Exercise:
A biased coin has a probability $p$ that it gives a tail when it is tossed. The random variable $T$ is the number of tosses up to and including the second tail.
Show that $\frac{1}{T-1}$ is an unbiased estimator of $p$.
My work so far:
I know that $P(T=t) = (t-1)(1-p)^{t-2}p^2$ for $t \ge 2$.
I know that $E(T) = \frac{2}{p}$.
I found a similar question at Finding an unbiased estimator for the negative binomial distribution, but I don't understand the first line (!) of the solution, which states:
$$E\left(\frac{r-1}{Y-r-1}\right)=\sum_{y=0}^\infty \frac{r-1}{y+r-1}\binom{y+r-1}{y} \theta^r(1-\theta)^y$$
Can someone please explain where the above expectation expression comes from?
Many thanks!
Best Answer
$Y$ takes nonnegative integer and
$$E(g(Y))=\sum_{y=0}^\infty g(y) P(Y=y)$$
Here $g(y)=\frac{r-1}{y+r-1}$ and $P(Y=y)=\binom{y+r-1}{y}\theta^r (1-\theta)^y$.