$k^{\text{th}}$ maxima of $n$ i.i.d chi-square random variables

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 $$P( \frac{X_{(k:n)}}{\log n} if $$a<2$$ and $$P( \frac{X_{(k:n)}}{\log n} if $$a>2$$. Any help will be appreciated.

• 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. Jul 27 '21 at 18:19
• Your order statistics are increasing (The most common convention goes the other way). Jul 28 '21 at 2:34

$$\newcommand{\ep}{\varepsilon} \newcommand{\pp}{\overset p\to}$$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

$$np< if $$a>2$$ and

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

we write $$A< and $$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.

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

• Thank you for the clarification. I was having a wrong intuition about how the order statistics behave asymptotically. Jul 28 '21 at 22:55