Quadratic variation of Brownian motion

What is left of the squared increments of a Brownian path when the steps get very small?

Split a time interval into many short steps, square the change of a path over each step, and add the squares. For the paths of ordinary calculus the answer is simple. For Brownian motion it is the reason Itô's formula has an extra term. Stage 1 asks for your prediction before the page shows the answer.

Before you start · 3 quick checks on Brownian increments, convergence in L² and the mean value theorem, about 3 min

Setup: Brownian motion, partitions and two sums

Brownian motion. \(W = (W_s)_{s \ge 0}\) is a standard Brownian motion (step 9 of the path): \(W_0 = 0\), its paths are continuous, and for \(0 \le r \lt s\) the increment \(W_s - W_r\) has the normal distribution \(N(0, s - r)\) and is independent of the path up to time \(r\). So increments over intervals that do not overlap are independent.
Partition. For \(t \gt 0\), a partition of \([0, t]\) into \(n \ge 1\) steps is \(0 = t_0 \lt t_1 \lt \dots \lt t_n = t\). Its steps are \(\Delta t_k = t_k - t_{k-1}\), and its mesh is \(\delta = \max_k \Delta t_k\). The equal partition has \(\Delta t_k = \Delta t = t/n\). The increments are \(\Delta W_k = W_{t_k} - W_{t_{k-1}}\). Every partition on this page is fixed in advance: its points are numbers chosen without looking at the path.
Two sums. Over a partition, the sum of squares is \(\sum_{k=1}^n (\Delta W_k)^2\) and the first variation is \(\sum_{k=1}^n \lvert \Delta W_k \rvert\). For any function \(f\) the same sums use \(\Delta f_k = f(t_k) - f(t_{k-1})\).
Quadratic variation. The quadratic variation of \(f\) on \([0, t]\) is the limit of \(\sum_k (\Delta f_k)^2\) as the mesh goes to 0, when that limit exists. For a random path the sums are random variables, so the limit is taken in a sense of convergence of random variables: in \(L^2\), in probability or almost surely. Stage 3 says which.

How the paths on this page are made

Each path is drawn from a fixed seed, so the numbers are the same on every visit. A path is simulated over 4096 equal time steps of \([0, 1]\): each of its 4096 increments is \(Z/64\) with \(Z\) a standard normal draw, since \(\sqrt{1/4096} = 1/64\). A coarser equal partition with \(n = 2^j\) steps uses every \((4096/n)\)th point of the same path, so splitting every step in two keeps the old points and adds the midpoints.

For an interval \([0, t]\) the page uses the scaled path \(B_s = \sqrt{t}\, W_{s/t}\), \(0 \le s \le t\). By Brownian scaling, \(B\) is again a standard Brownian motion, and its increments over the equal partition of \([0, t]\) are \(\sqrt{t}\) times those of \(W\) over the equal partition of \([0, 1]\).

1 See

Add up the squares

The explorer opens when you have checked your prediction.

2 Try

First variation, a smooth path and single terms

The questions use path 1 of stage 1 and the equal partition of \([0, 1]\), and the explorer at the end of this stage lets you change \(t\).

Explorer

Goals

3 Prove

The sum of squares converges to t in mean square

Theorem (quadratic variation of Brownian motion)

Assumptions.

\(W\) is a standard Brownian motion and \(t \gt 0\). The partition \(0 = t_0 \lt t_1 \lt \dots \lt t_n = t\) is fixed (not random, and not chosen after seeing the path), with \(n \ge 1\) steps \(\Delta t_k = t_k - t_{k-1}\) and mesh \(\delta = \max_k \Delta t_k\).
\(\Delta W_k = W_{t_k} - W_{t_{k-1}}\) and \(Q = \sum_{k=1}^n (\Delta W_k)^2\). Sums over \(k\) run from 1 to \(n\).
Facts used: the increments \(\Delta W_1, \dots, \Delta W_n\) are independent, and \(\Delta W_k\) has distribution \(N(0, \Delta t_k)\). So \(\Delta W_k = \sqrt{\Delta t_k}\, Z_k\) with \(Z_1, \dots, Z_n\) independent standard normal, and \(E[Z_k] = 0\), \(E[Z_k^2] = 1\), \(E[Z_k^4] = 3\).

Conclusion. \(E[Q] = t\) and \(E[(Q - t)^2] = \operatorname{Var}(Q) = 2\sum_k (\Delta t_k)^2 \le 2t\delta\). So along any sequence of fixed partitions whose mesh \(\delta \to 0\), \(Q \to t\) in \(L^2\), and hence in probability. For the equal partition, \(E[(Q - t)^2] = 2t^2/n\).

Scratch work, not part of the proof.

Almost sure convergence

The theorem gives convergence in \(L^2\) and in probability for any sequence of fixed partitions whose mesh goes to 0. Almost sure convergence, path by path, needs more, and it is a statement about one fixed sequence of partitions at a time: for that sequence, there is an event of probability 1 on which the sums converge to \(t\). It holds when each partition refines the one before it (for example the dyadic partitions of this page, where every step is split in two), and it holds when the meshes \(\delta_m\) of the sequence satisfy \(\sum_m \delta_m \lt \infty\). The second case follows from the bound above with the Borel–Cantelli lemma; the first needs a martingale argument. This page states both without proof.

The partitions must not depend on the path. If the points may be chosen after looking at the path, the sums of squares can be made as large as you like, even with the mesh going to 0: the supremum of \(\sum_k (\Delta W_k)^2\) over all partitions of \([0, t]\) is infinite with probability 1.

The table behind \((dW)^2 = dt\)

Itô calculus writes the rules below as a multiplication table. Each entry is shorthand for what a sum over a partition of \([0, t]\) does as the mesh \(\delta \to 0\), under the theorem's assumptions.

Sums over a partition of \([0, t]\) with mesh \(\delta\). The mean square error is \(E[(\text{sum} - \text{limit})^2]\).
Shorthand	Sum	Mean square error	Limit as \(\delta \to 0\)
\(dW \cdot dW = dt\)	\(\sum_k (\Delta W_k)^2\)	\(2\sum_k (\Delta t_k)^2 \le 2t\delta\)	\(t\), in \(L^2\)
\(dW \cdot dt = 0\)	\(\sum_k \Delta W_k\, \Delta t_k\)	\(\sum_k (\Delta t_k)^3 \le t\delta^2\)	\(0\), in \(L^2\)
\(dt \cdot dt = 0\)	\(\sum_k (\Delta t_k)^2\)	\(\big(\sum_k (\Delta t_k)^2\big)^2 \le t^2\delta^2\); the sum is not random	\(0\)

Why the second and third rows hold. The terms \(\Delta W_k\, \Delta t_k\) are independent with mean 0 and variance \((\Delta t_k)^2 \Delta t_k\), so the mean square of their sum is \(\sum_k (\Delta t_k)^3 \le \delta^2 \sum_k \Delta t_k = t\delta^2\). And \(\sum_k (\Delta t_k)^2 \le \delta \sum_k \Delta t_k = t\delta\). The same bound handles any \(C^1\) path \(f\): by the mean value theorem \(\lvert \Delta f_k \rvert \le M \Delta t_k\) with \(M = \max_{[0,t]} \lvert f' \rvert\), so \(\sum_k (\Delta f_k)^2 \le M^2 t\delta \to 0\). The quadratic variation of a \(C^1\) path is 0.

Where the table is used

Take a function \(g\) with two continuous derivatives. In a Taylor expansion of \(g(W_{t_k}) - g(W_{t_{k-1}})\) over each step, the second-order term is \(\tfrac12 g''(W_{t_{k-1}})(\Delta W_k)^2\), plus a remainder. Over a fine partition the sum of these terms behaves like \(\tfrac12 \sum_k g''(W_{t_{k-1}})\,\Delta t_k\), close to \(\tfrac12\int_0^t g''(W_s)\,ds\), which need not vanish; it is 0 for every path when \(g'' = 0\), for example \(g(x) = x\). For a \(C^1\) path in place of \(W\) the same terms would add up to nothing. That is the extra term of Itô's formula. Making this step rigorous needs the Itô integral of step 12; this page proves only the case \(g''= 2\), which is the theorem above.

Key point The proof needs only the mean \(\Delta t_k\) and the variance \(2(\Delta t_k)^2\) of each squared increment, and independence to add the variances. The variances add to \(2\sum_k (\Delta t_k)^2 \le 2t\delta\), which goes to 0 with the mesh, so the sum of squares converges to \(t\) in \(L^2\) and in probability.

4 Apply

Problems

\(W\) is a standard Brownian motion and every partition is equal, \(\Delta t = t/n\). A standard normal \(Z\) has \(E[Z^2] = 1\), \(E[Z^4] = 3\) and \(E\lvert Z \rvert = \sqrt{2/\pi}\).