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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.

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    $\begingroup$ Did you mean to write "that a sample median could often be"? $\endgroup$ – Juho Kokkala Nov 6 '14 at 13:55
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    $\begingroup$ 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
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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 $$ \mathbb{I}_{B^{n+1}_\theta}(\text{med}(x_{1:n+1}))=\mathbb{I}_{B^n_\theta}(\text{med}(x_{1:n}))\times \mathbb{I}_{A_\theta}(x_{n+1}) $$

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  • $\begingroup$ 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

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