How can I compute $P(W_{t_1} \in (a,b], W_{t_2} \in (a,b])$ where $(W_t)_{t \geq 0}$ is a 1-dimensional Brownian motion and $0 \leq t_1 <t_2$ and $a<b$? My idea was to somehow use the fact that the increments of a Brownian motion are independent
$P(W_{t_1} \in (a,b], W_{t_2} \in (a,b])=P(W_{t_1} \in (a,b], (W_{t_2}-W_{t_1}+W_{t_1}) \in (a,b])= \int_{a}^bP(W_{t_1} \in (a,b], (W_{t_2}-W_{t_1}+W_{t_1}) \in (a,b]\big\vert W_{t_1}=y)dy_1$
Is the last equality even true? If it is how can we explicitly compute it?
Yes, the equality is true because $W_{t_{2}}-W_{t_{1}}$ is independent of $W_{t_{1}}$. THe value of the integral equals $\int_a^{b} P\{W_{t_{2}}-W_{t_{1}} \in (a-y,b-y]\}dy$ Since $W_{t_{2}}-W_{t_{1}}$ is normally distributed with mean 0 and variance $t_2-t_1$ you get $\int_a^{b} [(\Phi (\frac {b-y} {\sqrt (t_2-t_1)}) -\Phi (\frac {a-y} {\sqrt (t_2-t_1})] e^{-y^{2}/2t_1} \frac 1 {\sqrt {2\pi t_1}} dy$.