Let $X \sim Gamma(\alpha,1)$ and $Y|X=x \sim Exp(\frac{1}{\theta x}), \alpha >1$ and $\theta >0$ are unknown. Let $\tau=E(Y)$. Suppose that based on the random sample $Y_1,...,Y_n$, we have MLEs, $\hat{\alpha}$ and $\hat{\theta}$. Use these MLEs to develop an asymptotic $1-\alpha$ confidence interval for $\tau$.
my work:
First, I need to find $\tau=E(Y)=E(\frac{1}{\theta x})=\frac{1}{\theta}E(\frac{1}{x})$. We use a transformation of $T=\frac{1}{X}$, where $f_T(t)=\frac{1}{\Gamma(\alpha)t^{\alpha+1}}e^{-1/t},t>0$. However, I am having trouble evaluating $E(T)=\int^\infty_0\frac{1}{\Gamma(\alpha)t^{\alpha}}e^{-1/t}dt$.
Assuming we have $\tau$, we can get the asymptotic $1-\alpha$ CI by using the asymptotic property of MLE. We know that $\hat{\alpha}\sim AN(\alpha,\frac{1}{ni(\alpha)})$ and $\hat{\theta} \sim AN(\theta,\frac{1}{ni(\theta)})$. However, I am failing to see how I can obtain the asymptotic CI for $\tau$.
updated work:
Thanks to Oriol, I get that $\tau=\frac{1}{\theta}\frac{\Gamma(\alpha-1)}{\Gamma(\alpha)}$. I now see that through the invariance property of MLE, $\hat{\tau}=\frac{1}{\hat{\theta}}\frac{\Gamma(\hat{\alpha}-1)}{\Gamma(\hat{\alpha})}$.
We can get our asymptotic $1-\alpha$ confidence interval for $\tau$ with
$\hat{\tau} \pm z_{\alpha/2}\frac{1}{\sqrt{ni(\hat{\tau})}} \implies \hat{\tau} \pm z_{\alpha/2}\sqrt{\hat{Var}(\tau(\hat{\theta},\hat{\alpha})|\theta,\alpha)}$.
To be quite honest, I do not see how to derive either $\sqrt{ni(\hat{\tau})}$ or $\sqrt{\hat{Var}(\tau(\hat{\theta},\hat{\alpha})|\theta,\alpha)}=\sqrt{\hat{V}}$. Up to this point, since regularity conditions hold, I have been using $i(\tau)$ to denote the Fisher information for a single observation and would prefer to see a solution using this form of Fisher information. Regarding the variance term, I know that
$\hat{V} \approx \frac{(\tau'(\hat{\theta},\hat{\alpha}))^2}{-\frac{\partial^2}{\partial \theta \partial \alpha}logL(\hat{\theta},\hat{\alpha}|X)}$,
but I do not know how to derive this term.