Expected averages of ordered draws from a uniform distribution.

Question

Suppose I take $n$ independent draws from a uniform distribution on $[0,1]$ and call the set of those draws $N$. Then define $X \subset N$ as the $K$ largest draws, and $Y \subset N$ as the $N-K$ smallest draws. What is the probability that the mean of $X$ is $k$ times larger than the mean of $Y$?

Answer 1

So I think, I can combine the order statistic of the uniform distribution and the Bates distribution as follows.

First, the order statistic of the $K^{th}$ largest draw is distributed via a Beta distribution with parameters $K, n-K+1$. So I can define a random variable $a$ based on that.

Then, because each draw is independent, I can use that a to bound intervals and apply the Bates distribution as follows:

$$pr(\mu(X) > k \mu(Y) ) = pr(Bates\left( [a,1], K\right) > k \; Bates\left( [0,a], n-K \right)) $$

This implies:

$$\int_{0}^{1}Beta(K, N-K+1)\int_{0}^{1}Bates\left( [0,a], n-K \right) \int_{xt}^{1} Bates\left( [a,1], K\right) dx dt da.$$

Does that look right?