Lecture 20. The Brownian motion as a rough path (1)

It is now time to give a fundamental example of rough path: The Brownian motion. As we are going to see, a Brownian motion is a $p$ -rough path for any $2 < p < 3$ .

We first remind the following basic definition.
Definition: Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space. A continuous $d$ -dimensional process $(B_t)_{t \ge 0}$ is called a standard Brownian motion if it is a Gaussian process with mean function
$m(t)=0$
and covariance function
$R(s,t)=\mathbb{E}(B_s \otimes B_t)=\min (s,t) \mathbf{I}_d.$

For a Brownian motion $(B_t)_{t \ge 0}$ , the following properties are easy to check:

$B_0=0$ a.s.;
For any $h \geq 0$ , the process $(B_{t+h} - B_h)_{t \ge 0}$ is a standard Brownian motion;
For any $t > s\geq 0$ , the random variable $B_t -B_s$ is independent of the $\sigma$ -algebra $\sigma(B_u, u \le s )$ .
For every $c > 0$ , the process $(B_{ct})_{t \geq 0}$ has the same law as the process $(\sqrt{c} B_t)_{t \geq 0}$ .

An easy computation shows that for $n \ge 0$ and $0 \le s \le t$ :
$\mathbb{E} \left( \|B_t - B_s\|^{2n} \right)=\frac{(2n)!}{2^n n!} (t-s)^n.$
Therefore, as a consequence of the Kolmogorov continuity theorem, for any $T \ge 0$ and $0 \le \varepsilon < 1/2$ , there exists a finite random variable $C_{T, \varepsilon}$ such that for $0 \le s \le t \le T$ ,
$\| B_t -B_s \| \le C_{T, \varepsilon} |t-s|^{1/2-\varepsilon }.$
We deduce in particular that for any $p > 2$ , we have almost surely
$\| B\|_{p-var,[0,T]} < +\infty.$

We now prove that for $1 \le p < 2$ , we have almost surely
$\| B\|_{p-var,[0,T]}=+\infty.$

In the sequel, 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, if $1 \le p < 2$ , for every $T \ge 0$ , almost surely,
$\| B\|_{p-var,[0,T]}=+\infty.$

Proof: We prove the result in dimension 1 and let the reader adapt it to the multidimensional setting.
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 $p$ -variation on the time interval $[0,t]$ if $1 \le p < 2$ . Reasoning by absurd, let us assume that $\| B \|_{p-var,[0,t]} \le M$ . 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
$\sum_{k=1}^{n} \left( B_{t^n_k} -B_{t^n_{k-1}}\right)^2 \le M^p \sup_{1\le k \le n} \mid B_{t^n_k} -B_{t^n_{k-1}} \mid^{2-p} \rightarrow_{n \rightarrow +\infty} 0,$
which is clearly absurd $\square$