1)})= \frac {1}{9}, \qquad Var(X_{(3)})= 1+\frac {1}{4}+\frac {1}{9}$

I also know that correlation = $$ \frac{cov(X_{(1)},X_{(3)})}{\sqrt{Var(X_{(1)})Var(X_{(3)})}}$$

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Update:

So from the implication of @NcH I figured the $E(X_{(1)}X_{(3)})=1$ But, when trying to compute the correlation coefficient I get 1 as the final answer as opposed to 2/7 suggested by the textbook.

$$r= \frac{1-11/18}{7/18}=1$$

Anyone has a different answer ?

Answer 1

I think that the simplest way is to use joint PDF of $X_{(1)}$ and $X_{(3)}$. For $0<x<y$ using formula for [joint pdf of two order ststistics]1 we get $$ f_{X_{(1)}, X_{(3)}}(x,y) = 3! \bigl((1-e^{-y})-(1-e^{-x})\bigr)e^{-x}e^{-y}=6(e^{-2x-y}-e^{-x-2y}) $$

Then $$\mathbb E[X_{(1)}X_{(3)}]=\iint_{0<x<y<\infty} xy f_{X_{(1)}, X_{(3)}}(x,y) \,dx\, dy.$$