I am having some difficulty finding the expected value of the MLE for the $\operatorname{Beta}(\theta,1)$ distribution.
Until now, I have found that the MLE for $\theta$ is: $$\hat{\theta} = -\frac{n}{\sum_{i=1}^n \ln X_i}$$
And I know that the expected value of $\hat{\theta}$ is: $$E[\hat{\theta}]=\frac{n}{n-1}\theta$$
But I really don't know how to calculate this expected value. I need to find the distribution of $\hat{\theta}$ to be able to calculate this expected value? Or exists another approach to this problem.
Furthermore, if I know that $T(X_1,...,X_n)=\sum_{i=1}^n\ln X_i$ is sufficient statistics for $\theta$, how can I find a non-biased estimator which is a function of the sufficient statistics for $\theta$? There is a method for that?
Thank you very much!
Observe that
$$Y=-\log X \sim\exp(\theta)$$
Thus $\Sigma_y Y_i\sim \text{Gamma}(n,\theta)$
Is this enough for you to derive your UMVU Estimator?
Here is how to calculate your expectation using gamma distribution
$$\mathbb{E}[\hat{\theta}]=n\int_0^{\infty}\frac{1}{y}\frac{\theta^n}{\Gamma(n)}y^{n-1}e^{-n\theta}dy=n\theta\frac{\Gamma(n-1)}{\Gamma(n)}\underbrace{\int_0^{\infty}\frac{\theta^{n-1}}{\Gamma(n-1)}y^{(n-1)-1}e^{-n\theta}dy}_{=1}=$$
$$=\frac{n}{n-1}\theta$$