# Infinite-dimensional Gaussian measure vs finite-dimensional Wiener measure

I'm trying to figure out the connections between two contructions of Gaussian measure.

Let $$(U, \langle\cdot,\cdot\rangle_U)$$ be a seprable Hilbert space, and $$\mathcal{B}(U)$$ be the Borel sigma-algebra.

Definition 2.1, page 10 gave the following definition of Gaussian measure:

A measure $$\mu$$ on $$(U, \mathcal{B}(u), \mu)$$ is Gaussian if for all $$v\in U$$, there are $$m_v\in \mathbb{R}$$, $$\sigma_v\in \mathbb{R}^+$$ such that if $$\sigma_v>0$$,

$$\mu(u \in U\colon \langle v,u\rangle_U \in A) = \frac{1}{\sqrt{2\pi\sigma_v}}\int_A\exp(-\frac{(s-m_v)^2}{2\sigma_v^2})ds,$$

for all $$A\in\mathcal{B}(\mathbb{R})$$.

However, I recall from when we construct the Brownian motion. See, page 23 of https://www.springer.com/gp/book/9780387287201

Let $$U$$ be the space of continuous funcitons $$u(t)$$ on $$[0,1]$$, and $$D$$ be a cylindrical set

$$D = \{u\in U: (u(t_1), u(t_2), \ldots, u(t_N))\in E\in \mathcal{B}(\mathbb{R}^N)\}$$

Then the measure is defined by

$$\mu(D) = \int_E \prod_{i=1}^N (\frac{1}{\sqrt{2\pi(t_i-t_{i-1})}} \exp(-\frac{(h_i-h_{i-1})}{2(t_i-t_{i-1})}))dh_1dh_2\cdots dh_N$$

on $$0.

My questions is: Is it possible to define the infintite Gaussian measure on finite-dimensional distributions similar to we did for Brownian motion? I don't know how to find the connection bewtween the inner product $$\langle v,u\rangle_U$$ and a finite-dimensional distribution.

• I'm not 100% sure what you're looking for, but you might recall the fact that a set of random variables is jointly Gaussian iff every linear combination is Gaussian. Each $v \in U$ can be viewed as a random variable on the probability space $(U, \mu)$, identified with $u \mapsto \langle u, v \rangle$. With this in hand, you can determine the covariance between the random variables corresponding to $v_1, \dots, v_n$, and thus write down their joint distribution. – Nate Eldredge Mar 1 at 19:51
• So you end up with a formula for $\mu(\{u \in U : (\langle u, v_1 \rangle, \dots \langle u, v_n \rangle) \in E\})$ for any Borel $E \subset \mathbb{R}^n$ which will look similar to what you have for Brownian motion. – Nate Eldredge Mar 1 at 19:52
• I think the problem is that I do't understand why $\langle u,v \rangle$ having Gaussian law induces the Gaussian measure. I can't find the intuition/motivation. To be the finite-dimensional construction is quite straightforward. – Nathan Explosion Mar 1 at 20:01
• At the risk of self-promotion, I have some notes that you may find helpful. – Nate Eldredge Mar 1 at 20:26
• You say construction though, and neither of these statements are really constructions. The Wiener measure paragraph states a certain property for a measure to satisfy; it's still difficult work to prove that there exists a (unique) measure on $C([0,1])$ satisfying this property. Likewise, the Hilbert space paragraph just states a property, and in this case one needs some fairly strong conditions on the $m_v, \sigma_v$ in order to be able to prove there exists a measure with that property. – Nate Eldredge Mar 1 at 20:34

If $$U$$ is a Banach space, then the natural thing to do is replacing the $$\langle v,\cdot\rangle_U$$ by any continuous form $$\phi\in U^*$$. Then the definition of a Gaussian measure is a measure $$\mu$$ such that the pushforward $$\phi^*\mu$$ is Gaussian for all $$\phi\in U^*$$ (note that this works in finite dimension too). In general, there are subtleties about the mean being in $$U$$ or $$U^{**}$$, but for centred examples everything works well.
Your definition of Brownian motion implies that the every form $$\phi:u\mapsto\alpha_1u(t_1)+\cdots+\alpha_n\phi(t_n)$$ is Gaussian (it is in fact equivalent, using the above remark). Now one can show that the subspace of those forms, $$\mathrm{Vect}(\mathrm{ev}_t,t\in[0,1])$$, is actually dense in $$U^*$$ in the weak-$$*$$ topology. Let us show that this proves that $$\phi^*\mu$$ is Gaussian for all $$\phi\in U^*$$.
Let $$\phi\in U^*$$. Let $$\phi_n\in U^*$$ be a sequence of functions in $$\mathrm{Vect}(\mathrm{ev}_t,t\in[0,1])$$ that converges to $$\phi\in U^*$$ in the weak-$$*$$ topology (here we use the fact that the topology is metrisable, which is true since $$U$$ is second countable). We show that $$\phi^*\mu$$ is the weak limit of $$\phi_n^*\mu$$ (in the sense of probability measures on $$\mathbb R$$), and that the limit is Gaussian.
The first point is a matter of unfolding the definitions: for every bounded continuous $$f:\mathbb R\to\mathbb R$$, $$f\circ\phi_n(u)\to f\circ\phi(u)$$ for all $$u$$, and since $$f$$ is bounded we apply Lebesgue's dominated convergence theorem to get $$\int f(x)\phi_n^*\mu(x) = \int f\circ\phi_n(u)\mu(du) \to \int f\circ\phi(u)\mu(du) = \int f(x)\phi^*\mu(dx).$$ It remains to show that a weak limit of Gaussian measures stays Gaussian. This can be seen by showing first that the means $$m_n$$ and variances $$\sigma^2_n$$ have to be bounded, then up to extraction we can assume they converge to $$m$$ and $$\sigma$$, and at this point one can see that the characteristic function of the limit is the limit of the characteristic functions $$t\mapsto\exp(im_nt-\sigma^2_nt^2/2)$$, which is of course the characteristic function of $$\mathcal N(m,\sigma^2)$$