# Under which conditions Mean Square Continuity implies Sample Continuity for Gaussian Processes?

First, let us give the setting.

Let $$(\Omega, \Sigma, \mathbf{P})$$ be a probability space, let $$T$$ be some interval of time, and let $$X: T \times \Omega \rightarrow S$$ be a stochastic process.

By Mean Square continuity I mean:

Given a time $$t \in T, X$$ is said to be continuous in mean-square at $$t$$ if $$\mathrm{E}\left[\left|X_t\right|^2\right]<+\infty$$ and $$\lim _{s \rightarrow t} \mathbf{E}\left[\left|X_s-X_t\right|^2\right]=0$$

One can check here also the definitions of continuity with probability 1 and sample continuity.

Q: I would like to know under which conditions for Gaussian Processes, continuity in mean square implies sample continuity.

I know that there exists the Kolmogorov Continuity Criterion that gives a sort of an answer. However, it requires an upper bound for the second moment to be bounded for all $$s,t \in T$$ where $$T$$ compact. A version for Gaussian Processes of this theorem is given here below:

Proposition. If ( $$W_t: t \in T$$ ) is a centered Gaussian process indexed by a compact set $$T \subset \mathbb{R}^d$$ with $$\mathrm{E}\left|W_s-W_t\right|^2 \leq\|s-t\|^{2 \alpha}$$, for all $$s, t \in T$$ and some $$\alpha \in(0,1]$$, then $$W$$ possesses a version with continuous sample paths such that $$\left|W_s-W_t\right|=O\left(\|s-t\|^\alpha \log (1 / \| s-\right.$$ $$t \|)$$ ), uniformly in $$(s, t)$$ with $$\|s-t\| \rightarrow 0$$, almost surely.

Now imagine that we have something of the form, $$\forall \: s,t \in T$$:

$$\mathrm{E}\left|W_s-W_t\right|^2 \leq |s-t| + o(|s-t|)$$

Q: Can we still conclude that the paths are sample continuous?

Another way to rephrase the question in an interesting way is by using the following property of Gaussian Processes:

Theorem A Gaussian process $$X$$ on $$T$$ has continuous sample paths with probability one if, and only if, it is continuous at each fixed $$t \in T$$ with probability one; i.e. $$P \left( \lim_{s \rightarrow t} X_s =X_t, \quad \forall t \in T \right) =1$$ if and only if $$P\left(\lim_{s \rightarrow t} X_s=X_t\right)=1, \quad \forall t \in T$$

Q: Therefore it would be enough to know when, for Gaussian Processes, mean square continuity implies continuity with probability 1.

## 1 Answer

I found an answer to this in "Weak Convergence and Empirical Processes" - van der Vaart, Wellner.

Sample path properties of a Gaussian process $$W=\left(W_t: t \in T\right)$$, such as boundedness, continuity or differentiability, are determined by the covariance kernel $$K(s, t)=$$ $$\operatorname{cov}\left(W_s, W_t\right)$$ of the process. Let $$T$$ be an index set and the square of the intrinsic semimetric be:

$$\rho^2(s, t)=\operatorname{var}\left(W_s-W_t\right)=K(t, t)+K(s, s)-2 K(s, t) \text {. }$$ Proposition (Continuity). If $$\left(W_t: t \in T\right)$$ is a separable, mean-zero Gaussian process with intrinsic metric $$\rho$$, then for any $$\delta>0$$, $$\mathrm{E}\left[\sup _{\rho(s, t) \leq \delta}\left|W_s-W_t\right|\right] \lesssim \int_0^\delta \sqrt{\log D(\epsilon, T, \rho)} d \epsilon .$$ Where $$D$$ is the $$\epsilon$$-packing number. Furthermore, for $$J(\delta)$$ the integral on the right side, $$M^* := \mathrm{E} \sup _{\rho(s, t)<\delta} \frac{\left|W_s-W_t\right|}{J(\rho(s, t))}<\infty .$$ Consequently, $$\left|W_s-W_t\right|=O(J(\rho(s, t)))$$, uniformly in $$(s, t)$$ with $$\rho(s, t) \rightarrow 0$$, almost surely.

$$M^*$$ is the modulus of continuity of subgaussian processes. An increasing function $$\omega$$ such that $$\omega(0)=0$$ is called a modulus of continuity for the random process $$\left\{X_t\right\}_{t \in T}$$ on the metric space $$(T, d)$$ if there is a random variable $$K$$ such that $$X_t-X_s \leq K \omega(d(t, s)) \quad \text { for all } t, s \in T \text {. }$$ Evidently the function $$\omega$$ controls the "degree of smoothness" of $$t \mapsto X_t$$. To show that $$\omega$$ is a modulus of continuity, it clearly suffices to prove that $$K=\sup _{t, s \in T} \frac{X_t-X_s}{\omega(d(t, s))}<\infty \quad \text { a.s. }$$

In the case prosed in my question one can verify that $$D(\epsilon,T,\rho) \sim \frac{c}{\epsilon}$$, for $$c>0$$ constant, so the process has indeed sample continuous paths.