I am struggling with understanding a certain result. I study poisson processes, and a solution to an exercise I am doing does not make sense to me.

Let $N_1(t)$ be a Poisson($\lambda$) process and $N_2(t)$ a Poisson(3$\lambda$) process that are independent. Then

$$P(N_1(h)=1,N_2(h)=0)=(\lambda h+o(h))(1-3\lambda h+o(h))=\lambda h+o(h)$$

What I don't understand is the last equality. How to we derive that? Thank you.


Expand: when $h\to 0$, $$\begin{align} (\lambda h+o(h))(1-3\lambda h+o(h)) &= \lambda h+o(h)-3\lambda^2h^2+o(h^2)+o(h^2) \\ &= \lambda h+o(h) \end{align}$$ since $h^2=o(h)$.

  • $\begingroup$ I see, thanks a lot! $\endgroup$ – user128836 Nov 9 '16 at 20:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.