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The Maximum of $X_1,\dots,X_n. \sim$ i.i.d. Standardnormals converges to the Standard Gumbel Distribution according to Extreme Value Theory.

How can we show that?

We have

$$P(\max X_i \leq x) = P(X_1 \leq x, \dots, X_n \leq x) = P(X_1 \leq x) \cdots P(X_n \leq x) = F(x)^n $$

We need to find/choose $a_n>0,b_n\in\mathbb{R}$ sequences of constants such that: $$\Phi\left(a_n x+b_n\right)^n\rightarrow^{n\rightarrow\infty} G(x) = e^{-\exp(-x)}$$

Can you solve it or find it in literature?

There are some examples pg.6/71, but not for the Normal case.

$$\Phi\left(a_n x+b_n\right)^n=\left(\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{a_n x+b_n} e^{-\frac{y^2}{2}}dy\right)^n\rightarrow e^{-\exp(-x)}$$

emcor
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1 Answers1

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Use the asymptotic expansion

$$\Phi(x) = 1-\frac{e^{-x^2/2}}{\sqrt{2\pi}}\left(\frac{1}{x}+ \ldots\right)$$

If you can select $a_n$ and $b_n$ such that as $n \rightarrow \infty,$

$$\frac{e^{-(a_n x + b_n)^2/2}}{(a_n x +b_n)\sqrt{2\pi}}\sim \frac{e^{-x}}{n}, $$

then, as desired,

$$\lim_{n \rightarrow \infty}\Phi(a_nx+b_n)^n =\lim_{n \rightarrow \infty}\left(1-\frac{e^{-x}}{n}\right)^n= \exp(-e^{-x}).$$

First select $a_n = 1/b_n$, leaving $b_n$ unspecified for now , but with the requirement $b_n \rightarrow \infty.$

Then

$$\frac{e^{-(a_nx+b_n)^2/2}}{(a_nx+b_n)\sqrt{2\pi}} = \frac{e^{-x^2/2b_n^2}e^{-x}}{x/b_n^2+1}\frac{e^{-b_n^2/2}}{b_n\sqrt{2\pi}}. $$

This a good start as

$$\lim_{n \rightarrow \infty} \frac{e^{-x^2/2b_n^2}e^{-x}}{x/b_n^2+1}=e^{-x}.$$

Now using the inverse normal distribution, choose $b_n = \Phi^{-1}(1 - 1/n)$, which satisfies the requirement, $b_n \rightarrow \infty$, and

$$\frac{e^{-b_n^2/2}}{b_n\sqrt{2\pi}} \sim 1-\Phi[\Phi^{-1}(1-1/n)]= \frac1{n}.$$

Putting everything together, we have as $n \rightarrow \infty$,

$$\frac{e^{-(a_n x + b_n)^2/2}}{(a_n x +b_n)\sqrt{2\pi}}\sim \frac{e^{-x}}{n}. $$

RRL
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