sampling distribution of order statistics

Let, $X_{1},X_{2},\ldots,X_{n}$ be an i.i.d. sample from $R(1-\theta,1+\theta)$. Show that, $(X_{(1)},\bar{X},X_{(n)})$ is sufficient for $\theta$.

Ans: $f(x_{1},x_{2},\ldots,x_{n})=\frac{1}{(2\theta)^n}I_{(1-\theta,1+\theta)}{(x_{1},x_{2},\ldots,x_{n})}=\frac{1}{(2\theta)^n}I_{\theta>\rm{max}(1-x_{(1)},x_{(n)}-1)}$.

Hence, evidently $X_{(1)}$ and $X_{(n)}$ are sufficient for $\theta$.But how $\bar{X}$ can be minimal sufficient??

• If $T_1$ is sufficient, then $(T_1, T_2)$ is also sufficient (for any statistic $T_2$). Adding more information won't eliminate sufficiency. The question becomes is $(X_{(1)}, \bar X, X_{(n)})$ minimal sufficient? Sep 14 '18 at 19:21
• @knrumsey - might want to expand that to a full answer, it wouldn't take much. Sep 14 '18 at 19:41
• "$R$" is not a particularly common notation for a uniform, it might help readers to make it explicit that this is what is meant here. Sep 15 '18 at 3:04
• The correct sentence is $(X_{(1)},X_{(n)})$ is sufficient rather than $X_{(1)}$ and $X_{(n)}$ are sufficient. Sep 16 '18 at 10:17

Sufficiency

Assume $$T_1$$ is sufficient for $$\theta$$. Then the statistic $$(T_1, T_2)$$ is also sufficient for $$\theta$$ (for any statistic $$T_2)$$.

Proof: Since $$T$$ is sufficient for $$\theta$$, there exists a factorization

$$f({\bf x}) = g(T_1|\theta)h(\theta)$$

Let $$A$$ be a set such that $$P(T_2 \in A) = 1%$$. Define $$g'((T_1, T_2)|\theta) = g(T_1|\theta)\cdot I(T_2 \in A)$$ Thus by Factorization Theorem $$(T_1, T_2)$$ is also sufficient for $$\theta$$.

$$\square$$

The intuition here is that sufficient statistics contain all information for $$\theta$$. If I add another statistic to the collection, I may not gain anything, but I certainly won't lose sufficiency. So in your case, the sufficiency of $$T_1 = (X_{(1)}, X_{(n)})$$ guarantees the sufficiency of $$(T_1, \bar X)$$.

Minimal Sufficiency

Now, assume that $$T_1$$ is minimal sufficient. This means that if $$T'$$ is another sufficient statistic, then $$T_1$$ is a function of $$T'$$.

Claim: If $$T_1$$ is minimal sufficient, then $$(T_1, T_2)$$ is minimal sufficient if and only if there exists a mapping $$T_2 = \phi(T_1)$$

Proof: Consider an arbitrary sufficient statistic $$T'$$. Since $$T_1$$ is min. suff., there must exist a function such that $$T_1 = h(T')$$.

First, assume that we can write $$T_2 = \phi(T_1)$$. Then $$[T_1, T_2] = [h(T'), \phi(h(T'))]$$ therefore $$(T_1, T_2)$$ is also minimal sufficient for $$\theta$$.

If there does not exist a $$\phi()$$ such that $$T_2 = \phi(T_1)$$, then the pair $$(T_1, T_2)$$ can not be written as a function of $$T_1$$. Since $$T_1$$ is sufficient, this implies (by definition) that $$(T_1, T_2)$$ is not minimal sufficient.

$$\square$$

Returning to your example, there is no way to write $$\bar X$$ as a function of $$X_{(1)}$$ and $$X_{(n)}$$, so the statistic $$(X_{(1)}, X_{(n)}, \bar X)$$ is not minimal sufficient for $$\theta$$.

Side Note: It turns out that the statistic $$(X_{(1)}, X_{(n)})$$ is not minimal sufficient here either. It is easy to see that the statistic $$T_{ms} = \max(1-X_{(1)}, X_{(n)} - 1)$$ is sufficient, and $$(X_{(1)}, X_{(n)})$$ cannot be expressed as a function of $$T_{ms}$$ and thus cannot be min suff.