Arbitrary Partitions and the Error in Quadratic Variation
A partition need not be uniform, nested, or chosen by a special rule for Brownian quadratic-variation sums to converge. What matters for convergence in probability is the largest time step: if that mesh tends to zero, the sum of squared increments approaches the length of the time interval. The exact variance quantifies this statement and shows how uneven partitions affect the error.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\) relative to the filtration. Fix a finite horizon \(T>0\). For a deterministic partition \(\Pi=\{0=t_0<t_1<\cdots<t_n=T\}\), define the interval lengths \(\Delta t_i=t_i-t_{i-1}\), increments \(\Delta B_i=B_{t_i}-B_{t_{i-1}}\), mesh \(|\Pi|=\max_i\Delta t_i\), and quadratic-variation sum \(Q_\Pi=\sum_{i=1}^n(\Delta B_i)^2\).
The deterministic qualification matters: it ensures each increment has a specified time length and that the Brownian increments in the sum are independent. No stochastic integral is needed here; the argument concerns the ordinary random variables formed by the partition sums.
Exact Mean and Variance on One Partition
Proof. Brownian independent increments give independent random variables \(\Delta B_i\), with \(\Delta B_i\sim N(0,\Delta t_i)\). The Gaussian moment formulas are \(\mathbb{E}[(\Delta B_i)^2]=\Delta t_i\) and \(\mathbb{E}[(\Delta B_i)^4]=3(\Delta t_i)^2\). Therefore $$ \mathbb{E}[Q_\Pi]=\sum_{i=1}^n\mathbb{E}[(\Delta B_i)^2] =\sum_{i=1}^n\Delta t_i=T. $$ For each term, the variance is $$ \operatorname{Var}((\Delta B_i)^2) =\mathbb{E}[(\Delta B_i)^4]-\big(\mathbb{E}[(\Delta B_i)^2]\big)^2 =3(\Delta t_i)^2-(\Delta t_i)^2 =2(\Delta t_i)^2. $$ The squared increments are independent, so the variance of their sum is the sum of their variances. This yields the stated variance. Since \(\mathbb{E}[Q_\Pi]=T\), the mean-square error around \(T\) equals the variance: \(\mathbb{E}[(Q_\Pi-T)^2]=\operatorname{Var}(Q_\Pi)\). \(\square\)
The exact formula reflects the sizes of all the intervals, not just their number. A useful bound follows by comparing every interval with the largest one: $$ \sum_{i=1}^n(\Delta t_i)^2 \le |\Pi|\sum_{i=1}^n\Delta t_i = T|\Pi|. $$ Thus \(\operatorname{Var}(Q_\Pi)\le 2T|\Pi|\). Equality holds in this bound when all intervals have the same length; for a nonuniform partition it can be strict.
Worked Example: An Uneven Partition of a Fixed Horizon
Take \(T=2\) and the partition \(0<\tfrac14<1<2\). Its three interval lengths are \(\tfrac14\), \(\tfrac34\), and \(1\). Compute the mean, variance, mean-square error, and mesh bound.
The expectation is the sum of the lengths: $$ \mathbb{E}[Q_\Pi]=\tfrac14+\tfrac34+1=2. $$ The variance is $$ \operatorname{Var}(Q_\Pi) =2\left((\tfrac14)^2+(\tfrac34)^2+1^2\right) =2\left(\tfrac1{16}+\tfrac9{16}+\tfrac{16}{16}\right) =2\cdot\tfrac{26}{16} =\tfrac{13}{4}. $$ The mean-square error is also \(\tfrac{13}{4}\), because the expectation is \(2=T\). The mesh is \(1\), so the general bound is \(2T|\Pi|=2\cdot2\cdot1=4\), and indeed \(\tfrac{13}{4}\le4\).
Check: Adding the individual variances gives the same result: the three contributions are \(2(\tfrac14)^2=\tfrac18\), \(2(\tfrac34)^2=\tfrac98\), and \(2(1)^2=2\). Their sum is \(\tfrac18+\tfrac98+2=\tfrac{10}{8}+2=\tfrac{13}{4}\).
Convergence for Every Vanishing-Mesh Sequence
Proof. Apply the exact mean-square formula to each partition: $$ \mathbb{E}[(Q_{\Pi_m}-T)^2] =2\sum_{i}(\Delta t_{m,i})^2 \le 2T|\Pi_m|. $$ The right-hand side tends to zero by the vanishing-mesh assumption. This proves mean-square convergence. For the probability bound, on the event \(\{|Q_{\Pi_m}-T|>\varepsilon\}\), the nonnegative random variable \((Q_{\Pi_m}-T)^2\) is greater than \(\varepsilon^2\). Consequently, $$ \varepsilon^2\mathbf{1}_{\{|Q_{\Pi_m}-T|>\varepsilon\}} \le (Q_{\Pi_m}-T)^2. $$ Taking expectations and dividing by \(\varepsilon^2>0\) gives $$ \mathbb{P}(|Q_{\Pi_m}-T|>\varepsilon) \le \frac{\mathbb{E}[(Q_{\Pi_m}-T)^2]}{\varepsilon^2} \le \frac{2T|\Pi_m|}{\varepsilon^2}. $$ The final bound tends to zero, which is the definition of convergence in probability. \(\square\)
This argument applies to each deterministic sequence separately and requires no relationship between its partitions. In particular, the partitions need not refine one another. The result establishes convergence in probability for every such sequence; it does not assert almost-sure convergence along every sequence.
Worked Example: Equal Partitions and a Quantitative Probability Bound
Partition \([0,3]\) into \(n\) equal intervals. Find the exact variance and use it to bound the probability that the sum differs from \(3\) by more than \(\tfrac12\). Evaluate the bound for \(n=6\) and \(n=24\).
Each interval has length \(\tfrac3n\), so $$ \operatorname{Var}(Q_\Pi) =2n\left(\tfrac3n\right)^2 =\frac{18}{n}. $$ The general probability bound at threshold \(\varepsilon=\tfrac12\) is $$ \mathbb{P}(|Q_\Pi-3|>\tfrac12) \le \frac{18/n}{(\tfrac12)^2} =\frac{18/n}{\tfrac14} =\frac{72}{n}. $$ For \(n=6\), the exact variance is \(18/6=3\), and the bound is \(72/6=12\). This exceeds \(1\), so it gives no useful restriction beyond the fact that a probability is at most \(1\). For \(n=24\), the exact variance is \(18/24=\tfrac34\), and the bound is \(72/24=3\), also uninformative at this threshold. For \(n=144\), the variance is \(18/144=\tfrac18\), and the bound is \(72/144=\tfrac12\).
Check: The mesh is \(3/n\), and the mesh-based bound gives \(\frac{2\cdot3\cdot(3/n)}{(\tfrac12)^2}=\frac{18/n}{\tfrac14}=\frac{72}{n}\), agreeing with the exact-variance calculation. The bound tending to zero as \(n\) grows is the convergence claim; for a fixed small \(n\), it may be numerically weak.
Worked Example: A Nonuniform Sequence with Vanishing Mesh
For each positive integer \(m\), divide \([0,1]\) into \(m\) equal intervals in the first half and \(2m\) equal intervals in the second half. Compute the mesh, the exact variance, and a probability bound at threshold \(\varepsilon=\tfrac14\).
The first half has \(m\) intervals, each of length \(\frac{1/2}{m}=\frac1{2m}\). The second half has \(2m\) intervals, each of length \(\frac{1/2}{2m}=\frac1{4m}\). The mesh is therefore \(|\Pi_m|=\frac1{2m}\). The sum of squared lengths is $$ m\left(\frac1{2m}\right)^2+2m\left(\frac1{4m}\right)^2 =\frac1{4m}+\frac1{8m} =\frac3{8m}. $$ Thus $$ \operatorname{Var}(Q_{\Pi_m})=2\cdot\frac3{8m}=\frac3{4m}. $$ At threshold \(\frac14\), the probability bound is $$ \mathbb{P}(|Q_{\Pi_m}-1|>\tfrac14) \le \frac{3/(4m)}{(1/4)^2} =\frac{3/(4m)}{1/16} =\frac{12}{m}. $$ This tends to zero as \(m\to\infty\), although intervals in the two halves have different lengths.
Check: The mesh estimate gives \(2T|\Pi_m|=2\cdot1\cdot\frac1{2m}=\frac1m\). The exact variance \(\frac3{4m}\) is no greater than \(\frac1m\), as required.
Why the Mesh Condition Matters
The estimate depends on the largest interval tending to zero, not merely on the number of partition points increasing. For contrast, consider partitions of \([0,1]\) with points \(0<\frac1m<1\), for \(m\ge2\). These have two intervals, of lengths \(\frac1m\) and \(1-\frac1m\). Their mesh is \(1-\frac1m\), which tends to \(1\), not to zero. Their exact variance is $$ 2\left(\left(\frac1m\right)^2+\left(1-\frac1m\right)^2\right) =2\left(\frac1{m^2}+1-\frac2m+\frac1{m^2}\right) =2-\frac4m+\frac4{m^2}, $$ which tends to \(2\), not to zero. This calculation shows why simply adding a point does not make the error vanish if a large interval remains.
Common mistake: A tempting but incorrect calculation is to replace each squared interval length by the interval length and conclude that the variance is \(2\sum_i\Delta t_i=2T\), independent of the partition. For the uneven partition \(0<\frac14<1<2\), this would give \(2T=4\). The correct calculation squares each length: the variance is \(2(\frac1{16}+\frac9{16}+1)=\frac{13}{4}\), not \(4\). The distinction is essential: it is the sum of squared lengths that shrinks as the mesh shrinks.
For a fixed horizon \(T>0\), the exact error variance and the mesh bound together give a practical way to assess a deterministic partition. A uniform partition minimizes the sum of squared lengths among partitions with a fixed number of intervals, while a coarse interval increases that sum. Regardless of how uneven the intervals are, vanishing mesh guarantees convergence in probability through the displayed bound.
Check Your Understanding
Use the exact variance formula and the mesh estimate to answer these questions. Probability events use the strict deviation condition shown.
- For a deterministic partition of \([0,4]\) with interval lengths \(\frac12,\frac32,2\), compute the expectation and variance of \(Q_\Pi\).
- For equal partitions of \([0,2]\) into \(n\) intervals, derive the exact variance and the bound for \(\mathbb{P}(|Q_\Pi-2|>\varepsilon)\).
- Explain why vanishing mesh implies that \(\sum_i(\Delta t_i)^2\to0\).
- For a sequence of partitions with a fixed interval of length \(\frac14\) in every partition, can the mesh tend to zero? Explain.
- State precisely which convergence conclusion follows from the variance estimate, and whether the estimate alone claims almost-sure convergence along every deterministic sequence.