Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 225 (Functional Data Analysis)

Write $Z=X-\mu$, so the covariance operator isFor $g,h\in L^2[0,1]$,so $C_X$ is self-adjoint. Moreover,so it is a positive operator.

Let $(e_j)$ be any orthonormal basis. Tonelli theorem and Parseval identity giveA positive operator with finite trace is a trace-class operator, completing the proof.

For every orthonormal basis $(e_j)$ of the infinite-dimensional Hilbert space $L^2[0,1]$, the identity operator satisfiesPart a shows that every covariance operator of a square-integrable Hilbert-space random element is trace-class. Therefore the identity cannot be a covariance operator.

Choose a unit vector $e$ and define the rank-one operatorIt is bounded, self-adjoint, and Hilbert-Schmidt, with $\Vert T{\Vert }_{\mathrm{HS}}=1$. Butso it is not positive. Since every covariance operator is positive, $T$ is the required counterexample.

Linearity of the Bochner integral givesIndependence and centering of $X_i-\mu$ imply

Since $C_X{\varphi }_k={\lambda }_k{\varphi }_k$ and the eigenfunctions can be chosen orthonormally, independence of the sample givesThis remains valid for repeated eigenvalues after choosing an orthonormal eigenbasis within each eigenspace.

Under $H_0$, the Hilbert-space central limit theorem giveswhere $G$ is a centered Gaussian random element with covariance $C_X$. Distinct leading eigenvalues imply consistency of the empirical eigenvalues and eigenfunctions, up to irrelevant signs. The first $K$ standardized functional principal component scores of $G$ are independent $N(0,1)$ variables, so Slutsky theorem yieldsAn asymptotic level-$\alpha$ test rejects above the $(1-\alpha )$ quantile of the chi-squared distribution.

Under a fixed alternative, put $\delta =\mu -{\mu }_0$. The weak law of large numbers and eigenpair consistency giveThus the statistic diverges and the test is consistent whenever at least one retained projection is nonzero. A fixed alternative orthogonal to the first $K$ eigenfunctions is invisible to this fixed-$K$ statistic.

Let $C^{\ast }=C_X+\delta \otimes \delta$, where $(\delta \otimes \delta )h=⟨\delta ,h⟩\delta$. Suppose for contradiction thatThen $C^{\ast }{\varphi }_l^{\ast }=C_X{\varphi }_l^{\ast }$, so every leading eigenpair $({\lambda }_l^{\ast },{\varphi }_l^{\ast })$ is also an eigenpair of $C_X$. Since $C^{\ast }\ge C_X$, the Courant–Fischer min-max principle gives ${\lambda }_l^{\ast }\ge {\lambda }_l$. If ${\lambda }_l^{\ast }={\lambda }_{j_l}$ in the strictly decreasing spectrum of $C_X$, this inequality implies $j_l\le l$. Orthogonality and distinctness forceHence the first $K$ eigenvectors of $C^{\ast }$ span the same space as ${\varphi }_1,\dots ,{\varphi }_K$. Their assumed orthogonality to $\delta$ would imply $⟨\delta ,{\varphi }_k⟩=0$ for every $k\le K$, contradicting the hypothesis. Therefore some $l\le K$ satisfies $⟨\delta ,{\varphi }_l^{\ast }⟩\ne 0$.

HereIt has the same eigenfunctions as $C_X$, but the $(K+1)$st eigenvalue is raised from ${\lambda }_{K+1}$ to ${\lambda }_{K+1}+b^2$. A test based on the first $K$ eigenfunctions of $C_X$ is blind to this alternative because $\delta =b{\varphi }_{K+1}$ is orthogonal to all of them. If the raised eigenvalue overtakes at least ${\lambda }_K$, then ${\varphi }_{K+1}$ enters the first $K$ eigenfunctions of $C^{\ast }$ and the mean shift acquires a retained nonzero coordinate. The $C^{\ast }$-based test can therefore detect alternatives that the original FPCA truncation misses.

For positive trace-class operators, the square-root distance between covariance operators isIf $C_i=L_i{L}_i^{\ast }$ are Hilbert-Schmidt factorizations, the Procrustes distance between covariance operators iswhere $\mathcal{O}(H)$ is the group of orthogonal operators on the real separable Hilbert space $H$.

For an orthogonal $R$, unitary invariance of the Hilbert-Schmidt norm givesThe product $L_2^{\ast }{L}_1$ is trace-class. Its polar decomposition of a bounded operator and trace duality givewhere $({\sigma }_k)$ are its singular values. Taking the infimum proves

Hilbert-Schmidt convergence impliesThe Schatten norm Hölder inequality givesThe sum of singular values is the trace norm, and the reverse triangle inequality givesSubstitution into part b proves

Choose a unit vector $e$ and distinct positive numbers $a,b$, and setThese are distinct rank-one covariance operators. Their positive square roots are $\sqrt{a}(e\otimes e)$ and $\sqrt{b}(e\otimes e)$, soUsing these square roots as the factors in the Procrustes definition, the identity alignment attains the same value. The singular-value formula gives the matching lower bound,Hence $d_P=d_R$ although $C_1\ne C_2$.

Letand define the functional principal component scoresUse empirical centered scores ${\hat{\xi }}_{ik},{\hat{\eta }}_{il}$ from the sample covariance eigenfunctions and formThe finite-dimensional statistic is

Under $H_0$, $Y=\epsilon$ is independent of $X$. The population cross-covariances vanish, andThe multivariate central limit theorem, consistency of the empirical eigenpairs, and Slutsky theorem therefore giveRejecting above the $(1-\alpha )$ quantile gives an asymptotic level-$\alpha$ test.

Under a fixed alternative, $\mathbb{E}[{\eta }_l{\xi }_k]$ is the $(l,k)$ coordinate of the cross-covariance operator $C_{YX}=B{C}_X$, where $B$ is the Hilbert-Schmidt operator with kernel $\beta$. ThusThe test is consistent whenever this retained block contains a nonzero cross-covariance; fixed truncation can miss alternatives outside the selected principal-component subspaces.