a) Suppose $N$ is a Poisson random variable with mean $\lambda$. Show that $E[(1+a)]^N]=e^{a\lambda}$ for any constant $a$.
b) Telephone calls enter a switchboard in a Poisson process of rate $\lambda$ every $10$ minutes. In a 10 minute period, $N$ calls enter the switchboard. In order to decide whether he can chance an unofficial tea-break, the operator wants to estimate the probability $\theta$ that there will be no incoming calls in the next $20$ minutes, using this single observation $N$.
Write down an expression for $\theta$ in terms of $\lambda$. Use this, together with the result in part (a), to suggest an unbiased estimator $T(N)$ of $\theta$. Is this estimator sensible?
I have done the first part of the question but I am unsure on how to find $\theta$ in the next part and how to find an estimator. I am thinking that I should set $\alpha$ to be something but I am unclear on how these estimators work.
The probability that there are no calls in the next $20$ minutes is simply $\Pr[N(2) = 0] = e^{-2\lambda}$, where $N(t) \sim \operatorname{Poisson}(\lambda t)$ is the random number of calls in $t$ time increments of $10$ minutes. So $$\theta = e^{-2\lambda}$$ is the desired relationship. This would seem to suggest letting $a = -2$ in part (a), which would give you $$\operatorname{E}[(-1)^N] = e^{-2\lambda} = \theta,$$ and so the estimator $$T(N) = (-1)^N$$ is unbiased for $\theta$. But when it comes to asking whether it's a good estimator, is it? It would mean that if $N$ is odd, you would choose $\theta = -1$, whereas if $N$ is even, you would choose $\theta = 1$, and since $\theta$ is a probability--remember, it is the probability of no calls in the next $20$ minutes--this is an awful choice for an estimator.
Is there a better estimator we can choose, even if it is possibly biased?