Stochastic Calculus

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Stochastic Calculus ——– an introduction to option pricing with martingales ——– Christophe Giraud1

Lecture notes IMAMIS Manila, august 2005

1 University

of Nice (France). E-mail: [email protected]

Contents 0 Goal of the lecture 0.1 Risk hedging - options . . . . . . 0.1.1 Problematic . . . . . . . . 0.1.2 Some examples of options 0.1.3 Option pricing - hedging . 0.2 Stochastic models . . . . . . . . . 0.3 ”One step, two states” model . .

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1 Conditional expectation 1.1 Discrete setting . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1.1 Conditioning against a random variable . . . . . . . . . 1.1.2 Conditioning against several random variables . . . . . 1.2 General case . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.1 Extension to the case where there exists a joint density 1.2.2 General case . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Martingales 2.1 Information, filtration, stopping time . . . 2.1.1 Information, filtration . . . . . . . 2.1.2 Stopping time . . . . . . . . . . . . 2.2 Martingales . . . . . . . . . . . . . . . . . 2.2.1 Definition . . . . . . . . . . . . . . 2.2.2 Optional stopping theorem . . . . . 2.3 Exercises . . . . . . . . . . . . . . . . . . . 2.3.1 Galton-Watson genealogy . . . . . 2.3.2 With the optional stopping theorem 3 The B-S market 3.1 The B-S market . . . . . . 3.1.1 Evolution . . . . . 3.1.2 Portfolio . . . . . . 3.2 Arbitrage and martingale .

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CONTENTS

3.3 3.4 3.5

2

3.2.1 Risk neutral probability . . . . . . . . . . . . . . . . . 3.2.2 Arbitrage . . . . . . . . . . . . . . . . . . . . . . . . . Complete market . . . . . . . . . . . . . . . . . . . . . . . . . Girsanov Lemma . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.1 Change of probability: Girsanov lemma . . . . . . . . . 3.5.2 Application: computation of a risk neutral probability

4 European option pricing 4.1 Problematic . . . . . . . . . . . . . . . . . . . . . 4.2 Pricing a european option in an arbitrage-free and 4.3 And what for an incomplete market? . . . . . . . 4.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . 4.4.1 Cox-Ross-Rubinstein model . . . . . . . . 4.4.2 Black-Scholes and Merton models . . . . .

7 Itˆ o 7.1 7.2 7.3 7.4

calculus Problematic . . . . . . . . . . Itˆo’s integral . . . . . . . . . . Itˆo’s processes . . . . . . . . . Girsanov formula . . . . . . . 7.4.1 Stochastic exponentials 7.4.2 Girsanov formula . . . 7.5 Exercises . . . . . . . . . . . .

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. . . . . . . . . . complete market . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

5 American option pricing 5.1 Problematic . . . . . . . . . . . . . . . . . . . . . . . 5.2 Pricing on american option . . . . . . . . . . . . . . . 5.3 Dynamic Programing Principle . . . . . . . . . . . . 5.4 Doob’s decomposition and predictable representation 5.5 Proof of theorem 5.1 . . . . . . . . . . . . . . . . . . 5.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 6 Brownian motion 6.1 Continuous-time processes . . . . . . . 6.2 Brownian motion . . . . . . . . . . . . 6.2.1 Gaussian law . . . . . . . . . . 6.2.2 Definition of Brownian motion . 6.2.3 Properties of Brownian motion 6.3 Continuous-time martingales . . . . . . 6.4 Exercises . . . . . . . . . . . . . . . . . 6.4.1 Basic properties . . . . . . . . . 6.4.2 Quadratic variation . . . . . . .

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CONTENTS 7.5.1 7.5.2

3 Scaling functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . Cameron-Martin formula . . . . . . . . . . . . . . . . . . . . . . . .

8 Black-Scholes model 8.1 Setting . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1.1 The Black-Scholes model . . . . . . . . . . . . 8.1.2 Portfolios . . . . . . . . . . . . . . . . . . . . 8.1.3 Risk neutral probability . . . . . . . . . . . . 8.2 Price of a european option in the Black-Scholes model 9 Appendix 9.1 Convergence of random variables . . . . . 9.1.1 Convergence a.s. . . . . . . . . . . 9.1.2 Convergence in L2 . . . . . . . . . 9.1.3 Convergence in probability . . . . . 9.1.4 Convergence in distribution . . . . 9.1.5 Relationships . . . . . . . . . . . . 9.2 Construction of Itˆo’s integral . . . . . . . . 9.2.1 Setting . . . . . . . . . . . . . . . . 9.2.2 Integration of elementary processes 9.2.3 Extension to L2T . . . . . . . . . . .

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69 69 69 69 69 69 69 70 70 70 72

CONTENTS

Abbreviations and notations

E(X) X⊥Y i.e. i.i.d. a.s. P∗ iif s.t. r.v.

expectation of X X is independent of Y namely independent and identically distributed almost surely (with probability 1) risk neutral probability if and only if stopping time random variable

4

Chapter 0 Goal of the lecture The goal: to give a comprehensive introduction to option pricing with martingale. The first two chapters give some mathematical stuff (conditional expectation, martingale). Chapter 3,4,5 focus on the completely discrete setting. We use the risk neutral probability approach to compute the price of european and american options.

0.1 0.1.1

Risk hedging - options Problematic

Some financial or industrial agents may wish to eliminate some risks, either as a commercial choice or because it does not enter into their fields of competence. Examples: 1. A european company manages its activity in euros, but signs its contracts in dollar, payable on receipt. Between today and the receipt, the euro/dollar exchange rate may fluctuate. The company thus have to face an exchange risk. If it does not want to take charge of it, the company will sign a contract that protects itself against this risk. 2. The market price of copper fluctuate dramatically. A copper mine may wish to protect itself against this fluctuation. The manager of the mine will then sign a contract, that warrants him a minimum price for its copper.

0.1.2

Some examples of options

European call : contract that gives the right (but not the obligation) to its owner to buy an asset at the fixed price K (strike) at time N (maturity). This contract has a price C (prime). We write Sn for the price of the asset at time n. Two cases may occurs at maturity:

5

CHAPTER 0. GOAL OF THE LECTURE

6

• when SN < K: the owner of the option has the right to buy at price K an asset that he could buy for less in the market. This right is not interesting. He does not exercise it : nothing happens. • when SN ≥ K : the owner of the call can buy an asset for less than the market price, which is interesting. The seller of the option, must then buy an action at price SN and sell it at price K to the owner of the option. Things goes as if the seller was giving SN − K to the owner. In short: at time n = 0 the owner gives C to the seller of the call. At time N , he receives the maximum between SN −K and 0, we denote by (SN −K)+ . We call payoff, the function f = (SN − K)+ .

Figure 1: Profit/loss of the owner of the call

European put: contract that gives the right (but not the obligation) to its owner to sell an asset at the fixed price K (strike) at time N (maturity). This contract has a price C (prime). In this case the payoff is f = (K − SN )+ .

Figure 2: Profit/loss of the owner of the put

CHAPTER 0. GOAL OF THE LECTURE

7

American call (resp. put): contract that gives the right (without obligation) to buy (resp. sell) an asset at any time before time N (maturity) at the fixed price K. This contract has a price.

exotic options: there exists many other options, called exotic options. Example given: the collar option, with payoff f = min(max(K1 , SN ), K2 ), the Boston option, with payoff f = (SN − K1 )+ − (K2 − K1 ), where K1 < K2 , etc.

0.1.3

Option pricing - hedging

The determination of the fair price C of an option and the way to hedge it is the main goal of the mathematical finance. The seller of the option will invest on the market in such a way that, whatever the evolution of the market, he will be able to face its engagement without loosing money. The fair price of the option will then correspond to the minimal initial investment needed to carry out this condition. The investment strategy of the seller is called hedging strategy. Fundamental remark: the hedging strategy of an option differs fundamentally from the hedging strategy of a classical insurance (against robbery, fire, etc). Indeed, in the present case the seller of the option must hedge the risk of a single contract. The risk must disappear. A contrario, the seller of a classical insurance sell many similar contracts. His strategy is then to face his engagement on average. He hopes that there will not be two many claims in the same time. Such an hedging strategy is so-called a hedging strategy via diversification.

0.2

Stochastic models

The future is uncertain, but there exit some models to describe its evolution. These model take into account the uncertainty of the future. They are so-called stochastic models (in opposition to deterministic models). A financial market may follow different evolutions, each evolution having a given probability to happen. For example, two basic parameters to describe the evolution of a stock is its trend and its volatility. A stochastic process is a ”value” that evolves randomly when time passes (NASDAC, quote of copper, exchange rates, etc). We will investigate in the next chapters some stochastic processes that appear when we model the financial market. A special attention will be given on martingales and brownian motions.

0.3

”One step, two states” model

The following model is the simplest one: there exists two dates (today and tomorrow) and two possible states for tomorrow. Its analysis uses (in this very simple setting) the methods

CHAPTER 0. GOAL OF THE LECTURE

8

that we will further develop for more involved cases. We focus on an option with maturity N = 1 and payoff of the form f = g(SN ) (ex: for a european call g(x) = (x − K)+ ). Today (n = 0), the quote of the underlying asset is S0 . Tomorrow (n = 1), the quote can take two possible values S− and S+ . The seller of the option creates a portfolio π = (β, γ) made of β units of bond and γ units of asset. Today, the value of the portfolio is X0 = β + γS0 . Tomorrow, it will be XN = βerN + γSN (where r represents the interest rate of the bond). To be able to face to its obligation, the value of the seller’s portfolio at maturity (time N ) must not be smaller than the payoff, namely XN ≥ g(SN ). Assume that either when SN = S+ , or when SN = S− , the value of the portfolio is larger than the payoff g(SN ). Then the seller could earn money with positive probability and no risk of loosing money. We assume that this is impossible (market at equilibrium). We say in this case that the market is arbitrage-free. As a consequence, we must have XN = g(SN ), which in turn enforces (β, γ) to satisfy the equations  rN βe + γS+ = g(S+ ) βerN + γS− = g(S− ). The unique solution to this equation is given by γ= and

1 β = e−rN 2



g(S+ ) − g(S− ) S + − S−

(1)

 S + + S− g(S+ ) + g(S− ) − (g(S+ ) − g(S− )) . S + − S−

Formula (1) is commonly called ”delta hedging formula”. The initial value of the portfolio is X0 = β + γS0 which is positive. This value corresponds to the cost of the contract, namely its fair price C. —————— In short: at time t = 0, the owner of the option gives C = X0 = β + γS0 to the seller. At time N , either the asset quotes S+ , and then the seller gives g(S+ ) to the owner, either the asset quotes S− , in which case the seller gives g(S− ) to the owner. Fundamental Remark: neither the price C nor the portfolio (β, γ) depend on the probability that the asset S takes value S+ or S− ! why? Because our hedging strategy works whatever the evolution of the market, not in average. If we want to face our obligation in every case, the probability of rise or fall does not

CHAPTER 0. GOAL OF THE LECTURE

9

matter. Our hedging must work both when the price rises and when it falls. There is no probabilistic arguments here (6= average hedging via diversification). Nevertheless, the price C of the option can be view as the expected value of the payoff under some artificial probability the risk neutral probability. This remark is the corner stone of the pricing methods developed below. Let us focus on a european call. We define the risk neutral probability P∗ by P∗ (SN = S+ ) =

erN S0 − S− =: p∗ S+ − S −

and P∗ (SN = S− ) = 1 − p∗ .

We then have S0 = E∗ e−rN SN



 and C = E∗ e−rN (SN − K)+ ,

where E∗ represents the expectation under probability P∗ . The first equality says that the discounted value1 Sn∗ of S, defined by Sn∗ = e−rn Sn , keeps a constant expected value under P∗ . This is the fundamental property of a risk neutral probability: it annihilates the trend of the discounted quote of the stock S. The second equality says that the fair price of the option suits with the expected value of the discounted payoff under P∗ . We shall see in the following, that this formula is very general. —————— Another remark: we have made two major hypotheses in the previous analysis. First, we assume that the market is arbitrage-free, which fits more or less with the reality. Second, we assume to be in an ideal market: no transaction costs, possibility to buy/sell any quantity of asset, possibility to borrow as much money as needed, etc. Such a market is so-called complete. In this case the fair price is unique, which falls without theses hypotheses. —————— Exercise: compute the price of a european call, when S0 = 120, K = 80, r = 0 and S+ = 180, S− = 60.

1

also called ”present value”

Chapter 1 Conditional expectation The goal: to formalize the notion of conditional expectation given some information I. Informally, the conditional expectation of a random variable (r.v.) X given I represents the ’average value expected for X when one knows the information I’. Example: throw two dices. Write X1 and X2 for the value of first and second dice, and set S = X1 + X2 . If you have no information, the average value expected for S is E(S) = E(X1 ) + E(X2 ) = 7. Assume now that you know the value of X1 . Then, the average value expected for S knowing the value of X1 (our information) is X1 + E(X2 ) = X1 + 3.5 . The latter quantity is what we call the ’conditional expectation of S knowing X1 ’. Notations: Ω will represent the universe of the possibilities, we will write F for the set of all the possible events (F is a so-called σ-algebra) and P(A) for the probability that the event A ∈ F occurs.

1.1

Discrete setting

We associate to an event A ∈ F and an event B ∈ F of positive probability P(A | B) :=

P(A ∩ B) P(B)

which represents ”the probability that the event A occurs, knowing that the event B has occurred”. In the same way, we define the conditional expectation of a real random variable X : Ω → {x1 , . . . , xm } knowing that the event B has occurred by E (X | B) :=

m X

xi P(X = xi | B).

i=1

10

(1.1)

CHAPTER 1. CONDITIONAL EXPECTATION

11

This quantity is a real number that represents ”the average value expected for X knowing that the event B has occurred”. - Example: throw a dice with 6 faces, and set X for the obtained value. Write B for the event ”X P is not smaller than 3”. Then, the expected value of X knowing B is E(X | B) = 6i=1 i P(X = i | B). Since P(B) = 2/3, we have 3 P({X = i} ∩ B) 2  0 if i = 1 or 2 = 1/4 else.

P(X = i | B) =

Finally, E(X | B) = 41 (3 + 4 + 5 + 6) = 9/2.

1.1.1

Conditioning against a random variable

We will introduce in this section the notion of conditioning against a random variable. To start with, let’s have look to an example. —————— - Example: we play at a game in two stages: first stage we throw a dice and call Z the obtained value. Second stage, we throw Z times the dice and multiply the Z obtained values. We write X for this product. Note that we are able to compute for example E(X | Z = 5). If Z(ω) = 5, we throw 5 times the dice, independently, so E(X | Z = 5) = m5 , where m = 61 (1 + 2 + 3 + 4 + 5 + 6) = 7/2 is the expected value for a throw. More generally, for i ∈ {1, . . . , 6}, we have E(X | Z = i) = mi . In this case, we will define the conditional expectation of X knowing Z as E(X | Z) := mZ . Then, E(X | Z) is a random variable such that: if i is the value of Z, i.e. if Z(ω) = i, then E(X | Z)(ω) = E(X | Z = i). —————— We now generalize this notion to two general random variables X : Ω → {x1 , . . . , xm } and Z : Ω → {z1 , . . . , zn }, with x1 , . . . , xm , z1 , . . . , zn ∈ R. Definition 1.1 Conditional expectation of X given Z. The conditional expectation of X given Z, we note E(X | Z), is a random variable, defined by E(X | Z)(ω) := h(Z(ω)), where h is the function defined by h(zj ) = E(X | Z = zj ) for any j ∈ {1, . . . , n}. The quantity E(X | Z = zj ) is the one defined by (1.1) (see also the following exercise). Warning: E(X | Z = zj ) is a real number, but E(X | Z) is a random variable. Indeed, E(X | Z)(ω) depends on ω, since the value of Z(ω) depends on ω.

CHAPTER 1. CONDITIONAL EXPECTATION -

12

Example: Let Z be a r.v. with uniform law on {1, . . . , n} (i.e. P(Z = i) = 1/n for i = 1, . . . , n) and ε be a r.v. independent of Z, such that P(ε = 1) = p and P(ε = −1) = 1 − p. We set X = εZ. Then, X is a r.v. with value in {−n, . . . , n}. Let us compute E (X | Z). For j ∈ {1, . . . , n}, E (X | Z = j) =

n X

i P(X = i | Z = j)

i=−n

and P(X = i | Z = j) = 0 when i 6= j or i 6= −j (since X = ±Z), so it remains E (X | Z = j) = −jP(X = −j | Z = j) + jP(X = j | Z = j) = −jP(ε = −1 | Z = j) + jP(ε = 1 | Z = j). Now, since the r.v. ε and Z are independent P(ε = −1 | Z = j) =

P(ε = −1, Z = j) P(ε = −1)P(Z = j) = = P(ε = −1) = 1 − p P(Z = j) P(Z = j)

and in the same manner P(ε = 1 | Z = j) = P(ε = 1) = p. Finally, E (X | Z = j) = −j(1 − p) + jp = j(2p − 1), which exactly means that E (X | Z) = (2p − 1)Z. —————— Exercise: Check that E (X | Z) can also be written in the following form: E (X | Z) (ω) = =

n X j=1 n X

1{Z(ω)=zj } E (X | Z = zj ) 1{Z(ω)=zj }

j=1

X m

 xi P(X = xi | Z = zj ) .

i=1

—————— Remark: we usually write {Z = zj } := {ω ∈ Ω : Z(ω) = zj }, for the event Z takes the value zj . We also usually write P(X = xi , Z = zj ) for P({X = xi } ∩ {Z = zj }).

1.1.2

Conditioning against several random variables

Let us consider n + 1 random variables X : Z1 : Zn :

.. .

Ω → {x1 , . . . , xm }, Ω → {z1 , . . . , zk }, Ω → {z1 , . . . , zk }.

CHAPTER 1. CONDITIONAL EXPECTATION

13

We will extend the previous definition to the case where we deal with several random P variables. In this case, formula (1.1) reads E (X | Z1 = zj1 , . . . , Zn = zjn ) = m x i=1 i P(X = xi | Z1 = zj1 , . . . , Zn = zjn ). Remind that this quantity is a real number. Definition 1.2 Conditional expectation of X given Z1 , . . . , Zn . We call conditional expectation of X given Z1 , . . . , Zn the random variable defined by: E (X | Z1 , . . . , Zn ) (ω) := h(Z1 (ω), . . . , Zn (ω)), where h : Rn → R is defined by h(zj1 , . . . , zjn ) = E (X | Z1 = zj1 , . . . , Zn = zjn ) .

1.2 1.2.1

General case Extension to the case where there exists a joint density

Reminder: Two random variables X : Ω → R and Z : Ω → R, are said to have a density with respect to the Lebesgue measure, if there exists a function f(X,Z) : R × R → R+ such that for any domain D ⊂ plane, Z  P (X, Z) ∈ D = f(X,Z) (x, z) dx dz . D

In particular, P(X ∈ dx, Z ∈ dz) := P(X ∈ [x, x + dx], Z ∈ [z, z + dz]) = f(X,Z) (x, z) dx dz . Note that the margins can be computed in the following way  Z Z f(X,Z) (x, z) dz dx and P(Z ∈ dz) = P(X ∈ dx) =

 f(X,Z) (x, z) dx dz .

x∈R

z∈R

—————— When we deal with random variables that have a density, then P(X = x) and P(Z = z) are equal to zero, so we cannot define the quantity P(X = x | Z = z). To bypass this difficulty we will define another quantity P(X ∈ dx | Z = z) that will represent ”the probability that X ∈ [x, x + dx] given  R Z = z.” In view of P(Z ∈ dz) = x∈R f(X,Z) (x, z)dx dz and P(X ∈ dx, Z ∈ dz) = f(X,Z) (x, z) dz dx, it is natural to define P(X ∈ dx | Z = z) by P(X ∈ dx | Z = z) :=

f(X,Z) (x, z) P(X ∈ dx, Z ∈ dz) =R dx P(Z ∈ dz) f (y, z)dy y∈R (X,Z)

Definition 1.3 We call conditional expectation of X given Z the random variable defined by E (X | Z) (ω) = h(Z(ω)) R where h : R → R is defined by h(z) = x∈R x P(X ∈ dx | Z = z). It is a ”continuous space” version of the formula of 1.1.1.

CHAPTER 1. CONDITIONAL EXPECTATION

1.2.2

14

General case

In the general case, we will give a more formal definition of the conditional expectation: we will define it by its properties, instead by a formula. —————— Reminder: let us consider n + 1 r.v. X : Ω → R, Z1 : Ω → R, . . . , Zn : Ω → R. The r.v. X is so-called ”σ(Z1 , . . . , Zn )-measurable” iif1 there exists a (measurable) function h : Rn → R such that X(ω) = h(Z1 (ω), . . . , Zn (ω)). Let us draw a little scheme to illustrate the case n = 1: a random variable X is σ(Z)measurable iif there exists a (measurable) function h : R → R such that X = h(Z). In short: X

−→

Ω & Z

R %

R

h

We are now ready for giving a general definition of the conditional expectation. Definition (and theorem) 1.1 Let us consider n + 1 random variables X : Ω → R, Z1 : Ω → R, . . . , Zn : Ω → R, and set Fn := σ(Z1 , . . . , Zn ). There exists a unique random variable, called conditional expectation of X given Z1 , . . . , Zn (or given Fn ), we note E (X | Z1 , . . . , Zn ) (or E (X | Fn )), fulfilling the two properties: a) E (X | Fn ) is Fn -measurable b) for any Y which is Fn -measurable, we have E (E (X | Fn ) Y ) = E (XY ). Comments: we admit the existence of the conditional expectation. Nevertheless, when X ∈ L2 (F), we can check that E (X | Fn ) corresponds to the orthogonal projection of X onto the subspace L2 (Fn ). See exercise 5 below for more details. Besides, modifying a random variable on a set of null probability, does not change its law. Therefore, the uniqueness claimed in the above definition is to be understood as ”uniqueness up to a modification on a set of null probability”. Remark 1: condition a) is equivalent to a’) there exists a (measurable) function h such that E (X | Fn ) = h(Z1 , . . . , Zn ). In the same way, condition b) is equivalent to b’) E (E (X | Fn ) f (Z1 , . . . , Zn )) = E (Xf (Z1 , . . . , Zn )) for any (measurable) function f : Rn → R. Remark 2: In the discrete setting E (X | Z) and E (X | Z1 , . . . , Zn ) are given by the formula of the section 1.1.1 and 1.1.2. In the same manner, if (X, Z) as a density with respect to the Lebesgue measure, E (X | Z) is given by the formula of section 1.2.1. —————— 1

if and only if

CHAPTER 1. CONDITIONAL EXPECTATION

15

Let us check for example that we find back the formula of Section 1.1.1 For X : Ω → {x1 , . . . , xm } and Z : Ω → {z1 , . . . , zk }, we define h by   h(z1 ) = E (X | Z = z1 )    .. .  h(zk ) = E (X | Z = zk )    h(z) = 0 if z ∈ / {z1 , . . . , zk } The formula of the section 1.1.1 defines the conditional expectation of X given Z by E (X | Z) (ω) := h(Z(ω)). Does this definition fits with the one given above (1.2.2)? Let us check that h(Z) fulfills conditions a’) and b’), (which implies that the two definitions fit, since there exists a unique random variable fulfilling this two conditions). Condition a’) is clearly satisfied by h(Z) ! ? Let us check condition b’): do we have E (h(Z)f (Z)) = E (Xf (Z)) for any function f ? Check that k X h(Z(ω)) = 1{Z(ω)=zj } E (X | Z = zj ), {z } | {z } | i=1

random variable

non−random

so that  Pk  (Z) E (h(Z)f (Z)) = E j=1 1{Z=zj } E (X | Z = zj ) f | {z } =f (zj )  Pk f (z ) E (X | Z = z ) E 1 = j j {Z=z } j=1 | {z j } =P(Z=zj )

| = E

hP

k j=1

{z

“ ” =E X1{Z=zj }

f (zj )X1{Z=zj }

}

i

P Note that kj=1 f (zj )1{Z(ω)=zj } = f (Z(ω)), which leads to E (h(Z)f (Z)) = E (Xf (Z)). Condition b’) is then satisfied. To conclude, we have check that the two definitions fit. —————— Informally, E (X | Fn ) represents the average value expected for X when one knows the values of Z1 , . . . , Zn . Let us come back to the example of the introduction. Example: We throw two dice : X1 = value of the first dice X2 = value of the second dice S = total value = X1 + X2 . Since X1 et X2 are independent we expect that E (S | X1 ) = X1 + E (X2 ), since ”X1 gives no information on X2 ”. Write h(x) = x + E (X2 ), and check that h(X1 ) fulfills conditions a) and b). • Condition a): no problem.

CHAPTER 1. CONDITIONAL EXPECTATION

16

• Condition b): let us check that E (h(X1 )f (X1 )) = E (Sf (X1 )):   E (h(X1 )f (X1 )) = E (X1 + E (X2 ))f (X1 ) = E (X1 f (X1 )) + E (X2 ) E (f (X1 )) | {z } = E (X2 f (X1 )) since X1 ⊥ X2  Remind that if X1 et X2 are independent, E (f (X1 )g(X2 )) = E (f (X1 )) E (g(X2 )) . Conclusion : E (h(X1 )f (X1 )) = E ((X1 + X2 )f (X1 )) = E (Sf (X1 )), so condition b) is fulfilled and thus 7 E (S | X1 ) = h(X1 ) = X1 + E (X2 ) = X1 + . 2 In Practice, to compute a conditional expectation, we use the following properties. Properties 1.1 (fundamental properties) 1. E (E (X | Fn )) = E (X), 2. When X is Fn -measurable, E (X | Fn ) = X, 3. When X is independent of Z1 , . . . , Zn then E (X | Z1 , . . . , Zn ) (ω) = E (X) a.s. 4. For any a, b ∈ R, E (aX + bY | Fn ) = a E (X | Fn ) + b E (Y | Fn ), 5. When Y is Fn -measurable (i.e. Y = f (Z1 , . . . , Zn )) then E (Y X | Fn ) = Y E (X | Fn )   6. E E (X | Fn+p ) | Fn = E (X | Fn ). Proof of some properties : 1. let Y be the random variable such that Y (ω) = 1, ∀ω ∈ Ω. Y is Fn -measurable since Y = h(Z1 , . . . , Zn ), where h(z1 , . . . , zn ) = 1. Condition b) gives E (E (X | Fn ) Y ) = E (XY ), and since Y = 1, we get E (E (X | Fn )) = E (X). 2. X satisfies a) and if Y is Fn measurable, we have E (XY ) = E (XY )(!), so X fulfills b), and X = E (X | Fn ). 6. We set W = E (X | Fn+p ). The random variable E (W | Fn ) is Fn -measurable, so it fulfills condition a). For any Fn -measurable random variable Y , we have E (Y E (W | Fn ))

(condition b)

=

E (Y W ) = E (Y E (X | Fn+p )) .

Now, since Y is Fn measurable, Y is Fn+p measurable (check it!), and condition b) gives E (Y E (X | Fn+p )) = E (Y X). As a consequence, the random variable E (W | Fn ) fulfills condition b), so that E (X | Fn ) = E (W | Fn ) = E (E (X | Fn+p ) | Fn ) . ——————

CHAPTER 1. CONDITIONAL EXPECTATION

17

Properties 1.2 (other properties) 1. When Y1 , . . . , Yp are independent of X and Z1 , . . . , Zn , then E (X | Z1 , . . . , Zn , Y1 , . . . , Yp ) = E (X | Z1 , . . . , Zn ) , 2. If X ≤ Y , E (X | Fn ) ≤ E (Y | Fn ). What can we say about E (f (X) | Fn ) ? • Finite setting : E (f (X) | Z1 , . . . , Zn ) (ω) = hf (Z1 (ω), . . . , Zn (ω)), where hf (z1 , . . . , zn ) =

n X

f (xi )P(X = xi | Z1 = z1 , . . . , Zn = zn ).

i=1

• Continuous setting : E (g(X) | Z) (ω) = hg (Z(ω)) where R g(x)f(X,Z) (x, z)dx hg (z) = R f(X,Z) (x, z)dx At last, when Φ : R → R is a convex function, we have the Jensen inequality: Φ(E (X | Fn )) ≤ E (Φ(X) | Fn ) , that ensures for example that E (X | Fn )2 ≤ E (X 2 | Fn ).

1.3

Exercises

1. We throw a dice and write N for the result (between 1 and 6). Then, we throw N 2 times the dice, and write S for the sum of the results (including the first throw). Compute E(S|N ) and then E(S). 2. Assume that the random variables (X, Y ) have a density f (x, y) := n(n − 1)(y − x)n−2 1(x,y)∈A where A := {(x, y) ∈ R2 /0 ≤ x ≤ y ≤ 1}. Check that E(Y |X) =

n−1+X . n

3. Let (Xn ; n ∈ N) be a sequence of independent random variables. We focus on the random walk Sn := X1 + . . . + Xn and set Fn = σ(S1 , . . . , Sn ). a) Compute E(Sn+1 |Fn ). b) For any z ∈ C, check that   E z Sn+1 |Fn = z Sn E z Xn+1 .

CHAPTER 1. CONDITIONAL EXPECTATION 4. With the Jensen inequality: a) For any p > 1 prove that E(|X||Fn )p ≤ E(|X|p |Fn ). b) Prove the Jensen inequality in the discrete setting. 5. We assume that X ∈ L2 (F). Check that   min E (X − h(Z))2 = E (X − E(X|Z))2 . h

Conclude that E(X|Z) is the orthogonal projection of X onto L2 (σ(Z)).

18

Chapter 2 Martingales Example: a gambler in a casino. Let us investigate the fortune of a gambler in a casino: • J = the set of all the possible games A game : j ∈ J ex: − to stake on 2 at the roulette − not to stake (n)

• we write Rj

for the return of game j at time n. This is a random variable.  36 if 2 occurs (n) ex: − if j = to stake on 2, Rj = 0 else (n) − if j = not to stake, Rj = 1 (n)

We also write R(n) := (Rj ; j ∈ J) for the random vector made of the return at time n of each game. (n)

• The gambler will stake at time n on each game: we write Mj zero) on game j at time n.

for its stake (possibly

• We write Xn for the fortune of the gambler after time n. X0 is then its initial fortune. Since, ”not to stake” is considered as a game, we have X (n) Xn−1 = Mj . j∈J

After time n its fortune is Xn =

P

(n)

j∈J

(n)

Mj R j .

• The information that have the gambler after time n is the value of the returns R(1) , . . . , R(n) . We write Fn := σ(R(1) , . . . , R(n) ) for this information.

19

CHAPTER 2. MARTINGALES

20

Hypotheses of the game: • The returns R(1) , . . . , R(n) , . . . are independent. (n)

• The stakes Mj are Fn−1 -measurable: this means that the gambler choose its stakes at time n only on the basis of the information he has collected after time n − 1. He is enable to predict the future!   (n) ≤ 1, ∀n ∈ N, ∀j ∈ J. • On average, the casino does not loose money, i.e. : E Rj Problematic: 1. Is there a way for the gambler to choose its stakes, so that he wins money on average, viz so that E (Xn ) > X0 ? 2. Is there a way to ”stop gambling” so that, if T is the (random) time at which ones stop, E (XT ) > X0 ? Answer to the first question: Let us compute the conditional expectation of the gambler’s fortune after time n, given Fn−1 : ! X (n) (n) E (Xn | Fn−1 ) = E Mj Rj | Fn−1 j∈J

=

X

=

X

  (n) (n) E Mj Rj | Fn−1

j∈J (n) Mj E



(n) Rj

| Fn−1



(n)

since Mj

is Fn−1 -measurable

j∈J

=

X



X

  (n) (n) Mj E Rj

(n)

since Rj

is independent of Fn−1

j∈J (n) Mj

= Xn−1

since E



(n) Rj



≤ 1.

j∈J

We thus have proved that E (Xn | Fn−1 ) ≤ Xn−1

(2.1)

and we say that (Xn )n is a surpermartingale Taking the expectation of inequality (2.1), leads to E (E (Xn | Fn−1 )) ≤ E (Xn−1 ). According to the property 1.1.1 of the conditional expectation, it follows that E (Xn ) ≤ E (Xn−1 ). Iterating this inequality gives E (Xn ) ≤ E (Xn−1 ) ≤ . . . ≤ X0 . The answer to the first question is thus ”no” (which does not surprise anybody). With regard to the second question, we need to define the notion of ”stop gambling”, which is the purpose of the next section.

CHAPTER 2. MARTINGALES

2.1 2.1.1

21

Information, filtration, stopping time Information, filtration

In the previous example, the gambler’s information after time n is the return of each game at times 1, . . . , n, viz R(1) , . . . , R(n) . We write σ(R(1) , . . . , R(n) ) for that information. In the general case, when we observe a temporal random phenomenon (ex: the quotation of a share), the information we have at time n is the value of X1 , . . . , Xn (ex: the daily quotation of a share). We write σ(X1 , . . . , Xn ) for this information. To get lighter notations, we usually set Fn := σ(X1 , . . . , Xn ). As time passes, we collect more and more information so that F0 ⊂ F1 ⊂ . . . ⊂ Fn ⊂ Fn+1 ⊂ . . . Comment: In mathematics, Fn corresponds to a σ-algebra, and a filtration corresponds to an increasing sequence (Fn )n∈N of sigma-algebra.

2.1.2

Stopping time

In this section, we will formalize the notion of ”stop gambling”. The gambler is not expected to be able to predict the future. As a consequence, when he decides to stop gambling at time n, he takes its decision only from the information Fn he has at this time. For example, he can decide to stop when his fortune has reached twice its initial fortune, or when 5 consecutive blacks occur in the roulette, etc. A contrario, he can’t decide to stop when his fortune is maximal (the ideal!), since he should be able to predict the future. Informally, a stopping time is a time where we decide to stop only from the information we have (at this time). Let us give another example of stopping time: a selling order to your banker. You own a share and you ask your banker to sell it at a given time T depending on the evolution of the market. Unfortunately, your banker can’t predict the future, so you can’t ask him to sell it when the quotation is at its maximum. The time T at which he will sell your share shall only depends on the information he has at this time. For example, T can be the first time where the quotation pass over 100 $, or the first time where the quotation has increased of 15% in the last 100 days, etc. In both cases, your banker does not need more information than the one he has at time T . What are the characteristic of a stopping time T ? Let us focus on the selling order and write Fn for the information we have at time n. Note that the time T is a random variable (it depends on the evolution of the market). The key idea is that we are able to say at time n (namely from the information Fn ) whether T = n or not. Before going further, let us give some notation. Notation 1: to a set A, we associate the indicator function 1A , which is a random variable defined by:

CHAPTER 2. MARTINGALES

22

1A : Ω → {0, 1} 

1 if ω ∈ Ω 0 if ω ∈ /Ω In the same way, if X : Ω → R is a random variable and B is a subset of R, we set 1{X∈B} for the random variable: 1{X∈B} : Ω → {0,  1} 1 if X(ω) ∈ B ω 7→ 0 if X(ω) ∈ /B ω 7→ 1A (ω) =

Notation 2: we write ”A ∈ Fn ” for ”1A is Fn -measurable”. For example ”{T = n} ∈ Fn ” means that ”1{T =n} is Fn -measurable”. We have the following properties (properties of a σ-algebra). Properties 2.1 (de Fn ) 1. If A ∈ Fn , then A ∈ Fn where A = complementary of A. T S 2. If A1 , A2 , . . . ∈ Fn , then i Ai ∈ Fn and i Ai ∈ Fn . —————— We are now ready for defining precisely a stopping time. Definition 2.1 A stopping time is a random variable T : Ω → N ∪ {+∞} such that {T = n} ∈ Fn for any n. Informally, this says that at time n, we are able to say whether T = n or not. Proposition 2.1 Let us consider a stochastic process (Xn ). Write Fn = σ(X1 , . . . , Xn ) for our information at time n and T = inf{n ≥ 1 such that Xn ≥ a}. The time T represents the first time where Xn passes over a. Then, the time T is a stopping time. This result still holds true when replacing ≥ by =, ≤, < or >. A contrario, The first time M where Xn reaches it maximum, namely M = inf{n ≥ 1 such thatXn = max{p∈N} Xp } is not a stopping time. Indeed, at time n, you do not know whether M = n or not (you need to known the future) Proof : Note that: {T = n} = {X1 < a} ∩ · · · ∩ {Xn−1 < a} ∩ {Xn ≥ a} . Moreover for k = 1 . . . n − 1, we have {Xk < a} ∈ Fk ⊂ Fn and also {Xn ≥ a} ∈ Fn . According to the property 2.1.2, we conclude to {T = n} ∈ Fn . 2 —————— We now define the information we have at time T . Definition 2.2 The information FT we have at time T is defined as follows: A ∈ FT ⇔ A ∩ {T = n} ∈ Fn , Next proposition enumerates the properties of FT .

∀n ∈ N.

CHAPTER 2. MARTINGALES

23

Proposition 2.2 1. If σ(X1 , . . . , Xn ) ⊂ Fn and T is a stopping time,

then the random variable XT defined by XT (ω) =

      

X1 (ω) .. .

if T (ω) = 1

Xn (ω)   ..   .   0

if T (ω) = n if T (ω) = +∞

is FT -measurable. 2. If T (ω) = n, then E (Y | FT ) (ω) = E (Y | Fn ) (ω). 3. If S ≤ T are two stopping times, then FS ⊂ FT . Proof : 1. we shall check that XT : Ω → R is FT -measurable, i.e. that {XT ∈ A} ∈ FT for any A⊂R: T {XT ∈ A} ∈ FT ⇔ {XT ∈ A} T {T = n} ∈ Fn ∀n ∈ N ⇔ {Xn ∈ A} {T = n} ∈ Fn ∀n ∈ N, which is true. | {z } | {z } ∈Fn

∈Fn

2. et 3. : exercise.

2.2 2.2.1

2

Martingales Definition

We observe a random process (Xn )n∈N . We write Fn for our information at time n. This information contains σ(X0 , X1 , . . . , Xn ), given by the values X0 , X1 , . . . , Xn , that is to say σ(X0 , X1 , . . . , Xn ) ⊂ Fn . Usually, our only information is X0 , X1 , . . . , Xn so that Fn = σ(X0 , X1 , . . . , Xn ). Definition 2.3 (Martingale) • When E (Xn+1 | Fn ) = Xn , ∀n ≥ 0, we say that (Xn )n∈N is a martingale (more precisely a Fn -martingale). • When E (Xn+1 | Fn ) ≥ Xn , ∀n ≥ 0, we say that (Xn )n∈N is a submartingale. • When E (Xn+1 | Fn ) ≤ Xn , ∀n ≥ 0, we say that (Xn )n∈N is a supermartingale.

CHAPTER 2. MARTINGALES

24

Informally, a martingale is a stochastic process, such that: ”the average value expected for tomorrow given the information we have today, is equal to the today’s value” (no rise nor fall in average) Exercise: Let (Mn )n∈N be a martingale. Check that for any positive n and p E(Mn+p |Fn ) = Mn . Example: • The gambler’s fortune in a casino is a supermartingale. In this case, σ(X1 , . . . , Xn )

⊂ 6 =

Fn

• The random walk Sn = Y1 + . . . + Yn − nm, where the (Yi )’s are iid and m = E (Y1 ). The process (Sn )n∈N is a martingale, with Fn = σ(S0 , S1 , . . . , Sn ). Indeed, E (Sn+1 | Fn ) = E (Y1 + . . . + Yn+1 − (n + 1)m | Fn ) = E (Sn + Yn+1 − m | Fn ) , with Sn Fn -measurable, and Yn+1 independent of Fn . Therefore E (Sn+1 | Fn ) = Sn + E (Yn+1 ) −m = Sn . | {z } =m

• Assume that X is a random variable fulfilling E (|X|) < ∞ and consider a given filtration (Fn )n∈N (ex: Fn = σ(Z0 , . . . , Zn )). Then the process defined by Mn = E (X | Fn ) is a martingale, since E (Mn+1 | Fn ) = E (E (X | Fn+1 ) | Fn ) = E (X | Fn ) = Mn . We now state the fundamental property of the martingale: Property 2.1 (fundamental) The expected value of a martingale is constant as time passes, viz for any n ∈ N E (Xn ) = E (X0 ) . (2.2) Proof : We check it by iterating the equality P1

E (Xn+1 ) = E (E(Xn+1 |Fn ))

martingale

=

E (Xn ) . 2

One may wonder if equality (2.2), which is true at a fixed time n, still holds true at a random time T , viz do we have E (XT ) = E (X0 ) ? For sure, this is not true at any random time T . For exemple, when T is the time where Xn reaches its maximum, we can’t have E (XT ) = E (X0 ), except if X0 = XT a.s. Indeed, XT ≥ X0 . Nevertheless, we shall see below, that this equality turns to be true when T is a stopping time (modulo some restrictions).

CHAPTER 2. MARTINGALES

2.2.2

25

Optional stopping theorem

Notations: we set n ∧ T := min(n, T ) and write Xn∧T for the random variable Xn∧T : Ω → R ω 7→ Xn∧T (ω) (ω). The next result due to the mathematician J.L. Doob, is one of the two fundamental results in martingale theory. Theorem 2.1 (Optional stopping theorem) (Doob) Assume that (Xn ) is a martingale (respectively a supermartingale) and T is a stopping time. Then : 1. The process (Xn∧T )n∈N is a martingale (resp. supermartingale), 2. When T is bounded a.s., i.e. when there exists N ∈ N such that P(T ≤ N ) = 1, E (XT ) = E (X0 )

(resp. ≤) ,

3. If P(T < ∞) = 1 and if there exists Y such that |Xn∧T | ≤ Y for any n ∈ N, with E(Y ) < ∞ then E (XT ) = E (X0 ) (resp. ≤) . Remarks: 1. Concerning the gambler, whatever his choice of T , E (XT ) ≤ E (X0 ) = X0 = its initial fortune. 2. The hypotheses of 3. are necessary. Indeed, assume that we stake on a coin: • When tail occurs, we double our fortune. • When head occurs, we loose everything. Write pn ∈ {tail, head} for the result at time n and Xn for our fortune. We have  2Xn if pn+1 = tail Xn+1 = 0 if pn+1 = head viz Xn+1 = 2Xn 1{pn+1 =tail} . Since X0 = 1, iterating previous equality gives Xn = 2n 1{p1 =tail,...,pn =tail} . Let us check that (Xn ) is a martingale:  E (Xn+1 | Fn ) = E 2Xn 1{pn+1 =tail} | Fn   2Xn is Fn -measurable = 2Xn E 1{pn+1 =tail} since 1{pn+1 =tail} is independent of Fn = Xn since P(pn+1 = tail) = 1/2

CHAPTER 2. MARTINGALES

26

The time T = inf{n ≥ 0 such that Xn = 0} is a stopping time fulfilling P(T < ∞) = 1. Since XT (ω) (ω) = 0, we have E (XT ) = 0, so that E (XT ) = 0 6= E (X0 ) = 1. Why? the preceding result cannot be applied since |Xn∧T | cannot be dominated by any Y with finite mean. Proof : of the optional stopping theorem 1. We focus on the case where (Xn ) is a martingale. We have  Xn if T > n − 1 Xn∧T = XT if T ≤ n − 1 or equivalently Xn∧T = Xn 1{T >n−1} + XT 1{T ≤n−1} , so that   E (Xn∧T | Fn−1 ) = E Xn 1{T >n−1} | Fn−1 + E XT 1{T ≤n−1} | Fn−1 {z } | {z } | ∈Fn−1 Fn−1 -measurable (prop 2.1)

= 1{T >n−1} E (Xn | Fn−1 ) +XT 1{T ≤n−1} {z } | = Xn−1 since (Xn ) martingale = Xn−1 1{T >n−1} + XT 1{T ≤n−1} = X(n−1)∧T . We have proved that Xn∧T is a martingale. 2. When T is bounded a.s.: let N ∈ N be such that T ≤ N a.s. The process (Xn∧T ) is a martingale, so at time N : E (XN ∧T ) = E (X0∧T ) k k E (XT ) E (X0 ) 3. If T < ∞ a.s. and Xn∧T is dominated. The process (Xn∧T ) is a martingale, so E (Xn∧T ) = E (X0 ). Let n goes to infinity. First, note that Xn∧T → XT . Second, we can apply the theorem of dominated convergence, so that E (Xn∧T ) → E (XT ). As a consequence E (XT ) = E (X0 ). 2 Exercise: With the help of the first part of the optional stopping theorem and Fatou’s lemma, prove that when (Xn ) is a positive (super)martingale and T a stopping time, then we always have E(XT ) ≤ E(X0 ) (without conditions on T ) —————— In practice: how to use it ?

CHAPTER 2. MARTINGALES

27

Usually, we choose some suitable martingale (Xn ) and stopping time T and then apply the equality E (XT ) = E (X0 ) (warning: check the hypotheses of the optional stopping theorem!). —————— Example: The bankruptcy problem. Here is a simple example of use of the optional stopping theorem. Two persons A and B stake 1$ on a coin. A the initial time A has a$ and B has b$. They play until one of them has been bankrupted.  1 if A wins at time i A’s fortune : Xn = a + ε1 + . . . + εn , with (εi ) iid, εi = −1 if A looses at time i (Xn ) is a martingale and the time T of bankruptcy, namely T = inf{n ≥ 0 such that Xn = 0 or a + b} is a stopping time. Furthermore, |Xn∧T | ≤ a + b, so we can apply the optional stopping theorem, which ensures that a = E(X0 ) = E (XT ) = 0 P(A bankrupted) + (a + b) P(B bankrupted). As a consequence P(A wins B’s fortune) =

2.3 2.3.1

a . a+b

Exercises Galton-Watson genealogy

We focus henceforth on a simple model of population. At the initial time n = 0, there is one individual (the ancestor). We write Zn for the number of individual at the n-th generation. We assume that each individual gives birth to children independently of the others. We also assume that the number of children of an individual follows some “universal” law µ. Universal means that it is the same for everybody. (n) Let us formalize this situation. We write Xk for the number of children of the individual k present at generation n. Previous hypothesis says that the random variables   (n) Xk ; k ∈ N, n ∈ N 

(n) Xk



are independent and that their law is µ, viz P = i = µ(i). Furthermore, note that we have the formula (n) (n) Zn+1 = X1 + · · · + XZn . Let X be a random variable distributed according to µ (i.e. P(X = i) = µ(i)). We set m := E(X) , and write G for its generative function, defined by  G(s) := E sX , for s ∈ [0, 1]. We assume henceforth that µ(0) > 0.

CHAPTER 2. MARTINGALES

28

1. Express G(s) in terms of the µ(k), k ∈ N. What is the value of G(1)? G(0)? 2. Check that G is non-decreasing and convex. What is the value of G0 (1)? Draw the functions x 7→ G(x) and x 7→ x in the case where m ≤ 1 and m > 1. Take care of the behavior of G around 1.   (1) (n−1) 3. We set Fn := σ Xk , . . . , Xk ; k ∈ N , which represents the information contained in the genealogical tree up to generation n−1. Compute E(sZn+1 |Fn ). Deduce that  · · ◦ G}(s). Gn (s) := E sZn = G | ◦ ·{z n times

4. Express with words what Gn (0) corresponds to. We focus now on the probability of extinction, viz on p := P(∃ n ∈ N : Zn = 0). Express p in terms of the Gn (0)’s. 5. With the help of the figure of 2., check that p = 1 when m ≤ 1 and p = G(p) < 1, when m > 1. 6. We set Mn := Zn /mn . Check that Mn is a martingale. What is the value of E(Mn )? 7. We admits that the limit M∞ (ω) := limn→∞ Mn (ω) exists a.s. When m ≤ 1, what is the value of M∞ ? E(M∞ )? Compare E(M∞ ) and limn→∞ E(Mn ). Give a heuristic explanation to this result. 8. Prove that: E(exp(−λMn )) = Gn (exp(−λ/mn )). 9. For any λ > 0, prove with the theorem of dominated convergence that lim E(exp(−λMn )) = E(exp(−λM∞ )) =: L(λ).

n→∞

Derive from 8. (and Gn = G ◦ Gn−1 ) the functional equation satisfied by L: L(λ) = G(L(λ/m)). Remark: This equation completely determine L and thus the distribution of M∞ .

2.3.2

With the optional stopping theorem

- Maximal inequality (Doob) Assume that (Xn )n∈N is a non-negative martingale. To any real a > 0, we associate τa := inf{k ∈ N : Xk ≥ a}, the first passage time of Xn above a. a) Is τa a stopping time? b) Fix an integer n. Prove that E(Xmin(τa ,n) ) = E(X0 ) = E(Xn )

CHAPTER 2. MARTINGALES

29

and derive the equality E(Xτa 1τa ≤n ) = E(Xn 1τa ≤n ). c) Derive from this last formula, the upper bound (so-called “Doob’s maximal inequality”)  P

 1 max Xk ≥ a ≤ E (Xn 1maxk=1...n Xk ≥a ) . k=1...n a

- Wald identity We focus on the random walk Sn = X1 + · · · + Xn , with (Xn )n∈N i.i.d. We set m := E(Xn ) and T := inf{n ∈ N : Sn ≥ a} for a given a > 0. a) What can you say about T ? b) Assume that m > 0. What can we say about (Sn − nm)n∈N ? Derive the equality m E(T ∧ n) = E(ST ∧n ) = E(ST 1T ≤n ) + E(Sn 1T >n ). c) Prove that E(Sn 1T >n ) ≤ a P(T > n), and then justify with the law of large numbers the equality (so-called Wald identity) E(T ) =

1 E(ST ) m

(focus only on the case where Xn ≥ 0 for all n).

d) We assume now that m = 0. Prove that (Sn ) is a martingale. Do we have E(ST ) = E(S0 )? Why? Fix  > 0, and set T = inf{k ∈ N : Sk + k ≥ a}. Prove that T ≥ T and (use the Wald identity) E(T ) ≥ E(T ) = What is the value of E(T )?

a 1 E(ST + T ) ≥ .  

Chapter 3 The B-S market Goals: to define the B-S market and to rely the mathematical concept of “neutral risk probability” to the economical concepts of “arbitrage” and “completeness”.

3.1 3.1.1

The B-S market Evolution

We focus henceforth on a market made of two assets: • a non risky asset B, or bond (with predictable evolution), • a risky asset S, or Stock (with unpredictable evolution). We write ∆Bn := Bn − Bn−1 and ∆Sn := Sn − Sn−1 for the variation of B ans S between the time n − 1 and n. The assets evolve according to the dynamic:  ∆Bn = rn Bn−1 (3.1) ∆Sn = ρn Sn−1 with rn (interest rate) and ρn (return) random in general. We write Fn for the information we have at time n. Since S0 , . . . , Sn are known at time n, we have σ(S0 , . . . , Sn ) ⊂ Fn . The asset B is said non risky, because its evolution is predictable: at time n − 1 we know the value of the interest rate rn for the time n. The variable rn is thus Fn−1 -measurable. A contrario, S is a risky asset: at time n − 1 we do not know the value of ρn . The random variable ρn is thus Fn -measurable, but not Fn−1 -measurable. Remark: Here the time n is discrete. Note that we can see ∆Xn = Xn − Xn−1 as the derivative of X in discrete time. We also assume henceforth that B and S can only take a finite number of value. In other words, we are in a completely discrete setting. Note that this corresponds to the reality. First the quotation are given with only a finite precision. Second, the quotations occur in discrete time. —————— 30

CHAPTER 3. THE B-S MARKET

31

We can rewrite the dynamic as follows: equation (3.1) gives Bn − Bn−1 = rn Bn−1 so that Bn = (1 + rn )Bn−1 . Iterating this formula gives : Bn = (1 + rn ) . . . (1 + r1 )B0 . Setting Un =

n X

rk

and

Vn =

k=1

we obtain



n X

ρk

k=1

Bn = B0 εn (U ) Sn = S0 εn (V )

with εn (U ) = (1 + ∆Un ) . . . (1 + ∆U1 ). In analogy with the continuous time setting, we will call the random variable εn (U ) “stochastic exponential”.

3.1.2

Portfolio

Let us consider a portfolio Π = (βn , γn )n≤N made of βn units of B and γn units of S at time n. Its value at time n is XnΠ = βn Bn + γn Sn . We manage the portfolio in the following way. At time n, we have βn units of asset B and γn units of asset S. We then decide to reinvest for the next quotation, namely we choose βn+1 and γn+1 . This choice occurs at time n. This means that βn+1 and γn+1 are Fn -measurable (or equivalently βn and γn are Fn−1 -measurable). At time n, the value of the portfolio is XnΠ = βn Bn + γn Sn . After the reinvestment, its value is βn+1 Bn + γn+1 Sn . It is natural to assume that when we reinvest money, no value is added or lost, viz βn Bn + γn Sn = βn+1 Bn + γn+1 Sn . Such a trading strategy is said self-financed. In the following definition, the self-financing condition is expressed at time n − 1 instead of n. Definition 3.1 A portfolio (or trading strategy) is self-financed when βn−1 Bn−1 + γn−1 Sn−1 = βn Bn−1 + γn Sn−1 which can also be written as ∆βn Bn−1 + ∆γn Sn−1 = 0. Warning: the value of the portfolio does not change during the reinvestment, but it changes between two consecutive time, due to the fluctuation of B and S. As an exercise, check that Π is self-financed if and only if the fluctuation of X Π between time n − 1 and time n is ∆XnΠ = βn ∆Bn + γn ∆Sn .

CHAPTER 3. THE B-S MARKET

3.2

32

Arbitrage and martingale

3.2.1

Risk neutral probability

We have seen on the model “one step, two states” that it could be judicious not to investigate the evolution of the market under the real world probability but under another artificial probability the so-called risk neutral probability. This probability tends to annihilate the average rise or fall of the stock. We generalize here this point of view. Definition 3.2 A probability P∗ is a risk neutral probability if • P∗ is equivalent to the real world probability P, namely P∗ (A) = 0 iif P(A) = 0, • (Sn /εn (U ))n≤N is a martingale under P∗ . The second condition is equivalent to ”(Sn /Bn )n≤N is a martingale under P∗ ”. Next proposition gives a condition on Un and Vn that ensures that a probability is “risk neutral”. Proposition 3.1 Assume that P∗ is equivalent to P. Then P∗ is a risk neutral probability iif (Vn − Un ) is a martingale under P∗ . To prove this result we need the following properties of the stochastic exponential. Lemma 3.1 Properties of εn (·): 1. εn (X)εn (Y ) = εn (X + Y + [X, Y ]), where [X, Y ]n = variation of X and Y . 2.

1 εn (X)

= εn (−X ∗ ), where ∆Xn∗ = ∆Xn −

Pn

k=1

∆Xk ∆Yk is the quadratic

(∆Xn )2 . 1+∆Xn

3. (εn (X))n≤N is a martingale iif (Xn )n≤N is a martingale. Proof : of the Lemma. 1. We have εn (X)εn (Y ) = εn−1 (X) (1 + ∆Xn ) εn−1 (Y ) (1 + ∆Yn ) = εn−1 (X)εn−1 (Y ) (1 + ∆Xn + ∆Yn + ∆Xn ∆Yn ) iterate = εn (X + Y + [X, Y ]) (∆Xn )2 ∗ 2. Define X by ∆Xn∗ = ∆Xn − 1+∆X . In the view of part 1.: n εn (X)εn (−X ∗ ) = εn (X − X ∗ − [X, X ∗ ]). Besides ∗



∆(X − X − [X, X ])n

  (∆Xn )2 (∆Xn )2 − ∆Xn ∆Xn − = ∆Xn − ∆Xn + 1 + ∆Xn 1 + ∆Xn = 0.

CHAPTER 3. THE B-S MARKET

33

P Since for any process Z, we have Zn = Z0 + nk=1 ∆Zk , it follows that (X −X ∗ −[X, X ∗ ])n = 0 and finally εn (X)εn (−X ∗ ) = εn (0) = 1. 3. Using that εn−1 (X) is Fn−1 -measurable: E(εn (X) | Fn−1 ) = E((1 + ∆Xn )εn−1 (X) | Fn−1 ) = εn−1 (X)(1 + E(∆Xn | Fn−1 )) so E(εn (X) | Fn−1 ) = εn−1 (X)

E(∆Xn | Fn−1 ) = 0 E(Xn | Fn−1 ) = Xn−1

iif iif

2

Proof : of the proposition. We only have to check that (Sn /εn (U )) is a martingale under P∗ if and only if (Vn − Un ) is a martingale under P∗ . We have Sn εn (U )

S0 εn (V ) εn (U )

= Lemma 2.

S0 εn (V ) εn (−U ∗ )

Lemma 1.

S0 εn (V − U ∗ − [V, U ∗ ])

=

=

so according to Lemma 3.1.3, the process (Sn /εn (U ))n≤N is a martingale under P∗ iif (εn (V − U ∗ − [V, U ∗ ])n≤N is a martingale under P∗ . A computation gives Vn −

Un∗



− [V, U ]n =

n X ∆Vk − ∆Uk k=1

1 + ∆Uk

so (εn (V − U ∗ − [V, U ∗ ])n≤N is a martingale under P∗ iif ! n−1 n X X ∆Vk − ∆Uk ∆V − ∆U k k E∗ | Fn−1 = . 1 + ∆Uk 1 + ∆Uk k=1 k=1 Moreover ∆Uk = rk is Fk−1 -measurable so ! n n X X ∆V − ∆U 1 k k ∗ E | Fn−1 = E∗ (∆Vk − ∆Uk | Fn−1 ) 1 + ∆U 1 + ∆U k k k=1 k=1 =

n−1 X ∆Vk − ∆Uk k=1

1 + ∆Uk

+

1 E∗ (∆Vn − ∆Un | Fn−1 ) 1 + ∆Un

and (εn (V − U ∗ − [V, U ∗ ])n≤N is a martingale iif E∗ (∆Vn − ∆Un | Fn−1 ) = 0, namely iif V − U is a martingale. Putting pieces together leads to the claimed result. 2 —————— Next lemma investigates the statistical evolution of the value of a portfolio based on a self-financed strategy under a risk neutral probability.

CHAPTER 3. THE B-S MARKET

34

Proposition 3.2 Assume that there exists a risk neutral probability P∗ . Then, if Π is a self-financed portfolio, its so-called discounted value (or present value) εn (U )−1 XnΠ is a martingale under P∗ . Proof : Since Bn = B0 εn (U ), the discounted value of the portfolio is given by XnΠ /εn (U ) = βn B0 + γn Sn /εn (U ). Taking conditional expectation on both side leads to  Π    Xn Sn ∗ ∗ ∗ E | Fn−1 = E (βn B0 | Fn−1 ) + E γn | Fn−1 εn (U ) εn (U )   Sn ∗ | Fn−1 since βn and γn are Fn−1 -measurable = βn B0 + γn E εn (U ) Sn−1 = βn B0 + γn since (Sn /εn (U )) is a martingale under P∗ εn−1 (U ) = (βn Bn−1 + γn Sn−1 ) /εn−1 (U ) since Bn−1 = B0 εn−1 (U ) = (βn−1 Bn−1 + γn−1 Sn−1 ) /εn−1 (U ) self-financing strategy Π = Xn−1 /εn−1 (U ). Finally (XnΠ /εn (U )) is thus a martingale under P∗ .

3.2.2

2

Arbitrage

Henceforth, we focus on the evolution of the market until a fixed time N . In economy, we say that there is an arbitrage opportunity if there is an opportunity to make a profit without any risk of loosing money. In mathematical words this become: Definition 3.3 There exists an arbitrage opportunity, when there exists a self-financed strategy Π such that X0Π = 0,

XnΠ ≥ 0

∀n ≤ N,

and P(XNΠ > 0) > 0.

Next fundamental theorem relies the economic notion of ”arbitrage opportunity” to the mathematical notion of ”neutral risk probability”. Theorem 3.1 There exists no arbitrage opportunity ⇐⇒ There exists at least one risk neutral probability.

CHAPTER 3. THE B-S MARKET

35

In this case one says that the market is arbitrage-free. Proof : (⇐) Assume that P∗ is a risk neutral probability and Π is a self-financed trading strategy such that X0Π = 0. According to Proposition 3.2 (XnΠ /εn (U ))n≤N is a martingale under P∗ . It follows that   Π   XNΠ X0 ∗ ∗ E =E = 0. εN (U ) ε0 (U ) This equality together with XNΠ /εN (U ) ≥ 0 imply that P∗ (XNΠ > 0) = 0. Moreover P and P∗ are equivalent, which implies in turns P(XNΠ > 0) = 0. In conclusion, there exists no arbitrage opportunities. (⇒) Admitted. See, example given [5] Chap. V Section 2.d. 2

3.3

Complete market

In economy, a so-called complete market corresponds to an ideal market, without any constraints, nor transaction costs, where every assets can be found at any time and any quantities. In mathematics, this translates as follows. Definition 3.4 The B-S market is so-called complete, when for any random variable f : Ω → R+ , there exists a self-financed trading strategy Π such that XNΠ (ω) = f (ω), ∀ω ∈ Ω. Next theorem links this concept to the uniqueness of risk neutral probability. Theorem 3.2 Let us consider an arbitrage-free B-S market. Then The market is complete ⇐⇒ There exists a unique risk neutral probability P∗ . Proof : (⇒) Assume that the market is complete and that there exists two different risk neutral probabilities P∗ and P0 . Since P∗ = 6 P0 , there exists an event A such that P∗ (A) 6= P0 (A) .

(3.2)

Set f (ω) = εN (U )(ω)1{A} (ω). Due to the completeness of the market there exists a trading strategy Π such that XNΠ (ω) = f (ω) whatever ω ∈ Ω. According to Proposition 3.2, the discounted value of the portfolio Π is a martingale under P∗ and P0 . As a consequence,     E∗ εN (U )−1 XNΠ = E∗ ε0 (U )−1 X0Π and E0 εN (U )−1 XNΠ = E0 ε0 (U )−1 X0Π . Now, on the one hand ε0 (U )−1 = 1 and on the other hand εN (U )−1 XNΠ = εN (U )−1 f = 1{A} . Therefore E∗ (1{A} ) = X0Π = E0 (1{A} ),

CHAPTER 3. THE B-S MARKET

36

i.e. P∗ (A) = P0 (A) which contradicts (3.2). There cannot exist two different risk neutral probabilities. Besides, since the market is arbitrage-free, there exists at least one risk neutral probability. (⇐) Admitted. See, example given [5] Chap. V Section 4. 2

3.4

Girsanov Lemma

We have seen that arbitrage and completeness are closely linked to risk neutral probabilities. Usually, it is hard to compute P∗ . Nevertheless, next lemma gives a way to compute P∗ from P. There exists a very simple way to construct a probability P0 from P. Let us consider a random variable Z > 0 such that E(Z) = 1. We define P0 in setting P0 (A) := E(Z1A ) for any A ∈ F.

(3.3)

It is easily checked that P0 is a probability measure. Furthermore, we can prove that any probability P0 equivalent to P is of the previous form (3.3). Besides, note that the expectation under P0 of any random variable X is given by E0 (X) = E(ZX). Assume that you have a process which is not a martingale under P. Next lemma gives (in some situations) a way to construct a probability P0 such that the process becomes a martingale under P0 . Lemma 3.2 Girsanov formula Let (Mn )n≤N be a Fn -martingale under P and ZN be a (FN -measurable) random variable, fulfilling E(ZN ) = 1 et ZN (ω) > 0, ∀ω ∈ Ω. Let us define P0 by P0 (A) := E(ZN 1A ), where E represents the expectation under the probability P. Then the process (Mn∗ )n≤N defined by   n X Zk ∗ ∆Mk Fk−1 , where Zn = E(ZN |Fn ), Mn = Mn − E Zk−1 k=1 is a martingale under P0 . See exercises below for a proof. Next example shows how to use this formula. Example: Assume that the interest rates rn are constant, viz rn = r > 0 for any n. Assume also that the ρn ’s are i.i.d. and set m = E(ρn ) ≥ 0,

and σ 2 = E((ρn − m)2 ) > 0.

CHAPTER 3. THE B-S MARKET

37

Then, Girsanov formula ensures that the probability P∗ defined by P∗ (A) := E(ZN 1A ) with ZN = EN (G) and Gn := exercises below).

3.5

r−m (ρ1 σ2

(3.4)

+ · · · + ρn − mn), is a risk neutral probability (see

Exercises

3.5.1

Change of probability: Girsanov lemma

Let us prove Girsanov formula. First note that for any random variable X, we have E0 (X) := E(ZN X). 1. Set Zn = E(ZN |Fn ). Prove that (Zn )n≤N is a martingale under P. 2. Check that for any Fp -measurable random variable W , we have: E0 (W ) = E(ZN W ) = E(Zp W ). 3. Check that for any Fn -measurable random variable Y , we have E0 (Y |Fn−1 ) =

1 Zn−1

E(Y Zn |Fn−1 ).

4. Set αn = Zn /Zn−1 . Check that the process (Mn∗ )n≤N defined by Mn∗

= Mn −

n X

E(αk ∆Mk |Fk−1 ),

k=1

is a martingale under P0 (Girsanov lemma).

3.5.2

Application: computation of a risk neutral probability

Example 3.4 (continued): assume that the interest rates rn are constant, viz rn = r > 0 for any n (and Un = rn) and that the ρn are i.i.d. We set m = E(ρn ) ≥ 0,

et σ 2 = E((ρn − m)2 ) > 0.

1. We set Vn = ρ1 + . . . + ρn and Mn = Vn − mn. Check that (Mn )n≤N is a martingale under P. 2. We set Gn := r−m Mn and ZN = EN (G). Check that (Gn )n≤N and (En (G))n≤N are σ2 martingales under P.

CHAPTER 3. THE B-S MARKET

38

3. Derive that Zn = En (G) (with the notations of 3.5.1). Compute αn . 4. We define P∗ according to (3.4). Check (with Girsanov formula) that (Vn − Un )n≤N is a martingale under P∗ . 5. Conclude that P∗ is a risk neutral probability (think to Proposition 3.1).

Chapter 4 European option pricing The goal: to compute the price of a european option in a B-S market.

4.1

Problematic

We want to compute the fair price of a european option with payoff f at maturity N . Examples of european options: • european call: f (ω) = (SN (ω) − K)+ • european put: f (ω) = (K − SN (ω))+ • Collar option: f (ω) = min(SN (ω) − K1 , K2 ) • ”look-back” option: f (ω) = SN (ω) − min(S0 (ω), . . . , SN (ω)) Definition 4.1 (Hedging portfolio) A hedging portfolio (or strategy) Π, is a portfolio such that its value at maturity is larger than the payoff f , namely XNΠ (ω) ≥ f (ω) ∀ω ∈ Ω. The fair price of an option will then corresponds to the minimal initial value X0Π that can have a self-financed hedging portfolio. In short: Definition 4.2 (Price of an option) The fair price of a european option with payoff f and maturity N is   − Π is self-financed Π C := inf X0 such that . − XNΠ (ω) ≥ f (ω), ∀ω ∈ Ω In next section, we solve this minimization problem in an arbitrage-free and complete market. 39

CHAPTER 4. EUROPEAN OPTION PRICING

4.2

40

Pricing a european option in an arbitrage-free and complete market

We assume in this section that the market is arbitrage-free and complete. According to the previous chapter, there exists in this case a unique risk neutral probability P∗ . Next result establishes the price of an option in terms of P∗ . Theorem 4.1 Price of a european option.

1. The price of a european option with payoff f at maturity N is  C = E∗ εN (U )−1 f . 2. There exists a self-financed hedging strategy Π∗ with initial value C. The value at time n of the portfolio Π∗ is  ∗ XnΠ = E∗ εN (U )−1 εn (U )f | Fn . Remarks: • There exist theoretical formulas giving the composition (βn∗ , γn∗ )n≤N of the portfolio Π∗ . But they are seldom useful. • Usually, the hard task is to compute P∗ . This may be done with the Girsanov formula. When P∗ is known, then we can compute C at least numerically (using Monte-Carlo methods). Proof : Let us consider a self-financed portfolio Π. Under P∗ , the discounted value of the portfolio (XnΠ /εn (U )) is a martingale (Proposition 3.2). As a consequence E∗ (εN (U )−1 XNΠ ) = X0Π . Moreover XNΠ (ω) ≥ f (ω) ∀ω ∈ Ω and thus X0Π ≥ E∗ (εN (U )−1 f ), and therefore  C := inf

X0Π

− Π is self-financed such that − XNΠ (ω) ≥ f (ω), ∀ω ∈ Ω



 ≥ E∗ εN (U )−1 f .

(4.1)

Besides, the market is complete, so there exists a self-financed strategy Π∗ such that ∗ ∗ XNΠ (ω) = f (ω) for any ω ∈ Ω. Since Π∗ is self-financed, (XnΠ /εn (U )) is a martingale. As a consequence  ∗ ∗ X0Π = E∗ εN (U )−1 XNΠ = E∗ εN (U )−1 f .

CHAPTER 4. EUROPEAN OPTION PRICING

41 ∗

Now, Π∗ is a self-financed portfolio, so due to the very definition of C we have X0Π ≥ C. It follows that  (4.2) C ≤ E∗ εN (U )−1 f . Combining (4.1) and (4.2) we get  C = E∗ εN (U )−1 f . We have proved the first part of the theorem. Concerning the second part, note that Π∗ ∗ suits. Moreover, since (XnΠ /εn (U )) is a martingale under P∗ we have ∗

  XnΠ ∗ = E∗ εN (U )−1 XNΠ | Fn = E∗ εN (U )−1 f | Fn . εn (U ) We get the value of Π∗ at time n in multiplying the equality by εn (U ), and then make εn (U ) enter into the conditional expectation (εn (U ) is Fn -measurable). 2

4.3

And what for an incomplete market?

What happens when the market is incomplete? In this case, there is no unique fair price, but a spread of fair price. Let us define the following two quantities   − Π is self-financed + Π ask : C := inf X0 such that − XNΠ (ω) ≥ f (ω), ∀ω ∈ Ω   − Π is self-financed − Π bid : C := sup X0 such that − XNΠ (ω) ≤ f (ω), ∀ω ∈ Ω First note that when P∗ is risk neutral:  C − ≤ E∗ εN (U )−1 f ≤ C + .

(4.3)

Indeed, if Π is a self-financed portfolio, (XnΠ /εn (U )) is a martingale under P∗ (Proposition  3.2), so X0Π = E∗ εN (U )−1 XNΠ . In particular: • if XNΠ ≥ f , then X0Π ≥ E∗ (εN (U )−1 f ) and thus C + ≥ E∗ (εN (U )−1 f ) , • if XNΠ ≤ f , then X0Π ≤ E∗ (εN (U )−1 f ) and thus C − ≤ E∗ (εN (U )−1 f ) . Note that when the market is complete, then C + = C − (= C). Indeed, we have seen the ∗ ∗ existence of a self-financed portfolio Π∗ such that X0Π = C + = E∗ (εN (U )−1 f ) et XNΠ = f . ∗ Besides Π∗ also fulfills the condition of C − , so C + = X0Π ≤ C − . Finally C + = C − , since we always have C − ≤ C + (see (4.3)). In view of (4.3), we have in an incomplete market  sup C − ≤ ∗ inf E∗ εN (U )−1 f ≤ P risk neutral

P∗ risk neutral

This means that usually C − < C + : the spread is positive. What are the meaning of C − and C + ?

 E∗ εN (U )−1 f ≤ C + .

CHAPTER 4. EUROPEAN OPTION PRICING

42

• imagine an option with price x > C + : the seller can then earn money without risk. Indeed, at the end, he will have at least (x − C + ) × E∗ (εN (U )) > 0. • imagine an option with price x < C − : the seller is sure to loose money. Indeed, he will loose at least (C − − x) × E∗ (εN (U )) > 0. We thus conclude that [C − , C + ] corresponds to the fair prices. It is actually interesting to develop other strategies in incomplete market, see for example [3, 4].

4.4 4.4.1

Exercises Cox-Ross-Rubinstein model

We assume that the interest rates are constant, i.e. rn = r ≥ 0 and that the return (ρn )n≤N are i.i.d. We also assume that they can only take two values a and b. Our aim is to compute the price and hedging of an option with payoff f := g(SN ) and maturity N . E∗ (ρn ) = r p∗ := P∗ (ρn = b) = r−a b−a  2. Compute the probability P∗ SN = S0 (1 + b)k (1 + a)N −k in terms of p∗ . Conclude that C = (1 + r)−N FN∗ (S0 ) where 1.

Prove that P∗ is risk neutral

FN∗ (x)

:=

N X

⇐⇒ ⇐⇒

 g x(1 + b)k (1 + a)N −k CNk p∗k (1 − p∗ )N −k .

k=0

3. Assume that π ∗ is the optimal hedging strategy. Check that its value at time n is ∗

Xnπ = (1 + r)−(N −n) FN∗ −n (Sn ). 4. Assume that we are at time n − 1. We know S0 , S1 , . . . , Sn−1 and we shall choose βn∗ et γn∗ . Check that they must satisfy:  ∗ βn B0 (1 + r)n + γn∗ Sn−1 (1 + a) = (1 + r)−(N −n) FN∗ −n (Sn−1 (1 + a)) βn∗ B0 (1 + r)n + γn∗ Sn−1 (1 + b) = (1 + r)−(N −n) FN∗ −n (Sn−1 (1 + b)). 5. Derive that γn∗ = (1 + r)−(N −n) ∗

and

βn∗

FN∗ −n (Sn−1 (1 + b)) − FN∗ −n (Sn−1 (1 + a)) (b − a)Sn−1

π (1 + r)−(N −n+1) FN∗ −n+1 (Sn−1 ) − γn∗ Sn−1 Xn−1 − γn∗ Sn−1 = = . B0 (1 + r)n−1 B0 (1 + r)n−1

CHAPTER 4. EUROPEAN OPTION PRICING

43

6. In the case of a european call, viz when g(SN ) = (SN − K)+ , check that C = S0 B(k0 , N, p0 ) − (1 + r)−N K B(k0 , N, p∗ ) 1+b ∗ p = p, 1+r 0

where

B(k0 , N, p) =

N X

CNk pk (1 − p)N −k

k=k0

  K 1+b and k0 := min{k ∈ N : S0 (1 + a)N −k (1 + b)k > K} = 1 + log S0 (1+a) N / log 1+a .

4.4.2

Black-Scholes and Merton models

We will now rely the Cox-Ross-Rubinstein model to the Black-Scholes and Merton models, in letting √ the time-step√tends to zero (continuous limit). Assume that N = T /∆, r = ρ∆, a = −σ ∆ and b = σ ∆. We investigate the limit ∆ → 0 with T = N ∆ constant. 1. Use the central limit theorem to prove

where Φ(x) =

√1 2π

B(k0 , N, p∗ )

∆→0

B(k0 , N, p0 )

∆→0

Rx

2 /2

e−t

−∞





N p∗ − k0

Φ

p

Φ

p

!

N p∗ (1 − p∗ ) ! N p0 − k 0 N p0 (1 − p0 )

dt.

2. Check then that k0 ∼ N p∗ ∼ N p0 ∼ p

N p∗ (1 − p∗ ) ∼

p

N p0 (1 − p0 ) ∼ (1 + r)−N ∼

√ log(K/S0 ) + T σ/ ∆ √ 2σ ∆ √ T (ρ − σ 2 /2) + T σ/ ∆ √ 2σ ∆ √ T (ρ + σ 2 /2) + T σ/ ∆ √ 2σ ∆ r 1 T 2r∆ 1 T 2 ∆ e−ρT .

3. Conclude that     T (ρ − σ 2 /2) + log(S0 /K) T (ρ + σ 2 /2) + log(S0 /K) ∆→0 −ρT √ √ C ∼ S0 Φ − Ke Φ . σ T σ T We find here the Black-Scholes formula (see Chapter 8).

CHAPTER 4. EUROPEAN OPTION PRICING

44

When one takes a = −σ∆ and b constant, then the limit model is the Merton model based on a Poisson process. In this case C → S0 P1 − Ke−ρT P2 where P1 = P2 =

∞ X [(1 + b)(ρ + σ)T /b]i i=k0 ∞ X i=k0

i!

exp[−(1 + b)(ρ + σ)T /b)]

[(ρ + σ)T /b]i exp[−(ρ + σ)T /b)] i!

Chapter 5 American option pricing The goal: to compute the price of an american option and the optimal time of exercise.

5.1

Problematic

In this chapter, we will consider an arbitrage-free and complete B-S market. According to the results of Chapter 3, this implies the existence of a unique risk neutral probability P∗ . We focus henceforth on an american option with maturity N and payoff fn at time n ≤ N . Example: • american call: fn (ω) = (Sn (ω) − K)+ • american put: fn (ω) = (K − Sn (ω))+ • russian option: fn (ω) = supk≤n Sk (ω) —————— The owner of the option can exercise his option at any time before time N . He then receives fn . For hedging, the seller thus must have a portfolio Π such that XnΠ (ω) ≥ fn (ω) ∀n ≥ N, ∀ω ∈ Ω.

(5.1)

Definition 5.1 An hedging portfolio (or strategy) is a portfolio Π fulfilling (5.1). As for european options, the fair price of an american option corresponds to the minimal initial value that can have a self-financed hedging portfolio. In short: Definition 5.2 The price of an american option is   − Π is self-financed Π C := inf X0 such that − XnΠ (ω) ≥ fn (ω), ∀n ≤ N, ∀ω ∈ Ω

45

(5.2)

CHAPTER 5. AMERICAN OPTION PRICING

46

The goal of this chapter is to • compute C, • hedge our portfolio in an optimal way, • determine the optimal (rational) time for exercising the option. Write τ exc for the exercise time. The time τ exc will depend on the evolution of the market, so it is a random variable. Besides, the owner of the option cannot predict the future. When he decides to exercise his option, he takes his decision only on the information available at that time. This exactly means that τ exc is a stopping time.

5.2

Pricing on american option

Assume that Π is a self-financed hedging portfolio. According to Proposition 3.2, (XnΠ /εn (U )) is a martingale under P∗ . We also have seen before that τ exc is a stopping time bounded by N . Optional stopping theorem thus ensures that   E∗ ετ exc (U )−1 XτΠexc = E∗ ε0 (U )−1 X0Π = X0Π . Moreover XnΠ ≥ fn for any n ≤ N , so  X0Π ≥ E∗ ετ exc (U )−1 fτ exc . Since this must be true for any exercise time τ exc , we must have  X0Π ≥ sup E∗ ετ (U )−1 fτ τ s.t. τ ≤N

where ”s.t.” means ”stopping time”. As a consequence the price C defined by (5.2) admits for lower bound  C ≥ sup E∗ ετ (U )−1 fτ . (5.3) τ s.t. τ ≤N

We will actually see that previous inequality is an equality. Theorem 5.1 Price of an american option. The price of an american option with maturity N and payoff fn is  C = sup E∗ ετ (U )−1 fτ τ s.t. τ ≤N

where ”s.t.” means ”stopping time” and where E∗ represents the expectation under P∗ .

CHAPTER 5. AMERICAN OPTION PRICING

47

Formula (5.3) already gives us the inequality  C ≥ sup E∗ ετ (U )−1 fτ . τ s.t. τ ≤N

Thus to prove the theorem, all we need is to find a self-financed hedging portfolio Π∗ with initial value  ∗ X0Π = sup E∗ ετ (U )−1 fτ . τ s.t. τ ≤N

This work will be done at section 5.5. Before, we will try to better understand the optimisation problem appearing in Theorem 5.1.

5.3

Dynamic Programing Principle

The dynamic programing principle will enable us to compute the price given in Theorem 5.1. We write henceforth Xn := εn (U )−1 fn . Theorem 5.2 Dynamic Programing Principle. We define the sequence (Yn )n≤N by a backward iteration YN := XN

and Yn := max (Xn , E∗ (Yn+1 | Fn )) for n going from N − 1 to 0.

We also set Tn := inf{k ∈ [n, N ], such that Xk = Yk }. Then, for any stopping time τ with value between n and N E∗ (XTn | Fn ) = Yn ≥ E∗ (Xτ | Fn ) .

(5.4)

Y0 = E∗ (XT0 ) = sup E∗ (Xτ ) .

(5.5)

In particular, τ s.t. τ ≤N

Proof : a) We first prove that (5.4) implies (5.5). Taking the expectation of (5.4) gives E∗ (XTn ) = E∗ (Yn ) ≥ E∗ (Xτ ) . Specifying previous bound for n = 0, gives E∗ (XT0 ) = E∗ (Y0 ) ≥ E∗ (Xτ ) for any τ ≤ N . Moreover E∗ (Y0 ) = Y0 , so that E∗ (XT0 ) = Y0 ≥ sup E∗ (Xτ ) . τ s.t. τ ≤N

CHAPTER 5. AMERICAN OPTION PRICING

48

To conclude, since T0 is himself a stopping time, the inequality turns to be an equality. b) Let us prove (5.4). First, we check that Yn = E∗ (XTn | Fn ). In view of the very definition of Tn , for any k ∈ [n, Tn (ω) − 1], we have Yk (ω) = E∗ (Yk+1 | Fk ) (ω). An iteration thus enforces Yn (ω) = E∗ (XTn | Fn ) (ω). We now prove Yn ≥ E∗ (Xτ | Fn ) by a backward iteration. First, note that it holds true for n = N . We now assume that (5.4) holds true for any n ≥ k + 1. For any stopping time τ taking value in [k, N ], we have E∗ (Xτ | Fk ) = 1τ =k E∗ (Xτ | Fk ) + 1τ >k E∗ (Xτ | Fk ) = 1τ =k Xk + E∗ (Xτ 1τ >k | Fk ) . Note that Xτ 1τ >k = Xτ 0 1τ >k , where  τ when τ ≥ k + 1 0 τ = k + 1 when τ = k is also a stopping time. We then have E∗ (Xτ | Fk ) = 1τ =k Xk + 1τ >k E∗ (Xτ 0 | Fk ) E∗ (Xτ | Fk ) = 1τ =k Xk + 1τ >k E∗ (E∗ (Xτ 0 | Fk+1 ) | Fk ) . Furthermore, we have assumed that (5.4) holds true for n = k+1, so that E∗ (Xτ 0 | Fk+1 ) ≤ Yk+1 . Putting pieces together gives E∗ (Xτ | Fk ) ≤ 1τ =k Xk + 1τ >k E∗ (Yk+1 | Fk ) ≤ max(Xk , E∗ (Yk+1 | Fk )) = Yk . We thus have checked that (5.4) holds true for n = k. A descending iteration ensures that (5.4) holds true for any n. 2 Remark: the definition of Yn involves the conditional expectation E∗ (Yn+1 | Fn ). When the risk neutral probability is of the form P∗ = ZN P, this conditional expectation is given by 1 E (Yn+1 Zn+1 | Fn ) E∗ (Yn+1 | Fn ) = Zn where Zn = E(ZN | Fn ), see Exercise 3.5.1 Chapter 3.

5.4

Doob’s decomposition and predictable representation

We will see in this section two results of the martingale theory, that we will use for the proof of Theorem 5.1. Let us first introduce the so-called ”non-decreasing predictable processes”.

CHAPTER 5. AMERICAN OPTION PRICING

49

Definition 5.3 A process (An )n≤N is a non-decreasing predictable process if 1. An is Fn−1 -measurable, for all n ≤ N , viz (An ) is predictable 2. ∆An (ω) = An (ω) − An−1 (ω) ≥ 0, for all n ≤ N , ω ∈ Ω, i.e. (An ) is non-decreasing. The first result gives the decomposition of a supermartingale as a martingale and a nondecreasing predictable process. Theorem 5.3 Doob’s decomposition. Assume that (Yn )n≤N is a supermartingale (i.e. Yn ≥ E(Yn+1 | Fn )). Then, there exists a unique martingale (Mn )n≤N and a unique non-decreasing predictable process (An )n≤N such that Yn = Mn − An and Y0 = M0 . Proof : • Existence of Mn and An : set M n = Y0 −

n X

[E (Yk | Fk−1 ) − Yk ]

k=1

An =

n X

[Yk−1 − E (Yk | Fk−1 )]

k=1

and check that they suit. • uniqueness: if Yn = Mn − An = Mn0 − A0n , then ∆A0n = ∆An + ∆Mn0 − ∆Mn .

(5.6)

Since An and A0n are Fn−1 -measurable E (∆An | Fn−1 ) = ∆An

and E (∆A0n | Fn−1 ) = ∆A0n .

In the same way, since Mn and Mn0 are martingales E (∆Mn | Fn−1 ) = 0 and E (∆Mn0 | Fn−1 ) = 0. So taking the conditional expectation E (· | Fn−1 ) of equalityP(5.6), gives ∆A0n = n 0 0 ∆A k=1 ∆Ak and An = Pn n , for 0 all n ≤ N . Since A0 = A0 0 = 0, we have An 0= k=1 ∆Ak . So, in view of ∆Ak = ∆Ak , we have An = An . As a consequence, we also have Mn = Mn0 . 2 —————— The second result gives a so-called predictable representation of a martingale under P∗ .

CHAPTER 5. AMERICAN OPTION PRICING

50

Theorem 5.4 Predictable representation. Let (Mn )n≤N be a martingale under the unique risk neutral probability P∗ . Then, there exists a predictable process (αn )n≤N (i.e. αn is Fn−1 -measurable) such that Mn (ω) = M0 +

n X

 αk (ω)∆ εk (U )−1 Sk (ω).

k=1

Proof : Set f = εN (U )MN . Since the market is complete, there exists a self-financed portfolio Π = (βn , γn )n≤N such that XNΠ = f . Moreover (Mn ) is a martingale under P∗ as well as (εn (U )−1 XnΠ ) (Proposition 3.2), so  E∗ (MN | Fn ) = Mn et E∗ εN (U )−1 XNΠ | Fn = εn (U )−1 XnΠ . Since XNΠ = f = εN (U )MN , we have Mn = εn (U )−1 XnΠ for all n ≤ N . Therefore, according to next lemma, we only need to choose αk = βk to conclude (remind that βk is Fk−1 -measurable). 2 Lemma 5.1 Assume that Π = (βn , γn )n≤N is a (generic) self-financed portfolio. Then   n X Sk XnΠ Π = X0 + . γk ∆ εn (U ) εk (U ) k=1 Proof : (of the lemma). Remind that a portfolio Π is self-financed if  Π Xn = βn Bn + γn Sn Π Xn−1 = βn−1 Bn−1 + γn−1 Sn−1 = βn Bn−1 + γn Sn−1 (self-financing) As a consequence  ∆

XnΠ εn (U )



Π Xn−1 XnΠ − εn (U ) εn−1 (U ) Sn Sn−1 = βn B0 + γn − βn B0 − γn εn (U ) εn−1 (U )   Sn Sn−1 = γn − . εn (U ) εn−1 (U )

=

2

This proves the lemma.

5.5

Proof of theorem 5.1

We set  Λ := sup E∗ ετ (U )−1 fτ . τ s.t. τ ≤N

CHAPTER 5. AMERICAN OPTION PRICING

51

All we need to prove Theorem 5.1, is to find a self-financed hedging portfolio with initial value Λ. Let us consider the sequence (Yn ) of the dynamic programing principle. In view of Yn := max (Xn , E∗ (Yn+1 | Fn )) , we have Yn ≥ E∗ (Yn+1 | Fn ), so (Yn )n≤N is a supermartingale. We write Yn = Mn − An for its Doob decomposition and Mn = M 0 +

n X

αk ∆ εk (U )−1 Sk



(5.7)

k=1

for the predictable representation of (Mn ). Note that M0 = Y0 = Λ. We will construct a self-financed hedging portfolio Π∗ = (βn∗ , γn∗ )n≤N with initial value Λ. First, we set γn∗ = αn for all n ≤ N . The initial value of the portfolio is β1∗ B0 + γ1∗ S0 . This value is equal to Λ, when we set β1∗ := (Λ − γ1∗ S0 ) /B0 . ∗

Π Then, the self-financing condition enforces that Xn−1 = βn∗ Bn−1 + γn∗ Sn−1 , which in turns enforces that  Π∗ βn∗ := Xn−1 − γn∗ Sn−1 /Bn−1 .

The sequence (βn∗ ) is thus defined by iteration. By construction, the portfolio Π∗ is selffinanced and its initial value is Λ. It remains to check that it is also an hedging portfolio. In view of Lemma 5.1 −1

εn (U )

∗ XnΠ

=Λ+

n X

 γk∗ ∆ εk (U )−1 Sk .

k=1

Keep in mind that γk∗ = αk and compare previous equality to (5.7). It turns out that ∗ Mn = εn (U )−1 XnΠ . In particular ∗

XnΠ = εn (U )Mn ≥ εn (U )Yn

since Mn = Yn + An with An ≥ 0.

Now according to the dynamic programing principle (5.4) εn (U )Yn =

sup

 E∗ ετ (U )−1 εn (U )fτ | Fn ≥ fn

τ s.t. n≤τ ≤N

(the latter equality comes from fn = E∗ (ετ (U )−1 εn (U )fτ | Fn ) for τ = n). Putting pieces ∗ together, we conclude that XnΠ ≥ fn for all n ≤ N . The portfolio Π∗ is then an hedging portfolio. Remark: T0 = inf{k ≤ N, such that εk (U )−1 fk = Yk } is the optimal time of exercise. Indeed, according to (5.5), for all stopping time τ bounded by N   E∗ ετ (U )−1 fτ ≤ E∗ εT0 (U )−1 fT0 .

CHAPTER 5. AMERICAN OPTION PRICING

5.6

52

Exercises

Assume that εn (U )−1 fn = g(Yn )Mn with Mn a martingale under P∗ , M0 = 1 and g such that g(y) ≤ g(y ∗ ) ∀y ∈ R. 1. check that εn (U )−1 fn ≤ g(y ∗ )Mn . Derive that for any stopping time τ ≤ N  E∗ ετ (U )−1 fτ ≤ g(y ∗ )E∗ (Mτ ) = g(y ∗ ). 2. Conclude that C ≤ g(y ∗ ). 3. Set τ ∗ := inf{n ≤ N : Yn = y ∗ }. We assume that with probability 1, Yn takes the value y ∗ before time N . Prove in this case that E∗ (ετ ∗ (U )−1 fτ ∗ ) = g(y ∗ ) and C = g(y ∗ ). What is the optimal exercise time? Remark: we usually do not have ”with probability 1, Yn takes the value y ∗ before time N ”. Nevertheless, when N goes to infinity the probability that Yn takes the value y ∗ before time N tends to 1. Therefore, for large value of N , g(y ∗ ) is a good approximation of C and τ ∗ is a good approximation of the optimal exercise time.

Chapter 6 Brownian motion The goal: to give a short introduction to continuous-time processes and especially to Brownian motions.

6.1

Continuous-time processes

A continuous-time process is a collection (Xt )t∈I of random variables indexed by some subinterval I of R, often I = [0, ∞[. We usually ask some regularity properties to the trajectories t 7→ Xt (ω). Definition 6.1 A process (Xt )t≥0 is so-called ”continuous” (respectively ”left-continuous”) when for any ω ∈ Ω the trajectories t 7→ Xt (ω) are continuous (resp. left-continuous). As in discrete time, we call filtration, a collection (Ft )t≥0 of nested σ-algebras F0 ⊂ · · · ⊂ Ft ⊂ · · · Typically, Ft will correspond to the information given by a process X up to time t, namely Ft = σ(Xs , s ≤ t). Definition 6.2 A process Y is so-called Ft -adapted, if for any t ≥ 0 the random variable Yt is Ft -measurable. . Example: consider a continuous process X. Set Yt = sups≤t Xs and Ft = σ(Xs , s ≤ t). Then, the process Y is Ft -adapted.

6.2

Brownian motion

Brownian motion is one of the basic process for modeling in continuous time. In this section we will define the Brownian motion and state its basic properties. No proof are given here, we refer the interested reader to [2] for (much) more details. 53

CHAPTER 6. BROWNIAN MOTION

6.2.1

54

Gaussian law

A Gaussian (or normal ) real random variable with mean m and variance t is a real random variable whose law admits a density against the Lebesgue measure given by   1 (x − m)2 √ exp − . 2t 2πt We write henceforth N (m, t) for the law of a Gaussian random variable with mean m and variance t. We remind some properties of Gaussian random variables. Properties 6.1 (of Gaussian laws) 1. the characteristic function of a N (m, t) random variable X is E(exp(izX)) = exp(izm − tz 2 /2),

∀ z ∈ C.

2. If X follows a N (m, t) law then a + bX follows a N (a + bm, b2 t) law. 3. If X and Y are two independent random variables with respectively N (a, t) and N (b, s) law then X + Y follows a N (a + b, t + s) law.

6.2.2

Definition of Brownian motion

Definition 6.3 A Brownian motion (or Wiener process) W = (Wt )t≥0 is a real-valued process such that 1. W is continuous and W0 = 0, 2. for any positive t, the random variable Wt follows a N (0, t) law, 3. For any s, t > 0, the increment Wt+s − Wt is independent of Ft and follows a N (0, s) law. We admit the existence of such a process. Warning! a continuous process X such that for any time t > 0, the variable Xt follows √ a N (0, t) law is not necessary a Brownian motion. Think, example given, to Xt = t N , with N distributed according to the N (0, 1) law. The process (Xt )t≥0 fulfills the two first conditions but not the last one. What does a Brownian motion looks like? Next result states that a Brownian motion is in some sense ”a continuous-time random walk”. Theorem 6.1 Donsker’s principle Assume that (Xi )i∈N is a sequence of i.i.d. random variables such that E(Xi ) = 0 and E(Xi2 ) = 1. Set Sn = X1 + · · · + Xn . Then the following convergence holds   S[nt] (d) √ −→ (Wt )t≥0 n t≥0

CHAPTER 6. BROWNIAN MOTION

55 (d)

where W is a Brownian motion, [x] represents the integer part of x and ”−→” means ”convergence in distribution” (see the Appendix for a reminder on the various types of convergence)

6.2.3

Properties of Brownian motion

We list below some basic properties of Brownian motion. Again, we refer the interested reader to [2] for proofs and finer properties. Properties 6.2 (of Brownian motion) Assume that W is a Brownian motion. Then, 1. Symmetry: the process (−Wt )t≥0 is also a Brownian motion.  2. Scaling: for any c > 0, the process c−1/2 Wct t≥0 is again a Brownian motion. 3. Roughness: the trajectories t 7→ Wt (ω) are nowhere differentiable. 4. Markov property: for any u > 0, the process (Wu+t − Wu )t≥0 is again a Brownian motion, independent of Fu . (Admitted) Some properties of the Brownian motion are rather unusual. Let us give an example. Fix a > 0 and write Ta for the first time where the Brownian motion W hits the value a. Then, whatever ε > 0, the Brownian motion W will hit again infinitely many times the value a during the little time interval [Ta , Ta + ε] !

6.3

Continuous-time martingales

As in the discrete setting, martingales will play a major role in the continuous-time theory of option pricing. Definition 6.4 A process (Mt )t≥0 is a martingale when for any s, t ≥ 0 E(Mt+s |Ft ) = Mt .

When a martingale is continuous, the properties we have seen in the discrete setting (fundamental property, optional stopping theorem, maximal inequality, Doob’s decomposition, etc) still hold true after replacing ”n ∈ N” by ”t ∈ [0, ∞[”.

CHAPTER 6. BROWNIAN MOTION

6.4

56

Exercises

6.4.1

Basic properties

√ 1. Assume that N follows a N (0, 1) law and set Xt = t N for any t ≥ 0. Check that the process X fulfills the two first properties of Brownian motion, but not the third one. 2. Check that a Brownian motion W is a martingale. 3. Set Ft = σ (Ws , s ≤ t) and Lt = Wt2 − t. Prove that L is a Ft -martingale. 4. Set Et = exp(Wt − t/2). Is Et a martingale? 5. Assume that M is a martingale such that E (Mt2 ) < +∞. Check that for any s ≤ t:   E (Mt − Ms )2 | Fs = E Mt2 − Ms2 | Fs .

6.4.2

Quadratic variation

Henceforth, W represents a Brownian motion. For a partition τ = (t1 , . . . , tn ), of [0, t] (i.e. 0 = t0 < t1 < · · · < tn = t), we set ∆tk = tk − tk−1 and ∆Wk = Wtk − Wtk−1 . 1. For s ≤ t, check that E (Wt Ws ) = E ((Wt − Ws )Ws ) + E (Ws2 ) = s.   2. Check that E (∆Wk )2 = ∆tk and E (∆Wk )4 = 3 (∆tk )2 . P 3. Write |τ | = maxk=1...n ∆tk and < W >τt (ω) = nk=1 (∆Wk (ω))2 . Check that n X n X      E (< W >τt −t)2 = E (∆Wj )2 − ∆tj (∆Wk )2 − ∆tk j=1 k=1 n X

(∆tk )2

= 2

k=1 |τ |→0

≤ 2|τ | × t −→ 0 . 4. Deduce from the inequality <W

>τt ≤

max |∆Wk | ×

k=1...n

n X

|∆Wk |

k=1

and the uniform continuity of the Brownian motion on [0, t] that n X

|τ |→0

|∆Wk | −→ +∞.

k=1

We say that the Brownian motion has no bounded variations.

Chapter 7 Itˆ o calculus The goal: to give a short introduction to Itˆo calculus and its special rules.

7.1

Problematic

Let us discuss briefly and informally the motivation for introducing the Itˆo calculus. We want to model a ”continuous-time noisy signal” X = (Xt )t≥0 . In analogy with the discrete time, we want the evolution on a short time interval δt to be given by ”δXt (ω) = a(t, ω)δt + σ(t, ω)t (ω)”

(7.1)

where t represents some ”noise” fulfilling the properties: • t is independent of s for s 6= t, and has the same law, • E(t ) = 0. P Therefore, setting ”Wt = s≤t s ” it is natural, in view of the Donsker’s principle, to assume that Wt is a Brownian motion. Equation (7.1) then turns to δXt (ω) = a(t, ω)δt + σ(t, ω)δWt (ω) . Unfortunately, the Brownian motion is nowhere differential, so we cannot give a meaning to this equality in terms of differentials. We will try instead to give a meaning to it in terms of integrals: Z t Z t Xt (ω) = X0 + a(s, ω) ds + σ(s, ω)dWs (ω). (7.2) 0

0

The first integral enters into the classical field of integration theory and is well-defined. But what is the meaning of the second integral? Should W be with bounded variations, could we define the second integral in terms of the Stieljes’ integral. Unfortunately, we have seen in Exercise 6.4.2 that a Brownian motion has no bounded variations. The goal of Itˆo’s integration theory is precisely to give a mathematical meaning to (7.2). 57

ˆ CALCULUS CHAPTER 7. ITO

7.2

58

Itˆ o’s integral

Henceforth, W is a Brownian motion and Ft = σ(Ws , s ≤ t). Definition (and theorem) 7.1 - Itˆ o’s integral Rt 2 Assume that H is a left-continuous Ft -adapted process fulfilling Hs ds < ∞ for any 0 R  t t > 0. Then, there exists a continuous process 0 Hs dWs such that t≥0

n X

P



Z

t

H(i−1)t/n Wit/n − W(i−1)t/n −→

Hs dWs

for any t > 0,

(7.3)

0

i=1 P

where ”−→” means ”convergence in probability” (see the Appendix for a reminder). This process fulfills the equality Z t 2 ! Z t  E Hs2 ds , = E Hs dWs 0

0

even when this quantity takes infinite value. Rt Furthermore, in the case where E (Hs2 ) ds < ∞ for any t > 0, the Itˆ o’s integral 0  R t Hs dWs is a Ft -martingale. 0 t≥0

R· We admit the existence of the process 0 Hs dWs and refer the interested reader to the Appendix 9.2 for the main lines of its construction. Before investigating the properties of Itˆo’s integral, let us stop for a comment.  P Question:Pdoes the above result still hold true when setting ni=1 Hit/n Wit/n − W(i−1)t/n  instead of ni=1 H(i−1)t/n Wit/n − W(i−1)t/n ? The answer is no in general. Indeed, takes H = W . According to the exercise 6.4.2 n X



Wit/n Wit/n − W(i−1)t/n −

i=1

and therefore

7.3

n X

 L2 W(i−1)t/n Wit/n − W(i−1)t/n −→ t

i=1

Pn

i=1

 P Rt Wit/n Wit/n − W(i−1)t/n → t + 0 Ws dWs .

Itˆ o’s processes

When K is R  a process whose path t 7→ Kt (ω) are left-continuous, we can define the process t Ks ds by 0 t≥0 Z t  Z t Ks ds (ω) := Ks (ω) ds , ∀ ω, t , 0

0

where in the right-hand side, the integral is a classic Riemann (or Lebesgue) integral.

ˆ CALCULUS CHAPTER 7. ITO

59

Definition 7.1 We call ”Ito process” a process X of the form Z t Z t Ks ds, Hs dWs + Xt = X0 + 0

0

where K and H are Ft -adapted and left-continuous, and where H fulfills for any t > 0.

Rt 0

Hs2 ds < ∞,

For short notations, we usually write dXt = Ht dWt + Kt dt. Next fundamental Lemma says that a smooth function of an Itˆo process is still an Itˆo process. Theorem 7.1 - Itˆ o’s formula - Assume that X is an Itˆ o process and g : R+ × R → R 2 belongs to C (twice continuously differentiable). Then Z t Z t Z ∂g ∂g 1 t 2 ∂2g g(t, Xt ) = g(0, X0 ) + (s, Xs ) dXs + (s, Xs ) ds + H (s, Xs ) ds . 2 0 s ∂x2 0 ∂x 0 ∂t Comments: 1. The term

Rt

∂g (s, Xs ) dXs 0 ∂x

Z 0

t

is a short notation for Z t ∂g ∂g (s, Xs ) Hs dWs + (s, Xs ) Ks ds . ∂x 0 ∂x

2. The last term in the right-hand side is unusual. This odd rule of calculus is due to the fact that Brownian motion has a quadratic variation. Let us inspect this. Proof : We only sketch the main lines of the proof of Itˆo’s formula: our goal is just to understand the appearance of the last integral. Set ti = it/n and for any process Y , we write ∆Yti = Yti − Yti−1 . The Taylor expansion ensures that X g(t, Xt ) = g(0, X0 ) + ∆g(ti , Xti ) i

= g(0, X0 ) +

X ∂g i

∂t

(ti−1 , Xti−1 )∆ti +

∂g (ti−1 , Xti−1 )∆Xti ∂x

1 X ∂2g + (ti−1 , Xti−1 ) (∆Xti )2 + residue, 2 2 i ∂x where the residue goes to 0 when n goes to infinity. Furthermore, note that (∆Xti )2 = Ht2i−1 (∆Wti )2 + ”residue”, and according to exercise 6.4.2 (∆Wti )2 ≈ ∆ti . So, the convergence of Riemann sums to Riemann integrals gives Z 1 t 2 ∂2g 1 X ∂2g 2 (ti−1 , Xti−1 ) (∆Xti ) ≈ H (s, Xs ) ds . 2 i ∂x2 2 0 s ∂x2

ˆ CALCULUS CHAPTER 7. ITO

60

Using again the convergence of Riemann sums to Riemann integrals for the first term of Taylor expansion and the approximation formula (7.3) for the second term leads to the claimed result: Z t Z t Z ∂g 1 t 2 ∂2g ∂g (s, Xs ) dXs + (s, Xs ) ds + g(t, Xt ) = g(0, X0 ) + H (s, Xs ) ds . 2 0 s ∂x2 0 ∂t 0 ∂x 2 Remark: The condition g ∈ C 2 is necessary. Indeed, set g(t, x) = |x|, which is not C 2 at 0. Let us check R t that Itˆo’s formula cannot hold true for this function. Otherwise, we would have |Wt | = 0 sgn(Ws ) dWs , which is impossible since the left-hand side is a submartingale (Jensen formula), whereas the right-hand side is a martingale. Indeed, the right-hand side Rt is a stochastic integral, with the integrand fulfilling 0 E (sgn(Ws )2 ) ds = t < ∞. Examples: 1. Check that Wt2 = 2

Rt 0

Ws dWs + t and Wt3 = 3

Rt 0

Ws2 dWs +

Rt 0

Ws ds.

2. Geometric Brownian motion - our goal is to find a process X such that dXt = µXt dt + σ Xt dWt , with µ, σ ∈ R. We will search this process in the form Xt = g(t, Wt ). Previous equation then reads dXt = µg(t, Wt ) dt + σg(t, Wt ) dWt . Besides, Itˆo’s formula says that ∂g (t, Wt ) dWt + dXt = ∂x



 ∂g 1 ∂2g (t, Wt ) + (t, Wt ) dt . ∂t 2 ∂x2

So, all we need is to solve ∂g , ∂x ∂g 1 ∂ 2 g and µg = + . ∂t 2 ∂x2 σg =

 First equation gives g(t, x) = h(t)eσx , and second equation leads to h0 (t) = µ − 21 σ 2 h(t). In conclusion, we find     σ2 Xt = X0 exp σWt + µ − t , 2 which is so-called ”geometric Brownian motion”.

ˆ CALCULUS CHAPTER 7. ITO

7.4 7.4.1

61

Girsanov formula Stochastic exponentials

Definition 7.2 We associate to an Itˆ o process dXt = Ht dWt , its stochastic exponential  Z t Z 1 t 2 H ds . εt (X) := exp Hs dWs − 2 0 s 0 Exercises: 1. Use Itˆo ’s formula to check that εt (X) is solution of the stochastic differential equation dYt = dXt . Yt

(7.4)

2. Derives of this equation that ε(X) is a martingale as soon as Z t  E (Hs εs (X))2 ds < ∞ , for any t > 0 . 0

Warning! the solution of (7.4) is Yt = εt (X) and not Yt = exp(Xt ) as you may have expected. This is due to the special rules of Itˆo calculus.

7.4.2

Girsanov formula

We state in this section a continuous-time Girsanov formula. We will not prove any result. Again, we refer the interested reader to [2] for further details. We have seen that under some technical conditions ε(X) is a martingale. In this case, E(εt (X)) = E(ε0 (X)) = 1. A sufficient and practical condition for this equality to hold true is the following criterion. Novikov’s Criterion Assume that X is an Itˆo process dXt = Ht dWt with H fulfilling    Z T 1 2 Hs ds <∞ E exp 2 0 for some positive T . Then E (εT (X)) = 1. Under the above hypotheses, we can define (as in the discrete setting) a probability measure Q by setting Q(A) = E (εT (X) 1A ) . (7.5)

ˆ CALCULUS CHAPTER 7. ITO

62

Theorem 7.2 - Girsanov formula - Assume that X fulfills the hypotheses of Novikov’s R ˜ t := Wt − t Hs ds, criterion (for any positive t) and define Q by (7.5). Then, the process W 0 t ≤ T , is a Brownian motion under Q. As a conseqence, for any measurable function g : R[0,T ] → R, we have the formula    Z T Z T    1 2 ˜ t; t ≤ T Hs ds , E g W = E g(Wt ; t ≤ T ) exp − Hs dWs − 2 0 0 where E represents the expectation under the original probability P. Exercise: under the hypotheses and notations of Girsanov formula, we set dYt = Kt Ht dt + Kt dWt , where K is a left-continuous adapted process. ˜ t. 1. Check that dYt = Kt dW RT 2. Assume furthermore that 0 E(Ks2 ) ds < ∞. Derives from Girsanov formula that (Yt )t≤T is martingale under Q.

7.5 7.5.1

Exercises Scaling functions

We consider a diffusion dXt = σ(Xt ) dWt + b(Xt ) dt, with b and σ continuous. We assume that b is bounded and that there exists  and M such that 0 <  ≤ σ(x) ≤ M < ∞, for any x ∈ R. We set also Ft = σ(Ws , s ≤ t). 1. Find a function s ∈ C 2 such that (s(Xt ))t≥0 is a martingale. This function is so-called ”scaling function”. 2. For any a < X0 < b, we set Ta = inf{t ≥ 0 : Xt = a} and Tb = inf{t ≥ 0 : Xt = b}. Admitting that the equality E(s(XTa ∧Tb )) = s(X0 ) holds true, show that P (Tb < Ta ) =

s(X0 ) − s(a) . s(b) − s(a)

3. Assume that limx→∞ s(x) < ∞. Show that in this case limb→∞ P (Tb < Ta ) > 0. Derive that with positive probability the diffusion X will never reach a.

ˆ CALCULUS CHAPTER 7. ITO

7.5.2

63

Cameron-Martin formula

We set Xt = µt + Wt , where Wt is a Brownian motion under P. Fix T > 0 and define     2T 1A . Q(A) = E exp −µWT − µ 2 1. Show that (Xt , t ≤ T ) is a Brownian motion under Q. 2. Check that for any measurable function g : R[0,T ] → R, we have E (g(Xt , t ≤ T )) = e−µ 3. Replace µt by µt = under Q.

Rt 0

2 T /2

 E eµWT g(Wt , t ≤ T ) .

ms ds. Find Q such that (Xt , t ≤ T ) is a Brownian motion

Chapter 8 Black-Scholes model The goal: to define the Black-Scholes model, and price a european option in this setting.

8.1 8.1.1

Setting The Black-Scholes model

As in the discrete setting, we focus on a market made of two assets: • a non risky asset B (bond), • a risky asset S (stock). The two assets B and S are assumed to evolves according to  dBt = rBt dt dSt = St (µ dt + σ dWt ). The parameter r corresponds to the interest rate of the bond B, whereas µ and σ corresponds to the trend and volatility of the asset S. We can give a closed form for the value of Bt and St . The evolution of B is driven by an ordinary differential equation, solved by Bt = B0 ert . The evolution of S is driven by the stochastic differential equation of a geometric Brownian motion (see Chapter 7, Section 3, Example 2). Therefore,     σ2 St = S0 exp σWt + µ − t . 2 Throughout this chapter, Ft refers to σ(Wu , u ≤ t). Note that according to the previous formula, we also have Ft = σ(Su , u ≤ t). 64

CHAPTER 8. BLACK-SCHOLES MODEL

8.1.2

65

Portfolios

A portfolio Π = (βt , γt )t≥0 made of βt units of B and γt units of S has value XtΠ = βt Bt + γt St . We will assume henceforth that βt and γt are left-continuous, and Ft -adapted. In the discrete setting, a portfolio was so-called ”self-financed” when its fluctuation between two consecutive times was given by ∆XnΠ = βn ∆Bn + γn ∆Sn . In continuous time, this condition turns to: Definition 8.1 A portfolio Π is self-financed when XtΠ solve dXtΠ = βt dBt + γt dSt . ˜ Π := e−rt X Π . We call discounted value (or present value) of a portfolio Π, the value X t t Next lemma gives a characterization of self-financed portfolios (compare with Lemma 5.1). ˜ tΠ fulfills Lemma 8.1 A portfolio Π is self-financed if and only if its discounted value X ˜ Π = γt dS˜t , with S˜t := e−rt St . dX t ˜ tΠ = −re−rt XtΠ dt + e−rt dXtΠ . Therefore, a portfolio Proof : Itˆo’s formula ensures that dX is self-financed if and only if  ˜ Π = −r βt e−rt Bt + γt e−rt St dt + e−rt (βt dBt + γt dSt ) dX t = βt e−rt (−rBt dt + dBt ) +γt (−re−rt St dt + e−rt dSt ) . | {z } | {z } =dS˜t

=0

2

The lemma follows.

8.1.3

Risk neutral probability

As in the discrete setting a risk neutral probability P∗ is a probability equivalent to P (i.e. P∗ (A) = 0 iif P(A) = 0) such that the discounted value of the stock (S˜t )t≤T is a martingale under P∗ . Let us focus on the evolution of S˜t :       σ2 σ2 ∗ ˜ St = exp σWt + µ − r − t = exp σWt − t , (8.1) 2 2 where we have set Wt∗ = Wt +

µ−r t. σ

CHAPTER 8. BLACK-SCHOLES MODEL

66

Now, according to the Girsanov-Cameron-Martin formula (see Exercise 7.5.2), the process (Wt∗ )t≤T is a Brownian motion under the probability P∗ defined by  2 ! µ−r µ−r T P∗ (A) := E (ZT 1A ) , with ZT := exp − WT − . σ σ 2   Since the discounted value of the stock S˜t

t≤T

is a stochastic exponential under P∗ , it is

a martingale under P∗ (as already checked in exercise 6.4.1.4). Conclusion: the probability P∗ is a risk neutral probability. Comment: Compare the probability P∗ with the risk neutral probability computed at the end of Section 3.4.

8.2

Price of a european option in the Black-Scholes model

We compute in this section, the price of a european option in the Black-Scholes model. To be more specific, we will focus on options with maturity T and payoff of the form f (ω) = g(ST (ω)). Definition 8.2 A portfolio Π is so-called an hedging portfolio when XTΠ ≥ g(ST ) and    Z T 2 ∗ ˜ tΠ E X dt < +∞ . (8.2) 0

The first condition is the same as in the discrete setting, while the second condition is technical. The price C of an option will again correspond to the minimal initial value X0Π that can have an hedging self-financed portfolio, namely   − Π is self-financed Π C := inf X0 such that . − Π is hedging Again, we will solve this minimization problem, with martingale methods. First, we will bound C from below by using that the discounted value of a self-financed hedging portfolio is a martingale under P∗ . Second, we will construct a self-financed hedging portfolio Π∗ , whose initial value fits with the lower bound found at the first step. In the next result, we assume that g is piecewise C 1 and fulfills E (g(ST )2 ) < ∞.

CHAPTER 8. BLACK-SCHOLES MODEL

67

Theorem 8.1 Price of a european option.

1. The price of a european option with payoff g(ST ) at maturity T is  C = E∗ e−rT g(ST ) = G(0, S0 ) where the function G is defined by  Z +∞   (r−σ 2 /2)(T −t)+σy −r(T −t) g xe exp G(t, x) := e −∞

−y 2 2(T − t)



dy p

2π(T − t)

.

2. There exists a self-financed hedging portfolio Π∗ with initial value C. The value at time t of the portfolio Π∗ is ∗

XtΠ = e−r(T −t) E∗ (g(ST ) | Ft ) = G(t, St ). Its composition is given by γt∗ =

∂G (t, St ) and ∂x

βt∗ =

G(t, St ) − γt∗ St . Bt

Proof : Let us first consider an arbitrary self-financed hedging portfolio Π. According to (8.1), we have dS˜t = σ S˜t dWt∗ . Combining with Lemma 8.1, thus leads to Z t Π Π ˜ ˜ Xt = X0 + σ γu S˜u dWu∗ , 0

˜ Π of Π is therefore a stochastic with W ∗ Brownian motion under P∗ . The discounted price X  t RT ∗ Π 2 ˜ ) dt < ∞, it is a martingale under integral, and since we have assumed that 0 E (X t ∗ P . In particular, we have    ˜ 0Π = E∗ X ˜ TΠ ≥ E∗ e−rT g(ST ) , X where the last inequality comes from the condition XTΠ ≥ g(ST ). It follows that C ≥ E∗ e−rT g(ST ) . Conversely, we will show that there exists a self-financed hedging portfolio Π∗ with initial   ∗ −rT ∗ −rT value E e g(ST ) . Set Mt = E e g(ST ) | Ft . Lemma 8.2 When M is defined by the above formula, its value is given by ˜ S˜t ), Mt = e−rt G(t, St ) = G(t, ˜ x) := e−rt G(t, xert ). with G(t,

CHAPTER 8. BLACK-SCHOLES MODEL

68

Proof : (of the lemma). The second equality is straightforward. Let us prove the first one. Due to the very definition of conditional expectation, all we need is to check that: 1. the random variable e−rt G(t, St ) is Ft -measurable (which is obvious!) 2. the equality E∗ (e−rT g(ST )h(St )) = E∗ (e−rt G(t, St )h(St )) holds for any measurable h. For the second point, note that  g(ST ) = g St exp (r − σ 2 /2)(T − t) + σ(WT∗ − Wt∗ ) with WT∗ − Wt∗ independent of St and N (0, T − t) distributed under P∗ . Therefore Z Z +∞  −y2 /2(T −t) dy   −rT (r−σ 2 /2)(T −t)+σy e −rT ∗ ∗ p e h(x) g x e P (St ∈ dx) E e g(ST )h(St ) = 2π(T − t) −∞  = E∗ e−rt G(t, St )h(St ) 2

This conclude the proof of the lemma. ˜ S˜t ), leads to Applying Itˆo’s formula to G(t, Z t ˜ Z t ∂G ˜ ˜ Mt = M0 + (u, Su ) dSu + 0 ∂x 0 |

˜ ˜ σ 2 x2 ∂ 2 G ∂G + ∂t 2 ∂x2 {z

! (u, S˜u ) du ,

=0

}

where the second integral turns to be 0, since ˜ σ 2 x2 ∂ 2 G ˜ ∂G + = 0. (check it!) (8.3) ∂t 2 ∂x2 We now define Π∗ by setting ˜ ∂G ∂G γt∗ = (t, S˜t ) = (t, St ) ∂x ∂x ∗ and βt∗ = (G(t, St ) − γt∗ St )/Bt . Then, the value of Π∗ is XtΠ = G(t, St ) and its discounted value is Z t Π∗ Π∗ ˜ ˜ ˜ ˜ Xt = G(t, St ) = X0 + γu∗ dS˜u . (Itˆo’s formula) 0

Therefore, according to Lemma 8.1, Π∗ is self-financed. Besides, Jensen inequality enforces   2  2   ∗ Π∗ ∗ ∗ −rT ˜ E Xt = E E e g(ST ) | Ft ≤ e−2rT E∗ g(ST )2 ∗

so that condition (8.2) holds. To conclude: since XTΠ = g(ST ), we have  constructed a self-financed hedging portfolio with initial value G(0, S0 ) = E∗ e−rT g(ST ) . 2 Comment: we derive from (8.3), that G is solution of  ∂G σ2 x2 ∂ 2 G + 2 ∂x2 + rx ∂G = rG in [0, T ] × R+ ∂t ∂x G(T, x) = g(x) for any x > 0. Exercise: compute the price of a call with maturity T and strike K.

Chapter 9 Appendix 9.1

Convergence of random variables

In the following, (Xn )n≥0 stands for a sequence of real random variables.

9.1.1

Convergence a.s.

(Xn ) is said to converge a.s. to X, when there exists a set Ω0 such that P(Ω0 ) = 1 and n→∞ Xn (ω) −→ X(ω) for any ω ∈ Ω0 .

9.1.2

Convergence in L2 n→∞

Xn is said to converge in L2 to X, when E (|Xn − X|2 ) −→ 0.

9.1.3

Convergence in probability

Xn is said to converge in probability to X, when for any  > 0 n→∞

P (|Xn − X| > ) −→ 0.

9.1.4

Convergence in distribution

Xn is said to converge in distribution to X, when for any bounded and continuous function f : R → R we have n→∞ E(f (Xn )) −→ E(f (X)).

9.1.5

Relationships

Convergence a.s. =⇒ Convergence in probability =⇒ Convergence in distribution ⇑ Convergence L2 69

CHAPTER 9. APPENDIX

9.2

70

Construction of Itˆ o’s integral

We sketch in this section the main lines of the construction of the Itˆo’s integral. For comprehensive proofs, we refer to [2].

9.2.1

Setting

Henceforth, W is a Brownian motion and Ft = σ(Ws , s ≤ t). Fix T ∈ [0, ∞] and write MT for the space of Ft -martingales which are continuous and start from 0 (i.e. M0 = 0) with probability one and fulfill E (MT2 ) < ∞. Proposition 9.1 MT endowed with the scalar product (M |N )M = E (NT MT ) is an Hilbert space. Proof : Checking that (M |N )M is a scalar product is straightforward. Proving the completeness needs both Doob’s inequality and Borel-Cantelli’s lemma, see [2]. 2 We will also introduce the linear space spanned by the left-continuous Ft -adapted processes (Ht )t≥0 , fulfilling Z T  2 Hs ds < ∞. E 0

L2T

for We  the closure of this space endowed with the scalar product (H|K)L = Rwrite T E 0 Hs Ks ds .

9.2.2

Integration of elementary processes

Elementary processes are left-continuous and bounded piecewise constant processes, i.e. processes H of the form n X Ht (ω) = hi (ω) 1]ti−1 ,ti ] (t) i=1

where the hi ’s are bounded Fti−1 -measurable random variables and 0 ≤ t0 < . . . < tn ≤ T . We write henceforth ET for the space of elementary processes endowed with the scalar product (·|·)L . For t ≤ T and a process H of the previous form, we define the stochastic integral as follows Z

t

 Hs dWs (ω) :=

0

n X

 hi (ω) Wt∧ti (ω) − Wt∧ti−1 (ω) ,

i=1

with s ∧ t := min(s, t). Proposition 9.2 The map (Ht )t≤T 7→

R

t 0

Hs dWs

 t≤T

is an isometry from ET to MT .

CHAPTER 9. APPENDIX

71

 Pn Proof : First, note that the process h W − W is continuous and i t∧t t∧t i i−1 i=1 t≤T takes value 0 at t = 0. Second, it is a martingale. Indeed, fix 0 ≤ tk ∧ t ≤ s < tk+1 ∧ t. Using the martingale property of Brownian motions (see exercise 6.4.1.2), we have ! n X  E hi Wt∧ti − Wt∧ti−1 | Fs i=1

=

k X

    E hi Wt∧ti − Wt∧ti−1 | Fs + E hk+1 Wt∧tk+1 − Wt∧tk | Fs

i=1

+

n X

  E hi Wt∧ti − Wt∧ti−1 | Fs

i=k+2

=

k X

   hi Wt∧ti − Wt∧ti−1 + hk+1 E Wt∧tk+1 | Fs − Wt∧tk

i=1

+

n X

   E E hi Wt∧ti − Wt∧ti−1 | Fti−1 ∧t | Fs

i=k+2

=

k X

 hi Wt∧ti − Wt∧ti−1 + hk+1 (Ws − Wt∧tk ) + 0

i=1

=

n X

 hi Ws∧ti − Ws∧ti−1 .

i=1 t 0



=

XX

R

belongs to MT , it remains to prove that the We thus have checked that Hs dWs t≤T R· map H 7→ 0 Hs dWs is an isometry. R· Let us compute the norm of 0 Hs dWs : Z · 2 Z T 2 ! Hs dWs = E Hs dWs 0

0

M

i

 E hi hj (Wti − Wti−1 )(Wtj − Wtj−1 ) .

j

For i < j, the variable hi hj (Wti − Wti−1 ) is Ftj−1 -measurable. Therefore, according to the third property of Brownian motion hi hj (Wti − Wti−1 ) is independent of Wtj − Wtj−1 . It follows that    E hi hj (Wti − Wti−1 )(Wtj − Wtj−1 ) = E hi hj (Wti − Wti−1 ) E Wtj − Wtj−1 = 0. | {z } =0

When i = j: since h2i is Fti−1 -measurable, it is independent of (Wti − Wti−1 )2 , so that    E h2i (Wti − Wti−1 )2 = E h2i (Wti − Wti−1 )2  = E h2i × (ti − ti−1 ).

CHAPTER 9. APPENDIX

72

Putting pieces together thus gives Z · 2 Z X  2 Hs dWs = E hi × (ti − ti−1 ) = 0

M

9.2.3

E(Hs2 )ds = ||H||2L .

0

i

We have check that the map H 7→

T

R· 0

2

Hs dWs is an isometry.

Extension to L2T

Thanks to the isometry property we can extend the previous integral to the closure of ET in L2T , which turns to be L2T itself. Proposition 9.3 The map ET → MT Z · H 7→ Hs dWs 0

can be extend in a unique way as an isometry from L2T to MT . RT Extending this integral to processes H fulfilling only 0 Hs2 ds < ∞, needs somewhat more technical arguments. We refer the interested reader to [2] for this extension as well as for the proof of the density of ET in L2T . Corollary 9.1 When H ∈ L2T and is left-continuous, the following convergence holds: n X



P

Z

H(i−1)t/n Wit/n − W(i−1)t/n −→

t

Hs dWs ,

for any t ≤ T.

0

i=1 (n)

Proof : Set Hs

=

Pn

i=1

Z 0

H(i−1)t/n 1](i−1)t/n,it/n] (s). On the one hand

t

Hs(n)

dWs =

n X

 H(i−1)t/n Wit/n − W(i−1)t/n .

i=1

R t (n) On the other hand H (n) converges to H in L2t : according to the isometry property 0 Hs dWs Rt converges to 0 Hs dWs in Mt and therefore in probability. 2 R· Remark: the Itˆo integral 0 Hs dWs constructed above belongs to the space MT and is therefore continuous with probability one. Since we can modify a Rprocess on a set of · probability 0 without changing its law, we can choose a version of 0 Hs dWs which is continuous everywhere.

Bibliography [1] Baxter M., Rennie A. Financial calculus: an introduction to derivative pricing. Cambridge University Press 1999, Cambridge. [2] Karatzas I., Shreve S.E. Brownian Motion and Stochastic Calculus. Springer-Verlag, New-York, 1988. [3] Mel’nikov A.V. Financial Markets: stochastic analysis and the pricing of derivative securities. Translations of Mathematical Monographs 1999, AMS edition, Providence. [4] Pliska S. Introduction to mathematical finance. Blackwell Publishers, Oxford UK. [5] Shiryaev A.N. Essentials of stochastic finance: facts, models, theory. Advanced Series on Statistical Science & Applied Probability 1999, World Scientific, Singapore. [6] Williams D. Probability with martingales. Cambridge Mathematical Textbooks. Cambridge University Press, Cambridge, 1991.

73

Index ∆Xn , 30 εn (U ), 31

market B-S, 30 complete, 35 incomplete, 41 martingale, 23, 55 measurable, 14 model Black-Scholes, 43 Merton, 43

arbitrage, 34 ask, 41 bankruptcy problem, 27 bid, 41 bond, 30 Brownian motion definition, 54 properties, 55

non-decreasing predictable process, 49 normal law, 54 Novikov’s criterion, 61

conditional expectation continuous setting, 13 discrete setting, 11 general setting, 14 Cox-Ross-Rubinstein model, 42

optional stopping theorem, 25 portfolio hedging, 39, 66 self-financed, 31, 65 predictable representation, 50 present value, 34, 65 price of an option american calculus, 46 definition, 45 european calculus, 40, 67 definition, 39 process adapted, 53 continuous, 53 elementary, 70 Itˆo, 59 left-continuous, 53

discounted value, 34, 65 Donsker’s principle, 54 Doob’s decomposition, 49 dynamic programing principle, 47 exercise time, 46 Galton-Watson, 27 gambler in a casino, 19 Gaussian law, 54 Girsanov formula, 36, 62 inequality maximal, 28 of Jensen, 17 Itˆo’s integral, 58 Ito’s formula, 59

risk neutral probability, 32, 65

74

INDEX spread, 41 stochastic exponential, 61 definition, 31 properties, 32 stock, 30 stopping time, 22 strategy hedging, 39 self-financed, 31 submartingale, 23 supermartingale, 23 Wald identity, 29 Wiener process, 54

75

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