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 $\theta$. 
 A: In this calculation, we assume that you are familiar with the moment generating function, and know in particular that the moment generating function of a Poissson $X$ with parameter (mean) $\mu$ is given by
$$M_X(t)=E(e^{tX})=e^{\mu(e^t-1)}.   \tag{1}$$
Temporarily, for typing ease, let $c=\frac{n-1}{n}$.  We want to find the expectation of $c^T$, that is, the expectation of 
$$c^{X_1+X_2+\cdots+X_n}.$$ 
Thus we want 
$$E(c^{X_1}c^{X_2}\cdots c^{X_n}).\tag{2}$$
By independence, this is the product of the expectations, which is
$$(E(c^{X_1})^n.\tag{3}$$
So now we go after $E(c^{X_1}$.
Putting $t=\ln c=\ln((n-1)/n))$ in (1) we find that
$$E(c^{X_1})=e^{\mu ((n-1)/n-1)}=e^{-\mu/n}.\tag{4}$$
Expression (3) tells us to take the $n$-th power of this. we get $e^{-\mu}$, which is $\theta$. This completes the proof.
Remark: For completeness, we sketch the calculation of the moment generating function. We have
$$E(e^{tX})=\sum_{k=0}^\infty e^{tk} e^{-\mu}\frac{\mu^k}{k!}.$$
Reorganize this as
$$e^{-\mu} \sum_{k=0}^\infty \frac{w^k}{k!},$$
where $w=\mu e^t$.
But we recognize $\sum_0^\infty \frac{w^k}{k!}$ as the series expansion of $e^w$, and now it is just a mater of putting pieces together.
A: 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
$$
\mathbb E\left( \left(\frac{n-1}{n}\right)^{X_1+\cdots+X_n} \right) = \theta.
$$
