Why $\mathbb P(B_{T_a+(t-T_a)}-B_{T_a}<0\mid T_a<t)=\frac{1}{2}$?

Question

Let $(B_t)$ a Brownian motion starting at $0$, and $T_a=\inf\{t\geq 0\mid B_t=a\}$. So, I know that $(B_{T_a+t}-B_{T_a})_t$ is a Brownian motion independent of $\mathcal F_{T_a}$ (where $\mathcal F_t$ is adapted at the brownian motion). So indeed $$\mathbb P(B_{T_a+s}-B_{T_a}<0)=\frac{1}{2}.$$

Now, I don't understand how to prove that $$\mathbb P(B_{T_a+(t-T_a)}-B_{T_a}<0\mid T_a<t)=\frac{1}{2}.$$

What I know is $$\mathbb P(B_{T_a+(t-T_a)}-B_{T_a}<0\mid T_a<t)=\frac{\mathbb P(B_{T_a+(t-T_a)}-B_{T_a}<0,T_a<t)}{\mathbb P(T_a<t)},$$ but I don't see why it should be equal to $\frac{1}{2}$.

Answer 1

Using the strong Markov property (e.g. Theorem 8.3.7 on page 314 here), \begin{align} \mathsf{P}_0(B_{T_a+(t-T_a)}<B_{T_a},T_a<t)&=\mathsf{E}_0\!\left[1\{T_a<t\}\mathsf{P}_0(B_{T_a+(t-T_a)}<a\mid \mathcal{F}_{T_a})\right] \\ &=\mathsf{E}_0\!\left[1\{T_a<t\}f(B_{T_a},T_a)\right]=\frac{1}{2}\mathsf{P}_0(T_a<t), \end{align} where $f(x,s)=\mathsf{P}_x(B_{t-s}<a)$ for $s<t$.