Verifying Covariance in a Gaussian-Series Construction

Covariance from Independent Gaussian Coefficients

The Schauder representation expresses a Brownian path as a linear term plus a sum of dyadic tent functions with independent Gaussian coefficients. To verify its covariance, the key is to compute what each coefficient contributes at a pair of times, then sum those contributions. Each tent is supported on one dyadic interval, so a coefficient contributes only when both times lie in that same interval.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let the time index be \(t\in[0,1]\) and the state space be \(S=\mathbb{R}\). Take \(Z_*\) and \(Z_{n,k}\), for \(n\ge 0\) and \(0\le k<2^n\), to be mutually independent standard normal random variables. At level \(n\), the dyadic intervals have length \(h_n=2^{-n}\). On the interval \([a,b]=[k2^{-n},(k+1)2^{-n}]\), define the tent \(\phi_{n,k}\) to be zero outside \([a,b]\), linear from \(0\) at \(a\) to \(1\) at \((a+b)/2\), and linear from \(1\) there back to \(0\) at \(b\). In particular, its values at both endpoints are zero.

The covariance contribution of a coefficient is the product of its values at the two times, multiplied by the coefficient’s variance. Independence makes the covariances between distinct coefficients zero. Thus the linear term contributes \(st\), and the tents contribute the sum \(\sum_{n,k}2^{-(n+2)}\phi_{n,k}(s)\phi_{n,k}(t)\). The main task is to identify this sum exactly.

The Dyadic Refinement Identity

Separate the desired covariance into the endpoint contribution \(st\) and the remaining kernel \(K(s,t)=\min(s,t)-st\). This is the covariance kernel of a Brownian bridge, but here it is used simply as a function to be recovered from the tent contributions. We show that refining a dyadic interpolation adds exactly one level of tents to its covariance.

For \(m\ge0\), let \(x_{m,r}=r2^{-m}\), \(0\le r\le2^m\). For any \(s\in[0,1]\), write \(\lambda_{m,r}(s)\) for the piecewise-linear interpolation weights on this grid: if \(s\in[x_{m,r},x_{m,r+1}]\), only the two weights at the endpoints can be nonzero, and they are the linear weights summing to \(1\). Define the interpolated kernel \(K_m(s,t)=\sum_{r,q}\lambda_{m,r}(s)\lambda_{m,q}(t)K(x_{m,r},x_{m,q})\). At level \(m=0\), the grid consists of \(0,1\), and \(K\) vanishes at all pairs of these endpoints. Hence \(K_0(s,t)=0\).

Proof. Fix an old grid interval \([a,b]\) at level \(m\), let \(h=b-a=2^{-m}\), and let \(c=(a+b)/2\). Refinement adds the midpoint \(c\). The new midpoint value, relative to linear interpolation between the old endpoints, is represented by the residual \(D=Y(c)-\tfrac12(Y(a)+Y(b))\), where \(Y\) denotes values with covariance \(K\). The covariance calculations below are algebraic calculations with \(K\), so no process \(Y\) needs to be assumed to exist.

First, \(K(c,c)=c-c^2\), \(K(a,a)=a-a^2\), \(K(b,b)=b-b^2\), and \(K(a,b)=a-ab\), since \(a\le b\). Therefore

For completeness, the last equality follows by substituting \(c=(a+b)/2\) and \(h=b-a\) into \(K(u,v)=\min(u,v)-uv\); the terms involving \(a\) and \(b\) cancel, leaving \((b-a)/4\). Also, for any old grid point \(x\), \(\operatorname{Cov}(D,Y(x))=K(c,x)-\tfrac12K(a,x)-\tfrac12K(b,x)=0\). If \(x\le a\) or \(x\ge b\), the function \(u\mapsto K(u,x)\) is linear on \([a,b]\), so its midpoint value is the average of its endpoint values. For two distinct old grid intervals, the same piecewise-linearity calculation gives zero covariance between their midpoint residuals.

Refining the interpolation therefore adds an independent-in-covariance midpoint residual on each old interval. Its coefficient in the interpolated value at \(s\) is \(\phi_{m,k}(s)\). Its variance is \(h/4=2^{-(m+2)}\). The covariance increase is consequently the sum over intervals of \(2^{-(m+2)}\phi_{m,k}(s)\phi_{m,k}(t)\), as claimed. \(\square\)

Summing the lemma from \(m=0\) through \(m=M\) telescopes: the total tent contribution through level \(M\) is \(K_{M+1}(s,t)-K_0(s,t)=K_{M+1}(s,t)\). As the grid spacing decreases to zero, \(K_{M+1}(s,t)\) converges to \(K(s,t)\). Indeed, \(K\) is continuous on the compact square \([0,1]^2\), and the interpolation uses values at grid points within one mesh width of each argument. Thus the infinite sum of tent contributions is \(K(s,t)\).

Proof. Every coefficient has mean zero, and distinct coefficients are independent, so their cross-covariances vanish. The linear term contributes \(\operatorname{Cov}(Z_*s,Z_*t)=st\). By the refinement identity and its telescoping limit, the tent terms contribute \(K(s,t)=\min(s,t)-st\). Adding these contributions gives \(st+\min(s,t)-st=\min(s,t)\). The \(L^2\) convergence at each time justifies taking the covariance limit: for \(L^2\)-convergent centered random variables, the products of pairs converge in expectation by the Cauchy–Schwarz inequality. \(\square\)

Worked Covariance Calculations

Why the Coefficient Scaling Matters

Each coefficient is multiplied by \(2^{-(n+2)/2}\), but covariance uses the square of that multiplier. At level \(n\), the variance is \(2^{-(n+2)}\). This is precisely the scale needed for the midpoint refinement: an interval of length \(2^{-n}\) contributes a midpoint residual with variance \(2^{-n}/4\).

Common mistake. At \(t=\tfrac12\), the linear term contributes variance \((\tfrac12)^2=\tfrac14\). The level-\(0\) tent equals \(1\) there, and its coefficient variance is \(2^{-(0+2)}=\tfrac14\). All finer tents vanish at this dyadic endpoint. Thus the correct variance is \(\tfrac14+\tfrac14=\tfrac12\), matching \(\min(\tfrac12,\tfrac12)\). A common incorrect calculation treats the coefficient multiplier \(2^{-(0+2)/2}=\tfrac12\) as the variance, rather than squaring it. That gives \(\tfrac14+\tfrac12=\tfrac34\), which fails the covariance check. Whenever independent Gaussian coefficients are summed, multiply their values at the two times by the coefficient variance, not by the coefficient’s standard deviation.

The refinement argument also explains why the computation is more than a collection of special cases. Each level supplies the difference between two successive interpolations of the kernel, and those differences telescope to the full bridge part. Adding the independent linear endpoint term then gives the covariance of the complete series. The answer is a covariance identity; by itself it is not a substitute for checking the construction’s other properties.