Suppose that $\{N_1(t),t\geq0\}$ and $\{N_2(t),t\geq0\}$ are independent
Poisson Process with rates $\lambda_1$ and $\lambda_2$. Show that
$\{N_1(t)+N_2(t),t\geq0\}$ is a Poisson process with rate
$\lambda_1+\lambda_2$. Also, show that the probability that the first
event of the combined process comes from $\{N_1(t),t\geq0\}$ is
$\frac{\lambda_1}{\lambda_1+\lambda_2}$, independently of the time of
the event.
If $N_1(t)\sim Poisson(\lambda_1t)$ and $N_2(t)\sim Poisson(\lambda_2t)$ then $$N_1(t)+N_2(t)\sim Poisson((\lambda_1+\lambda_2)t)$$
For the probability I did
$$P(N_1(t)=1\mid N_1(t)+N_2(t)=0)=\frac{P(N_1(t)=1\cap N_1(t)+N_2(t)=1)}{P(N_1(t)+N_2(t)=1)}=\frac{P(N_1(t)=1\cap N_2(t)=0)}{P(N_1(t)+N_2(t)=1)}=\frac{P(N_1(t)=1)\cdot P(N_2(t)=0)}{P(N_1(t)+N_2(t)=1)}=\frac{(e^{-\lambda_1 t}\lambda_1 t)(e^{-\lambda_2 t})}{e^{-t(\lambda_1+\lambda_2)}(\lambda_1+\lambda_2)t}=\frac{\lambda_1}{\lambda_1+\lambda_2}$$
I'm not too sure about the first part, I need to show that $N_1(0)+N_2(0)=0$
and the increments are independent?
Best Answer
You need to show independence of increments, i.e. if $0\le a<b<c<d$ then $(N_1(d)+N_2(d)) - (N_1(c)+N_2(c))$ is independent of $(N_1(b)+N_2(b)) - (N_1(a)+N_2(a))$, and similarly for more than two intervals. You can prove that by using independence of increments of each of the two processes separately plus independence of $N_1$ and $N_2$.
You also need to show $(N_1(b)+N_2(b)) - (N_1(a)+N_2(a))$ has a Poisson distribution with expected value $(\lambda_1+\lambda_2)(b-a)$.
If $X,Y$ are independent with respective distributions $\mathrm{Poisson}(\alpha)$ and $\mathrm{Poisson}(\beta)$ then \begin{align} & \Pr(X+Y= w) = \sum_{v=0}^w \Pr(X=v\ \&\ Y=w-v) = \sum_{v=0}^w \Pr(X=v)\cdot\Pr(Y=w-v) \\[8pt] = {} & \sum_{v=0}^w \frac{e^{-\alpha}\alpha^v}{v!} \cdot \frac{e^{-\beta} \beta^{w-v}}{(w-v)!} = \frac{e^{-(\alpha+\beta)}}{w!} \sum_{v=0}^w \frac{w!}{v!(w-v)!} \alpha^v \beta^{w-v} = \frac{e^{-(\alpha+\beta)}}{w!} (\alpha+\beta)^w. \end{align}