4
$\begingroup$

I'm working on Extreme Values Theory, and I found the following sufficient condition to find the domain of attraction of a distribution and the corresponding normalizing constants:

For sufficiently smooth distribution $F$ with density $f$, define $\displaystyle h(x) = \frac{1-F(x)}{f(x)}$; let $\displaystyle b_n = F^{-1} \left( 1 - \frac{1}{n} \right)$, $a_n = h(b_n)$ and $\displaystyle \xi = \lim_{x \rightarrow +\infty}{h'(x)}$. Then the distribution of the maxima $F^n(a_nx+b_n)$ converges to a GEV distribution with shape parameter $\xi$.

My questions:

  1. What are the exact conditions for this result? I guess "sufficiently smooth" means that there must exist a density; there is also a definition problem for $h(x)$ if $f(x)=0$: what are exactly the assumptions here?

  2. I've got a theoretical book on Extreme Values (Resnick 87) but I haven't found this result; so how do you prove it?

Thanks

$\endgroup$
1
  • $\begingroup$ The inverse of the hazard rate is used in sufficient conditions for the Gumbel case $\xi = 0$ only. In the general case the constants $a_n$ and $b_n$ are different, see my answer to this question. $\endgroup$
    – Yves
    Feb 17, 2023 at 10:09

1 Answer 1

4
$\begingroup$

The result you stated is also known as the von Mises condition in Extreme Value Theory. You can find the proof as well as the exact technical conditions in theorem 1.1.8. of the book "Extreme value theory: An introduction" by Laurens de Haan and Ana Ferreira.

Here's the exact statement:

von Mises condition

Note that the right endpoint $x^*$ of a distribution function $F$ is defined as $x^*=\sup\{x: F(x)<1\}$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.