Almost-Sure Quadratic Variation on Dyadic Grids

From Mean-Square Control to Almost-Sure Convergence

For Brownian motion, the sum of squared increments over a fine partition has mean equal to the length of the time interval. A mean-square error bound shows that these sums become close to that value in \(L^2\). Almost-sure convergence is stronger: it says that, with probability one, the sums converge along the specified sequence of partitions. The dyadic grids provide a particularly direct route from the error bound to this pathwise conclusion, because the variances decrease fast enough to be summable.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), and let \((B_t)_{t\ge0}\) be standard Brownian motion with continuous sample paths. Fix \(T>0\). For \(n\ge0\), divide \([0,T]\) into \(2^n\) equal intervals, each of length \(h_n=T2^{-n}\). For \(t\in[0,T]\), define the squared-increment sum, including the possibly incomplete final increment, by

If \(t\) is a grid point, the final term is zero. At \(t=T\), the sum contains all \(2^n\) full increments. We will first study the sum at fixed times, then use dyadic endpoints and continuity to obtain a simultaneous pathwise statement for all \(t\in[0,T]\).

The Variance Bound and a Summability Principle

Each increment on a grid of width \(h_n\) has distribution \(N(0,h_n)\), and the increments on that grid are independent. If \(Z\) has distribution \(N(0,h)\), then \(\mathbb{E}[Z^2]=h\) and \(\mathbb{E}[Z^4]=3h^2\), so \(\operatorname{Var}(Z^2)=3h^2-h^2=2h^2\). Adding the means and variances of the independent squared increments gives, for \(m\le 2^n\),

In particular, at the right endpoint \(T\), there are \(2^n\) increments. Thus the mean is \(2^n h_n=T\), and the variance is \(2\cdot 2^n h_n^2=2T^2 2^{-n}\). The key feature is not just that this variance tends to zero: its sum over \(n\) is finite.

Proof. For each \(m\), the event that at least one \(E_n\) occurs for \(n\ge m\) is contained in \(\bigcup_{n\ge m}E_n\). By countable subadditivity, its probability is at most \(\sum_{n\ge m}\mathbb{P}(E_n)\). These tail sums tend to zero as \(m\to\infty\). The event that infinitely many \(E_n\) occur is contained in \(\bigcup_{n\ge m}E_n\) for every \(m\), so its probability is zero. \(\square\)

Proof. First consider the completed-increment sum \(A_n(t)=\sum_{k=0}^{m_n(t)-1}(B_{(k+1)h_n}-B_{kh_n})^2\). For fixed \(t\), its number of increments is \(m_n(t)\), and its mean is \(m_n(t)h_n\). Its variance is \(2m_n(t)h_n^2\le 2Th_n=2T^2 2^{-n}\). Also \(0\le t-m_n(t)h_n<h_n\), so its mean tends to \(t\). For any \(\varepsilon>0\), once \(h_n<\varepsilon/2\), Chebyshev’s inequality gives \(\mathbb{P}(|A_n(t)-t|>\varepsilon)\le 8T^2\varepsilon^{-2}2^{-n}\). These bounds are summable. Applying the lemma for each threshold \(\varepsilon=1/j\), \(j\ge1\), and taking the countable intersection of the resulting probability-one events proves \(A_n(t)\to t\) almost surely. The remaining term in \(Q_n(t)\) is at most the square of the largest Brownian increment over a grid interval. By continuity of the path on the compact interval \([0,T]\), that largest increment tends to zero. Hence \(Q_n(t)-A_n(t)\to0\) almost surely, proving the claim. \(\square\)

Worked Variance Calculations

One Probability-One Event for Every Time

Proof. First restrict attention to dyadic endpoints \(r=jT/2^q\), where \(q\ge0\) and \(0\le j\le2^q\). There are countably many such endpoints. At each fixed endpoint, the proposition gives \(A_n(r)\to r\) almost surely; for \(n\ge q\), \(r\) is a grid point, so \(Q_n(r)=A_n(r)\). Intersecting these countably many probability-one events gives one event of probability one on which convergence holds at every dyadic endpoint.

Fix a path in this event and let \(q\) be fixed. For \(n\ge q\), \(A_n(t)\) is nondecreasing in \(t\), since it is a sum of nonnegative completed squared increments. If \(r_j=jT/2^q\le t\le r_{j+1}=(j+1)T/2^q\), then \(A_n(r_j)\le A_n(t)\le A_n(r_{j+1})\). Since also \(r_j\le t\le r_{j+1}\), it follows that \[ |A_n(t)-t| \le \max_{0\le i\le2^q}|A_n(r_i)-r_i|+\frac{T}{2^q}. \] For fixed \(q\), the maximum on the right tends to zero as \(n\to\infty\), because it is over finitely many endpoints. The remaining bound \(T/2^q\) can be made arbitrarily small by choosing \(q\) large. Therefore \(\sup_{t\in[0,T]}|A_n(t)-t|\to0\).

Finally, \(Q_n(t)-A_n(t)\) is the squared increment from the left endpoint of the grid interval containing \(t\) to \(t\). Its supremum over \(t\) is bounded by the square of the largest oscillation of the continuous path over any interval of length at most \(h_n\). Uniform continuity on \([0,T]\) makes this bound tend to zero. Thus \(\sup_{t\in[0,T]}|Q_n(t)-t|\to0\) on the same probability-one event. \(\square\)

Why the Details Matter

The conclusion is about this particular sequence of dyadic partitions. The variance estimate alone gives a probability bound at each level; summability is what lets the first Borel–Cantelli implication control the sequence almost surely. The extension from countably many grid points to all times uses both monotonicity of the completed sums and continuity of Brownian paths. The incomplete final increment is not itself monotone as a function of time, so it must be controlled separately.

Common mistake: omitting the factor of two in the variance. A tempting but incorrect calculation is \(\operatorname{Var}(Z^2)=h^2\) for \(Z\sim N(0,h)\). For \(T=1\), \(n=2\), this would give a total variance of \(4(\tfrac14)^2=\tfrac14\). The correct fourth moment is \(\mathbb{E}[Z^4]=3h^2\), so \(\operatorname{Var}(Z^2)=3h^2-(\mathbb{E}[Z^2])^2=3h^2-h^2=2h^2\). The correct total variance is therefore \(4\cdot2(\tfrac14)^2=\tfrac12\), also obtained from \(2T^2 2^{-n}=2\cdot1^2\cdot2^{-2}=\tfrac12\).