FactorizableCovariance - crowlogic/arb4j GitHub Wiki
Canonical Representation of Centered Gaussian Process with Factorizable Covariance
Definitions
- Let $(T, \Sigma)$ be a measurable space.
- Let $\mu$ be a positive measure on $(T, \Sigma)$.
- 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
- Define $W$ as a Gaussian random measure on $(T, \Sigma)$ with variance measure $\mu$.
- For each $t \in T$, define $Y(t) = \int_T f(s) W(ds)$.
- Show that $Y$ is a centered Gaussian process with covariance function $K(s,t) = f(s)f(t)$.
- Use the uniqueness of finite-dimensional distributions for Gaussian processes to conclude that $X$ and $Y$ have the same distribution.