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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?

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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}

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  • $\begingroup$ I don't need to show that $N_1(0)+N_2(0)=0$? $\endgroup$ – Roland Jul 21 '15 at 23:56
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    $\begingroup$ $N_1(0)=N_2(0)=0$ by definition of Poisson process, so... $\endgroup$ – Math1000 Jul 22 '15 at 0:00
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For sake of compact notation, we let $\Delta N_i \mathop{:=} N_i(t+\Delta t)-N_i(t)$, denote the change in event count over a period $\Delta t$, and make note of the memoryless property of Poisson distribution's waiting times.   Meaning, $\mathsf P(\Delta N_i=n)$ depends only on $\Delta t, \lambda_i,$ and $n$; it is irrespective of $t$.   Indeed: $$\mathsf P(\Delta N_i=n) =\dfrac{ \mathsf e^{-\lambda_i \Delta t}(\lambda_i \Delta t)^n}{n!}$$

Now, investigate $\;\mathsf P(\Delta N_1 +\Delta N_2 = n)$ and see if it has the same form.

$$\begin{align} \mathsf P(\Delta N_1 +\Delta N_2 = n) \; & = \; \sum_{x=0}^n \mathsf P(\Delta N_1 =x , \Delta N_2 =n-x) \\ & \mathop{=}^\text{ind}\; \sum_{x=0}^n \mathsf P(\Delta N_1 =x )\,\mathsf P(\Delta N_2 =n-x) \\ & = \ldots \\ & \vdots \end{align}$$

Can you complete?

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  • $\begingroup$ Being quite honest, I do not quite understand your idea. $\Delta$ is time variation? $\endgroup$ – Roland Jul 21 '15 at 23:27
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    $\begingroup$ Poisson distributions are not memoryless; rather, the distributions of the waiting times in in Poisson processes are memoryless. $\endgroup$ – Michael Hardy Jul 21 '15 at 23:39
  • $\begingroup$ @JohnSheridan Yes, the change in value over period $\Delta t$. In particular: $$\mathsf P(\Delta N_1 + \Delta N_2= n) \quad = \quad \mathsf P(N_1(\Delta t) + N_2(\Delta t) = n \mid N_1(0)=0,N_2(0)=0)$$ ... due to memorylessness of waiting times, giving us a much more compact notation. $\endgroup$ – Graham Kemp Jul 21 '15 at 23:42
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First, $N_1(0) + N_2(0) = 0 + 0 = 0$.

Second, we need to show $\{N_1(t) + N_2(t)\}$ has independent increments, namely, for $0 < t_1 < t_2 < \cdots < t_k$, $N_1(t_1) + N_2(t_1), (N_1(t_2) + N_2(t_2)) - (N_1(t_1) + N_2(t_1)), \ldots, (N_1(t_k) + N_2(t_k)) - (N_1(t_{k - 1}) + N_2(t_{k - 1}))$ are independent, which follows from that $\{N_1(t)\}$ and $\{N_2(t)\}$ are independent and each of $\{N_i(t)\}$ has independent increments.

Third, it needs to be shown $\{N_1(t) + N_2(t)\}$ has stationary increments, namely, for $0 \leq s < t$ and every $n \geq 0$, $$P((N_1(t) + N_2(t)) - (N_1(s) + N_2(s)) = n) = P(N_1(t - s) + N_2(t - s) = n). $$ Indeed, by the stationary increments property of $N_i(t)$ and the law of total probability, it follows that

\begin{align*} & P((N_1(t) + N_2(t)) - (N_1(s) + N_2(s)) = n) \\ = & P(N_1(t) - N_1(s) + N_2(t) - N_2(s) = n) \\ = & \sum_{k = 0}^n P(N_1(t) - N_1(s) + N_2(t) - N_2(s) = n|N_2(t) - N_2(s) = k) P(N_2(t) - N_2(s) = k) \\ = & \sum_{k = 0}^n P(N_1(t) - N_1(s) = n - k|N_2(t) - N_2(s) = k) P(N_2(t) - N_2(s) = k) \\ = & \sum_{k = 0}^n P(N_1(t) - N_1(s) = n - k) P(N_2(t) - N_2(s) = k) \text{ by independence of } N_1(t) \text{ and } N_2(t)\\ = & \sum_{k = 0}^n P(N_1(t - s) = n - k)P(N_2(t - s) = k) \\ = & P(N_1(t - s) + N_2(t - s) = n). \end{align*}

Finally, by the law of total probability, for every $n \in \{0, 1, \ldots\}$, \begin{align*} & P(N_1(t) + N_2(t) = n) = \sum_{k = 0}^n P(N_1(t) + N_2(t) = n, N_1(t) = k) \\ = & \sum_{k = 0}^n P\left(N_1(t) + N_2(t) = n\big | N_1(t) = k\right) P(N_1(t) = k) \\ = & \sum_{k = 0}^nP(N_2(t) = n - k|N_1(t) = k)P(N_1(t) = k) \\ = & \sum_{k = 0}^nP(N_2(t) = n - k) P(N_1(t) = k) \quad \text{by independence of } N_1(t) \text{ and } N_2(t) \\ = & \sum_{k = 0}^n e^{-{\lambda_2t}}\frac{(\lambda_2t)^{n - k}}{(n - k)!}\times e^{-\lambda_1t}\frac{(\lambda_1t)^k}{k!} \\ = & \frac{e^{-(\lambda_1 + \lambda_2)}}{n!}\sum_{k = 0}^n \binom{n}{k}(\lambda_2t)^{n - k}(\lambda_1t)^k \\ = & e^{-(\lambda_1 + \lambda_2)t}\frac{((\lambda_1 + \lambda_2)t)^n}{n!} \end{align*} Hence $N_1(t) + N_2(t) \sim \text{Poisson}((\lambda_1 + \lambda_2)t))$.

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