Shared-minimum coupling of two Poisson counts

If $A\sim \mathrm{Pois}(\alpha )$ and $B\sim \mathrm{Pois}(\beta )$ are independent, then for $\alpha ,\beta >0$ and $n\in {\mathbb{N}}_0$ their convolution is $\mathbb{P}(A+B=n)=e^{-(\alpha +\beta )}{\sum }_{k=0}^n\frac{{\alpha }^k{\beta }^{n-k}}{k!(n-k)!}=e^{-(\alpha +\beta )}\frac{(\alpha +\beta {)}^n}{n!},$ by the binomial theorem. If either parameter is zero, the corresponding count is identically zero by [F1], and the same conclusion follows. Thus an independent sum has Poisson law with the sum of its parameters.

Apply step 1.1 to $C+R_1$ and $C+R_2$. Their parameters are respectively $h\min ({\lambda }_1,{\lambda }_2)+h({\lambda }_1-{\lambda }_2{)}^{+}=h{\lambda }_1$ and $h\min ({\lambda }_1,{\lambda }_2)+h({\lambda }_2-{\lambda }_1{)}^{+}=h{\lambda }_2$. This proves the marginal laws.

The shared count cancels in $K_1-K_2=R_1-R_2$. Independence and [F2] give its mean as $h({\lambda }_1-{\lambda }_2{)}^{+}-h({\lambda }_2-{\lambda }_1{)}^{+}=h({\lambda }_1-{\lambda }_2)$ and its variance as $h({\lambda }_1-{\lambda }_2{)}^{+}+h({\lambda }_2-{\lambda }_1{)}^{+}=h\mathrm{\mid }{\lambda }_1-{\lambda }_2\mathrm{\mid }$. Also, since $R_1,R_2$ are independent of $C$ and of each other, $\mathrm{Cov}(K_1,K_2)=\mathrm{Var}(C)=h\min ({\lambda }_1,{\lambda }_2)$ by [F2]. This proves all identities.