Order statistic - Rate of convergence of a p-quantile to the expectation Fix some $k\in\mathbb N$ and some probability $p\in[0,1]$. Denote with $F_n$ the cdf of the k-th highest oder statistic (i.e. the distribution of the k-th highest draw) of $n$ draws from a uniform distribution on $[0,1]$. Obviously, for $n\to\infty$, the $p$-quantile of $F_n$ as well as the expectation of the lower $p$-quantile approach $1$. I am interested in the speed they converge to each other. More precisely, for $X_k^n$ being the $k$-th highest order statistic of $n$ draws I need to determine
$ \lim_{n\to\infty} n\big(F_n^{-1}(p)-\mathbb E[X^n_k|X^n_q\leq F_n^{-1}(p)]\big)$
It seems related to the question The behavior of a uniform order statistic near zero, but I don't see how I can solve it.
 A: $\newcommand{\E}{\operatorname{\mathsf E}} 
\newcommand{\eD}{\overset{\text{D}}\to} 
\newcommand{\D}{\overset{\text{D}}=}$ 
Let $U_1,U_2,\dots$ be iid random variables, each uniformly distributed on $[0,1]$. For a fixed natural $k$, let $Y_{n,k}$ be the $k$th largest value among $U_1,\dots,U_n$. For a fixed $p\in(0,1)$, let $q_{n,k}(p)$ be the $p$-quantile of $Y_{n,k}$. The problem then is to find 
\begin{equation}
 \lim_{n\to\infty}n(q_{n,k}(p)-\E(Y_{n,k}|Y_{n,k}\le q_{n,k}(p)). 
\end{equation}
Note that $Y_{n,k}$ has the beta distribution with parameters $n+1-k,k$. So (see Sections Derived_from other distributions and Summation), 
\begin{equation}
 1-Y_{n,k}\D\frac{S_k}{S_{n+1}},
\end{equation}
where $\D$ denotes the equality in distribution, $S_j:=X_1+\dots+X_j$, and $X_1,X_2,\dots$ are iid standard exponential r.v.'s. So, by the law of large numbers,
\begin{equation}
 S_{n,k}:=n(1-Y_{n,k})\eD S_k, 
\end{equation}
where $\eD$ denotes the convergence in distribution. 
(Note also that $S_k$ has the gamma distribution with parameters $k$ and $1$.)  
So, $n(1-q_{n,k}(p))=\tilde q_{n,k}(1-p)\to \tilde q_k(1-p)$, where $\tilde q_{n,k}(1-p)$ and $\tilde q_k(1-p)$ denote the $(1-p)$-quantiles of $S_{n,k}$ and $S_k$, respectively. Hence,
\begin{align}
 n(1-\E(Y_{n,k}|Y_{n,k}\le q_{n,k}(p))
 &=\E(S_{n,k}|Y_{n,k}\le q_{n,k}(p)) \\
  &=\E(S_{n,k}|S_{n,k}\ge \tilde q_{n,k}(1-p)) \\ 
  &\to\E(S_k|S_k\ge \tilde q_k(1-p)),  
\end{align}
by an appropriate uniform integrability involving (say) second moments. 
Thus,
\begin{equation}
 \lim_{n\to\infty}n(q_{n,k}(p)-\E(Y_{n,k}|Y_{n,k}\le q_{n,k}(p))
 =\E(S_k|S_k\ge \tilde q_k(1-p))-\tilde q_k(1-p). 
\end{equation}
