Constructing Brownian Motion from Consistent Gaussian Laws

From Finite-Time Laws to a Process

A stochastic process is a family of random variables on one probability space, indexed here by continuous time \(t\ge0\). One way to construct such a family is to specify the joint distribution of its values at every finite collection of times, then check that these distributions fit together. For Brownian motion, the proposed values are jointly Gaussian, have mean zero, and have covariance \(\min(s,t)\). The essential work is to check that these prescriptions define valid finite-dimensional probability laws and that taking a marginal at fewer times gives the law prescribed for those times.

We use the state space \(S=\mathbb{R}\) and the time index set \(I=[0,\infty)\). For a finite ordered list \(t_1,\ldots,t_m\) of nonnegative times, define the \(m\times m\) matrix \(C\) by \(C_{ij}=\min(t_i,t_j)\). We propose that the random vector \((X_{t_1},\ldots,X_{t_m})\) have the centered Gaussian distribution with covariance matrix \(C\). A centered Gaussian vector with covariance \(C\), including a possibly degenerate one, is specified by its characteristic function \(u\mapsto\exp(-\tfrac12 u^\mathsf{T}Cu)\), for \(u\in\mathbb{R}^m\).

Before calling the proposed distributions probability laws, we must show that \(C\) is positive semidefinite. This is necessary for a covariance matrix. It is also sufficient to define a centered Gaussian vector: by the spectral theorem, a symmetric positive semidefinite matrix can be written \(C=AA^\mathsf{T}\); if \(Z\) has independent standard normal coordinates, then \(AZ\) has the required centered Gaussian distribution and covariance \(C\). This argument applies even when \(C\) is singular.

Validity of the Gaussian Covariance Matrices

Proof. Symmetry follows from \(\min(t_i,t_j)=\min(t_j,t_i)\). Fix any real numbers \(a_1,\ldots,a_m\). For \(u\ge0\), the indicator \(\mathbf{1}_{[0,t_i]}(u)\) equals one precisely when \(u\le t_i\). The length of the intersection of \([0,t_i]\) and \([0,t_j]\) is \(\min(t_i,t_j)\), so \[ \min(t_i,t_j)=\int_0^\infty \mathbf{1}_{[0,t_i]}(u)\mathbf{1}_{[0,t_j]}(u)\,du. \] Therefore \[ \sum_{i=1}^m\sum_{j=1}^m a_iC_{ij}a_j =\int_0^\infty\left(\sum_{i=1}^m a_i\mathbf{1}_{[0,t_i]}(u)\right)^2\,du\ge0. \] The integral is finite because its integrand vanishes for \(u>\max_i t_i\), and it is nonnegative because it is the integral of a square. This proves positive semidefiniteness. If every \(t_i=0\), the integral and the matrix are both zero, so the degenerate case is included. \(\square\)

Thus every finite list has a well-defined centered Gaussian law. In particular, the specification is meaningful at \(t=0\): its variance is \(\min(0,0)=0\), so \(X_0=0\) with probability one under that one-time law. Repeated times also produce valid, possibly singular covariance matrices; repeated coordinates are required to agree rather than to behave like independent observations.

Why the Finite-Time Laws Are Consistent

For a centered Gaussian vector with covariance \(C\), the characteristic function at \(u\in\mathbb{R}^m\) is \(\exp(-\tfrac12 u^\mathsf{T}Cu)\). This formula gives a direct way to check marginals: to retain only selected coordinates, set the other coordinates of \(u\) equal to zero. The resulting quadratic form uses just the covariance entries of the retained coordinates.

Proof. First retain coordinates indexed by a subset \(J\subseteq\{1,\ldots,m\}\). In the characteristic function of the full vector, set \(u_i=0\) for each \(i\notin J\). The exponent becomes \[ -\tfrac12\sum_{i\in J}\sum_{j\in J}u_i\min(t_i,t_j)u_j, \] which is exactly the characteristic-function exponent prescribed for the retained times. Since characteristic functions determine probability distributions, the marginal law is the specified law on those times. This includes deleting all coordinates, whose law is the unique probability measure on the one-point space \(\mathbb{R}^0\).

Next, permute the time coordinates by a permutation \(\sigma\). The covariance entry in the new row \(i\), column \(j\) is \(\min(t_{\sigma(i)},t_{\sigma(j)})\), exactly the entry obtained by applying the same permutation to the original covariance matrix. Substituting the permuted coordinates into the characteristic-function formula therefore gives the distribution of the correspondingly permuted vector. Adding a coordinate does not change the marginal on the original coordinates, by the deletion calculation just given. Hence the family satisfies the stated consistency requirements. \(\square\)

Applying the Kolmogorov Extension Theorem

Here \(I=[0,\infty)\), \(S=\mathbb{R}\) with its Borel sets, and the finite-dimensional distributions are the centered Gaussian laws just checked. The state space \(\mathbb{R}\) is standard Borel, and the propositions establish that the family is well-defined and consistent. The extension theorem therefore gives a probability measure on \(\mathbb{R}^{[0,\infty)}\). On this canonical space, define \(X_t(\omega)=\omega(t)\), the coordinate map at time \(t\). With the probability measure supplied by the theorem, \((X_t)_{t\ge0}\) is a stochastic process whose every finite collection of coordinates has the specified Gaussian law.

This proves the existence of a process with the Brownian finite-dimensional distributions. It does not, by itself, prove that its sample paths are continuous. The extension theorem controls the joint distributions at finitely many fixed times; a statement about continuity concerns the behavior of a path across infinitely many times. Thus constructing the process with these laws and establishing a continuous-path version are distinct tasks.

The covariance prescription also explains the familiar increment structure. For \(0\le s\le t\), the increment \(X_t-X_s\) is Gaussian because it is a linear combination of a Gaussian vector. Its mean is zero and its variance is \[ \operatorname{Var}(X_t-X_s)=t+s-2s=t-s. \] For disjoint successive intervals \([s,t]\) and \([u,v]\), where \(s\le t\le u\le v\), expanding the covariance using \(\operatorname{Cov}(X_a,X_b)=\min(a,b)\) gives zero covariance between \(X_t-X_s\) and \(X_v-X_u\). These increments are jointly Gaussian, so zero covariance implies independence. This is a consequence of the finite-dimensional Gaussian laws; it does not supply path continuity.

A Covariance Pitfall and What the Construction Delivers

Common mistake. Replacing \(\min(s,t)\) by \(\max(s,t)\) may seem plausible because both expressions are symmetric in the times. But at times \(1\) and \(3\), that incorrect choice gives \[ C_{\mathrm{wrong}}=\begin{bmatrix}1&3\\3&3\end{bmatrix}. \] Its determinant is \(1\cdot3-3\cdot3=-6\), so it is not positive semidefinite. For example, the quadratic form at \((a_1,a_2)=(3,-1)\) equals \(9-18+3=-6<0\). A covariance matrix cannot have a negative quadratic form. By contrast, the correct matrix has off-diagonal entries \(\min(1,3)=1\), and its quadratic form is nonnegative by the integral proof above. The failure is not merely a different model: the proposed wrong entries cannot be the covariance of any random vector.

The construction has three logically separate checks. First, each proposed covariance matrix is positive semidefinite, so its Gaussian law exists, possibly in a degenerate form. Second, the distributions agree under permutations and marginalization, so no finite collection contradicts another. Third, the extension theorem turns this consistent family into one process on a common probability space. Keeping these checks separate prevents two common overstatements: a covariance formula alone is not enough until positivity is verified, and existence of the finite-dimensional laws alone does not assert regularity of sample paths.