Lecture 15. Quadratic variation of the Brownian motion paths

Posted on August 24, 2012 by Fabrice Baudoin

If \Delta_n [0,t]=\left\{ 0=t^n_0 \le t^n_1 \le ...\le t^n_n=t\right\}, is a subdivision of the time interval [0,t], we denote by \mid\Delta_n [0,t] \mid=\max \{ \mid t^n_{k+1}-t^n_k \mid , k=0,...,n-1 \}, the mesh of this subdivision.

Proposition. Let (B_t)_{t\ge 0} be a standard Brownian motion. Let t \ge 0. For every sequence \Delta_n [0,t] of subdivisions such that \lim_{n \rightarrow +\infty}\mid\Delta_n [0,t]\mid=0, the following convergence takes place in L^2 (and thus in probability),
\lim_{n \rightarrow +\infty} \sum_{k=1}^{n} \left( B_{t^n_k}-B_{t^n_{k-1}}\right)^2=t.
As a consequence, almost surely, Brownian paths have an infinite variation on the time interval [0,t].

Proof.

Let us denote

V_n=\sum_{k=1}^{n} \left( B_{t^n_k} -B_{t^n_{k-1}}\right)^2.

Thanks to the stationarity and the independence of Brownian increments, we have:
\mathbb{E} \left( (V_n-t)^2\right)=\mathbb{E} \left(V_n^2\right)-2t\mathbb{E} \left( V_n\right)+t^2
=\sum_{j,k=1}^n\mathbb{E} \left( \left( B_{t^n_j} -B_{t^n_{j-1}}\right)^2\left( B_{t^n_k}-B_{t^n_{k-1}}\right)^2\right)-t^2
=\sum_{k=1}^n\mathbb{E} \left( \left( B_{t^n_j}-B_{t^n_{j-1}}\right)^4\right)+2\sum_{1\le j<k\le n}^n\mathbb{E} \left( \left( B_{t^n_j} -B_{t^n_{j-1}}\right)^2\left( B_{t^n_k}-B_{t^n_{k-1}}\right)^2\right)-t^2
=\sum_{k=1}^n (t^n_k-t^n_{k-1})^2 \mathbb{E} \left( B_1^4\right)+2\sum_{1\le j<k\le n}^n (t^n_j-t^n_{j-1})(t^n_k-t^n_{k-1})-t^2
=3\sum_{k=1}^n (t^n_j-t^n_{j-1})^2+2\sum_{1\le j<k\le n}^n (t^n_j-t^n_{j-1})(t^n_k-t^n_{k-1})-t^2
=2\sum_{k=1}^n (t^n_k-t^n_{k-1})^2
\le 2t\mid\Delta_n [0,t]\mid \rightarrow_{n \rightarrow +\infty} 0.

Let us now prove that, as a consequence of this convergence, the paths of the process (B_t)_{t\ge 0} almost surely have an infinite variation on the time interval [0,t]. It suffices to prove that there exists a sequence of subdivisions \Delta_n [0,t] such that almost surely

\lim_{n \rightarrow +\infty} \sum_{k=1}^n \mid B_{t^n_k} -B_{t^n_{k-1}}\mid=+\infty.

Reasoning by absurd, let us assume that the supremum on all the subdivisions of the time interval [0,t] of the sums

\lim_{n \rightarrow +\infty} \sum_{k=1}^n \mid B_{t^n_k} -B_{t^n_{k-1}}\mid

may be bounded from above by some positive M . From the above result, since the convergence in probability implies the existence of an almost surely convergent subsequence, we can find a sequence of subdivisions \Delta_n [0,t] whose mesh tends to 0 and such that almost surely,

\lim_{n \rightarrow +\infty} \sum_{k=1}^{n} \left( B_{t^n_k}-B_{t^n_{k-1}}\right)^2=t.

We get then
\sum_{k=1}^{n} \left( B_{t^n_k} -B_{t^n_{k-1}}\right)^2 \le M \sup_{1\le k \le n} \mid B_{t^n_k} -B_{t^n_{k-1}} \mid \rightarrow_{n \rightarrow +\infty} 0,
which is clearly absurd \square

Exercise. Let (B_t)_{t\ge 0} be a Brownian motion.

  1. Show that for t \ge 0, almost surely
    \lim_{n \rightarrow +\infty}\sum_{k=1}^{2^n} \left(B_{\frac{kt}{2^n}} -B_{\frac{(k-1)t}{2^n}}\right)^2=t.
  2. Show that there exists a sequence of subdivisions \Delta_n [0,t] whose mesh tends to 0 and such that almost surely

    \lim_{n \rightarrow +\infty} \sum_{k=1}^{n} \left( B_{t^n_k} -B_{t^n_{k-1}}\right)^2=+\infty.

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