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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?

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    $\begingroup$ 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. $\endgroup$
    – Surb
    Commented Jan 21, 2021 at 13:54
  • $\begingroup$ 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. $\endgroup$ Commented Jan 21, 2021 at 14:26
  • $\begingroup$ @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. $\endgroup$
    – 123 456
    Commented Jan 21, 2021 at 18:36
  • $\begingroup$ @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 $\mathbb{R}^{[0,\infty)}$. $\endgroup$ Commented Jan 22, 2021 at 10:23

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