Prove sum of two independent Poisson processes is another Poisson process

Question

I was trying to prove that the sum of two independent Poisson processes is another Poisson process. I know how to prove that the sum of the Poisson distributions is another Poisson distribution. But I think that is not enough. How can I continue from there?

Answer 1

If ${N_1(t):t\geqslant 0}$ and ${N_2(t):t\geqslant 0}$ are independent Poisson processes with rates $\lambda_1$ and $\lambda_2$, then for any $0\leqslant t_1 < \cdots < t_m$ \begin{align} N_1(t_1), N_1(t_2)-N_2(t_1),\ldots, N_2(t_m)-N_2(t_{m-1})\\ N_2(t_1), N_2(t_2)-N_2(t_1),\ldots, N_2(t_m)-N_2(t_{m-1}) \end{align} are independent, and hence $$ N(t_1), N(t_2)-N(t_1),\ldots, N(t_m)-N(t_{m-1}) $$ are independent. Moreover, for each $s,t\geqslant 0$ \begin{align} N(s+t)-N(s) &= N_1(s+t)+N_2(s+t)-(N_1(s)+N_2(s))\\ &= N_1(s+t)-N_1(s) + N_2(s+t)-N_2(s), \end{align} so as $N_1(s+t)-N_1(s)\sim\mathsf{Pois}(\lambda_1 t)$ and $N_2(s+t)-N_2(s)\sim\mathsf{Pois}(\lambda_2 t)$, it follows that $$ N(s+t)-N(s)\sim\mathsf{Pois}((\lambda_1+\lambda_2)t), $$ and hence the superposition $N(t)=N_1(t)+N_2(t)$ is a Poisson process with rate $\lambda_1+\lambda_2$.