$1/E(x)$ vs. $E(1/x)$ Question concerning MVUE and geometric distribution, trying to apply Rao-Blackwell Theorem here.
We know that the geometric distribution is a regular exponential class with $$Y = \sum x$$ as our sufficient and complete statistic. However $$ E(Y) = \frac{n(1- \theta)}{\theta} $$ is not an unbiased estimator because it does not equal theta. Since 1/E(X) is usually not E(1/X), I tried $$ E(\frac{1}{y}) = \sum\frac{1}{y}\theta(1-\theta)^\frac{1}{y} $$ but that's where I got royally stuck. Is there a way around this summation so that I can get $$ E(Y) = \theta $$
 A: The definition of the Geometric distribution you're using is one for which $X_1, \dots, X_n$ have the probability mass function
$$p_\theta(k) = (1-\theta)^k\theta\text{, }\quad k = 0, 1, 2, \dots\text{.}$$
Now we have that
$$f_{X_1, \dots, X_n}(x_1, \dots, x_n) =\prod_{i=1}^{n}p_\theta(x_i) = (1-\theta)^{\sum x_i}\theta^n$$
and as you stated, this is of the exponential family, and $Y = \sum X_i$ is a complete and sufficient statistic. Notice that $Y$ is a sum of iid Geometric random variables, hence $Y$ is negative binomial with
$$f_{Y}(y) = \binom{y + n - 1}{n-1}(1-\theta)^y\theta^n\text{, }\quad y = 0, 1, 2, \dots\text{.}$$
Consider the estimator
$$T = \begin{cases}
1, & X_1 = 0 \\
0, & X_1 \neq 0\text{.}
\end{cases}$$
$T$ is an unbiased estimator of $\theta$. Consider, through Rao-Blackwellization,
$\mathbb{E}[T \mid Y]$.
We have that
$$\begin{align}
\mathbb{P}(T = 1 \mid Y = y) &= \dfrac{\mathbb{P}(T = 1 \text{ and }Y = y)}{\mathbb{P}(Y = y)} \\
&= \dfrac{\mathbb{P}(X_1 = 0 \text{ and }\sum_{i=1}^{n}X_i = y)}{\binom{y + n - 1}{n-1}(1-\theta)^y\theta^n} \\
&= \dfrac{\mathbb{P}(X_1 = 0 \text{ and }\sum_{i=2}^{n}X_i = y)}{\binom{y + n - 1}{n-1}(1-\theta)^y\theta^n} \\
&= \dfrac{\mathbb{P}(X_1 = 0)\mathbb{P}(\sum_{i=2}^{n}X_i = y)}{\binom{y + n - 1}{n-1}(1-\theta)^y\theta^n} \\
&= \dfrac{\theta\cdot \mathbb{P}(\sum_{i=2}^{n}X_i = y)}{\binom{y + n - 1}{n-1}(1-\theta)^y\theta^n}
\end{align}$$
Now we observe $\sum_{i=2}^{n}X_i$ is a sum of $n - 1$ iid Geometric random variables, hence
$$\mathbb{P}\left(\sum_{i=2}^{n}X_i = y\right) = \binom{y + (n-1)-1}{(n-1)-1}(1-\theta)^y\theta^{n-1} = \binom{y + n -2}{n - 2}(1-\theta)^y\theta^{n-1}\text{.}$$
Through some algebra, we obtain
$$\dfrac{\theta\cdot \mathbb{P}(\sum_{i=2}^{n}X_i = y)}{\binom{y + n - 1}{n-1}(1-\theta)^y\theta^n} = \dfrac{\theta^n}{\theta^{n}}\cdot \dfrac{(y+n-2)!}{(n-2)!y!} \cdot \dfrac{(n-1)!y!}{(y+n-1)!} = \dfrac{n-1}{y+n-1}$$
hence (note that $T = 0$ adds nothing to this expected value)
$$\mathbb{E}[T \mid Y] = 1 \cdot \dfrac{n-1}{Y+n-1} = \dfrac{1}{\frac{\sum_{i=1}^{n}X_i}{n-1} + 1} = \left(\dfrac{\sum_{i=1}^{n}X_i}{n-1} + 1 \right)^{-1}\text{.}$$
Because we have that $T$ is unbiased for $\theta$, we have $$\theta = \mathbb{E}[T] = \mathbb{E}[\mathbb{E}[T \mid Y]] = \mathbb{E}\left[\left(\dfrac{\sum_{i=1}^{n}X_i}{n-1} + 1 \right)^{-1} \right]\text{.}$$
A: The probability of success on each trial is $\theta,$ and $X_1$ is the number of failures before the first success, so $X_1\in \{0,1,2,3,\ldots\}.$
$Y= X_1+\cdots+X_n$ is a sufficient statistic for $\theta,$ i.e. the conditional probability distribution of $(X_1,\ldots,X_n)$ given $Y$ does not depend on $\theta.$
Now let $W= \begin{cases} 1 & \text{if } X_1=0, \\ 0 & \text{if } X_1>0. \end{cases}$
Then $\operatorname E(W) = \theta.$ So here we have an unbiased estimator of $\theta.$
The corresponding Rao–Blackwell estimator is $\operatorname E(W\mid Y).$
\begin{align}
\operatorname E(W\mid Y=y) = {} & \Pr(W=1\mid Y=y) \\[8pt]
= {} & \Pr(X_1=0\mid X_1+\cdots + X_n=y) \\[8pt]
= {} & \frac{\Pr(X_1=0\ \&\ X_1+\cdots +X_n=y)}{\Pr(X_1+\cdots + X_n=y)} \\[8pt]
= {} & \frac{\Pr(X_1=0\ \&\ X_2+\cdots + X_n=y)}{\Pr(X_1+\cdots + X_n=y)} \\[8pt]
= {} & \frac{\Pr(X_1=0)\Pr(X_2+\cdots + X_n=y)}{\Pr(X_1+\cdots+X_n=y)}
\end{align}
And $\theta$ will cancel out.
Can you finish this>
