Lecture 10. Estimating iterated integrals (Part 1)

In the previous lecture we introduced the signature of a bounded variation path $x$ as the formal series
$\mathfrak{S} (x)_{s,t} =1 + \sum_{k=1}^{+\infty} \int_{\Delta^k [s,t]} dx^{\otimes k}.$
If now $x \in C^{p-var}([0,T],\mathbb{R}^d)$ , $p \ge 1$ the iterated integrals $\int_{\Delta^k [s,t]} dx^{\otimes k}$ can only be defined as Young integrals when $p < 2$ . In this lecture, we are going to derive some estimates that allow to define the signature of some (not all) paths with a finite $p$ variation when $p \ge 2$ . These estimates are due to Terry Lyons in his seminal paper and this is where the rough paths theory really begins.

For $P \in \mathbb{R} [[X_1,...,X_d]]$ that can be writen as
$P=P_0+\sum_{k = 1}^{+\infty} \sum_{I \in \{1,...,d\}^k}a_{i_1,...,i_k} X_{i_1}...X_{i_k},$
we define
$\| P \| =|P_0|+\sum_{k = 1}^{+\infty} \sum_{I \in \{1,...,d\}^k}|a_{i_1,...,i_k}| \in [0,\infty].$
It is quite easy to check that for $P,Q \in \mathbb{R} [[X_1,...,X_d]]$
$\| PQ \| \le \| P \| \| Q\|.$
Let $x \in C^{1-var}([0,T],\mathbb{R}^d)$ . For $p \ge 1$ , we denote
$\left\| \int dx^{\otimes k}\right\|_{p-var, [s,t]}=\left( \sup_{ \Pi \in \mathcal{D}[s,t]} \sum_{i=0}^{n-1} \left\| \int_{\Delta^k [t_i,t_{i+1}]} dx^{\otimes k} \right\|^p \right)^{1/p},$
where $\mathcal{D}[s,t]$ is the set of subdivisions of the interval $[s,t]$ . Observe that for $k \ge 2$ , in general
$\int_{\Delta^k [s,t]} dx^{\otimes k}+ \int_{\Delta^k [t,u]} dx^{\otimes k} \neq \int_{\Delta^k [s,u]} dx^{\otimes k}.$
Actually from the Chen’s relations we have
$\int_{\Delta^n [s,u]} dx^{\otimes n}= \int_{\Delta^n [s,t]} dx^{\otimes k}+ \int_{\Delta^n [t,u]} dx^{\otimes k} +\sum_{k=1}^{n-1} \int_{\Delta^k [s,t]} dx^{\otimes k }\int_{\Delta^{n-k} [t,u]} dx^{\otimes (n-k) }.$
It follows that $\left\| \int dx^{\otimes k}\right\|_{p-var, [s,t]}$ needs not to be the $p$ -variation of $t \to \int_{\Delta^k [s,t]} dx^{\otimes k}$ .
The first major result of rough paths theory is the following estimate:

Proposition: Let $p \ge 1$ . There exists a constant $C \ge 0$ , depending only on $p$ , such that for every $x \in C^{1-var}([0,T],\mathbb{R}^d)$ and $k \ge 0$ ,
$\left\| \int_{\Delta^k [s,t]} dx^{\otimes k} \right\| \le \frac{C^k}{\left( \frac{k}{p}\right)!} \left( \sum_{j=1}^{[p]} \left\| \int dx^{\otimes j}\right\|^{1/j}_{\frac{p}{j}-var, [s,t]} \right)^k, \quad 0 \le s \le t \le T.$

By $\left( \frac{k}{p}\right)!$ , we of course mean $\Gamma \left( \frac{k}{p}+1\right)$ . Some remarks are in order before we prove the result. If $p=1$ , then the estimate becomes
$\left\| \int_{\Delta^k [s,t]} dx^{\otimes k} \right\| \le \frac{C^k}{k!} \| x \|_{1-var, [s,t]}^k,$
which is immediately checked because
$\left\| \int_{\Delta^k [s,t]} dx^{\otimes k} \right\|$
$\le \sum_{I \in \{1,...,d\}^k} \left\| \int_{\Delta^{k}[s,t]}dx^{I} \right\|$
$\le \sum_{I \in \{1,...,d\}^k} \int_{s \le t_1 \le t_2 \le \cdots \le t_k \le t} \| dx^{i_1}(t_1) \| \cdots \| dx^{i_k}(t_k)\|$
$\le \frac{1}{k!} \left( \sum_{j=1}^ d \| x^j \|_{1-var, [s,t]} \right)^k.$

We can also observe that for $k \le p$ , the estimate is easy to obtain because
$\left\| \int_{\Delta^k [s,t]} dx^{\otimes k} \right\| \le \left\| \int dx^{\otimes k}\right\|_{\frac{p}{k}-var, [s,t]}.$
So, all the work is to prove the estimate when $k >p$ . The proof is split into two lemmas. The first one is a binomial inequality which is actually quite difficult to prove:

Lemma: For $x,y >0$ , $n \in \mathbb{N}, n \ge 0$ , and $p \ge 1$ ,
$\sum_{j=0}^n \frac{x^{j/p}}{\left( \frac{j}{p}\right)!} \frac{y^{(n-j)/p}}{\left( \frac{n-j}{p}\right)!} \le p \frac{(x+y)^{n/p}}{ {\left( \frac{n}{p}\right)!}}.$

Proof: See Lemma 2.2.2 in the article by Lyons or this proof for the sharp constant $\square$

The second one is a lemma that actually already was essentially proved in the Lecture on Young’s integral, but which was not explicitly stated.

Lemma: Let $\Gamma: \{ 0 \le s \le t \le T \} \to \mathbb{R}^N$ . Let us assume that:

There exists a control $\tilde{\omega}$ such that
$\lim_{r \to 0} \sup_{(s,t), \tilde{\omega}(s,t) \le r } \frac{\| \Gamma_{s,t} \|}{r}=0;$

There exists a control $\omega$ and $\theta >1, \xi >0$ such that for $0 \le s \le t \le u \le T$ ,
$\| \Gamma_{s,u} \| \le \| \Gamma_{s,t} \|+ \| \Gamma_{t,u} \| +\xi \omega(s,u)^\theta.$

Then, for all $0 \le s \le t \le T$ ,
$\| \Gamma_{s,t} \| \le \frac{\xi}{1-2^{1-\theta}} \omega(s,t)^\theta.$

Proof:
See the proof of the Young-Loeve estimate or Lemma 6.2 in the book by Friz-Victoir $\square$

We can now turn to the proof of the main result.

Proof:
Let us denote
$\omega(s,t)=\left( \sum_{j=1}^{[p]} \left\| \int dx^{\otimes j}\right\|^{1/j}_{\frac{p}{j}-var, [s,t]} \right)^p.$
We claim that $\omega$ is a control. Indeed for $0 \le s \le t \le u \le T$ , we have from Holder’s inequality
$\omega(s,t)+\omega(t,u)$
$= \left( \sum_{j=1}^{[p]} \left\| \int dx^{\otimes j}\right\|^{1/j}_{\frac{p}{j}-var, [s,t]} \right)^p+\left( \sum_{j=1}^{[p]} \left\| \int dx^{\otimes j}\right\|^{1/j}_{\frac{p}{j}-var, [t,u]} \right)^p$
$\le \left( \sum_{j=1}^{[p]}\left( \left\| \int dx^{\otimes j}\right\|^{p/j}_{\frac{p}{j}-var, [s,t]} + \left\| \int dx^{\otimes j}\right\|^{p/j}_{\frac{p}{j}-var, [t,u]}\right)^{1/p} \right)^p$
$\le \left( \sum_{j=1}^{[p]} \left\| \int dx^{\otimes j}\right\|^{1/j}_{\frac{p}{j}-var, [s,u]} \right)^p =\omega(s,u).$

It is clear that for some constant $\beta > 0$ which is small enough, we have for $k \le p$ ,
$\left\| \int_{\Delta^k [s,t]} dx^{\otimes k} \right\| \le \frac{1}{\beta \left( \frac{k}{p}\right)!} \omega(s,t)^{k/p}.$

Let us now consider
$\Gamma_{s,t}= \int_{\Delta^{[p]+1} [s,t]} dx^{\otimes ([p]+1)}.$
From the Chen’s relations, for $0 \le s \le t \le u \le T$ ,
$\Gamma_{s,u}= \Gamma_{s,t}+ \Gamma_{t,u}+\sum_{j=1}^{[p]} \int_{\Delta^j [s,t]} dx^{\otimes j }\int_{\Delta^{[p]+1-j} [t,u]} dx^{\otimes ([p]+1-j) }.$
Therefore,
$\| \Gamma_{s,u}\|$
$\le \| \Gamma_{s,t} \| + \| \Gamma_{t,u} \| +\sum_{j=1}^{[p]} \left\| \int_{\Delta^j [s,t]} dx^{\otimes j }\right\| \left\| \int_{\Delta^{[p]+1-j} [t,u]} dx^{\otimes ([p]+1-j) }\right\|$
$\le \| \Gamma_{s,t} \| + \| \Gamma_{t,u} \| +\frac{1}{\beta^2} \sum_{j=1}^{[p]} \frac{1}{ \left( \frac{j}{p}\right)!} \omega(s,t)^{j/p}\frac{1}{ \left( \frac{[p]+1-j}{p}\right)!} \omega(t,u)^{([p]+1-j)/p}$
$\le \| \Gamma_{s,t} \| + \| \Gamma_{t,u} \| +\frac{1}{\beta^2} \sum_{j=0}^{[p]+1} \frac{1}{ \left( \frac{j}{p}\right)!} \omega(s,t)^{j/p}\frac{1}{ \left( \frac{[p]+1-j}{p}\right)!} \omega(t,u)^{([p]+1-j)/p}$
$\le \| \Gamma_{s,t} \| + \| \Gamma_{t,u} \| +\frac{1}{\beta^2} p \frac{(\omega(s,t)+\omega(t,u))^{([p]+1)/p}}{ {\left( \frac{[p]+1}{p}\right)!}}$
$\le \| \Gamma_{s,t} \| + \| \Gamma_{t,u} \| +\frac{1}{\beta^2} p \frac{\omega(s,u)^{([p]+1)/p}}{ {\left( \frac{[p]+1}{p}\right)!}}.$
On the other hand, we have
$\| \Gamma_{s,t} \| \le A \| x \|_{1-var,[s,t]}^{[p]+1}.$
We deduce from the previous lemma that
$\| \Gamma_{s,t} \| \le \frac{1}{\beta^2} \frac{p}{1-2^{1-\theta}} \frac{\omega(s,t)^{([p]+1)/p}}{ {\left( \frac{[p]+1}{p}\right)!}},$
with $\theta=\frac{[p]+1}{p}$ . The general case $k \ge p$ is dealt by induction. The details are let to the reader $\square$

4 Responses to Lecture 10. Estimating iterated integrals (Part 1)

Taras says:

May 10, 2013 at 3:30 pm

There is a misprint in the definition of the p-variation of iterated integral.
Could you please clarify, how you use Holder inequality in the part of proof the main theorem, that corresponds to the statement that smth is control? I believe, you want to prove that $(\sum a_j)^p+(\sum b_j)^p\leq (\sum (a_j^p+b^p_j)^{1/p})^p$ , however I do not understand how it follows from Holder inequality. Thanks!

- Fabrice Baudoin says:
  
  June 25, 2013 at 7:36 pm
  
  Thanks. This inequality actually follows from the reverse Minkowski inequality, $\| x+ y \|_{1/p} \ge \| x \|_{1/p} +\|y \|_{1/p}$ where $x_i=a_i^p$ and $y_i=b_i^p$ .
  
Michael Hartig says:

January 21, 2014 at 4:55 pm

First of all, thanks a lot for this lecture, it’s helping me quite a bit.
Can you please verify, that in the definition of the p-variation the running index of the sum on the right hand side should be i instead of k?
\left\| \int dx^{\otimes k}\right\|_{p-var, [s,t]}=\left( \sup_{ \Pi \in \mathcal{D}[s,t]} \sum_{k=0}^{n-1} \left\| \int_{\Delta^k [t_i,t_{i+1}]} dx^{\otimes k} \right\|^p \right)^{1/p},

Fabrice Baudoin says:

January 21, 2014 at 7:43 pm

Thanks. This is now corrected.