Let $\{N(t), t \geq 0\}$ be the Poisson process with $\lambda=1$. We know $$N(a,b) = N(b)-N(a)$$ $$\mathbb{P}(N(a,b) = n) = \frac{(\lambda(b-a))^n}{n!}e^{-\lambda (b-a)}$$
Now I want to compute $\mathbb{P}(N(0,3) = 6 | N(0,1) = 2)$. I know the solution is given by $$\mathbb{P}(N(0,3) = 6 | N(0,1) = 2) = \frac{\mathbb{P}(N(0,3) = 6 \cap N(0,1) = 2)}{\mathbb{P}(N(0,1) = 2)} = \frac{\mathbb{P}(N(1,3) = 4 \cap N(0,1) = 2)}{\mathbb{P}(N(0,1) = 2)}$$ $$= \frac{\mathbb{P}(N(1,3) = 4) \mathbb{P}(N(0,1) = 2)}{\mathbb{P}(N(0,1) = 2)} = \mathbb{P}(N(1,3) = 4) = \frac{2^4}{4!}e^{-2}$$
However I need help to understand why $$\mathbb{P}(N(0,3) = 6 \cap N(0,1) = 2) = \mathbb{P}(N(1,3) = 4 \cap N(0,1) = 2) = \mathbb{P}(N(1,3) = 4) \mathbb{P}(N(0,1) = 2)$$ Can someone explain this to me?
First you need to understand:
If $a < b < c$, we have
$$ N(a, c) = N(a, b) + N(b, c)$$
Then you need to understand the equivalence of the two system: $$ \begin{cases} x + y &= r \\ x &= s\end{cases} \iff \begin{cases} y &= r - s \\ x &= s \end{cases}$$
So these $2$ will explain the first equation.
The independence increment property (for non-overlapping time interval) of Poisson Process will give you the second equation.