$X_1, \dots, X_n$ iis geometric: $P(X=x) = (1-p)p^{x-1}$, $x=1,2, \dots$
Find UMVUE for $p$ when $n=1$ and in general.
Attempt
$T=\sum_{i=1}^n X_i$ is a sufficient statistic
$w = \begin{cases}1 & X_i\neq 1,\\ 0 & X_i=1 \end{cases}$
then $EW = P(X_i\neq 1) = p$ so $W=I[X_i \neq 1]$ is an unbiased estimator of $p$
For $n=1, E[W|T=t] = \frac{P(X_1\neq 1, X_1 = t) }{P(X_1=t)} = \frac{p}{(1-p)p^{t-1}}$
For general $n$ $E[W|T] = \frac{P(X_1\neq 1, \sum_{i=1}^nX_i = t) }{P(\sum_{i=1}^nX_i = t)} = \frac{\sum_{x_1=2}P(X_1=x_1)P(\sum_{i=2}^n X_i = t-x_1)}{\sum_{i=1}^n X_i = t}$
where $P(T=t) = \binom{t-1}{n-1}(1-p)^np^{t-n}$
Is this correct? Is there a closed form for UMVUE in the general case?
Best Answer
No. If $E[W|T=t]=g(p)$, i.e., its expected value depends on the unknown parameter $p$ then $T$ is not a sufficient statistic. It should be \begin{align} E[W|T=t] &= \frac{P(X_1 = 1, T=t)}{P(T=t)}\\[5pt] &= \frac{P(X_1 = 1)P(X_2+\cdots +X_n=t-1)}{P(T=t)}\\[5pt] & = \frac{p\binom{t-2}{n-2}p^{n-1}(1-p)^{t-n}}{\binom{t-1}{n-1}p^{n}(1-p)^{t-n}}\\[5pt] &= \frac{n-1}{t-1}\end{align} and therefore $$E[W\mid T] = \frac{n - 1}{T - 1} =\frac{n-1}{\sum_{i=1}^n X_i-1}\xrightarrow{P}\frac{1}{EX}=p.$$