Note: I don't know if this is the best place for this question, because it is very specific. However, I'm not sure of a better place to go (apart from one of the other SE's). If you have a recommendation, I'd love to hear it.
Anyway, here goes:
I'm reading a paper (Arxiv) in which a critical part is the derivation of a limiting form of a distribution---the relevant equations are (7) and (8). Specifically, given the negative binomial distribution
$$P(s,t|s_0) = \binom{s-1}{s_0-1}(e^{-\kappa t})^{s_0}(1-e^{-\kappa t})^{(s-s_0)},$$
with $s,s_0,\kappa,t>0$, the authors claim that in the limit as $t\to\infty$,
$$P(s,t\to\infty|s_0) = \frac{1}{\Gamma(s_0)s_0^{-s_0}}s^{s_0-1}\exp(-s_0s),$$
where $\Gamma(x)$ is the usual Gamma function. This is precisely a $\mathrm{Gamma}(s_0,1/s_0)$ density in the location-shape parameterization.
I am flabbergasted how they arrived at that. If I try putting the exact distribution $P(s,t|s_0)$ into Mathematica for a given value of $s_0$ and let $t\to\infty$, I decidedly do not observe that asymptotic result; rather, the mean and width of the distribution grow without limit as $t$ does. The result I obtain from taking a limit on the first expression recapitulates what I see in Mathematica, but it looks like
$$P(s,t\to\infty|s_0) = \frac{1}{\Gamma(s_0)(e^{-\kappa t})^{-s_0}}s^{s_0-1}\exp({-se^{-\kappa t}})$$
i.e., a $\mathrm{Gamma}(s_0,e^{\kappa t})$ density. If I continue to take the limit to $t\to\infty$, I won't get a distribution at all.
This is a pretty central result of their paper, the paper's now been published in PRL, and it would be a pretty severe typo to make, so my first instinct is that I'm misunderstanding what they mean by "asymptotic limit."
Can you provide any clarification as to why the authors claim the second equation is a limit of the first equation? In what sense is this true? Or what am I missing?
If you agree with me, then I'll feel more confident going to the paper authors and asking for clarification.
