Given a Bernoulli r.v., $W$, which is derived from r.v. $T$ (Poisson)
(a) if $T=0$ then $W=1$ and
(b) if $T>0$ then $W=0$.
One has to show that the sample mean (the proportion of $0$s in the sample), is an unbiased estimate of $\phi=e^\lambda$. Also, how does one find the variance of the sample mean and show that this variance exceeds the CRLB?
I am unsure how to make the function to have a second derivative in order to solve the rest of the question.
At the moment based on the rules of the Bernoulli the function equals to $e^{-\lambda}$. How do I proceed?
I suspect you mean: $T$ is a Poisson random variable with parameter $\lambda$, and $$W = \cases{1 & if $T = 0$\cr 0 & if $T \ne 0$\cr}$$ and you want an unbiased estimator of $e^{-\lambda}$. It couldn't be $e^\lambda$ because $e^\lambda > 1$ if $\lambda > 0$, while the proposed estimator is always $\le 1$. Hint: What is $P(T = 0)$?