Expectation associated with the Kth largest value of iid variables We have $\{x_i\}$, $i=1,2,...,N$ randomly and independently drawn from uniform distribution U(0,1). Let $x'_K$ $(K<N)$ denote the Kth number, from the smallest to the largest, among them. Is it possible to calculate the following expectation?
$E[1_{\{x_i\leq x'_K\}} \cdot x'_{K+1}]$
 A: From Balakrishnan and Cohen:

If $X_i$ are iid random variables from a uniform $(0,1)$ distribution, then $X_{i:n}$ and $X_{j:n}$ have the joint density function (for $1 \leq i < j \leq n$)
  \begin{align}
f_{i,j:n}(x_{i:n}, x_{j:n}) = \frac{n!~ x_{i:n}^{i-1} (x_{j:n}-x_{i:n})^{j-i-1}(1-x_{j:n})^{n-j}}{(i-1)!(j-i-1)!(n-j)!} , \qquad 0 < x_{i:n} < x_{j:n} < 1
\end{align}

Therefore, setting $a := n-K$ and $b := n+1-J$ for some $J$; using $s = x_{i:n}$ and $t = x_{j:n}$ to declutter notation, we can look at the slightly simpler problem of
\begin{align*}
\mathbb E[X_{k+1}' \mid X_i = X_J', X_i = w] 
&= \mathbb E[X_{a:n} \mid X_i = X_{b:n}, X_i = w]
= \int_w^1 \int_0^t s f_{a,b:n}(s,t) \,\mathrm{d}s\,\mathrm{d}t
\\
&= \frac{a n!}{b!(n-b)!}\int_w^1 t^b (1-t)^{n-b} \,\mathrm{d}t
= a(1 - I_w(b+1,n-b+1))\\
& =: e(w,a,b,n)
\end{align*}
where $I_x(\alpha, \beta)$ is the regularized incomplete beta function (i.e. the CDF of the beta distribution with parameters $\alpha, \beta$). 
Next, use the fact that $X_i = X_J'$ with probability $1/n$ and the fact that $J < K+1$ which means that $b > a$. 
\begin{align*}
\mathbb E[X_{k+1}' \mid X_i \leq X_K'] 
&= \mathbb E_{X_i}\big[\frac{1}{n}\sum_{1\leq J<K}\mathbb E[X_{k+1}' \mid X_i = X_J', X_i = w] \big]
= \frac{1}{n}\sum_{a < b \leq n}\mathbb E_{X_i} \big[e(X_i,a,b,n)\big]
\end{align*}
Lastly, there might be some hope of an exact solution since $X_i \sim \mathrm{Unif}(0,1)$, but I haven't gotten that far. Since it's just integration, you should be able to take it from here!
