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From http://math.stackexchange.com/a/1605995/321264

I have understood that any θ within the range defined by first and last order statistics can be MLE. Of these, how can we choose the one which is unbiased?

SrikRaja
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  • Take MLE as the average of the end points $X_{(n)}-1,X_{(1)}$ – StubbornAtom Jan 30 '21 at 14:52
  • But $$ E(1/2)(X_{(n)} - 1 + X_{(1)}) = 0 $$. How to show that this is equal to θ? May be I'm missing something here. Little help is very much appreciated – SrikRaja Jan 30 '21 at 17:20
  • @StubbornAtom, For Uniform disribution and expectation of order statistics, $$ E(X_{(n)}) = n/(n+1) $$ and $$ E(X_{(1)}) = 1/(n+1) $$. So,$$ E(X_{(n)} + X_{(1)} - 1) = 0 $$
    Did I do anything wrong?     – SrikRaja Jan 31 '21 at 10:21
  • This would have been correct if the common distribution was $\text{Uniform}(0,1)$. Here $X_i\sim \text{Uniform}(\theta,\theta+1)$, so that $X_i-\theta\sim \text{Uniform}(0,1)$. This means $\operatorname E\left[X_{(1)}\right]=\frac1{n+1}+\theta$ and $\operatorname E\left[X_{(n)}\right]=\frac n{n+1}+\theta$. – StubbornAtom Jan 31 '21 at 10:51
  • @StubbornAtom, thank you for correcting my understanding! – SrikRaja Jan 31 '21 at 12:04

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