2
$\begingroup$

I dont' really know how to realize the following: Let $\lambda \sim \Gamma(\alpha,\beta)$ and let $X$ conditional on $\lambda$ be Poisson$(\lambda)$. Argue that for $n=0,1,2,\ldots,$ $$P(X=n)=\int_0^\infty \frac{\lambda^n}{n!}e^{-\lambda}\frac{\beta^\alpha}{\Gamma(\alpha)}\lambda^{\alpha-1}e^{-\beta\lambda} \, d\lambda.$$ I don't know how to deal with this type of conditional probability. I'd appreciate some help.

Fuente

  • 1
    I think instead of thinking about conditional probability, what you need here is to know how to clear away the clutter and camouflage from the integral to see that it's really just like another integral you know how to do, but with different numbers. See my answer below.2012-10-09

3 Answers 3