Let $X_1...X_n \sim \Gamma(\alpha,\beta)$, what unbiased estimator of $\frac{1}{\beta}$ has minimum variance? Given that $\alpha$ is known and $\beta$ is an unknown rate parameter, how do I find the unbiased estimator of $\frac{1}{\beta}$? Does it involve finding $\hat\theta_{\text{MLE}}$ first? 
I honestly don't know where to start.
 A: Assuming $f(x|\beta) = \frac{\beta^\alpha}{\Gamma(\alpha)}x^{(\alpha-1)}e^{-\beta x}$, I proceeded with the MLE method and found the following estimator:
$$\hat{\frac{1}{\beta}} = \frac{1}{n\alpha}\sum_{i=1}^n X_i$$
And found that $$E\left[\hat{\frac{1}{\beta}}\right] = \frac{1}{\beta}$$ and $$Var\left(\hat{\frac{1}{\beta}}\right) = \frac{1}{n\alpha\beta^2}$$
Also, because the distribution is an exponential family when $\alpha$ is fixed, the MLE estimator must attain the lower bound of the Cramer-Rao inequality:
$$Var\left(\hat{\frac{1}{\beta}}\right) \geq \frac{\left(\frac{d}{d\beta}E\left[\hat{\frac{1}{\beta}}\right]\right)^2}{E\left[\left(\frac{d}{d\beta}ln\left(f(X|\beta)\right)\right)^2\right]}$$
For the numerator:
$$\left(\frac{d}{d\beta}E\left[\hat{\frac{1}{\beta}}\right]\right)^2 = \frac{1}{\beta^4}$$
For the denominator:
\begin{align}
E\left[\left(\frac{d}{d\beta}ln\left(f(X|\beta)\right)\right)^2\right] &= E\left[\left(\frac{d}{d\beta}n \alpha ln(\beta) -n ln(\Gamma(a)) + (\alpha-1) \sum_{i=1}^nln(X_i)- \beta \sum_{i=1}^n X_i\right)^2\right] \\
&= E\left[\left(\frac{n\alpha}{\beta} - \sum_{i=1}^n X_i\right)^2\right] \\
&= E\left[\left(\sum_{i=1}^n E[X_i] - \sum_{i=1}^n X_i\right)^2\right] \\
&= \sum_{i=1}^n Var(X_i) \\
&= \frac{n\alpha}{\beta^2} \\
\end{align}
Now, when I plug in and simplify on the right hand side: 
$$ \frac{1}{n\alpha\beta^2} = \frac{1}{n\alpha\beta^2}$$
