Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 225 / 2

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.