Poisson Estimators Consider a simple random sample of size $n$ from a Poisson distribution with mean $\mu$. Let $\theta=P(X=0)$. 
Let $T=\sum X_{i}$. Show that $\tilde{\theta}=[(n-1)/n]^{T}$ is an unbiased estimator of $\theta$. 

We have $\Pr(X_1=0)=e^{-\mu}=\theta$.
Therefore
$$
\theta=\mathbb E(\Pr(X_1=0\mid X_1+\cdots+X_n)).
$$
So what is
$$
\Pr(X_1=0\mid X_1+\cdots+X_n=x)\text{ ?}
$$
It is
$$
\begin{align}
& {}\qquad \frac{\Pr(X_1=0\text{ and } X_1+\cdots+X_n=x)}{\Pr(X_1+\cdots+X_n=x)} = \frac{\Pr(X_1=0)\cdot\Pr(X_2+\cdots+X_n=x)}{e^{-n\mu}(n\mu)^x/(x!)} \\[10pt]
& = \frac{\left(e^{-\mu}\right)\cdot\left(e^{-(n-1)\mu}((n-1)\mu)^x/(x!)\right)}{e^{-n\mu}(n\mu)^x/(x!)} = \left(\frac{n-1}{n}\right)^x \\[10pt]
\end{align}
$$
Therefore
$$
\mathbb E\left( \left(\frac{n-1}{n}\right)^{X_1+\cdots+X_n} \right) = \theta.
$$
