## HW6. MA3160 Fall 2017

Exercise 1. Patricia receives  an average of two texts every 2 hours. If we assume that the number of texts is Poisson distributed, what is the probability that she receives five or more texts in a 9 hours period?

Exercise 2.  A UConn student claims that she can distinguish Dairy Bar ice cream from Friendly’s ice cream. As a test, she is given ten samples of ice cream (each sample is either from the Dairy Bar or Friendly’s) and asked to identify each one. She is right eight times. What is the probability that she would be right exactly eight times if she guessed randomly for each sample?

## HW5. MA3160 Fall 2017

Exercise 1. Three balls are randomly chosen with replacement from an urn containing 5 blue, 4 red, and 2 yellow balls. Let X denote the number of red balls chosen.

(a) What are the possible values of X?
(b) What are the probabilities associated to each value?

Exercise 2. Suppose X is a random variable such that E[X] = 50 and Var(X) = 12. Calculate the following quantities.

(a) $E[X^2]$
(b) E [3X + 2]

(c) $E [(X+2)^2]$

(d) Var[−X]

## Annales de la faculte des sciences de Toulouse

Annales de la Faculte des Sciences de Toulouse is a peer-reviewed international  journal with a long tradition of excellence (going back to 1887 and Thomas Stieltjes). The journal periodically publishes surveys by the recipients of the Fermat Prize.  The Editorial Board encourages high level submissions.

Submissions in all areas of mathematics are accepted and decisions are usually made within 3 months. The electronic version is free and accessible without subscription.

## MA3160. Fall 2017. Midterm 1 sample

Practice midterm 1

We will do the correction in class on 09/28.

## HW4. MA360 Fall 2017

Exercise 1. Two dice are rolled. Consider the events A = {sum of two dice equals 3}, B = {sum of two dice equals 7 }, and C = {at least one of the dice shows a 1}.

(a) What is P(A | C)?

(b) What is P(B | C)?

(c) Are A and C independent? What about B and C?

Exercise 2. Suppose you roll two standard, fair, 6-sided dice. What is the probability that the sum is at least 9 given that you rolled at least one 6?

Exercise 3.  Color blindness is a sex-linked condition, and 5% of men and 0.25% of women are color blind. The population of the United States is 51% female. What is the probability that a color-blind American is a man?

## Lecture 7. Rough paths. Fall 2017

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 $\square$

Let $x \in C^{1-var}([0,T],\mathbb{R}^d)$. Since
$\omega(s,t)=\left( \sum_{j=1}^{[p]} \left\| \int dx^{\otimes j}\right\|^{1/j}_{\frac{p}{j}-var, [s,t]} \right)^p$
is a control, the estimate
$\left\| \int_{\Delta^k [s,t]} dx^{\otimes k} \right\| \le \frac{C^k}{\left( \frac{k}{p}\right)!} \left( \sum_{j=1}^{[p]} \left\| \int dx^{\otimes j}\right\|^{1/j}_{\frac{p}{j}-var, [s,t]} \right)^k, \quad 0 \le s \le t \le T.$
easily implies that for $k > p$,
$\left\| \int dx^{\otimes k} \right\|_{1-var, [s,t]} \le \frac{C^k}{\left( \frac{k}{p}\right)!} \omega(s,t)^{k/p}.$
We stress that it does not imply a bound on the 1-variation of the path $t \to \int_{\Delta^k [0,t]} dx^{\otimes k}$. What we can get for this path, are bounds in $p$-variation:

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 $k \ge 0$,
$\left\| \int_{\Delta^k [0,\cdot]} dx^{\otimes k} \right\|_{p-var, [s,t]} \le \frac{C^k}{\left( \frac{k}{p}\right)!} \omega(s,t)^{1/p} \omega(0,T)^{\frac{k-1}{p}}$
where
$\omega(s,t)= \left( \sum_{j=1}^{[p]} \left\| \int dx^{\otimes j}\right\|^{1/j}_{\frac{p}{j}-var, [s,t]} \right)^p, \quad 0 \le s \le t \le T.$

Proof: This is an easy consequence of the Chen’s relations. Indeed,

$\left\| \int_{\Delta^k [0,t]} dx^{\otimes k} - \int_{\Delta^k [0,s]} dx^{\otimes k} \right\|$
$=\left\| \sum_{j=1}^k \int_{\Delta^j [s,t]} dx^{\otimes j} \int_{\Delta^{j-k} [0,s]} dx^{\otimes (k-j)} \right\|$
$\le \sum_{j=1}^k \left\| \int_{\Delta^j [s,t]} dx^{\otimes j} \right\| \left\| \int_{\Delta^{j-k} [0,s]} dx^{\otimes (k-j)} \right\|$
$\le C^k \sum_{j=1}^k \frac{1}{\left( \frac{j}{p}\right)!} \omega(s,t)^{j/p} \frac{1}{\left( \frac{k-j}{p}\right)!} \omega(s,t)^{(k-j)/p}$
$\le C^k \omega(s,t)^{1/p} \sum_{j=1}^k \frac{1}{\left( \frac{j}{p}\right)!} \omega(0,T)^{(j-1)/p} \frac{1}{\left( \frac{k-j}{p}\right)!} \omega(0,T)^{(k-j)/p}$
$\le C^k \omega(s,t)^{1/p} \omega(0,T)^{(k-1)/p}\sum_{j=1}^k \frac{1}{\left( \frac{j}{p}\right)!} \frac{1}{\left( \frac{k-j}{p}\right)!}.$
and we conclude with the binomial inequality $\square$

We are now ready for a second major estimate which is the key to define iterated integrals of a path with $p$-bounded variation when $p \ge 2$.

Theorem: Let $p \ge 1$, $K > 0$ and $x,y \in C^{1-var}([0,T],\mathbb{R}^d)$ such that
$\sum_{j=1}^{[p]} \left\| \int dx^{\otimes j}- \int dy^{\otimes j} \right\|^{1/j}_{\frac{p}{j}-var, [0,T]} \le 1,$
and
$\left( \sum_{j=1}^{[p]} \left\| \int dx^{\otimes j}\right\|^{1/j}_{\frac{p}{j}-var, [0,T]} \right)^p+ \left( \sum_{j=1}^{[p]} \left\| \int dy^{\otimes j}\right\|^{1/j}_{\frac{p}{j}-var, [0,T]} \right)^p \le K.$
Then there exists a constant $C \ge 0$ depending only on $p$ and $K$ such that for $0\le s \le t \le T$ and $k \ge 1$
$\left\| \int_{\Delta^k [s,t]} dx^{\otimes k}- \int_{\Delta^k [s,t]} dy^{\otimes k} \right\| \le \left( \sum_{j=1}^{[p]} \left\| \int dx^{\otimes j}- \int dy^{\otimes j} \right\|^{1/j}_{\frac{p}{j}-var, [0,T]} \right) \frac{C^k}{\left( \frac{k}{p}\right)!} \omega(s,t)^{k/p} ,$
$\left\| \int_{\Delta^k [s,t]} dx^{\otimes k}\right\| +\left\| \int_{\Delta^k [s,t]} dy^{\otimes k} \right\| \le \frac{C^k}{\left( \frac{k}{p}\right)!} \omega(s,t)^{k/p}$
where $\omega$ is the control
$\omega(s,t)= \frac{ \left( \sum_{j=1}^{[p]} \left\| \int dx^{\otimes j}\right\|^{1/j}_{\frac{p}{j}-var, [s,t]} \right)^p+ \left( \sum_{j=1}^{[p]} \left\| \int dy^{\otimes j}\right\|^{1/j}_{\frac{p}{j}-var, [s,t]} \right)^p } { \left( \sum_{j=1}^{[p]} \left\| \int dx^{\otimes j}\right\|^{1/j}_{\frac{p}{j}-var, [0,T]} \right)^p+ \left( \sum_{j=1}^{[p]} \left\| \int dy^{\otimes j}\right\|^{1/j}_{\frac{p}{j}-var, [0,T]} \right)^p }$
$+\left( \frac{\sum_{j=1}^{[p]} \left\| \int dx^{\otimes j} - \int dy^{\otimes j}\right\|^{1/j}_{\frac{p}{j}-var, [s,t]} }{\sum_{j=1}^{[p]} \left\| \int dx^{\otimes j} - \int dy^{\otimes j}\right\|^{1/j}_{\frac{p}{j}-var, [0,T]} } \right)^p$

Proof: We prove by induction on $k$ that for some constants $C,\beta$,
$\left\| \int_{\Delta^k [s,t]} dx^{\otimes k}- \int_{\Delta^k [s,t]} dy^{\otimes k} \right\| \le \left( \sum_{j=1}^{[p]} \left\| \int dx^{\otimes j}- \int dy^{\otimes j} \right\|^{1/j}_{\frac{p}{j}-var, [0,T]} \right) \frac{C^k}{\beta \left( \frac{k}{p}\right)!} \omega(s,t)^{k/p},$
$\left\| \int_{\Delta^k [s,t]} dx^{\otimes k}\right\| +\left\| \int_{\Delta^k [s,t]} dy^{\otimes k} \right\| \le \frac{C^k}{\beta \left( \frac{k}{p}\right)!} \omega(s,t)^{k/p}$

For $k \le p$, we trivially have
$\left\| \int_{\Delta^k [s,t]} dx^{\otimes k}- \int_{\Delta^k [s,t]} dy^{\otimes k} \right\|$ $\le \left( \sum_{j=1}^{[p]} \left\| \int dx^{\otimes j}- \int dy^{\otimes j} \right\|^{1/j}_{\frac{p}{j}-var, [0,T]} \right)^k \omega(s,t)^{k/p}$
$\le \left( \sum_{j=1}^{[p]} \left\| \int dx^{\otimes j}- \int dy^{\otimes j} \right\|^{1/j}_{\frac{p}{j}-var, [0,T]} \right) \omega(s,t)^{k/p}.$
and
$\left\| \int_{\Delta^k [s,t]} dx^{\otimes k}\right\| +\left\| \int_{\Delta^k [s,t]} dy^{\otimes k} \right\| \le K^{k/p} \omega(s,t)^{k/p}$.
Not let us assume that the result is true for $0 \le j \le k$ with $k > p$. Let
$\Gamma_{s,t}=\int_{\Delta^k [s,t]} dx^{\otimes (k+1)}- \int_{\Delta^k [s,t]} dy^{\otimes (k+1)}$
From the Chen’s relations, for $0 \le s \le t \le u \le T$,
$\Gamma_{s,u}= \Gamma_{s,t}+ \Gamma_{t,u}$
$+\sum_{j=1}^{k} \int_{\Delta^j [s,t]} dx^{\otimes j }\int_{\Delta^{k+1-j} [t,u]} dx^{\otimes (k+1-j) }-\sum_{j=1}^{k} \int_{\Delta^j [s,t]} dy^{\otimes j }\int_{\Delta^{k+1-j} [t,u]} dy^{\otimes (k+1-j) }.$
Therefore, from the binomial inequality
$\| \Gamma_{s,u}\|$
$\le \| \Gamma_{s,t} \| + \| \Gamma_{t,u} \| +\sum_{j=1}^{k} \left\| \int_{\Delta^j [s,t]} dx^{\otimes j }- \int_{\Delta^j [s,t]} dy^{\otimes j } \right\| \left\| \int_{\Delta^{k+1-j} [t,u]} dx^{\otimes (k+1-j) }\right\|$
$+\sum_{j=1}^{k} \left\| \int_{\Delta^{j} [s,t]} dy^{\otimes j }\right\| \left\| \int_{\Delta^{k+1-j} [t,u]} dx^{\otimes (k+1-j) }- \int_{\Delta^{k+1-j} [t,u]} dy^{\otimes (k+1-j) } \right\|$
$\le \| \Gamma_{s,t} \| + \| \Gamma_{t,u} \| +\frac{1}{\beta^2}\tilde{\omega}(0,T) \sum_{j=1}^{k} \frac{C^j}{\left( \frac{j}{p}\right)!} \omega(s,t)^{j/p} \frac{C^{k+1-j}}{\left( \frac{k+1-j}{p}\right)!} \omega(t,u)^{(k+1-j)/p}$
$+\frac{1}{\beta^2}\tilde{\omega}(0,T) \sum_{j=1}^{k} \frac{C^j}{\left( \frac{j}{p}\right)!} \omega(s,t)^{j/p} \frac{C^{k+1-j}}{\left( \frac{k+1-j}{p}\right)!} \omega(t,u)^{(k+1-j)/p}$
$\le \| \Gamma_{s,t} \| + \| \Gamma_{t,u} \| +\frac{2p}{\beta^2} \tilde{\omega}(0,T) C^{k+1} \frac{ \omega(s,u)^{(k+1)/p}}{\left( \frac{k+1}{p}\right)! }$
where
$\tilde{\omega}(0,T)=\sum_{j=1}^{[p]} \left\| \int dx^{\otimes j}- \int dy^{\otimes j} \right\|^{1/j}_{\frac{p}{j}-var, [0,T]} .$
We deduce
$\| \Gamma_{s,t} \| \le \frac{2p}{\beta^2(1-2^{1-\theta})} \tilde{\omega}(0,T) C^{k+1} \frac{ \omega(s,t)^{(k+1)/p}}{\left( \frac{k+1}{p}\right)! }$
with $\theta= \frac{k+1}{p}$. A correct choice of $\beta$ finishes the induction argument $\square$

Posted in Rough paths theory | 3 Comments

## Lecture 6. Rough paths. Fall 2017

In this lecture we introduce the central notion of the signature of a path $x \in C^{1-var}([0,T],\mathbb{R}^d)$ which is a convenient way to encode all the algebraic information on the path $x$ which is relevant to study differential equations driven by $x$. The motivation for the definition of the signature comes from formal manipulations on Taylor series.

Let us consider a differential equation
$y(t)=y(s)+\sum_{i=1}^d \int_s^t V_i (y(u) )dx^i(u),$
where the $V_i$‘s are smooth vector fields on $\mathbb{R}^n$.

If $f: \mathbb{R}^{n} \rightarrow \mathbb{R}$ is a $C^{\infty}$ function, by the change of variable formula,
$f(y(t))=f(y(s))+\sum^{d}_{i=1}\int^{t}_{s}V_{i}f(y(u))dx^{i}(u).$

Now, a new application of the change of variable formula to $V_{i}f(y(s))$ leads to
$f(y(t))=f(y(s))+\sum^{d}_{i=1}V_{i}f(y(s))\int^{t}_{s}dx^{i}(u)+\sum^{d}_{i,j=1}\int^{t}_{s}\int^{u}_{s} V_{j}V_{i}f(y(v))dx^{j}(v)dx^{i}(u).$

We can continue this procedure to get after $N$ steps
$f(y(t))=f(y(s))+\sum^{N}_{k=1}\sum_{I=(i_1,\cdots,i_k)}(V_{i_1}\cdots V_{i_k}f)(y(s))\int_{\Delta^{k}[s,t]}dx^{I}+R_{N}(s,t)$
for some remainder term $R_{N}(s,t)$, where we used the notations:

• $\Delta^{k}[s,t]=\{(t_1,\cdots,t_k)\in[s,t]^{k}, s\leq t_1\leq t_2\cdots\leq t_k\leq t\}$
• If $I=\left(i_1,\cdots,i_k\right)\in\{1,\cdots,d\}^k$ is a word with length $k$, $\int_{\Delta^{k}[s,t]}dx^{I}=\displaystyle \int_{s \le t_1 \le t_2 \le \cdots \le t_k \le t}dx^{i_1}(t_1)\cdots dx^{i_k}(t_k).$

If we let $N\rightarrow +\infty$, assuming $R_{N}(s,t) \to 0$ (which is by the way true for $t-s$ small enough if the $V_i$‘s are analytic), we are led to the formal expansion formula:
$f(y(t))=f(y(s))+\sum^{+\infty}_{k=1}\sum_{I=(i_1,\cdots,i_k)}(V_{i_1}\cdots V_{i_k}f)(y(s))\int_{\Delta^{k}[s,t]}dx^{I}.$
This shows, at least at the formal level, that all the information given by $x$ on $y$ is contained in the iterated integrals $\int_{\Delta^{k}[s,t]}dx^{I}$.

Let $\mathbb{R} [[X_1,...,X_d]]$ be the non commutative algebra over $\mathbb{R}$ of the formal series with $d$ indeterminates, that is the set of series
$Y=y_0+\sum_{k = 1}^{+\infty} \sum_{I \in \{1,...,d\}^k} a_{i_1,...,i_k} X_{i_1}...X_{i_k}.$

Definition: Let $x \in C^{1-var}([0,T],\mathbb{R}^d)$. The signature of $x$ (or Chen’s series) is the formal series:
$\mathfrak{S} (x)_{s,t} =1 + \sum_{k=1}^{+\infty} \sum_{I \in \{1,...,d\}^k} \left( \int_{\Delta^{k}[s,t]}dx^{I} \right) X_{i_1} \cdots X_{i_k}, \quad 0 \le s \le t \le T.$

As we are going to see in the next few lectures, the signature is a fascinating algebraic object. At the source of the numerous properties of the signature lie the following so-called Chen’s relations

Lemma: Let $x \in C^{1-var}([0,T],\mathbb{R}^d)$. For any word $(i_1,...,i_n) \in \{ 1, ... , d \}^n$ and any $0 \le s \le t \le u \le T$,
$\int_{\Delta^n [s,u]} dx^{(i_1,...,i_n)}=\sum_{k=0}^{n} \int_{\Delta^k [s,t]} dx^{(i_1,...,i_k)}\int_{\Delta^{n-k} [t,u]} dx^{(i_{k+1},...,i_n)},$
where we used the convention that if $I$ is a word with length 0, then $\int_{\Delta^{0} [0,t]} dx^I =1$.

Proof: It follows readily by induction on $n$ by noticing that
$\int_{\Delta^n [s,u]} dx^{(i_1,...,i_n)}=\int_s^u \left( \int_{\Delta^{n-1} [s,t_n]} dx^{(i_1,...,i_{n-1})} \right) dx^{i_n}(t_n)$ $\square$

To avoid heavy notations, it will be convenient to denote
$\int_{\Delta^k [s,t]} dx^{\otimes k} =\sum_{I \in \{1,...,d\}^k} \left( \int_{\Delta^{k}[s,t]}dx^{I} \right) X_{i_1} \cdots X_{i_k}.$

This notation actually reflects a natural algebra isomorphism between $\mathbb{R} [[X_1,...,X_d]]$ and $1\oplus_{k=1}^{+\infty} (\mathbb{R}^d)^{\otimes k}$. With this notation, observe that the signature writes then
$\mathfrak{S} (x)_{s,t} =1 + \sum_{k=1}^{+\infty} \int_{\Delta^k [s,t]} dx^{\otimes k},$
and that the Chen’s relations become
$\int_{\Delta^n [s,u]} dx^{\otimes n}=\sum_{k=0}^{n} \int_{\Delta^k [s,t]} dx^{\otimes k }\int_{\Delta^{n-k} [t,u]} dx^{\otimes (n-k) }.$
The Chen’s relations imply the following flow property for the signature:

Proposition: Let $x \in C^{1-var}([0,T],\mathbb{R}^d)$. For any $0 \le s \le t \le u \le T$,
$\mathfrak{S} (x)_{s,u} =\mathfrak{S} (x)_{s,t}\mathfrak{S} (x)_{t,u}$

Proof: Indeed,
$\mathfrak{S} (x)_{s,u}$
$=1 + \sum_{k=1}^{+\infty} \int_{\Delta^k [s,u]} dx^{\otimes k}$
$=1 + \sum_{k=1}^{+\infty}\sum_{j=0}^{k} \int_{\Delta^j [s,t]} dx^{\otimes j }\int_{\Delta^{k-j} [t,u]} dx^{\otimes (k-j) }$
$=\mathfrak{S} (x)_{s,t}\mathfrak{S} (x)_{t,u}$
$\square$

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