The paths of $UX_t$ are continuous and start at zero. Its increments are independent because they are deterministic functions of the independent increments of $X$. They are centered Gaussian, and orthogonality gives
$$
\operatorname{Cov}(U(X_t-X_s))
=U((t-s)I_d)U^T
=(t-s)I_d.
$$
Thus $UX$ is Brownian motion. This is the <orthogonal invariance of Brownian motion>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 201 / 4 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-201/4/b/solution