User talk:Drone Better

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March 2019

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Thank you. Drm310 🍁 (talk) 15:39, 17 March 2019 (UTC)

further comment on Coupon collector's problem

Continuing from my revision, I realized that it should be quite trivial to get the moment generating function for the Coupon collector's problem: Let $\mu _{k}$ be the k-th moment of the Coupon collector problem, that is, $\mu _{n,k}=(n-1)!((D_{x}x)^{k}f_{n-1}(x)){\Big |}_{x=1/n}$. Then the moment generating function is $F(t;n)=\sum _{k=0}^{\infty }{\frac {\mu _{n,k}}{k!}}t^{k}=(n-1)!\sum _{k=0}^{\infty }\left({\frac {(tD_{x}x)^{k}}{k!}}f_{n-1}(x)\right){\Big |}_{x=1/n}=(n-1)!e^{t}\left(e^{txD_{x}}f_{n-1}(x)\right){\Big |}_{x=1/n}$

By umbral calculus formula, it is $(n-1)!e^{t}f_{n-1}(e^{t}/n)$. The cumulant generating function would have a simpler expression, as a sum of terms:

$K(t;n)=\ln F(t;n)=nt+\ln n!-n\ln n-\sum _{r=1}^{n-1}\ln(1-re^{t}/n)$

which I believe is equivalent to ${\frac {n^{\{x\}}}{n^{x}}}$. I didn't check, but I calculated the first 3 cumulants and they did match your result.

This allows some quick derivations. For example, direct calculation gives $\partial _{t}|_{t=0}K(t;n)=nH_{n}$, and $\kappa _{l}=-\sum _{r=0}^{n-1}\partial _{t}^{l}|_{t=0}\ln \left(1-re^{t}/n\right)$ for all $l\geq 2$, which is computer-evaluatable.

As for how this may connect to Stirling numbers, I believe this combinatorial formula would be useful: $\left(D_{x}x\right)^{n}=\sum _{k=0}^{n}S(n,k)x^{k}D_{x}^{k}$.

pony in a strange land (talk) 18:23, 17 July 2025 (UTC)