The process $M$ is continuous and Gaussian. Since $A$ has finite variation,
$$
[M]_t=[X]_t=t.
$$
The <Lévy characterization of Brownian motion> makes $M$ a Brownian motion in the enlarged filtration.

Moreover,
$$
\operatorname{Cov}(M_t,X_1)
=t-\int_0^t\frac{\operatorname{Cov}(X_1-X_s,X_1)}{1-s}\,ds
=t-\int_0^t1\,ds=0.
$$
Every finite vector from $M$ is jointly Gaussian with $X_1$, so zero covariance implies independence. Thus the whole process $M$ is independent of $X_1$.

Solution

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

 Solution

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