Let $N(t)$ be a Poisson process. Calculate $P(N(s)\neq N(t))$ and $P(N(s)=0, N(t)=1)$

Question

Let $N(t)$ be a Poisson process with parameter $\lambda$ . Calculate

a) $P(N(s)=0, N(t)=1)$

b) $P(N(s)\neq N(t))$

My try:

Let $0\leq s < t$

a) \begin{alignat*}{1} P(N(s)=0, N(t)=1) &= P(N(s)=0, N(t)-N(s)=1)\\ &=P(N(s)=0)P(N(t)-N(s)=1)\\ &=e^{-\lambda s}*e^{-\lambda(t-s)}(\lambda(t-s))=e^{-\lambda t} \lambda(t-s) \end{alignat*}

I'm not completely sure if this is the correct way.

b) I understand the meaning of this probability, but I'm stuck on how to calculate it. Any suggestions would be great!

Answer 1

For part a), what you did is correct. However, you have to justify the step by the properties of a Poisson process (independence of increments and the fact that $N(t)-N(s)$ follows a Poisson distribution with parameter $\lambda(t-s)$).

For the second part, note that $$ \mathbb P\left(N(s)\neq N(t)\right)=1-\mathbb P\left(N(s)= N(t)\right) =1-\mathbb P\left( N(t)-N(s)=0\right). $$