Context, name or explanation for a formula in the "law of rare events" The formula is
$$\psi(m) = \frac{m^m\,e^{-m}}{m!}$$
and it is described as la probabilité de l'espérance mathématique dans la loi des événements rares in the original paper by E.J. Gumbel's Les valeurs extrêmes des distributions statistiques.
In it I tend to think that $m$ stands for the highest $m\in \mathbb N$ number of draws in a samples $N>m$ from a symmetrical distribution:
Quand on fait $N $observations ou quand on prend $N$ échantillons
sur une distribution initiale illimitée dans les deux sens, on peut
classer les valeurs numériques qui constituent ces $N$ observations
par ordre de grandeur. Elles seront toutes finies. Il y en aura une
qui sera la plus grande, la dernière valeur, $x_1,$ une autre immédiatement inférieure $x_2,$ que l’on peut considérer comme la plus petite
parmi les deux plus grandes, en général une $m^{\text{ième}}$ qui sera la plus
petite parmi les $\mathbf m$ plus grandes. Considérons ces valeurs en partant
de l’autre extrémité de la suite.
I don't know if there is a connection to the gamma or the digamma function (psi symbol), but I see that it is approximated one line below in the paper by
$$\frac 1{\sqrt{2\pi\, m}}$$
clearly the re-arrangement of the Stirling approximation:
$$m! \sim \frac{\sqrt{2\pi \,m}\;m^m}{e^m}$$
 A: Let $X$ be a Poisson random variable with rate $\lambda$: the limit of binomial random variables in which the the number of trials $n \to \infty$ but the probability of success $p = \frac{\lambda}{n}$ goes to $0$, keeping the expected value at $np = \lambda$. Hence the "law of rare events" in the paper you're reading: this is the law (probability distribution) we get for counting the number of these rare (probability tending to $0$) events in a pool of very many trials.
Then $\Pr[X=k]$ is given by the formula $e^{-\lambda} \cdot \frac{\lambda^k}{k!}$. The formula you're looking at is the probability that a Poisson random variable with rate $m$ is equal to $m$. (Here, $m$ is also the expected value of this random variable; I'm not fluent in French, but I bet that this is what l'espérance mathematique is.)
You can estimate $\frac{m^m e^{-m}}{m!}$ as $\frac1{\sqrt{2\pi m}}$ by Stirling's formula. But this is also the estimate provided by the central limit theorem, since for large $m$, we can approximate a Poisson random variable with rate $m$ by a normal random variable with mean $m$ and variance $m$. So actually, this gives us a probabilistic proof of Stirling's formula that $m! \sim \frac1{\sqrt {2\pi m}} (\frac me)^m$, starting from the central limit theorem.
