FactorizableCovariance

Canonical Representation of Centered Gaussian Process with Factorizable Covariance

Definitions

1. Let $(T, \Sigma)$ be a measurable space.
2. Let $\mu$ be a positive measure on $(T, \Sigma)$.
3. Let $f: T \to \mathbb{R}$ be a measurable function.

Theorem

Let $X = {X(t): t \in T}$ be a centered Gaussian process on a probability space $(\Omega, \mathcal{F}, \mathbb{P})$ with covariance function $K(s,t) = f(s)f(t)$ for $s,t \in T$, where $f$ is square-integrable with respect to $\mu$. Then there exists a Gaussian random measure $W$ on $(T, \Sigma)$ with variance measure $\mu$ such that:

$$X(t) = \int_T f(s) W(ds) \quad \text{(a.s.)}$$

for all $t \in T$, where the integral is in the $L^2(\Omega, \mathcal{F}, \mathbb{P})$ sense.

Lemma

For the Gaussian process $X$ defined in the theorem, we have:

$$E[X(s)X(t)] = \int_T f(s)f(t) \mu(du) = f(s)f(t)$$

Proof of Lemma

$$\begin{align*} E[X(s)X(t)] &= E\left[\left(\int_T f(u) W(du)\right)\left(\int_T f(v) W(dv)\right)\right] \\\ &= \int_T \int_T f(u)f(v) E[W(du)W(dv)] \\\ &= \int_T f(u)f(v) \mu(du) \quad \text{(by properties of $W$)} \\\ &= f(s)f(t) \end{align*}$$

This completes the lemma.

Proof Sketch of Theorem

1. Define $W$ as a Gaussian random measure on $(T, \Sigma)$ with variance measure $\mu$.
2. For each $t \in T$, define $Y(t) = \int_T f(s) W(ds)$.
3. Show that $Y$ is a centered Gaussian process with covariance function $K(s,t) = f(s)f(t)$.
4. Use the uniqueness of finite-dimensional distributions for Gaussian processes to conclude that $X$ and $Y$ have the same distribution.