A first result which is a consequence of the theorem proved in the previous lecture is the following continuity of the iterated iterated integrals with respect to a convenient topology. The proof uses very similar arguments to the previous two lectures, so we let it as an exercise to the student.

**Theorem:** * Let , and such that
and
Then there exists a constant depending only on and such that for
*

This continuity result naturally leads to the following definition.

**Definition:*** Let and . We say that is a -rough path if there exists a sequence such that in -variation and such that for every , there exists such that for ,
The space of -rough paths will be denoted .*

From the very definition, is the closure of inside for the distance

If and is such that in -variation and such that for every , there exists such that for ,

then we define for as the limit of the iterated integrals . However it is important to observe that may then depend on the choice of the approximating sequence . Once the integrals are defined for , we can then use the previous theorem to construct all the iterated integrals for . It is then obvious that if , then

implies that

In other words the signature of a -rough path is completely determinated by its truncated signature at order :

For this reason, it is natural to present a -rough path by this truncated signature at order in order to stress that the choice of the approximating sequence to contruct the iterated integrals up to order has been made. This will be further explained in much more details when we will introduce the notion of geometric rough path over a rough path.

The following results are straightforward to obtain from the previous lectures by a limiting argument.

**Lemma:** * Let , . For , and ,
*

**Theorem:*** Let . There exists a constant , depending only on , such that for every and ,
*

If , the space is not a priori a Banach space (it is not a linear space) but it is a complete metric space for the distance

The structure of will be better understood in the next lectures, but let us remind that if , then is the closure of inside for the variation distance it is therefore what we denoted . As a corollary we deduce

**Proposition:*** Let . Then if and only if
where is the set of subdivisions of . In particular, for ,
*

We are now ready to define solutions of linear differential equations driven by -rough paths, and present the Lyons’ continuity theorem in this setting. Let be a -rough path with truncated signature and let be an approximating sequence such that

Let us consider matrices . We have the following theorem:

**Theorem:** * Let be the solution of the differential equation
Then, when , converges in the -variation distance to some . is called the solution of the rough differential equation
*

**Proof:** It is a classical result that the solution of the equation

can be expanded as the convergent Volterra series:

Therefore, in particular, for ,

which implies that

with . From the theorems of the previous lectures, there exists a constant depending only on and

such that for and big enough:

As a consequence, there exists a constant such that for big enough:

This already proves that converges in the supremum topology to some . We now have

and we can bound

Again, from the theorems of the previous lectures, there exists a constant , depending only on and

such that for and big enough

where is a control such that . Consequently, there is a constant , such that

This implies the estimate

and thus gives the conclusion

With just a little more work, it is possible to prove the following stronger result whose proof is let to the reader.

**Theorem:** * Let be the solution of the differential equation
and be the solution of the rough differential equation:
Then, and when ,
*

We can get useful estimates for solutions of rough differential equations. For that, we need the following analysis lemma:

**Proposition:*** For and ,
*

**Proof:** For , we denote

This is a special function called the Mittag-Leffler function. From the binomial inequality

Thus we proved

Iterating this inequality, times we obtain

It is known (and not difficult to prove) that

By letting we conclude

This estimate provides the following result:

**Proposition:** * Let be the solution of the rough differential equation:
Then, there exists a constant depending only on such that for ,
where .
*

**Proof:** We have

Thus we obtain

,

and we conclude by using estimates on iterated integrals of rough paths together with the previous lemma

Filed under: Rough paths theory ]]>

**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?

Filed under: MA3160 ]]>

**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)

(b) E [3X + 2]

(c)

(d) Var[−X]

Filed under: Uncategorized ]]>

*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.

Filed under: Mathematicians ]]>

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

Filed under: Uncategorized ]]>

**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?

Filed under: MA3160 ]]>

If now , the iterated integrals can only be defined as Young integrals when . In this lecture, we are going to derive some estimates that allow to define the signature of some (not all) paths with a finite variation when . These estimates are due to Terry Lyons in his seminal paper and this is where the rough paths theory really begins.

For that can be writen as

we define

It is quite easy to check that for

Let . For , we denote

where is the set of subdivisions of the interval . Observe that for , in general

Actually from the Chen’s relations we have

It follows that needs not to be the -variation of .

The first major result of rough paths theory is the following estimate:

**Proposition:** *Let . There exists a constant , depending only on , such that for every and ,
*

By , we of course mean . Some remarks are in order before we prove the result. If , then the estimate becomes

which is immediately checked because

We can also observe that for , the estimate is easy to obtain because

So, all the work is to prove the estimate when . The proof is split into two lemmas. The first one is a binomial inequality which is actually quite difficult to prove:

**Lemma:*** For , , and ,
*

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

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 . Let us assume that:*

- There exists a control such that

- There exists a control and such that for ,

Then, for all ,

**Proof:**

See the proof of the Young-Loeve estimate or Lemma 6.2 in the book by Friz-Victoir

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

**Proof:**

Let us denote

We claim that is a control. Indeed for , we have from Holder’s inequality

It is clear that for some constant which is small enough, we have for ,

Let us now consider

From the Chen’s relations, for ,

Therefore,

On the other hand, we have

We deduce from the previous lemma that

with . The general case is dealt by induction. The details are let to the reader

Let . Since

is a control, the estimate

easily implies that for ,

We stress that it does not imply a bound on the 1-variation of the path . What we can get for this path, are bounds in -variation:

**Proposition:*** Let . There exists a constant , depending only on , such that for every and ,
where
*

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

and we conclude with the binomial inequality

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

**Theorem:*** Let , and such that
and
Then there exists a constant depending only on and such that for and
where is the control
*

**Proof:** We prove by induction on that for some constants ,

For , we trivially have

and

.

Not let us assume that the result is true for with . Let

From the Chen’s relations, for ,

Therefore, from the binomial inequality

where

We deduce

with . A correct choice of finishes the induction argument

Filed under: Rough paths theory ]]>

Let us consider a differential equation

where the ‘s are smooth vector fields on .

If is a function, by the change of variable formula,

Now, a new application of the change of variable formula to leads to

We can continue this procedure to get after steps

for some remainder term , where we used the notations:

- If is a word with length ,

If we let , assuming (which is by the way true for small enough if the ‘s are analytic), we are led to the formal expansion formula:

This shows, at least at the formal level, that all the information given by on is contained in the iterated integrals .

Let be the non commutative algebra over of the formal series with indeterminates, that is the set of series

**Definition:** *Let . The signature of (or Chen’s series) is the formal series:
*

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 . For any word and any ,
where we used the convention that if is a word with length 0, then . *

**Proof:** It follows readily by induction on by noticing that

To avoid heavy notations, it will be convenient to denote

This notation actually reflects a natural algebra isomorphism between and . With this notation, observe that the signature writes then

and that the Chen’s relations become

The Chen’s relations imply the following flow property for the signature:

**Proposition:*** Let . For any ,
*

**Proof:** Indeed,

Filed under: Rough paths theory ]]>

The basic existence and uniqueness result is the following. Throughout this lecture, we assume that .

**Theorem:** *Let and let be a Lipschitz continuous map, that is there exists a constant such that for every ,
For every , there is a unique solution to the differential equation:
Moreover . *

**Proof:** The proof is of course based again of the fixed point theorem. Let and consider the map going from the space into itself, which is defined by

By using basic estimates on the Young’s integrals, we deduce that

If is small enough, then , which means that is a contraction of the Banach space endowed with the norm .

The fixed point of , let us say , is the unique solution to the differential equation:

By considering then a subdivision

such that , we obtain a unique solution to the differential equation:

As for the bounded variation case, the solution of a Young’s differential equation is a function of the initial condition,

**Proposition:** * Let and let be a Lipschitz continuous map. Let be the flow of the equation
Then for every , the map is and the Jacobian is the unique solution of the matrix linear equation
*

As we already mentioned it before, solutions of Young’s differential equations are continuous with respect to the driving path in the -variation topology

**Theorem:** * Let and let be a Lipschitz and bounded continuous map such that for every ,
Let be the solution of the differential equation:
If converges to in -variation, then converges in -variation to the solution of the differential equation:
*

**Proof:** Let . We have

Thus, if is such that , we obtain

In the very same way, provided , we get

Let us fix and pick a sequence such that . Since , for with big enough, we have

We deduce that for ,

and

For with and big enough, we have

which implies

Filed under: Uncategorized ]]>

**Exercise 1. **Two dice are simultaneously rolled. For each pair of events defined below, compute if they are independent or not.

(a) A1 ={thesumis7},B1 ={thefirstdielandsa3}.

(b) A2 = {the sum is 9}, B2 = {the second die lands a 3}.

(c) A3 = {the sum is 9}, B3 = {the first die lands even}.

(d) A4 = {the sum is 9}, B4 = {the first die is less than the second}.

(e) A5 = {two dice are equal}, B5 = {the sum is 8}.

(f) A6 = {two dice are equal}, B6 = {the first die lands even}.

(g) A7 = {two dice are not equal}, B7 = {the first die is less than the second}.

**Exercise 2**. Are the events A1, B1 and B3 from Exercise 1 independent?

**Exercise 3.** Suppose you toss a fair coin repeatedly and independently. If it comes up heads, you win a dollar, and if it comes up tails, you lose a dollar. Suppose you start with $20. What is the probability you will get to $150 before you go broke?

Filed under: MA3160 ]]>