Let $X\_1$,$X\_2$,…,$X\_n$ be $n$ i.i.d. random variables with $f(x)$ as the pdf and $F(x)$ as the cdf in interval $[0,1]$. Let $F$ be uniformly distributed. Let $X\_{i:n}$ be the $i^{th}$ order statistic such that $X\_{1:n} \leq X\_{2:n} \leq ... \leq X\_{n:n}$. I wish to compute the expected value $\mathbb{E} [\frac{X\_{(k-1):n} X\_{i:n}}{X\_{k:n}} ]$ for any $ k < i \leq n$. So the question is are $\frac{X\_{(k-1):n}}{X\_{k:n}}$ and $X\_{i:n}$ independent? Because if they are not, then the problem is non-trivial. Due to a standard result in theory of order statistics, we already know that for any $i \leq n$, $\frac{X\_{(i-1):n}}{X\_{i:n}}$ and $X\_{i:n}$ are independent.