Snell envelope and the pricing of American-style contingent claims

Snell envelope and the pricing of -style contingent claims 80-646-08 Stochastic calculus I Geneviève Gauthier HEC Montréal

Notation The The riskless security Market model The following text draws heavily on Isabelle Cormier s (UQAM) master s thesis.

Notation I Notation The Let s rst establish the notation that will be used in this text: the triple (Ω, F, P) represents a nite probability space (Card (Ω) < ) on which the ltration F = ff t : t = 0, 1,..., T g is built.

Notation The Notation II We will use the market model introduced in the article written in 1981 by J. Michael Harrison and Stanley R. Pliska: so, there are on the market K + 1 nancial securities, whose prices at time t are given by a random vector! > S t = St 0, St 1,..., St K each component of which is a random variable taking a nite number of strictly positive n values. Thus, such a!s o stochastic process as S = t : t = 0, 1,..., T models the evolution of the security prices on the market.

Notation The Notation III The rst security has a particular status riskless investment (less risky!), i.e. 8ω 2 Ω, 8t = 1,..., T, S 0 t (ω) S 0 t 1 (ω). We may also assume, without loss of generality, that 8ω 2 Ω, S 0 0 (ω) = 1.

Notation The Notation IV Since the market value of one dollar is not the same at di erent times, investment gains and losses cannot be directly compared if they don t occur at the same time. For that reason, we must study the discounted price process β! S where the adapted, strictly positive-valued, stochastic process β = fβ t : t = 0, 1,..., T g, de ned as β t = 1 St 0, represents our discount factor at each time point. From Harrison and Pliska s Theorem 2.7 (1981, p. 228), we know that, if our market model contains no arbitrage opportunities, there exists at least one risk-neutral probability measure Q according to which the discounted price processes for the securities are martingales.

Notation The The I Mathematically speaking, a European-style contingent claim is a non-negative random variable, F T measurable since the said contingent claim can only be exercised at maturity, that is at time T. As a consequence, the holder of such a contingent claim is passive in the sense that he or she has no decision to make during the lifetime of the contingent claim.

Notation The The II An -style contingent claim, by contrast, can be represented as an F adapted stochastic process X = fx t : t = 0, 1,..., T g where X t represents the contingent claim value at time t if it is exercised at that time. The holder of an -style contingent claim must therefore ask himself or herself, at each time period during the lifetime of the aforementioned contingent claim, whether it is better to exercise his or her claim or to wait.

Notation The The III The random time τ : Ω! f0,..., T g, representing the time when the holder of the contingent claim exercises his or her claim, must be such that 8t 2 f0,..., T g, fω 2 Ω jτ (ω) = t g 2 F t since the decision to exercise the claim at time t can only be made based on the information available at that time. It is therefore a stopping time. In what follows, the set of random times representing the time of exercise is de ned as Λ 0 = fτ : Ω! f0,..., T g jτ is a stopping timeg.

Inequalities Proof Pricing an -style contingent claim I Similar to the European-style contingent claim, the price of an -style contingent claim can be viewed from two angles : the contingent claim seller s perspective, and the buyer s perspective.

Inequalities Proof Pricing an -style contingent claim II In what follows, Φ denotes the set of admissible investment strategies (Harrison and Pliska, 1981, p. 226), a strategy! ϕ = n!ϕ t = ϕ 0 t, ϕ1 t,..., ϕk t o : t = 1,...T being a predictable stochastic process indicating, at each time and for each security, the number of shares being held. Reminder. An investment strategy is said to be admissible if it is self- nancing and its market value is never negative. An admissible strategy thus is such that the investor is never in a debt position. That doesn t mean however that short sales are prohibited.

Inequalities Proof Pricing an -style contingent claim III Seller s perspective. The seller s rst goal is to ensure that, if he adequately invests the amount x obtained from selling the contingent claim X, then, at time τ when the buyer exercises his claim, he is able to meet his obligation, i.e. to pay the amount X τ. The minimum price acceptable to the seller of contingent claim X is therefore x sup = inf x 0 9! ϕ 2 Φ such that V 0! ϕ = x and Vτ! ϕ Xτ, 8τ 2 Λ 0.

Inequalities Proof Pricing an -style contingent claim IV Buyer s perspective. If the buyer borrows an amount x at time t = 0 in order to buy the contingent claim X then, at time τ when he exercises his claim, he wants to be able to pay back his debt, i.e. there must exist an investment strategy φ!φ!φ such that V 0 = x and V τ + X τ 0. Thus, the maximum price that the buyer of the contingent claim X is ready to pay out is 8 < x inf = sup : x 0 9!!φ 9 φ 2 Φ such that V 0 = x and =!φ V τ + X τ 0 for some τ 2 Λ 0 ;

Inequalities Proof Inequalities Pricing an -style contingent claim Theorem If the market model contains no arbitrage opportunities, then, for any martingale measure Q, X 0 x inf sup τ2λ 0 E Q [β τ X τ ] x sup. Remark. In solving, we obtain a stopping time τ satisfying E Q [β τ X τ ] = sup τ2λ0 E Q [β τ X τ ], which allows us to determine the latter quantity without the need to know every stopping time in the set Λ 0

Inequalities Proof Proof - First inequality Pricing an -style contingent claim The rst inequality comes from the fact that X 0 belongs to the set 8 < : x 0 9!!φ 9 φ 2 Φ such that V 0 = x =!φ and V τ + X τ 0, for some τ 2 Λ 0 ;. Indeed, let s choose the strategy! φ that consists, throughout time interval [0, T ], of holding portfolio ( X 0, 0,..., 0) (a debt of X 0 shares of the riskless security) and let s set τ = 0. Then V 0!φ = ( X 0, 0,..., 0)! S 0 = X 0 and V τ!φ + X τ = V 0!φ + X 0 = 0.

Inequalities Proof Proof - Third inequality I Pricing an -style contingent claim Let s choose arbitrarily x 0 2 x 0 9! ϕ 2 Φ such that V! 0 ϕ = x and Vτ! ϕ Xτ, 8τ 2 Λ 0. That choice is such that there exists a strategy! ϕ 2 Φ such that V 0! ϕ = x0 and V τ! ϕ Xτ for any stopping time τ 2 Λ 0. Thus, based on the latter inequality, we state that 8τ 2 Λ 0, E Q [β τ X τ ] E Q β τ V τ! ϕ. (1)

Inequalities Proof Proof - Third inequality II Pricing an -style contingent claim Since the market contains no arbitrage opportunities, the stochastic process βv! ϕ is a (Q, F) martingale (Harrison and Pliska, 1981, Proposal 2.8, p. 230). The Optional Stopping Theorem (Revuz and Yor, Theorem 3.2, p. 65) implies that, for any bounded stopping time τ, E Q β τ V τ! ϕ = E Q β 0 V 0! ϕ which leads to the following: 8τ 2 Λ 0, E Q β τ V τ! ϕ = E Q β 0 V 0! ϕ = β 0 x 0 = x 0.

Inequalities Proof Proof - Third inequality III Pricing an -style contingent claim By substitution into inequality (1) E Q [β τ X τ ] E Q β τ V! τ ϕ, we obtain hence 8τ 2 Λ 0, E Q [β τ X τ ] x 0 sup E Q [β τ X τ ] x 0. τ2λ 0 Since the choice of x 0 was arbitrary, then sup E Q [β τ X τ ] τ2λ 0 inf x 0 0 9! ϕ 2 Φ such that V! 0 ϕ = x0 and V! τ ϕ Xτ, 8τ 2 Λ 0 = x sup.

Inequalities Proof Proof - Second inequality I Pricing an -style contingent claim Let s choose arbitrarily 8 < x 0 2 : x 0 9!!φ φ 2 Φ such that V 0 = x!φ and V τ + X τ 0 for some τ 2 Λ 0 thus establishing the existence of an admissible strategy! φ 2 Φ satisfying V0!φ = x 0 and V τx0!φ + X τx0 0, for some τ x0 2 Λ 0. Thus, given that β τx0 > 0, 0 E Q h β τx0 V τx0!φ + X τx0 i = E Q h β τx0 V τx0!φ i + E Q h β τx0 X τx0 i. 9 = ;

Inequalities Proof Proof - Second inequality II Pricing an -style contingent claim Besides, since βv! ϕ is a (Q, F) martingale and τ x0 is a bounded stopping time, we can use the Optional Stopping Theorem and obtain E Q h β τx0 V τx0!φ i = E Q h β 0 V 0!φ i = β 0 x 0 = x 0.

Inequalities Proof As a consequence, Proof - Second inequality III Pricing an -style contingent claim 0 E Q h β τx0 V τx0!φ i + E Q h β τx0 X τx0 i = x 0 + E Q h β τx0 X τx0 i x 0 + sup τ2λ 0 E Q [β τ X τ ] hence sup E Q [β τ X τ ] x 0. τ2λ 0

Inequalities Proof Proof - Second inequality IV Pricing an -style contingent claim Since the choice of x 0 was arbitrary, sup E Q [β τ X τ ] τ2λ 0 8 >< 9!!φ φ 2 Φ such that V 0 = x 0 and sup x 0 0!φ V τ + X τ 0, >: for some τ 2 Λ 0 = x inf. 9 >= >;

Inequalities Proof I Pricing an -style contingent claim Let s consider the following three-period binomial model ω S 0 0 (ω) S0 1 (ω) ω 1 1 80 S 0 1 (ω) S1 1 (ω) 1.115 100 S 0 2 (ω) S2 1 (ω) 1.115 2 125 S 0 3 (ω) S3 1 (ω) 1.115 3 156, 25 Q (ω) 0.343 ω 2 1 80 1.115 100 1.115 2 125 1.115 3 100 0.147 ω 3 1 80 1.115 100 1.115 2 80 1.115 3 100 0.147 ω 4 1 80 1.115 100 1.115 2 80 1.115 3 64 0.063

Inequalities Proof ω S 0 0 (ω) S0 1 (ω) ω 5 1 80 II Pricing an -style contingent claim S 0 1 (ω) S1 1 (ω) 1.115 64 S 0 2 (ω) S2 1 (ω) 1.115 2 80 S 0 3 (ω) S3 1 (ω) 1.115 3 100 Q (ω) 0.147 ω 6 1 80 1.115 64 1.115 2 80 1.115 3 64 0.063 ω 7 1 80 1.115 64 1.115 2 51.20 1.115 3 64 0.063 ω 8 1 80 1.115 64 1.115 2 51.20 1.115 3 40.96 0.027

III Pricing an -style contingent claim Inequalities Proof So, the ltration is F 0 = f?, Ωg F 1 = σ ffω 1, ω 2, ω 3, ω 4 g, fω 5, ω 6, ω 7, ω 8 gg F 2 = σ ffω 1, ω 2 g, fω 3, ω 4 g, fω 5, ω 6 g, fω 7, ω 8 gg F 3 = all events in Ω

Inequalities Proof IV Pricing an -style contingent claim Exercise. It follows that Card (Λ 0 ) = 26. (Check by yourself... if you have time!) Partial answer. If we represent the stopping time τ as a point in R Card (Ω), (τ (ω 1 ), τ (ω 2 ), τ (ω 3 ), τ (ω 4 ), τ (ω 5 ), τ (ω 6 ), τ (ω 7 ), τ (ω 8 )), then the 26 stopping times in Λ 0 are (0, 0, 0, 0, 0, 0, 0, 0) (1, 1, 1, 1, 1, 1, 1, 1) (2, 2, 2, 2, 2, 2, 2, 2) (3, 3, 3, 3, 3, 3, 3, 3) (1, 1, 1, 1, 2, 2, 2, 2) (1, 1, 1, 1, 2, 2, 3, 3) (1, 1, 1, 1, 3, 3, 2, 2) (1, 1, 1, 1, 3, 3, 3, 3) (2, 2, 2, 2, 1, 1, 1, 1) (2, 2, 3, 3, 1, 1, 1, 1) (3, 3, 2, 2, 1, 1, 1, 1) (3, 3, 3, 3, 1, 1, 1, 1) (2, 2, 2, 2, 2, 2, 3, 3) (2, 2, 2, 2, 3, 3, 2, 2) (2, 2, 3, 3, 2, 2, 2, 2) (3, 3, 2, 2, 2, 2, 2, 2) (2, 2, 2, 2, 3, 3, 3, 3) (2, 2, 3, 3, 2, 2, 3, 3) (2, 2, 3, 3, 3, 3, 2, 2) (3, 3, 2, 2, 2, 2, 3, 3) (3, 3, 2, 2, 3, 3, 2, 2) (3, 3, 3, 3, 2, 2, 2, 2) (3, 3, 3, 3, 3, 3, 2, 2) (3, 3, 3, 3, 2, 2, 3, 3) (3, 3, 2, 2, 3, 3, 3, 3) (2, 2, 3, 3, 3, 3, 3, 3)

V Pricing an -style contingent claim Inequalities Proof Exercise. Check that the probability measure Q given in the table is the only measure that makes the discounted price of the risky security a martingale. Exercise. Justify why Q is also called the risk-neutral measure.

Inequalities Proof VI Pricing an -style contingent claim Let s consider, as an -style contingent claim, a put option with an 80-dollar strike price. The value of such a put at time t, if the put is exercised, is X t = max 80 S 1 t, 0 and its discounted value at time t, if again the put is exercised, is Y t = β t X t = 1.115 t max 80 S 1 t, 0

Inequalities Proof VII Pricing an -style contingent claim Y t = β t X t = 1.115 t max 80 S 1 t, 0 ω Y 0 Y 1 Y 2 Y 3 ω 1 0 0 0 0 ω 2 0 0 0 0 ω 3 0 0 0 0 16 ω 4 0 0 0 1.115 = 11.5424 3 16 ω 5 0 1.115 = 14.3498 0 0 16 ω 6 0 16 1.115 = 14.3498 0 1.115 = 11.5424 3 16 ω 7 0 28.80 1.115 = 14.3498 16 1.115 = 23.1656 2 1.115 = 11.5424 3 16 ω 8 0 28.80 1.115 = 14.3498 39.04 1.115 = 23.1656 2 1.115 = 28.1634 3

Inequalities Proof VIII Pricing an -style contingent claim Note that one of the random times representing, for each ω, a time when the option value is the greatest is (τ (ω 1 ), τ (ω 2 ), τ (ω 3 ), τ (ω 4 ), τ (ω 5 ), τ (ω 6 ), τ (ω 7 ), τ (ω 8 )) = (3, 3, 3, 3, 1, 1, 2, 3). But such a random time is not a stopping time, since fω 2 Ω : τ (ω) = 2g = fω 7 g /2 F 2 i.e., in order to be able to maximize our gains, we would need, at the time of deciding whether to exercise, to know the future. By contrast, it will be possible for us, using the stopping times, to maximize our expected gain conditionally on the information available at decision times (whether to exercise the contingent claim or not). This idea will be made more precise later on.

General Formulation of I Let Y = fy t : t = 0, 1,..., T g be a stochastic process, F adapted. For all t 2 f0, 1,..., T g, we can de ne the set of stopping times taking their values in the set ft,..., T g : Λ t = fτ : Ω! ft,..., T g jτ is a stopping timeg. Note that Λ T Λ T 1... Λ 0. is as follows: can we determine a stopping time τ 2 Λ 0 satisfying E [Y τ ] = sup τ2λ 0 E [Y τ ] (2) In other words, we are looking to determine, for each of the ω 2 Ω, the time τ (ω) when we should stop the stochastic process Y in order to maximize the expected value of the random variable Y τ.

General I The ltration contains no information Let s assume that 8t 2 f0, 1,..., T g, F t = f?, Ωg. Under such conditions, since Y t is F t measurable, then Y t is constant, i.e. there exists a real number y t for which 8ω 2 Ω, Y t (ω) = y t.

General II The ltration contains no information Moreover, that special ltration is such that Λ 0 = fτ 0, τ 1,..., τ T g (3) where 8ω 2 Ω, τ t (ω) = t. Indeed, if τ is any stopping time (with respect to the ltration F) then 8t 2 f0, 1,..., T g, fω 2 Ω : τ (ω) = tg 2 F t = f?, Ωg. Thus, if there exists a k 2 f0, 1,..., T g for which τ (ω) = k, then Ω fω 2 Ω : τ (ω) = tg = if t = k? otherwise, which means that any stopping time existing in such a ltered probability space is constant. If we restrict ourselves to the stopping times taking their values in the set f0, 1,..., T g then there are T + 1 possible values for k, hence equation (3).

III The ltration contains no information General If we rewrite equation (2) E [Y τ ] = sup τ2λ0 E [Y τ ] keeping in mind this special only, we observe that resolving, in this context, amounts to determine t such that y t = max y t. (4) t2f0,1,...,t g It s obviously a trivial, but it is not useless to explain the steps of the solution, because there will be similar steps, less obvious, in solving the general case.

General IV The ltration contains no information The basic idea is to introduce an auxiliary sequence fz t : t = 0, 1,..., T g de ned as z t = max y u. u2ft,...,t g Note that this sequence is decreasing (i.e. 8t 2 f1,..., T g, z t z t 1 ) and z t = max fy t, z t+1 g. (5) Let s set t = min ft 2 f0, 1,..., T g j z t = y t g (6) and let s show that t satis es equation (4).

General V The ltration contains no information From the de nition of t, we observe that z 0 = z 1 =... = z t z t +1... z T. (7) Indeed, if there existed t 2 f0, 1,..., t 1g such that z t is striclty greater than z t+1, then, from equation (5), z t = max fy t, z t+1 g > z t+1. It would therefore follow that z t = y t, which would contradict the de nition of t. As a consequence, z t z t+1. But, since the sequence of z t is decreasing, then z t = z t+1. Besides, expression (7) shows that t truly is the index we are after, since y t = z t = z 0 = max y t. t2f0,1,...,t g

General General I Now, let s see how to adapt the ideas from the previous section to any ltration. The key to the solution lies in building a proper version of the decreasing sequence satisfying equation (5) z t = max fy t, z t+1 g. Let s set 8 < Y T if t = T Z t = : max fy t, E [Z t+1 jf t ]g if t 2 f0,..., T 1g.

General General II The stochastic process Z = fz t : t = 0, 1,..., T g is adapted to the ltration F. Indeed, E [Z t+1 jf t ] and Y t being F t measurable, max fy t, E [Z t+1 jf t ]g is also measurable. It is to be noted that, thus de ned, the sequence fz t : t = 0, 1,..., T g is not necessarily decreasing ω by ω, but it is decreasing in terms of conditional expectations.

I General ω Y 0 Y 1 Y 2 Y 3 Q ω 1 0 0 0 0 0.343 ω 2 0 0 0 0 0.147 ω 3 0 0 0 0 0.147 ω 4 0 0 0 = 11.5424 0.063 ω 5 0 = 14.3498 0 0 0.147 ω 6 0 = 14.3498 0 = 11.5424 0.063 ω 7 0 = 14.3498 = 23.1656 = 11.5424 0.063 ω 8 0 = 14.3498 = 23.1656 = 28.1634 0.027

II General ω Z 0 Z 1 Z 2 Z 3 ω 1 = 5. 0321 = 1.0388 0 0 ω 2 = 5. 0321 = 1.0388 0 0 ω 3 = 5. 0321 = 1.0388 = 3.4627 0 ω 4 = 5. 0321 = 1.0388 = 3.4627 = 11.5424 ω 5 = 5. 0321 = 14.3498 = 3.4627 0 ω 6 = 5. 0321 = 14.3498 = 3.4627 = 11.5424 ω 7 = 5. 0321 = 14.3498 = 23.1656 = 11.5424 ω 8 = 5. 0321 = 14.3498 = 23.1656 = 28.1634

General Calculation of the preceding table entries : 8ω 2 fω 1, ω 2 g, E [Z 3 jf 2 ] (ω) = 0 III Z 2 (ω) = max fy 2 (ω), E [Z 3 jf 2 ] (ω)g = max f0; 0g = 0 8ω 2 fω 3, ω 4 g, E [Z 3 jf 2 ] (ω) = 16 0.063 1.115 3 = 3.4627 0.21 Z 2 (ω) = max fy 2 (ω), E [Z 3 jf 2 ] (ω)g = max f0; 3.4627g = 3.4627

General IV 8ω 2 fω 5, ω 6 g, E [Z 3 jf 2 ] (ω) = 16 0.063 1.115 3 = 3.4627 0.21 Z 2 (ω) = max fy 2 (ω), E [Z 3 jf 2 ] (ω)g = max f0; 3.4627g = 3.4627 8ω 2 fω 7, ω 8 g, E [Z 3 jf 2 ] (ω) = 16 0.063 1.115 3 0.09 + 39.04 0.027 1.115 3 = 16.5287 0.09 Z 2 (ω) = max fy 2 (ω), E [Z 3 jf 2 ] (ω)g = max f23.1656; 16.5287g = 23.1656

General V 8ω 2 fω 1, ω 2, ω 3, ω 4 g, E [Z 2 jf 1 ] (ω) = 0.21 3.4627 = 1.0388 0, 7 Z 1 (ω) = max fy 1 (ω), E [Z 2 jf 1 ] (ω)g = max f0; 1.0388g = 1.0388 8ω 2 fω 5, ω 6, ω 7, ω 8 g, E [Z 2 jf 1 ] (ω) = 0.21 3.4627 0.3 + 23.16560.09 = 7.9885 0.3 Z 1 (ω) = max fy 1 (ω), E [Z 2 jf 1 ] (ω)g = max f14.3498; 7.9885g = 14.3498 8ω 2 Ω E [Z 1 jf 0 ] (ω) = 1.0388 0.7 + 14.3498 0.3 = 5. 0321

VI General Interpretation. Z t represents, conditionally to the information available at time t, the maximum between the discounted value of the option if it is exercised at that time and its expected discounted value if it is exercised subsequently, at a time judiciously chosen. Z t is thus the discounted value of the option at time t.

I General Theorem. For all t 2 f0,..., T g, i.e. and, more particularly, 8τ 2 Λ t, Z t E [Y τ jf t ], (8) Z t sup τ2λ t E [Y τ jf t ] Z 0 sup τ2λ 0 E [Y τ jf 0 ].

General Proof of. II We will proceed by backward induction on t. When t = T, then Λ T = fτ : Ω! ft g jτ is a stopping timeg, i.e. Λ T contains one stopping time only, the one identically equal to T. Since Z T = Y T = E [Y T jf T ], inequality (8) 8τ 2 Λ T, Z T E [Y τ jf T ] is clearly satis ed. Now, let s assume that inequality (8) 8τ 2 Λ t, Z t E [Y τ jf t ] is veri ed for some t 2 f1,..., T g and let s show that is also veri ed for index t 1.

General III Let s choose arbitrarily a stopping time τ 2 Λ t 1. We will evaluate the expectation on the right side of inequality (8) 8τ 2 Λ t, Z t E [Y τ jf t ] by breaking it down according to the values taken by τ : E [Y τ jf t 1 ] = E Y τ I fτ=t 1g jf t 1 + E Yτ I fτ>t 1g jf t 1 = E Y t 1 I fτ=t 1g jf t 1 + E Yτ_t I fτ>t 1g jf t 1.(9) First, we re going to tackle the second term in the equality stated above.

IV General Given that τ _ t, or the maximum of the two stopping times t and τ, is itself a stopping time, which besides belongs to Λ t since, by its very de nition, it takes its values in the set ft,..., T g, the induction hypothesis then allows us to state that Z t E [Y τ_t jf t ]. (10)

General V On another front, fτ > t 1g 2 F t 1 since fτ > t 1g c 2 F t 1. Indeed, fτ t 1g = t 1 [ fτ = jg {z } j=0 2F j F t 1 {z } 2F t 1 2 F t 1 (11) implies that I fτ>t 1g is F t 1 measurable.

General It follows that E Y τ_t I fτ>t 1g jf t 1 = I fτ>t 1g E [Y τ_t jf t 1 ] VI = I fτ>t 1g E [E [Y τ_t jf t ] jf t 1 ] I fτ>t 1g E [Z t jf t 1 ] I fτ>t 1g max fy t 1, E [Z t jf t 1 ]g {z } =Z t 1 = I fτ>t 1g Z t 1 where the rst inequality comes from expression (10) Z t E [Y τ_t jf t ].

VII General What about the rst term in equation (9) E [Y τ jf t 1 ] = E Y t 1 I fτ=t 1g jf t 1 + E Yτ_t I fτ>t 1g jf t 1? From the de nition of Z t hence 1, we have that Z t 1 = max fy t 1, E [Z t jf t 1 ]g Y t 1, E Y t 1 I fτ=t 1g jf t 1 E Z t 1 I fτ=t 1g jf t 1 = Z t 1 I fτ=t 1g.

General VIII Let s go back to equation (9) i i E [Y τ jf t 1 ] = E hy t 1 I fτ=t 1g jf t 1 + E hy τ_t I fτ>t 1g jf t 1 : E [Y τ jf t 1 ] = E Y t 1 I fτ=t 1g jf t 1 + E Yτ_t I fτ>t 1g jf t 1 Z t 1 I fτ=t 1g + Z t 1 I fτ>t 1g = Z t 1 I fτt 1g = Z t 1, since τ 2 Λ t 1 leads to I fτt 1g = I Ω = 1. The proof is complete.

IX General By de nition, Z t dominates Y t and coincides with it at t = T. This incites us to introduce, for each t 2 f0,..., T g, the random time τ t (ω) = min fs 2 ft,..., T g jz s (ω) = Y s (ω) g. (12) We nd here, with τ 0, the anolog of the t introduced, for the purpose of solving the simpli ed, in equation (6)

General ω Y 0 Y 1 Y 2 Y 3 Z 0 Z 1 Z 2 Z 3 ω 1 0 0 0 0 5.0321 1.0388 0 0 ω 2 0 0 0 0 5.0321 1.0388 0 0 ω 3 0 0 0 0 5.0321 1.0388 3.4627 0 ω 4 0 0 0 11.5424 5.0321 1.0388 3.4627 11.5424 ω 5 0 14.3498 0 0 5.0321 14.3498 3.4627 0 ω 6 0 14.3498 0 11.5424 5.0321 14.3498 3.4627 11.5424 ω 7 0 14.3498 23.1656 11.5424 5.0321 14.3498 23.1656 11.5424 ω 8 0 14.3498 23.1656 28.1634 5.0321 14.3498 23.1656 28.1634 ω τ 0 (ω) τ 1 (ω) τ 2 (ω) τ 3 (ω) ω 1 2 2 2 3 ω 2 2 2 2 3 ω 3 3 3 3 3 ω 4 3 3 3 3 ω 5 1 1 3 3 ω 6 1 1 3 3 ω 7 1 1 2 3 ω 8 1 1 2 3

General I General The next two lemmas show, respectively, that τ t is a stopping time belonging to the set Λ t and that E [Y τt jf t ] = Z t.

I General Theorem. For all t 2 f0,..., T g, the random time τ t is a stopping time, element of Λ t. Proof of. In order to establish that τ t is a stopping time, it is su cient to show that, for all s 2 ft,..., T g, fω 2 Ω : τ t (ω) = sg 2 F s.

General But fω 2 Ω jτ t (ω) = s g II = fω 2 Ω j8u 2 ft,..., s 1g, Z u (ω) > Y u (ω) and Z s (ω) = Y s (ω) g! = = s 1 \ u=t 0 s B \ 1 @ u=t fω 2 Ω jz u (ω) > Y u (ω) g fω 2 Ω jy u (ω) {z } 2F u F s \ fω 2 Ω jz s (ω) = Y s (ω) g 1 C Z u (ω) < 0 ga \ fω 2 Ω jy s (ω) Z s (ω) = 0 g 2 F s. {z } 2F s Now, since τ t is a stopping time taking its values in the set ft,..., T g, it is an element of Λ t, which completes the proof.

I General Theorem. For all t 2 f0,..., T g, E [Y τt jf t ] = Z t. (13) We are going to proceed by backward induction on t.

II General Let s begin with the simple case: t = T. For all ω, τ T (ω) = min fs 2 ft,..., T g jz s (ω) = Y s (ω) g = T. As a consequence, E [Y τt jf T ] = E [Y T jf T ] = Y T = Z T (14) where the last equality comes from the very de nition of Z T.

General III Now, let s assume there exists a t 2 f1,..., T g for which equation (13) is satis ed, and let s show that, in such as case, we can establish the result for t 1. Let s break down the left-side member of equation (13) according to the values taken by τ t 1. E Y τt 1 jf t 1 i = E hy τt 1 I fτt 1 =t 1g jf t 1 i = E hy t 1 I fτt 1 =t 1g jf t 1 + E i + E hy τt 1 I fτt 1 >t 1g jf t 1 hy τt 1 _ti fτt 1 >t 1g jf t 1 i.

General IV Since, on the set fω 2 Ω : τ t 1 (ω) > t 1g, we have the following equality: τ t 1 _ t = min fs 2 ft 1,..., T g jz s (ω) = Y s (ω) g _ t = min fs 2 ft,..., T g jz s (ω) = Y s (ω) g = τ t,

General V then, using the induction hypothesis to justify the equation before the last, we can write E Y τt 1 _ti fτt 1 >t 1g jf t 1 = E Y τt I fτt 1 >t 1g jf t 1 = I fτt 1 >t 1gE [Y τt jf t 1 ] = I fτt 1 >t 1gE [E [Y τt jf t ] jf t 1 ] = I fτt 1 >t 1gE [Z t jf t 1 ] = I fτt 1 >t 1gZ t 1 (see next page)

VI General Justi cation for the last equality: it is explained by the fact that fω 2 Ω : τ t 1 (ω) > t 1g = fω 2 Ω : Z t 1 (ω) > Y t 1 (ω)g implies that, on the set fω 2 Ω : τ t 1 (ω) > t 1g, Z t 1 (ω) = max fy t 1 (ω), E [Z t jf t 1 ] (ω)g = E [Z t jf t 1 ] (ω).

VII General Besides, since fω 2 Ω : τ t 1 (ω) = t 1g = fω 2 Ω : Z t 1 (ω) = Y t 1 (ω)g, then E Y t 1 I fτt 1 =t 1g jf t 1 = Y t 1 I fτt 1 =t 1g = Z t 1 I fτt 1 =t 1g.

General Thus, VIII E [Y τt 1 jf t 1 ] = E Y t 1 I fτt 1 =t 1g jf t 1 +E Y τt 1 _ti fτt 1 >t 1g jf t 1 = Z t 1 I fτt 1 =t 1g + Z t 1 I fτt 1 >t 1g = Z t 1 I fτt 1 t 1g = Z t 1. Which completes the proof.

I General Z t sup τ2λt E [Y τ jf t ] and (E [Y τt jf t ] = Z t ), applied to time t = 0, give E [Y τ0 jf 0 ] = Z 0 sup τ2λ 0 E [Y τ jf 0 ] E [Y τ jf 0 ], 8τ 2 Λ 0. By taking the expectation on both sides, we nd E [Y τ0 ] E [Y τ ], 8τ 2 Λ 0.

II General As a consequence, hence E [Y τ0 ] sup τ2λ 0 E [Y τ ] E [Y τ0 ] E [Y τ0 ] = sup τ2λ 0 E [Y τ ]. The stopping time τ 0 is therefore the stopping time τ we were looking for.

III General Interpretation. The stopping time τ 0, which represents the time chosen to exercise the contingent claim, can be interpreted as follows: if, conditionally on the information available at that time, and according to the risk-neutral measure, our expected gain from exercising at a subsequent time judiciously chosen is no greater than the gain realized by exercising now, we must exercise now.

General I ω Y 0 Y 1 Y 2 Y 3 τ 0 Y τ0 Q ω 1 0 0 0 0 2 0 0.343 ω 2 0 0 0 0 2 0 0.147 ω 3 0 0 0 0 3 0 0.147 ω 4 0 0 0 11.5424 3 11.5424 0.063 ω 5 0 14.3498 0 0 1 14.3498 0.147 ω 6 0 14.3498 0 11.5424 1 14.3498 0.063 ω 7 0 14.3498 23.1656 11.5424 1 14.3498 0.063 ω 8 0 14.3498 23.1656 28.1634 1 14.3498 0.027 E Q [Y τ0 ] = 16 1.115 3 {z } =11.5424 0.063 + 16 1.115 {z } =14.3498 0.3 = 5.0321

General II Note that a stopping time solving is not unique, i.e. there may exist several stopping times, elements of the set Λ 0, say τ and τ such that τ 6= τ and E Q [Y τ ] = sup τ2λ 0 E Q [Y τ ] and E Q [Y τ ] = sup τ2λ 0 E Q [Y τ ]. Thus, in our example, we can verify that (τ (ω 1 ), τ (ω 2 ), τ (ω 3 ), τ (ω 4 ), τ (ω 5 ), τ (ω 6 ), τ (ω 7 ), τ (ω 8 )) = (3, 3, 3, 3, 1, 1, 1, 1) is a stopping time for which E Q [Y τ ] = sup τ2λ 0 E Q [Y τ ] = 5.0321.

I CORMIER, Isabelle (1997). L évaluation des droits conditionnels dans les marchés discrets, Thesis submitted in partial ful llment of the requirements for the Master s degree in Mathematics, Université du Québec à Montréal. HARRISON, J. Michael and PLISKA, Stanley R. (1981). Martingales and Stochastic Integrals in the Theory of Continuous Trading, Stochastic Processes and their Applications, vol. 11, no 3, pages 215-260. NEVEU, Jacques (1970). Martingales à temps discret, Masson, Paris. REVUZ, Daniel and YOR, Marc (1980). Continuous Martingales and Brownian Motion, Springer-Verlag.