# PMF of Poisson processes [closed]

if i have 2 intervals on the same PP

[15,25] [16,36]

the 2 pmf are still independent ?

• $\textbf{WHAT IS PMF?}$ ${}{}{}{}{}{}{}{}$ – mjw Sep 24 at 21:09
• @mjw : "probablity mass function". Each discrete probability distribution is characterized by one of these. – Michael Hardy Sep 24 at 21:22
• @mjw : And one can define "discrete probability distribution" as one that is fully characterized by a pmf. Many books say it is one for which the set of its possible values is finite or countably infinite, but I have long viewed that as a definition by non-essentials. – Michael Hardy Sep 24 at 21:24
• Overlapping how, exactly? There are two possibilities - either a partial overlap or one interval is a subset of the other. – Math1000 Sep 24 at 23:03
• @Math1000 : Both of the possibilities you mention are included in the case in which $X$ and $Y$ have expectations $\alpha+\beta$ and $\beta+\gamma,$ where $\beta$ is the expected number of arrivals in the overlap. The case of one of them being included in the other is the case in which either $\alpha=0$ or $\gamma=0.$ – Michael Hardy Sep 25 at 16:04

$$(X,Y) = (U+V,V+W)$$ where $$V$$ is the count in the overlap, and $$U,V,W$$ are independent and Poisson-distributed.
\begin{align} & \Pr(X=x\ \&\ Y=y) \\[8pt] = {} & \sum_{z\,=\,0}^{\min\{\,x,\,y\,\}} \Pr(V=z)\Pr(U = x-z)\Pr(W=y-z) \\[8pt] = {} & \sum_{z\,=\,0}^{\min\{\,x,\,y\,\}} \frac{\beta^z e^{-\beta}}{z!} \cdot\frac{\alpha^{x-z} e^{-\alpha}}{(x-z)!} \cdot \frac{\gamma^{y-z} e^{-\gamma}}{(y-z)!} \end{align}
In case $$\gamma=0,$$ this has a closed form: \begin{align} & \Pr(X=x\ \&\ Y=y) \\[8pt] = {} & \Pr(V=y\ \&\ U=x-y) = \frac{\gamma^y e^{-\gamma}}{y!} \cdot \frac{\alpha^{x-y} e^{-\alpha}}{(x-y)!}. \end{align}
(And if $$\beta=0$$ then $$X,Y$$ are independent and the sum has only the one term in which $$z=0.$$)