Find $\sum_{k=0}^{\infty}(1-1/n)^{2k}\frac{e^{-n\theta}(n\theta)^{k}}{k!}$ (the variance of $(1-1/n)^{X_1+\cdots+X_n}$)

Question

Given a random sample $X_1,\ldots,X_n$ from Poisson distribution with an unknown parameter $\theta>0$.$T:=(1-1/n)^{X_1+\cdots+X_n}$. Find $\operatorname{var}(T)$.

My work:

I find $T$ is a UMVUE of $e^{-\theta}$. But I cannot use C-R lower bound since not all UMVUE achieve lower bound. My scheme is using transformation to get the pdf of $T$. And then get the variance. Is there any other easier to solve it?

Added:

Let $a=(1-1/n)$ and $Y=a^{X_1+\cdots+X_n}$. Then, one has $f_{Y}(y)=\frac{e^{-n\theta}(n\theta)^{\log_a y}}{(y \ln a)(\log_a y!)}$. But it seems too complicated for me.

Answer 1

First show that $E(T)=e^{-\theta}$ use the fact that sum of $n$ poisson with parameter $\lambda$ is $\mathrm{poisson}(n\lambda)$.

Now show $\sum_{I=1}^nX_i$ is complete sufficient statistic.

Next use the fact that any unbiased estimator based on a complete sufficient statistic is a UMVUE.

Reference : Rao -Blackwell theorem.

To get the variance. $E(T^2)=E((1-1/n)^{2\sum_{I=1}^nX_i})$

Now $S=\sum_{I=1}^nX_i$ follows $\mathrm{poisson}(n\theta)$. so, \begin{align} E(T^2) & =\sum_{i=1}^{\infty}(1-1/n)^{2s} e^{-n\theta}(n\theta)^s/s! \\ & =e^{-n\theta} \sum_{i=1}^{\infty}((n-1^2)\theta/n)^s/s! \\ & = e^{-n\theta}e^{(n-1^2)\theta/n} =e^{(2n-1)\theta/n} \end{align}

Now you can calculate the variance.