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Given a sample $X_1, \ldots, X_n$, which is Poisson distributed, i.e. $X_i\sim P_\theta$, how can one prove that $\overline{X}_n = \frac 1 n \sum_{i=1}^n X_i$ is complete for $\theta$?

I've found this, but since the statistics here is $T=\sum_{i=1}^n X_i$, that doesn't help, does it?

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  • $\begingroup$ A.E. answer is fine, also if you recognize that the Poisson distribution is a one-parameter exponential family, then you easily see $\overline{X}_n$ is complete & sufficient for $\theta$ $\endgroup$ – user365239 Nov 27 '16 at 20:06
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The properties of "sufficiency" and "completeness" preserved under one-to-one transformations as $g(x)=x/n$, $n\in \mathbb{N}$. So you can use the linked proof for $\bar{X}_n$ as well.

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