Let $X_{1},\ldots,X_{n}$ be i.i.d. $U[\alpha,\beta]$ r.v.s., and let $X_{(1)}$ denote the $\min$, and $X_{(n)}$ the $\max$. Show that $$ \mathbb{E}\left(\overline{X}_{n}\mid X_{(1)},X_{(n)}\right) = \frac{X_{(1)}+X_{(n)}}{2}. $$
I know that $\displaystyle\mathbb{E}\left(\overline{X}_{n}\mid X_{(1)},X_{(n)}\right)=\mathbb{E}\left({X}_{1}\mid X_{(1)},X_{(n)}\right)$ but not much more.
On
Assume (because of symmetry, it doesn't matter; and it simpifies notation) that $x_{(1)}=x_1=A$ and $x_{(n)}=x_n=B$ ; and $n\ge 3$.
Then the variables $x_2, x_3 \cdots x_{n-1}$ are iid uniform on $[A,B]$
Hence $$E\left(\frac{\sum x_i}{n} \mid x_{(1)},x_{(2)}\right) =\frac{1}{n}\left(A+ (n-2)\frac{(A+B)}2+B \right)=\frac{A+B}{2}$$
After having used a translation and rescaling, we can assume that $\alpha=0$ and $\beta=1$. Assume that $1\lt i\lt n$. Denote $X_{(i)}$ the $i$th greater element among $X_1,\dots,X_n$ (which is almost surely well-defined, as the vector $\left(X_1,\dots,X_n\right)$ has a continuous distribution). Then for each Borel subset $B$ of $\mathbb R^2$, we have, by symmetry, $$ \mathbb E\left[\left(\frac{X_{(1)}+X_{(n)}}2-X_{(i)} \right)\mathbf 1\left\{\left(X_{(1)},X_{(n)}\right)\in B\right\} \right]\\ = n!\mathbb E\left[\left(\frac{X_{1}+X_{n}}2-X_{i} \right)\mathbf 1\left\{\left(X_{1},X_{n}\right)\in B\right\} \mathbf 1\left\{X_1\lt X_2\lt \dots X_i\lt \dots \lt X_n \right\} \right]. $$
Now, we use the fact that if $a$ and $b$ are two fixed real numbers such that $0\leqslant a\lt b\leqslant 1$, then $$\int_0^1\mathbf 1\left\{a\lt x_2\lt\dots \lt x_{i-1}\lt b\right\}\mathrm dx_2\dots dx_{i-1}=\frac{\left(b-a\right)^{i-2}}{(i-2)!} \mbox{ and } $$ $$\int_0^1\mathbf 1\left\{a\lt x_{i+1} \lt\dots \lt x_{n-1}\lt b\right\}\mathrm dx_{i+1} \dots dx_{n-1}=\frac{\left(b-a\right)^{n-i-1}}{(n-i-1)!} . $$ We get, using independence and then the fact that $\left(X_1,X_i,X_n\right)$ has the same distribution as $\left(X_1,X_2,X_3\right)$, $$ \mathbb E\left[\left(\frac{X_{(1)}+X_{(n)}}2-X_{(i)} \right)\mathbf 1\left\{\left(X_{(1)},X_{(n)}\right)\in B\right\} \right]\\ = \frac{ n!}{\left(i-2\right)!\left(n-i-1\right)!} \mathbb E\left[\left(\frac{X_{1}+X_{n}}2-X_{i} \right)\mathbf 1\left\{\left(X_{1},X_{n}\right)\in B\right\}\left(X_i-X_1\right)^{i-2}\left(X_n-X_i\right)^{n-i-1} \mathbf 1\left\{X_1\lt X_i\lt X_n \right\} \right]\\ = \frac{ n!}{\left(i-2\right)!\left(n-i-1\right)!} \mathbb E\left[\left(\frac{X_{1}+X_{3}}2-X_{2} \right)\mathbf 1\left\{\left(X_{1},X_{3}\right)\in B\right\}\left(X_2-X_1\right)^{i-2}\left(X_3-X_2\right)^{n-i-1} \mathbf 1\left\{X_1\lt X_2\lt X_3 \right\} \right]. $$
Define $A :=\sum_{i=2}^{n-1} \mathbb E\left[\left(\frac{X_{(1)}+X_{(n)}}2-X_{(i)} \right)\mathbf 1\left\{\left(X_{(1)},X_{(n)}\right)\in B\right\} \right]$. In view of the previous computations, we have $$A=\sum_{j=0}^{n-3}\frac{n!}{j!(n-3-j)!} \mathbb E\left[\left(\frac{X_{1}+X_{3}}2-X_{2} \right)\mathbf 1\left\{\left(X_{1},X_{3}\right)\in B\right\}\left(X_2-X_1\right)^{j}\left(X_3-X_2\right)^{n-3-j} \mathbf 1\left\{X_1\lt X_2\lt X_3 \right\} \right] \\ =n(n-1) (n-2) \mathbb E\left[\left(\frac{X_{1}+X_{3}}2-X_{2} \right)\mathbf 1\left\{\left(X_{1},X_{3}\right)\in B\right\} \left(X_3-X_1\right)^{n-3} \mathbf 1\left\{X_1\lt X_2\lt X_3 \right\} \right] .$$ Starting to integrate the last expectation with respect to $X_2$, we derive that $A=0$. We therefore showed that $$\mathbb E\left[\sum_{i=2}^{n-1}X_{(i)}\mid X_{(1) },X_{(n)} \right] =\frac{n-2}2\left(X_{(1)} + X_{(n)} \right).$$ To conclude the wanted result, it suffices to notice that $\sum_{i=1}^nX_i= \sum_{i=1}^nX_ { (i)}$.