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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?

ithisa
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1 Answers1

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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 $1$.

André Nicolas
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