I am studying on UMVUE, and I'm struggling to find that conditional expectation
Let $X_1,\ldots,X_n$ random sample of $X\sim U[0,\theta]$. i) Show that $2X_1$ is a unbiased estimator for $\theta$ and use the Rao-BlackWell Theorem for found the UMVUE for $\theta$.
ii)Calculate $E[X_{(n)}]$ and explicitly find UMVUE for $\theta$
I already show that $2X_1$ is unbiased and also found that $X_{(n)}=\max(X_1,\ldots,X_n)$ is a complete and sufficient statistic for $\theta$, but I am having trouble finding the conditional, how I can calulate $$E[2X_1\mid X_{(n)}]=2E[X_1\mid X_{(n)}]$$ How do I calculate the conditional distribution and the expectation?
$\newcommand{\E}{\operatorname{E}}$I've always found this particular problem mildly icky in the way in which it mixes the discrete and the continuous and apparently has to be done piecewise.
You need \begin{align} & \E(X_1\mid X_{(n)}) \\[10pt] = {} & \E(\E(X_1 \mid X_1=X_{(n)})\mid X_{(n)})\Pr(X_1=X_{(n)}\mid X_{(n)}) \\[2pt] & {} + \E(\E(X_1 \mid X_1\ne X_{(n)})\mid X_{(n)})\Pr(X_1\ne X_{(n)}\mid X_{(n)}) \end{align}
Observe that $\Pr(X_1=X_{(n)}) = 1/n$ because every one of the observations is equally likely to be the maximum, and $\E(\E(X_1\mid X_1=X_{(n)})\mid X_{(n)}) = X_{(n)}$. So you get $$ \frac{X_{(n)}}n + \frac{n-1} n \E(\E(X_1 \mid X_1\ne X_{(n)})\mid X_{(n)}). $$ If you can show that $\E(\E(X_1 \mid X_1\ne X_{(n)})\mid X_{(n)})= X_{(n)}/2$ then you get $$ X_{(n)}\cdot\left( \frac 1 n + \frac{n-1}{2n} \right) = X_{(n)} \cdot\frac{n+1}{2n} $$ and your best unbiased estimator will be $2$ times that.