Bayes estimate for loss function $\ell(t,\alpha)=\frac{1}{\alpha^2}(t-\alpha^2)^2$? 
I am given the following info for $\{X_i\}_{i=1}^{n}$:
$$X\sim f(X|\alpha)=\alpha X^{-(\alpha+1)}I(X>1).$$
Propose a convenient family of priors and find the Bayes estimate for the loss function $\ell(t,\alpha)=\frac{1}{\alpha^2}(t-\alpha^2)^2$.

The joint distribution is of the form
$$f(\underline{X}|\alpha)=\alpha^n\prod_{i=1}^{n}X_i^{-(\alpha+1)}I(X_i>1)=\alpha^n e^{-(\alpha+1)\sum_{i=1}^{n}\log(X_i)}I(X_{(1)}>1).$$
Then, $\text{Gamma}(a,b)$ seem like good priors with a posterior $\text{Gamma}(a+n,\sum_{i = 1}^{n}\log(X_i)+b)$.
To find the posterior:
$$P(\alpha|X)\propto P(X|\alpha)P(\alpha)\propto\alpha^n e^{-\alpha\sum_{i=1}^{n}\log(X_i)}e^{-b\alpha}\alpha^{a-1}$$.
Adding the like terms gives us the posterior.
To find the Bayes estimate we minimize with respect to $t$, we consider the following Bayes risk function
$$\int \ell(t,\alpha)f(\alpha|\underline{X})d\alpha=\int \frac{1}{\alpha^2}(t-\alpha^2)^2f(\alpha|\underline{X})d\alpha.$$
Since $\frac{\partial}{\partial t}\frac{1}{\alpha^2}(t-\alpha^2)^2=\frac{2t}{\alpha^2}-2$ and
$\int (\frac{2t}{\alpha^2}-2)f(\alpha|\underline{X})d\alpha<\infty$, we have
$$\frac{\partial}{\partial t}\int \ell(t,\alpha)f(\alpha|\underline{X})d\alpha=2\int \left(\frac{t}{\alpha^2}-1\right)f(\alpha|\underline{X})d\alpha
= 2tE\bigg[\frac{1}{\alpha^2}\bigg|\underline{X}\bigg]-2.$$
Setting this equal to $0$ implies the Bayes estimate is $$\hat{t}=\frac{1}{E\bigg[\frac{1}{\alpha^2}\bigg|\underline{X}\bigg]}.$$
Define $R=\sum_{i = 1}^{n}\log(X_i)$.
Then
\begin{align}
E\bigg[\frac{1}{\alpha^2}\bigg|\underline{X}\bigg]
& =\int \alpha^{-2}\frac{(R+b)^{n+a}}{\Gamma(n+a)}\alpha^{n+a-1}e^{-\alpha(R+b)}d\alpha \\
& = \frac{(R+b)^{n+a}}{\Gamma(n+a)}\frac{\Gamma(n+a-2)}{(R+b)^{n+a-2}}\int \frac{(R+b)^{n+a-2}}{\Gamma(n+a-2)}\alpha^{n+a-2-1}e^{-\alpha(R+b)}d\alpha \\
& =\frac{(R+b)^{n+a}}{\Gamma(n+a)}\frac{\Gamma(n+a-2)}{(R+b)^{n+a-2}} \\
& =\frac{(R+b)^2}{(n+a)(n+a-1)}
\end{align}
Then $\hat{t}=\frac{(n+a)(n+a-1)}{(\sum_{i = 1}^{n}\log(X_i)+b)^2}$.

I am asked to find the asymptotic distribution as well.

We have $\frac{(n+a)(n+a-1)}{(\sum_{i = 1}^{n}\log(X_i)+b)^2}\stackrel{p}\to\alpha^2$ so it is a consistent estimator. I am trying to use the Central Limit Theorem, but need to find the variance, is my work up to now correct? If so, how do I find the asymptotic distribution?
 A: For the conceptual understanding of the task it is important to keep in mind that, given $X_1, \ldots, X_n$, we are trying to estimate $\alpha > 0$.


*

*A mistake you made is that the parameters, $\alpha = a + n$ and $\beta = \sum_{k = 1}^{n} \log(X_k) + b$, you gave for the posterior distribution depend on the sample $(X_1, \ldots, X_n)$, which shouldn't be the case - the parameters should be independent of the input sample.
For the computation of the posterior, see this Math.SE question.

*You also seem to be claiming that
$$
\int f(X, \alpha) \; \text{d}(\alpha)
= 1,
$$
which is not true (if you integrated over $X$, it would be true, because $f$ is a density): consider $n = 1$.
Then (with the convention $0 \cdot \infty = 0$) we have
$$
\int_{0}^{\infty} \alpha X^{- \alpha - 1} 1_{\{ X > 1 \}}(x) \; \text{d}(\alpha)
= 1_{\{ X > 1 \}}(x) \cdot \int_{0}^{\infty} \alpha X^{- \alpha - 1} \; \text{d}(\alpha)
= \frac{1}{x \log^2(x)} 1_{\{ X > 1 \}}(x),
$$
in particular this always depends on $x$.

*Also you seem to be claiming that
$$
\int \frac{2}{\alpha^2} f(\alpha | \underline{X}) \; \text{d}\alpha
= E\left[\frac{1}{\alpha^2} | \underline{X}\right]
= \int \alpha^{-2}\frac{(R+b)^{n+a}}{\Gamma(n+a)}\alpha^{n+a-1}e^{-\alpha(R+b)}\; \text{d}\alpha
$$
and I don't see how the term on the left and the term on the right should be equal, in particular because the term on the right depends on $a$ and $b$, while the term on the left does not.
Lastly, I think those integrals should be definite ones over $(0, \infty)$ and not indefinite ones.

