Let $X$ be a random variable of the negative binomial distribution, of parameters $r\in \mathbb{N},~p\in (0,1)$. Evaluate $E\left(\dfrac{1}{X}\right)$.
Attempt. Of course
$$E\left(\dfrac{1}{X}\right)=\sum_{k=r}^{\infty}\frac{1}{k} P(X=k)$$
but the substitution of the pmf of $X$ seems to make the calculation difficult. Am I on the right path, or should we work on X being the sum of independent geometric rvs?
Thanks for the help.
*Edit. Writing down the calculations, we would go like:
$$E\left(\frac{1}{X}\right)=\sum_{x=r}^{\infty}\frac{1}{x}\binom{x−1}{r-1}p^r(1-p)^{x-r},$$
where the term $\frac{1}{x}$ seems not to be absorbed by the binomial coefficient $\binom{x−1}{r-1}=\frac{(x-1)!}{(r-1)!(x-r)!}$.
Best Answer
Hint: By differentiating the equation with respect to $p$ we obtain:$${dE\{{1\over X}\}\over dp}=r\sum_{x=r}^{\infty}\frac{1}{x}\binom{x−1}{r-1}p^{r-1}(1-p)^{x-r}-\sum_{x=r}^{\infty}(x-r)\frac{1}{x}\binom{x−1}{r-1}p^r(1-p)^{x-r-1}$$by using the definition of $E\{{1\over X}\}$ and plugging it in the above equation we can conclude:$${dE\{{1\over X}\}\over dp}={r\over p}E\{{1\over X}\}-{1\over 1-p}+{r\over 1-p}E\{{1\over X}\}$$can you finish now?