I am trying to solve the following problem.
"Find the asymptotic relative efficiency of the MLE of $a$ with respect to the UMVUE of $a$". Where,
$$X_1, ... , X_n \sim f(x) = \theta a^\theta x^{-(\theta +1)}, \quad x\in (a,\infty)$$ is the pdf and both $a$ and $\theta$ are positive.
So far I have found that
$$\hat{a}_{MLE}=X_{(1)}$$
and that the unbiased estimator based on this statistics is
$$\hat{a}^*=\frac{n\theta - 1}{n\theta}X_{(1)}$$.
I found the Fisher information is
$$nI(a) = \frac{(n\theta)^2}{a^2}$$
and
$$Var[\hat{a}_{MLE}]=a^2 \frac{n\theta}{(n\theta-2)(n\theta-1)^2}\quad , \quad Var[\hat{a}^*]=a^2 \frac{1}{n\theta(n\theta-2)}$$
There are a couple of things I am not sure here, so I would like some help on the following.
1), Is $\hat{a}^*$ a UMVUE? From the problem it seems like I need to find the UMVUE. So, I know that I at least found an unbiased sufficient statistic, so as long as I can show that $X$ is a complete distribution then I am done constructing the UMVUE. But I cannot use the regular exponential class argument for the support depends on the parameter $a$.
The other approach is to use $E[u(X)]=0$ then show $u(X)=0$. A common approach is to take the derivative of
$$\int_{a}^{\infty} u[x]\theta a^\theta x^{-(\theta+1)}dx$$ with respect to $x$ but I am not sure how to proceed, so maybe I can get some help here.
2), If I am understanding the question correctly, then the asymptotic efficiency is the efficiency of a statistic when $n \rightarrow \infty$. Both statistics $\hat{a}_{MLE}$ and $\hat{a}^*$ seem to approach the $RCLB = \frac{1}{nI(a)}$ so they are asymptotically efficient. However, I am not 100% sure if this is really what is being asked since this argument doesn't really depend on he UMVUE.