Gaussian Processes characterised by their co variance.

Question

Let $x_{t}$ be a Gaussian Process. Let $E[x_{t}]=0$, I have seen it said it many places that $x_{t}$ is then characterised by its co variance $E[x_{t}x_{t}]$.

Why is this so? What do they mean by characterised?

Answer 1

Comes from the fact that a Gaussian distribution can be fully defined by its first two moments. Moreover, in a Gaussian process the unconditional probability density function is just

$$ f_{X_t}(x) = \frac{1}{\sqrt{2\pi t}}e^{-x^2/2t} $$

with $\mathbb{E}[X_t] = 0$ and $\mathbb{V}{\rm ar}[X_t] = t = \mathbb{C}{\rm ov}[X_t, X_t]$. It is easy to show that

$$ \mathbb{C}{\rm ov}[X_t, X_s] = \min\{t, s\} $$