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Timeline for Find the unique MVUE

Current License: CC BY-SA 4.0

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Jan 14 at 16:29 comment added whuber Your are right, but that error was corrected in my answer, which you haven't addressed.
Jan 14 at 8:56 comment added Juan Hortez I think that the last equality in question formulation $ f(x_1;\theta)f(x_2;\theta)...f(x_n;\theta)=\frac{1}{(3\theta)^n}\prod_{i}^{n} I(-\theta<x_i<2\theta)=\frac{1}{(3\theta)^n}I(max(x_i)<2\theta)\times 1 $ is not true, because we just get rid of one of the condition (this one about min(x_i)). From factorization theorem, we should get (max(x_i), min(x_i)) as sufficient statistic. It doesn't mean that (max(x_i)) and (min(x_i)) are sufficient statistics separately.
Jan 13 at 20:48 comment added whuber You are still just asserting the conclusion without deriving it. In the notation of my answer, the likelihood equals $(3\theta)^{-n}\mathcal{I}(\hat\theta\le\theta).$ That depends on a single statistic $\hat \theta.$
Jan 13 at 19:43 comment added Juan Hortez No, I'm talking about b)
Jan 13 at 19:29 comment added kjetil b halvorsen This might be an answer to another question! Did you post at the wrong place?
Jan 13 at 17:09 comment added Juan Hortez In question formulation above we can conclude that (-X(1),X(n)/2) is minimal sufficient statistic. It is two dimentional, so there are no one-dimensional sufficient statistics
Jan 13 at 16:39 comment added whuber In what way, specifically, is this a "similar" case and why would that imply lack of sufficiency?
Jan 13 at 16:30 review Low quality posts
Jan 17 at 19:27
Jan 13 at 13:36 review Late answers
Jan 13 at 14:05
S Jan 13 at 13:23 review First answers
Jan 13 at 14:05
S Jan 13 at 13:23 history edited Juan Hortez CC BY-SA 4.0
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S Jan 13 at 13:19 review First answers
Jan 13 at 13:21
S Jan 13 at 13:19 history answered Juan Hortez CC BY-SA 4.0