Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 225 / 2

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.