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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 $\theta$?

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    $\begingroup$ Since you seem to have a lot of very closely related questions, it's time to take advantage of our site search capabilities. For this one, "UMVUE Poisson" are good keywords. Some of the first hits include stats.stackexchange.com/questions/55377 and stats.stackexchange.com/questions/143086, both of which appear to answer this question as well as several other questions you have posted. $\endgroup$ – whuber May 15 '15 at 19:43
  • $\begingroup$ @whuber My main question is in the second part, and I found nothing about. $\endgroup$ – user72621 May 15 '15 at 19:51
  • $\begingroup$ @askazy There exists no unbiased estimator for case $ii$ $\endgroup$ – rightskewed May 16 '15 at 9:32
  • $\begingroup$ @rightskewed But how I can show it? $\endgroup$ – user72621 May 16 '15 at 12:40
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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

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