Gaussian process

Remarks

Equivalently, every finite linear combination $\sum_{i=1}^{k} c_i X_{t_i}$ with real $c_i$ is normal or constant.

The mean function is $m(t) = E[X_t]$ and the covariance function is $K(s, t) = \operatorname{Cov}(X_s, X_t)$. They determine the finite-dimensional distributions, hence the law of $X$. The function $K$ is positive semidefinite: the matrix $(K(t_i, t_j))_{i, j}$ is positive semidefinite for all $t_1, \dots, t_k$. Conversely, by the Kolmogorov extension theorem, for every function $m$ and every symmetric positive semidefinite $K$ there is a Gaussian process with mean $m$ and covariance $K$.

Example: the Wiener process is a Gaussian process on $[0, \infty)$ with $m = 0$ and $K(s, t) = \min(s, t)$. Check: $(W_{t_1}, \dots, W_{t_k})$ for $t_1 < \cdots < t_k$ is a linear image of the independent normal increments $W_{t_1}, W_{t_2} - W_{t_1}, \dots$, hence multivariate normal.

Non-example: let $Z$ be standard normal and let $\varepsilon$ be independent of $Z$ with $P(\varepsilon = 1) = P(\varepsilon = -1) = 1/2$. Let $X_1 = Z$ and $X_2 = \varepsilon Z$ on $T = \set{1, 2}$. Each $X_t$ is standard normal, but $X_1 + X_2 = (1 + \varepsilon) Z$ equals $0$ with probability $1/2$ and is not constant, so it is neither normal nor constant. Thus $(X_1, X_2)$ is not multivariate normal, and $X$ is not a Gaussian process.