# The expectation of the inverse of a negative binomial random variable?

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

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.