Poisson-Gamma mixture is Negative Binomially Distributed
Let
\[\begin{align*} X & \sim Poisson(\lambda) \\ \lambda & \sim Gamma\left(r,\frac{p}{1-p}\right) \end{align*}\]We will show that the Poisson-Gamma mixture above follows a Negative Binomial Distribution.
First, recall the gamma function:
\[\Gamma(\alpha) = \int_{0}^{\infty} x^{\alpha-1}e^{-x}dx\]This function, which is a generalisation of the factorial function to the complex numbers, has the property:
\[\Gamma(\alpha)=(\alpha - 1)\Gamma(\alpha - 1)\tag{$\star$}\]This will prove useful in a moment.
Proof
\[\begin{align*} P(X=x) & = \int_{0}^{\infty} \left(\frac{\lambda^{x}}{x!}e^{-\lambda}\right)\left(\frac{1}{\Gamma(r) (\frac{p}{1-p})^r} \lambda^{r- 1} e^{-\frac{\lambda}{\frac{p}{1-p}}}\right)d\lambda\\ & = \frac{1}{\Gamma(r)x!(\frac{p}{1-p})^r}\int_{0}^{\infty} \lambda^{x+r-1}e^{-\frac{\lambda}{p}}d\lambda \\ & = \frac{p^{x+r}}{\Gamma(r)x!(\frac{p}{1-p})^r}\int_{0}^{\infty} u^{x+r-1}e^{-u}du \\ & = \frac{\Gamma(r+x)}{\Gamma(r)x!}p^{x}(1-p)^r \\ & = {x+r-1\choose x}p^{x}(1-p)^r \tag{By $\star$}\\ \end{align*}\]Therefore \(X\sim NegBin(r,p)\), i.e. \(X\) is the random variable of the number of successes before \(r\) number of failures, where \(p\) is the probability of success.