# Extreme Value Theory - Show: Normal to Gumbel

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.

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

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## 1 Answer

An indirect way, is as follows:
For absolutely continuous distributions, Richard von Mises (in a 1936 paper "La distribution de la plus grande de n valeurs", which appears to have been reproduced -in English?- in a 1964 edition with selected papers of his), has provided the following sufficient condition for the maximum of a sample to converge to the standard Gumbel, $G(x)$:

Let $F(x)$ be the common distribution function of $n$ i.i.d. random variables, and $f(x)$ their common density. Then, if

$$\lim_{x\rightarrow F^{-1}(1)}\left (\frac d{dx}\frac {(1-F(x))}{f(x)}\right) =0 \Rightarrow X_{(n)} \xrightarrow{d} G(x)$$

Using the usual notation for the standard normal and calculating the derivative, we have

$$\frac d{dx}\frac {(1-\Phi(x))}{\phi(x)} = \frac {-\phi(x)^2-\phi'(x)(1-\Phi(x))}{\phi(x)^2} = \frac {-\phi'(x)}{\phi(x)}\frac {(1-\Phi(x))}{\phi(x)}-1$$

Note that $\frac {-\phi'(x)}{\phi(x)} =x$. Also, for the normal distribution, $F^{-1}(1) = \infty$. So we have to evaluate the limit

$$\lim_{x\rightarrow \infty}\left (x\frac {(1-\Phi(x))}{\phi(x)}-1\right)$$

But $\frac {(1-\Phi(x))}{\phi(x)}$ is Mill's ratio, and we know that the Mill's ratio for the standard normal tends to $1/x$ as $x$ grows. So

$$\lim_{x\rightarrow \infty}\left (x\frac {(1-\Phi(x))}{\phi(x)}-1\right) = x\frac {1}{x}-1= 0$$

and the sufficient condition is satisfied.

The associated series are given as $$a_n = \frac 1{n\phi(b_n)},\;\;\; b_n = \Phi^{-1}(1-1/n)$$

ADDENDUM

This is from ch. 10.5 of the book H.A. David & H.N. Nagaraja (2003), "Order Statistics" (3d edition).

$\xi_a = F^{-1}(a)$. Also, the reference to de Haan is "Haan, L. D. (1976). Sample extremes: an elementary introduction. Statistica Neerlandica, 30(4), 161-172." But beware because some of the notation has different content in de Haan -for example in the book $f(t)$ is the probability density function, while in de Haan $f(t)$ means the function $w(t)$ of the book (i.e. Mill's ratio). Also, de Hann examines the sufficient condition already differentiated.

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I'm not quite sure I understood your solution. So you took $F$ to be the standard normal CDF. I followed through and agree that the sufficient condition is satisfied. But how is the associated series $a_n$ and $b_n$ all of the sudden given by those? –  renrenthehamster Jul 7 at 14:16
@renrenthehamster I think these two parts are independently stated (no direct connection). –  emcor Jul 7 at 15:12
And so how might the associated series be obtained? Anyway, I opened a question about this issue (and more generally, for other distributions beyond the standard normal) –  renrenthehamster Jul 7 at 15:29
@renrenthehamster I have added relevant material. I don't believe there is a standard recipe for all cases, to find these series. –  Alecos Papadopoulos Jul 7 at 16:10