6
$\begingroup$

Question

Does there exist a (non-degenerate) distribution that does NOT belong to any maximum domain of attraction?

That is: Does there exist any non-degenerate probability distribution function $F$ such that if $X_1,X_2,\dots \overset{\text{iid}}{\sim} F$, then there do not exist any sequences $(a_n) \subset \mathbb R_{>0}$, $(b_n) \subset \mathbb R$ such that $$ \frac{\max\{ X_1, \dots, X_n\} - b_n}{a_n} $$ converges in distribution to a non-degenerate distribution?

Note: By "degenerate distribution", I mean one that takes a certain value with probability $1$, i.e. $\mathbb P(X = c) = 1$ for some $c \in \mathbb R$. Or equivalently for a distribution function $F(x) = \begin{cases} 0, &x<c \\ 1, &x \geq c\end{cases}$ for some $c \in \mathbb R$.

Thoughts

The Fisher-Tippett-Gnedenko theorem tells us that if a distribution function $F$ belongs to the maximum domain of attraction (MDA) of any non-degenerate probability distribution, then it belongs to the MDA of a generalized extreme value distribution.

The lecture notes and books I've seen covering this topic take care to emphasize that this result only hold if $F$ belongs to the MDA of some non-degenerate $G$. However, the treatments I've seen do not then offer counterexamples of non-degenerate $F$'s that don't lie in any MDA.

Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ A Poisson distribution would be a good candidate for $F.$ $\endgroup$
    – whuber ♦
    Feb 10, 2022 at 18:22
  • 1
    $\begingroup$ The exercises of chap 1 in the book Extreme Value Theory by Laurens de Hann and Ana Ferreira contain some (counter)examples. $\endgroup$
    – Yves
    Feb 11, 2022 at 10:43
  • 2
    $\begingroup$ Thanks for the reference! Exercise 1.13 in de Haan shows that the geometric distribution $F(x) = 1 - \text e^{[x]}$, $x > 0$, and the Poisson distribution are not in any maximum domain of attraction. Exercise 1.18 is to show that $F(x) = 1 - \text e^{-x-\sin x}$, $x > 0 $ is not in any maximum domain of attraction. $\endgroup$
    – zxmkn
    Feb 11, 2022 at 12:20

1 Answer 1

Reset to default
0
$\begingroup$

Does there exist any non-degenerate probability distribution function $F$ such that if $X_1,X_2,\dots \overset{\text{iid}}{\sim} F$, then there do not exist any sequences $(a_n) \subset \mathbb R_{>0}$, $(b_n) \subset \mathbb R$ such that $$ \frac{\max\{ X_1, \dots, X_n\} - b_n}{a_n} $$ converges in distribution to a non-degenerate distribution?

Any discrete distribution whose maximum value in the domain has non-zero probability is an example for a distribution $F$ that is not degenerate but $\max\{ X_1, \dots, X_n\}$ converges to the maximum value of the domain and becomes a degenerate distribution. Hence, we can not find $a_n$ and $b_n$ such that there is convergence to a non-degenerate distribution.

An other example, for continuous distributions, that does not converge are the distributions with Super-Heavy Tails as described here: How do we call a more extreme case of fat tails than a power law? (they are distributions for which any order statistic of a sample will have infinite expectation values).

(Edit note: I got to that idea of super heavy tails by thinking of neccesary condition of the tail behaviour. In my edits you can see a line of thought about it, but it is incorrect and I have to refine it, so I deleted it.)

Improve this answer
$\endgroup$
5
  • $\begingroup$ Re "non-zero slope at the maximum of the domain:" why would that be? For instance, let the CDF of $X$ be $F(x)=1-(1-x)^2$ with domain $[0,1]$ (where the slope at its maximum value $1$ is equal to $0$) and observe that with $a_n=\sqrt{n}$ and $b_n = -\sqrt{n}$ the standardized maximum converges to a nondegenerate distribution with CDF given by $e^{-x^2}$ for $x\le 0.$ (Thus, the negative of such a variable follows a Fréchet distribution, suggesting it would be advisable to allow negative values of the $a_n$.) $\endgroup$
    – whuber ♦
    Apr 20 at 16:55
  • $\begingroup$ BTW, I have (for the first time) seen the comments to the question you reference. I kept the informative ones but deleted the nasty ones. You should have flagged them for moderator attention at the time! $\endgroup$
    – whuber ♦
    Apr 20 at 17:06
  • 1
    $\begingroup$ @whuber you are right, my description of non-zero slope doesn't accurately describe the image that I have about it, and I have to reconsider it. I was thinking of 'a piece that is constant', such that the maximum will concentrate with probability one in that piece. But indeed, if you have this zero slope only for a infinitely small piece, then it will also work. $\endgroup$
    – Sextus Empiricus
    Apr 20 at 17:43
  • $\begingroup$ Maybe it is the quantile distribution that has the non-zero slope. $\endgroup$
    – Sextus Empiricus
    Apr 20 at 18:26
  • $\begingroup$ Possibly--that looks right. I also believe there are bounded discrete distributions whose scaled maxima do have limiting distributions. Consider the distribution of the random variable $X$ given by $\Pr(X=1-1/n)=2^{-n},$ $n=1,2,3\ldots.$ They must, of course, have countable support, thereby not contradicting your assertions about distributions with finite support. But they are interesting examples nonetheless! $\endgroup$
    – whuber ♦
    Apr 20 at 19:00

Your Answer

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

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