Let $X$ be a $\Gamma(1, \theta)$ random variable. Derive the variance bound for estimating $\theta$:
My attempt:
We seek a random variable $S$ s.t. $1=E_\theta(WS)$ for all estimators $W$ with $E_\theta(W)=\theta$.
So $$ \begin{split} 1 &= \frac{d}{d\theta}E(W) = \frac{d}{d\theta}\int_{\mathcal{R}^+}\frac{w}{\theta}e^{-t/\theta}dt = \int_{\mathcal{R}^+}\frac{d}{d\theta}\frac{w}{\theta}e^{-t/\theta}dt \\ &= \int_{\mathcal{R}^+} \left[w\left(\frac{t}{\theta^2}-\theta^{-1} \right) \right] \theta^{-1}e^{-t/\theta}dt \\ &= E_\theta\left[W\left(\frac{X}{\theta^2}-\theta^{-1}\right)\right] \end{split} $$
So $S=\frac{X}{\theta^2}-\theta^{-1}$ and the Variance bound is given by $Var_\theta S=\theta^{-4}Var_\theta X=\theta^{-2}$?
By Cramér–Rao_bound in exponential family, for all unbiased estimator, $\hat{\theta}$ ,
$Var(\hat{\theta})\geq \frac{1}{\theta^2}$
Since in the exponential distribution ($f(x)=\frac{1}{\theta} e^{-\frac{x}{\theta}}$) $$I(\theta)=\theta^2$$ Exponential_distribution#Fisher_Information
and $Var(\hat{\theta})\geq \frac{\{1+ \frac{d}{d\theta} \left( E(\hat{\theta})-\theta \right)\}^2}{\theta^2}$