Integration of Gaussian process

Question

Let $\textbf{G}(t)$ be a zero-mean tight Gaussian process and $f(t)$ be a deterministic function. What theorem can be used to prove that $\int_0^\tau \textbf{G}(t)df(t)$ is a zero-mean Gaussian variable? Thanks.

Answer 1

I think establishing convergence in distribution of the sums whos limit defines your integral is relatively straightforward; just work with the characteristic function of the appropriate sums (and the fact that $G$ is Gaussian) then take the limit. If you're just going for convergence in distribution you won't have to worry about subtleties in commuting a limit with an integral. I'm also not aware of the meaning of what a "tight" process is but presumably it will tell you something about the first and second moment properties of $G(t)$ which you will need once you take the limit. You almost certainly need $f$ to have bounded variation I think because you will have to do Riemann-Stieltjes integrals of the type $\int \mu(t) \mathrm{d}f(t)$ and similar for the second moment.

EDIT: A little more details. We know that $ \operatorname{E}\left(\exp\left(i \ \sum_{\ell=1}^k (f(t_\ell)-f(t_{\ell-1})) \ \mathbf{G}(t_\ell)\right)\right) \\ = \exp \left(-\frac{1}{2} \, \sum_{\ell, j} \sigma(t_\ell,t_j) (f(t_\ell)-f(t_{\ell-1})) (f(t_j)-f(t_{j-1})) + i \sum_\ell \mu(t_\ell) (f(t_\ell)-f(t_{\ell-1}))\right).$

Since $G$ is Gaussian. Take the limit as $k \to 0$ (such that the partition of $t_\ell$ is refining), then the sums in the exponential turn into Riemann-Stieltjes integrals, the one with $\sigma$ is a double integral.