So there is this problem
(Truncated Poisson) Let $X \sim POI(\mu)$, and suppose we cannot observe $X=0$ so the observed random variable, $Y$, has discrete pdf
$\displaystyle {f(y; \mu)= \begin{cases} \frac{e^{-\mu} \mu^y}{y! (1-e^{-\mu})} & y = 1,2,... \\ 0 & \text{otherwise} \end{cases}}$
We desire to estimate $P[X>0] = 1 - e^{-\mu}$. Show that an unbiased (but unreasonable) estimate of $1 - e^{-\mu}$ is given by $u(Y)$ where $u(y) = 0$ if $y$ is odd, and $u(y) = 2$ if $y$ is even. Hint: Consider the power series expansion of $\frac{(e^\mu + e^{-\mu})}{2}$
So we have never gone over this kind of problem before in class and none of the people I talked to know how to do this problem. Can anyone help?
You need to show that $E[u(Y)]=1-e^{-\mu}$. They gave you the hint to find the power series of $\frac{e^{\mu}+e^{-\mu}}{2}$. That is the function $\cosh(\cdot)$ and its power series is (show this yourself!): \begin{align} \frac{e^{\mu}+e^{-\mu}}{2}=\sum_{k=0}^\infty\frac{\mu^{2k}}{(2k)!} \end{align} Its radius of convergence is the whole real line.
Let's show unbiasedness: \begin{align} E[u(Y)]&=\sum_{k=1}^\infty 2\frac{e^{-\mu} \mu^{2k}}{(2k)!(1-e^{-\mu})}\\ &=\frac{2e^{-\mu}}{1-e^{-\mu}}\sum_{k=1}^\infty\frac{\mu^{2k}}{(2k)!}\\ &=\frac{2e^{-\mu}}{1-e^{-\mu}}\left(\frac{e^{\mu}+e^{-\mu}}{2}-1\right)\\ &=\frac{e^{-\mu}}{1-e^{-\mu}}\left( e^{\mu}+e^{-\mu}-2\right)\\ &=\frac{1+e^{-2\mu}-2e^{-\mu}}{1-e^{-\mu}}\\ &=\frac{(1-e^{-\mu})^2}{1-e^{-\mu}}\\ &=1-e^{-\mu} \end{align} Done!