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](https://mathoverflow.net/questions/346157/the-behavior-of-a-uniform-order-statistic-near-zero), but I don't see how I can solve it.



Order statistic - Rate of convergence of a p-quantile to the expectation
$\newcommand{\E}{\operatorname{\mathsf E}} 
\newcommand{\eD}{\overset{\text{D}}\to} 
\newcommand{\D}{\overset{\text{D}}=}$ 
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](https://en.wikipedia.org/wiki/Beta_distribution#Derived_from_other_distributions) and [Summation](https://en.wikipedia.org/wiki/Gamma_distribution#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}


