Past exam of the mathematics course of the University of Cambridge / 2002 / ib / Paper 1 / 3H / Solution

For a dominated family of sample distributions, a statistic $T$ is a sufficient statistic precisely when the probability mass function or density factors aswhere $h$ does not depend on the parameter. This is the Fisher-Neyman factorization theorem.

Here is the discrete proof in both directions. Suppose the factorization holds, and write $H(t)={\sum }_{x:T(x)=t}h(x)$. Then ${\mathbb{P}}_{\theta }(T=t)=g_{\theta }(t)H(t)$. Whenever this probability is positive, $H(t)$ is finite and positive andis independent of $\theta$. Thus the conditional distribution of the whole sample given $T$ is parameter-free, which is the definition of a sufficient statistic.

Conversely, if $T$ is sufficient, let $q_t(x)$ denote this common conditional mass function on each possible fiber. ThenTake $g_{\theta }(t)={\mathbb{P}}_{\theta }(T=t)$ and $h(x)=q_{T(x)}(x)$ to obtain the factorization. Fibers impossible for every parameter can be assigned $h=0$; conditional probabilities are required only for fibers with positive probability.

For the independent Poisson sample the joint mass function isAll parameter dependence is throughso this is a one-dimensional sufficient statistic. The sample mean contains the same information, since $n$ is known.