The <square-root distance between covariance operators> is
$$
d_R(C_1,C_2)=\lVert C_1^{1/2}-C_2^{1/2}\rVert_{\rm HS}.
$$
Writing $A_i=C_i^{1/2}$, minimizing $\sum_i\lVert A_i-A\rVert_{\rm HS}^2$ gives $A=n^{-1}\sum_iA_i$. Since this average is positive,
$$
\boxed{\widehat C_R=
\left(\frac1n\sum_{i=1}^nC_i^{1/2}\right)^2},
$$
the <square-root barycenter of covariance operators>.

Solution

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

 Solution

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