1
$\begingroup$

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?

$\endgroup$
3
  • 3
    $\begingroup$ 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. $\endgroup$ Commented May 10, 2022 at 0:22
  • $\begingroup$ 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? $\endgroup$
    – Alfonso_MA
    Commented May 10, 2022 at 6:57
  • $\begingroup$ 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. $\endgroup$ Commented May 10, 2022 at 13:28

1 Answer 1

1
$\begingroup$

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 $\lambda_1+\lambda_2$.

$\endgroup$
2
  • $\begingroup$ Hello @Math1000. In my opinion, OP's question should be handled using characteristic functions for both the distribution and the independence part $\endgroup$
    – Snoop
    Commented Dec 3, 2022 at 11:58
  • $\begingroup$ That would be another approach, but I don't see any error in mine. I answered a question very similar to this in the same manner here: math.stackexchange.com/questions/4146699/… $\endgroup$
    – Math1000
    Commented Dec 8, 2022 at 13:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .