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

  • $\begingroup$ 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 Aug 6 '12 at 1:05
  • 8
    $\begingroup$ @MichaelChernick: Reference for your claims? (Both are false.) The sample mean is not even a function of the sufficient statistic! $\endgroup$ – cardinal Aug 6 '12 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 Aug 6 '12 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 Aug 6 '12 at 15:56
  • $\begingroup$ Never mind. I see it in Henry's answer. $\endgroup$ – Michael R. Chernick Aug 6 '12 at 16:02

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.

  • 1
    $\begingroup$ (+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 Aug 6 '12 at 11:37
  • 1
    $\begingroup$ 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 Aug 6 '12 at 15:59
  • 1
    $\begingroup$ @Michael: Perhaps almost always would be a better way of stating it. It is effectively the maximum likelihood estimate. $\endgroup$ – Henry Aug 6 '12 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 Aug 6 '12 at 18:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.