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?
$E[1_{\{x_i\leq x'_K\}} \cdot x'_{K+1}]$
From Balakrishnan and Cohen:
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!