Let $B$ be the <Hilbert-Schmidt operator> with kernel $\beta$, so the <function-on-function linear model> is $Y=BX+\varepsilon$. Independence and centering give $\mathbb E[Y\mid X]=BX$. Let
$$
C_X\phi_j=\lambda_j\phi_j,
\qquad
C_Y\psi_k=\gamma_k\psi_k,
$$
and define the <functional principal component scores>
$$
\xi_j=\langle X,\phi_j\rangle,
\qquad
\eta_k=\langle Y,\psi_k\rangle.
$$
The <cross-covariance operator> identity $C_{YX}=BC_X$ gives, for every $\lambda_j>0$,
$$
\mathbb E[\eta_k\xi_j]
=\langle C_{YX}\phi_j,\psi_k\rangle
=\lambda_j\langle B\phi_j,\psi_k\rangle.
$$

Since $BX$ is centered, the requested integrated variance is $\mathbb E\lVert BX\rVert^2$. Applying the <Karhunen–Loève expansion> to $X$ and the <Parseval identity> in the $\psi_k$ basis yields
$$
\begin{aligned}
\int_0^1\operatorname{Var}(\mathbb E[Y(t)\mid X])dt
&=\mathbb E\lVert BX\rVert^2\\
&=\sum_{j:\lambda_j>0}\lambda_j\lVert B\phi_j\rVert^2\\
&=\sum_{j:\lambda_j>0}\sum_{k\geq1}
\frac{\{\mathbb E(\xi_j\eta_k)\}^2}{\lambda_j}.
\end{aligned}
$$
Equivalently, if $\rho_{jk}=\operatorname{Corr}(\xi_j,\eta_k)$, the expression is $\sum_k\gamma_k\sum_{j:\lambda_j>0}\rho_{jk}^2$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 225 / 4 / Solution

{title2=$\xi_k$}

The functional principal component score of a centered function $X$ along $\phi_k$ is $\xi_k=\langle X,\phi_k\rangle$. Its variance is the corresponding covariance eigenvalue $\lambda_k$.


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