Let X|α have a single-parameter Pareto distribution with parameters θ and α, where theta is a known and fixed constant and α has an exponential distribution with parameter λ>0. Find the density function of X.
Below is my attempt thus far:
I believe you have to integrate the pdf of a single parameter pareto distribution * the pdf of an exponential distribution:
$$\int^∞_{−∞}\dfrac{(αθ^α)}{(x^{(α+1)})} λe^{(−λx)} d\lambda$$ I rewrote this to equal: $$\int \dfrac{( α θ^α)}{(x^{(α+1)}) } \dfrac{(e^{(-x/θ)})}{θ}$$
and then simplified that to equal $$\int\dfrac{( α θ^{(α-1)})}{x^{(α+1)} } e^{-(1/θx)}$$
Would this be the final answer or can it be simplified? Or am I completely not on the right track? Any guidance would be greatly appreciated!
(Also I apologize in advance if this is a duplicate, I tried to post this before but was not logged in and am unsure if it went through.)
You are off to a rocky start. Do over!
You know $f_{X\mid A}(x\mid \alpha) =\dfrac{\alpha\theta^\alpha}{x^{\alpha-1}}\mathbf 1_{x\geqslant \theta}$ and $f_A(\alpha)=\lambda e^{-\alpha\lambda}\mathbf 1_{\alpha\geqslant 0}$, so: $$\begin{align}f_X(x) &= \int_\Bbb R f_{X\mid A}(x\mid \alpha)f_A(\alpha)~\mathsf d \alpha \\ &=\mathbf 1_{x\geqslant\theta}\int_0^\infty \dfrac{\alpha\theta^\alpha~\lambda e^{-\alpha\lambda}}{x^{\alpha-1}\qquad~~~}~\mathsf d \alpha \\ &= \dfrac \lambda x\mathbf 1_{x\geqslant\theta}\int_0^\infty \alpha~(\theta/x)^\alpha~e^{-\lambda \alpha}\mathsf d \alpha \\ &~~\vdots\end{align}$$