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When a covariance operator on a function space is an integral operator, its covariance kernel satisfies $(C_{X}f)(s)=\int c_X(s,t)f(t)dt$. For a centered process, $c_X(s,t)=\mathbb{E}[X(s)X(t)]$ whenever point evaluation is meaningful.

Functional data analysis treats each observation as a function or another infinite-dimensional object. Means, covariance operators, spectral coordinates, and regression operators replace their finite-dimensional vector and matrix counterparts.

Functional principal component analysis diagonalizes a compact covariance operator. Its eigenfunctions are principal component functions, and projecting a centered observation onto them gives uncorrelated functional principal component scores.

A random element $G$ of a Hilbert space is Gaussian when every continuous linear functional of $G$ has a normal distribution. Its mean and covariance operator determine its distribution.

For independent identically distributed centered random variables in a separable Hilbert space with finite second moment, $n^{-1/2}{\sum }_{i=1}^n{X}_i$ converges in distribution to a centered Gaussian random element with the same 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.

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 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.

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

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.

A principal component function is a normalized eigenfunction ${\varphi }_k$ of a covariance operator, ordered by decreasing eigenvalue ${\lambda }_k$.