What is the difference between a Gaussian process and Brownian motion?

Question

A Gaussian process is defined to be a stochastic process $X_t$ such that for every finite collection $(t_1, \ldots, t_k)$, the random variable $(X_{t_1}, \ldots, X_{t_k})$ is jointly Gaussian.

A Brownian motion is a stochastic process $B_t$ such that

1. $B_0 = 0$ a.s.
2. $B_t$ has independent increments
3. $B_t - B_s \sim \mathcal{N}(0, t-s)$.

Additionally it can be shown that every Brownian motion is a Gaussian process. It appears to me that these two are almost describing the same thing and I cannot figure out where the difference is.

Are these two objects the same? If not, what is an example of a Gaussian process that is not a Brownian motion.

Answer 1

A Brownian motion is a specific type of Gaussian process, but it is not the only one. Brownian motion can be characterized as a centered Gaussian process $X$ having the covariance function $\Gamma(s,t) := \text{Cov}(X_s,X_t) = \min(s,t)$, but any positive semidefinite function $\Gamma$ can be used to define a centered Gaussian process. One easy way to define another Gaussian process is to just start from a Brownian motion $W$ and define, for example, $X_t := 2 W_t$ or $X_t := W_{2t}$. Both of these are Gaussian processes, but they don't fit the requirement that $X_t - X_s \sim N(0,t-s)$ or $\text{Cov}(X_s,X_t) = \min(s,t)$.

Another commonly used Gaussian process is the Brownian Bridge defined on $[0,1]$ by $X_t := B_t - t B_1$. This is again a centered Gaussian process, but with $\text{Cov}(X_s,X_t) = s(1-t)$ for $s \le t$.

For a somewhat trivial example, we could also take the constant process $X_t = X_0 \sim N(0,1)$.

If you are willing to accept the existence of uncountably many independent random variables, we could also define a process $X$ by $X_t$ being i.i.d. $N(0,1)$ random variables.