# Prove sum of two independent Poisson processes is another Poisson process

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?

• Presumably, when you say "sum of two Poisson processes" you mean the sum of two independent Poisson counting processes, $\ N_1(t)\sim\frac{(\lambda_1t)^ne^{-\lambda_1t}}{n!}\$ and $\ N_2(t)\sim$$\,\frac{(\lambda_2t)^ne^{-\lambda_2t}}{n!}\$, say. If so, and you already know how to show that $\ N_1(t)+N_2(t)\sim\frac{((\lambda_ 1+\lambda_2)t)^ne^{-(\lambda_1+\lambda_2)t}}{n!}\$, then that's all you would need to do. Commented May 10, 2022 at 0:22
• Yes, independent processes, I just updated the question thanks. Yes, that's the part that I know about, but don't I need to say (prove) something about the independent and stacionary increments of the resulting proccess to complete the proof? Commented May 10, 2022 at 6:57
• Yes, you're quite correct. I had misremembered the properties needed for a Poisson process. I don't think the proof of independent increments (using the fact that the two summand processes have that property) should be particularly difficult, if a little tedious. I'll have a go at it and post some hints if I don't run into any serious snags. Commented May 10, 2022 at 13:28

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 $$\lambda_1+\lambda_2$$.