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?
Best Answer
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.