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(1) How can we prove that the integral i.e. $\int_{a}^{b} X(t) dt$ (or any linear functional) of a Gaussian process $X(t)$ has Gaussian distribution?

(2) And how can we find that distribution in the case of integral?

Appreciate any insight, thanks.

Edit: (per comment) $$\lim_{\Delta t \to 0}\sum_{a}^{b}X(t)\Delta t$$

triomphe
  • 3,848

1 Answers1

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As @Did wrote in his comment we can derive it using the Riemann sum, here is a much longer derivation:
Let $X(t)$ a Gaussian Process: $$X(t) = GP(m(t), k(t,t'))$$

Then, writing the integral as a Riemann sum: $$\int_{a}^{b} X(t) dt = \lim_{\Delta t \to 0}\sum_{a}^{b}X(t)\Delta t$$

By the affine property of Multivariate Gaussian random variables we have that: $$X = \mathcal{N}(\mu, \Sigma) \implies AX = \mathcal{N}(A\mu, A\Sigma A')$$

Hence, $$\int_{a}^{b} X(t) dt = \lim_{\Delta t \to 0}\sum_{a}^{b}GP(\Delta t \ m(t), \Delta t\ k(t,t')\ \Delta t')$$

Also, we know that the sum of two Gaussians random variables is Gaussian:
If $X \sim \mathcal{N}(\mu_X, \sigma_X^2), Y \sim \mathcal{N}(\mu_Y, \sigma_Y^2)$ and $Z = X + Y$,
then $Z \sim \mathcal{N}(\mu_X + \mu_Y, \sigma_X^2 + \sigma_Y^2)$.

Hence, we obtain:
$$\int_{a}^{b} X(t) dt = \lim_{\Delta t \to 0}GP(\sum_{a}^{b}\Delta t \ m(t), \sum_{a}^{b}\sum_{a}^{b}\Delta t\ k(t,t')\ \Delta t')$$

Which can also be written as,
$$\int_{a}^{b} X(t) dt = GP(\int_{a}^{b}m(t)\ dt, \int_{a}^{b}\int_{a}^{b}k(t,t')\ dtdt')$$