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 $0<t_1<t_2<\cdots<t_N<1$.

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.

  • $\begingroup$ 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. $\endgroup$ Mar 1, 2021 at 19:51
  • $\begingroup$ 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. $\endgroup$ Mar 1, 2021 at 19:52
  • $\begingroup$ 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. $\endgroup$
    – null
    Mar 1, 2021 at 20:01
  • 2
    $\begingroup$ At the risk of self-promotion, I have some notes that you may find helpful. $\endgroup$ Mar 1, 2021 at 20:26
  • 1
    $\begingroup$ 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. $\endgroup$ Mar 1, 2021 at 20:34

1 Answer 1


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 $\mathcal N(m,\sigma^2)$

  • $\begingroup$ I guess this might not be what you are interested in. In hindsight, I show that Brownian motion is an example of a Gaussian measure using your definition, and you're asking for a construction à la Brownian motion for all Gaussian measure. Is that it? $\endgroup$
    – Pierre PC
    Mar 1, 2021 at 20:39
  • $\begingroup$ My answer is more or less the exercise that Nate Eldredge suggests in his comments. $\endgroup$
    – Pierre PC
    Mar 1, 2021 at 20:40

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