In Wikipedia:


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.

  • 4
    $\begingroup$ This is not clashing with the definition of sufficiency. $\endgroup$
    – Xi'an
    Jan 25, 2019 at 18:59
  • 2
    $\begingroup$ 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). $\endgroup$
    – whuber
    Jan 25, 2019 at 20:12

2 Answers 2


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 $x\text{s}$ change. Thus the proof is not erroneous.

If the number of independent identically distribued Poisson observations were a random variable in its own right, then one would have a different family of distributions and the sum of the observations would probably not be a sufficient statistic.


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.

Most "answers"/"comments" here focus on claiming "yes"/"obvious", without scientific basis, but this provides a more formal answer.


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