I came across a 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.

  • 1
    $\begingroup$ Did you mean to write "that a sample median could often be"? $\endgroup$ – Juho Kokkala Nov 6 '14 at 13:55
  • 8
    $\begingroup$ It's an interesting question; the double exponential has the median for a ML estimator of its location parameter, but it's not sufficient. $\endgroup$ – Glen_b Nov 6 '14 at 15:57

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 $S$, which is impossible: modifying an extreme in the observations $x_1,\ldots,x_n$, $n>2$, modifies $S$ but does not modify the median.

In the alternative case when the support of the distribution does depend on the unknown parameter θ, we can consider the case when $$ f(x|\theta) = h(x) \mathbb{I}_{A_\theta}(x) \tau(\theta) $$ where the set $A_\theta$ indexed by θ is the support of $f$. In that case, the factorisation theorem implies that $$ \prod_{i=1}^n \mathbb{I}_{A_\theta}(x_i) $$ is a 0-1 function of the sample median $$ \prod_{i=1}^n \mathbb{I}_{A_\theta}(x_i) = \mathbb{I}_{B^n_\theta}(\text{med}(x_{1:n})) $$ 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 $$ \mathbb{I}_{B^{n+1}_\theta}(\text{med}(x_{1:n+1}))=\mathbb{I}_{B^n_\theta}(\text{med}(x_{1:n}))\times \mathbb{I}_{A_\theta}(x_{n+1}) $$

  • $\begingroup$ What is set $B_\theta^n$? $\endgroup$ – 3x89g2 May 3 '17 at 5:56
  • $\begingroup$ It is the support of the median. $\endgroup$ – Xi'an May 3 '17 at 6:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.