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Let $X_1,...X_n$ $U(-\theta , \theta)$ I want to find the UMVUE of $\theta$ if it is exists.

My answer is , there is no UMVUE in this case.

Because there is no complete sufficient statistic that exists for $\theta$ . So although there exist an unbiased estimator of $\theta$, it is not a function of complete sufficient statistic.

So there is no there exist no UMVUE for $\theta$.

Am I correct in this situation?

kjetil b halvorsen
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student_R123
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  • @StubbornAtom Hi . I think here we need to consider the relationship between UMVUE and unbiased estimator of zero. – student_R123 Jul 19 '18 at 19:58
  • Hi, did you actually verify that a complete sufficient statistic does not exist? By my calculations, $\max_{1\le i\le n}|X_i|$ is a complete sufficient statistic for $\theta$. If so, UMVUE of $\theta$ would naturally exist. – StubbornAtom Jul 24 '18 at 16:23
  • Hi . No i got a different answer. I will post my answer in a later day. – student_R123 Jul 24 '18 at 16:41
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    Using factorisation theorem, one would arrive at $\max(-X_{(1)},X_{(n)})$ as a sufficient statistic, which is the same as $\max |X_i|$. Since $ |X_i|\sim U(0,\theta)$, $\max |X_i|$ is complete which can be shown as a separate problem. – StubbornAtom Jul 24 '18 at 16:51
  • here you can find an unbiased estimator of $\theta $ . then you can find the UMVUE by considering the fact that UMVUE is uncorrelated with all unbiased estimators of zero. – student_R123 Jul 24 '18 at 17:00
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    The UMVUE I am getting is $\frac{n+1}{n}\max |X_i|$. I presume you are getting the same answer. Your original question no longer remains I guess. – StubbornAtom Jul 24 '18 at 17:25
  • what i got was a function of sufficient statistic . so it involved both min X and Max X. For my point of view it is reasonable because then it is a function of a sufficient statistic as well . – student_R123 Jul 24 '18 at 17:33
  • @StubbornAtom Hi , Based on your calculation ,the Minimum sufficient statistic for $\theta$ is $max|X_i| $ isnt it ? – student_R123 Aug 04 '18 at 14:25
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    Yes, it is .... – StubbornAtom Aug 04 '18 at 14:27
  • @StubbornAtom But when we have a case like $ X$ ~ $ uni [\theta , 2\theta] $ or $X$ ~ $ uni [\theta -1/2 , \theta + 1/2 ] $ then the sufficient and minimum sufficient is $ min (X_i) $ and $ max (X_i) $ isnt it ? – student_R123 Aug 04 '18 at 14:35

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