The common eigenbasis gives
$$
C_1^{1/2}\phi_k=\sqrt{\lambda_k}\phi_k,\qquad
C_2^{1/2}\phi_k=\sqrt{\lambda_k^*}\phi_k.
$$
Consequently
$$
\boxed{d_R(C_1,C_2)^2
=\sum_{k=1}^\infty(\sqrt{\lambda_k}-\sqrt{\lambda_k^*})^2}.
$$
For the positive square-root factors, the singular values of $C_2^{1/2}C_1^{1/2}$ are $\sqrt{\lambda_k\lambda_k^*}$, so
$$
d_P(C_1,C_2)^2
=\sum_k\lambda_k+\sum_k\lambda_k^*
-2\sum_k\sqrt{\lambda_k\lambda_k^*}
=d_R(C_1,C_2)^2.
$$
Thus the distances coincide when the covariance operators commute and share an eigenbasis.


Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 225 / 3 / iv / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-225/3/iv/solution