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 $$X_q^n$$ being the $$q$$-th highest order statistic of $$n$$ draws I need to determine

$$\lim_{n\to\infty} n\big(F_n^{-1}(p)-\mathbb E[X^n_q|X^n_q\leq F_n^{-1}(p)]\big)$$

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.

• True, I fixed it, thanks! – jonasvw Nov 25 at 14:49
• What is $\mathbb E_{F_n}[X|X\leq F_n^{-1}(p)]$? In particular, what is $X$ there? Is $\mathbb E_{F_n}[X|X\leq F_n^{-1}(p)]$ the conditional expectation of the $q$th largest order statistic (say $Y_q$) given that $Y_q\le F_n^{-1}(p)$? – Iosif Pinelis Nov 25 at 15:04
• Once there's an answer to the question from @IosifPinelis, I think this will reduce to a problem about Beta functions where Mathematica will be able to run easy numerical simulations and probably provide a limit. – Matt F. Nov 25 at 15:14
• Yes, I clarified the notation. Indeen, it boils down to a Beta function and solvable numerically (if I can get my hands on mathematica), I just hoped there is a nice argument here, "proved by mathematica" doesn't sound too nice in a paper :-) – jonasvw Nov 25 at 15:26

$$\newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\eD}{\overset{\text{D}}\to} \newcommand{\D}{\overset{\text{D}}=}$$ 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 $$$$\lim_{n\to\infty}n(q_{n,k}(p)-\E(Y_{n,k}|Y_{n,k}\le q_{n,k}(p)).$$$$
Note that $$Y_{n,k}$$ has the beta distribution with parameters $$n+1-k,k$$. So (see Sections Derived_from other distributions and Summation), $$$$1-Y_{n,k}\D\frac{S_k}{S_{n+1}},$$$$ where $$\D$$ denotes the equality in distribution, $$S_j:=X_1+\dots+X_j$$, and $$X_1,X_2,\dots$$ are iid standard exponential r.v.'s. So, by the law of large numbers, $$$$S_{n,k}:=n(1-Y_{n,k})\eD S_k,$$$$ where $$\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$$, respectively. Hence,
\begin{align} n(1-\E(Y_{n,k}|Y_{n,k}\le q_{n,k}(p)) &=\E(S_{n,k}|Y_{n,k}\le q_{n,k}(p)) \\ &=\E(S_{n,k}|S_{n,k}\ge \tilde q_{n,k}(1-p)) \\ &\to\E(S_k|S_k\ge \tilde q_k(1-p)), \end{align} by an appropriate uniform integrability involving (say) second moments. Thus, $$$$\lim_{n\to\infty}n(q_{n,k}(p)-\E(Y_{n,k}|Y_{n,k}\le q_{n,k}(p)) =\E(S_k|S_k\ge \tilde q_k(1-p))-\tilde q_k(1-p).$$$$
• Thanks a million, this is amazing! The result is as I hoped: neither zero nor infinity but a positive constant. Some steps I have to digest a bit (statistic is not my main field). Does standard exponential r.v. mean distributed according to an exponential distribution with $\lambda=1$? Is the used equality in distribution a standard result? Best, Jonas – jonasvw Nov 26 at 16:20
• @jonasvw : I am glad this was of use. The equality in distribution is indeed a standard result; I have now added references to it. The standard exponential distribution is indeed the exponential distribution with mean $1$. – Iosif Pinelis Nov 26 at 16:37