Dyadic Midpoint Construction of Brownian Motion

Why Midpoints Determine the Construction

A Brownian path can be specified progressively on finer and finer time grids. Begin with its values at times \(0\) and \(1\); then fill in the midpoint, then the quarter points, and continue. The key is not to draw each new value independently of the values already assigned. Given the two endpoints of an interval, the midpoint has a particular Gaussian conditional distribution. Sampling from that distribution preserves the joint laws required of Brownian motion.

We construct the process on \([0,1]\). Let \((\Omega,\mathcal F,\mathbb P)\) carry a countable family of independent standard normal random variables \(Z_*\) and \(Z_{n,k}\), where \(n\ge0\) and \(0\le k<2^n\). For example, these can be the coordinate variables on a countable product of standard normal probability spaces. Set \(X_0=0\) and \(X_1=Z_*\). At level \(n\), the grid points are \(k2^{-n}\), for \(0\le k\le2^n\). Let \(\mathcal G_n\) be the sigma-field generated by the values assigned on that grid. For each grid interval, assign its midpoint by

The two endpoints on the right have already been assigned at level \(n\). Since \(Z_{n,k}\) is independent of all the normals used to form \(\mathcal G_n\), the new midpoint values are conditionally independent given \(\mathcal G_n\). Each has conditional mean equal to the average of its endpoints and conditional variance \(2^{-(n+2)}\).

The Conditional Gaussian Rule

Here is why the midpoint formula has the stated mean and variance. Consider Brownian motion over an interval \([a,a+h]\), with \(h>0\), and write \(x\) and \(y\) for its endpoint values. The increments on the two halves, \(A=X_{a+h/2}-X_a\) and \(B=X_{a+h}-X_{a+h/2}\), are independent \(N(0,h/2)\) variables. The total increment is \(A+B=y-x\). Define \(R=(A-B)/2\). Then

Because \(R\) and \(A+B\) are jointly Gaussian and have zero covariance, they are independent. Thus, given the total increment, the first half-increment has conditional mean half the total and conditional variance \(h/4\). Adding the left endpoint gives the midpoint rule.

Proof. Use the variables \(A,B,R\) above. Their means are zero, and independence of \(A\) and \(B\), with variance \(h/2\) each, gives \(\operatorname{Var}(R)=(h/2+h/2)/4=h/4\) and \(\operatorname{Cov}(R,A+B)=(h/2-h/2)/2=0\). The pair \((R,A+B)\) is jointly Gaussian, so zero covariance implies independence. Since \(X_{a+h/2}=x+A=x+(y-x)/2+R=(x+y)/2+R\), its conditional mean and variance are the stated ones. Increments on disjoint intervals are independent; applying the same decomposition on each interval gives independent residuals, also independent of the endpoint increments. This proves both assertions. \(\square\)

For a level-\(n\) interval, \(h=2^{-n}\), so the midpoint variance is \(h/4=2^{-(n+2)}\). Its standard deviation is \(2^{-(n+2)/2}\), exactly the multiplier in the recursive rule.

Grid Laws and a Continuous Limit

The construction has the right Brownian distribution at every finite dyadic grid. To see the role of this fact, recall that a centered Gaussian vector with covariance entries \(\min(s,t)\) has independent increments with increment variance equal to the interval length. We now verify that the recursively constructed values have exactly these grid laws.

Proof. First prove the grid-law assertion by induction. At level \(0\), \(X_0=0\) and \(X_1=Z_*\sim N(0,1)\), which are the Brownian grid values at \(0\) and \(1\). Suppose the assertion holds at level \(n\). On each interval of length \(h=2^{-n}\), the midpoint formula adds a centered Gaussian residual of variance \(h/4\), independent of the level-\(n\) values. The conditional midpoint law proved above is exactly the law obtained by splitting a Gaussian increment of variance \(h\) into two independent Gaussian increments of variance \(h/2\). Residuals for distinct intervals are independent. Consequently, conditional on the old grid values, the new values have the same joint distribution as Brownian midpoints conditional on those values. Combining this conditional law with the induction hypothesis gives the Brownian joint law on the refined grid. This proves the assertion at every level.

For continuity, let \(D_{n,k}\) be the level-\(n\) midpoint minus the average of its two endpoints. The definition gives \(D_{n,k}=2^{-(n+2)/2}Z_{n,k}\). On each interval, \(L_{n+1}-L_n\) is linear on each half, is zero at the old endpoints, and has value \(D_{n,k}\) at the midpoint. Hence \(\lVert L_{n+1}-L_n\rVert_\infty=\max_{0\le k<2^n}|D_{n,k}|\).

For a standard normal \(Z\) and \(u>0\), its moment generating function gives \(\mathbb P(Z>u)\le e^{-u^2/2}\): for \(\theta>0\), Markov's inequality gives \(\mathbb P(Z>u)\le e^{-\theta u}\mathbb E e^{\theta Z}=e^{-\theta u+\theta^2/2}\), and choosing \(\theta=u\) gives the bound. Applying it to \(Z\) and \(-Z\), and then using the union bound over \(2^n\) variables, gives

The right-hand side is summable in \(n\). Indeed, for \(n\ge4\), \(n\log 2\le n^2/4\), so it is at most \(2e^{-n^2/4}\), a summable sequence. By the first Borel–Cantelli lemma, only finitely many of the events in the display occur almost surely. Thus, almost surely, for all sufficiently large \(n\), \(\lVert L_{n+1}-L_n\rVert_\infty\le n2^{-(n+2)/2}\). The series of these bounds converges, so the sequence \((L_n)\) is uniformly Cauchy almost surely. Its uniform limit is continuous, since each \(L_n\) is continuous. Every dyadic value is preserved at all finer levels, so the limit agrees with the recursive assignments there.

It remains to verify the law at arbitrary times. For dyadic \(s,t\), the grid-law assertion gives \(\mathbb E[(X_t-X_s)^2]=|t-s|\). Approximate any finite collection of times by dyadic times. Their constructed values converge almost surely by continuity and in \(L^2\), since the displayed second-moment identity makes the approximations \(L^2\)-Cauchy. The approximating vectors are centered Gaussian with covariance entries \(\min(s_n,t_n)\), which converge to \(\min(s,t)\). Their characteristic functions therefore converge to those of a centered Gaussian vector with covariance entries \(\min(s,t)\); the \(L^2\) limits have this same distribution. The limit process is thus centered Gaussian with covariance \(\min(s,t)\), starts at zero, and has continuous paths. For disjoint increments the covariance is zero, so joint Gaussianity makes those increments independent; their variances are their time lengths. This is Brownian motion on \([0,1]\). \(\square\)

What the Construction Does—and Does Not—Say

The recursion is local: each new value uses the two endpoints of its own interval and a fresh independent normal input. The conditional law is what makes the local assignments compatible with the global Brownian covariance. A finite collection of grid values has the correct joint law, and the uniform-limit argument supplies continuous paths. The same construction can be carried out on any finite time interval after rescaling time and the Gaussian variances. Independent copies on successive unit intervals, with each new interval started at the preceding endpoint, extend the construction to \([0,\infty)\).

Common mistake: using the unconditioned half-increment variance. On an interval of length \(h=\tfrac12\), each unconditioned half-increment has variance \(h/2=\tfrac14\). It is wrong to use \(\tfrac14\) as the conditional midpoint variance after both endpoints are fixed. The endpoint fixes the sum of the two half-increments; once that sum is known, the first half-increment has conditional variance \(h/4=(\tfrac12)/4=\tfrac18\). Equivalently, the midpoint residual is \((A-B)/2\), whose variance is \((\tfrac14+\tfrac14)/4=\tfrac18\). Using \(\tfrac14\) would make the midpoint fluctuate too much and would not preserve the Brownian grid law.

The construction is useful because it turns the specification of a continuous Gaussian process into explicit, countably many Gaussian draws. It also separates two issues that must both be checked: the conditional distributions must give the correct finite-grid laws, and the successive interpolants must converge to continuous paths. Matching midpoint means alone would not establish either conclusion.