$X_1, \dots, X_n$ 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.

  • $\begingroup$ can you please explain me how could you get this result? 1/(1+∑Xi/n−1) $\endgroup$ 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



\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.

  • $\begingroup$ Is the answer, $\frac{1}{1+\frac{\sum_{i=1}^n X_i}{n-1}}$ $\endgroup$
    – Harry
    Apr 8 '19 at 11:13
  • $\begingroup$ That is what I get, yes. $\endgroup$ Apr 8 '19 at 13:34
  • $\begingroup$ 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? $\endgroup$
    – Harry
    Apr 9 '19 at 2:33
  • $\begingroup$ That looks like a 'method of moment' estimator, found by equating sample mean with population mean. Do you think UMVUE is the same? $\endgroup$ Apr 9 '19 at 20:29
  • 1
    $\begingroup$ @ZeroPancakes No, $E(W)=P(X_1=0)=p$. $\endgroup$ May 10 '20 at 6:46

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