Fisher–Tippett–Gnedenko theorem

Conditions of convergence

The Fisher–Tippett–Gnedenko theorem is a statement about the convergence of the limiting distribution ${\displaystyle G(x)}$, above. The study of conditions for convergence of ${\displaystyle G}$ to particular cases of the generalized extreme value distribution began with Mises (1936) and was further developed by Gnedenko (1943).

Let ${\displaystyle F}$ be the distribution function of ${\displaystyle X}$, and ${\displaystyle X_1,\dots ,X_n}$ be some i.i.d. sample thereof. Also let ${\displaystyle x_{\mathsf{m}\mathsf{a}\mathsf{x}}}$ be the population maximum: ${\displaystyle x_{\mathsf{m}\mathsf{a}\mathsf{x}}\equiv sup\{x\mid F(x)<1\}}$.

Then the limiting distribution of the normalized sample maximum, given by ${\displaystyle G}$ above, will then be one of the following three types:

• Fréchet distribution (${\displaystyle \gamma >0}$): For strictly positive ${\displaystyle \gamma >0}$, the limiting distribution converges if and only if ${\displaystyle x_{\mathsf{m}\mathsf{a}\mathsf{x}}=\mathrm{\infty }}$ and

${\displaystyle \underset{t\to \mathrm{\infty }}{lim}\frac{1-F(ut)}{1-F(t)}=u^{-1/\gamma }}$ for all ${\displaystyle u>0}$.

In this case, possible sequences that will satisfy the theorem conditions are ${\displaystyle b_n=0}$ and ${\displaystyle a_n=F^{-1}\left(1-\frac{1}{n}\right)}$. Strictly positive ${\displaystyle \gamma }$ corresponds to what is called a heavy tailed distribution.

• Gumbel distribution (${\displaystyle \gamma =0}$): For trivial ${\displaystyle \gamma =0}$, and with ${\displaystyle x_{\mathsf{m}\mathsf{a}\mathsf{x}}}$ either finite or infinite, the limiting distribution converges if and only if

${\displaystyle \underset{t\to x_{\mathsf{m}\mathsf{a}\mathsf{x}}}{lim}\frac{1-F(t+u\tilde{g}(t))}{1-F(t)}={\mathrm{e}}^{-u}}$ for all ${\displaystyle u>0}$ with ${\displaystyle \tilde{g}(t)\equiv \frac{{\int }_t^{x_{\mathsf{m}\mathsf{a}\mathsf{x}}}(1-F(s))\mathrm{d}s}{1-F(t)}}$.

Possible sequences here are ${\displaystyle b_n=F^{-1}\left(1-\frac{1}{n}\right)}$ and ${\displaystyle a_n=\tilde{g}(F^{-1}\left(1-\frac{1}{n}\right))}$.

• Weibull distribution (${\displaystyle \gamma <0}$): For strictly negative ${\displaystyle \gamma <0}$, the limiting distribution converges if and only if ${\displaystyle x_{\mathsf{m}\mathsf{a}\mathsf{x}}<\mathrm{\infty }}$ (is finite) and

${\displaystyle \underset{t\to 0^{+}}{lim}\frac{1-F(x_{\mathsf{m}\mathsf{a}\mathsf{x}}-ut)}{1-F(x_{\mathsf{m}\mathsf{a}\mathsf{x}}-t)}=u^{-1/\gamma }}$ for all ${\displaystyle u>0}$.

Note that for this case the exponential term ${\displaystyle -1/\gamma }$ is strictly positive, since ${\displaystyle \gamma }$ is strictly negative.

Possible sequences here are ${\displaystyle b_n=x_{\mathsf{m}\mathsf{a}\mathsf{x}}}$ and ${\displaystyle a_n=x_{\mathsf{m}\mathsf{a}\mathsf{x}}-F^{-1}\left(1-\frac{1}{n}\right)}$.

Note that the second formula (the Gumbel distribution) is the limit of the first (the Fréchet distribution) as ${\displaystyle \gamma }$ goes to zero.

Examples

Fréchet distribution

The Cauchy distribution's density function is:

${\displaystyle f(x)=\frac{1}{\pi (1+x^2)},}$

and its cumulative distribution function is:

${\displaystyle F(x)=\frac{1}{2}+\frac{1}{\pi }\arctan \left(\frac{x}{\pi }\right).}$

A little bit of calculus show that the right tail's cumulative distribution ${\displaystyle 1-F(x)}$ is asymptotic to ${\displaystyle \frac{1}{x},}$ or

${\displaystyle \ln F(x)\to \frac{-1}{x}\mathsf{a}\mathsf{s}x\to \mathrm{\infty },}$

so we have

${\displaystyle \ln \left(F(x{)}^n\right)=n\ln F(x)\sim \frac{-n}{x}.}$

Thus we have

${\displaystyle F(x{)}^n\approx \exp \left(\frac{-n}{x}\right)}$

and letting ${\displaystyle u\equiv \frac{x}{n}-1}$ (and skipping some explanation)

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}(F(n(u+1){)}^n)=\exp \left(\frac{-1}{1+u}\right)=G_1(u)}$

for any ${\displaystyle u.}$

Gumbel distribution

Let us take the normal distribution with cumulative distribution function

${\displaystyle F(x)=\frac{1}{2}\mathrm{erfc}\left(\frac{-x}{\sqrt{2}}\right).}$

We have

${\displaystyle \ln F(x)\to -\frac{\exp \left(-\frac{1}{2}{x}^2\right)}{\sqrt{2\pi }x}\mathsf{a}\mathsf{s}x\to \mathrm{\infty }}$

and thus

${\displaystyle \ln \left(F(x{)}^n\right)=n\ln F(x)\to -\frac{n\exp \left(-\frac{1}{2}{x}^2\right)}{\sqrt{2\pi }x}\mathsf{a}\mathsf{s}x\to \mathrm{\infty }.}$

Hence we have

${\displaystyle F(x{)}^n\approx \exp \left(-\frac{n\exp \left(-\frac{1}{2}{x}^2\right)}{\sqrt{2\pi }x}\right).}$

If we define ${\displaystyle c_n}$ as the value that exactly satisfies

${\displaystyle \frac{n\exp \left(-\frac{1}{2}{c}_n^2\right)}{\sqrt{2\pi }{c}_n}=1,}$

then for ${\displaystyle x\approx c_n,}$ we find that

${\displaystyle \frac{n\exp \left(-\frac{1}{2}{x}^2\right)}{\sqrt{2\pi }x}\approx \exp \left(c_n(c_n-x)\right).}$

As ${\displaystyle n}$ increases, this becomes a good approximation for a wider and wider range of ${\displaystyle c_n(c_n-x)}$ so letting ${\displaystyle u\equiv c_n(x-c_n)}$ we find that

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}(F{\left(\frac{u}{c_n}+c_n\right)}^n)=\exp (-\exp (-u))=G_0(u).}$

Equivalently,

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\mathbb{P}(\frac{max\{X_1,\dots ,X_n\}-c_n}{\left(\frac{1}{c_n}\right)}\le u)=\exp (-\exp (-u))=G_0(u).}$

With this result, we see retrospectively that we need ${\displaystyle \ln c_n\approx \frac{\ln \ln n}{2}}$ and then

${\displaystyle c_n\approx \sqrt{2\ln n},}$

so the maximum is expected to climb toward infinity ever more slowly.

Weibull distribution

We may take the simplest example, a uniform distribution between 0 and 1, with cumulative distribution function

${\displaystyle F(x)=x}$ for any x value from 0 to 1 .

For values of ${\displaystyle x\to 1}$ we have

${\displaystyle \ln (F(x{)}^n)=n\ln F(x)\to n(1-x).}$

So for ${\displaystyle x\approx 1}$ we have

${\displaystyle F(x{)}^n\approx \exp (n-nx).}$

Let ${\displaystyle u\equiv 1+n(1-x)}$ and get

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}(F\left(\frac{u}{n}+1-\frac{1}{n}\right){)}^n=\exp (-(1-u))=G_{-1}(u).}$

Close examination of that limit shows that the expected maximum approaches 1 in inverse proportion to n .