Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 225 (Functional Data Analysis)

This is the Brownian covariance kernel, so its integral operator is a covariance operator. To obtain its eigendecomposition, suppose $Uf=\lambda f$ with $\lambda \ne 0$. Splitting the integral at $s$ givesDifferentiation yields $\lambda f^{\prime }(s)={\int }_s^{1}f(t)dt$ and $\lambda f^{\prime \prime }(s)=-f(s)$, with boundary conditions $f(0)=0$ and $f^{\prime }(1)=0$. Hence the normalized eigenpairs areThe eigenvalues are positive and summable, consistently with positivity and the trace-class operator property of a covariance operator.

The identity operator is positive and self-adjoint, but on the infinite-dimensional Hilbert space $L^2[0,1]$ it has eigenvalue one with infinite multiplicity. ConsequentlyA square-integrable Hilbert-space random variable has a trace-class covariance operator with trace $\mathbb{E}\Vert X{\Vert }^2<\infty$. The identity therefore cannot be such a covariance operator.

This symmetric finite-rank integral operator is not positive. For $f(t)=1-\frac{3}{2}t$, set $A_j={\int }_0^1{t}^{j}f(t)dt$. Directly,The associated quadratic form isEvery covariance operator is a positive operator, so this operator is not a covariance operator.

Write $x\otimes y$ for the rank-one operator $h\mapsto ⟨h,y⟩x$. The estimator is the kernel representation of the empirical covariance operatorSince $\mathbb{E}X=0$,Thus the estimator has the fixed bias of an estimator ${\mu }_0\otimes {\mu }_0$ for $C_X$.

The fourth-moment assumption makes $(X_i-{\mu }_0)\otimes (X_i-{\mu }_0)$ square-integrable in the Hilbert space of Hilbert-Schmidt operators. The weak law of large numbers therefore givesConsequently consistency for $C_X$ holds exactly when ${\mu }_0=0$. More precisely, if $Y=(X-{\mu }_0)\otimes (X-{\mu }_0)$, thenThe Hilbert-space central limit theorem also yieldswhere $G$ is a centered Gaussian random element in the Hilbert-Schmidt operator space with covariance determined by $Y$. Relative to $C_X$, the same fluctuation is displaced by $\sqrt{n}({\mu }_0\otimes {\mu }_0)$ and hence does not have a finite centered limit when ${\mu }_0\ne 0$.

Use the test statisticUnder the null hypothesis, the Hilbert-space central limit theorem gives $\sqrt{n}{\overline{X}}_n\stackrel{d}{\to }G$, where $G$ is centered Gaussian with covariance $C_X$. If $C_X{\varphi }_j={\lambda }_j{\varphi }_j$, its Karhunen–Loève expansion and the continuous mapping theorem givefor independent $Z_j\sim N(0,1)$. Reject for $T_n$ above the $(1-\alpha )$ quantile of this weighted chi-squared law; replacing the ${\lambda }_j$ by empirical covariance eigenvalues gives a plug-in estimator of the critical value.

Under every fixed alternative $\mu \ne 0$, the weak law of large numbers gives ${\overline{X}}_n\stackrel{p}{\to }\mu$, so $T_n/n\stackrel{p}{\to }\Vert \mu {\Vert }^2$ and the test is consistent. Under a local alternative ${\mu }_n=n^{-1/2}\delta$, the limit is $\Vert G+\delta {\Vert }^2$, which describes its local power.

Let $({\hat{\lambda }}_k,{\hat{\varphi }}_k)$ be the leading eigenpairs of the sample covariance operator. For fixed $K$ with ${\lambda }_1>\cdots >{\lambda }_K>{\lambda }_{K+1}$ and ${\lambda }_K>0$, the FPCA mean test usesUnder the null, consistency of the empirical eigenpairs and the multivariate central limit theorem implyThe level-$\alpha$ test therefore rejects above the $(1-\alpha )$ quantile of the chi-squared distribution with $K$ degrees of freedom.

For a fixed mean $\mu$, if at least one leading coordinate $⟨\mu ,{\varphi }_k⟩$, $k\le K$, is nonzero, then $T_{K,n}\to \infty$ in probability and the test is consistent. It has only null-level asymptotic power against means orthogonal to the first $K$ principal component functions. Under ${\mu }_n=n^{-1/2}\delta$, the limit is noncentral chi-squared with noncentrality

Assume the null distribution is a centrally symmetric probability distribution, so $X_i$ and $-X_i$ have the same law. A sign-flip randomization test draws signs $s_i\in \{-1,1\}$ independently and recomputes, for example,The exact p-value averages over all $2^n$ sign vectors:With $B$ random sign vectors, including the observed configuration, the standard Monte Carlo version is $(1+\#\{b:T_b\ge T_0\})/(B+1)$, where $T_0$ is the observed statistic. Joint sign invariance under the null makes this finite-sample valid.

The squared-norm test is omnibus: every fixed nonzero mean eventually changes $\Vert {\overline{X}}_n\Vert$. Its null law, however, is an infinite weighted chi-squared distribution and requires accurate estimation of enough covariance eigenvalues; noisy low-variance directions can also make calibration inefficient.

The FPCA mean test has the simple ${\chi }_K^2$ limit and standardizes retained directions by their variances. It is effective when the signal lies in the leading principal component subspace, but choosing $K$ introduces a tuning decision and truncation makes the test blind to alternatives orthogonal to that subspace. Close or repeated eigenvalues also make individual empirical eigenfunctions unstable.

The sign-flip randomization test can provide finite-sample calibration and avoids estimating a limiting covariance spectrum. Its exactness requires central symmetry, which is stronger than merely having zero mean, and exhaustive enumeration costs $2^n$ evaluations; Monte Carlo sign flips introduce simulation error. Its power still depends on the statistic used inside the randomization scheme.

The minimizer is the arithmetic meanIt remains a positive self-adjoint trace-class operator and hence a covariance operator. The Hilbert-Schmidt inner product gives, for every Hilbert-Schmidt operator $C$,because ${\sum }_k(C_k-\overline{C})=0$. Thus $\overline{C}$ is the unique minimizer, including when the minimization is restricted to covariance operators.

For factorizations $C_i=L_i{L}_i^{\ast }$ and $C_j=L_j{L}_j^{\ast }$, the Procrustes distance between covariance operators is the following infimum over unitary operators $R$:Unitary invariance of the Hilbert-Schmidt norm givesThe polar decomposition of a bounded operator and trace duality implywhere the last equality expresses the trace norm as the sum of the singular values. Taking the infimum proves

The two rank-one operators have orthogonal ranges. Their Hilbert-Schmidt inner product is zero, while $\Vert C_i{\Vert }_{\mathrm{HS}}={\lambda }_i$, so the Hilbert-Schmidt distance between covariance operators isFor each $C_i$, the positive square root of an operator is $C_i^{1/2}=\sqrt{{\lambda }_i}{e}_i\otimes e_i$. Orthogonality therefore gives the square-root distance between covariance operatorsFinally $C_2^{1/2}{C}_1^{1/2}=0$, so every singular value in the Procrustes cross-term vanishes. Hence

Let $B$ be the Hilbert-Schmidt operator with kernel $\beta$, so the function-on-function linear model is $Y=BX+\epsilon$. Independence and centering give $\mathbb{E}[Y\mid X]=BX$. Letand define the functional principal component scoresThe cross-covariance operator identity $C_{YX}=B{C}_X$ gives, for every ${\lambda }_j>0$,

Since $BX$ is centered, the requested integrated variance is $\mathbb{E}\Vert BX{\Vert }^2$. Applying the Karhunen–Loève expansion to $X$ and the Parseval identity in the ${\psi }_k$ basis yieldsEquivalently, if ${\rho }_{jk}=\mathrm{Corr}({\xi }_j,{\eta }_k)$, the expression is ${\sum }_k{\gamma }_k{\sum }_{j:{\lambda }_j>0}{\rho }_{jk}^2$.