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!


\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}

  • 1
    $\begingroup$ 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

$Y$ 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 $P(Y=y)=\binom{y+r-1}{y}\theta^r (1-\theta)^y$.

  • $\begingroup$ 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

Sample over all possibilities using your $P(T = t)$:

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.