I searched for "rao blackwell poisson", "umvu poisson", and "umvue poisson" on this website but didn't find anything that specifically answered my question. I also searched on Google for "umvu poisson" and found a bunch of lecture notes, but nothing answering my question.
Setup
I am studying An Introduction to Probability and Statistical Inference, second edition, written by George G. Roussas. I am trying to solve this problem from page 313:
Example 17. Determine the UMVU estimate of $\theta$ on the basis of the random sample $X_1, \dots, X_n$ from the distribution $P(\theta)$.
For context, I quote Theorem 7 from page 312:
Theorem 7 (Rao-Blackwell and Lehmann-Scheffé). Let $X_1, \dots, X_n$ be a random sample with p.d.f. $f(\cdot ; \theta)$, $\theta \in \Omega \subseteq \mathbb{R}$, and let $T = T(X_1, \dots, X_n)$ be a sufficient statistic for $\theta$ (which is complete). Let $U = U(X_1, \dots, X_n$) be any unbiased estimate of $\theta$, and define the statistic $\varphi (T)$ by the relation
\begin{align} \varphi (T) = \text{E}_{\theta} (U \mid T) \tag{11}\label{eq11}. \end{align}
Then $\varphi (T)$ is unbiased, $\text{Var}_{\theta} \big( \varphi (T) \big) \leq \text{Var}_{\theta}(U)$ for all $\theta \in \Omega$, and, indeed, $\varphi (T)$ is a UMVU estimate of $\theta$. If $U$ is already a function of $T$ only, then the conditioning in $(11)$ is superfluous.
My approach
I need an unbiased estimator $U$ and sufficient (and complete) statistic $T$. It's fairly straightforward to figure out that $U = \bar{X}$ is unbiased and $T = \sum_{i = 1}^n X_i$ is sufficient. (I did not verify completeness.)
Since I can write $U = \frac{1}{n}T$, Theorem 7 tells me that I'm done: I don't need to bother with the conditional expectation. However, I wanted to do it anyway, and this is where I run into trouble.
Is computing $\text{E} \big( \bar{X} \mid \sum_{i = 1}^n X_i \big) = \text{E}(U \mid T)$ tractable? The first problem I see is that $\bar{X}$ can take non-integer values, so that makes me think I'm going to need to integrate from zero to infinity. The second problem is figuring out $f_{U \mid T} (u \mid t)$, i.e. the conditional distribution needed to compute the expectation.
Going back to the definition of conditional probability, I know that $f_{U \mid T} (u \mid t) = \frac{f_{U, T} (u, t)}{f_T(t)}$. The joint probability in the numerator looks "redundant" in the sense that $(U = u \cap T = t) = (U = u) = (T = t)$.
If I eliminate the $U$ part, I am left with $\frac{f_T(t)}{f_T(t)} = 1$, which doesn't make sense. If I eliminate the $T$ part, I am left with $\frac{f_U(u)}{f_T(t)}$ and then I need to figure out how $U$ and $T$ are distributed. $T$ is easy, but $U$ is probably harder.
Am I way off track here?
George Roussas' approach
Roussas lets $U = X_1$ and uses the same $T$ as I did. He then uses the fact that $X_1 \mid T = t \sim B(t, \frac{1}{n})$ so that $\text{E}(X_1 \mid T) = \frac{T}{n} = \bar{X}$. I am supposed to verify completeness of $T$ in Exercise 3.17 to conclude that $\bar{X}$ is UMVU for $\theta$.
Summary
Is the moral of the story here that, if you want to compute the Rao-Blackwell conditional expectation, then you should pick a simple unbiased estimator for which the conditional distribution is easily found?
Thanks.
