# Sufficient statistic for Poisson in wiki?

In Wikipedia:

https://en.wikipedia.org/wiki/Sufficient_statistic#Poisson_distribution

it says that $$X_1+\cdots+X_n$$ is a sufficient statistic for the parameter of the Poisson distribution and its proof follows by using the factorization theorem. However, the expression they obtain is:

$$e^{-n\lambda}\lambda^{(x_1+x_2+\cdots +x_n)}\cdot {1 \over x_1!x_2!\cdots x_n!}$$

which also depends on $$n$$. Am I correct by assuming that this proof is flawed?

I know another proof using another result, but I wonder about the correctness of this claim in wiki.

• This is not clashing with the definition of sufficiency. – Xi'an Jan 25 '19 at 18:59
• You misread Wikipedia: the formula you quote is the probability for the vector $(x_1,x_2,\ldots,x_n).$ Please read on from that statement through the end of their demonstration that the likelihood depends only on the mean of the $x_i$ (or, equivalently, their sum, because $n$ is known). – whuber Jan 25 '19 at 20:12

The proof of the factorization theorem is based on the sample size (the conditional distribution of the sample given the test statistic, of size $$n$$, for each $$n$$), thus, it is specific to each $$n$$. For this reason, the sample size $$n$$ is considered fixed/known.
A sufficient statistic is sufficient for a particular family of probability distributions, and in this case that family is actually not the family of Poisson distributions, each member of which is supported on the set $$\{0,1,2,3,\ldots\},$$ but rather it is a family of distributions supported on the set $$\{0,1,2,3,\ldots\}^n$$ of all $$n$$-tuples of finite cardinal numbers. Thus the random variable involved is an $$n$$-tuple $$(X_1,\ldots,X_n)$$ in which the components are independent and each has a $$\operatorname{Poisson}(\lambda)$$ distribution.
Within the set of all such $$n$$-tuples one finds the observable data $$(x_1,\ldots,x_n)$$. If the observable data are altered within that set, the number $$n$$ does not change; only the $$x$$s change. Thus the proof is not erroneous.