UMVUE for pareto distribution Let $X_1,\ldots,X_n$ random sample with $f(x;\theta,a)=\frac{\theta}{a}(\frac{a}{x})^{(\theta+1)}I_{(a,\infty)}(x),a>0,\theta>0$. Find the UMVUE for $\theta$ when $a$ is fixed.
My attempt
$$f(x;\theta,a)=\frac{\theta}{a}\left(\frac{a}{x}\right)^{(\theta+1)} I_{(a,\infty)}(x) = \frac{\theta}{a}e^{(\theta+1)[\log a-\log x]}I_{(a,\infty)}(x)$$
Then I found that $T(x)=-\sum \log X_i$ is a sufficient and complete statistics. Now that the problems started 
$$y=-\log x\Rightarrow x=e^{-y}\Rightarrow \frac{\partial x}{\partial y}=-e^{-y}$$
Then I applied the transformation of density
$$f(y)=\frac{\theta}{a}e^{(\theta+1)[\log a-\log(e^{-y})]}e^{-y} = \frac{\theta}{a} e^{\theta y+(\theta+1)\log a}$$
But I make no idea how to proceed, perhaps I made a mistake as my statistics.
 A: You have already established that $T(\mathbf{x}) = - \sum_{i=1}^n \ln(x_i)$ is a complete sufficient statistic for the parameter $\theta$.  For reasons made apparent below, it is simpler to work with the equivalent statistic:
$$T^*(\mathbf{x}) \equiv \sum_{i=1}^n \ln(x_i) - n \ln(a).$$
From here, you can use the Lehmann–Scheffé theorem, which says that there is only one unbiased estimator of $\theta$ that is a function of $T^*$ (or equivalently, $T$), and this is the is UMVUE of $\theta$.  So all you need to do now is find some function of $T^*$ that is an unbiased estimator of $\theta$.
To do this, we will first find the distribution of the statistic $T^*$.  We let $T_i^*(\mathbf{X}) \equiv \ln (X_i) - \ln(a)$ and find its distribution function as follows:
$$\begin{equation} \begin{aligned}
F_{T_i^*}(t) = \mathbb{P}(T_i^* \leqslant t) 
&= \mathbb{P}( \ln (X_i) - \ln(a) \leqslant t) \\[6pt]
&= \mathbb{P}( \ln (X_i) \leqslant t + \ln(a)) \\[6pt]
&= \mathbb{P}( X_i \leqslant a \exp(t)) \\[6pt]
&= \begin{cases} 
1- \exp(-\theta t) & & \text{if } t \geqslant 0, \\[6pt]
0 & & \text{if } t < 0. \\[6pt]
\end{cases} \\[6pt]
\end{aligned} \end{equation}$$
From this result we see that $T_i^* \sim \text{Exp}(\theta)$ (this simple distributional form is why we have decided to work with $T^*$ instead of $T$) so we have the distribution:
$$T^*(\mathbf{X}) = \sum_{i=1}^n T_i^* \sim \text{Gamma} (n, \theta).$$
From here we can form the estimator:
$$\hat{\theta}(\mathbf{X}) \equiv \frac{n-1}{T^*(\mathbf{X})} = \frac{n-1}{\sum_{i=1}^n \ln(X_i) - n \ln(a)}.$$
Now, it can easily be shown that $\hat{\theta}(\mathbf{X}) \sim (n-1) \cdot \text{Inv-Gamma} (n, \theta)$, and from the moments of the inverse-gamma distribution this means it is an unbiased estimator for $\theta$.  Since we now have an unbiased estimator for $\theta$ that is a function of the complete sufficient statistic, it follows from the Lehmann–Scheffé theorem that this estimator is the UMVUE.
