My stats class is studying sufficient statistics. So far, every example we have done has assumed $h(\mathbf{X}) = 1$ and so all our examples have involved really trivial uses of the factorization theorem in this regard.

Fisher-Neyman factorisation theorem: If the probability density function is $ƒ_θ(x)$, then $T$ is sufficient for $θ$ if and only if nonnegative functions $g$ and $h$ can be found such that $$ f_\theta(x)=h(x) \, g_\theta(T(x)), \,\! $$

I am a little worried this might come up on an exam where $h(\mathbf{X}) \neq 1$.

My Question

Can someone provide a basic example of using the factorization theorem to show a statistic is sufficient when $h(\mathbf{X}) \ne 1$?


1 Answer 1


As explained in this answer to another question on sufficiency, if you consider an iid sample $(X_1,\ldots,X_n)$ from an arbitrary distribution with density $f$, the order statistic $S(\mathbf{X})=(X_{(1)},\ldots,X_{(n)})$ is sufficient. The distribution of the sample $(X_1,\ldots,X_n)$ conditionally on $S(\mathbf{X})$ is then uniform over all permutations of $S(\mathbf{X})$, with density $h(\mathbf{X})=1/n!$.

Note however that the function $h$ appearing in the factorisation theorem is somewhat arbitrary in that it is a density against a particular dominating measure. For instance the counting measure in the above example. If one incorporates $h$ into the measure, then the corresponding density becomes $1$, which means that we can always assume $h(x)\equiv 1$.


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