Stochastic Processes – Problem 35: Prove: (1) is a Poisson process with rate

Let $N_{1}(t),N_{2}(t)$ be independent Poisson processes with rates $\lambda_{1},\lambda_{2}$ respectively. Prove: (1) $N_{1}(t)+N_{2}(t)$ is a Poisson process with rate $\lambda_{1}+\lambda_{2}$. (2) For $N_{1}(t)+N_{2}(t)$, the probability that the first event comes from $N_{1}(t)$ is $\frac{\lambda_{1}}{\lambda_{1}+\lambda_{2}}$, and prove that this event is independent of the time of occurrence.

Step 1. Define the superposed process $$N(t):=N_1(t)+N_2(t),\qquad t\ge 0,$$ where $N_1,N_2$ are independent Poisson processes with rates $\lambda_1,\lambda_2$. Clearly $N(0)=0$.

Step 2. Independent increments: for disjoint intervals, increments of $N_1$ are independent, increments of $N_2$ are independent, and the two processes are mutually independent. Therefore the corresponding increments of $N$ are independent.

Step 3. Stationary increment law: for $h>0$, $$N(t+h)-N(t)\stackrel{d}=\bigl(N_1(t+h)-N_1(t)\bigr)+\bigl(N_2(t+h)-N_2(t)\bigr),$$ which is a sum of independent Poisson variables with means $\lambda_1 h$ and $\lambda_2 h$. Hence $$N(t+h)-N(t)\sim\operatorname{Poisson}\bigl((\lambda_1+\lambda_2)h\bigr).$$ Thus $N$ is a Poisson process of rate $\lambda_1+\lambda_2$.

Step 4. Let $$T_1:=\inf\{t>0:N_1(t)=1\},\quad T_2:=\inf\{t>0:N_2(t)=1\},\quad T:=\min(T_1,T_2).$$ Then $T_1\sim\operatorname{Exp}(\lambda_1)$, $T_2\sim\operatorname{Exp}(\lambda_2)$, independent. Therefore $$P(T>t)=P(T_1>t,T_2>t)=e^{-\lambda_1 t}e^{-\lambda_2 t}=e^{-(\lambda_1+\lambda_2)t},$$ so $T\sim\operatorname{Exp}(\lambda_1+\lambda_2)$.

Step 5. Probability that the first jump comes from process 1: $$P(T_1<T_2)=\int_0^\infty P(T_2>t)\,\lambda_1 e^{-\lambda_1 t}\,dt =\int_0^\infty \lambda_1 e^{-(\lambda_1+\lambda_2)t}dt =\frac{\lambda_1}{\lambda_1+\lambda_2}.$$

Step 6. Independence between label and first-jump time. For any $t\ge0$, $$P(T_1<T_2,\,T\le t) =\int_0^t \lambda_1 e^{-(\lambda_1+\lambda_2)s}ds =\frac{\lambda_1}{\lambda_1+\lambda_2}\bigl(1-e^{-(\lambda_1+\lambda_2)t}\bigr).$$ Since $$P(T_1<T_2)=\frac{\lambda_1}{\lambda_1+\lambda_2},\qquad P(T\le t)=1-e^{-(\lambda_1+\lambda_2)t},$$ we have $$P(T_1<T_2,\,T\le t)=P(T_1<T_2)P(T\le t).$$ Hence the indicator $\mathbf 1_{\{T_1<T_2\}}$ is independent of $T$.

Therefore: 1) $N_1+N_2$ is Poisson with rate $\lambda_1+\lambda_2$; 2) the first jump comes from $N_1$ with probability $\lambda_1/(\lambda_1+\lambda_2)$, and this event is independent of the first-jump time.

QED.