Variance of $\frac{1}{Y-1}$ where Y is Negative Binomial Let $Y\sim NB(2,\theta)$ represent the trial on which the 2nd success is made. 
From a previous question I posed in which many people here helped, I got the UMVUE for $\theta$ as $\frac{1}{Y-1}$. Can someone point me towards the right direction in getting the variance of this estimator? Thanks.
 A: The Variance is $\mathbb{E}(\left(\frac{1}{Y-1}\right)^2)-\mu^2$. We already know by UMVUE (the UE part) that the expectation of this quantity is $p$ so the only thing you need to find is the second moment. To find this write: 
$\mathbb{E}[\left(\frac{1}{Y-1}\right)^2]=\sum_{y=2}^{\infty}\left(\frac{1}{y-1}\right)^2{y-1\choose1}p^2(1-p)^{y-2}$. One of the $y-1$s will cancel with the binomial coefficient and you can factor out the constants $p^2$ and $(1-p)^2$ to get: 
$$\frac{p^2}{(1-p)^2}\sum_{y=2}^{\infty}\left(\frac{1}{y-1}\right)(1-p)^{y}.$$
Now change to a variable $z=y-1$ to get rid of the pesky $y-1$ piece in the denominator. The new equation will be: 
$$\frac{p^2}{(1-p)^2}\sum_{z=1}^{\infty}\left(\frac{1}{z}\right)(1-p)^{z+1}.$$
Again factor out the $(1-p)$ constant term and note the Taylor series expansion: 
$$-\log(1-x)=x+\frac{x^2}{2}+\frac{x^3}{3}+\dots.$$
Here your $x$ is $1-p$. When you put this in and reduce you should get:
$$\mathbb{E}\left[\left(\frac{1}{Y-1}\right)^2\right] = \frac{-\log(p)\,p^2}{(1-p)}$$
now using the variance formula above you should see that the variance will be: 
$$\mathbb{V}ar\left[\frac{1}{y-1}\right]= \frac{-\log(p)\,p^2}{(1-p)} - p^2.$$ 
