Probability of achieving the maximum among absolute value of Gaussians Yesterday the following question was asked by user sigmatau:   

I'm interested in the following question: 
given $n$ i.i.d. random variables $X_i \sim \mathcal{N}(0,\sigma^2_1), i=1,\ldots,n$ and $n(n-1)/2$ i.i.d. random variables (independent w.r.t. the first set of random variables)  with $X_{n+j} \sim \mathcal{N}(0, \sigma^2_2), j = 1, \ldots, n(n-1)/2$, what is the probability that 
  \begin{equation}
\text{
argmax}_{i=1,\ldots,(n^2+n)/2} |X_i| \leq n
\end{equation}
In particular I'm interested in the case where $\sigma^2_1=2, \sigma^2_2=1$. In this case some simple numerical simulations suggest that the event that I want to know the probability of happens with high probability, but I was not able to give a closed form estimate of it. Any help is much appreciated. 

The question was deleted soon afterwards by the user. I think the question might be of interest to some other users and am therefore recreating it here, accompanied with an answer. 
 A: $\newcommand{\si}{\sigma}$
Let $M_n$ be the maximum of $n$ iid standard normal random variables (r.v.'s) $Z_1,\dots,Z_n$, and let $L_N$ be the maximum of $N$ iid standard normal r.v.'s $W_1,\dots,W_N$, where $N:=n(n-1)/2$. Assume also the r.v.'s $M_n$ and $L_N$ are independent. By rescaling, the question can be restated as follows: 

Given a positive real $\si$, how does the probability 
  $P(M_n>\si L_N)$ behave? 

The answer to this questions is this: 

$$P(M_n>\si L_N)\longrightarrow
\begin{cases}
1&\text{ if }\si<1/\sqrt2, \\
0&\text{ if }\si\ge1/\sqrt2;
\end{cases} \tag{1}
$$
  the convergence everywhere here is as $n\to\infty$. Moreover, it does not matter that the r.v.'s $M_n$ and $L_N$ are independent. 

Indeed, it is a well-known fact of the so-called extreme value theory (see e.g. Theorem 18.4) that 
$$M_n=d_n+\frac{G_n}{\sqrt{2\ln n}}, \tag{2}
$$
where 
$$d_n:=\sqrt{2\ln n}-\frac{\ln\ln n+\ln(4\pi)}{2\sqrt{2\ln n}}$$ 
and $G_n\to G$ in distribution, where in turn $G$ is a Gumbel r.v. 
So, $M_n\sim d_n\sim\sqrt{2\ln n}$ and $L_N\sim d_N\sim\sqrt{2\ln N}\sim\sqrt{4\ln n}$ in probability, which immediately implies (1) for $\si\ne1/\sqrt2$. 
The case $\si=1/\sqrt2$ (which was of special interest in the original question), a bit more effort shows that 
$$\frac1{\sqrt2}\,d(N)-d(n)\sim\frac{\ln\ln n}{4\sqrt{2\ln n}}
>>\frac1{\sqrt{\ln n}}, 
$$
where $A>>B$ means $B=o(A)$. Now, in view of (2), the case $\si=1/\sqrt2$ follows as well. 

Somehow, I have just noticed that the original question was, not about the maximum of $n$ iid standard normal r.v.'s, but about the maximum of the absolute values of $n$ iid standard normal r.v.'s. Anyway, (1) holds as well with $M_n^{|\;|}:=\max_1^n|Z_i|$ and $L_N^{|\;|}:=\max_1^N|W_j|$ in place of $M_n=\max_1^n Z_i$ and $L_N=\max_1^N W_j$, respectively. Here one only needs to make small adjustments. Indeed, the essential fact used in the above derivation is that 
$$M_n=d_n+\frac{\ln\ln n}{\sqrt{\ln n}}\,o_P(1), \tag{3}
$$
where $o_P(1)$ stands for a r.v. depending on $n$ and converging to $0$ in probability. But (3) holds with $M_n^-:=\max_1^n(-Z_i)$ in place of $M_n$, because $M_n^-$ equals $M_n$ in distribution. So, (3) holds with $M_n^{|\;|}$ in place of $M_n$, because every value of $M_n^{|\;|}$ is either the corresponding value of $M_n$ or the corresponding value of $M_n^-$. 
