$k^{\text{th}}$ 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 $P( \frac{X_{(k:n)}}{\log n} <a  ) \to 0$ if $a<2$ and $P( \frac{X_{(k:n)}}{\log n} <a  ) \to 1$ if $a>2$. Any help will be appreciated.
 A: $\newcommand{\ep}{\varepsilon}
\newcommand{\pp}{\overset p\to}$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.
A: 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 $r>1-b$.
