For $0\leq s\leq t$, write $B_t=B_s+(B_t-B_s)$. The increment is independent of $\mathcal F_s$ and is normally distributed with variance $t-s$. Its moment generating function gives
$$
\begin{aligned}
\mathbb E[M_\lambda(t)\mid\mathcal F_s]
&=e^{\lambda B_s-\lambda^2s/2}
\mathbb E\left[e^{\lambda(B_t-B_s)-\lambda^2(t-s)/2}\right]\\
&=M_\lambda(s).
\end{aligned}
$$
The process is integrable for every real $\lambda$, so it is the <exponential Brownian martingale>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 201 / 4 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-201/4/a/solution