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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.

zxmkn
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    A Poisson distribution would be a good candidate for $F.$ – whuber Feb 10 '22 at 18:22
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    The exercises of chap 1 in the book Extreme Value Theory by Laurens de Hann and Ana Ferreira contain some (counter)examples. – Yves Feb 11 '22 at 10:43
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    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. – zxmkn Feb 11 '22 at 12:20

1 Answers1

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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 potential* 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.)

*edit note: I have added 'potential' in an edit. While re-reading this old post, I do not understand anymore why I considered the infinite expectation as a problem for convergence. It can still have converge to one of the distributions from the Fisher–Tippett–Gnedenko theorem theorem, namely the Fréchet distribution can have finite expectation.

Sextus Empiricus
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  • 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 $a_n$.) – whuber Apr 20 '23 at 16:55
  • 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! – whuber Apr 20 '23 at 17:06
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    @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. – Sextus Empiricus Apr 20 '23 at 17:43
  • Maybe it is the quantile distribution that has the non-zero slope. – Sextus Empiricus Apr 20 '23 at 18:26
  • 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! – whuber Apr 20 '23 at 19:00