Let $X_1, · · · , X_n$ be i.i.d. $\mathrm{Uniform}[\alpha, \beta]$, where $\alpha$ and $\beta$ are unknown.
Show that $$ E(X_\ast|X(1), X(n)) = \frac{X(1) + X(n)}{2} $$
where $X_\ast$ is the sample mean, and $X(n)$ is the $n$th order statistic.
I know that $X(1),X(n)$ are the sufficient statistics for this family (minimal in fact) and intuitively it makes sense that the expected value of the sample mean will lie in the middle of the max and min. But how do I show it rigorously?
Let $S:=\{X_1,\dots,X_n\}\setminus\{X(1),X(2)\}$ and let $f:\{1,\dots,n-2\}\to S$ be an injective function (such functions exists a.s.).
Denoting $Y_i=f(i)$ it can be shown that under condition $X(1)=a<b=X(2)$ the random variables $Y_1,\dots,Y_{n-2}$ are iid and $\sim\mathsf{Uniform}(a,b)$.
Then denoting their sample mean as $Y_*$: $$\mathbb E[X_*\mid X(1)=a,X(n)=b]=\mathbb E\frac1n\left[a+b+(n-2)Y_*\right]=\frac1n\left[a+b+(n-2)\left(\frac{a+b}2\right)\right]=\frac{a+b}2$$
This justifies the conclusion that $$\mathbb E[X_*\mid X(1),X(n)]=\frac{X(1)+X(n)}2$$