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Suppose that $X_1,\cdots, X_n$ are independent random variables from uniform distribution on interval $(\theta_1,\theta_2)$, $\theta_2>\theta_1>0$. It is know that $T(X)=(X_{(1)},X_{(n)})$ is the complete sufficient statistic for $(\theta_1,\theta_2)$.

(a) Find the UMVUE of the range parameter $\tau(\theta)=\theta_2-\theta_1$.

(b) Find the UMVUE of the mean parameter $\tau(\theta)=(\theta_2+\theta_1)/2$.

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  • $\begingroup$ This is phrased quite a lot like a homework problem. Can you tell us specifically what difficulties you've had with it. For part (a) common sense suggests that $X_{(n)}-X_{(1)}$ should be looked at, but obviously its expected value is less than $\theta_2-\theta_1$. But if the amount you need to multiply it by in order to get expectation $\theta_2-\theta_1$ does not depend on $\theta_1$ or $\theta_2$, then you can do something. For part (b), what function of $(X_{(1)},X_{(n)})$ would common sense suggest? ${}\qquad{}$ $\endgroup$ – Michael Hardy Mar 25 '15 at 22:58
  • $\begingroup$ Why $T(X)$ is complete statistic? I can just show it is a sufficient statistic. $\endgroup$ – username Nov 19 '18 at 6:34
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You have $T(X)=(X_{(1)}=\min \mathbf{X},X_{(n)}=\max \mathbf{X})$ which is the sufficient statistic for the uniform distribution.

Then, the first UMVUE for $\theta_2-\theta_1$ is $((n+1)/n)(X_{(n)}-X_{(1)})$ and for the second $((n+1)/n)((X_{(n)}+X_{(1)})/2)$

The idea is as follows: Given two random variable with $X$ and $Y$ if they have minimum variance for the same true mean, the linear combinations of them is also minimum variance.

Why they are unbiased? have a look at this one

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    $\begingroup$ From order statistics I get $\mathrm{E}[X_{(i)}] = \theta_1 + (\theta_2 - \theta_1)\cdot i/(n+1)$ and thus $\mathrm{E}[X_{(n)} - X_{(1)}] = (\theta_2 - \theta_1)(n-1)/(n+1)$ and $\mathrm{E}[X_{(n)} + X_{(1)}] = \theta_2 + \theta_1$. Thus, I think the MVUE are $(n+1)/(n-1)(X_{(n)} - X_{(1)})$ and just $(X_{(n)} + X_{(1)})/2$. Wikipedia also states the MVUE of the mean is just the average of max and min sample. Care to comment? $\endgroup$ – GDumphart Mar 19 '18 at 19:06
  • $\begingroup$ Your final answers are incorrect, as rightly pointed out by the above comment. $\endgroup$ – StubbornAtom May 27 '18 at 22:14
  • $\begingroup$ @GDumphart please feel free to correct it. It seems I wrote it fast. I just took care of the bias correction with respect to the end point estimators as stated here: en.wikipedia.org/wiki/Minimum-variance_unbiased_estimator please see "other examples". $\endgroup$ – Seyhmus Güngören May 27 '18 at 23:58

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