# Distribution of arrival times of Poisson point processes

Let $$(M_{t})_{t\geq 0}$$ and $$(N_t)_{t\geq0}$$ be two independent Poisson point processes with rate $$\lambda$$ and $$\mu$$ respectively. Let $$\tau$$ be the first arrival time for the process $$N_{t}$$. Find:

a) the distribution of $$M_{\tau }$$

b) the distribution of $$N_{2\tau} - N_{\tau}$$

I was hoping for some clarification as to how the distribution of $$M_\tau$$ changes based on the arrival time of $$N_t$$, as I just don't understand how it could if the processes are independent?

Thanks in advance

• the distributions are asked so they are related. independency condition is not bad here. – Seyhmus Güngören Apr 22 at 23:01

## 1 Answer

Hints: $$P(M_{\tau}=k)=\sum_j P(\tau =j, M_j=k)$$ and the events $$(\tau =j), (M_j=k)$$ are independent. Can you compute this now?

For the second part $$P(N_{2\tau}-N_{\tau} =k)=\sum_j P(\tau =j,N_{2j}-N_{j} =k)$$. Use the fact that $$(\tau =j)$$ and $$(N_{2j}-N_{j} =k)$$ are independent (because poissson process has independent increments).

• I think I can now. Thank you – David Stevens Apr 23 at 0:56