If $N_1(t) + N_2(t) = N(t)$ is a split Poisson Process, which we can realize as events coming in and going into the first category with probability $p$ and the second with probability $(1-p)$ why is $(N_1(t), N(t))$ jointly INDEPENDENT of $N(s)-N(t)$ for $s \geq t$?
I know that $N_1(t)$ and $N_2(t)$ are independent themselves, and that in general, $N(t)$ is independent of $N(s)-N(t)$ for $s \geq t$, but why is it further that $(N_1(t), N(t))$ jointly INDEPENDENT of $N(s)-N(t)$ for $s \geq t$? Thanks!
We know that ${N_1}(t)$, ${N_2}(t)$ are independent themselves and they come from Poisson Processes with parameters ${\lambda _1} = \lambda p,{\lambda _2} = \lambda (1 - p)$ respectively, just because the $N(t) = {N_1}(t) + {N_2}(t)$ is the split Poisson process. Also, we know that $N(t)$ has independent increments and the memoryless property. So, $N(t)$ is independent of $N(s) - N(t)$ which is $N(s - t)$ for $s \ge t$. While, $N(t) = {N_1}(t) + {N_2}(t)$ we know that ${N_1}(t) + {N_2}(t)$ is independent of $N(s) - N(t)$ for $s \ge t$. Equivalently, we can claim that both ${N_1}(t),{N_2}(t)$ are independent of $N(s) - N(t)$ for $s \ge t$. Finally, we have the result you want that $({N_1}(t),N(t))$ is jointly independent of $N(s) - N(t)$ for $s \ge t$.