This is related to one of my previous questions here.

Let $(Z_1, Z_2, \ldots, Z_n)\sim N(0, \Omega)$, where $\Omega = (1-\mu) I_{n\times n} + \mu \boldsymbol{1}_n\boldsymbol{1}_n^\top $. Here $\boldsymbol{1}_n$ denotes the vector of 1's of length $n$. Let us define $X_i = Z_i^2$ and I am trying to investigate the asymptotic of $k^{th}$ order statistic $X_{(k:n)}$ where $k/n \to 1$ as $n \to \infty$.

As a special case let $k=n$. When $\mu =0$, it is known that $T_n:=X_{(n:n)}/\log n \overset{p}{\to} 2$.

What can we say about $T_n$ when $0\leq \mu<1$? I have some conjectures about $T_n$. If I am not wrong then one can use Slepian's lemma to conclude that $P(X_{(n:n)}>t)\leq P(Y_{(n:n)}>t)$ for $t\in \mathbb{R}$, where $\{Y_i\}_{i=1}^n$ are i.i.d $\chi^2_1$. Though I am not completely sure of this fact. Here is the idea I am thinking about. Let $(W_1, \cdots,W_n)\sim N(0, I_{n\times n})$. By Slepians' lemma we know $P(Z_{(n:n)}> t)\leq P(W_{(n:n)}>t)$. Now I am trying to take square of $Z_{(n:n)}$ and $W_{(n:n)}$. But squaring is not monotone and I am stuck. But I believe in asymptotic regime both of them are positive and hence my claim should work.

If this is true then I believe that $P(T_n>2)\to 0$, i.e. heuristically speaking $T_n \leq 2$ in limiting sense.

Any help will be appreciated. Also is any result for general $\Omega$ known? Thanks.