Let $X_1, \cdots, X_n$ be iid from a uniform distribution $U[\theta-\frac{1}{2}, \theta+\frac{1}{2}]$ with $\theta \in \mathbb{R}$ unknown. Take for granted that $T(\mathbf{X}) = (X_{(1)}, X_{(n)})$ is minimal sufficient statistic for $\theta$ and consider the unbiased estimator $\overline{X}$. Apply Rao-Blackwell's Theorem and find the estimator that improves upon $\overline{X}$ by conditioning on $T$.
My attempt: Let $\phi(T) = E(\overline{X}|T)$. By Rao-Blackwell's Theorem, $\phi(T)$ is a uniformly better unbiased estimator of $\theta$ than $\overline{X}$. However, how can I find an expression for this estimator?
My initial idea: To compute the expectation, I will need to find the conditional distribution of $\overline{X}$ given $T$ (which ofcourse does not depend on $\theta$ since $T$ is sufficient). I have found the joint distribution of $T= (X_{(1)}, X_{(n)})$ which is given by
\begin{align*} f_{X_{(1)}, X_{(n)}}(x_{(1)}, x_{(n)}) = \begin{cases} n(n-1)(x_{(n)} - x_{(1)})^{n-2} & \text{ if } \theta-\frac{1}{2} < x_{(1)} < x_{(n)} < \theta+\frac{1}{2} \\ 0 & \text{ otherwise } \end{cases}. \end{align*}
If I can somehow find the joint distribution of $\overline{X}$ and $T$, then I should be able to find the conditional distribution and then compute the conditional expectation.
Is this the right approach? If so, how can I find the joint distribution of $\overline{X}$ and $T$?
Here the first step is to choose an arbitrary $X_i$.
Note that
\begin{equation*} \begin{split} & \hspace{15pt} E(X_i \ | \ X_{(1)} = l, \ X_{(n)} = u) \\ & = E(X_i \ |\ X_{(1)} = l, \ X_{(n)} = u, \ X_{i} = l) * P(X_i = X_{(1)}) \\ & + \ E(X_i \ | \ X_{(1)} = l, \ X_{(n)} = u, \ X_i = u) * P(X_i = X_{(n)}) \\ & + \ E(X_i \ | \ X_{(1)} = l, \ X_{(n)} = u, \ l < X_i < u) * P (X_i \neq X_{(1)}, X_i \neq X_{(n)}) \\ & = l*\frac{1}{n}+u*\frac{1}{n}+\frac{l+u}{2}*\frac{n-2}{n} \\ &= \frac{(l + u)}{2} \end{split} \end{equation*}
Now just apply the linearity of expectation to get that $E(\bar{X} \ | \ X_{(1)}, X_{(n)}) = \frac{X_{(1)} + X_{(n)}}{2}$
Edit: tweaked to make it clear that we are consider the probability $X_i$ is equal to the order statistic.