Poisson Process: indepedent increment

Let $\{N(t): t\geq0\}$ be a Poisson process of rate $\lambda$, and let $S_n$ denote the time until the $n_{th}$ event occurs.

compute $P(S_3>5|N(2)=1)$

Attempt:

Notice that $P(S_3>5)=P(N(5)<3)$. Therefore, we write $P(N(5)<3|N(2)=1)$. Using indepedent increment, this is equivalent as $P(N(5)-N(2)\leq1)=P(N(3)\leq1)=e^{-3\lambda}(1+3\lambda)$ .

What do you guys think?

• Seems fine. Good job.
– Em.
Mar 9 '16 at 6:49