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.

  • $\begingroup$ True, I fixed it, thanks! $\endgroup$ – jonasvw Nov 25 at 14:49
  • $\begingroup$ 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 $Y_q\le F_n^{-1}(p)$? $\endgroup$ – Iosif Pinelis Nov 25 at 15:04
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    $\begingroup$ 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. $\endgroup$ – Matt F. Nov 25 at 15:14
  • $\begingroup$ 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 :-) $\endgroup$ – 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 \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}

  • $\begingroup$ 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 $\endgroup$ – jonasvw Nov 26 at 16:20
  • $\begingroup$ @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 $1$. $\endgroup$ – Iosif Pinelis Nov 26 at 16:37

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