Brownian Bridge Covariance and Finite-Dimensional Laws

From a Brownian Motion to a Bridge

A Brownian bridge is pinned to specified values at the ends of a time interval. That pinning changes the dependence among its interior values: knowing the process must return to its endpoint affects its covariance at every pair of times. Once the covariance is known, the Gaussian-process property gives the full joint law at any finite collection of times.

Fix \(T>0\). On a probability space \((\Omega,\mathcal{F},\mathbb{P})\), let \((W_t)_{t\ge 0}\) be a standard Brownian motion, with \(W_0=0\), mean zero, and covariance \(\operatorname{Cov}(W_s,W_t)=\min(s,t)\). Define the standard Brownian bridge on \([0,T]\) by \(X_t=W_t-\frac{t}{T}W_T\), for \(0\le t\le T\). This representation gives the same Gaussian-process law as Brownian motion conditioned to have value zero at time \(T\). Since the event \(W_T=0\) has probability zero, that conditioning is understood through a regular conditional Gaussian law, not ordinary conditioning on a positive-probability event. The earlier material on Brownian bridges establishes this connection.

The representation immediately pins both endpoints: \(X_0=W_0=0\), and \(X_T=W_T-W_T=0\), on every path. It also gives the covariance by expanding the covariance of the two differences. The following result collects the resulting finite-dimensional law.

Proof. Each \(X_t\) is a linear combination of \(W_t\) and \(W_T\), so any finite vector of bridge values is a linear transformation of a finite Gaussian vector of Brownian-motion values. It is therefore Gaussian. Its mean is zero because both Brownian-motion terms have mean zero. Expanding the covariance gives

where \(s,t\le T\), so \(\min(s,T)=s\), \(\min(T,t)=t\), and \(\operatorname{Var}(W_T)=T\). Applying this formula to every pair \(t_i,t_j\) gives the stated covariance matrix. At the endpoints, \(K(0,t)=0\) and \(K(T,t)=t-t=0\); these agree with the pathwise endpoint values.

It remains to prove positive definiteness at distinct interior times. Order them as \(0<t_1<\cdots<t_m<T\), and take real coefficients \(a_1,\ldots,a_m\), not all zero. Put \(c=\sum_{i=1}^m a_i t_i/T\), \(t_0=0\), and \(t_{m+1}=T\). Then

where the sum \(\sum_{i=m+1}^m a_i\) for \(k=m+1\) is defined to be zero. The Brownian increments in this expression are independent and have variances \(t_k-t_{k-1}>0\). Thus the variance of the expression is the sum, over \(k\), of the squared coefficients multiplied by \(t_k-t_{k-1}\). If that variance were zero, every coefficient would have to be zero. The last coefficient gives \(c=0\); the coefficient for \(k=m\) then gives \(a_m=0\), and working backwards gives \(a_{m-1}=\cdots=a_1=0\). This contradicts the choice of the \(a_i\). Hence every nonzero linear combination has positive variance, which is exactly positive definiteness of \(\Sigma\). If an endpoint is included, its bridge coordinate is identically zero; if a time is repeated, two coordinates coincide. Either case makes the covariance matrix singular. \(\square\)

Computing a Joint Law

For distinct interior times, the theorem gives a nonsingular multivariate normal distribution. To write its density, use the standard density formula: if the mean is zero and the covariance is \(\Sigma\), then the density at a column vector \(x\) is \((2\pi)^{-m/2}(\det\Sigma)^{-1/2}\exp(-\tfrac12x^{\mathsf T}\Sigma^{-1}x)\). The examples below calculate the covariance matrix first, then its determinant and inverse. Each density is with respect to ordinary Lebesgue measure on the corresponding Euclidean space.

Endpoint Values, Degeneracy, and a Common Error

The covariance formula also handles finite-dimensional laws that include endpoints. For example, if the selected times are \(0,\tfrac12,T\), the first and last coordinates are zero with probability one, and their rows and columns in the covariance matrix are zero. The joint Gaussian vector is then degenerate: it is supported on the set of vectors whose first and last coordinates are zero. This is not a failure of the Gaussian-process description. It means that the nonsingular density formula does not apply to that full vector. Repeated observation times likewise create duplicate coordinates and a singular covariance matrix.

A common mistake is to compute the variance of \(X_t=W_t-\frac{t}{T}W_T\) by treating its two Brownian-motion terms as independent. They are not independent: \(\operatorname{Cov}(W_t,W_T)=t\) for \(t\le T\). For \(T=1\) and \(t=\tfrac12\), the incorrect calculation gives \(\operatorname{Var}(W_{1/2})+(\tfrac12)^2\operatorname{Var}(W_1)=\tfrac12+\tfrac14=\tfrac34\). The correction includes the covariance terms: \(\operatorname{Var}(X_{1/2})=\tfrac12+\tfrac14-2(\tfrac12)(\tfrac12)=\tfrac14\). This agrees with the kernel, \(K(\tfrac12,\tfrac12)=\tfrac12-\tfrac14=\tfrac14\), and with the fact that the bridge variance must decrease to zero at the terminal time.

The covariance kernel is useful because a Gaussian process is determined in finite-dimensional law by its means and covariances. In this case, the kernel provides the joint law at any finite set of times without requiring a separate conditional calculation for every collection. It also makes the endpoint constraint visible: every covariance with either endpoint is zero, and the endpoint values themselves are fixed rather than random.