Finite-Dimensional Laws and the Brownian Covariance Kernel

From the Covariance Kernel to a Joint Law

A covariance formula specifies how pairs of random variables vary together, but by itself it does not generally determine their joint distribution. Brownian motion is special: its values at any finite list of times have a multivariate Gaussian law, and the covariance kernel determines that law completely. This lets us pass from the pathwise definition of Brownian motion to explicit characteristic functions and, when the covariance matrix is nonsingular, a density.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\). Thus \(B_0=0\), its paths are continuous, and for \(0\le s<t\), \(B_t-B_s\) is independent of \(\mathcal{F}_s\) and has distribution \(N(0,t-s)\). In particular, Brownian motion is centered and has variance \(\mathbb{E}[B_t^2]=t\). No stochastic integral is needed here.

For \(s\le t\), write \(B_t=B_s+(B_t-B_s)\). The increment is independent of \(B_s\), is centered, and \(B_s\) is centered. Therefore \(\mathbb{E}[B_sB_t]=\mathbb{E}[B_s^2]=s\). Interchanging \(s\) and \(t\) when \(t<s\) gives the Brownian covariance kernel

This pairwise calculation does not alone establish that a vector of Brownian values is Gaussian. To show that, we use the independent normal increments in the definition and calculate the vector’s characteristic function.

The Finite-Dimensional Gaussian Law

Proof. Set \(t_0=0\), \(\Delta t_k=t_k-t_{k-1}>0\), and \(\Delta B_k=B_{t_k}-B_{t_{k-1}}\). These increments are independent, with \(\Delta B_k\sim N(0,\Delta t_k)\), and \(B_{t_j}=\sum_{k=1}^j\Delta B_k\). For any \(u=(u_1,\ldots,u_n)^{\mathsf T}\), rearranging this finite sum gives \(u^{\mathsf T}X=\sum_{k=1}^n(\sum_{j=k}^n u_j)\Delta B_k\). The characteristic function of a centered normal variable of variance \(v\) at argument \(a\) is \(\exp(-a^2v/2)\). Independence of the increments therefore gives

Expanding the squares, the coefficient of \(u_i^2\) in the sum in the exponent is \(\sum_{k=1}^i\Delta t_k=t_i\). For \(i\ne j\), the coefficient of \(u_i u_j\) is \(2\sum_{k=1}^{\min(i,j)}\Delta t_k=2t_{\min(i,j)}\), because the product occurs once in each of the two orders in each square. Equivalently, the matrix in the quadratic form has entries \(C_{ij}=t_{\min(i,j)}=\min(t_i,t_j)\); its off-diagonal entries occur twice when \(u^{\mathsf T}Cu\) is expanded. This proves the displayed characteristic function and the stated covariance matrix. By the standard uniqueness theorem for characteristic functions, this characteristic function determines the joint law. \(\square\)

The same covariance matrix can also be read directly from the kernel. It is positive definite for strictly increasing positive times: the representation as cumulative sums of independent normal increments has an invertible triangular change of variables, and each increment has positive variance. That representation also supplies a useful density formula.

Proof. The vector of increments \((\Delta B_1,\ldots,\Delta B_n)\) has density equal to the product of its independent normal densities. The map from increments \(y_k\) to values \(x_j=\sum_{k=1}^j y_k\) is triangular with diagonal entries \(1\), so its determinant is \(1\); its inverse is \(y_k=x_k-x_{k-1}\), with \(x_0=0\). The change-of-variables formula therefore gives the displayed density, including its normalizing factor. In matrix form \(X=A Y\), where \(Y\) is the increment vector and \(A\) is the lower triangular matrix of ones on and below the diagonal. Since \(\det A=1\) and \(\operatorname{Cov}(Y)\) is diagonal with entries \(\Delta t_k\), \(C=A\operatorname{Cov}(Y)A^{\mathsf T}\), so \(\det C=(\det A)^2\prod_k\Delta t_k=\prod_k\Delta t_k\). \(\square\)

Worked Calculations

Repeated Times, Zero Times, and the Density Question

The theorem’s ordered, strictly positive times guarantee a nonsingular covariance matrix and an ordinary density. For an arbitrary finite list of nonnegative times, sort the distinct positive times, apply the theorem to those values, and then restore the original order and any repetitions. Since \(B_0=0\) almost surely, a coordinate at time zero is deterministic; repeated times give equal coordinates almost surely. These constraints make the full vector singular whenever the list includes a zero time or a repetition. Its characteristic function is still given by the same covariance-kernel formula, but it has no density with respect to Lebesgue measure in the full coordinate space.

A Common Quadratic-Form Pitfall

When expanding \(u^{\mathsf T}Cu\), an off-diagonal covariance contributes twice: once as \(C_{ij}u_i u_j\) and once as \(C_{ji}u_j u_i\). For the two-time covariance matrix \(C=\begin{pmatrix}1&1\\1&2\end{pmatrix}\), the correct expansion is \(u_1^2+2u_1u_2+2u_2^2\). A common incorrect calculation writes \(u_1^2+u_1u_2+2u_2^2\), counting the off-diagonal term only once. The correction is to retain both terms, or to write the quadratic form as \(\sum_i C_{ii}u_i^2+2\sum_{i<j}C_{ij}u_i u_j\). The coefficient of the mixed monomial is twice the matrix entry; the covariance matrix entry itself remains \(C_{ij}=\min(t_i,t_j)\).

The covariance kernel is useful because it turns the dependence among Brownian values into finite-dimensional linear algebra. The characteristic function gives expectations of complex exponentials, while the density formula gives explicit probabilities and likelihoods for sampled values at distinct times. For ordered positive times, the time gaps determine both the density’s normalizing constant and the covariance determinant. For degenerate time lists, the same kernel still describes the law, but the support must be treated as a lower-dimensional set rather than assigned a full-dimensional density.