# Find probability of a Poisson process.

Given that $N=\{N(t)\mid t\geq 0\}$ is a Poisson process with parameter $\lambda>0$

I need to find $P(N(3)=2\mid N(1)=0, N(5)=4)$

So this is a conditional probability (can anyone clarify if this is correct) and I assume that the mass function for a Poisson process is:

$p(N(t)=k)=e^{-\lambda t}\frac{(\lambda t)^k}{k!}$

But if I think in the way $P(A\mid B)=\frac{P(A\cap B)}{P(B)}$, I get wrong answer (the answer should not depend on $\lambda$).

Can anyone clarify how this calculation should be done? (note my coursebook is extremely abstract with no examples whatsoever) :(

• Please post the steps of your calculation, the approach should work. Jan 1 '14 at 20:43
• $P(N(3)=2)=e^{-3\lambda}\frac{(3\lambda)^2}{2!}$, $P(N(1)=0)=e^{-\lambda}$, $P(N(5)=4)=e^{-5\lambda}\frac{(5\lambda)^4}{4!}$ Jan 1 '14 at 20:56
• OK, that was not helpful. What is $P(N(3)=2|N(1)=0)$? What is $P(N(5)=4|N(1)=0)$? What is $P(N(5)=4|N(3)=2)$? Jan 1 '14 at 21:00
• I am aiming for a very basic property of Poisson processes. Jan 1 '14 at 21:06
• No, the events $\{N(3)=2\}$ and $\{N(1)=0\}$ are not independent. Jan 1 '14 at 21:17

Here, three disjoint intervals are involved, namely $(0,1]$, $(1,3]$ and $[3,5]$, hence one should try to write down everything in terms of $X=N(1)$, $Y=N(3)-N(1)$ and $Z=N(5)-N(3)$. Then, $X$, $Y$ and $Z$ are Poisson with the parameters you know and independent, and you are interested in the events $$B=[N(1)=0,N(5)=4]=[X=0,Y+Z=4],$$ and $$A=[N(1)=0,N(3)=2,N(5)=4]=[X=0,Y=2,Z=2].$$ Thus, $$P[B]=p_\lambda(0)p_{4\lambda}(4),\qquad P[A]=p_\lambda(0)p_{2\lambda}(2)p_{2\lambda}(2),$$ where $p_\mu(n)$ denotes the probabiolity that a Poisson random variable with parameter $\mu$ is $n$. Thus, you are looking for $$\frac{P[A]}{P[B]}=\frac{p_\lambda(0)p_{2\lambda}(2)p_{2\lambda}(2)}{p_\lambda(0)p_{4\lambda}(4)}.$$ If you compute this ratio, you should see that $\lambda$ disappears...