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:


where the term $\frac{1}{x}$ seems not to be absorbed by the binomial coefficient $\binom{x−1}{r-1}=\frac{(x-1)!}{(r-1)!(x-r)!}$.

  • 2
    $\begingroup$ You seems too early to judge that the calculation is difficult before actually substitute and see what happen. Note that there is a Binomial coefficient in the pmf which can "absorp" the $1/x$ part and form another negative binomial pmf kernel. Then you find out the normalizing constant. $\endgroup$ – BGM Nov 9 '18 at 8:07
  • $\begingroup$ Thank you. How can $1/x$ be absorbed in $\frac{1}{x}\binom{x-1}{r-1} = \frac{1}{x} \frac{(x-1)!}{(r-1)!(x-r)!}$? $\endgroup$ – Nikolaos Skout Nov 9 '18 at 9:03
  • $\begingroup$ Sorry I overlooked the parametrization. You may take a look at wolframalpha.com/input/?i=sum+x+from+r+to+inf,+1%2Fx*(x-1)!%2F(r-1)!%2F(x-r)!*(1-p)%5Er*p%5E(x-r) $\endgroup$ – BGM Nov 10 '18 at 16:58
  • 1
    $\begingroup$ I doubt that has a simple form. In case that $E[1/(X-1)]$ is more tractable (as it seems), then we could get some bounds using $a/(X-1)\le 1/X < 1/(X-1)$ with $a=(r-1)/r$ $\endgroup$ – leonbloy Nov 11 '18 at 20:27

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?

  • $\begingroup$ This differential equation doesn't seem very trivial. One would assume a bounty calls for more than a hint that may or may not be helpful. $\endgroup$ – Lee David Chung Lin Nov 14 '18 at 0:06

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