Suppose $X \tilde{} NB(n,p)$ and $\mathbb{P}(X=x) = \binom{x-1}{n-1}p^n(1-p)^{x-n}$. Then what is $\mathbb{E}(\frac{1}{X}) = \sum_{x=n}^\infty \frac{1}{x}\binom{x-1}{n-1}p^n(1-p)^{x-n}$?
Many thanks
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These negative moments of random variables are in general difficult to obtain closed-form expressions for. You already have $\text{E}[X^{-1}]$ in the form of an infinite series, but what you ultimately want is a representation as a finite sum or product. As I learned from my mentor, sometimes it is possible to get this using the generating function of $X$ as follows.
Let $G(z)=\text{E}[z^X]$ be the probability generating function of $X$. In this case, $X$ is negative binomial which yields $$ G(z) = \text{E}[z^X] = \Big(\frac{pz}{1-\bar{p}z}\Big)^n $$ with the short notation $\bar{p}=1-p$. Now observe that $$ \int_0^1 \frac{G(z)}{z}\text{d}z = \int_0^1 \text{E}[z^{X-1}]\text{d}z = \text{E}\Big[\int_0^1 z^{X-1}\text{d}z\Big]= \text{E}\Big[\frac{1}{X}\Big] $$ where switching the expectation operator and integral requires some regularity conditions to be fulfilled. I am quite sure in this case they are. So $$ \text{E}\Big[\frac{1}{X}\Big] = \int_0^1 \Big(\frac{pz}{1-\bar{p}z}\Big)^n \frac{\text{d}z}{z} $$ For any finite value of $n$, the integrand is a rational function, so at least in principle it should be possible to integrate this analytically, giving a closed-form result.
