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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

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2 Answers 2

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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.

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  • $\begingroup$ Is there a book that includes the above steps in your proof? $\endgroup$ Jul 30, 2019 at 11:40
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    $\begingroup$ @ErdoganCEVHER The normal distribution is one of the distributions and discussed in most books on probability theory... shouldn't be much trouble to find a discussion of its Fourier transform. Difficult to give a precise recommendation without knowing anything about your background. Also you find plenty of material online; just search for "multivariate normal distribution characteristic function" or something like this... $\endgroup$
    – saz
    Jul 30, 2019 at 13:14
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Because these determine the characteristic functions of the process.

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