The Schauder Series Representation of Brownian Motion

From Dyadic Midpoints to Tent Functions

Brownian motion can be built by filling in its values on successively finer dyadic grids. At each step, a new midpoint equals the average of its two endpoints plus an independent Gaussian innovation. The Schauder series writes all of those midpoint corrections at once: each correction is a random multiple of a tent-shaped function supported on the interval where that midpoint is inserted.

Work on a probability space $(\Omega,\mathcal{F},\mathbb{P})$, and consider the time interval $[0,1]$. The earlier dyadic midpoint construction uses a standard normal endpoint value $Z_*$ and independent standard normal innovations $Z_{n,k}$, for levels $n\ge 0$ and indices $0\le k<2^n$. These variables are mutually independent. At level $n$, the dyadic intervals are $[k2^{-n},(k+1)2^{-n}]$; the innovation $Z_{n,k}$ is used to set the value at that interval's midpoint.

Definition: For $n\ge0$ and $0\le k<2^n$, put $a_{n,k}=k2^{-n}$ and $b_{n,k}=(k+1)2^{-n}$. The dyadic tent function $\phi_{n,k}$ is zero outside $[a_{n,k},b_{n,k}]$, is linear on each half of that interval, and has value $1$ at its midpoint. Equivalently, on its support, $$ \phi_{n,k}(t)=2^{n+1}\min(t-a_{n,k},b_{n,k}-t). $$ Its maximum is $1$, attained at $t=(a_{n,k}+b_{n,k})/2$.

The interval length at level $n$ is $2^{-n}$. In the dyadic midpoint construction, conditional on the values at the endpoints, the midpoint value is their average plus a centered normal variable with variance $2^{-n}/4$. Thus its standard deviation is $2^{-(n+2)/2}$. This fixes the coefficient scale: the midpoint correction is $2^{-(n+2)/2}Z_{n,k}$. The level-zero tent has scale $1/2$, as required for the midpoint of $[0,1]$ to have conditional variance $1/4$.

Definition (Schauder series): The level-$m$ partial sum and its proposed limit are $$ S_m(t)=Z_*t+\sum_{n=0}^{m}\sum_{k=0}^{2^n-1}2^{-(n+2)/2}Z_{n,k}\phi_{n,k}(t), \qquad W(t)=\lim_{m\to\infty}S_m(t),\quad 0\le t\le1, $$ when the limit exists. The term $Z_*t$ linearly interpolates the endpoint values $0$ and $Z_*$; every subsequent term adds one level of midpoint corrections.

Why the Series Converges Uniformly

At any fixed level $n$, the interiors of the tents have disjoint supports. Consequently, at a given $t$, at most one level-$n$ tent is nonzero, and $$ \sup_{0\le t\le1}\left|\sum_{k=0}^{2^n-1}2^{-(n+2)/2}Z_{n,k}\phi_{n,k}(t)\right| \le 2^{-(n+2)/2}\max_{0\le k<2^n}|Z_{n,k}|. $$ We show that these bounds are summable almost surely. This proves uniform convergence without needing any path-regularity result about Brownian motion.

Proposition: If the $Z_{n,k}$ are independent standard normal random variables, then the Schauder series converges uniformly on $[0,1]$ almost surely. Its limit $W$ is therefore a continuous function almost surely.

Proof. For a standard normal variable $Z$, its moment generating function is $\mathbb{E}e^{\theta Z}=e^{\theta^2/2}$. For $x\ge0$ and $\theta>0$, Markov's inequality gives $\mathbb{P}(Z>x)\le e^{-\theta x}\mathbb{E}e^{\theta Z}=e^{-\theta x+\theta^2/2}$. Choosing $\theta=x$ gives $\mathbb{P}(Z>x)\le e^{-x^2/2}$; applying the same argument to $-Z$ yields $\mathbb{P}(|Z|>x)\le2e^{-x^2/2}$. By the union bound, for each integer $n\ge1$, $$ \mathbb{P}\left(\max_{0\le k<2^n}|Z_{n,k}|>n\right) \le 2^n\cdot 2e^{-n^2/2}. $$ The series $\sum_{n\ge1}2^{n+1}e^{-n^2/2}$ is finite: for all sufficiently large $n$, $n\log 2-n^2/2\le -n^2/4$, and $\sum_{n\ge1}e^{-n^2/4}\le\sum_{n\ge1}e^{-n/4}<\infty$. The first Borel–Cantelli lemma therefore implies that, almost surely, $\max_{k}|Z_{n,k}|\le n$ for all sufficiently large $n$. On that event the supremum norm of the level-$n$ contribution is eventually at most $n2^{-(n+2)/2}$. The series of these bounds converges, since it is a constant multiple of $\sum_{n\ge1}n(2^{-1/2})^n$. Hence the series of level functions converges uniformly by the uniform Cauchy criterion. Each partial sum is continuous, and a uniform limit of continuous functions is continuous. $\square$

The Midpoint Recursion in the Series

Write $S_{-1}(t)=Z_*t$. The new term at level $n$ vanishes at all endpoints of the level-$n$ intervals. At the midpoint of $[a_{n,k},b_{n,k}]$, its tent has value $1$, while every other level-$n$ tent is zero. The earlier levels together form the straight line between the two endpoint values on this interval. Thus the midpoint value becomes the endpoint average plus precisely the level-$n$ innovation.

Proposition (midpoint identity): Let $m_{n,k}=(a_{n,k}+b_{n,k})/2$. For the partial sums, the midpoint inserted at level $n$ satisfies $$ S_n(m_{n,k})=\frac{S_{n-1}(a_{n,k})+S_{n-1}(b_{n,k})}{2} +2^{-(n+2)/2}Z_{n,k}. $$ All values already assigned at coarser dyadic grid points remain unchanged.

Proof. Every tent from a level $r<n$ is linear on the level-$n$ interval $[a_{n,k},b_{n,k}]$, because its possible change of slope occurs only at a coarser dyadic midpoint, which is an endpoint of this level-$n$ interval or lies outside it. Therefore $S_{n-1}$, the sum of the linear term and all levels below $n$, is linear on this interval. Its midpoint value is the average of its endpoint values. At $m_{n,k}$, $\phi_{n,k}=1$; every other level-$n$ tent vanishes there. The level-$n$ addition is consequently $2^{-(n+2)/2}Z_{n,k}$, proving the identity. At any coarser dyadic grid point, each level-$n$ tent is zero, so earlier values are unchanged. $\square$

This identity connects the series to Brownian motion. In the dyadic midpoint construction, the midpoint corrections on intervals of length $2^{-n}$ are independent centered normals with variance $2^{-n}/4$, and hence have the form $2^{-(n+2)/2}Z_{n,k}$ with independent standard normals. The identity shows that the series performs exactly the same recursive construction, not just a similar-looking interpolation. Conversely, starting with a Brownian motion on $[0,1]$, its endpoint $B_1$ and its successive standardized midpoint deviations give these coefficients. The piecewise linear interpolants on dyadic grids converge uniformly to each continuous Brownian path, and their successive differences are exactly the tent levels. This gives the Schauder series representation of Brownian motion.

Worked Example: The First Two Midpoint Levels

Take $Z_*=2$, $Z_{0,0}=2$, and $Z_{1,0}=\sqrt{2}$. The other level-one coefficient does not affect the value at $1/4$. Compute the partial-sum values at $1/4$ and $1/2$, then use the level-two innovation $Z_{2,1}=-1$ to compute the value at $3/8$. These are particular coefficient values for checking the algebra, not claims about positive-probability events of exact normal values.

At $t=1/4$, the linear term is $Z_*t=2(1/4)=1/2$. The level-zero tent has value $\phi_{0,0}(1/4)=2\min(1/4,3/4)=1/2$, so its contribution is $(1/2)(2)(1/2)=1/2$. The level-one tent on $[0,1/2]$ has value $\phi_{1,0}(1/4)=1$, and its contribution is $2^{-3/2}\sqrt{2}=1/2$. Therefore $S_1(1/4)=1/2+1/2+1/2=3/2$. At $1/2$, the linear term is $1$, the level-zero contribution is $(1/2)(2)(1)=1$, and the level-one tents vanish at this coarser grid point. Thus $S_1(1/2)=2$.

The endpoints of $[1/4,1/2]$ are now $3/2$ and $2$. Their average is $(3/2+2)/2=7/4$. The level-two scale is $2^{-(2+2)/2}=1/4$, so the innovation at $3/8$ is $(1/4)(-1)=-1/4$. The recursion gives $$ S_2(3/8)=\frac{7}{4}-\frac{1}{4}=\frac{3}{2}. $$ Check by direct expansion: the linear term at $3/8$ is $2(3/8)=3/4$; the level-zero tent value is $3/4$, contributing $(1/2)(2)(3/4)=3/4$; the level-one tent on $[0,1/2]$ has value $1/2$, contributing $2^{-3/2}\sqrt{2}(1/2)=1/4$; and the level-two tent on $[1/4,1/2]$ has value $1$, contributing $-1/4$. The sum $3/4+3/4+1/4-1/4=3/2$ agrees with the recursion.

Worked Example: Reading a Coefficient from a Midpoint

Suppose the level-zero endpoint values are $S_{-1}(0)=0$ and $S_{-1}(1)=2$, and the observed midpoint value is $S_0(1/2)=3/2$. Find the corresponding standard normal coefficient and check the scale.

The endpoint average is $(0+2)/2=1$, so the midpoint deviation is $3/2-1=1/2$. At level $n=0$, the coefficient scale is $2^{-(0+2)/2}=1/2$. Therefore $(1/2)Z_{0,0}=1/2$, giving $Z_{0,0}=1$. Substitution checks the result: the midpoint is $1+(1/2)(1)=3/2$. The deviation has variance $(1/2)^2=1/4$ when $Z_{0,0}$ is standard normal, as required by the midpoint construction.

Worked Example: Support and Scale at a Fine Level

At level $n=2$, consider the interval $[1/4,1/2]$, corresponding to $k=1$. Find its tent function at $t=3/8$, its coefficient scale, and the effect of coefficient $Z_{2,1}=4$.

The interval length is $2^{-2}=1/4$, and $3/8$ is its midpoint. The tent value is $$ \phi_{2,1}(3/8)=2^{3}\min(3/8-1/4,1/2-3/8)=8\min(1/8,1/8)=1. $$ The scale is $2^{-(2+2)/2}=1/4$. Hence this term contributes $(1/4)(4)(1)=1$ at $3/8$. At $t=1/4$ and $t=1/2$, the tent vanishes, so this innovation changes the midpoint without changing either endpoint. Its midpoint deviation has variance $(1/4)^2=1/16$ when the coefficient is standard normal, matching an interval of length $1/4$ with conditional midpoint variance $(1/4)/4=1/16$.

What the Representation Does—and Does Not—Say

The series is useful because it separates Brownian motion into independent standard normal coordinates and deterministic functions. Its first term sets the endpoint at time $1$; each subsequent level refines the path on a dyadic grid. At a fixed level, only one tent can affect an interior point, and its height is scaled by the square root of the interval length. The coefficient scale is therefore not arbitrary: it is exactly the standard deviation of the corresponding midpoint innovation.

This tutorial establishes the series representation and its convergence, but does not use the series to verify the covariance between arbitrary times. That requires tracking how tents from different levels contribute to pairs of times. In particular, independent coefficients do not by themselves mean that the process values at different times are independent: the same coefficient can contribute to multiple time points. The covariance calculation is a separate step.

Common pitfall: omitting an earlier innovation. In the first worked example, calculating $S_1(1/4)$ as only the linear term plus the level-zero contribution gives $1/2+1/2=1$, which is wrong: the level-one innovation at $1/4$ must also be included. Its contribution is $2^{-3/2}\sqrt2\cdot1=1/2$, so $S_1(1/4)=3/2$. The level-two midpoint calculation must therefore use the endpoints $3/2$ and $2$, whose average is $7/4$, not the average obtained from the incomplete value. Adding the level-two innovation $-1/4$ gives $3/2$.

Check Your Understanding

Use the tent definition and midpoint identity to answer the following questions.

For the level-$n$ interval length $2^{-n}$, derive the midpoint coefficient scale from the conditional midpoint variance $2^{-n}/4$.
At level $n=1$, what is the support and peak location of $\phi_{1,1}$? What is its value at $t=3/4$?
If the endpoint values on an interval are $1$ and $3$, and its midpoint innovation is $-1/2$, what is the midpoint value?
Why does a uniform limit of the partial sums have continuous paths on $[0,1]$?
At a coarse dyadic grid point, why do the tent terms from all finer levels leave its already assigned value unchanged?

The Schauder Series Representation of Brownian Motion

What You'll Learn