A stochastic process $X_t$ is called Gaussian if the random vectors $(X_{t_1},…,X_{t_n})$
are multivariate normal.
Why are the finite dimensional distributions of a Gaussian process determined by its mean and covariance functions:
$m(t) = EX_t$,
$\rho(s,t) = E[(X_s-m(s))(X_t – m(t))^T]$?
Thank you
Best Answer
Any distribution is uniquely determined by its Fourier transform. If $Y \sim N(m,C)$ is a normal distributed random variable with mean vector $m$ and covariance matrix $C$, then the Fourier transform $\phi(\xi) := \mathbb{E}e^{\imath \, Y\xi}$ equals
$$\phi(\xi) := \exp \left( \imath \, m \cdot \xi - \frac{1}{2} \langle \xi, C \xi \rangle \right).$$
In particular, we see that the Fourier transform is uniquely determined by $m$ and $C$. Using this representation, it is not difficult to see that
$$m = \mathbb{E}Y \qquad \qquad C = (\text{cov}(Y_i,Y_j))_{i,j}.$$
If we apply this to the random vector $Y:= (X_{t_1},\ldots,X_{t_n})$, the claim follows.