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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 $v_i$?

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  • $\begingroup$ If you have similar questions, you might like to know that this is called a compound distribution. $\endgroup$ – Neil G Feb 27 '14 at 7:22
  • $\begingroup$ (Added a tag for compound-distribution. There are about 50 questions that might benefit from that tag if it makes sense to keep it.) $\endgroup$ – Neil G Feb 27 '14 at 7:27
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$$ f(t)= \frac1{\theta^k\Gamma(k)}\int_0^\infty s e^{-st}\cdot s^{k-1}\cdot e^{-s/\theta}\cdot ds $$ 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...

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    $\begingroup$ Hey, thanks for following me here and answering my question!! $\endgroup$ – Paul Feb 27 '14 at 6:47
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    $\begingroup$ It's a generalized beta prime distribution. $\endgroup$ – Neil G Feb 27 '14 at 7:20

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