3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
3
$\begingroup$

$\newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\eD}{\overset{\text{D}}\to} \newcommand{\D}{\overset{\text{D}}=}$ 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}

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.