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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!

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    $\begingroup$ You first state you do know the distribution of $Y_1+\cdots+Y_n = n\bar Y$, so how could you not know the distribution of $\bar Y$? BTW, something seems to be missing after "this can be done by," because the criterion you give is incomplete. $\endgroup$
    – whuber
    Commented Feb 22, 2016 at 18:58
  • $\begingroup$ Hi, thanks for your response. I know that the distribution of the mean is $\frac{1}{n}Pois(n\lambda)$, but scaling a poisson results in a binomial, is this what you meant? Thanks! $\endgroup$ Commented Feb 22, 2016 at 19:04
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    $\begingroup$ "scaling a poisson results in a binomial" that's for large n $\endgroup$
    – Aksakal
    Commented Feb 22, 2016 at 19:14
  • $\begingroup$ Your definition of completeness is a bit off. Completeness of a statistic $T$ means that $\text{E}[g(T)] = 0$ implies that $g(T) = 0$ with probability one. $\endgroup$
    – dsaxton
    Commented Feb 22, 2016 at 21:33
  • $\begingroup$ "This can be done by showing that $\operatorname E[h(\overline Y)]=0$ for all $\lambda$." That makes no sense without saying what $h$ is. I presume you meant "This can be done by showing that $\operatorname E[h(\overline Y)]=0$ for all $\lambda$ ONLY IF $h$ is the identically zero function." $\endgroup$ Commented Jan 25, 2019 at 20:06

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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.

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