# Bounding the variance of an unbiased estimator for a uniform-distribution parameter

$X_1,\ldots,X_6$ is a sample from a uniform distribution $\left[ 0, \theta \right]$, $\theta$ is $[1,2]$. Find an unbiased estimator for $\theta$ with variance less than $\dfrac{1}{10}$.

I thought the M.L.E is $\max \left( X_i \right)$,and the unbiased estimator without other restriction shoud be $\hat\theta_N=\dfrac{N+1}N\max(X_i)$, (N=6).

But, I have no idea how to make the variance be less than $\dfrac{1}{10}$. I know that $$\mathrm{Var}\left(\hat\theta\right) = \theta^2\dfrac{N}{(N+1)^2(N+2)}\,,$$ so $\mathrm{Var}\left(\hat\theta\dfrac{N+1}{N}\right) = \dfrac{\theta^2}{(N+2)N}$.

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Dear nina, Please don't use only the homework tag. I've added the statistics one as well. An outline on how to enter equations can be found near the bottom of this section in the FAQ. Proper punctuation and capitalization will help make your question more legible and, hence, easier to answer. Welcome to the site! Cheers. :) –  cardinal Jan 22 '13 at 13:33
Note that $\hat\theta = \max_i X_i$ is not (quite) the MLE in this case since you have a restricted parameter space $\Theta = [1,2]$ instead of $\Theta = (0,\infty)$. –  cardinal Jan 22 '13 at 13:37
thanks for your help:) –  nina li Jan 22 '13 at 13:37
If $\theta\lt2$ and $N=6$, then $\dfrac{\theta^2}{(N+1)N}\lt\dfrac{2^2}{(6+1)6}\lt\dfrac1{10}$. –  Did Jan 22 '13 at 14:11
yes,I think you are right.thank you. –  nina li Jan 22 '13 at 14:47

If $\theta\lt2$ and $N=6$, then $\dfrac{\theta^2}{(N+2)N}\lt\dfrac{2^2}{(6+2)6}\lt\dfrac1{10}$.