3

The standard generalised extreme value (GEV) distribution is given by $H_{\xi}$ which is

$exp(-(1+\xi x)^{-1/\xi}$ if $\xi<>0$ and

$exp(-e^{-x})$ if $\xi=0$

In the lecture notes it is stated

$1-H_\xi (x)$ approximatle equals $e^{-x}$ for $\xi=0$ for $x_{H_\xi}$ going to infinity which is the Gumbel distribution.

$1-H_\xi (x)$ approximatle equals $(\xi x)^{-1/\xi}$ for $\xi=0$ for $x_{H_\xi}$ going to infinity which is the Fréchet distribution.

I would like to do the math to derive the Gumbel and the Fréchet from the GEV to understand deriving limits better (I seem to have some deficits). I would be grateful for a solution or a textbook hint.

Many thanks.

Florian
  • 75

1 Answers1

5

To restate your question, the form of the generalized extreme value cumulative distribution function is given for the cases $\xi\not=0$ and $\xi=0$. The case $\xi=0$ is known as the Gumbel distribution. You are asking for a demonstration of how the formula for the $\xi=0$ cases arises as a limit of the $\xi\not=0$ case.

By inspection, this amounts to the assertion $$ \lim_{\xi\to0} (1+\xi x)^{-1/\xi}=e^{-x}. $$ Taking negative $\log$ of both sides, it is equivalent to $$ \lim_{\xi\to0} \frac{\log(1+\xi x)}{\xi}=x.\qquad(\star) $$ The left side is recognized as the partial derivative of $\log(1+\xi x)$ with respect to $\xi$, evaluated at $\xi=0$. Using basic rules of differentiation, we have $$ \frac{\partial }{\partial \xi}\log(1+\xi x)=\frac{x}{1+\xi x}. $$ Therefore $$ \lim_{\xi\to0} \frac{\log(1+\xi x)}{\xi}=\left.\frac{\partial }{\partial \xi}\log(1+\xi x)\right|_{\xi=0}=x, $$ establishing $(\star)$ and therefore proving that the Gumbel CDF is the $\xi=0$ limit of the GEV CDF.

pre-kidney
  • 30,884
  • Many thanks, pre-kidney. My question would rather have been how can I derive the approximate distributions from the GEV survival functions taking the limit of $x$ to infinity. How do I get $e^{-x}$? – Florian Jun 04 '19 at 08:42
  • @Florian Since you are a new user of the website, let me just point out that gratitude is best expressed by an upvote and clicking "accept" on the answer (green checkbox) and comments saying thank you and nothing else are discouraged (since they don't add to the discussion). – pre-kidney Jun 04 '19 at 08:44
  • Sorry, I pressed save to quickly. I tried to restate my question into another direction. – Florian Jun 04 '19 at 08:52
  • 2
    That second part of the question is the more straightforward part. In both cases (Gumbel and GEV) it follows from the observation that $1-e^{-y}$ is approximately equal to $y$ as $y\to 0$. This, in turn, follows from the first term in the Taylor series for $e^y$, namely $e^y\approx 1+y$ when $y$ is small. – pre-kidney Jun 04 '19 at 08:56