= Solution
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 $X_1$.
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