If $Y_1, \ldots, Y_n$ are iid from $Pois(\lambda)$, how to show that the sample mean is a complete sufficient statistic? If $Y_1, \ldots, Y_n$ are iid from $Pois(\lambda)$, then I know that the sum of iid poisson random variables has the $Pois(n\lambda)$ distribution. However, I am trying to show that $\bar{Y}$ is a complete sufficient statistic. This can be done by showing that $E[h(\bar{Y})]=0$ for all $\lambda$. But, without knowing the distribution of $\bar{Y}$, how is it possible to show this? Thanks!
 A: To show that a statistic $T = \sum_i y_i$ is sufficient you can appeal to the factorization theorem, which says that $T$ is sufficient if and only if the likelihood can be written in the form $L_\lambda(y) = h(y) g_\lambda(T)$ for nonnegative functions $h$ and $g$.  In the Poisson case the likelihood takes the form
$$
L_\lambda(y) = \frac{1}{\prod_{i} y_i!} e^{-n \lambda} \lambda^{\sum_i y_i}
$$
and so the condition is satisfied with $h(y) = \left ( \prod_i y_i! \right )^{-1}$ and $g_\lambda(T) = e^{-n \lambda} \lambda^{\sum_i y_i}$.  Now if we're sampling from an exponential family then $T$ will be complete if the natural parameter space contains an open set (this is a convenient property of exponential families that allows us to avoid direct reference to the somewhat unusual definition of completeness).  The Poisson mass function can be written
$$
f_\lambda(y) = \frac{e^{-\lambda}}{y!} \exp \left [ y \log(\lambda) \right ] 
$$
and so it is in fact an exponential family with natural parameter space $\{ \log(\lambda) : \lambda > 0 \} = \mathbb{R}$.  Since this space contains an open set (it is an open set) $T$ is also complete.
