Show that $(1-\frac{1}{n})^{\sum_{i=1}^n X_i}$ is an unbiased estimator of $\tau(\theta)=e^{-\lambda}$ Let $S=\sum_{i=1}^n X_i$ than show that $T(X_1,\ldots,X_n)= \left(1-\frac{1}{n}\right)^{\sum_{i=1}^n X_i}$ is an unbiased estimator of $\tau(\theta)=e^{-\lambda}$ where $X_1,\ldots,X_n$ are IID from POI($\lambda$) with $n>1$. 
 A: Hint: Try using the fact that $S= \sum X_i$ is a Poisson random variable with parameter $n\lambda$ so that $$E\left[\left(1-\frac{1}{n}\right)^S\right]=\sum_{i=0}^\infty \left(1-\frac{1}{n}\right)^i e^{-n\lambda}\frac{(n\lambda)^i}{i!}=e^{-n\lambda}\sum_{i=0}^\infty \frac{(n\lambda - \lambda)^i}{i!}$$
A: You have $e^{-\lambda} = \Pr(X_1=0)$, so one unbiased estimator of $e^{-\lambda}$ is
$$
Y = \begin{cases} 1 & \text{if }X_1=0, \\  0 & \text{if } X_1>0. \end{cases}
$$
Lemma:
The conditional distribution of $X_1,\ldots,X_n$ given $S$ does not depend on $\lambda$.
Proof:
\begin{align}
& \Pr(X_1=x_1\ \&\ \cdots \ \&\ X_n=x_n \mid S=s) = \frac{\Pr(X_1=x_1\ \&\ \cdots \ \&\ X_n=x_n\ \& \ S=s)}{\Pr(S=s)} \\[10pt]
= {} & \frac{\Pr(X_1=x_1\ \&\ \cdots \ \&\ X_n=x_n)}{\Pr(S=s)} = \frac{\dfrac{e^{-\lambda} \lambda^{x_1}}{x_1!} \cdots \dfrac{e^{-\lambda} \lambda^{x_n} }{x_n!}}{\left(\dfrac{(n\lambda)^s e^{-n\lambda}}{s!}\right)} = \frac{s!}{x_1!\cdots x_n!} \cdot \frac 1 {n^s}. \qquad \blacksquare
\end{align}
Corollary: $\Pr(Y=1\mid S=s)$ does not depend on $\lambda$.
Therefore $\Pr(Y=1\mid S=s)$ is just a function of $s$, and which function it is can be known without knowing $\lambda$.  So let us evaluate that function.
\begin{align}
& \Pr(Y=1\mid S=s) = \frac{\Pr(Y=1\ \&\ S=s)}{\Pr(S=s)} = \frac{\Pr(X_1=0\ \&\ X_2+\cdots+X_n = s)}{\Pr(S=s)} \\[10pt]
= {} & \frac{e^{-\lambda} \cdot \left(\dfrac{((n-1)\lambda)^s e^{-(n-1)\lambda}}{s!} \right)}{\left(\dfrac{(n\lambda)^s e^{-n\lambda}}{s!}\right)} = \left( \frac{n-1} n \right)^s.
\end{align}
By the law of total probability,
$$
\operatorname{E}\left( \left(\frac{n-1} n \right)^S \right) = \operatorname{E}\left( \Pr\left( Y= 1 \mid S \right) \right) = \Pr(Y=1) = e^{-\lambda}.
$$
Therefore $\left( \dfrac{n-1} n \right)^S$ is an unbiased estimator of $e^{-\lambda}$.
A: So basically, you can say that since you're looking for an unbiased estimator, you need
$$E[\hat{\tau(\theta)}]=\tau(\theta)=e^{-\lambda}$$
So we know
$$E[\hat{\tau(\theta)}]=E\left[\left(1-\frac{1}{n}\right)^S\right]$$
And the sum of n random IID Poisson variables is just another Poisson r.v with parameter $n\lambda$.  Then using the definition of expectation you get your answer.
