The squared chord distance is
$$
\lVert u(t)-u(t')\rVert^2
=2-2\cos\{\omega(t-t')\}
=4\sin^2\left(\frac{\omega(t-t')}{2}\right).
$$
Hence the restricted <Gaussian process> has covariance
$$
\boxed{
k_\theta(t,t')
=A\exp\left[-\frac{2}{l^2}
\sin^2\left\{\frac{\omega(t-t')}{2}\right\}\right]}.
$$
It depends only on $t-t'$, so it is stationary. It is periodic in either argument with period $T=2\pi/\omega$, and Gaussian-process realizations inherit that period almost surely because $f(t+T)-f(t)$ has zero variance.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 219 / 2 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-219/2/a/solution