Let
$$
\overline C=\frac12(C_X+C_{X^*}),
\qquad
\overline C\phi_k=\lambda_k\phi_k,
$$
and estimate it by the average of the two within-sample <empirical covariance operator>[empirical covariance operators]. Let $(\widehat\lambda_k,\widehat\phi_k)$ be its leading empirical eigenpairs and let $\overline X_n,\overline X_n^*$ be the sample means. Use the <Two-sample FPCA mean statistic>
$$
\boxed{
T_{n,K}=\frac n2\sum_{k=1}^K
\frac{\langle\overline X_n-\overline X_n^*,\widehat\phi_k\rangle^2}
{\widehat\lambda_k}}.
$$

Under $H_0$, the <Hilbert-space central limit theorem> gives
$$
\sqrt{\frac n2}(\overline X_n-\overline X_n^*)
\xrightarrow dG,
$$
where $G$ is a centered <Gaussian random element> with covariance $\overline C$. Distinct eigenvalues give consistent empirical eigenpairs, up to signs, and the standardized leading scores are independent standard normal variables. Hence
$$
T_{n,K}\xrightarrow d\chi_K^2.
$$
Rejecting above the $(1-\alpha)$ quantile gives an asymptotic level-$\alpha$ test.

Under a fixed alternative $\delta=\mu-\mu^*$,
$$
\frac{T_{n,K}}n\xrightarrow p
\frac12\sum_{k=1}^K\frac{\langle\delta,\phi_k\rangle^2}{\lambda_k}.
$$
The test is consistent whenever one retained projection is nonzero. Alternatives orthogonal to the first $K$ eigenfunctions are invisible at fixed $K$. Under local alternatives $\delta=h/\sqrt n$, the limit is noncentral chi-squared with noncentrality $\frac12\sum_{k\leq K}\langle h,\phi_k\rangle^2/\lambda_k$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 225 / 2 / Solution

Let $(\lambda_j,\phi_j)$ be the ordered eigenpairs of the common <covariance operator> $C_X$, and write $\delta=\mu-\mu^*$. Estimate $C_X$ by the <pooled covariance operator>
$$
\widehat C
=\frac1{2n}\sum_{i=1}^n\left[
(X_i-\overline X)\otimes(X_i-\overline X)
+(X_i^*-\overline X^*)\otimes(X_i^*-\overline X^*)
\right]
$$
and denote its first $K$ eigenpairs by $(\widehat\lambda_j,\widehat\phi_j)$. The sign ambiguity of each eigenfunction disappears after squaring. Consider the <Two-sample FPCA mean statistic>
$$
T_{n,K}=\frac n2\sum_{j=1}^K
\frac{\langle\overline X-\overline X^*,\widehat\phi_j\rangle^2}
{\widehat\lambda_j}.
$$

Under $H_0$, the <Hilbert-space central limit theorem> gives
$$
\sqrt{\frac n2}(\overline X-\overline X^*)
\xrightarrow{d}G,
$$
where $G$ is a centered <Gaussian random element> with covariance $C_X$. The assumed eigenvalue gaps give consistency of the estimated eigenvalues and eigenfunctions, so <Slutsky theorem>[Slutsky's theorem] yields
$$
T_{n,K}\xrightarrow{d}\sum_{j=1}^K
\frac{\langle G,\phi_j\rangle^2}{\lambda_j}
\sim\chi_K^2.
$$
An asymptotic level-$\alpha$ test therefore rejects when $T_{n,K}$ exceeds the $(1-\alpha)$-quantile of the <chi-squared distribution> with $K$ degrees of freedom.

Under a fixed alternative,
$$
\frac{T_{n,K}}n\xrightarrow{p}
\frac12\sum_{j=1}^K\frac{\langle\delta,\phi_j\rangle^2}{\lambda_j}.
$$
The test is consequently consistent whenever the mean difference has a nonzero projection onto one of the retained principal components. A difference orthogonal to their span is invisible to this fixed-$K$ test, so $\mu\ne\mu^*$ alone does not guarantee consistency. Increasing $K$ with $n$ can recover such alternatives, but requires additional eigenvalue and approximation control.

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 225 / 2 / Solution

{c}

For two independent samples of size $n$ with a common covariance eigenbasis, the two-sample FPCA mean statistic is
$$
\frac n2\sum_{j=1}^K\frac{\langle\overline X-\overline Y,\widehat\phi_j\rangle^2}{\widehat\lambda_j}.
$$
Under equality of means and consistent pooled eigencomponent estimates, it converges to a chi-squared distribution with $K$ degrees of freedom.


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