# Displaced Poisson distribution

(Redirected from Displaced Poisson)
${\displaystyle P(X=n)={\begin{cases}e^{-\lambda }{\dfrac {\lambda ^{n+r}}{\left(n+r\right)!}}\cdot {\dfrac {1}{I\left(r,\lambda \right)}},\quad n=0,1,2,\ldots &{\text{if }}r\geq 0\\[10pt]e^{-\lambda }{\dfrac {\lambda ^{n+r}}{\left(n+r\right)!}}\cdot {\dfrac {1}{I\left(r+s,\lambda \right)}},\quad n=s,s+1,s+2,\ldots &{\text{otherwise}}\end{cases}}}$
where ${\displaystyle \lambda >0}$ and r is a new parameter; the Poisson distribution is recovered at r = 0. Here ${\displaystyle I\left(\cdot ,\cdot \right)}$ is the incomplete gamma function and s is the integral part of r. The motivation given by Staff[1] is that the ratio of successive probabilities in the Poisson distribution (that is ${\displaystyle P(X=n)/P(X=n-1)}$) is given by ${\displaystyle \lambda /n}$ for ${\displaystyle n>0}$ and the displaced Poisson generalizes this ratio to ${\displaystyle \lambda /\left(n+r\right)}$.