Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 1 / 19H / a / Solution

A statistic $T$ is a sufficient statistic for $\theta$ if the conditional distribution of the data given $T$ does not depend on $\theta$. For a dominated family with density or mass function $f_{\theta }(x)$, the Fisher-Neyman factorization theorem states that this is equivalent towhere $h$ is nonnegative and independent of $\theta$, and $g_{\theta }$ is nonnegative and depends on the data only through $T$.

For a discrete sample space, suppose first that the factorization holds. If ${\mathbb{P}}_{\theta }(T=t)>0$, summing over the fibre gives ${\mathbb{P}}_{\theta }(T=t)=g_{\theta }(t){\sum }_{T(x)=t}h(x)$, soThis conditional distribution is independent of $\theta$, proving that $T$ is a sufficient statistic. Conversely, let $q_t(x)$ be the common conditional mass function guaranteed by the sufficient statistic property, for every fibre attainable under at least one parameter. Set $h(x)=q_{T(x)}(x)$ and $g_{\theta }(t)={\mathbb{P}}_{\theta }(T=t)$. Then ${\mathbb{P}}_{\theta }(X=x)=g_{\theta }(T(x))h(x)$. Fibres never attained under any parameter can be assigned any conditional distribution, since their multiplying probability is always zero. This proves both directions of the discrete criterion, including possible parameter-dependent supports.