Let $X_1\dots,X_n$ be $Poisson(\lambda)$. Show that $$T= \left( \frac{n-1}{n} \right)^{\sum_{i=1}^{n}X_i}$$ is an UMVUE for $e^{-\lambda}$
I know that $\sum_{i=1}^{n}X_i$
it is a sufficient and complete statistic since the distribution belongs to the exponential family. But I don't know how to go on to show that $T$ is an UMVUE.
Any hint?
Best Answer
Since $Y = \sum_{i=1}^nX_i \sim$ Poisson$(n\lambda)$, we can compute $E[T] = E\left[\left(\frac{n-1}{n}\right)^Y\right]$ directly,
\begin{eqnarray} E[T] &=& E\left[\left(\frac{n-1}{n}\right)^Y\right]\\ &=& \sum\limits_{k=0}^\infty\left(\frac{n-1}{n}\right)^k\frac{e^{-n\lambda}(n\lambda)^k}{k!}\\ &=& e^{-n\lambda}\underbrace{\sum\limits_{k=0}^\infty\frac{\left(n\lambda\left(\frac{n-1}{n}\right)\right)^k}{k!}}_{e^{n\lambda\frac{n-1}{n}}}\\ &=& e^{-n\lambda}e^{n\lambda\frac{n-1}{n}} \\ &=& e^{-\lambda} \end{eqnarray}
So $T$ is unbiased. By Lehmann-Scheffe, it is UMVUE.