Find the distribution of $\theta$, bayesian statistics Given the prior distribution of $\theta$:
$$s(\theta)=\theta e^{-\theta}1_{[0,\infty)}(\theta).$$
and
$$f_{x|\theta}(x|\theta)=\frac{\theta a^\theta}{x ^{\theta+1}}1_{[a,\infty)}(x).$$
Find the posterior distribution:
$$p_{\Theta|X}=\frac{p(x,\theta)}{\int p(x,\theta)d\theta}$$
My attempt:
$$p(x,\theta)=\Pi_{i=1}^nf_{x|\theta}(x|\theta)s(\theta)=\frac{\theta^n a^{n\theta}\theta e^{-\theta}}{\Pi_{i=1}^nx_i^{\theta+1}}{}$$
How can I simplify the integral:$\int\frac{\theta^{n+1} a^{n\theta}e^{-\theta}}{\Pi_{i=1}^nx_i^{\theta+1}}d\theta$?
 A: The integrand in question is essentially a gamma likelihood with respect to $\theta$.  However, I would proceed more simply as follows: the posterior distribution of $\theta$ given the sample $\boldsymbol x = (x_1, \ldots, x_n)$ is proportional to the likelihood; $$f(\theta \mid \boldsymbol x) \propto f(\boldsymbol x \mid \theta) f(\theta).$$  Note the likelihood of $\theta$ from a single observation in the sample can be written as $$\mathcal L(\theta \mid x) = \frac{\theta a^\theta}{x^{\theta+1}} \mathbb 1_{[a,\infty)}(x) \propto \theta(a/x)^\theta \mathbb 1_{[a,\infty)}(x).$$  One simply removes a factor of $x$ that does not depend on $\theta$.  Consequently,
$$ f(\boldsymbol x \mid \theta) f(\theta) \propto \theta^n \left(\frac{1}{a} \prod_{i=1}^n x_i \right)^{-\theta} \theta e^{-\theta} \mathbb 1_{[a,\infty)}(x_{(1)}) \mathbb 1_{[0,\infty)}(\theta).$$  Ignoring the indicator functions for the sake of simplicity, we get a likelihood that is proportional to $$\theta^{n+1} c^{-\theta} = \theta^{n+1} e^{-\theta \log c},$$ where $c = \frac{e}{a} \prod_{i=1}^n x_i$.  This is clearly proportional to a gamma posterior density with shape $n+2$ and rate $$\log c = 1 + n \, \overline{\log x} - \log a.$$  Integration is not necessary because the form of posterior is familiar.
