Let $X$ be a topological affine space. A Gaussian measure on $X$ is characterized by the property that its finite-dimensional projections are multivariate Gaussian distributions.
Is there a direct characterization of a Gaussian measure which does not rely on finite-dimensional projections? This definition is analogous to describing a duck as the animal whose shadows look like $2$-dimensional ducks. The definition is sufficient for doing analysis, but to me it misses the essence of what a Gaussian measure is as a mathematical object in and of itself.
Here is the precise definition of a Gaussian measure that I usually work with, which relies on the fact that Gaussians are entirely described by their covariance structure.
For $X$ a topological affine space as above, let $X^*$ denote its dual space of affine functionals. The dual space is a linear space, since there there is a natural zero functional $0 \in X^*$.
Let $K : X^* \to X$ be a continuous affine operator which is symmetric and non-negative-definite. i.e., $f'(Kf) = f(Kf')$ and $f(Kf) \ge 0$ for all $f, f' \in X^*$. Let $m_K := K(0)$ denote the image of the zero functional.
There is a unique Gaussian measure $P_K$ on $X$ with mean point $m_K \in X$ and covariance operator $K : X^* \to X$. That is, if $\pi : X \to \mathbb R^n$ denotes a finite-dimensional projection, then the push-forward measure $\pi_* P_K := P_K \circ \pi^{-1}$ is an $n$-dimensiona Gaussian distribution with mean vector $\pi(m_K) \in \mathbb R^n$ and covariance matrix $\pi K \pi^*$, where $\pi^* : (\mathbb R^n)^* \to K^*$ denotes the formal adjoint operator.
Furthermore, the structure theorem for Gaussian measures states that all Gaussian measures arise in this way. Consequently, we may parametrize the space of Gaussian measures by the space $\mathcal K(X)$ of symmetric, non-negative operators from $X^*$ to $X$.
This provides a weak answer to the question stated at the top of this post: yes, Gaussian measures can be directly characterized by their covariance structure. Consequently, here is the stronger form of my question:
- Is there a geometric description of the space $\mathcal K(X)$ of Gaussian covariance operators?
For example, is the space $\mathcal K(X)$ an infinite-dimensional manifold? What is its symmetry group?
Edit: My above post implicitly defines the covariance form incorrectly. In the affine setting, the covariance form is defined by $\langle f', f \rangle_K := f'(Kf) – f'(0)$, and the conditions of symmetry and non-negative-definiteness are $\langle f', f \rangle_K = \langle f, f' \rangle_K$ and $\langle f, f \rangle_K \ge 0$, respectively. It is an easy exercise to verify that this defines a bilinear form on the dual space $X^*$ of affine functionals.
Best Answer
You could alternatively try defining Gaussian measures as $2$-stable distributions. This does remove any reliance on finite dimensional projections, and even removes reference to topology. Let $V$ be a measurable vector space (by which, I mean a real vector space $V$ with sigma-algebra $\mathcal{F}$ with respect to which addition and multiplication are measurable).
A probability measure $\mu$ on $V$ is then a centered Gaussian iff, for any independent pair $X,Y$ of $V$-valued random variables each with measure $\mu$, then $aX+bY$ also has measure $\mu$ for all $a,b\in\mathbb{R}$ with $a^2+b^2=1$.
If $A$ is a (measurable) affine space with underlying vector space $V$, then we could similarly say that $\mu$ is Gaussian iff there exists an $m\in A$ such that $X-m$ is a centered Gaussian on $V$ for a random variable $X$ with measure $\mu$.
Several facts should then follow quickly from this:
This seems to give an answer first paragraph of the question, and does not depend on projections. I'm not sure if it is going in the direction that the question was asking for though, as it says nothing about the stronger form of the question further down and didn't mention covariance operators at all.