Since $C_X\phi_k=\lambda_k\phi_k$ and the eigenfunctions can be chosen orthonormally, independence of the sample gives
$$
\begin{aligned}
\operatorname{Cov}(\langle\widehat\mu,\phi_k\rangle,
\langle\widehat\mu,\phi_{k'}\rangle)
&=\frac1n\langle C_X\phi_k,\phi_{k'}\rangle\\
&=\frac{\lambda_k}{n}\mathbf1_{\{k=k'\}}.
\end{aligned}
$$
This remains valid for repeated eigenvalues after choosing an orthonormal eigenbasis within each eigenspace.


Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 225 / 1 / e / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-225/1/e/solution