# Regarding sample continuity of Gaussian Processes

Suppose we have a Gaussian Process $$X_t$$ on $$\mathbb{R}^n$$ with mean function $$m(t)$$ and covariance function $$K(t,s)$$. Then is $$X_t$$ being sample continuous (i.e. the sample paths of $$X_t$$ are almost surely continuous everywhere) equivalent to $$X_t$$ having a modification that is sample continuous (via for example Kolmogorov's continuity theorem)? I've seen sources use them interchangeably but I don't quite understand why?

If there is a difference, does there exist a condition to ensure $$X_t$$ itself is sample continuous, and not just having a sample continuous modification? If so, can those conditions be purely expressed in terms of the moments of $$X_t$$ (like with Kolmogorov's continuity theorem)?

If $$X_t$$ is sample continuous, does there always exist a modification (which has the same mean and covariance as $$X_t$$) that is not sample continuous?

• A Gaussian process is continuous by definition... Or in other word, if $(X_t)$ is s.t. all finite vector is a Gaussian vector, then there is a modification of (X_t)$that is continuous. The continuous modification will be called "Gaussian process", not the non-continuous modification. – Surb Commented Jan 21, 2021 at 13:54 • Just as a remark on the second question: If$X_t$is sample continuous and Gaussian and$\tau$is any positive, random variable with a continuous distribution which is independent of$X$, then defining$Y_t=X_t$for$t\neq \tau$and$0$for$t=\tau$makes$Y$a version of$X$(it's easy to see that they have the same finite dimensional distributions) and$Y$almost surely has discontinuous sample paths. Commented Jan 21, 2021 at 14:26 • @WoolierThanThou Is$Y$a separable stochastic process? Basically I'm trying to understand the difference between results like Kolmogorov's continuity theorem (which only shows there's a continuous modification), and ones like equation (2) on page 2 in this paper core.ac.uk/download/pdf/82123257.pdf that just says the original process is sample continuous. Commented Jan 21, 2021 at 18:36 • @123456 Y is very much not separable. But ultimately, whether your process has continuous sample paths ultimately just comes down to how you specifically constructed it. You cannot in general tell whether a process has continuous sample paths from its distribution alone. This is closely related to the fact that the space of continuous functions is not a measurable subset of$\mathbb{R}^{[0,\infty)}\$. Commented Jan 22, 2021 at 10:23