Weighted Sums of Squared Increments
Quadratic variation records the cumulative size of Brownian increments after squaring them. A deterministic weight lets us ask where along the time interval that quadratic variation is being accumulated. For a continuous function \(w\), the natural limit of the weighted sum is the ordinary time integral of \(w\), not a random integral involving Brownian increments. The result makes precise how the small squared increments contribute across different parts of the interval.
Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space carrying a standard Brownian motion \((B_t)_{t\ge0}\), adapted to a filtration \((\mathcal{F}_t)_{t\ge0}\). In particular, Brownian increments on disjoint time intervals are independent, and \(B_{t+h}-B_t\) is normal with mean \(0\) and variance \(h\). Fix \(T>0\). A tagged partition of \([0,T]\) consists of times \(0=t_0<t_1<\cdots<t_n=T\) and tags \(\xi_i\in[t_i,t_{i+1}]\), for \(0\le i<n\). Write \(\Delta t_i=t_{i+1}-t_i\), \(\Delta B_i=B_{t_{i+1}}-B_{t_i}\), and \(\|\Pi\|=\max_i\Delta t_i\) for the mesh.
The tags may be left endpoints, right endpoints, or other points in their respective intervals. The key conditions are that \(w\) is continuous and deterministic, and that the mesh tends to zero. We first prove convergence in mean square, which means that the expectation of the squared error tends to zero. This also implies convergence in probability.
Proof. Fix one tagged partition and abbreviate \(Z_i=\Delta B_i\). Each \(Z_i\) is normal with mean \(0\) and variance \(\Delta t_i\), so \(\mathbb{E}[Z_i^2]=\Delta t_i\) and \(\operatorname{Var}(Z_i^2)=2(\Delta t_i)^2\). The latter follows from the fourth moment of a centered normal variable: \(\mathbb{E}[Z_i^4]=3(\Delta t_i)^2\). The \(Z_i\) are independent because the intervals are disjoint. Therefore $$ \mathbb{E}[S_{\Pi}(w)]=\sum_i w(\xi_i)\Delta t_i,\qquad \operatorname{Var}(S_{\Pi}(w))=2\sum_i w(\xi_i)^2(\Delta t_i)^2. $$ For any square-integrable random variable \(Y\) and constant \(c\), \(\mathbb{E}[(Y-c)^2]=\operatorname{Var}(Y)+(\mathbb{E}Y-c)^2\); applying this with \(Y=S_{\Pi}(w)\) gives the displayed identity. Since \(w\) is continuous on the compact interval, it is bounded, and \(|w|\le M\). Thus $$ 2\sum_i w(\xi_i)^2(\Delta t_i)^2 \le 2M^2\|\Pi\|\sum_i\Delta t_i =2M^2T\|\Pi\|. $$ The other term in the identity tends to zero because \(\sum_iw(\xi_i)\Delta t_i\) is a tagged Riemann sum for the continuous function \(w\). Both terms therefore tend to zero as \(\|\Pi\|\to0\), proving mean-square convergence. Finally, for every \(\varepsilon>0\), Markov's inequality gives $$ \mathbb{P}\left(\left|S_{\Pi}(w)-\int_0^T w(t)\,dt\right|\ge\varepsilon\right) \le \frac{\mathbb{E}\left[(S_{\Pi}(w)-\int_0^T w(t)\,dt)^2\right]}{\varepsilon^2}, $$ so the same limit gives convergence in probability. \(\square\)
The formula separates two sources of error. The mean of the random sum is a Riemann sum, so its difference from the integral is deterministic. The variance measures the remaining random fluctuation. Independent increments make that variance a sum of terms of order \((\Delta t_i)^2\), which vanishes as the partition is refined.
Worked Example: The Linear Weight \(w(t)=t\)
Take \([0,T]\), use the uniform partition \(t_i=iT/n\) for \(0\le i\le n\), and use left endpoints as tags. Then \(\Delta t_i=T/n\), \(w(t_i)=iT/n\), and $$ S_n(t)=\sum_{i=0}^{n-1}\frac{iT}{n}(\Delta B_i)^2. $$ The expectation is $$ \mathbb{E}[S_n(t)] =\sum_{i=0}^{n-1}\frac{iT}{n}\frac{T}{n} =\frac{T^2}{n^2}\sum_{i=0}^{n-1}i =\frac{T^2}{n^2}\frac{n(n-1)}2 =\frac{T^2(n-1)}{2n}. $$ The target integral is \(\int_0^T t\,dt=T^2/2\), so the bias is \(-T^2/(2n)\). The variance is $$ \begin{aligned} \operatorname{Var}(S_n(t)) &=2\sum_{i=0}^{n-1}\left(\frac{iT}{n}\right)^2\left(\frac{T}{n}\right)^2\\ &=\frac{2T^4}{n^4}\sum_{i=0}^{n-1}i^2 =\frac{2T^4}{n^4}\frac{n(n-1)(2n-1)}6\\ &=\frac{T^4(n-1)(2n-1)}{3n^3}. \end{aligned} $$ This tends to zero, as does the squared bias. For a finite-partition check, when \(n=2\) the only nonzero weight is \(w(T/2)=T/2\). Thus \(\mathbb{E}[S_2(t)]=(T/2)(T/2)=T^2/4\) and \(\operatorname{Var}(S_2(t))=(T/2)^2\,2(T/2)^2=T^4/8\). The formulas give \(T^2(2-1)/(2\cdot2)=T^2/4\) and \(T^4(2-1)(4-1)/(3\cdot2^3)=T^4/8\), respectively.
Worked Example: The Quadratic Weight \(w(t)=t^2\)
On \([0,1]\), take the uniform partition \(t_i=i/n\) and left tags. The sum is \(S_n(t^2)=\sum_{i=0}^{n-1}(i/n)^2(\Delta B_i)^2\). Its expectation is $$ \mathbb{E}[S_n(t^2)] =\frac1{n^3}\sum_{i=0}^{n-1}i^2 =\frac{n(n-1)(2n-1)}{6n^3} =\frac{(n-1)(2n-1)}{6n^2}. $$ The target is \(\int_0^1t^2\,dt=1/3\). For the variance, use \(\operatorname{Var}((\Delta B_i)^2)=2/n^2\): $$ \begin{aligned} \operatorname{Var}(S_n(t^2)) &=2\sum_{i=0}^{n-1}\left(\frac{i}{n}\right)^4\left(\frac1n\right)^2\\ &=\frac{2}{n^6}\sum_{i=0}^{n-1}i^4\\ &=\frac{(n-1)(2n-1)(3n^2-3n-1)}{15n^5}. \end{aligned} $$ Here the finite sums use \(\sum_{i=0}^{n-1}i^2=n(n-1)(2n-1)/6\) and \(\sum_{i=0}^{n-1}i^4=(n-1)n(2n-1)(3n^2-3n-1)/30\). The expectation tends to \(1/3\), and the variance tends to zero. For a check at \(n=2\), only the interval starting at \(1/2\) has nonzero weight. Its weighted expectation is \((1/2)^2(1/2)=1/8\), and its variance is \((1/2)^4\,2(1/2)^2=1/32\). Substitution into the formulas gives \(1\cdot3/(6\cdot4)=1/8\) and \(1\cdot3\cdot5/(15\cdot32)=1/32\), as required.
Worked Example: A Polynomial Weight \(w(t)=1+2t\)
On \([0,1]\), again use the uniform partition and left tags. Now \(w(t_i)=1+2i/n\), and $$ S_n(1+2t)=\sum_{i=0}^{n-1}\left(1+\frac{2i}{n}\right)(\Delta B_i)^2. $$ The expectation is $$ \begin{aligned} \mathbb{E}[S_n(1+2t)] &=\frac1n\sum_{i=0}^{n-1}\left(1+\frac{2i}{n}\right)\\ &=\frac1n\left(n+\frac2n\frac{n(n-1)}2\right) =1+\frac{n-1}{n} =2-\frac1n. \end{aligned} $$ The target integral is \(\int_0^1(1+2t)\,dt=1+1=2\). For completeness, the variance is $$ \begin{aligned} \operatorname{Var}(S_n(1+2t)) &=\frac{2}{n^2}\sum_{i=0}^{n-1}\left(1+\frac{2i}{n}\right)^2\\ &=\frac{2}{n^2}\left(n+2(n-1)+\frac{2(n-1)(2n-1)}{3n}\right). \end{aligned} $$ In the last equality, expand the square and use \(\sum i=n(n-1)/2\) and \(\sum i^2=n(n-1)(2n-1)/6\). The expression tends to zero, while the expectation tends to \(2\), so the theorem applies. At \(n=1\), the sum is \(B_1^2\): its expectation is \(1\), matching \(2-1=1\), and its variance is \(2\), matching the displayed variance formula at \(n=1\). This check also shows that the finite sum is still random even though its expectation is simple.
Almost-Sure Convergence Along Dyadic Partitions
Mean-square convergence implies convergence in probability, but it does not by itself imply almost-sure convergence along every sequence of partitions. For dyadic partitions, a stronger conclusion follows from a summable variance bound. Take the left-tagged partition of \([0,T]\) into \(2^m\) equal intervals and write its sum as \(S_m(w)\).
Proof. Set \(Y_m=S_m(w)-\mathbb{E}[S_m(w)]\). The theorem's variance calculation, with \(\|\Pi\|=T/2^m\), gives $$ \operatorname{Var}(Y_m)\le \frac{2M^2T^2}{2^m}. $$ For any \(\varepsilon>0\), Chebyshev's inequality therefore gives $$ \mathbb{P}(|Y_m|\ge\varepsilon)\le\frac{2M^2T^2}{\varepsilon^2\,2^m}. $$ The sum of these upper bounds over \(m\ge0\) is finite. The first Borel–Cantelli lemma says that if the sum of the probabilities of events is finite, then with probability one only finitely many of those events occur. Applying it for each \(\varepsilon=1/r\), where \(r\) ranges over the positive integers, and taking the intersection of these countably many probability-one events gives \(Y_m\to0\) almost surely. The expectations \(\mathbb{E}[S_m(w)]\) are deterministic left Riemann sums, so they tend to \(\int_0^T w(t)\,dt\). Adding these two limits proves the corollary. \(\square\)
The mesh condition matters. If it is dropped, even the unweighted sum need not approach \(T\). Keep the single-interval partition \([0,T]\) at every stage and take \(w(t)=1\). Every sum is then \(B_T^2\), which has variance \(2T^2>0\); this constant sequence of random variables cannot converge in probability to the constant \(T\). Thus refinement is essential to the stated convergence result. The theorem does not require the individual squared increments to equal their interval lengths; it controls their accumulated error as the mesh decreases.
Common mistake. Replacing \((\Delta B_i)^2\) by \(\Delta t_i\) in each finite sum treats a random quantity as if it were exactly its expectation. For example, on \([0,1]\) with one interval and \(w(t)=1\), the actual sum is \(B_1^2\), not \(1\). It has expectation \(1\) and variance \(2\). The correction is to use \(\mathbb{E}[(\Delta B_i)^2]=\Delta t_i\) only when calculating the expectation; the variance calculation and the mesh limit are what show that the random sum itself converges.
Check Your Understanding
For each question, distinguish the finite-partition sum from its limiting integral.
- For \(w(t)=1\) on \([0,T]\), what are the expectation and variance of the sum on a uniform partition into \(n\) intervals?
- Why does independence of Brownian increments allow the variance of the weighted sum to be written as a sum?
- On \([0,1]\), use left tags and two equal intervals to calculate the expectation of the sum for \(w(t)=t\). Compare it with the limiting integral.
- Which step in the mean-square proof uses continuity of \(w\), and which step uses independent increments?
- Why does the dyadic almost-sure argument not establish almost-sure convergence along every possible sequence of partitions?