For $s<r<1$, part a gives
$$
\mathbb E\left(\left.\frac{X_1-X_r}{1-r}\right|\mathcal F_s\right)
=\frac{X_1-X_s}{1-s}.
$$
Conditional Fubini then yields
$$
\mathbb E(A_t-A_s\mid\mathcal F_s)
=\frac{t-s}{1-s}(X_1-X_s)
=\mathbb E(X_t-X_s\mid\mathcal F_s).
$$
Hence $\mathbb E(M_t\mid\mathcal F_s)=M_s$, so $M$ is a martingale.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 201 / 4 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-201/4/c/solution