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??

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    $\begingroup$ 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? $\endgroup$
    – knrumsey
    Sep 14 '18 at 19:21
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    $\begingroup$ @knrumsey - might want to expand that to a full answer, it wouldn't take much. $\endgroup$
    – jbowman
    Sep 14 '18 at 19:41
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    $\begingroup$ "$R$" is not a particularly common notation for a uniform, it might help readers to make it explicit that this is what is meant here. $\endgroup$
    – Glen_b
    Sep 15 '18 at 3:04
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    $\begingroup$ The correct sentence is $(X_{(1)},X_{(n)})$ is sufficient rather than $X_{(1)}$ and $X_{(n)}$ are sufficient. $\endgroup$
    – Xi'an
    Sep 16 '18 at 10:17


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 $\theta$.


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.


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.


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