Generalized Extreme Value Distribution

Study guide: Exam ASTAM

The Generalized Extreme Value (GEV) Distribution is the only possible limit of a suitably normalized block maximum $M_n = \max(X_1, \ldots, X_n)$ of independent, identically distributed losses, much as the normal is the limit of a normalized sum. Its shape parameter $\xi$ classifies the tail of the loss distribution into three cases, the Gumbel ( $\xi = 0$ ), the Fréchet ( $\xi > 0$ ) and the Weibull ( $\xi < 0$ ). It is fitted to block maxima, such as the largest loss of each year, to estimate the probabilities and quantiles of extreme losses.

$\lim_{n \to \infty} \Pr\left[\frac{M_n - d_n}{c_n} \le x\right] = H(x)$

$H_{\xi,\mu,\theta}(x) = \exp\left\{-\left[1 + \xi\,\frac{x - \mu}{\theta}\right]^{-1/\xi}\right\}$

$H_{0,\mu,\theta}(x) = \exp\left\{-e^{-(x - \mu)/\theta}\right\}$

$Q_p(M) = \mu + \frac{\theta}{\xi}\left[(-\ln p)^{-\xi} - 1\right]$

Symbols: Following Hardy and Saunders (the ASTAM study note on extreme value theory), $\mu$ is a location parameter, $\theta > 0$ a scale parameter and $\xi$ the shape parameter. The second block is the case $\xi \neq 0$ and holds where $1 + \xi(x - \mu)/\theta > 0$ , and the third is its limit as $\xi \to 0$ . The constants $c_n > 0$ and $d_n$ normalize the maximum, and the last block is the $p$ -quantile of a block maximum that follows the GEV, which becomes $\mu - \theta \ln(-\ln p)$ when $\xi = 0$ . When the first block holds, the loss distribution $F$ is said to be in the maximum domain of attraction (MDA) of $H$ .
Fisher–Tippett–Gnedenko theorem: If $F$ is in the MDA of a non-degenerate distribution $H$ , then $H$ must be a GEV. For a finite block, $\Pr(M_n \le m) = F(m)^n$ exactly, so the theorem matters when $F$ is unknown in its tail and the maxima themselves are modelled.
The three cases: The Fréchet case, $\xi > 0$ , is bounded below and has a fat tail, with moments only of order below $1/\xi$ , and its MDA includes the Pareto (with $\xi = 1/\alpha$ ), Student's $t$ and Burr distributions. The Gumbel case, $\xi = 0$ , is unbounded, and its MDA includes the normal, exponential, gamma and lognormal distributions, all of which have every moment. The Weibull case, $\xi < 0$ , is bounded above at $\mu - \theta/\xi$ , and its MDA includes the uniform and beta distributions.
Fitting: The data are split into blocks, the maximum of each block is taken, and the GEV is fitted to the maxima by maximum likelihood. Larger blocks put the maxima further into the tail but leave fewer of them, so $\xi$ is estimated with a large standard error. The fitted $\mu$ and $\theta$ change with the block size, but $\xi$ should not.

Pareto Maxima and the Fréchet Limit A book has 100 independent large claims a year, each Pareto with $\alpha = 2.5$ and scale 1,000, so $S(x) = \left(\frac{1{,}000}{1{,}000 + x}\right)^{2.5}$ . For the Pareto, the normalizing constants are $c_n = 1{,}000\,n^{1/\alpha}/\alpha$ and $d_n = 1{,}000\,(n^{1/\alpha} - 1)$ , and $\xi = 1/\alpha$ . Use the GEV to approximate the probability that the year's largest claim exceeds 20,000, and compare it with the exact value.
Answer With $n = 100$ , $n^{0.4} = 6.309573$ , so $c_{100} = 2{,}523.83$ and $d_{100} = 5{,}309.57$ , and $\xi = 0.4$ .
 $\begin{align*} z &= \frac{20{,}000 - 5{,}309.57}{2{,}523.83} \\ &= 5.82069 \\ 1 + 0.4z &= 3.32828 \\ H(z) &= \exp\left(-3.32828^{-2.5}\right) \\ &= \exp(-0.049483) \\ &= 0.951722 \end{align*}$ 
The approximation is $\Pr(M_{100} > 20{,}000) \approx 0.048278$ . Exactly, $S(20{,}000) = 0.000494825$ , and $1 - (1 - 0.000494825)^{100} = 0.048290$ . For a Pareto, $(1 + \xi z)^{-1/\xi}$ equals $n\,S(x)$ , so the approximation is $1 - e^{-nS(x)}$ against the exact $1 - (1 - S(x))^n$ , and the two agree closely whenever $S(x)$ is small.

Tail Estimates From a Fitted GEV A GEV fitted to the annual maximum catastrophe loss (in $ millions) has $\xi = 0.2$ , $\mu = 10$ and $\theta = 4$ . Find the probability that next year's largest loss exceeds 30, and the 99th percentile of the annual maximum. Compare the percentile with a Gumbel fit that has the same $\mu$ and $\theta$ .
Answer At $x = 30$ , $1 + 0.2(30 - 10)/4 = 2$ , and $2^{-1/0.2} = 2^{-5}$ .
 $\begin{align*} \Pr(M > 30) &= 1 - \exp\left(-2^{-5}\right) \\ &= 1 - e^{-0.03125} \\ &= 0.030767 \\ Q_{0.99}(M) &= 10 + \frac{4}{0.2}\left[(0.0100503)^{-0.2} - 1\right] \\ &= 10 + 20(1.509365) \\ &= 40.19 \end{align*}$ 
A loss of 30 million is exceeded in about one year in 33, and the annual maximum exceeds 40.19 million in one year in 100. The Gumbel quantile is $10 - 4\ln(0.0100503) = 28.40$ , so ignoring the positive shape understates the one-in-100-year maximum by almost 30%.

How it appears on the exam

ASTAM asks candidates to understand the derivation and characteristics of the GEV. A question gives a loss distribution and normalizing constants and asks for the limiting distribution of the maximum, as in showing that the exponential is in the Gumbel MDA, or asks which case a distribution belongs to from its support and its moments.

Tail estimates: ASTAM also asks candidates to apply the GEV to tail probabilities and risk measures. A question gives fitted values of $\xi$ , $\mu$ and $\theta$ and asks for the probability that a block maximum exceeds a level, or for a quantile of the block maximum.

Written comments: The reading's exercises also ask for the trade-off in choosing the block size, between the bias of blocks that are too small and the variance of too few maxima.

Exam traps

The sign of $\xi$ : The Fréchet case is $\xi > 0$ and has the heavy tail, and the Weibull case is $\xi < 0$ and is bounded above. The Weibull extreme value distribution is not the Weibull severity of the exam tables, to which it is related by a change of sign.

The support: The $\xi \neq 0$ formula holds only where $1 + \xi(x - \mu)/\theta > 0$ . Outside that range, $H$ is 0 below the Fréchet's lower bound and 1 above the Weibull case's upper bound.

A maximum, not a single loss: The GEV describes the largest loss in a block. A 1% chance that the annual maximum exceeds $x$ is not a 1% chance that a single loss does, and for a single loss in a block of $n$ , $S(x) \approx -\ln H(x)/n$ .

Assuming a Gumbel tail: Setting $\xi = 0$ because its estimate is uncertain understates every extreme quantile when the true $\xi$ is positive, by almost 30% in the second example.