Refining Brownian Paths with Brownian Bridges

Why Refine a Brownian Path?

A Brownian path simulated on a coarse time grid gives values only at the selected times. If a later calculation needs an intermediate value, linear interpolation may look like a natural choice, but it removes the random fluctuation Brownian motion has between the endpoints. Brownian bridges provide the correct conditional distribution for inserting such values. Drawing from that distribution refines the grid while preserving the Brownian law at all grid points.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). A standard Brownian motion \((B_t)_{t\ge0}\) has \(B_0=0\), continuous paths, and independent increments, with \(B_t-B_s\) normally distributed with mean \(0\) and variance \(t-s\) for \(0\le s<t\). Its natural filtration is \(\mathcal{F}_t=\sigma(B_u:0\le u\le t)\). For a fixed interval \([a,a+h]\), where \(h>0\), we will condition on its two endpoint values. Conditioning on random endpoint values is understood through the regular conditional distribution, which is defined here by the Gaussian densities.

The Conditional Midpoint Distribution

Proof. First take the two half-interval increments \(X=B_{a+h/2}-B_a\) and \(Y=B_{a+h}-B_{a+h/2}\). By the independent-increment property, \(X\) and \(Y\) are independent \(N(0,h/2)\) random variables. Set \(U=X+Y\) and \(V=(X-Y)/2\). These are jointly normal because they are linear combinations of the jointly normal vector \((X,Y)\). Their means are zero, and direct calculation gives $$ \operatorname{Var}(U)=h,\qquad \operatorname{Var}(V)=\frac{h}{4},\qquad \operatorname{Cov}(U,V)=\frac{1}{2}\bigl(\operatorname{Var}(X)-\operatorname{Var}(Y)\bigr)=0. $$ For jointly normal variables, zero covariance implies independence; in this case their joint normal density factors into the product of the \(N(0,h)\) and \(N(0,h/4)\) densities. Thus \(V\) is independent of \(U\). Given the endpoint values, \(U=B_{a+h}-B_a=y-x\) is fixed, while $$ B_{a+h/2}=x+X=x+\frac{U}{2}+V=\frac{x+y}{2}+V. $$ Since \(V\sim N(0,h/4)\) and is independent of \(U\), the stated conditional normal distribution follows. \(\square\)

The same argument also explains how to refine more than one interval at a time. On disjoint intervals, the pairs of half-interval increments are independent. The residual \(V\) on each interval is independent of that interval's total increment, and the residuals from different intervals are independent of each other. Therefore, given all values on a coarse grid, midpoint values on its distinct intervals can be drawn independently using the theorem.

A Refinement Algorithm and Why It Works

Suppose a Brownian path has already been simulated at deterministic times \(0=t_0<t_1<\cdots<t_m\). Write \(h_i=t_{i+1}-t_i>0\) for the length of interval \(i\). For each interval, generate a standard normal variable \(Z_i\), independently across intervals and independently of the simulated coarse-grid values. Insert the midpoint value $$ \widetilde B_{(t_i+t_{i+1})/2} =\frac{B_{t_i}+B_{t_{i+1}}}{2} +\frac{\sqrt{h_i}}{2}Z_i. $$ The old values are retained. If still finer values are needed, apply the same rule separately to each new interval, using fresh independent standard normal variables.

Proof. Let \(\mathcal{G}=\sigma(B_{t_0},\ldots,B_{t_m})\), the information in the coarse-grid values. On interval \([t_i,t_{i+1}]\), the midpoint theorem gives the conditional distribution \(N((B_{t_i}+B_{t_{i+1}})/2,h_i/4)\). The independent-increment argument above shows that, conditional on \(\mathcal{G}\), the residuals on the distinct intervals are independent: each is independent of its interval's total increment, and increments on different intervals are independent. Thus the conditional joint distribution of all the inserted midpoints is the product of the normal distributions generated by the algorithm.

For any event \(C\) specified by the coarse and inserted values, the conditional distribution just identified is the same as the conditional distribution of the corresponding Brownian midpoint values given the coarse values. Averaging over the coarse values gives the same unconditional probability for \(C\), by the law of total probability. Hence the joint distribution on the refined grid agrees with Brownian motion's distribution there. Applying this argument again to the refined grid proves the same claim at the next refinement. Induction gives the claim after any finite number of refinement steps. \(\square\)

This result is about the distribution of values on the refined grid. It does not say that the finitely many simulated values determine every point of a continuous path. Rather, each additional refinement supplies further correctly distributed values, conditional on those already generated.

Why Linear Interpolation Fails

Linear interpolation sets each midpoint equal to the average of its endpoints. But the bridge theorem shows that this average is only the conditional mean: the midpoint still has variance \(h/4\) for an interval of positive length \(h\). Omitting the normal term gives a deterministic midpoint conditional on the endpoints and therefore changes the conditional distribution. It also suppresses fluctuations at every inserted time.

Common mistake. On \([0,4]\) with both endpoint values equal to \(0\), linear interpolation assigns midpoint \(0\) with probability \(1\). The bridge formula instead gives conditional mean \((0+0)/2=0\), variance \(4/4=1\), and midpoint \(Z\) for \(Z\sim N(0,1)\). For example, an input \(Z=\tfrac12\) produces midpoint \(\tfrac12\), not \(0\). The correction is to retain the normal residual \(\sqrt{h}Z/2\), even when the endpoints coincide.

For implementation, generate a new independent standard normal for every interval at every refinement level. Reusing the same normal input, or treating the midpoint as fixed once the endpoints are known, does not implement the conditional product distribution in the theorem. The formula assumes standard Brownian motion; if the process has a different diffusion scale, the variance must be adjusted to match that process.