 Solution

By the Reflection invariance of Brownian motion, $-B$ is Brownian motion, and the first time $B_t-t$ reaches $b<0$ is the first time $-B_t+t$ reaches $-b>0$. Part b therefore gives

Now $L_t=\exp (B_t-t/2)$ is the exponential Brownian martingale. At $S_b$, the identity $B_{S_b}=b+S_b$ gives $L_{S_b}=e^{b+S_b/2}$, so $\mathbb{E}{L}_{S_b}=1$. The nonnegative stopped martingale $(L_{t\wedge S_b})$ therefore loses no mass at infinity and is uniformly integrable. The optional sampling theorem at any stopping time $R$ gives