# 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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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 $\dfrac{\theta^2}{(N+2)N}\lt\dfrac{2^2}{(6+2)6}\lt\dfrac1{10}$.