I am trying to study the asymptotic behavior of $k^{th}$ order statistic of $n$ i.i.d chisquare distribution. Let $X_1, \cdots , X_n$ be i.i.d $\chi^2_1$ random variables and $X_{(k:n)}$ be the $k^{th}$ order statistic of these random variables. I do know that, $$ \frac{X_{(n:n)}}{\log n} \overset{p}{\to} 2\; \text{as $n \to \infty$}. $$ But now I am interested in studying the behavior of $X_{(k:n)}$ where $\frac{k}{n}\to 1$ as $n \to \infty$. My conjecture is that, $$ \frac{X_{(k:n)}}{\log n} \overset{p}{\to} 2 \; \text{as $n \to \infty$}. $$ My idea is to show that $P( \frac{X_{(k:n)}}{\log n} <a ) \to 0$ if $a<2$ and $P( \frac{X_{(k:n)}}{\log n} <a ) \to 1$ if $a>2$. Any help will be appreciated.

$\begingroup$ My first instinct was to look at the $k/n$ quantile, which evaluates to 2 InverseGammaRegularized[n/2, 0, k/n] in Mathematica, or alternatively InverseGammaRegularized[n/2, 1  k/n]. Unfortunately, Mathematica knows only the information at functions.wolfram.com/GammaBetaErf/InverseGammaRegularized about these functions, which is not enough to compute any asymptotics. Getting asymptotics for the actual order statistics would be even more difficult. $\endgroup$– user44143Commented Jul 27, 2021 at 18:19

$\begingroup$ Your order statistics are increasing (The most common convention goes the other way). $\endgroup$– Yuval PeresCommented Jul 28, 2021 at 2:34
2 Answers
$\newcommand{\ep}{\varepsilon} \newcommand{\pp}{\overset p\to}$Let $Y_k:=X_{(k:n)}$, where $n1\ge k\sim n$. The correct asymptotics for $Y_k$ is as follows: \begin{equation*} Y_k/l_{n,k}\pp2,\tag{1} \end{equation*} where \begin{equation*} l_{n,k}:=\ln\frac n{nk}, \end{equation*} so that $l_{n,k}\to\infty$. In particlular, if $nk=O(n^\ep)$ for each $\ep>0$, then (1) does imply $Y_k/\ln n\pp2$.
To prove (1), note first that for all real $x>0$ \begin{equation*} P(Y_k\le x)=P(N_x\le nk),\tag{1.5} \end{equation*} where \begin{equation*} N_x:=\sum_{i=1}^n 1(X_i>x), \end{equation*} so that the random variable $N_x$ is binomial with parameters $n$ and \begin{equation*} p:=P(X_1>x)=2(1\Phi(\sqrt x)), \end{equation*} where $\Phi$ is the standard normal cdf, so that \begin{equation*} p=e^{x/(2+o(1))}\to0 \tag{2} \end{equation*} as $x\to\infty$.
Fix any real $a>0$. Choosing then $x=a\,l_{n,k}$ and using (2), we see that
$np<<nk$ if $a>2$ and
$np>>nk$ if $a<2$;
we write $A<<B$ and $B>>A$ if $A/B\to0$.
So, if $a>2$, then, by Markov's inequality, \begin{equation*} P(N_x\le nk)=1P(N_x>nk)\ge1\frac{np}{nk}\to1. \end{equation*} If now $a<2$, then, by Chebyshev's inequality, \begin{equation*} P(N_x\le nk)=P(N_xnp\le(np(nk)) \le\frac{np(1p)}{(np(nk))^2}\sim\frac{np}{(np)^2}\to0. \end{equation*}
So, for $x=a\,l_{n,k}$, in view of (1.5), $P(Y_k\le x)$ converges to $1$ or $0$ depending on whether $a>2$ or $a<2$. Thus, (1) is proved.

$\begingroup$ @losif Pinelis Thank you very much for your help. The proof is awesome. Just one question. When you define $l_{n,k}$, what is the meaning of $\to \infty$ inside the square bracket? $\endgroup$– De vinciCommented Jul 27, 2021 at 21:31

1$\begingroup$ @Devinci : I am glad you liked the proof. I have now spelled out the meaning of →∞ inside the square brackets. $\endgroup$ Commented Jul 27, 2021 at 21:54

$\begingroup$ @Devinci : So, to have a closure here, are you satisfied with this answer? $\endgroup$ Commented Aug 2, 2021 at 19:25


1$\begingroup$ @Devinci : All right, I'll look at your other question. For now, let us have a closure with this one. As a new contributor, you might not know about these guidelines: mathoverflow.net/help/someoneanswers and mathoverflow.net/help/acceptedanswer $\endgroup$ Commented Aug 3, 2021 at 3:06
The behavior of the order statistics is different than you predicted: Suppose that $k=n\ell$, where $\ell=\ell(n)$ satisfies $\ell/n \to 0$. Denote by $Z$ a standard normal variable. Then for $b \in (0,1)$ we have \begin{align} P[X_{(k:n)}>2b \log(n)] &\le {n \choose \ell} P[Z^2> 2b \log(n)]^{\ell} \\ &\le (\frac{ne}{\ell})^{\ell} \exp(\ell b \log n) \\ &=\exp(\ell \cdot [\log (ne/\ell)b\log n])\,, \end{align} which tends to zero if $\ell>n^r$ with $r>1b$.

$\begingroup$ Thank you for the clarification. I was having a wrong intuition about how the order statistics behave asymptotically. $\endgroup$– De vinciCommented Jul 28, 2021 at 22:55