# Distribution that doesn't belong to any maximum domain of attraction?

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 $$F(x) = \begin{cases} 0, &x for some $$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.

• A Poisson distribution would be a good candidate for $F.$
– whuber
Feb 10, 2022 at 18:22
• 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, 2022 at 10:43
• 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. Feb 11, 2022 at 12:20

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.)

• 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 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 at 17:06
• @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. Apr 20 at 17:43
• Maybe it is the quantile distribution that has the non-zero slope. Apr 20 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 at 19:00