When does a stochastic process have its sample paths a.s. in the reproducing kernel hilbert space (RKHS) induced by its covariance function?

Question

Let $T$ be a compact metrizable space. Consider a centered second order measurable process $(X_t\colon t\in T)$ with continuous covariance function $c(t,s):= \mathbb{E}X_t X_s$.

Are there any known sufficient conditions ensuring that $t \mapsto X_t$ lies almost surely in the reproducing kernel Hilbert space $\mathcal{H}(c)$ generated by $c$? What about necessary conditions on that matter?

The question is motivated by the Karhunen-Loève representation for $X_t$ (endow $T$ with a strictly positive Borel measure $\nu$ for this to make sense), \begin{equation} \sup_{t\in T} \mathbb{E}\left(X_t - \sum_{j=1}^n\langle X_\cdot , \varphi_j\rangle_{L^2(\nu)}\varphi_j(t)\right)^2\rightarrow 0 \end{equation} and the fact that $\{\sqrt{\lambda_j}\varphi_j\}$ is an orthonormal basis for $\mathcal{H}(c)$ so that one may wonder whether the (formal) series \begin{equation} \sum_{j=1}^\infty a_j\,\sqrt{\lambda_j}\varphi_j, \qquad a_j:=\lambda_j^{-1/2}\langle X_\cdot , \varphi_j\rangle_{L^2(\nu)} \end{equation} is a well defined element of $\mathcal{H}(c)$.

In the above, $(\lambda_j,\varphi_j)$ are the eigenvalues / eigenfunctions as given in Mercer's Theorem.

Answer 1

If $T$ is an infinite set and the Gaussian measure is non-degenerate, then the RKHS (Cameron Martin space, in Bogachev's language) is infinite dimensional; hence $\operatorname{Prob}(X \in \mathcal H(c)) = 0$. If $T$ is a finite set, then $\operatorname{Prob}(X \in \mathcal H(c)) = 1$.

For both results, see Bogachev, Gaussian Measures, Theorem 2.4.7.

As an example, think about Brownian motion; sample paths are nowhere differentiable with probability 1, whereas the RKHS consists of differentiable 'paths'.

For more information on Gaussian series, including Karhunen-Loeve expansion, see Bogachev, Section 3.5.

Answer 2

The Gaussianity assumption mentioned in the answer of Joris is a particular case of a more general condition:

The family of random variables $\{\langle X, \varphi \rangle^2 \mid \varphi \in S\}$ is uniformly integrable, where $S$ is the set of linear functionals of the form $\langle X, \varphi \rangle := \sum_{k=1}^n \varphi_k X(t_k)$ whose $\mathcal{H}(c)^\ast$-norm is $\le 1$ (i.e. $\forall \xi \in \mathcal{H}(c): \sum_k \varphi_k \xi(t_k) \le \Vert \xi \Vert_{\mathcal{H}(c)}$).

In the Gaussian case this condition is satisfied because the $L^2(\Omega,\mathsf{P})$ norm of the linear functionals $\langle X, \varphi \rangle$ (which is the same as $\Vert \varphi \Vert_{\mathcal{H}(c)^\ast}$) bounds an $L^p(\Omega,\mathsf{P})$ norm for $p>2$. More generally, it is also satisfied for processes that depend polynomially on a Gaussian system, or on Bernoulli random signs, or basically on anything where hypercontractivity works.

In general the condition above is enough to conclude that the sample paths of $X$ do not lie in $\mathcal{H}(c)$ (unless it's finite-dimensional).

To prove this, note that by the Lebesgue's theorem, uniform integrability implies that the $L^2$ and $L^0$ topologies (the latter is convergence in probability) coincide on the space of linear functionals of $X$. Note that the $L^0$ topology won't change if we change the probability measure $\mathsf{P}$ by an equivalent measure $\mathsf{Q}$.

Now argue towards a contradiction. Let $X$ be a random element of $\mathcal{H}(c)$. Then $\Vert X \Vert_{\mathcal{H}(c)}^2$ is a well-defined, finite random variable, so there exists a measure $\mathsf{Q}$, equivalent to $\mathsf{P}$, such that $\mathsf{E}_{\mathsf{Q}} \Vert X \Vert_{\mathcal{H}(c)}^2 < \infty$. On the other hand, $\mathsf{E}_{\mathsf{Q}} \Vert X \Vert^2_{\mathcal{H}(c)}$ is the trace of the covariance operator of $X$ in $\mathcal{H}(c)$ (computed for the measure $\mathsf{Q}$), so the covariance must be trace class, and therefore compact. Thus there is a sequence of linear functionals $\varphi_n \in S$, such that $\Vert \varphi_n \Vert_{\mathcal{H}(c)^\ast} = 1$, and $\langle \varphi_n, X \rangle \to 0$ in $L^2(\mathsf{Q})$. Thus we also have $\langle \varphi_n, X \rangle \to 0$ in $L^0(\mathsf{P})$, and, by uniform integrability, also in $L^2(\mathsf{P})$. This contradicts the fact that $\Vert \langle \varphi, X \rangle \Vert_{L^2(\mathsf{P})} = \Vert \varphi \Vert_{\mathcal{H}(c)^\ast} = 1$.