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If $X=(X_t)_{t \geq 0}$ is a one-dimensional Gaussian real process I know that it means that $$ (X_{t_1}, X_{t_2}, \ldots , X_{t_n}) $$ has $n$-dimensional Gaussian distribution for any $0 \leq t_i< t_{i+1}$ for $i=1,2,....,n-1$ and $n \geq 1$. What if $(X,Y)$ is a two-dimensional Gaussian process. Does it mean that $$ (X_{t_1}, X_{t_2}, \ldots , X_{t_n}, Y_{t_1}, Y_{t_2}, \ldots , Y_{t_n}) $$ has $2n$-dimensional Gaussian distribution for any $0 \leq t_i< t_{i+1}$ for $i=1,2,....,n-1$ and $n \geq 1$?

Note An answer is given in the post Definition of a $\mathbb{R}^d$-valued Gaussian process.

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  • $\begingroup$ I think a two-dimensional Gaussian process is more commonly understood to be a process where the index set is two-dimensional. This is contrasted with your one-dimensional process indexed by $\mathbb{R}$. $\endgroup$ May 3, 2021 at 14:18
  • $\begingroup$ I don't agree. Gaussian processes on higher dimensional index sets are rather referred to as Gaussian fields. An $n$-dimensional Gaussian process, clearly is a Gaussian process valued $\mathbb{R}^{n}$. $\endgroup$
    – Tobsn
    May 3, 2021 at 17:15
  • $\begingroup$ Now in order to also provide a hint towards a possible answer. One way to characterise a multivariate Gaussian RV $X$ is by imposing that the inner product $a\cdot X$ shall be a (scalar) Gaussian for every $a\in\mathbb{R}^{n}$. Can you cook up an answer yourself now? $\endgroup$
    – Tobsn
    May 3, 2021 at 17:17
  • $\begingroup$ @Tobsn Thank you. I know that definition of multivariate Gaussian random variables. I think I found an answer to my question in a related post, math.stackexchange.com/questions/3313305/… $\endgroup$
    – J. Goles
    May 5, 2021 at 5:39
  • $\begingroup$ It's not a good answer. I may repeat myself: try to cook up a reasonable definition using the same philosophy as used in the definition of a multivariate Gaussian RV. $\endgroup$
    – Tobsn
    May 5, 2021 at 6:03

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