There was an exercise in class asking to solve for the expectation E[1/(Y+1)] when Y is a negative binomial random variable (number of failures).
How do I solve this? Whatever transformation or substitution I try I just can't seem to derive a constant number that looks like an expectation.
I would appreciate an explanation. Thanks!
$$\mathbb{E}\Bigg[\frac{1}{Y+1}\Bigg]=\sum_{y=0}^{\infty}\frac{1}{y+1}\frac{(r+y-1)!}{y!(r-1)!}\theta^r(1-\theta)^y=$$
$$=\frac{\theta}{(1-\theta)(r-1)}\underbrace{\sum_{(y+1)=0}^{\infty}\frac{[(r-1)+(y+1)-1]!}{(y+1)!(r-2)!}\theta^{r-1}(1-\theta)^{y+1}}_{=1}=\frac{\theta}{(1-\theta)(r-1)}$$
The expression over the brace is 1 because it is the sum of
$$P(Z=r-1)=\binom{(r-1)+z-1}{z}\theta^{r-1}(1-\theta)^{z}$$
over all Z. (it is still a negative binomial counting the number of failures before the $(r-1)$th success