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I am trying to study the asymptotic behavior of $k^{th}$ order statistic of $n$ i.i.d chi-square 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.

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  • $\begingroup$ 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. $\endgroup$
    – Matt F.
    Jul 27 '21 at 18:19
  • $\begingroup$ Your order statistics are increasing (The most common convention goes the other way). $\endgroup$ Jul 28 '21 at 2:34
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$\newcommand{\ep}{\varepsilon} \newcommand{\pp}{\overset p\to}$Let $Y_k:=X_{(k:n)}$, where $n-1\ge k\sim n$. 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{n-k}, \end{equation*} so that $l_{n,k}\to\infty$. In particlular, if $n-k=O(n^\ep)$ 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 n-k),\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

$np<<n-k$ if $a>2$ and

$np>>n-k$ 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 n-k)=1-P(N_x>n-k)\ge1-\frac{np}{n-k}\to1. \end{equation*} If now $a<2$, then, by Chebyshev's inequality, \begin{equation*} P(N_x\le n-k)=P(N_x-np\le-(np-(n-k)) \le\frac{np(1-p)}{(np-(n-k))^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.

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  • $\begingroup$ @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 vinci
    Jul 27 '21 at 21:31
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    $\begingroup$ @Devinci : I am glad you liked the proof. I have now spelled out the meaning of →∞ inside the square brackets. $\endgroup$ Jul 27 '21 at 21:54
  • $\begingroup$ @Devinci : So, to have a closure here, are you satisfied with this answer? $\endgroup$ Aug 2 '21 at 19:25
  • $\begingroup$ @losif Pinelis: I am very satisfied. Thank you for asking. I am also currently thinking about a related problem. Can you share your view? $\endgroup$
    – De vinci
    Aug 2 '21 at 19:30
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    $\begingroup$ @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/someone-answers and mathoverflow.net/help/accepted-answer $\endgroup$ Aug 3 '21 at 3:06
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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 $r>1-b$.

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  • $\begingroup$ Thank you for the clarification. I was having a wrong intuition about how the order statistics behave asymptotically. $\endgroup$
    – De vinci
    Jul 28 '21 at 22:55

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