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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?

hmakholm left over Monica
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Roland
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    None of this is necessary to compute the UMVUE $\hat\theta_n$ since one knows that $\hat\theta_n=X_{(n)}/x_n$, with $E(X_{(n)})=x_n\theta$. Can you compute $x_n$? (The conditional distribution of $X_1$ conditionally on $X_{(n)}$ happens to be an interesting beast, being partially discrete and partially absolutely continuous, but it is not needed to compute $\hat\theta_n$.) – Did May 19 '15 at 21:15
  • @Did I'm having difficulty understanding when I need to find the conditional and when it is not needed. I'll edit the exercise because it has two items – Roland May 19 '15 at 21:24
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    I've posted an answer that does it in the pedestrian way. Now let me take a shot at understanding "Did"'s way. If you can show that $X_{(n)}$ is complete and sufficient, and that $\operatorname{E}(X_{(n)})=(n-1)\theta/n$, then you can find and estimator that depends on the data only through $X_{(n)}$ and that is unbiased. Since it's complete, sufficient, and unbiased, the Lehmann–Scheffé theorem theorem tells you it's the best unbiased estimator. ${}\qquad{}$ – Michael Hardy May 19 '15 at 21:32
  • https://math.stackexchange.com/questions/261530/finding-ex-1-max-x-i – StubbornAtom May 06 '19 at 10:33
  • https://math.stackexchange.com/q/60497/321264, https://math.stackexchange.com/q/2941489/321264 – StubbornAtom Jun 12 '20 at 19:05

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$\newcommand{\E}{\operatorname{E}}$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.

Michael Hardy
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  • When you have a sufficient and complete statistics, it is better to use Lehmann-Scheffé theorem, calculate the expectation of statistics and manipulate it to find the UMVUE, than finding the conditional? – Roland May 19 '15 at 21:50
  • It's simpler in cases where it can be done. – Michael Hardy May 19 '15 at 22:16