{title2=$R\sim\operatorname{Poisson}(rm),\quad G\sim\operatorname{Poisson}((1-r)m)$}

For a finite-intensity set of mean $m$, independent red marking with <probability> $r$ gives counts $R,G$ with
$$
\mathbb P(R=a,G=b)=e^{-m}\frac{m^{a+b}}{(a+b)!}\binom{a+b}{a}r^a(1-r)^b=\left[e^{-rm}\frac{(rm)^a}{a!}\right]\left[e^{-(1-r)m}\frac{((1-r)m)^b}{b!}\right].
$$
Disjoint source sets give independent pairs of these counts. Refining overlapping test sets into disjoint cells then proves independence of the two whole <Poisson point processes>, not only their counts on one set.

Joint red and green Poisson count formula

Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 39 / 1 / Solution

Joint red and green Poisson count formula ( $R \sim Poisson (r m), G \sim Poisson ((1 - r) m)$ )

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Joint red and green Poisson count formula (R∼Poisson(rm),G∼Poisson((1−r)m))

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Joint red and green Poisson count formula ( $R \sim Poisson (r m), G \sim Poisson ((1 - r) m)$ )

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