Mean-Square Convergence of Brownian Quadratic Variation

From Squared Increments to a Quantitative Limit

Brownian increments are small on short time intervals, but their squares accumulate rather than disappearing. The useful fact is not that each squared increment equals the length of its interval: it does not. Instead, the sum over a deterministic partition has a mean equal to the total elapsed time, and its mean-square error is controlled by the largest interval in the partition. This gives a precise sense in which Brownian quadratic variation emerges from increasingly fine observations.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$, adapted to $(\mathcal{F}_t)$. Thus, for $0\le s<t$, $B_t-B_s$ is independent of $\mathcal{F}_s$ and has distribution $N(0,t-s)$. Fix $T\ge0$. A deterministic partition of $[0,T]$ is a finite sequence $0=t_0<t_1<\cdots<t_n=T$, with interval lengths $h_i=t_i-t_{i-1}$ and mesh $|\Pi|=\max_{1\le i\le n}h_i$. Define $\Delta_iB=B_{t_i}-B_{t_{i-1}}$.

Definition: The squared-increment sum on $\Pi$ is $Q_\Pi=\sum_{i=1}^n(\Delta_iB)^2$. Mean-square convergence of $Q_\Pi$ to $T$ means $\mathbb{E}[(Q_\Pi-T)^2]\to0$ as the mesh tends to zero.

The intervals need not have equal lengths. That matters: the exact error depends on all the interval lengths, while a simpler bound depends only on the mesh. We first compute the exact quantities and then deduce the convergence statement.

Exact Mean and Variance

Proposition: For every deterministic partition $\Pi$ of $[0,T]$, $$ \mathbb{E}[Q_\Pi]=T,\qquad \operatorname{Var}(Q_\Pi)=2\sum_{i=1}^n h_i^2,\qquad \mathbb{E}[(Q_\Pi-T)^2]=2\sum_{i=1}^n h_i^2. $$

Proof. For $h_i>0$, the Brownian increment $\Delta_iB$ has the same distribution as $\sqrt{h_i}Z$, where $Z\sim N(0,1)$. The standard Gaussian moments are $\mathbb{E}[Z^2]=1$ and $\mathbb{E}[Z^4]=3$. They follow, for example, by integrating the standard normal density by parts: its derivative is $-z$ times the density, which gives $\mathbb{E}[Z^{2k}]=(2k-1)\mathbb{E}[Z^{2k-2}]$ for $k\ge1$. Hence $\mathbb{E}[(\Delta_iB)^2]=h_i$ and $\mathbb{E}[(\Delta_iB)^4]=3h_i^2$, so $$ \operatorname{Var}((\Delta_iB)^2) =\mathbb{E}[(\Delta_iB)^4]-\big(\mathbb{E}[(\Delta_iB)^2]\big)^2 =3h_i^2-h_i^2=2h_i^2. $$ A zero-length interval, if included, has increment zero and satisfies the same formulas with $h_i=0$. Brownian increments on disjoint successive intervals are independent; therefore their squares are independent, and the variance of their sum is the sum of their variances. Linearity of expectation gives $\mathbb{E}[Q_\Pi]=\sum_i h_i=T$, while independence gives $\operatorname{Var}(Q_\Pi)=2\sum_i h_i^2$. Since $\mathbb{E}[Q_\Pi]=T$, the identity $\mathbb{E}[(Q_\Pi-T)^2]=\operatorname{Var}(Q_\Pi)$ proves the final equality. If $T=0$, every increment and both sides of each equality are zero. $\square$

This proposition identifies the full mean-square error, not just an upper bound. It also shows why the mean alone is not enough: for every partition the mean is exactly $T$, even when the squared-increment sum still has substantial variability.

The Mesh Bound and Mean-Square Convergence

Theorem: For every deterministic partition $\Pi$ of $[0,T]$, $$ \mathbb{E}[(Q_\Pi-T)^2] =2\sum_{i=1}^n h_i^2 \le 2T|\Pi|. $$ Consequently, for any sequence of deterministic partitions $(\Pi_m)$ of a fixed interval $[0,T]$ satisfying $|\Pi_m|\to0$, $Q_{\Pi_m}\to T$ in mean square.

Proof. Each $h_i\le|\Pi|$, so $h_i^2\le|\Pi|h_i$. Summing these inequalities and using $\sum_i h_i=T$ gives $\sum_i h_i^2\le|\Pi|T$. The proposition therefore gives $\mathbb{E}[(Q_\Pi-T)^2]\le2T|\Pi|$. Along any sequence with mesh tending to zero, the right-hand side tends to zero, which is exactly the definition of mean-square convergence. When $T=0$, the sums are identically zero, so the conclusion holds as well. $\square$

The bound applies to every deterministic partition, whether or not its interval lengths are equal. For $T>0$ and $\varepsilon>0$, Chebyshev’s inequality gives the associated strict-threshold estimate $$ \mathbb{P}(|Q_\Pi-T|>\varepsilon) \le \frac{\mathbb{E}[(Q_\Pi-T)^2]}{\varepsilon^2} \le\frac{2T|\Pi|}{\varepsilon^2}. $$ Thus, at a fixed positive threshold, the probability of an error larger than that threshold tends to zero as the mesh tends to zero. The estimate can exceed one for a coarse partition; in that case it remains a valid, but uninformative, upper bound.

The variance formula also records the effect of the partition’s shape. For equal intervals, it is especially simple. For uneven intervals, the exact sum of squared lengths retains information that the mesh bound discards.

Worked Example: Equal Intervals on an Interval of Length Three

Divide $[0,3]$ into $n=6$ equal intervals. Compute the mean, variance, and mean-square error of $Q_\Pi$.

Each interval has length $h_i=3/6=1/2$, so the sum has six squared increments. Its mean is $$ \mathbb{E}[Q_\Pi]=\sum_{i=1}^6h_i=6\cdot\frac12=3. $$ Its variance, and therefore its mean-square error about $3$, is $$ \mathbb{E}[(Q_\Pi-3)^2] =2\sum_{i=1}^6h_i^2 =2\cdot6\cdot\left(\frac12\right)^2 =12\cdot\frac14=3. $$

Check: the mesh bound gives $2T|\Pi|=2\cdot3\cdot(1/2)=3$, equal to the exact error because all intervals have the same length. More generally, $n$ equal intervals on $[0,3]$ have length $3/n$, so the exact error is $2n(3/n)^2=18/n$, which tends to zero as $n$ tends to infinity.

Worked Example: An Uneven Partition

For the partition $0<\tfrac14<\tfrac34<2$ of $[0,2]$, calculate the exact mean-square error and compare it with the mesh bound.

The three interval lengths are $h_1=1/4$, $h_2=1/2$, and $h_3=5/4$. Their sum is $1/4+1/2+5/4=2$, so $\mathbb{E}[Q_\Pi]=2$. For the error, $$ \begin{aligned} \mathbb{E}[(Q_\Pi-2)^2] &=2\left(\left(\frac14\right)^2+\left(\frac12\right)^2+\left(\frac54\right)^2\right)\\ &=2\left(\frac1{16}+\frac4{16}+\frac{25}{16}\right) =2\cdot\frac{30}{16} =\frac{15}{4}. \end{aligned} $$ The mesh is $5/4$, so the general bound is $2\cdot2\cdot(5/4)=5$, which is larger than the exact error $15/4$.

Check: the exact variance is positive, as it must be for a sum of nonconstant squared Gaussian increments. Also, $15/4\le5$, in agreement with the mesh bound. The exact formula uses all three interval lengths; the bound uses only the largest one.

Worked Example: A Strict-Threshold Probability Bound

Let $\Pi$ divide $[0,1]$ into $32$ equal intervals. Bound the probability of the strict event $|Q_\Pi-1|>1/2$.

Each interval has length $1/32$. The exact mean-square error is $$ \mathbb{E}[(Q_\Pi-1)^2] =2\cdot32\cdot\left(\frac1{32}\right)^2 =\frac{64}{1024} =\frac1{16}. $$ Since $1/2>0$, Chebyshev’s inequality applies and yields $$ \mathbb{P}(|Q_\Pi-1|>1/2) \le\frac{1/16}{(1/2)^2} =\frac{1/16}{1/4} =\frac14. $$

Check: the mesh bound gives $2\cdot1\cdot(1/32)=1/16$, the same mean-square error for this equal partition. Dividing by the squared threshold $1/4$ again gives $1/4$. This is an upper bound on the probability, not an assertion that the probability equals $1/4$.

What the Convergence Says—and What It Does Not Say

Mean-square convergence is a statement about the expectation of the squared error. In particular, it implies convergence in probability: for every $\varepsilon>0$, the probability of the strict event $|Q_\Pi-T|>\varepsilon$ is bounded by the mean-square error divided by $\varepsilon^2$. It does not say that any one squared increment equals its interval length. For a single interval of length $h>0$, the random variable $(\Delta B)^2$ has mean $h$ and variance $2h^2$; it is not a constant.

Nor does the theorem, by itself, assert convergence for every sample path. Its conclusion is convergence in mean square along any sequence of deterministic partitions whose meshes tend to zero. Almost-sure convergence along particular partition sequences is a distinct question. Likewise, the identity $[B]_t=t$ is a quadratic-variation statement, while the shorthand $dB_t\,dB_t=dt$ is not an equality between individual increments. The ordinary differential rules $dB_t\,dt=0$ and $dt\,dt=0$ are not needed in this proof; the convergence here follows directly from Gaussian moments and independence.

A common error is to assume that unequal intervals can be handled by the equal-interval variance formula $2T^2/n$. For the partition $0<\tfrac14<\tfrac34<2$, that shortcut would give $2\cdot2^2/3=8/3$, which is not the correct mean-square error. Substituting the actual interval lengths gives $2(1/16+1/4+25/16)=15/4$. The shortcut applies only when all $n$ intervals have length $T/n$; for a general partition, use $2\sum_i h_i^2$.

The practical value of the theorem is its explicit error control. For a target threshold $\varepsilon>0$, the estimate $2T|\Pi|/\varepsilon^2$ shows how refining the largest interval reduces the probability bound uniformly over deterministic partitions with that mesh. The exact variance can be smaller, particularly when the interval lengths are uneven but most are short. In either form, the result makes the passage from finite sums of observations to Brownian quadratic variation quantitative.

Check Your Understanding

Use the exact variance formula first, then compare with the mesh bound where appropriate.

For a partition of $[0,2]$ into four equal intervals, compute the mean and mean-square error of $Q_\Pi$.
Explain why independence of Brownian increments is needed to add the variances of the squared increments.
For interval lengths $1/3$, $2/3$, and $1$, compute the exact mean-square error and the mesh bound.
State the conclusion of mean-square convergence using its defining expectation, and distinguish it from an almost-sure convergence claim.
For $T>0$ and $\varepsilon>0$, derive the upper bound on $\mathbb{P}(|Q_\Pi-T|>\varepsilon)$ from the mean-square error.

Mean-Square Convergence of Brownian Quadratic Variation

What You'll Learn