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

**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?

Posted in MA3160
Leave a comment

In the previous lecture we introduced the signature of a bounded variation path as the formal series

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

Posted in Rough paths theory
3 Comments

In this lecture we introduce the central notion of the signature of a path which is a convenient way to encode all the algebraic information on the path which is relevant to study differential equations driven by . The motivation for the definition of the signature comes from formal manipulations on Taylor series.

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,

Posted in Rough paths theory
2 Comments

In the previous lecture we defined the Young’s integral when and with . The integral path has then a bounded -variation. Now, if is a Lipschitz map, then the integral, is only defined when , that is for . With this in mind, it is apparent that Young’s integration should be useful to solve differential equations driven by continuous paths with bounded -variation for . If , then the Young’s integral is of no help and the rough paths theory later explained is the correct one.

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

Posted in Uncategorized
Leave a comment

**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?

Posted in MA3160
Leave a comment

In this lecture we define the Young‘s integral when and with . The cornerstone is the following Young-Loeve estimate.

**Theorem:** * Let and . Consider now with . The following estimate holds: for ,
*

**Proof:** For , let us define

We have for ,

As a consequence, we get

Let now . We claim that is a control. The continuity and the vanishing on the diagonal are obvious to check, so we just need to justify the superadditivity. Let , we have from Holder’s inequality,

We have then

For , consider then the control

Define now

If is such that , we can find a such that , . Indeed, the continuity of forces the existence of a such that . We obtain therefore

which implies by maximization,

By iterating times this inequality, we obtain

It is now clear that:

Since

We conclude

and thus

Sending , finishes the proof

It is remarkable that the Young-Loeve estimate only involves and . As a consequence, we obtain the following result whose proof is let to the reader:

**Proposition:*** Let and with . Let us assume that there exists a sequence such that in and a sequence such that in , then for every , converges to a limit that we call the Young’s integral of against on the interval and denote .
The integral does not depend of the sequences and and the following estimate holds: for ,
*

The closure of in is and we know that . It is therefore obvious to extend the Young’s integral for every and with and the Young-Loeve estimate still holds

From this estimate, we easily see that for and with the sequence of Riemann sums

will converge to when the mesh of the subdivision goes to 0. We record for later use the following estimate on the Young’s integral, which is also an easy consequence of the Young-Loeve estimate (see Theorem 6.8 in the book for further details).

**Proposition:** * Let and with . The integral path is continuous with a finite -variation and we have
*

Posted in Rough paths theory
1 Comment