poisson gamma mixture

Question

Let $N$ be Poisson distributed with parameter $10B$, where $B \sim \Gamma(3,1)$ (i.e. $f(b)=\frac{b^2 e^{-b}}{2}$). Find the p.m.f of $N$. How should I manipulate $10B$ in the integration? What is its pdf?

Answer 1

$\newcommand{\E}{\operatorname{E}}$We are given this conditional distribution: $$ \Pr(N=n\mid B) = \frac{(10B)^ne^{-10B}}{n!}. $$ We need to find a marginal distribution: \begin{align} \Pr(N=n) & = \E(\Pr(N=n\mid B)) = \E\left( \frac{(10B)^ne^{-10B}}{n!} \right) \\[8pt] & = \int_0^\infty \frac{(10b)^n e^{-10b}}{n!} \cdot \frac{b^2 e^{-b}}2\,db \\[8pt] & = \frac{10^n}{n!2} \int_0^\infty b^{n+2} e^{-11b} \, db \\[8pt] & = \frac{10^n}{n!2}\cdot\frac{1}{11^{n+3}} \int_0^\infty (11b)^{n+2} e^{-11b} (11\, db) \\[8pt] & = \frac{10^n}{n!2}\cdot\frac{1}{11^{n+3}} \int_0^\infty c^{n+2} e^{-c}\,dc \\[8pt] & = \frac{10^n}{n!2}\cdot\frac{1}{11^{n+3}} \cdot (n+2)! \\[8pt] & = \frac 1 {2\cdot11^3}\left(\frac{10}{11}\right)^n (n+2)(n+1) \\[8pt] & = \binom{n+3-1}{n} \left(\frac{10}{11}\right)^n \left(1-\frac{10}{11}\right)^3. \end{align} So this is a negative binomial distribution: the distribution of the number of failures before third success in i.i.d. Bernoulli trials with probability $10/11$ of failure on each trial.