Let $X_{1}, . . . , X_{n} > 0$ be a random sample from $U(0, \theta)$. Suppose $\theta$ has the prior $\pi(θ) = e^{-\theta} ; \theta > 0$. Find the Bayes estimate of $\frac{1}{\theta}$ with respect to squared error loss function.
Here is my approach:
Let $D = \{X_{1}, . . . , X_{n}\}$
First, we find the expression for the likelihood $p(D|\theta)$
Now $X_{(n)}$ cannot be greater than $\theta$. So, likelihood is $0$ for $\theta < X_{(n)}$
\begin{equation} p(D|\theta) = \begin{cases} 0, & \theta < X_{(n)} \\ 1/\theta^{n}, & \text{otherwise} \end{cases} \end{equation}
The posterior is proportional to prior times likelihood
\begin{equation} p(\theta | D) \propto \begin{cases} 0, & \theta < X_{(n)} \\ \theta^{-n} e^{-\theta}, & \text{otherwise} \end{cases} \end{equation}
Let $\frac{1}{\theta} = \beta$ \begin{equation} p(\beta | D) \propto \begin{cases} 0, & \beta > X_{(n)} \\ \theta^{-n} e^{-\theta}, & 0 < \beta \leq X_{(n)} \end{cases} \end{equation} Bayes estimate of $\beta$, \begin{equation} \hat{\beta} = \mathbb{E}_{p(\beta| D)}[\beta] = \int_{0}^{X_{(n)}} \beta p(\beta|D) d\beta = \int_{0}^{X_{(n)}} \theta^{-(n + 1)}e^{-\theta} d\beta = \int_{X_{(n)}}^{\infty} \theta^{-(n + 3)}e^{-\theta} d\theta \end{equation}
First, I don't know if what I have done till now is correct or not. Second, I have no idea how to solve this integral. Third, is it correct to just find the Bayes estimate of $\theta$ and invert it to get Bayes estimate of $1/\theta$?
First you have to normalize you posterior which results
$$\pi(\theta|\mathbf{x})=\frac{1}{\Gamma[1-n,x_{(n)}]}\theta^{-n}e^{-\theta}\cdot\mathbb{1}_{(x_{(n)};+\infty)}(\theta)$$
Then, given that the loss function is quadratic, Bayes estimation is the posterior expectation, that is
$$\mathbb{E}\left[\frac{1}{\theta};\mathbf{x}\right]=\frac{1}{\Gamma[1-n,x_{(n)}]}\int_{x_{(n)}}^{+\infty}\theta^{-(n+1)}e^{-\theta} d\theta=\frac{\Gamma[-n,x_{(n)}]}{\Gamma[1-n,x_{(n)}]}$$