# Does $\mathbb E X = \infty$ imply that $\mathbb P \circ X^{-1}$ belongs to Fréchet domain of attraction?

Question

Let $$F$$ be a distribution function and $$X \sim F$$ a random variable such that $$\mathbb E X = \infty$$. That is $$\mathbb E X^+ = \infty$$ and $$\mathbb E X^- \in \mathbb R$$.

Does it follow that $$F$$ belongs to the Fréchet maximum domain of attraction (MDA)? If not, what is an example?

Thoughts

I'm trying to understand extreme value theory and how distributions with infinite expectation fit into it. Here are a few related results that I know:

The Fisher-Tippett-Gnedenko theorem says that if $$F$$ belongs to an MDA, then it belongs to the Fréchet, Gumbel, or Weibull MDA.

The Generalized Extreme Value (GEV) distribution generalizes the Fréchet, Gumbel, and Weibull distributions and contains exactly these distributions. Denote by $$H_\xi$$ the distribution function of a GEV distribution with shape parameter $$\xi \in \mathbb R$$, location parameter $$0$$, and scale parameter $$1$$. We can then summarize the previous theorem by:

If $$F$$ belongs to an MDA, then $$F \in \text{MDA}(H_\xi)$$ for some $$\xi \in \mathbb R$$.

Additionally I know that $$F \in \text{MDA}(H_\xi)$$ with $$\xi > 0$$, i.e. the Fréchet MDA, if and only if $$\overline F(x) := 1- F(x) = L(x) x^{-1/\xi},$$ where $$L$$ is a slowly varying function.

Moreover, if $$X \sim F \in \text{MDA}(H_\xi)$$ with $$\xi > 0$$, then $$\mathbb E X^k = \infty$$ if $$k > 1/\xi$$. In particular, $$\mathbb E X = \infty$$ if $$\xi > 1$$.

Up to this point, though, these are only sufficient conditions for $$\mathbb E X = \infty$$.

I won't explicitly construct a complete counter-example here but here's the idea: for two different values of $$\xi_i > 0$$, with $$\xi_1 < 1$$, let 1 - F(x) interpolate between $$c_{n,1} \xi_1^{-1}$$ on $$(n,n+1/2]$$ and $$c_{n,2}\xi_2^{-1}$$ on $$(n+1/2,n+1]$$, where $$c_{n,i}$$ are chosen so that the function $$1-F(x)$$ is continuous and therefore decreasing. Thus, the mean of $$X$$ with such a distribution does not exist, and is in fact infinite, but it will not be in a domain of attraction, as the distribution cannot "decide" between attraction to $$G_{1/\xi_1}$$ and $$G_{1/\xi_2}$$.