When if ever is a median statistic a sufficient statistic?

Question

I came across a casual remark on The Chemical Statistician that a sample median could often be a choice for a sufficient statistic but, besides the obvious case of one or two observations where it equals the sample mean, I cannot think of another non-trivial and iid case where the sample median is sufficient.

Answer 1

In the case when the support of the distribution does not depend on the unknown parameter θ, we can invoke the (Fréchet-Darmois-)Pitman-Koopman theorem, namely that the density of the observations is necessarily of the exponential family form, $$ \exp\{ \theta T(x) - \psi(\theta) \}h(x) $$ to conclude that, since the natural sufficient statistic $$ S=\sum_{i=1}^n T(x_i) $$ is also minimal sufficient, then the median should be a function of $S$, and the other way as well, which is impossible: modifying an extreme in the observations $x_1,\ldots,x_n$, $n>2$, modifies $S$ but does not modify the median. Therefore, the median cannot be sufficient when $n>2$.

In the alternative case when the support of the distribution does depend on the unknown parameter $θ$, I am less happy with the following proof: first, we can wlog consider the simple case when $$ f(x|\theta) = h(x) \mathbb{I}_{A_\theta}(x) \tau(\theta) $$ where the set $A_\theta$ indexed by $θ$ denotes the support of $f(\cdot|\theta)$. In that case, assuming the median is sufficient, the factorisation theorem implies that we have that $$ \prod_{i=1}^n \mathbb{I}_{A_\theta}(x_i) $$ is a binary ($0-1$) function of the sample median $$ \prod_{i=1}^n \mathbb{I}_{A_\theta}(x_i) = \mathbb{I}_{B^n_\theta}(\text{med}(x_{1:n})) $$ Indeed, there is no extra term in the factorisation since it should also be (i) a binary function of the data and (ii) independent from $\theta$. Adding a further observation $x_{n+1}$ which value is such that it does not modify the sample median then leads to a contradiction since it may be in or outside the support set, while $$ \mathbb{I}_{B^{n+1}_\theta}(\text{med}(x_{1:n+1}))=\mathbb{I}_{B^n_\theta}(\text{med}(x_{1:n}))\times \mathbb{I}_{A_\theta}(x_{n+1}) $$