Let $X_1,...X_n$ be independent standard normal random variables ~$N(0,1$).
How does the maximum of these random variables, $M_n$, which is also a random variable, behave as a function of $n$ when $n\rightarrow+\infty$ ? An exercise suggests an asymptotic form with some coefficients $a_n$ and $b_n$, but I don't get how they were derived... Please, see below.
Attempt (through some guided exercise):
We have that $P(M_n\leq x)=P(X_1\leq x \space \text{AND} \space X_2\leq x ...\text{AND} \space X_n\leq x)=P(X\leq x)^n$ where $X$~$N(0,1)$.
Therefore, $F_{M_n}(x)=F_X(x)^n$ ($F_X$ is the C.D.F of r.v. $X$)
We admit that, modulo some shift $a_n>0$ and rescaling $b_n>0$, the C.D.F. of $\bar{M_n}=\frac{M_n-a_n}{b_n}$ converges to the Gumbel distribution, i.e. $F_{\bar{M_n}}(x)=e^{-e^{-x}}$.
Since $P(\bar{M_n}\leq x) \Leftrightarrow P(M_n \leq b_nx+a_n)$, we have that:
$F_{\bar{M_n}}(x)=F_{M_n}(a_n+b_nx)=F_X(a_n+b_nx)^n$
Assuming $F_X(a_n+b_nx)=1-\frac{\gamma(x)}{n}+o(\frac1n)$, we can deduce:
$F_{\bar{M_n}}(x)=e^{-e^{-x}}=e^{-\gamma(x)}$, so $\gamma(x)=e^{-x}$ by identification.
From there, the problem states that the use of the following values for $a_n$ and $b_n$ is consistent, allowing one to deduce that $M_n$~$a_n+b_nG(x)$ with $G$ the Gumbel distribution.
$a_n=\sqrt{2\ln(n)}-\frac{\ln\ln n+\ln 4\pi}{2\sqrt{2\ln n}}$ and $b_n=\frac{1}{\sqrt{2\ln n}}$
with the indication :
$F_X(x)=1-\frac1x \frac{1}{\sqrt{2\pi}}e^{\frac{-x^2}{2}}$ as $x\rightarrow+\infty$ (which is straightforward, doing integration by parts)
In other words, if I understood correctly, the coefficients can be obtained solving:
$F_X(a_n+b_nx)=1-\frac{e^{-x}}{n}$ for $a_n$ and $b_n$
I think that since we are interested in the $n\rightarrow +\infty$ limit, it is wise to use the asymptotic formula since apparently, $a_n\rightarrow +\infty$ (which makes sense).
However, I can't solve it, since the exponent in the LHS involves $x^2$ but it does not in the RHS, leading to $b_n=0$...
$x+\ln n = \frac{(a_n+b_nx)^2}{2}+\ln{(a_n+b_nx)}+\ln{\sqrt{2\pi}}$
has no solution in $a_n$ and $b_n$ as far as I know...
$\color{red}{\text{So, where do the two values}}$ $a_n=\sqrt{2\ln(n)}-\frac{\ln\ln n+\ln 4\pi}{2\sqrt{2\ln n}}$ and $b_n=\frac{1}{\sqrt{2\ln n}}$ $\color{red}{\text{come from ?}}$
I have tried to expand the logarithm to get two terms in $x^2$ and get rid of them, but it still wouldn't yield the correct values for $a_n$ and $b_n$. (I got $a_n=1$ doing so)
Any advice on how to proceed further ? Or maybe I made a mistake somewhere.
