UMVUE explanations

Question

Let $X_1,...,X_n$ a random sample where $X$~Poisson$(\theta)$.

i)Find UMVUE for $\theta$

ii)Exists UMVUE for $\frac{1}{\theta}$

For i) I found that $T=\overline{X}$ is UMVUE for $\theta$, but for ii) I tried a few things and I could not get anything, anyone can help me?

Another question I have is suposse I found the UMVUE for $\theta$ for some distribution, and suppose that is asked me to check for the UMVUE for one function $g(\theta)$. Is there any easy way to find out UMVUE for any particular function knowing the UMVUE of $\theta$?

Answer 1

Naive way to solve $ii$:

Observation 1: $\sum X_i$ is complete and sufficient statistic for $\theta$

Observation 2: $\sum X_i \sim Poisson(n\theta)$

We need to look for an unbiased estimator of $\frac{1}{\theta}$ in order to utilise Rao-Blackwell theorem

Let $\delta(x)$ be such an estimator:

$E[\delta(x)] = \frac{1}{\theta}$

Then, $\sum_{k=0}^{\infty}\frac{\delta(k)e^{-\theta}\theta^k}{k!} = \frac{1}{\theta}$ $\implies$ $\sum_{k=0}^{\infty}\frac{\delta(k)\theta^k}{k!} = \frac{e^\theta}{\theta}$

By Taylor expansion: $e^\theta = \sum_{k=0}^{\infty}\frac{\theta^k}{k!}$ $\implies$ $\frac{e^\theta}{\theta} = \sum_{k=0}^{\infty}\frac{\theta^{k-1}}{k!}$

Thus, $\sum_{k=0}^{\infty}\frac{\delta(k)\theta^k}{k!} =\sum_{k=0}^{\infty}\frac{\theta^{k-1}}{k!} $ $\implies$ $\delta(k) = \frac{1}{\theta} $ which is a 'useless' estimator