An L2 Error Bound for Brownian Quadratic Variation

From Squared Increments to an Error Bound

Brownian motion has random increments, but the sum of their squares over a partition has a simple average: it equals the length of the time interval. The sum is not usually equal to that length on each sample path. Its fluctuations can be quantified exactly, and a bound in terms of the partition mesh tells us how those fluctuations shrink as the partition is refined.

Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, and let \((B_t)_{t\ge0}\) be standard Brownian motion with state space \(S=\mathbb{R}\). Thus \(B_0=0\), its paths are continuous, and its increments over disjoint time intervals are independent; for \(0\le s<t\), \(B_t-B_s\) has distribution \(N(0,t-s)\). Fix \(T>0\) and a deterministic partition \(\Pi=\{0=t_0<t_1<\cdots<t_n=T\}\). Write \(\Delta_i=t_i-t_{i-1}\) for the \(i\)th interval length, and define the mesh \(|\Pi|=\max_{1\le i\le n}\Delta_i\).

The key calculation comes from a single Gaussian increment. If \(Z\) has distribution \(N(0,h)\), then \(\mathbb{E}[Z^2]=h\) and \(\mathbb{E}[Z^4]=3h^2\). Consequently, the variance of its square is \(2h^2\), not \(2h\). We first justify the fourth moment and then use independence of Brownian increments to calculate the error for the whole partition.

Proof. Write \(Z=\sqrt{h}\,Y\), where \(Y\sim N(0,1)\). The standard normal density is \(\phi(x)=(2\pi)^{-1/2}e^{-x^2/2}\), and \(\phi'(x)=-x\phi(x)\). Integration by parts, with boundary terms vanishing because \(x^k\phi(x)\to0\) as \(|x|\to\infty\), gives \(\int_{\mathbb{R}}x^2\phi(x)\,dx=-\int_{\mathbb{R}}x\phi'(x)\,dx=\int_{\mathbb{R}}\phi(x)\,dx=1\). Applying integration by parts again, \(\int_{\mathbb{R}}x^4\phi(x)\,dx=-\int_{\mathbb{R}}x^3\phi'(x)\,dx=3\int_{\mathbb{R}}x^2\phi(x)\,dx=3\). Scaling by \(\sqrt h\) yields \(\mathbb{E}[Z^2]=h\) and \(\mathbb{E}[Z^4]=3h^2\). Therefore \(\operatorname{Var}(Z^2)=\mathbb{E}[Z^4]-(\mathbb{E}[Z^2])^2=3h^2-h^2=2h^2\). \(\square\)

The Exact Error Identity and Mesh Bound

Proof. Define \(Z_i=B_{t_i}-B_{t_{i-1}}\). By the Brownian increment assumptions, the \(Z_i\) are independent and \(Z_i\sim N(0,\Delta_i)\). Their squares are therefore independent as well. The Gaussian square-moment result gives \(\mathbb{E}[Z_i^2]=\Delta_i\) and \(\operatorname{Var}(Z_i^2)=2\Delta_i^2\). Linearity of expectation now yields \(\mathbb{E}[Q_\Pi]=\sum_{i=1}^n\Delta_i=T\). Independence gives \(\operatorname{Var}(Q_\Pi)=\sum_{i=1}^n\operatorname{Var}(Z_i^2)=2\sum_{i=1}^n\Delta_i^2\). Because \(\mathbb{E}[Q_\Pi]=T\), the definition of variance gives \(\mathbb{E}[(Q_\Pi-T)^2]=\operatorname{Var}(Q_\Pi)\). Finally, every interval length satisfies \(\Delta_i\le|\Pi|\), so \(\Delta_i^2\le|\Pi|\Delta_i\). Summing these inequalities gives \(\sum_i\Delta_i^2\le|\Pi|\sum_i\Delta_i=T|\Pi|\), proving the bound. Taking nonnegative square roots gives the stated \(L^2\) bound. \(\square\)

The exact identity contains more information than the bound: it records how every interval length contributes to the variance. The mesh bound is useful when the partition is specified only through its largest interval. In particular, for a sequence of deterministic partitions of the same interval with mesh tending to zero, the bound implies that \(Q_\Pi\) converges to \(T\) in \(L^2\). It also implies convergence in probability: for every \(\varepsilon>0\), the pointwise inequality \(\mathbf{1}_{\{|Q_\Pi-T|\ge\varepsilon\}}\le (Q_\Pi-T)^2/\varepsilon^2\) gives \(\mathbb{P}(|Q_\Pi-T|\ge\varepsilon)\le 2T|\Pi|/\varepsilon^2\), which tends to zero with the mesh. This estimate alone is a mean-square and probability statement; it does not assert almost-sure convergence.

Using the Bound to Choose a Partition

For a uniform partition of \([0,T]\) into \(n\) intervals, each length and the mesh equal \(T/n\). Substituting into the exact formula gives \(\operatorname{Var}(Q_\Pi)=2n(T/n)^2=2T^2/n\). The resulting \(L^2\) error is \(\sqrt{2}\,T/\sqrt n\). Thus, if a target error \(\varepsilon>0\) is prescribed, the exact formula guarantees an \(L^2\) error at most \(\varepsilon\) whenever \(n\ge 2T^2/\varepsilon^2\). The mesh bound gives the same sufficient condition for this uniform partition.

Why Independence and the Squared Length Matter

The variance calculation relies on the increments being independent, which is part of the Brownian motion assumptions for disjoint intervals. In general, the variance of a sum includes covariance terms; those terms vanish here because the squared increments are independent. The other essential detail is the square on each interval length: the variance contributed by an increment over an interval of length \(h\) is \(2h^2\). The interval length itself is its contribution to the mean, not to the variance.

Common mistake. On the four-interval partition of \([0,1]\), an incorrect calculation might use \(2\Delta_i=\tfrac12\) as the variance of each squared increment, giving a total variance of \(4\cdot\tfrac12=2\). This omits the square on the interval length. The correct calculation is \(2\Delta_i^2=2(\tfrac14)^2=\tfrac18\) for each increment square, so the total variance is \(4\cdot\tfrac18=\tfrac12\). Substituting into the exact identity confirms that the mean-square error is \(\tfrac12\), not \(2\).

A second distinction is between an exact mean and a nonrandom value of the sum. The identity \(\mathbb{E}[Q_\Pi]=T\) says that the average over outcomes is \(T\); it does not say that \(Q_\Pi=T\) on every outcome. For a finite partition the variance \(2\sum_i\Delta_i^2\) is positive, since every \(\Delta_i>0\), so the sum is not almost surely constant. The mesh bound quantifies the size of its error in \(L^2\), and the exact variance explains how the individual partition intervals contribute.