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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!

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\begin{align} & \operatorname E\left( \frac 1 {T-1} \right) = \sum_{t=2}^\infty \frac 1 {t-1} \Pr(T=t) = \sum_{t=2}^\infty \frac 1 {t-1} (t-1)(1-p)^{t-2}p^2 \\[10pt] = {} & p^2 \sum_{t=2}^\infty (1-p)^{t-2} = p^2 \times \text{sum of a geometric series} = p^2 \times \frac{\text{first term}}{1 - \text{common ratio}} \\[10pt] = {} & p^2 \times \frac 1 {1-(1-p)} \end{align}

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    $\begingroup$ This makes sense to me. It's actually just the first step that wasn't occurring to me; the rest is straightforward. Thanks! $\endgroup$ – Kevin Frederick Mar 24 '18 at 6:31
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$Y$ 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 $P(Y=y)=\binom{y+r-1}{y}\theta^r (1-\theta)^y$.

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  • $\begingroup$ Could you be more informative? Why does $E(g(Y))$ take this value? $\endgroup$ – Taroccoesbrocco Mar 23 '18 at 5:22
  • $\begingroup$ it is Definition 1.1 in this document $\endgroup$ – Siong Thye Goh Mar 23 '18 at 5:28
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Sample over all possibilities using your $P(T = t)$:

$$\sum_{t=2}^\infty (t-1)(1-p)^{t-2}p^2 \frac{1}{t - 1} = $$ $$\sum_{t=2}^\infty (1-p)^{t-2}p^2 = $$ $$p^2\sum_{t=0}^\infty (1-p)^{t} = $$ $$p^2\times\frac{1}{p} = $$ $$p$$

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