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Let $X_1,X_2,\dots ,X_n$ follows $U(n_1,θ)$ where $θ>n_1$ and $n_1\le n$. Find the UMVUE of $θ$.

My answer is: $$\frac{(n+1)X_n}n - \frac{n_1}n$$ Is that correct?

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  • $\begingroup$ Try creating a sample and see if your estimator makes sense. For example, set $n_1 = 100$, $n=2$, and $x_1 = 102, x_2 = 110$. What do you get? $\endgroup$ – jbowman Dec 14 '16 at 17:11
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    $\begingroup$ This question has the [self-study] tag, & states the OP's attempt to solve the question. I think this meets our standards & can stay open. $\endgroup$ – gung - Reinstate Monica Dec 14 '16 at 17:40
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The UMVUE of $U(0,θ)$ is $(n+1)x_n/n$ where $X_1,X_2,\ldots,X_n$ follows $U(0,θ)$. Let $Y_i=X_i-n_1$, then $Y_i$ follows $U(0,ɸ)$. Then the UMVUE of $ɸ$ is $(n+1)Y_n/n$ where $ɸ=θ-n_1$.

\begin{align} \newcommand{\or}{{\rm or}}\newcommand{\(}{{\big{[}}}\newcommand{\)}{{\big{]}}} & &E\((n+1)Y_n/n\) &= ɸ \\[5pt] &\or, &E\((n+1)(X_n-n_1)/n\) &= θ-n1 \\[5pt] &\or, &E\((n+1)(X_n-n_1)/n +n_1\) &= θ \\[5pt] &\or, &E\((n+1)X_n/n-((n+1)n_1)/n+n_1\) &= θ \\[5pt] &\or, &E\((n+1)X_n/n -n_1/n\) &= θ \\[5pt] \end{align}

Hence the UMVUE of $θ$ is $(n+1)X_n/n - n_1/n$.

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  • $\begingroup$ I edited your post to use the $\LaTeX$ formatting the site affords. I think it makes it more readable. Please ensure it still says what you wanted it to. $\endgroup$ – gung - Reinstate Monica Mar 3 '17 at 16:49

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