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

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.

• Gaussian processes are characterized by their covariance function Cov$(X_t,X_s)\,.$ That of BM is $t\wedge s\,.$ This answer contains three other important covariance functions. Commented Feb 29 at 18:04

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.
• Thank you very much! What do you mean that we may assume $(X_{t_1},\cdots,X_{t_k})$ to be Gaussian rather than jointly Gaussian? Since it's a multidimensional vector wouldn't it being Gaussian be synonymous with it being jointly Gaussian? Commented Feb 29 at 18:25