Recognizing Brownian Motion from Gaussian Finite-Dimensional Laws

From Covariance to Increment Laws

A covariance function can encode more than the variability of a process at individual times. For a centered Gaussian process, the covariance \(C(s,t)=\min(s,t)\) determines the distribution of every finite collection of values. In particular, it determines the joint distribution of increments. The crucial point is that increments over disjoint time intervals have zero covariance; Gaussianity then turns that calculation into independence.

Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, let the time index be \(t\ge 0\), and let the state space be \(S=\mathbb{R}\). A process \((X_t)_{t\ge0}\) is a centered Gaussian process if, for every finite set of times \(t_1,\ldots,t_m\), the vector \((X_{t_1},\ldots,X_{t_m})\) has a multivariate Gaussian distribution with mean vector zero. The Gaussian distribution is allowed to be degenerate. Assume throughout that \(\mathbb{E}[X_t]=0\) and \(\operatorname{Cov}(X_s,X_t)=\min(s,t)\) for all \(s,t\ge0\).

Here an interval of zero length is permitted: its increment is zero almost surely, with the degenerate distribution \(N(0,0)\). Pairwise disjoint intervals may share endpoints, as consecutive intervals in a partition do. Independence of a finite collection means that their joint distribution is the product of their marginal distributions. The theorem concerns finite collections; it does not assert path continuity from the covariance assumptions.

Proof. Since \(\mathbb{E}[X_0]=0\) and \(\operatorname{Var}(X_0)=\operatorname{Cov}(X_0,X_0)=\min(0,0)=0\), we have \(\mathbb{E}[X_0^2]=0\). A nonnegative random variable with expectation zero is zero almost surely, so \(X_0=0\) almost surely.

Take an interval \([a,b]\) with \(a\le b\). Its increment is a linear combination of two coordinates of a Gaussian vector, so it is Gaussian and has mean zero. Its variance is

Thus \(X_b-X_a\) has distribution \(N(0,b-a)\), including the zero-variance case \(a=b\).

Now consider two disjoint intervals ordered in time, \([a,b]\) and \([c,d]\), with \(a\le b\le c\le d\). Their increments are jointly Gaussian because they are linear combinations of the finite Gaussian vector \((X_a,X_b,X_c,X_d)\). Bilinearity of covariance gives

The last equality uses \(b\le c\le d\), so the four covariances are respectively \(b,b,a,a\). This calculation also covers \(b=c\), when the intervals share an endpoint.

For a finite collection of pairwise disjoint intervals, order them in time and apply the same calculation to every pair. The vector of increments is jointly Gaussian, and its covariance matrix is diagonal, with diagonal entries \(b_r-a_r\). The characteristic function of a centered Gaussian vector with covariance matrix \(\Sigma\) is \(\exp(-u^{\mathsf T}\Sigma u/2)\). When \(\Sigma\) is diagonal, this expression is the product of the characteristic functions of its coordinates. By uniqueness of characteristic functions, the coordinates are independent. This argument includes zero-length intervals: a coordinate with variance zero is constant almost surely and is independent of the others. We have proved the increment claim.

Finally, standard Brownian motion is characterized by starting at zero almost surely, having independent increments with \(X_t-X_s\sim N(0,t-s)\) for \(0\le s\le t\), and having almost surely continuous paths. The first two properties have just been proved. If the additional continuity assumption holds, all defining properties are satisfied. \(\square\)

Working with the Covariance Calculation

Why the Gaussian Assumption Matters

The covariance calculation alone proves that increments on disjoint intervals are uncorrelated. The step from uncorrelated to independent uses joint Gaussianity. Without that assumption, zero covariance does not generally imply independence. It is therefore important to check both parts of the argument: the covariance function gives the zero cross-covariances, and the Gaussian process assumption makes those cross-covariances sufficient.

Another important distinction is between finite-dimensional laws and path properties. The theorem identifies the distribution of every finite collection of increments and, with continuity added, identifies a standard Brownian motion. The discontinuous process in the third example has precisely the same finite-dimensional distributions as Brownian motion. No check involving only finitely many fixed times can rule out that kind of path modification.

Common mistake. When computing the covariance of two increments, it is incorrect to keep only the covariance of their right endpoints and subtract the covariance of their left endpoints. For the intervals \([1/4,3/4]\) and \([3/4,1]\), that shortcut gives \(\operatorname{Cov}(X_{3/4},X_1)-\operatorname{Cov}(X_{1/4},X_{3/4})=3/4-1/4=1/2\), incorrectly suggesting dependence. The increment covariance has four terms, because each increment has two terms. Including all four gives

These increments are jointly Gaussian, so the corrected calculation establishes their independence. The practical rule is to expand both increments before applying the covariance function.