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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.

I think the idea is because Gaussian Processes are only uniquely defined up to the mean and covariance and therefore not unique (is this correct?), you can 'choose' to define your process to be the version/modification that is sample continuous, but is there a more formal way of saying this?

In general (i.e. not necessarily just for Gaussian Processes) when is having a sample continuous modification equivalent to sample continuity? Clearly if $X_t$ was sample continuous, you can easily find a modification $Y_t$ that is sample discontinuous (see for example https://math.stackexchange.com/questions/2608352/discontinuous-brownian-motion).

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  • $\begingroup$ You may have answered your own question in your final paragraph. The sample discontinuous process $Y_t$ admits a sample-continuous modification (namely, $X_t$), but is not itself sample continuous. $\endgroup$ Jan 22 at 18:58
  • $\begingroup$ Is there a condition which ensures the existence of a sample continuous modification is equivalent to the process itself being sample continuous? I suspect separability might work but I'm not sure and I cannot find a source which explicitly states this. $\endgroup$
    – 123 456
    Jan 23 at 7:27
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I think your suspicion is correct. Suppose $X=(X_t)_{t\ge 0}$ is separable and $Y=(Y_t)_{t\ge 0}$ is a continuous modification of $X$. Thus there is a countable dense set $D\subset[0,\infty)$ and a null set $\Omega_0$ that witness the separability of $X$. Let $\Omega_1$ be a null set such that $\omega\notin\Omega_1$ implies that $X_t(\omega) = Y_t(\omega)$ for all $t\in D$. Finally put $\Omega_2=\Omega_0\cup \Omega_1$. Then for $\omega\notin\Omega_2$ and $t\ge 0$, we have $$ \eqalign{ \lim_{s\to t}X_s(\omega) &=\lim_{s\to t,s\in D} X_s(\omega)\cr &=\lim_{s\to t,s\in D} Y_s(\omega)\cr &=Y_t(\omega),\cr } $$ the very existence of the initial limit being part of the assertion. The existence of this limit implies that $X_t(\omega) =\lim_{s\to t} X_s(\omega)$. It follows that $X$ and $Y$ coincide on $\Omega_2$, and in particular $X$ is a.s. path continuous

Special case: If $X$ is cadlag and admits a continuous modification, then it will be sample continuous. Ditto for caglad.

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  • $\begingroup$ Thank you, that was very helpful. I'm guessing $X$ is cadlag or caglad is an easy way to ensure $\lim_{s\to t} X_s(\omega)$ exists? $\endgroup$
    – 123 456
    Jan 24 at 19:11
  • $\begingroup$ Or at least a one-sided limit, yes. $\endgroup$ Jan 24 at 23:26
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Usually a stochastic process is described via finite-dimensional distributions (In the Gaussian case, these are determined by mean and covariance). But the finite-dimensional distributions do not determine the process uniquely, as the example you cite shows. If there is a continuous version, then that is unique (and preferred). More generally, if there is no continuous version, one needs to settle for a cadlag version of the process, see [1], [2]

[1] https://en.wikipedia.org/wiki/C%C3%A0dl%C3%A0g

[2] Billingsley, P. Convergence of Probability Measures. New York: Wiley.

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  • $\begingroup$ Thank you. Do you know of any citable examples of this being explicitly said? That you can somehow 'pick' the continuous version/modification of a process that's only defined in terms of its mean square properties (such as a Gaussian Process defined in terms of its mean and covariance) without losing generality of whatever it is you're trying to do. $\endgroup$
    – 123 456
    Jan 25 at 11:43

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