Why is $P^b(B_{T-T_b} \in (-\infty,b))=1/2$

Question

Why is $P^b(B_{T-T_b} \in (-\infty,b))=1/2$ on the set $\{T_b<t\} $ where $T_b=\inf\{t \ge 0 :B_t=b\}$ and $T=t 1_{\{T_b<t\}}+\infty 1_{\{T_b \ge 0\}}$.

I am trying to understand Proposition 2.6.19 in Shreve and trying to figure out why $$(U_{T(\omega)-T_b(\omega)}1_\Gamma)(B_{S(\omega)}(\omega))=1/2$$ My attempt I write $$(U_{T(\omega)-T_b(\omega)}1_\Gamma)(B_{S(\omega)}(\omega))=P^b(B_{t-T_b)} \in (-\infty,b))=P^0(B_{t-T_b)} \in (-\infty,0))$$

But then can I somehow argue that the distribution of $B_{t-T_b}$ is normal with mean $0$? I cant see why this should be true? Or am I totally off track

Answer 1

Fix $t>0$. By the very definition of $S=T_b$ we have $B_{S(\omega)}(\omega)=b$, i.e.

$$(U_{t} 1_{(-\infty,b)})(B_{S(\omega)}(\omega)) = (U_{t} 1_{(-\infty,b)})(b).$$

Since $$ (U_t 1_{(-\infty,b)})(x) = \mathbb{P}^x(B_t < b)$$

for any $x \in \mathbb{R}$, we find

$$(U_{t} 1_{(-\infty,b)})(B_{S(\omega)}(\omega)) = \mathbb{P}^b(B_t < b).$$

As $\mathbb{P}^b(B_t < b) = \mathbb{P}^0(b+B_t < b)$ this implies

$$(U_{t} 1_{(-\infty,b)})(B_{S(\omega)}(\omega)) = \mathbb{P}^0(B_t < 0) = \frac{1}{2}.$$

Since $t>0$ is arbitrary, we can choose in particular $t:=T(\omega)-S(\omega)$ for $\omega \in \{T<\infty\} \subseteq \{S<T\}$.