# UMVUE for $\theta$ in Weibull distribution

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).

• The way you can find it is by studying hard, then doing your own homework. – Mark L. Stone Mar 14 '16 at 23:04
• @MarkL.Stone Thank you for your reply! I can calculate its estimator while have no idea to find a complete sufficient statistics for it. – purod Mar 14 '16 at 23:10
• How did you get the mean here? It should only be a function of $\theta$. – dsaxton Mar 15 '16 at 2:30
• @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... – purod Mar 15 '16 at 2:41
• $\theta$ 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. – dsaxton Mar 15 '16 at 2:52

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 $\theta$.