Suppose a random variable $X$ is distributed as $\operatorname{NB}(\mu, \theta)$, and its mass is as follows
$$ \mathrm{P}(X = y) = \binom{y + \theta - 1}{y} \left(\frac{\mu}{\mu + \theta}\right)^{y}\left(\frac{\theta}{\mu + \theta}\right)^{\theta}.$$
Does anyone know how to calculate the expectation of $1 / X$ in this kind of definition of negative binomial distribution?


If there is no explicit expression for $\mathrm{E} (1 / X)$, I wonder whether exists some lower bounds for
$$\mathrm{E} \left[ \frac{X}{(X + c)^n}\right],$$
where $c$ is some positive constant and $n \in \mathbb{N}$.


Thanks so much!



Finding the expectation $\mathrm{E} (1/ X)$ for a negative binomial random variable $X$
Mathematica answers your second question for concrete values of $n$ (e.g. $n=3$) by



```
Mean[TransformedDistribution[X/(c + X)^3, X \[Distributed] NegativeBinomialDistribution[\[Mu], \[Theta]],Assumptions->c>0]]

```

$$-\frac{(\theta -1) \mu \theta ^{\mu } \, \_4F\_3(c+1,c+1,c+1,\mu +1;c+2,c+2,c+2;1-\theta )}{(c+1)^3} $$


