# How to find the $(a_n,b_n)$ for extreme value theory

In the solution to this question (Extreme Value Theory - Show: Normal to Gumbel), the OP asked for the sequence $(a_n, b_n)$ such that $\Phi(a_nx+b_n)$ converges to the Gumbel CDF. Not only did I not able to understand the derivation in the accepted answer (I don't see the derivation for the last paragraph), I am also curious to know how one would go about deriving the sequence for distributions other than the standard normal.

For example, the sequences in examples 7.5 and 7.6 in this document (page 6) seems to be pulled out of thin air.

• The link on "this document" no longer works. Moved?
– Yves
Commented Feb 16, 2017 at 10:41

To discuss about the constants $a_n$ and $b_n$ in the Fisher-Tippett-Gnedenko theorem, let us assume that $X_k$ is a sequence of i.i.d. r.vs with distribution $F(x)$; we consider $M_n:= \max\{X_1,\,X_2,\,\dots,\,X_n\}$ so that $F_{M_n}(x) = F(x)^n$. The constants are such that $(M_n - b_n) / a_n$ converges to a non-degenerate distribution, say $G(x)$, or in other words: $F(a_n \,x + b_n)^n \to G(x)$ for all $x$. Note that the constants are not uniquely determined: sequences $a_n'$ and $b_n'$ with $a_n' \sim a_n$ and $(b_n - b_n') / a_n \to 0$ can be used as well, with an unchanged non-degenerate limit distribution.
As a general rule, one makes use of the the so-called tail-quantile function $U(t)$. When $F(x)$ is continuous, $U(t)$ is defined for $t > 1$ with its value given by $$1 - F(U) = 1 / t.$$ In the vocabulary of applications, $U(t)$ is nothing but the $t$-years return level; the scalar $U$ then has the same physical dimension as the r.vs $X_i$ (length, time, temperature, ...). Now $U(n)$ gives an order of magnitude of $M_n$. Indeed with $U_n:= U(n)$ $$F_{M_n}(U_n) = F(U_n)^n = \left\{ 1 - [1 - F(U_n)] \right\}^n = \left\{ 1 - 1 / n \right\}^n \approx e^{-1},$$ so clearly $U(n)$ is in the bulk of the distribution of $M_n$ for large $n$. By either subtracting $U(n)$ or by dividing by $U(n)$ we can hope to scale $M_n$ so that it nearly has a fixed distribution for large $n$. But the choice depends on the type of tail i.e. of the domain of attraction of $X$.
The two simple cases are Weibull and Fréchet. Indeed with $\omega$ being the upper end-point of $F$, it can be proved that $$\text{Weibull, type III} \qquad \frac{M_n - \omega}{U(n) - \omega} \to G(x),$$ and $$\text{Fréchet, type II} \qquad \frac{M_n - 0}{U(n)} \to G(x).$$ The Gumbel case is more complicated and quite subtle. The value $U(n)$ is then subtracted to $M_n$, i.e. used as $b_n$ and we need the scale $a_n$. If $X$ turns out to have a density $f(x)$ at least near $\omega$, we can use the hazard rate $h(x)$ and the mean residual life $e(x)$ $$h(x) := \frac{f(x)}{1 - F(x)}, \quad e(x) := \frac{\int_x^\omega [1 - F(t)] \, \text{d}t}{1 - F(x)}.$$ Both $1/h(x)$ and $e(x)$ have the same physical dimension as $x$. Under some mild restrictions we have then $$\label{eq:Gum} \text{Gumbel, type I} \qquad \frac{M_n - U(n)}{e(U(n))} \to G(x)$$ One condition ensuring that this convergence holds is one of Von Mises' conditions: the derivative of $1 / h(x)$ exists and tends to $0$ for $x \to \omega$. For many application cases, $h(x)$ is positive and monotonic for $x$ close enough to $\omega$. Assuming this, it can be shown that $F$ is in the Gumbel domain of attraction if and only if the product $h(x)\times e(x)$ tends to $1$ when $x \to \omega$, and then $e(U_n)$ can be replaced by $1 / h(U_n)$.
For many classical distributions such as normal or gamma, neither $F(x)$ nor $U(t)$ are available in closed form. Equivalent quantities can be used, but this requires some math.
As a final remark, note that we can have a finite upper end-point $\omega < \infty$ and yet $F$ in the Gumbel domain of attraction. An example is provided by the reversed Fréchet distribution, for which the determination of the constants is a good exercise.