if i have 2 intervals on the same PP

[15,25] [16,36]

the 2 pmf are still independent ?

  • 1
    $\begingroup$ $\textbf{WHAT IS PMF?}$ ${}{}{}{}{}{}{}{}$ $\endgroup$ – mjw Sep 24 '20 at 21:09
  • 1
    $\begingroup$ @mjw : "probablity mass function". Each discrete probability distribution is characterized by one of these. $\endgroup$ – Michael Hardy Sep 24 '20 at 21:22
  • $\begingroup$ @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. $\endgroup$ – Michael Hardy Sep 24 '20 at 21:24
  • $\begingroup$ Overlapping how, exactly? There are two possibilities - either a partial overlap or one interval is a subset of the other. $\endgroup$ – Math1000 Sep 24 '20 at 23:03
  • $\begingroup$ @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 $\gamma=0.$ $\endgroup$ – Michael Hardy Sep 25 '20 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 $z=0.$)


Not the answer you're looking for? Browse other questions tagged or ask your own question.