# UMVUE of $\mathbb{P}[X=x_0]$ where $X_1,\cdots,X_n$ is Poisson

I came up with this problem and tried to solve it myself. Please check my solution, I am a bit unsure it is correct because it says the UMVUE for the probability where $$x_0>\text{sum of observations}$$ is actually $$0$$. I'm also interested in how to manually verify that the final expression is indeed unbiased (i.e. without knowing that it was created by Rao-Blackwellization of an unbiased estimator). It's easy to verify in the $$x_0=0$$ case since we can use the MGF of $$S$$. $$\newcommand{\P}{\mathbb{P}}\newcommand{\E}{\mathbb{E}}$$

Let $$X_1,\cdots,X_n$$ be a random sample from a $$\mathrm{Po}(\lambda)$$ distribution. What is the UMVUE of $$p(x_0|\lambda)=\P[X=x_0]=\frac{e^{-\lambda}\lambda^{x_0}}{x_0!}?$$

I got the answer $$T=\begin{cases}\binom{S}{x_0}(1/n)^{x_0}(1-1/n)^{S-x_0}&x_0\leqq S,\\0&x_0>S,\end{cases}$$where $$S=\sum X_i$$, via the following method:

Consider the rudimentary estimator $$W(X_1,\cdots,X_n)=\begin{cases}1&X_1=x_0,\\0&X_1\neq x_0.\end{cases}$$Then $$\E[W]=\P[X_1=x_0]=p(x_0|\lambda)$$, so $$W$$ is unbiased. The exponential family pmf of the Poisson distribution tells us that $$S=\sum_{i=1}^n X_i$$is a complete sufficient statistic for $$\lambda$$. Define the Rao-Blackwell estimator $$T=\E[W|S].$$Notice we can write $$S=\underbrace{X_1}_{\mathrm{Po}(\lambda)}+\underbrace{X_2+\cdots+X_n}_{\mathrm{Po}((n-1)\lambda)}.$$It is a well known property that if $$A\sim\mathrm{Po}(\alpha),B\sim\mathrm{Po}(\beta)$$, where both are independent, then the conditional distribution of $$A$$ given $$A+B=s$$ is $$A|_{A+B=s}\sim\mathrm{Bin}\left(s,\frac{\alpha}{\alpha+\beta}\right).$$It follows from Bayes' theorem. Applying this property to our problem, we have $$X_1|_{S}\sim\mathrm{Bin}(S,1/n).$$Finally, we have $$T=\E[W|S]=\P[X_1=x_0|S]=\begin{cases}\binom{S}{x_0}(1/n)^{x_0}(1-1/n)^{S-x_0}&x_0\leqq S,\\0&x_0>S.\end{cases}$$Since this unbiased estimator only depends on the data through a complete sufficient statistic, it is the UMVUE by Lehmann-Scheffe.

• Is my solution correct?
• If it is correct, can it be simplified?
• How can we directly verify that this final estimator $$T$$ is unbiased without appealing to the fact that it is the Rao-Blackwellization of an unbiased estimator?
• Testing some cases, it seems to work correctly and be unbiased. You can show $\sum\limits_{S=x_0}^\infty e^{-n\lambda}\frac{(n\lambda)^S}{S!} \, {S \choose x_0}\frac{(n-1)^{S-x_0}}{n^S}$ $=e^{-\lambda} \frac{\lambda^{x_0}}{x_0!}\sum\limits_{S=x_0}^\infty e^{-(n-1)\lambda}\frac{((n-1)\lambda)^{S-x_0}}{(S-x_0)!}$ $= e^{-\lambda}\frac{\lambda^{x_0}}{x_0!}$ with some careful cancellation and factoring Commented May 24 at 0:48
• There shouldn't be any doubt regarding the answer. Also see math.stackexchange.com/q/3268755/321264. Commented May 24 at 6:04