# Finding the expectation $\mathrm{E} (1/ X)$ for a negative binomial random variable $X$

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!

• I believe this question would be more appropriate in math.stackexchange.com. Moreover, I think the title does not represent well your final question. Nov 2, 2020 at 9:43
• Isn't $X = 0$ with positive probability? Nov 2, 2020 at 9:51
• @Kernel Thanks. My final question is finding a lower bound for $\mathrm{E} \left[ X/ (X + c)^n\right]$. The reason for my first part in the question is that: by Jensen inequality, $\mathrm{E} \left[ X / (X + c)^2\right] \geq \left[ \mathrm{E} X + 2c + c^2 \mathrm{E} (1/ X)\right]^{-1}$, then finding a upper bound for $\mathrm{E} (1/X)$ is enough.
– 香结丁
Nov 3, 2020 at 5:54
• @Kernel Oh! You said the title. Got it! Thanks so much~
– 香结丁
Nov 3, 2020 at 5:56

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]]


$$-\frac{(\theta -1) \mu \theta ^{\mu } \, _4F_3(c+1,c+1,c+1,\mu +1;c+2,c+2,c+2;1-\theta )}{(c+1)^3}$$

• Thanks. But it seems that it is also involved to find a lower bound of the expectation from this formula.
– 香结丁
Nov 3, 2020 at 6:01