# Conditional probability of Brownian motion

I am given the following exercise, where I don't see how to do this

Here $B_t$ is a Brownian motion. The expectation value $E^x$ means that we take the expectation value w.r.t. a Brownian motion starting in $x$ instead of zero. $\theta_s$ maps a Brownian motion $B_t$ to $B_{t+s}.$

What I don't understand here is now how to use this Markov property.

I mean sure we have $P(\tau_{a+b}< \infty; \tau_a < \infty) = E(1_{\tau_{a+b}< \infty} ; \tau_a < \infty),$ but how can I apply then the Markov property to actually show this relation for the conditional probability?

If anything is unclear, please let me know.

Assume $B(0)=0$. On $\{\tau_a<\infty\}$,
$$\{\omega:\tau_{a+b}<\infty\}=\{\omega: \tau_{a+b}-\tau_a<\infty\}=\{\omega: \tau_{a+b}\circ \theta_{\tau_a}<\infty\}.$$
$$\mathbb{E}_0\left[1\{\tau_{a+b}<\infty \}\times 1\{\tau_a<\infty\}\right]$$ $$=\mathbb{E_0}\left[\mathbb{E_0}[1\{\tau_{a+b}<\infty\}\circ \theta_{\tau_a}\mid \mathcal{F}_{\tau_a}]\times1\{\tau_a<\infty\}\right]$$ $$=\mathbb{E_0}\left[\mathbb{E_{B(\tau_a)}}[1\{\tau_{a+b}<\infty\}]\times1\{\tau_a<\infty\}\right]$$ $$=\mathbb{E_0}\left[\mathbb{E_0}[1\{\tau_{b}<\infty\}]\times1\{\tau_a<\infty\}\right]=P_0\{\tau_b<\infty\}P_0\{\tau_a<\infty\},$$
$$P_0\{\tau_{a+b}<\infty\mid \tau_a<\infty\}$$ $$=\mathbb{E}_0\left[1\{\tau_{a+b}<\infty \}\times 1\{\tau_a<\infty\}\right] / P_0\{\tau_a<\infty\}=P_0\{\tau_b<\infty\}.$$