# Calculating a marginal distribution for the joint density distribution of an exponential distribution with a rate given by a Gamma distribution

This is a follow-up question from one I asked over at MathOverflow: https://mathoverflow.net/questions/158806/is-there-a-simple-closed-form-solution-for-the-joint-density-distribution-of-an

The previous question was: I have an exponential distribution with rate $\lambda$, where $\lambda$ is drawn from a Gamma distribution with shape and scale parameters $(k,\theta)$. I'd like to calculate an exact PDF for values, $v_i$, drawn from the exponential distribution if, for each sampling event, we randomly sample a value of $\lambda$ from the aforementioned Gamma distribution. Is there a simple closed-form solution for the PDF of the $v_i$?

My question here is: Is it possible to calculate a marginal distribution for the PDF of the $v_i$?

• If you have similar questions, you might like to know that this is called a compound distribution. – Neil G Feb 27 '14 at 7:22
• (Added a tag for compound-distribution. There are about 50 questions that might benefit from that tag if it makes sense to keep it.) – Neil G Feb 27 '14 at 7:27

$$f(t)= \frac1{\theta^k\Gamma(k)}\int_0^\infty s e^{-st}\cdot s^{k-1}\cdot e^{-s/\theta}\cdot ds$$ The integral being $$\int_0^\infty s^k e^{-s(t+1/\theta)}ds$$ Now $t+1/\theta = 1/\psi$ where $\psi=\frac{\theta}{\theta t+1}$, so we get $\psi^{k+1}\Gamma(k+1)$ and hence $$f(t)=\frac{\psi^{k+1}\Gamma(k+1)}{\theta^k\Gamma(k)} = \frac{k\theta}{(\theta t+1)^{(k+1)}}$$ I forget what this distribution is called...