For some discrete random variables, is possible calculate probabilities using a recursive formula. Bionomial and Poisson random variables have these property. Lets check if the same apply for the Generalized Poisson (GP) distribution.

$$ \begin{align} \Pr(X = x) &= \Pr(X = x - 1) \times h(x)\ h(x) &= \frac{\Pr(X = x)}{\Pr(X = x - 1)}\ &= \frac{ \left(\frac{\lambda}{1+\alpha\lambda}\right)^{y} (1 + \alpha y)^{y - 1} \exp\left{-\lambda\frac{1+\alpha y}{1+\alpha\lambda}\right} (y!)^{-1}}{ \left(\frac{\lambda}{1+\alpha\lambda}\right)^{y - 1} (1 + \alpha (y - 1))^{y - 2} \exp\left{-\lambda\frac{1+\alpha (y - 1)}{1+\alpha\lambda}\right} ((y - 1)!)^{-1}}\ &= \frac{\left(\frac{\lambda}{1+\alpha\lambda}\right)^{y}}{ \left(\frac{\lambda}{1+\alpha\lambda}\right)^{y-1}} \times \frac{(1 + \alpha y)^{y - 1}}{ (1 + \alpha (y - 1))^{y - 2}} \times \frac{\exp\left{-\lambda\frac{1+\alpha y}{ 1+\alpha\lambda}\right}}{ \exp\left{-\lambda\frac{1+\alpha (y - 1)}{ 1+\alpha\lambda}\right}} \times \frac{(y!)^{-1}}{((y - 1)!)^{-1}}\ &= \left(\frac{\lambda}{1+\alpha\lambda}\right) \times (1 + \alpha y)^{y - 1}(1 + \alpha (y - 1))^{2 - y} \times \exp\left{-\frac{\alpha\lambda}{1+\alpha\lambda}\right} \times y^{-1} \end{align} $$

These property is valid for GP distribution. Note that is recursive formula is only valid for values of $x > 1$. For $\Pr(X=0)$ and $\Pr(X=1)$ should be used the probability function.

$$ \begin{align} \Pr(X = 0) &= \left(\frac{\lambda}{1+\alpha\lambda}\right)^{0} (1 + \alpha 0)^{0 - 1} \exp\left{-\lambda\frac{1+\alpha 0}{1+\alpha\lambda}\right} = \exp\left{-\frac{\lambda}{1+\alpha\lambda}\right} \ \Pr(X = 1) &= \left(\frac{\lambda}{1+\alpha\lambda}\right)^{1} (1 + \alpha 1)^{1 - 1} \exp\left{-\lambda\frac{1+\alpha 1}{1+\alpha\lambda}\right} = \left(\frac{\lambda}{1+\alpha\lambda}\right) \exp\left{-\lambda\frac{1+\alpha}{1+\alpha\lambda}\right} \end{align} $$



JrEduardo/gammacount documentation built on May 8, 2019, 4:41 p.m.