0
$\begingroup$

How can I find the UMVUE for $\theta$ in the following Weibull distribution? $$f\left(y|\theta\right) = \frac{2y}{\theta}\, \exp\left(-\frac{y^2}{\theta}\right),\: y>0, \theta>0$$

I want to use the Cramér-Rao thereom or the Rao-Blackwell thereom.

For the first method, I need to find an estimator. By calculating its mean, I get the answer is $\frac{4}{\pi}\, \bar{y}^2$. The difficult part for me is to get its variance.

The other thing I should do is to get the lower bound of estimator's variance. I think this would be fine for me. The second method(RB), I know that the complete sufficient statistic is $\sum_i y_i^2$ and I already get an unbiased estimator in method one. The difficult part is to calculate E(estimator|complete sufficient statistic).

$\endgroup$
4
  • $\begingroup$ How did you get the mean here? It should only be a function of $\theta$. $\endgroup$
    – dsaxton
    Mar 15, 2016 at 2:30
  • $\begingroup$ @dsaxton What I mean is that theta can be expressed as function of y bar. I think I have got some ideas for this problem. Express this distribution as a form of exponential family and I get the complete sufficient statistics(sum of y^2). The work next is to construct an unbiased estimator which is function of the complete sufficient statistics. Using ML, I get theta = 2*(sum of y^2). The only thing I need to prove is that this is unbiased. Still thinking about this... $\endgroup$
    – purod
    Mar 15, 2016 at 2:41
  • $\begingroup$ $\theta$ isn't a function of the data. A good general approach is to take the expectation of the complete sufficient statistic and try different transformations until you find one that's unbiased. $\endgroup$
    – dsaxton
    Mar 15, 2016 at 2:52
  • $\begingroup$ @dsaxton Yes, your way is pretty useful and I have solved it. Thank you! $\endgroup$
    – purod
    Mar 15, 2016 at 3:15

1 Answer 1

1
$\begingroup$

Basically, both ways are correct.

Nonetheless I'd suggest, instead of using the Cramer-Rao Lower Bound (CRLB), to use Rao-Blackwell combined with Lehman-Scheffé theorem.

First of all take your sufficient and complete statistic for the unknown parameter $\theta$, $T=\sum_{i=1}^n y_i^2$. Compute its expected value exploiting the fact that $Z=X^2$ follows an exponential distribution if $X$ has a Weibull distribution and correct it if necessary. In this way you will get an unbiased estimator $U=g(T)$ that, for Lehman-Scheffè theorem is UMVUE for $\theta$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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