Finding UMVUE of function of poisson parameter

Question

I am to estimate $\exp(-\lambda)\lambda^2/2$ from the distribution $Exp(\lambda) \sim \frac{e^{-\lambda}\lambda^x}{x!}$

I used the indicator function $W=\mathbb I_{2}(X_1)$ as an initial unbiased estimator and the sufficient statistic $T=\sum{X_i}$. Using the Rao-Blackwell Theorem, $\hat{W} = \mathbb E(W|T) = \frac{t(t-1)(n-1)^{t-2}}{2n^t}$ is the UMVUE.

However, when I check$$\mathbb E_\lambda(\hat{W}) = \sum_{t=0}^{n}\frac{t(t-1)(n-1)^{t-2}e^{-n\lambda}(n\lambda)^t}{2n^tt!}=\frac{1}{2}\exp(-n\lambda)\lambda^2$$where is the n coming from and how do i get rid of it? Did I make a mistake anywhere?

Answer 1

While the conditional expectation is correct, the series is not properly identified: \begin{array} {}\displaystyle\sum_{t=2}^{\infty}&\displaystyle\frac{t(t-1)(n-1)^{t-2}e^{-n\lambda}(n\lambda)^t}{2n^tt!}\\ &\displaystyle= \frac{\lambda^2 e^{-n\lambda}}{2}\,\sum_{t=2}^{\infty}\frac{t(t-1)(n-1)^{t-2}n^t\lambda^{t-2}}{n^t\,t!}\\ &\displaystyle= \frac{\lambda^2 e^{-n\lambda}}{2}\,\sum_{t=2}^{\infty}\frac{(n-1)^{t-2}\lambda^{t-2}}{(t-2)!}\\ &\displaystyle= \frac{\lambda^2 e^{-n\lambda}}{2}\,\sum_{k=0}^{\infty}\frac{(n-1)^{k}\lambda^{k}}{k!}\\ &\displaystyle= \frac{\lambda^2 e^{-n\lambda}}{2}\,e^{(n-1)\lambda}\\ &\displaystyle=\frac{\lambda^2}{2}e^{-\lambda} \end{array}