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
$$
\mathbb E e^{S_b/2}=e^{-b}.
$$

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 <uniform integrability>[uniformly integrable]. The <optional sampling theorem> at any stopping time $R$ gives
$$
\mathbb E\exp\!\left(B_{R\wedge S_b}-\frac12(R\wedge S_b)\right)=1.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 202 / 3 / c / Solution

{c}

If $B$ is standard Brownian motion, then $-B$ is also standard Brownian motion. More generally, applying an orthogonal transformation to a multidimensional Brownian motion preserves its law.


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 Incoming links: Reflection invariance of Brownian motion