Given a Poisson process $N(t)$ with rate $2$, find $P( N(3) = 4 | N(1) = 1 )$.

Question

Given a Poisson process $N(t)$ with rate $2$, find
a. $P( N(3) = 4 | N(1) = 1 )$
b. $P( N(1) = 1 | N(3) = 4 )$

Using the method, I have

$$\frac{P( N(3) = 4 \cap N(1) = 1 )}{P( N(1) = 1)}$$
$$P( N(1) = 1) = 2 / \exp(-2)$$

Then because $P( N(3) = 4 \cap N(1) = 1 )$ is independent, therefore $$\frac{P( N(3) = 4 ) P( N(1) = 1 )}{P( N(1) = 1)} = P( N(3) = 4 )$$

Am I correct in saying this for the first a?

Answer 1

That's false, they are not independent. Notice that $$\{N(3)= 4, N(1) = 1\}\iff\{N(1) = 1,N(3)-N(1) = 3\}$$ and $N(1)$ is independent of $N(3)-N(1)$. Hence \begin{align*} P(N(3) = 4|N(1) = 1) &= \frac{P(N(3) = 4,N(1) = 1)}{P(1) = 1)}\\ &=\frac{P(N(3)-N(1) = 3, N(1) = 1)}{P(N(1) = 1)}\\ &=\frac{P(N(3)-N(1) = 3)P(N(1) = 1)}{P(N(1) = 1)}\\ &=P(N(3)-N(1) = 3) \end{align*}