I am trying to understand NB process, and how it can be derived from Poisson process.
Zhou & Carin states that:
"By placing a gamma prior with shape $r$ and scale $\frac{p}{1−p}$ on $λ$ as $m \sim Pois(λ)$, $λ \sim Gamma(r, \frac{p}{1−p})$ and marginalizing out $λ$, an NB distribution $m \sim NB(r,p)$ is obtained, with PMF
$f_{M} (m|r,p) = \frac{Γ(m|r,p)}{m!Γ(r)}(1 - p)^rp^m, m \in Z_{+}$
Now my question is:
Does that mean that to construct a NB process model, I should specify:
$m \sim Pois(λ)$ (1)
or
$f_{M} (m|r,p) = \frac{Γ(m|r,p)}{m!Γ(r)}(1 - p)^rp^m, m \in Z_{+}$ (2)
or
both (1) and (2)?
Thanks in advance.
That part of the article has nothing to do with Poisson process or negative binomial process. What they are doing is deriving the negative binomial distribution from a gamma prior for $\lambda$ like so.
$$\begin{split}f(m,\lambda|r,p)&=f(\lambda|r,p)f(m|\lambda)\\ &=\frac{\left(\frac{p}{1-p}\right)^r}{\Gamma(r)}\lambda^{r-1}e^{-\frac p{1-p}\lambda}\frac{e^{-\lambda}}{\lambda ^m}m!\end{split}$$
Now marginalize out $\lambda$:
$$\begin{split}\int_0^\infty e^{-\frac\lambda{1-p}}\lambda^{m+r-1}d\lambda&=(1-p)^{m+r-1}\int_0^\infty e^{-\frac\lambda{1-p}}\left(\frac\lambda{1-p}\right)^{m+r-1}d\lambda\\ &=(1-p)^{m+r}\int_0^\infty e^{-u}u^{m+r-1}du\\ &=(1-p)^{m+r}\Gamma(m+r)\end{split}$$
so that $$\begin{split}f(m|r,p)&=\frac{\Gamma(m+r)}{\Gamma(r)m!}p^r(1-p)^m\end{split}$$
This is the pdf of a $\text{Negative Binomial}(r, p)$ distribution.
But if you were interested in a negative binomial process, its relation to the poisson process is as follows. Suppose $N_t$ is the number of arrivals in a period of time $t$. We know that $Negative Binomial(r, p)\approx Poisson(\lambda)$ if $r(1-p)=\lambda$, and further that $r$ is large and $p$ is close to $1$. So if a Poisson process has rate $\lambda$, then the negative binomial process that models arrivals in an interval of length $t$ as $Negative Binomial(rt, p)$ will be approximately the same as the poisson process subject to said condition on $r$ and $p$.