Suppose that $\{N_1(t)\,t \ge 0\}$ and $\{N_1(t)\,t \ge 0\}$ are two independent Poisson processes with rates of $\lambda_1$ and $\lambda_2$ respectively, then prove the sum process where $N(t) = N_1(t)+N_2(t)$ for all $t \ge 0$ is a Poisson process with rate of $\lambda_1+\lambda_2$.

The proof on my textbook includes the step of proving that $N(t)$ has independent increments, of which the reason is that $N_1(t)$ and $N_2(t)$ are independent and both possess stationary and independent increments properties. But I couldn't understand such an obvious step.

Let $I_1$ and $I_2$ be two disjoint intervals, then our goal is to show that $N_1(I_1)+N_2(I_1)$ and $N_1(I_1)+N_2(I_1)$ are independent. Obviously, we have:

  • $N_1(I_1)$, $N_1(I_2)$, $N_2(I_1)$, $N_2(I_1)$ are independent of each other;
  • $(N_1(I_1),N_1(I_2))$ and $(N_2(I_1),N_2(I_2))$ are independent random vectors since $\{N_1(t)\,t \ge 0\}$ and $\{N_1(t)\,t \ge 0\}$ are two independent processes.

Next, I only need to prove that $(N_1(I_1),N_2(I_1))$ and $(N_1(I_2),N_2(I_2))$ are two independent vectors to achieve our goal. But I was stuck here. It seems that I'm complicating the question. Does someone have a better idea? Please help me out here.

  • $\begingroup$ Take a look at youtube.com/watch?v=0J5acxn3y5c $\endgroup$ – John Douma Mar 28 '18 at 14:03
  • $\begingroup$ @JohnDouma Main problem of OP is independence of increments of summation of processes. Not so much the fact that the sum of independent Poissons is again Poisson. $\endgroup$ – drhab Mar 28 '18 at 14:34
  • $\begingroup$ @drhab Yeah, I’ve updated the title. $\endgroup$ – Mengfan Ma Mar 28 '18 at 14:42

$$\begin{aligned} & P\left(N_{1}\left(I_{1}\right)+N_{2}\left(I_{1}\right)=r\wedge N_{1}\left(I_{2}\right)+N_{2}\left(I_{2}\right)=s\right)\\ & =\sum_{r_{1}+r_{2}=r}\sum_{s_{1}+s_{2}=s}P\left(N_{1}\left(I_{1}\right)=r_{1}\wedge N_{2}\left(I_{1}\right)=r_{2}\wedge N_{1}\left(I_{2}\right)=s_{1}\wedge N_{2}\left(I_{2}\right)=s_{2}\right)\\ & =\sum_{r_{1}+r_{2}=r}\sum_{s_{1}+s_{2}=s}P\left(N_{1}\left(I_{1}\right)=r_{1}\right)P\left(N_{2}\left(I_{1}\right)=r_{2}\right)P\left(N_{1}\left(I_{2}\right)=s_{1}\right)P\left(N_{2}\left(I_{2}\right)=s_{2}\right)\\ & =\sum_{r_{1}+r_{2}=r}\left[P\left(N_{1}\left(I_{1}\right)=r_{1}\right)P\left(N_{2}\left(I_{1}\right)=r_{2}\right)\sum_{s_{1}+s_{2}=s}P\left(N_{1}\left(I_{2}\right)=s_{1}\right)P\left(N_{2}\left(I_{2}\right)=s_{2}\right)\right]\\ & =\left[\sum_{s_{1}+s_{2}=s}P\left(N_{1}\left(I_{2}\right)=s_{1}\right)P\left(N_{2}\left(I_{2}\right)=s_{2}\right)\right]\times\left[\sum_{r_{1}+r_{2}=r}P\left(N_{1}\left(I_{1}\right)=r_{1}\right)P\left(N_{2}\left(I_{1}\right)=r_{2}\right)\right]\\ & =P\left(N_{1}\left(I_{2}\right)+N_{2}\left(I_{2}\right)=s\right)\times P\left(N_{1}\left(I_{1}\right)+N_{2}\left(I_{1}\right)=r\right) \end{aligned} $$ showing that $N_{1}\left(I_{1}\right)+N_{2}\left(I_{1}\right)$ and $N_{1}\left(I_{2}\right)+N_{2}\left(I_{2}\right)$ are independent.

  • $\begingroup$ I guess it’s not as obvious as in the textbook? $\endgroup$ – Mengfan Ma Mar 28 '18 at 14:48
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    $\begingroup$ Essential is that the $N_i(I_j)$ are $4$ independent random variables. From that then you can immediately conclude that $N_1(I_1)+N_2(I_1)$ and $N_1(I_2)+N_2(I_2)$ are also independent. In my answer that conclusion is justified. $\endgroup$ – drhab Mar 28 '18 at 14:55
  • $\begingroup$ Can we extend that to any $n$ independent processes? $\endgroup$ – Mengfan Ma Mar 28 '18 at 14:57
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    $\begingroup$ Sure, I don't see why not. You can use induction on $n$ for a proof. $\endgroup$ – drhab Mar 28 '18 at 14:58
  • $\begingroup$ I was still wondering if $(N_1(I_1),N_2(I_1))$ and $(N_1(I_2),N_2(I_2))$ are two independent vectors. Could you shed more light on it? :) $\endgroup$ – Mengfan Ma Mar 30 '18 at 1:25

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