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Let $X_1, X_2,\cdots,X_n$ be a random sample from the distribution with the p.d.f $$f(x)=\frac{1}{\beta -\alpha},\alpha<x<\beta $$ where $0<\alpha<\beta<\infty$. Obtain the minimum variance unbiased estimators of $\frac{\alpha+\beta}{2}$ and $\beta-\alpha$.

Here I try to use Rao Blackwell Method but I am not able to solve with that. Please help

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  • $\begingroup$ The sample mean is minimum variance unbiased estimate for (α+β)/2. I don't think there is a UMVU estimator for β-α. $\endgroup$
    – Michael R. Chernick
    Commented Aug 6, 2012 at 1:05
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    $\begingroup$ @MichaelChernick: Reference for your claims? (Both are false.) The sample mean is not even a function of the sufficient statistic! $\endgroup$
    – cardinal
    Commented Aug 6, 2012 at 1:49
  • $\begingroup$ For what it is worth, if $n=3$ then taking the sample mean has a variance of $(\beta-\alpha)^2/36$ while taking the mean of the largest and smallest value has a slightly smaller variance of $(\beta-\alpha)^2/40$. $\endgroup$
    – Henry
    Commented Aug 6, 2012 at 12:51
  • $\begingroup$ @cardinal What is an unbiased estimator for β-α? The sample mean is unbiased for (α+β)/2 I did not know that it did not have minimum variance. Also I didn't think about sufficient statistic for the mean of uniform. Thanks for the correction. $\endgroup$
    – Michael R. Chernick
    Commented Aug 6, 2012 at 15:56
  • $\begingroup$ Never mind. I see it in Henry's answer. $\endgroup$
    – Michael R. Chernick
    Commented Aug 6, 2012 at 16:02

1 Answer 1

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The sufficient statistic is $(\min X_i, \max X_i)$ so you might expect these minimum variance unbiased estimators to be something related to $$\frac{\max X_i + \min X_i}{2}$$ and $$\max X_i - \min X_i$$ respectively.

The first of these turns out to be the minimum variance unbiased estimator for $\frac{\alpha+\beta}{2}$ while the second is a biased estimator for $\beta-\alpha$ as it is usually too small: you can calculate its expectation to be $(\beta-\alpha)\frac{n-1}{n+1}$, and so multiply it by $\frac{n+1}{n-1}$ to get an unbiased estimator which turns out to be the minimum variance unbiased estimator.

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    $\begingroup$ (+1) And, you've left a little bit for the OP to do as well! :-) (Sufficiency, ironically enough, does not suffice for the conclusion.) $\endgroup$
    – cardinal
    Commented Aug 6, 2012 at 11:37
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    $\begingroup$ It is not usually too small, it is always too small. Max X$_i$ < β with probability 1 and Min X$_i$>α with probability 1. $\endgroup$
    – Michael R. Chernick
    Commented Aug 6, 2012 at 15:59
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    $\begingroup$ @Michael: Perhaps almost always would be a better way of stating it. It is effectively the maximum likelihood estimate. $\endgroup$
    – Henry
    Commented Aug 6, 2012 at 16:41
  • $\begingroup$ @Henry I think you have a UMVUE for β-α which is also mle but it inflates Max X$_i$ -Min X$_i$ by the factor (n+1)/(n-1) to make it unbiased. β is an upper limit on the uniform and Max X$_i$ increasingly approaches it from below as n increases but it never equals it. Similarly Min X$_i$ decreases to α but never equals it. $\endgroup$
    – Michael R. Chernick
    Commented Aug 6, 2012 at 18:42

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