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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.


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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. – Michael Hardy Oct 9 '12 at 17:13

Use the tower rule: $$ \mathbb{P}\left(X=n\right) = \mathbb{E}_\lambda\left(\underbrace{\mathbb{P}\left(X=n | \lambda\right)}_{X|\lambda \sim \operatorname{Poi}\left(\lambda\right)}\right) $$ Write out the probability for the Poisson random variable, and then write the definition of the expectation.

The final expectation can also be computed easily: $$ \mathbb{P}\left(X=n\right) = \int_0^\infty \frac{\lambda^n}{n!} \mathrm{e}^{-\lambda} \frac{\beta^\alpha}{\Gamma(\alpha)} \lambda^{\alpha-1} \mathrm{e}^{-\beta \lambda} \mathrm{d} \lambda = \binom{n+\alpha-1}{n} \left(\frac{\beta}{\beta+1} \right)^\alpha \frac{1}{(\beta+1)^n} $$ This shows that $X$ is a negative binomial random variable, $X \sim \operatorname{NB}\left(\alpha, \frac{\beta}{\beta+1}\right)$.

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this question deals with Compound Probability distribution.

I'll give you some hints.

If $f(x|\theta)$ and $\pi(\theta | \phi)$ are the distribution functions (with parameters $\theta$ and $\phi$ respectively and you want to find distribution of $X|\phi$ then use the following formula: $$F(x|\phi)= \int_\Theta f(x|\theta) \pi(\theta|\phi) d\theta$$

These type of problems belong to Bayesian Statics so you can read more if you are interested.

I suppose in your case the answer would be negative binomial as a gamma mixture of poissons gives negative binomial but I am not very sure,so verify it first.

Hope this helps.

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$$ \int_0^\infty \frac{\lambda^n}{n!}e^{-\lambda}\frac{\beta^\alpha}{\Gamma(\alpha)}\lambda^{\alpha-1}e^{-\beta\lambda} \, d\lambda = \frac{\beta^\alpha}{n!\Gamma(\alpha)} \int_0^\infty \lambda^{n+\alpha-1} e^{-\lambda(\beta+1)} \, d\lambda.\tag{1} $$ The second integral is $$ \int_0^\infty \lambda^{\text{something}-1} e^{-\text{something}\cdot\lambda} \, d\lambda. $$ Such an integral evaluates to $$ \frac{\Gamma(\text{something})}{\text{something}^\text{something}}, $$ where, of course, you have to be careful about which "something" is which. Hence, it is $$ \frac{\Gamma(n+\alpha)}{(\beta+1)^{n+\alpha}}. $$ And then of course, you have to multiply it by the fraction in front of the integral in $(1)$. You get $$ \frac{\beta^\alpha}{n!\Gamma(\alpha)}\cdot \frac{\Gamma(n+\alpha)}{(\beta+1)^{n+\alpha}}. $$ Then of course $\Gamma(n+\alpha)/\Gamma(\alpha)$ reduces to $$ (n+\alpha-1)(n+\alpha-2)\cdots(3+\alpha)(2+\alpha)(1+\alpha)\alpha. $$

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