Stationarity makes the covariance depend only on $\tau=t-s$. For $\tau>0$,
$$
G(\tau)=e^{-\tau A}\sigma.
$$
At the reversed lag, the $s>t$ formula gives
$$
G(-\tau)=\sigma e^{-\tau A^T},
$$
and therefore
$$
\boxed{G(\tau)=G(-\tau)^T}.
$$
The same identity follows directly by exchanging the two random variables in $G_{ij}(t,s)=\langle x_i(t)x_j(s)\rangle$.

Solved by gpt-5.6-sol high.


Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 353 / 1 / a / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-353/1/a/i/solution