I have the following problem:
The prior distribution for $\theta$ is distributed $\pi(\theta) = \frac{aP^a}{\theta^{a+1}}$, $\theta >P$
The likelihood for X is uniformly distributed, i.e. $f(x|\theta)=\frac{1}{\theta}$, $0 \leq x \leq \theta$
How do I determine the marginal distribution of Y, and how do I show whether the Pareto the natural conjugate for $\theta$?
I am stuck on calculating the marginal with integral $\int_P^\infty\frac{aP^a}{\theta^{a+2}}d\theta$
Best Answer
$$\int_P^\infty\frac{aP^a}{\theta^{a+2}}d\theta = \left. -\frac{1}{a+1}\frac{aP^a}{\theta^{a+1}} \right|_P^\infty = \frac{a}{(a+1)P}$$
but your posterior distribution and indeed your integral should take different forms depending on whether $x \le P$ or $x \gt P$ as you know $\theta \ge x$.
Added:
If $x \gt P$ then $\Pr(\theta \lt x|X=x)=0$ so your integral should have $x$ as its lower limit rather than $P$, giving $$\int_{\max(x,P)}^\infty\frac{aP^a}{\theta^{a+2}}d\theta = \frac{a P^a}{(a+1)\max\left(P^{a+1},x^{a+1}\right)}$$ and a posterior density of $$\frac{(a+1)\max\left(P^{a+1},x^{a+1}\right)}{\theta^{a+2}}$$ when $\theta \ge \max(x,P)$.