Quadratic Variation

The increments of Brownian motion are so jagged that the path has no derivative and no finite length, yet a different accumulated quantity, the sum of squared increments, is perfectly regular and equals the elapsed time. This quadratic variation is the reason a stochastic integral cannot be defined pathwise as a Riemann-Stieltjes integral and the reason the chain rule for Brownian functions carries a second-order correction. This post proves that Brownian motion has quadratic variation equal to time, building on the probability-space and independence results developed earlier [1], [2]. Throughout, $W$ is a standard Brownian motion and partitions of $[0,t]$ are $0=t_0<t_1<\cdots<t_m=t$ with mesh $\max_i(t_i-t_{i-1})$ .

#The quadratic variation of a path

Definition1

The quadratic variation of a process $X$ over $[0,t]$ along a partition is $\sum_{i=1}^m(X_{t_i}-X_{t_{i -1}})^2$ . The process has quadratic variation $\qv{X}_t$ when these sums converge to $\qv{X}_t$ in $L^2$ as the mesh tends to $0$ (which in particular forces convergence in probability).

The convergence mode is part of the definition, so the value of $\qv{X}_t$ depends a priori on the sequence of partitions. The results below settle this for the two cases we need. A deterministic $C^1$ path has $\qv{X}_t=0$ pathwise (immediately below); Brownian motion has $\qv{W}_t=t$ in $L^2$ along every mesh- $\to 0$ sequence (Theorem 2), with the convergence upgrading to almost sure along the dyadic sequences (Corollary 3).

For a continuously differentiable path the quadratic variation is zero. With $f\in C^1$ on $[0,t]$ each increment obeys $\abs{f(t_i)-f(t_{i-1})}\le\|f'\|_\infty(t_i-t_{i-1})$ , so $\max_i\abs{\Delta f}\le\|f'\|_ \infty\cdot\text{mesh}$ and

\sum_i(\Delta f)^2\le\big(\max_i\abs{\Delta f}\big)\sum_i\abs{\Delta f}\le\|f'\|_\infty\cdot\text{mesh}\cdot V, \tag{1}

where $V=\int_0^t\abs{f'}\le\|f'\|_\infty\,t$ is the (finite) total variation. The product vanishes because the mesh tends to $0$ while $V$ stays finite. A nonzero quadratic variation is therefore a signature of nonsmoothness, and Brownian motion has the simplest possible one.

#Brownian motion accumulates quadratic variation

Theorem2

For a standard Brownian motion $W$ and any partition of $[0,t]$ , the sum $S=\sum_{i=1}^m(W_{t_i}-W_{t_{i-1}} )^2$ has $\E[S]=t$ and $\Var(S)=\sum_{i=1}^m 2(t_i-t_{i-1})^2$ . Consequently $S\to t$ in mean square as the mesh tends to $0$ , so $\qv{W}_t=t$ .

Proof

Write $\Delta_i=W_{t_i}-W_{t_{i-1}}$ , which by the increment law is $N(0,t_i-t_{i-1})$ and independent across $i$ . The second moment is $\E[\Delta_i^2]=t_i-t_{i-1}$ , so $\E[S]=\sum_i(t_i-t_{i-1})=t$ by telescoping. For the variance, the $\Delta_i^2$ are uncorrelated (independence of the increments across $i$ suffices), so $\Var(S)=\sum_i\Var(\Delta_i^2)$ . A centred Gaussian of variance $\sigma^2$ has fourth moment $3\sigma^4$ , read from the moment expansion of its characteristic function, so $\Var(\Delta_i^2)=\E[ \Delta_i^4]-\E[\Delta_i^2]^2=3(t_i-t_{i-1})^2-(t_i-t_{i-1})^2=2(t_i-t_{i-1})^2$ . Hence

\E[(S-t)^2]=\Var(S)=\sum_{i=1}^m 2(t_i-t_{i-1})^2\le 2\,\big(\max_i(t_i-t_{i-1})\big)\sum_{i=1}^m(t_i-t_{i-1} )=2t\cdot\text{mesh}, \tag{2}

which tends to $0$ as the mesh does. So $S\to t$ in mean square, hence $\qv{W}_t=t$ .

Along the dyadic partitions, where $[0,t]$ is cut into $2^n$ equal pieces, the convergence is almost sure, not merely in mean square.

Corollary3

For the dyadic partitions of $[0,t]$ , the quadratic-variation sums $S_n$ converge to $t$ almost surely.

Proof

With $2^n$ pieces each of length $t2^{-n}$ , the bound Equation (2) gives $\E[(S_n-t)^2]=2\cdot 2^n (t2^{-n})^2=2t^2 2^{-n}$ . By Chebyshev's inequality, $\P(\abs{S_n -t}>\varepsilon)\le 2t^2 2^{-n}/\varepsilon^2$ , a convergent series in $n$ , so by the Borel-Cantelli lemma the event $\abs{S_n-t}>\varepsilon$ occurs only finitely often. Taking $\varepsilon=1/k$ and intersecting the countably many resulting probability-one events gives $S_n\to t$ almost surely.

#Infinite total variation

The roughness that produces a nonzero quadratic variation forbids a finite length.

Corollary4

Almost surely, Brownian motion has infinite total variation on every interval $[0,t]$ with $t>0$ .

Proof

Suppose the total variation $V=\sup\sum_i\abs{W_{t_i}-W_{t_{i-1}}}$ were finite on a path. Along the dyadic partitions,

S_n=\sum_i(W_{t_i}-W_{t_{i-1}})^2\le\Big(\max_i\abs{W_{t_i}-W_{t_{i-1}}}\Big)\sum_i\abs{W_{t_i}-W_{t_{i-1}}} \le V\max_i\abs{W_{t_i}-W_{t_{i-1}}}. \tag{3}

Since $W$ is the continuous version of Brownian motion fixed in the construction, the event $A$ that the path is continuous has $\P(A)=1$ ; let $B$ be the event $S_n\to t$ from Corollary 3, also of probability one, so $\P(A\cap B)=1$ . Fix $\omega\in A\cap B$ with finite total variation $V(\omega)$ . Continuity on the compact $[0,t]$ is uniform, so the maximal increment over the dyadic partition tends to $0$ as $n\to\infty$ , forcing $S_n(\omega)\to 0$ by Equation (3). This contradicts $S_n(\omega)\to t>0$ , so $V(\omega)=\infty$ for every $\omega\in A\cap B$ , a set of probability one.

The infinite total variation is exactly why the stochastic integral cannot be built as a pathwise Stieltjes integral against $W$ , since that construction requires the integrator to have finite variation. The stochastic integral instead uses the finiteness of the quadratic variation, isometrically matching the second moment of the integral with an ordinary integral against $d\qv{W}_s=ds$ .

#Covariation

Polarising the quadratic variation gives a bilinear pairing of two processes.

Definition5

The covariation of $X$ and $Y$ over $[0,t]$ is $\qv{X,Y}_t=\lim\sum_i(X_{t_i}-X_{t_{i-1}})(Y_{t_i}-Y_{t _{i-1}})$ , the $L^2$ limit along partitions of vanishing mesh. When the three quadratic variations on the right exist, linearity of the $L^2$ limit gives $\qv{X,Y}_t=\tfrac12(\qv{X+Y}_t-\qv X_t-\qv Y_t)$ .

For a second standard Brownian motion $W'$ independent of $W$ the covariation is zero. Then $W+W'$ has independent increments with $(W+W')_{t_i}-(W+W')_{t_{i-1}}\sim N(0,2(t_i-t_{i-1}))$ , so its quadratic- variation sum has mean $\sum_i 2(t_i-t_{i-1})=2t$ and variance $8\sum_i(t_i-t_{i-1})^2\le 8t\cdot\text{mesh} \to 0$ , giving $\qv{W+W'}_t=2t$ in $L^2$ . Since the algebraic identity $(\Delta W)(\Delta W')=\tfrac12[( \Delta W+\Delta W')^2-(\Delta W)^2-(\Delta W')^2]$ and all three quadratic variations converge in $L^2$ over the same partitions, the cross sum has $L^2$ limit $\qv{W,W'}_t=\tfrac12(2t-t-t)=0$ . The covariation of a Brownian motion with any continuous finite-variation process $A$ is likewise zero by the same estimate as Equation (3). The cross sum obeys $\abs{\sum_i(\Delta W)(\Delta A)}\le(\max_i\abs{ \Delta W})\sum_i\abs{\Delta A}\le V_A\max_i\abs{\Delta W}\to 0$ , since continuity of $W$ on $[0,t]$ sends the maximal increment to $0$ while the total variation $V_A$ stays finite. So smooth drifts do not interact with Brownian noise at second order. These facts assemble into the multiplication rule $dW\,dW=dt$ , $dW\,dt=0$ , $dt\,dt=0$ . This bookkeeping drives Ito's formula, whose second-order term is exactly the quadratic variation that smooth calculus never sees.

[1]

I. Karatzas and S. E. Shreve, Brownian Motion and Stochastic Calculus, 2nd ed. Springer, 1991.

[2]

J.-F. Le Gall, Brownian Motion, Martingales, and Stochastic Calculus. Springer, 2016.

Explore connections

see in the atlas

referenced by (1)

Ito's Formula

cite

@misc{quadratic-variation,
  author = {Zac Kienzle},
  title  = {Quadratic Variation},
  year   = {2026},
  month  = {06},
  url    = {https://zackienzle.com/blog/quadratic-variation}
}

#The quadratic variation of a path

Definition1

\sum_i(\Delta f)^2\le\big(\max_i\abs{\Delta f}\big)\sum_i\abs{\Delta f}\le\|f'\|_\infty\cdot\text{mesh}\cdot V, \tag{1}

#Brownian motion accumulates quadratic variation

Theorem2

Proof

\E[(S-t)^2]=\Var(S)=\sum_{i=1}^m 2(t_i-t_{i-1})^2\le 2\,\big(\max_i(t_i-t_{i-1})\big)\sum_{i=1}^m(t_i-t_{i-1} )=2t\cdot\text{mesh}, \tag{2}

which tends to $0$ as the mesh does. So $S\to t$ in mean square, hence $\qv{W}_t=t$ .

Along the dyadic partitions, where $[0,t]$ is cut into $2^n$ equal pieces, the convergence is almost sure, not merely in mean square.

Corollary3

For the dyadic partitions of $[0,t]$ , the quadratic-variation sums $S_n$ converge to $t$ almost surely.

Proof

#Infinite total variation

The roughness that produces a nonzero quadratic variation forbids a finite length.

Corollary4

Almost surely, Brownian motion has infinite total variation on every interval $[0,t]$ with $t>0$ .

Proof

Suppose the total variation $V=\sup\sum_i\abs{W_{t_i}-W_{t_{i-1}}}$ were finite on a path. Along the dyadic partitions,

S_n=\sum_i(W_{t_i}-W_{t_{i-1}})^2\le\Big(\max_i\abs{W_{t_i}-W_{t_{i-1}}}\Big)\sum_i\abs{W_{t_i}-W_{t_{i-1}}} \le V\max_i\abs{W_{t_i}-W_{t_{i-1}}}. \tag{3}

#Covariation

Polarising the quadratic variation gives a bilinear pairing of two processes.

Definition5

[1]

I. Karatzas and S. E. Shreve, Brownian Motion and Stochastic Calculus, 2nd ed. Springer, 1991.

[2]

J.-F. Le Gall, Brownian Motion, Martingales, and Stochastic Calculus. Springer, 2016.

Explore connections

see in the atlas

referenced by (1)

Ito's Formula

cite

@misc{quadratic-variation,
  author = {Zac Kienzle},
  title  = {Quadratic Variation},
  year   = {2026},
  month  = {06},
  url    = {https://zackienzle.com/blog/quadratic-variation}
}