The paths of $UX_t$ are continuous, and linear transformation preserves independence of increments. Moreover,
$$
U(X_t-X_s)\sim N\!\left(0,(t-s)UI_dU^T\right)
=N(0,(t-s)I_d)
$$
because $U$ is an <orthogonal matrix>. Thus all defining properties are preserved, proving the <orthogonal invariance of Brownian motion>:
$$
\boxed{(UX_t)_{t\geq0}\text{ is Brownian motion in }\mathbb R^d.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 201 / 4 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-201/4/b/solution