Expectation of product of two order statistics

Question

Let $X_1, X_2,\ldots, X_n$ be $n$ i.i.d. random variables with $f(x)$ as the pdf and $F(x)$ as the cdf in interval $[a,b]$. Let $X_{i:n}$ be the $i^\text{th}$ order statistic such that $X_{1:n}\leq X_{2:n}\leq \cdots \leq X_{n:n}$. How does one compute the expected value $E [X_{k:k} X_{i:n}]$ for any $k< i \leq n$ ($X_{k:k}$ is the highest order statistic if there are only $k$ i.i.d. random variables)? Also, are $X_{k:k}$ and $X_{i:n}$ independent?

Answer 1

Suppose we're looking at the random variable $x_{k:n}x_{i:n}$, $i>k$, from the same sample. The joint density is \begin{multline} f(x_{k:n},x_{i:n}) = \dfrac{n!}{(k-1)!(i-k-1)!(n-i)!}F(x_{k:n})^{k-1}\\ [F(x_{i:n})-F(x_{k:n})]^{i-k-1}[1-F(x_{i:n})]^{n-i}f(x_{i:n})f(x_{k:n}) \end{multline} Then the expectation $\mathbb{E}[x_{k:n}x_{i:n}]$ can be computed as $$ \int_{x_{k:n} = -\infty}^{\infty} \int_{x_{i:n} = x_{k:n}}^\infty x_{k:n}x_{i:n} f(x_{k:n},x_{i:n})dx_{i:n} dx_{k:n}. $$

If the two variables are from two independent but identical samples, then \begin{multline} \mathbb{E}[x_{k:n} x_{i:n}] = \mathbb{E}[x_{k:n}]\mathbb{E}[x_{i:n}] = \int_{\infty}^\infty z\dfrac{(n-1)!}{(n-1-k)!(k-1)!}F^{k-1}(z)[1-F(z)]^{n-k}f(z)dz \\ *\int_{\infty}^\infty z\dfrac{(n-1)!}{(n-1-i)!(i-1)!}F^{i-1}(z)[1-F(z)]^{n-i}f(z)dz. \end{multline}