Poisson Estimators

Question

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 $\theta$.

Answer 1

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.

Answer 2

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 $$ \mathbb E\left( \left(\frac{n-1}{n}\right)^{X_1+\cdots+X_n} \right) = \theta. $$