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??
 A: 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.
