# UMVUE- geometric distribution where $X$ is the number of failures preceding the first success

$$X_1, \dots, X_n$$ iis geometric: $$P(X=x) = (1-p)^{x}p$$, $$x=0,1,2, \dots$$

My Attempt:

$$T=\sum_{i=1}^n X_i$$ is a sufficient statistic

$$W= \begin{cases}1 & X_1= 0,\\ 0 & X_1\neq 0\end{cases}$$ W is an unbiased estimator of $$p$$

To find UMVUE, \begin{align} E[W|T=t] &= \frac{P(X_1 = 0, T=t)}{P(T=t)}\\[5pt] &= \frac{P(X_1 = 0)P(X_1+\cdots +X_n=t)}{P(T=t)}\\[5pt] \end{align} Can somebody please help me expand this step. It's confusing whether I should use $$t-1$$ or $$t-2$$ in the combination part of negative binomial pdf.

• can you please explain me how could you get this result? 1/(1+∑Xi/n−1) Sep 18 '21 at 9:52

Since $$X_j$$ is the number of failures preceding the first success for each $$j$$, $$T=\sum\limits_{j=1}^n X_j$$ is the number of failures before the $$n$$th success. Therefore pmf of $$T$$ is

$$P(T=t)=\binom{n+t-1}{t}\theta^n(1-\theta)^{t}\,\mathbf1_{t\in\{0,1,2,\ldots\}}$$

Now,

\begin{align} E\left[W\mid T=t\right]&=\frac{P\left[X_1=0,\sum\limits_{i=2}^n X_i=t\right]}{P(T=t)} \\&=\frac{P(X_1=0)P\left[\sum\limits_{j=1}^{n-1} X_j=t\right]}{P(T=t)}\qquad,\,\small j=i-1 \end{align}

So the '$$t$$' remains as it is; it is a matter of '$$n$$' and '$$n-1$$' in the pmf of negative binomial.

• Is the answer, $\frac{1}{1+\frac{\sum_{i=1}^n X_i}{n-1}}$ Apr 8 '19 at 11:13
• That is what I get, yes. Apr 8 '19 at 13:34
• I also tried solving this question using the fact that geometric distribution belongs to exponential family and therefore $\sum_{i=1}^n X_i$ is complete sufficient. Taking \begin{align}E(\sum_{i=1}^n X_i) &= \frac{n(1-p)}{p} \end{align} from this I got the umvue of $p$ as $\frac{1}{X\bar+1}$. Is this method wrong? Apr 9 '19 at 2:33
• That looks like a 'method of moment' estimator, found by equating sample mean with population mean. Do you think UMVUE is the same? Apr 9 '19 at 20:29
• @ZeroPancakes No, $E(W)=P(X_1=0)=p$. May 10 '20 at 6:46