# Relation between regularities of the trajectory of a mean zero gaussian process and its covariance operator

Let $\xi_t$ be a zero-mean gaussian process on $[0,1]$ with covariance operator $C$. I would like to better understand the relation between the covariance operator and the regularity of the trajectories.

Theorem (Kolmogorov) If there exists $\alpha>1,C\geq 0$ and $\epsilon>0$ such that $$E[|\xi_t-\xi_s|^{\alpha}] \leq C > |t-s|^{1+\epsilon}$$ then there exists a modification of the process that is almost surely Hölder-$\delta$ for $\delta\in ]0,\epsilon/\alpha[$.

I would like to know if there are other results in this direction (with other spaces than Holder ?) and especially one that relates directly a norm (spectral norm under stationnarity assumption ?) of the covariance operator and the regularities of the trajectories.

note: if I remember there is a link between the closure of the cameron martin space and the support of the gaussian measure associated to the process... how can I reformulate this to answer my question ? )

-

Suppose we already know that the process is continuous with probability one, so that the process takes values in the Banach space $X = C([0,1])$ with distribution $\mathbb P$. The covariance operator $C : X^* \to X$ is then a map from the dual space $X^*$ to $X$. The support of the Gaussian measure $\mathbb P$ is then the closure of the image of $X^*$ under $C$: $$\operatorname{supp} \mathbb P = \overline{ CX^* }.$$ (This is the main theorem of [Vakhania 1975])

Now let's construct the Cameron-Martin space. The operator $C$ defines an inner product on the dual space $X^*$ by $$\langle f, g \rangle = f(Cg)$$ for $f, g \in X^*$. The space $X^*$ isn't necessarily closed under the topology induced by the inner product, so let $H$ be the Hilbert space completion, and let $\iota : X^* \hookrightarrow H$ be the inclusion map. Define a map $\iota^* : H \hookrightarrow X$ first on the dense subspace $\iota X^* \subseteq H$ by $$\iota^*(\iota f) = Cf,$$ and extend continuously to all of $H$. Thus the covariance operator factors as $C = \iota^* \circ \iota$.

The Hilbert space $\iota^* H$ is a subspace of $X$, and is called the Cameron-Martin space of the process. Interpreting this in the context of the support of the Gaussian measure $\mathbb P$, we have $$\operatorname{supp} \mathbb P = \overline{\iota^* H},$$ so that the closure of the Cameron-Martin space (with respect to the original norm of $X$) is exactly the support of $\mathbb P$.

I go into these ideas in more detail in Section 2 of my preprint [LaGatta 2010].

Suppose your covariance operator is an integral operator with kernel $c(s,t)$, called the covariance function of the process. That is, if $\mu$ is a Radon measure on $[0,1]$, then $$(C\mu)(s) = \int_0^1 c(s,t) \, d\mu(t).$$

If we write $c_s(t) = c(s,t)$, then the support of $\mathbb P$ is the closure of the span of the functions $c_s$ in $C([0,1])$. So to answer your question: if you know that the closure $\overline{\operatorname{span}\{c_s\}}$ is a space with regularity property $P$, then the process $\xi_t$ satisfies property $P$ with probability one.

-
Thanks Tom this is very clearly developping my note to answer the question ! I could have a further question such as "how far can we extend the idea with the kernel replacing the kernel function by a kernel distribution" but I'll read the preprint first. –  robin girard Aug 13 '10 at 11:26
Glad to help, Robin. I always enjoy your questions. Don't focus too much on the last paragraph I wrote, discussing covariance functions. I included that only to make the abstract formalism a little more concrete. All the functional analysis works as long as you have a covariance operator C. The important statement that you want to focus on is the first equation I posted: $\operatorname{supp} \mathbb P = \overline{CX^*}$. –  Tom LaGatta Aug 13 '10 at 13:19

Hi Robin

You should have a look to the following reference :

Adler - Introduction to Continuity, Extrema, and other Topics of Gaussian Processes

Best Regards

-
thanks ! I will take a look at it. –  robin girard Aug 11 '10 at 10:53