Lecture 28. Signature of the Brownian rough path

Since a $d$ -dimensional Brownian motion $(B_t)_{t \ge 0}$ is a $p$ -rough path for $p > 2$ , we know how to give a sense to the signature of the Brownian motion.
In particular, the iterated integrals at any order of the Brownian motion are well defined using rough path theory. It turns out that these iterated integrals do not coincide with iterated Ito’s integrals but with iterated Stratonovitch integrals.
We start with some reminders about Stratonovitch integration. Let $(B_t)_{t \ge 0}$ be a one dimensional Brownian motion defined on a filtered probability space $(\Omega, (\mathcal{F}_t)_{t \ge 0}, \mathbb{P})$ . Let $(\Theta_t)_{0 \le t \le T}$ be a $\mathcal{F}$ adapted process such that $\mathbb{E} \left( \int_0^T \Theta_s^2 ds \right)< +\infty$ . The Stratonovitch integral of $\Theta$ against $B$ can be defined as the limit in probability of the sums
$\sum_{k=0}^{n-1} \frac{\Theta_{t_{k+1}^n} +\Theta_{t_{k}^n}}{2} (B_{t_{k+1}^n}-B_{t_{k}^n}),$
where $0=t_0^n \le t_1^n \le \cdots \le t_n^n=T$ is a sequence of subdivisions whose mesh goes to 0. This limit is denoted $\int_0^T \Theta_s \circ dB_s$ and does not depend on the choice of the subdivision. It is an easy exercise to see that the relation between Ito’s integral and Stratonovitch’s is given by:
$\int_0^T \Theta_s \circ dB_s=\int_0^T \Theta_s dB_s+\frac{1}{2} \langle \Theta, B \rangle_T,$
where $\langle \Theta, B \rangle_T$ is the quadratic covariation between $\Theta$ and $B$ .
If $(B_t)_{t \ge 0}$ is $d$ dimensional Brownian motion, we can then inductively define the iterated Stratonovitch integrals $\int_{0 \leq t_1 \leq ... \leq t_k \leq t} \circ dB^{i_1}_{t_1} \cdots \circ dB^{i_k}_{t_k}$ . The next theorem proves that the signature of the Brownian rough path is given by multiple Stratonovitch integrals.

Theorem: If $(B_t)_{t \ge 0}$ is a $d$ -dimensional Brownian motion, the signature of $B$ as a rough path is the formal series:
$\mathfrak{S} (B)_t = 1+ \sum_{k=1}^{+\infty} \int_{\Delta^k[0,t]} \circ dB ^{\otimes k}$
$=1 + \sum_{k=1}^{+\infty} \sum_{I \in \{1,...,d\}^k} \left( \int_{0 \leq t_1 \leq ... \leq t_k \leq t} \circ dB^{i_1}_{t_1} \cdots \circ dB^{i_k}_{t_k} \right) X_{i_1} \cdots X_{i_k}.$

Proof: Let us work on a fixed interval $[0,T]$ and consider a sequence $D_n$ of subdivisions of $[0,T]$ such that $D_{n+1} \subset D_n$ and whose mesh goes to 0 when $n \to +\infty$ .
As in the previous lecture, we denote by $B^n$ the piecewise linear process which is obtained from $B$ by interpolation along the subdivision $D_n$ , that is for $t_i^n \le t \le t_{i+1}^n$ ,
$B^n_t= \frac{t_{i+1}^n -t}{ t_{i+1}^n-t_i^n} B_{t_i} + \frac{t-t_i^n}{ t_{i+1}^n-t_i^n} B_{t_{i+1}}.$
We know from the previous lecture that $B^n$ converges to $B$ in the $p$ -rough paths topology $2 < p < 3$ . In particular all the iterated integrals $\int_{\Delta^k [s,t]} dB^{n,\otimes k}$ converge. We claim that actually,
$\lim_{n \to \infty} \int_{\Delta^k [s,t]} dB^{n,\otimes k}= \int_{\Delta^k[0,t]} \circ dB ^{\otimes k}.$
Let us denote
$\int_{\Delta^k[s,t]} \partial B ^{\otimes k}=\lim_{n \to \infty} \int_{\Delta^k [s,t]} dB^{n,\otimes k}.$
We are going to prove by induction on $k$ that $\int_{\Delta^k[s,t]} \partial B ^{\otimes k} =\int_{\Delta^k[s,t]} \circ dB ^{\otimes k}$ . We have
$\int_0^T B_s^n \otimes dB_s^{n} =\sum_{i=0}^{n-1} \int_{t_i^n}^{t_{i+1}^n} B_s^n \otimes dB_s^{n}$
$= \sum_{i=0}^{n-1} \int_{t_i^n}^{t_{i+1}^n} \left( \frac{t_{i+1}^n -s}{ t_{i+1}^n-t_i^n} B_{t^n_i} + \frac{s-t_i^n}{ t_{i+1}^n-t_i^n} B_{t^n_{i+1}}\right)ds \otimes \frac{ B_{t^n_{i+1}}- B_{t^n_{i}}}{ t_{i+1}^n-t_i^n}$
$= \frac{1}{2} \sum_{i=0}^{n-1} \left(B_{t^n_{i+1}}- B_{t^n_{i}} \right) \otimes \left( B_{t^n_{i+1}}+ B_{t^n_{i}}\right)$
By taking the limit when $t \to \infty$ , we deduce therefore that $\int_{\Delta^2[0,T]} \partial B ^{\otimes 2} =\int_{\Delta^2[0,T]} \circ dB ^{\otimes 2}$ . In the same way, we have for $0 \le s < t \le T$ , $\int_{\Delta^2[s,t]} \partial B ^{\otimes 2} =\int_{\Delta^2[s,t]} \circ dB ^{\otimes 2}$ . Assume now by induction, that for every $0 \le s \le t \le T$ and $1 \le j \le k$ , $\int_{\Delta^k[s,t]} \partial B ^{\otimes k} =\int_{\Delta^k[s,t]} \circ dB ^{\otimes k}$ . Let us denote
$\Gamma_{s,t}= \int_{\Delta^{k+1}[s,t]} \partial B ^{\otimes (k+1)} -\int_{\Delta^{k+1}[s,t]} \circ dB ^{\otimes (k+1)}.$
From the Chen’s relations, we immediately see that
$\Gamma_{s,u}=\Gamma_{s,t}+\Gamma_{t,u}.$
Moreover, it is easy to estimate
$\|\Gamma_{s,t} \| \le C \omega(s,t)^{\frac{k+1}{p}},$
where $2<p<3$ and $\omega (s,t)= \| \mathbf{B} \|_{p-var, [s,t]}$ , $\mathbf{B}$ being the lift of $B$ in the free Carnot group of step 2. Indeed, the bound
$\int_{\Delta^{k+1}[s,t]} \partial B ^{\otimes (k+1)} \le C_1 \omega(s,t)^{\frac{k+1}{p}},$
comes from the continuity of Lyons' lift and the bound
$\int_{\Delta^{k+1}[s,t]} \circ dB ^{\otimes (k+1)} \le C_2 \omega(s,t)^{\frac{k+1}{p}},$
easily comes from the Garsia-Rodemich-Rumsey inequality. As a conclusion, we deduce that $\Gamma_{s,t}=0$ which proves the induction $\square$

We finish this lecture by a very interesting probabilistic object, the expectation of the Brownian signature.
If
$Y=y_0+\sum_{k = 1}^{+\infty} \sum_{I \in \{1,...,d\}^k} a_{i_1,...,i_k} X_{i_1}...X_{i_k}.$
is a random series, that is if the coefficients are real random variables defined on a probability space, we will denote
$\mathbb{E}(Y)=\mathbb{E}(y_0)+\sum_{k = 1}^{+\infty} \sum_{I \in \{1,...,d\}^k} \mathbb{E}(a_{i_1,...,i_k}) X_{i_1}...X_{i_k}.$
as soon as the coefficients of $Y$ are integrable, where $\mathbb{E}$ stands for the expectation.

Theorem: For $t \ge 0$ ,
$\mathbb{E} \left( \mathfrak{S} (B)_t \right)=\exp \left( t \left(\frac{1}{2}\sum_{i=1}^d X_i^2 \right)\right).$

Proof:
An easy computation shows that if $\mathcal{I}_n$ is the set of words with length $n$ obtained by all the possible concatenations of the words $\{ (i,i) \}, \quad i \in \{1,...,d\}$ , then, if $I \notin \mathcal{I}_n$ then
$\mathbb{E} \left( \int_{\Delta^n [0,t]} \circ dB^I \right) =0$
and if $I \in \mathcal{I}_n$ then
$\mathbb{E} \left( \int_{\Delta^n [0,t]} \circ dB^I \right) =\frac{t^{\frac{n}{2}}}{2^{\frac{n}{2}}\left(\frac{n}{2} \right) ! }$
Therefore,
$\mathbb{E} \left( \mathfrak{S} (B)_t \right) =1+\sum_{k = 1}^{+\infty} \sum_{I \in \mathcal{I}_k} \frac{t^{\frac{k}{2}}}{2^{\frac{k}{2}}\left(\frac{k}{2} \right) ! } X_{i_1}...X_{i_k} = \exp \left( t\left(\frac{1}{2}\sum_{i=1}^d X_i^2 \right)\right)$
$\square$