Finding an unbiased estimator for the negative binomial distribution Consider a negative binomial random variable Y
  as the number of failures that occur before the r
 th success in a sequence of independent and identical success/failure trials. The pmf of $Y$
  is $$nb(y;r,\theta)=\begin{cases}
{y+r-1 \choose y}\theta^{r}(1-\theta)^{y} & y=0,1,2,\dots\\
0 & \text{otherwise }
\end{cases}$$
 Suppose that $r\geq2$
 .
(a) Show that $\tilde{\theta}=\frac{r-1}{Y+r-1}$
  is an unbiased estimator for $\theta$
 ; i.e. show that $E(\tilde{\theta})=\theta$
 . 
How woudl I proceed? How do I even calculate $E(\tilde{\theta})$? This might be a stupid question since the teacher hasn't covered this yet (i'm learning more quickly) but how would this thing work?
 A: Note that 
$$E\left(\frac{r-1}{Y-r-1}\right)=\sum_{y=0}^\infty \frac{r-1}{y+r-1}\binom{y+r-1}{y} \theta^r(1-\theta)^y.$$
First we do a bit of binomial coefficient manipulation. In general 
$$\binom{n}{k}=\frac{n}{k}\binom{n-1}{k-1}.$$ 
This can be proved easily by expressing binomial coefficients in terms of factorials. (There are also combinatorial proofs.)
Since $r\ge 2$,  our expectation is
$$\theta\sum_{y=0}^\infty \binom{y+r-2}{y} \theta^{r-1}(1-\theta)^y.\tag{1}$$
The expression (1) is $\theta$ times a certain sum. We will be finished if we show that sum is equal to $1$. Note that
$$\binom{y+r-2}{y} \theta^{r-1}(1-\theta)^y$$
is the probability that the $r-1$-th success occurs on the $y+r-1$-th trial. 
For we must choose $y$ places among the first $y-r-2$ to have failure. The probability we have failures in the chosen places and success in the remaining $r-2$ places is $\binom{y+r-2}{y}\theta^{r-2}(1-\theta)^y$. finally, multiply by $\theta$ to take account of the fact we have success in the $y+r-1$th place. 
Summing from $y=0$ to $\infty$ therefore must yield $1$. 
