The Quadratic Variation Process of Brownian Motion

From Partition Sums to a Process

A partition sum on one fixed interval identifies a number; the quadratic variation process records the corresponding limit at every time. For Brownian motion, that process is $t\mapsto t$. Thus the increase in quadratic variation over $[s,t]$ is exactly $t-s$, and the path of the quadratic variation process is continuous and increasing. This is a statement about limits of sums of squared increments, not an ordinary chain rule.

Throughout, 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 $(\mathcal{F}_t)$. No stochastic integral is needed here. For a partition $\Pi=\{0=t_0<t_1<\cdots<t_n=T\}$, write $\Delta t_i=t_i-t_{i-1}$, $\Delta B_i=B_{t_i}-B_{t_{i-1}}$, and $|\Pi|=\max_i\Delta t_i$.

Definition: The quadratic variation sum up to time $t$ is the sum of squared increments over the partition intervals completed by $t$. The quadratic variation process $[B]_t$ is the limit in probability of these sums as the partition mesh tends to zero. Equivalently, one may include the squared increment from the last partition point before $t$ to $t$; for Brownian motion, this convention gives the same limit.

The process is defined by considering all times, not just the endpoint of one fixed interval. To make that precise, the next result establishes convergence uniformly in probability on a bounded time interval. This means that, for every $\varepsilon>0$, the probability that the largest error over $0\le t\le T$ exceeds $\varepsilon$ tends to zero.

The Brownian Quadratic Variation Process

Theorem: Let $\Pi_n$ be any deterministic partitions of $[0,T]$ with $|\Pi_n|\to0$, and let $Q_n(t)$ be the sum of $(\Delta B_i)^2$ over intervals completed by time $t$. Then $$ \sup_{0\le t\le T}|Q_n(t)-t|\longrightarrow 0 $$ in probability. Consequently, $[B]_t=t$ on $[0,T]$, and this process has continuous, strictly increasing paths.

Proof. Fix $u\in[0,T]$, and let $\tau_n(u)$ be the last partition point not exceeding $u$. The increments included in $Q_n(u)$ are independent centered normal variables with variances equal to their interval lengths. A centered normal variable of variance $h$ has square with mean $h$ and variance $2h^2$. Therefore $$ \mathbb{E}[Q_n(u)]=\tau_n(u),\qquad \operatorname{Var}(Q_n(u))=2\sum_{t_i\le u}(\Delta t_i)^2 \le 2T|\Pi_n|. $$ Since $0\le u-\tau_n(u)\le|\Pi_n|$, we obtain $$ \mathbb{E}\big[(Q_n(u)-u)^2\big] =\operatorname{Var}(Q_n(u))+(\tau_n(u)-u)^2 \le 2T|\Pi_n|+|\Pi_n|^2. $$ Thus $Q_n(u)\to u$ in probability at every fixed $u$.

To pass from fixed times to a uniform statement, choose a finite grid $0=u_0<u_1<\cdots<u_m=T$ whose largest spacing is at most $\eta$, where $\eta>0$. The process $Q_n$ is nondecreasing. For $u_j\le t\le u_{j+1}$, this gives $Q_n(u_j)\le Q_n(t)\le Q_n(u_{j+1})$, so $$ |Q_n(t)-t|\le \max\{|Q_n(u_j)-u_j|,\ |Q_n(u_{j+1})-u_{j+1}|\}+\eta. $$ For fixed $\eta$, the grid has finitely many points, and the fixed-time convergence just proved implies that the maximum error at the grid points tends to zero in probability. Given $\varepsilon>0$, first choose $\eta<\varepsilon/2$; then the probability that the supremum error exceeds $\varepsilon$ tends to zero. This proves uniform convergence in probability. The limit process is $t\mapsto t$, which is continuous and strictly increasing on $[0,T]$. $\square$

Some definitions use a partial final interval, setting $$ \widetilde Q_n(t)=Q_n(t)+(B_t-B_{\tau_n(t)})^2. $$ On each continuous Brownian path, uniform continuity on $[0,T]$ implies that $\sup_{t\le T}|B_t-B_{\tau_n(t)}|\to0$ as the mesh tends to zero. Hence $\sup_{t\le T}|\widetilde Q_n(t)-Q_n(t)|\to0$ almost surely, so this convention has the same limit.

Increments and Scaling

Because the process is identified at every time, its increments follow by subtraction. For $0\le s\le t$, the quadratic variation accumulated between $s$ and $t$ is $[B]_t-[B]_s=t-s$. The interval length, rather than the location of the interval, determines the result.

Worked Example: An Increment on a Nonzero Interval

Find the quadratic variation accumulated by $B$ between $s=\tfrac14$ and $t=\tfrac34$.

The process is $[B]_u=u$, so the increment is $$ [B]_{3/4}-[B]_{1/4}=\frac34-\frac14=\frac12. $$ The interval length is also $t-s=\tfrac34-\tfrac14=\tfrac12$, confirming the result. The starting time does not add any extra term: only the length of the interval enters.

Proposition: For any constant $c\in\mathbb{R}$, the quadratic variation process of $cB$ is $[cB]_t=c^2t$. In particular, if $c=0$, the process is identically zero.

Proof. On each partition interval, the increment of $cB$ is $c\Delta B_i$. Its squared increment is $c^2(\Delta B_i)^2$, so the completed-interval sum for $cB$ is exactly $c^2Q_n(t)$. Since $Q_n$ converges uniformly in probability to $t$, multiplication by the fixed constant $c^2$ gives uniform convergence in probability to $c^2t$. When $c=0$, every partition sum is zero and the conclusion still holds. $\square$

Worked Example: A Negative Scaling Constant

Let $Y_t=-3B_t$. Find its quadratic variation at $t=2$, and its increment between $t=1$ and $t=2$.

Here $c=-3$, so $c^2=(-3)^2=9$. Thus $[Y]_t=9t$, giving $[Y]_2=9\cdot2=18$. The increment is $$ [Y]_2-[Y]_1=18-9\cdot1=9. $$ This agrees with $9(2-1)=9$. The negative sign disappears on squaring: reversing the sign of every Brownian increment does not change its squared value.

Adding a Deterministic Linear Drift

A related calculation distinguishes the effect of the Brownian part from that of a deterministic linear drift. The following result is specific to this form of process; it does not assume a general rule for arbitrary finite-variation paths.

Proposition: Let $X_t=a+bt+cB_t$, where $a,b,c$ are constants. Along deterministic partitions whose mesh tends to zero, the quadratic variation sums for $X$ converge in probability to $c^2t$ on every bounded interval $[0,T]$.

Proof. On an interval of length $\Delta t_i$, $$ \Delta X_i=b\Delta t_i+c\Delta B_i,\qquad (\Delta X_i)^2=b^2(\Delta t_i)^2+2bc\,\Delta t_i\Delta B_i+c^2(\Delta B_i)^2. $$ The first term summed over intervals up to $t\le T$ is bounded by $b^2T|\Pi|$, because $\sum_i(\Delta t_i)^2\le|\Pi|\sum_i\Delta t_i\le T|\Pi|$. It therefore tends to zero.

For the cross term, the Cauchy–Schwarz inequality gives $$ \left|\sum_i\Delta t_i\Delta B_i\right| \le \left(\sum_i(\Delta t_i)^2\right)^{1/2} \left(\sum_i(\Delta B_i)^2\right)^{1/2}. $$ The first factor is at most $(T|\Pi|)^{1/2}$, which tends to zero. The second factor is bounded in probability because the Brownian quadratic variation sums converge in probability to a value no greater than $T$. Hence the product tends to zero in probability. The remaining sum, $c^2\sum_i(\Delta B_i)^2$, converges to $c^2t$. These bounds also apply to sums stopped at any $t\le T$; the uniform convergence of the Brownian sums gives the asserted process limit. The constants $a$ and $b$ contribute no limiting quadratic variation, while $c$ contributes its square. $\square$

Worked Example: Linear Drift and Brownian Noise

For $X_t=5+2t-2B_t$, compute the quadratic variation at $t=2$ and the increment from $t=\tfrac12$ to $t=2$.

The coefficients are $a=5$, $b=2$, and $c=-2$, so $c^2=(-2)^2=4$. The proposition gives $[X]_t=4t$, and therefore $[X]_2=4\cdot2=8$. At the earlier time, $[X]_{1/2}=4\cdot\tfrac12=2$; the increment is $8-2=6$. Directly from the interval length, $c^2(2-\tfrac12)=4\cdot\tfrac32=6$, which checks the calculation.

For the partition-sum check, each increment is $\Delta X_i=2\Delta t_i-2\Delta B_i$. Squaring gives $4(\Delta t_i)^2-8\Delta t_i\Delta B_i+4(\Delta B_i)^2$. The first two sums vanish in probability as shown in the proof, and the last tends to $4\cdot2=8$ on $[0,2]$.

What the Process Tells Us—and a Common Pitfall

The statement $[B]_t=t$ means that the limiting squared-increment accumulation is deterministic, continuous, and strictly increasing. Its increase over a time interval of length $h$ is $h$. This is the quadratic variation content of the shorthand rule $dB_t\,dB_t=dt$; the rules $dB_t\,dt=0$ and $dt\,dt=0$ are also shorthand for vanishing mixed and time-only contributions in the relevant partition calculations. They are not identities between ordinary real-valued increments.

Common mistake: treating a Brownian increment like a time increment. A wrong calculation assumes $\Delta B_i$ is of size $\Delta t_i$, then concludes that $n$ squared increments on $[0,T]$ total roughly $n(T/n)^2=T^2/n$, which tends to zero. But Brownian increments have variance $\Delta t_i$, and their squares have mean $\Delta t_i$. For $n$ equal intervals, the expected sum is therefore $n(T/n)=T$, not a quantity tending to zero. The theorem verifies that these sums converge to $T$ in probability.

A related algebraic check shows where the nonzero limit enters. For a partition of $[0,T]$, expansion of each square gives $$ B_T^2=2\sum_i B_{t_{i-1}}\Delta B_i+\sum_i(\Delta B_i)^2. $$ The final sum converges to $T$. Omitting it would incorrectly apply the ordinary chain rule to the squared Brownian path. The partition limit, rather than a differentiability assumption, determines the quadratic variation process.

Check Your Understanding

Use the process identity and the partition-sum definition to justify each answer.

What is $[B]_4-[B]_{3/2}$, and which interval length does it represent?
For $Y_t=5B_t$, compute $[Y]_t$ and its increment over $[s,t]$.
For $X_t=7-3t+B_t$, find $[X]_3$. Which coefficient determines the result?
Explain why the completed-interval sums are nondecreasing in time and how this helps prove uniform convergence in probability.
Why does the estimate $n(T/n)^2\to0$ fail to describe Brownian quadratic variation?