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For any real orthonormal basis $(G_i)$, the argument of (b)(ii) gives almost sure convergence and convergence in L2 of $\tilde{W}h$. Finite linear combinations have normal distributions, and the Parseval identity for a Hilbertian basis givesThus $W$ and $\tilde{W}$ are centered Gaussian processes with the same covariance function. More explicitly, their finite-dimensional characteristic functions are bothThis proves the basis invariance of an isonormal Gaussian process: equality of all finite-dimensional distributions, and hence equality in law of the indexed collections.

For the specified cosine orthonormal basis, put ${\lambda }_i=\pi (i-\frac{1}{2})$. Its coefficient against an interval indicator isConsequently the Brownian half-integer sine expansion, a Karhunen–Loève expansion, isFor every fixed $t$, the sum converges almost surely and in L2 space because the coefficient squares sum to $\Vert 1_{[0,t]}{\Vert }^2=t$. Its continuous modification is a standard Brownian motion by (b)(v). The expansion also holds in $L^2(\Omega \times [0,1])$: the expectation of the squared tail is the integral of the coefficient-square tail, which tends to zero by the dominated convergence theorem, with bound $t$.

To realize this expansion for the original $B$, choose $b_i=W{G}_i$. These are Jointly Gaussian with covariance matrix equal to the identity, hence are independent standard normal random variables. The finite orthogonal projections of $1_{[0,t]}$ onto $G_1,\dots ,G_N$ converge in $\mathcal{H}$, and the identity $\mathbb{E}|Wh{|}^2=\Vert h{\Vert }^2$ transfers that convergence to the displayed expansion for $B_t$. With an independently chosen sequence $(b_i)$ the resulting Brownian motion has the same law; a pathwise identification uses the coupling just given.

Let $V=(X_1(t_1),X_1(t_2){)}^T$. The Karhunen–Loève expansion and Gaussianity giveandThe conditional multivariate normal distribution formula yieldsIf $\Sigma$ is singular, the same formula uses its Moore-Penrose pseudoinverse.

The Karhunen–Loève expansion has the following Hilbert-space form. Let $X$ be a square-integrable Random element of a Hilbert space $H$, with mean $\mu$ and covariance operator $C$. There are nonnegative eigenvalues ${\lambda }_1\ge {\lambda }_2\ge \cdots$ and orthonormal eigenvectors ${\varphi }_k$ spanning the closure of the range of $C$ such thatin $L^2(\Omega ;H)$. The functional principal component scoressatisfyIf $X$ is a Gaussian random element, the scores are independent normal random variables.

For the proof, $C$ is positive, self-adjoint, and trace-class, withIt is therefore compact, so the spectral theorem for compact Hermitian operators supplies the eigenpairs. Their score covariance isFor the residual $R_m=X-\mu -{\sum }_{k=1}^m{\xi }_k{\varphi }_k$, Parseval identity and the trace formula giveThe centered variable's projection onto $\ker C$ has zero second moment and is therefore zero almost surely, which completes the mean-square expansion. When $C$ has a continuous covariance kernel, Mercer's theorem additionally expands that kernel as $c(s,t)={\sum }_k{\lambda }_k{\varphi }_k(s){\varphi }_k(t)$.

The stated kernel is the Brownian bridge covariance kernel. Its eigenvalue equation isThe right-hand side vanishes at $s=0$ and $s=1$, and differentiating it twice givesThus the normalized eigenfunctions and eigenvalues areThe Karhunen–Loève expansion is consequentlywith convergence in $L^2(\Omega ;L^2[0,1])$, where $\mathbb{E}{Z}_k=0$ and $\mathbb{E}[Z_j{Z}_k]=1_{\{j=k\}}$. Covariance alone does not imply that the $Z_k$ are independent or normal; they are independent standard normal variables when $X$ is Gaussian.

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 $B$ 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 ${\sum }_k{\gamma }_k{\sum }_{j:{\lambda }_j>0}{\rho }_{jk}^2$.