Snell envelope and the pricing of American-style contingent claims

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Introduction Tari…cation Snell’s problem Reference

Snell envelope and the pricing of American-style contingent claims

80-646-08 Stochastic calculus I

Geneviève Gauthier

HEC Montréal

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Introduction Notation The problem

Tari…cation Snell’s problem Reference

The riskless security

Market model

The following text draws heavily on Isabelle Cormier’s (UQAM) master’s thesis.

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Notation I

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, ...,Tg^{is built.}

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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 marketK +1 …nancial securities, whose prices at time t are given by a random vector

!_S

t = S_t⁰,S_t¹, ...,S_t^K ^>

each component of which is a random variable taking a

…nite number of strictly positive values. Thus, such a stochastic process as S =ⁿ!_S

t :t=0,1, ...,To

models the evolution of the security prices on the market.

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Notation III

The …rst security has a particular status riskless investment (less risky!), i.e.

8^ω 2Ω,8^t =1, ...,T,S_t⁰(ω) S_t⁰ ₁(ω). We may also assume, without loss of generality, that

8^ω 2Ω,S₀⁰(ω) =1.

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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=f^βt :t =0,1, ...,Tg, de…ned as

β_t = ¹ S_t⁰,

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.

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The problem 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.

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The problem II

An American-style contingent claim, by contrast, can be represented as an F adapted stochastic process X =f^X^t ^:^t =0,1, ...,Tg^where^X^t represents the contingent claim value at time t if it is exercised at that time.

The holder of an American-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.

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The problem III

The random time τ:Ω ! f^{0, ...,}^Tg^,representing the time when the holder of the contingent claim exercises his or her claim, must be such that 8^t 2 f^{0, ...,}^Tg^,

f^ω 2Ωj^τ(ω) =tg 2 F^t since the decision to exercise the claim at timet 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

Λ0 =f^τ^:^Ω ! f^{0, ...,}^Tg j^τ is a stopping timeg^.

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Introduction Tari…cation

Inequalities Proof Example Snell’s problem Reference

Pricing an American-style contingent claim I

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

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Pricing an American-style contingent claim II

In what follows, Φdenotes the set of admissible investment strategies (Harrison and Pliska, 1981, p.

226), a strategy

!_ϕ =ⁿ!_ϕ_t = ϕ⁰_t,ϕ¹_t, ...,ϕ^K_t :t =1, ...To 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.

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Pricing an American-style contingent claim III

Seller’s perspective. The seller’s …rst goal is to ensure that, if he adequately invests the amountx obtained from selling the contingent claimX, then, at time τ when the buyer exercises his claim, he is able to meet his obligation, i.e. to pay the amountX_τ. The minimum price acceptable to the seller of contingent claimX is therefore

xsup =inf x 0 9!^ϕ 2 Φsuch that V0 !_ϕ =x andV_τ !_ϕ _X

τ,8^τ2Λ0 .

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Pricing an American-style contingent claim IV

Buyer’s perspective. If the buyer borrows an amount x at timet =0 in order to buy the contingent claim X then, at timeτ when he exercises his claim, he wants to be able to pay back his debt, i.e. there must exist an investment strategyφ such thatV₀ !

φ = x and V_τ !

φ +X_τ 0. Thus, the maximum price that the buyer of the contingent claimX is ready to pay out is

x_inf =sup 8<

:x 0 9!

φ 2^Φ ^{such that}^V⁰ !

φ = x and V_τ !

φ +X_τ 0 for some τ2Λ0

9=

;

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Inequalities

Pricing an American-style contingent claim

Theorem

If the market model contains no arbitrage opportunities, then, for any martingale measureQ,

X0 xinf sup

τ2Λ0

E^Q[β_τX_τ] xsup.

Remark. In solving Snell’s problem, we obtain a stopping time τ satisfying E^Q[β_τ X_τ ] =sup_τ₂_Λ₀E^Q[β_τX_τ], which allows us to determine the latter quantity without the need to know every stopping time in the set Λ0

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Proof - First inequality

The …rst inequality comes from the fact thatX₀ belongs to the set

8<

:x 0 9!

φ 2^Φ^{such that} ^V⁰ !

φ = x andV_τ !

φ +X_τ 0, for some τ2Λ0

9=

;. Indeed, let’s choose the strategy !

φ that consists, throughout time interval[0,T], of holding portfolio ( X₀,0, ...,0)(a debt ofX0 shares of the riskless security) and let’s setτ=0. Then

V₀ !

φ = ( X₀,0, ...,0) !_S

0 = X₀ and

V_τ !

φ +X_τ =V0 !

φ +X0 =0.

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Proof - Third inequality I

Let’s choose arbitrarily

x₀ 2 ^x ⁰ 9!^ϕ 2^Φ^{such that}

V₀ !_ϕ =x andV_τ !_ϕ _X

τ,8^τ2 Λ0

.

That choice is such that there exists a strategy !_ϕ 2 Φsuch thatV₀ !_ϕ =x₀ andV_τ !_ϕ _X

τ for any stopping time τ2^Λ⁰. Thus, based on the latter inequality, we state that

8^τ2^Λ⁰^,^E^Q[β_τX_τ] E^Q β_τV_τ !_ϕ _. ₍₁₎

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Proof - Third inequality II

Since the market contains no arbitrage opportunities, the stochastic processβV !_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_τV_τ !_ϕ =E^Q β₀V₀ !_ϕ which leads to the following:

8^τ2Λ0,E^Q β_τV_τ !_ϕ = E^Q β₀V₀ !_ϕ

= β₀x₀

= x₀.

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Proof - Third inequality III

By substitution into inequality (1) E^Q[β_τX_τ] E^Q β_τV_τ !_ϕ , we obtain

8^τ2 ^Λ⁰^,^E^Q[β_τX_τ] x₀ hence

sup

τ2Λ⁰E^Q[β_τX_τ] x0. Since the choice ofx0 was arbitrary, then

sup

τ2Λ0

E^Q[β_τX_τ]

inf x0 0 9!^ϕ 2 Φsuch that V0 !_ϕ =x0

andV_τ !_ϕ _X

τ,8^τ2^Λ⁰

= x_sup.

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Proof - Second inequality I

Let’s choose arbitrarily

x0 2 8<

:x 0 9!

φ 2^Φ ^{such that}^V⁰ !

φ = x andV_τ !

φ +X_τ 0 for someτ2 Λ0

9=

; thus establishing the existence of an admissible strategy

!φ 2^Φ^satisfying ^V⁰ !

φ = x₀ and V_τ_x

0

!φ +X_τ_x

0 0,for some τx0 2 ^Λ⁰^. Thus, given that β_τ

x0 >0,

0 E^Qh

β_τ_x

0 V_τ_x

0

!φ +X_τ_x

0

i

= E^Qh β_τ

x0V_τ_x

0

!φ

i+E^Qh β_τ

x0X_τ_x

0

i .

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Proof - Second inequality II

Besides, sinceβV !_ϕ _{is a} (Q,F) martingale and τx0 is a bounded stopping time, we can use the ”Optional Stopping Theorem” and obtain

E^Qh β_τ_x

0V_τ_x

0

!φ i

= E^Qh

β₀V₀ ! φ

i

= β₀x₀

= x₀.

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Proof - Second inequality III

As a consequence,

0 E^Qh

β_τ

x0V_τ_x

0

!φ

i+E^Qh β_τ

x0X_τ_x

0

i

= x₀+E^Qh β_τ_x

0X_τ_x

0

i x0+ sup

τ2Λ⁰E^Q[β_τX_τ] hence

sup

τ2Λ0

E^Q[β_τX_τ] x0.

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Proof - Second inequality IV

Since the choice ofx0 was arbitrary, sup

τ2Λ⁰E^Q[β_τX_τ]

sup 8>

><

>>

: x₀ 0

9!

φ 2Φsuch that V₀ !

φ = x₀ and V_τ !

φ +X_τ 0, for some τ2Λ0

9>

>=

>>

;

= x_inf.

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Example I

Let’s consider the following three-period binomial model

ω S₀⁰(ω) S₀¹(ω)

S₁⁰(ω) S₁¹(ω)

S₂⁰(ω) S₂¹(ω)

S₃⁰(ω)

S₃¹(ω) ^Q(ω)

ω1 1

80

1.115 100

1.115² 125

1.115³

156,25 0.343

ω2

1 80

1.115 100

1.115² 125

1.115³

100 0.147

ω3 1

80

1.115 100

1.115² 80

1.115³

100 0.147

ω4 1

80

1.115 100

1.115² 80

1.115³

64 0.063

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Example II

ω S₀⁰(ω) S₀¹(ω)

S₁⁰(ω) S₁¹(ω)

S₂⁰(ω) S₂¹(ω)

S₃⁰(ω)

S₃¹(ω) ^Q(ω)

ω5 1

80

1.115 64

1.115² 80

1.115³

100 0.147

ω6 1

80

1.115 64

1.115² 80

1.115³

64 0.063

ω7 1

80

1.115 64

1.115² 51.20

1.115³

64 0.063

ω8 1

80

1.115 64

1.115² 51.20

1.115³

40.96 0.027

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Example III

So, the …ltration is F⁰ = f?^,^Ωg

F¹ = σff^ω¹^,^ω²^,^ω³^,^ω⁴g^,f^ω⁵^,^ω⁶^,^ω⁷^,^ω⁸gg F² = σff^ω¹^,^ω²g^,f^ω³^,^ω⁴g^,f^ω⁵^,^ω⁶g^,f^ω⁷^,^ω⁸gg F³ = all events inΩ

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Example IV

Exercise. It follows that Card(Λ0) =26. (Check by yourself...

if you have time!)

Partial answer. If we represent the stopping timeτ as a point inR^Card⁽^Ω⁾,

(τ(ω1), τ(ω2), τ(ω3), τ(ω4), τ(ω5), τ(ω6), τ(ω7),τ(ω8)), then the 26 stopping times inΛ0 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)

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Example V

Exercise. Check that the probability measureQ 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.

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Example VI

Let’s consider, as an American-style contingent claim, a put option with an 80-dollar strike price. The value of such a put at timet, if the put is exercised, is

Xt =max 80 S_t¹,0

and its discounted value at timet, if again the put is exercised, is

Y_t =β_tX_t =1.115 ^tmax 80 S_t¹,0

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Example VII

Y_t = β_tX_t =1.115 ^tmax 80 S_t¹,0

ω Y₀ Y₁ Y₂ Y₃

ω1 0 0 0 0

ω2 0 0 0 0

ω3 0 0 0 0

ω4 0 0 0 _1.115¹⁶3 =11.5424

ω5 0 _1.115¹⁶ =14.3498 0 0

ω6 0 _1.115¹⁶ =14.3498 0 _1.115¹⁶3 =11.5424

ω7 0 _1.115¹⁶ =14.3498 _1.115^28.802 =23.1656 _1.115¹⁶3 =11.5424 ω8 0 _1.115¹⁶ =14.3498 _1.115^28.802 =23.1656 _1.115^39.043 =28.1634

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Example VIII

Note that one of the random times representing, for each ω, a time when the option value is the greatest is

(τ(ω1), τ(ω2), τ(ω3), τ(ω4), τ(ω5), τ(ω6), τ(ω7),τ(ω8))

= (3,3,3,3,1,1,2,3).

But such a random time is not a stopping time, since

f^ω2^Ω^:^τ(ω) =2g=f^ω7g2 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.

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Introduction Tari…cation Snell’s problem

Simpli…ed situation General situation

Example Lemma 1 Example Lemma 2 Lemma 3 Summary Example Reference

Formulation of Snell’s problem I

Let Y =f^Y^t ^:^t =0,1, ...,Tgbe a stochastic process, F adapted.

For all t 2 f^0,^{1, ...,}^Tg, we can de…ne the set of stopping times taking their values in the set f^{t, ...,}^Tg^:

Λt =f^τ^:^Ω ! f^{t, ...,}^Tg j^τ is a stopping timeg^. Note thatΛT ΛT 1 ... Λ0.

Snell’s problem is as follows: can we determine a stopping time τ 2^Λ⁰ ^satisfying

E[Y_τ ] = sup

τ2Λ0

E[Y_τ] (2) In other words, we are looking to determine, for each of the ω2Ω, the time τ (ω) when we should stop the stochastic processY in order to maximize the expected value of the random variable Y_τ.

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Simpli…ed situation I

The …ltration contains no information

Let’s assume that 8^t2 f^0,^{1, ...,}^Tg^,F^t =f?^,Ωg^. Under such conditions, sinceYt is F^t measurable, then Y_t is constant, i.e. there exists a real numbery_t for which

8^ω2 Ω,Yt(ω) =yt.

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Simpli…ed situation II

Moreover, that special …ltration is such that

Λ0 =f^τ⁰^,^τ¹^{, ...,}^τ^Tg ⁽³⁾

where 8^ω2^Ω^,^τ^t(ω) =t.

Indeed, if τ is any stopping time (with respect to the

…ltration F) then8^t 2 f^0,^{1, ...,}^Tg^,

f^ω 2^Ω^:^τ(ω) =tg 2 F^t =f?^,^Ωg. Thus, if there exists a k 2 f^0,^{1, ...,}^Tg^{for which} ^τ(ω) =k,then

f^ω2 ^Ω^:^τ(ω) =tg= ^Ω ^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 f^0,^{1, ...,}^Tgthen there areT +1 possible values for k, hence equation (3).

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Simpli…ed situation III

If we rewrite equation (2) E[Y_τ ] =sup_τ₂_Λ₀E[Y_τ]

keeping in mind this special situation only, we observe that resolving Snell’s problem, in this context, amounts to determine t such that

y_t = max

t2f^0,1,...,Tgy_t. (4) It’s obviously a trivial problem, 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.

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Simpli…ed situation IV

The basic idea is to introduce an auxiliary sequence f^z^t ^:^t =0,1, ...,Tg^{de…ned as}

zt = max

u2f^t,...,Tgyu. Note that this sequence is decreasing (i.e.

8^t 2 f^{1, ...,}^Tg^,^z^t ^z^t ¹^{) and}

zt =maxf^y^t^, ^z^t⁺¹g^. ⁽⁵⁾ Let’s set

t =minf^t 2 f^0,^{1, ...,}^Tg j ^z^t =y_tg ⁽⁶⁾ and let’s show thatt satis…es equation (4).

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Simpli…ed situation V

From the de…nition of t , we observe that

z₀ =z₁=...=z_t z_t +1 ... z_T. (7) Indeed, if there existed t 2 f^0,^{1, ...,}^t ¹g^{such that}^z^t is striclty greater thanz_t₊₁, then, from equation (5), zt =maxf^y^t^, ^z^t⁺¹g>zt+1. It would therefore follow that z_t =y_t, which would contradict the de…nition oft . As a consequence, z_t z_t+1.

But, since the sequence ofz_t is decreasing, thenz_t =z_t+1. Besides, expression (7) shows that t truly is the index we are after, since

y_t =z_t =z₀= max

t2f^0,1,...,Tgy_t.

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General situation I

Snell’s problem

Now, let’s see how to adapt the ideas from the previous section to any …ltration.

The key to the problem solution lies in building a proper version of the decreasing sequence satisfying equation (5) zt =maxf^y^t^, ^z^t⁺¹g^.

Let’s set Zt =

8<

:

Y_T ift =T

maxf^Y^t^,^E[Z_t+1jF^t]g ^if^t 2 f^{0, ...,}^T ¹g^.

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General situation II

Snell’s problem

The stochastic process Z =f^Z^t ^:^t =0,1, ...,Tg^is adapted to the …ltration F.

Indeed, E[Zt+1jF^t]andYt beingF^t measurable, maxf^Y^t^,^E[Z_t+1jF^t]gis also measurable.

It is to be noted that, thus de…ned, the sequence

f^Z^t ^:^t =0,1, ...,Tgis not necessarily decreasing ω byω, but it is decreasing in terms of conditional expectations.

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Example I

Snell’s problem

ω Y0 Y1 Y2 Y3 Q

ω1 0 0 0 0 0.343

ω2 0 0 0 0 0.147

ω3 0 0 0 0 0.147

ω4 0 0 0 =^11.5424 ^0.063

ω5 0 =14.3498 0 0 0.147

ω6 0 =^14.3498 ⁰ =^11.5424 ^0.063

ω7 0 =^14.3498 =^23.1656 =^11.5424 ^0.063 ω8 0 =14.3498 =23.1656 =28.1634 0.027

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Example II

Snell’s problem

ω Z₀ Z₁ Z₂ Z₃

ω1 =5.0321˙ =1.0388 0 0

ω2 =^5.⁰³²¹^˙ =^1.0388 ⁰ ⁰

ω3 =^5.⁰³²¹^˙ =^1.0388 =^3.4627 ⁰ ω4 =5.0321˙ =1.0388 =3.4627 =11.5424 ω5 =^5.⁰³²¹^˙ =^14.3498 =^3.4627 ⁰ ω6 =^5.⁰³²¹^˙ =^14.3498 =^3.4627 =^11.5424 ω7 =5.0321˙ =14.3498 =23.1656 =11.5424 ω8 =^5.⁰³²¹^˙ =^14.3498 =^23.1656 =^28.1634

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Example III

Snell’s problem

Calculation of the preceding table entries:

8^ω 2 f^ω¹^,^ω²g^, E[Z3jF²] (ω) = 0

Z₂(ω) = maxf^Y²(ω),E[Z₃jF²] (ω)g

= maxf^0;⁰g=0 8^ω 2 f^ω³^,^ω⁴g^, E[Z₃jF²] (ω) = ¹⁶

1.115³ 0.063

0.21 =3.4627 Z2(ω) = maxf^Y²(ω),E[Z3jF²] (ω)g

= maxf^0;^3.4627g=3.4627

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Example IV

Snell’s problem

8^ω 2 f^ω⁵^,^ω⁶g^, E[Z₃jF²] (ω) = ¹⁶

1.115³ 0.063

0.21 =^3.4627 Z₂(ω) = maxf^Y²(ω),E[Z₃jF²] (ω)g

= ^maxf^0;^3.4627g=3.4627 8^ω 2 f^ω⁷^,^ω⁸g^,

E[Z3jF²] (ω) = ¹⁶ 1.115³

0.063

0.09 + ^39.04 1.115³

0.027

0.09 =16.5287 Z2(ω) = maxf^Y²(ω),E[Z3jF²] (ω)g

= maxf^23.1656;^16.5287g=23.1656

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Example V

Snell’s problem

8^ω 2 f^ω¹^,^ω²^,^ω³^,^ω⁴g^, E[Z₂jF¹] (ω) = 3.46270.21

0,7 =^1.0388

Z₁(ω) = maxf^Y¹(ω),E[Z₂jF¹] (ω)g

= maxf^0;^1.0388g=1.0388 8^ω 2 f^ω⁵^,^ω⁶^,^ω⁷^,^ω⁸g^,

E[Z2jF¹] (ω) = 3.46270.21

0.3 +23.16560.09

0.3 =7.9885 Z1(ω) = maxf^Y¹(ω),E[Z2jF¹] (ω)g

= ^maxf^14.3498;^7.9885g=14.3498 8^ω 2 ^Ω

E[Z1jF⁰] (ω) = 1.0388 0.7+14.3498 0.3=5.0321˙

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Example VI

Snell’s problem

Interpretation. Zt represents, conditionally to the

information available at timet, 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 American option at time t.

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Lemma 1 I

Snell’s problem

Theorem

Lemma 1. For all t 2 f^{0, ...,}^Tg^,

8^τ2 Λt, Zt E[Y_τjF^t], (8) i.e.

Zt sup

τ2Λt

E[Y_τjF^t] and, more particularly,

Z0 sup

τ2Λ⁰E[Y_τjF⁰].

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Lemma 1 II

Snell’s problem

Proof of Lemma 1.

We will proceed by backward induction on t. When t =T,then

ΛT =f^τ^:^Ω ! f^Tg j^τ is a stopping timeg^, i.e. ΛT contains one stopping time only, the one identically equal to T. Since ZT =YT =E[YT jF^T], inequality (8)8^τ2 ^Λ^T^,^Z^T ^E[Y_τjF^T] is clearly satis…ed.

Now, let’s assume that inequality (8)8^τ2 Λt, Zt E[Y_τjF^t] is veri…ed for somet 2 f^{1, ...,}^Tg^and let’s show that is also veri…ed for index t 1.

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Lemma 1 III

Snell’s problem

Let’s choose arbitrarily a stopping time τ2^Λ^t ¹^{. We will} evaluate the expectation on the right side of inequality (8) 8^τ2 ^Λ^t^,^Z^t ^E[Y_τjF^t]by breaking it down according to the values taken by τ:

E[Y_τjF^t ¹]

= E Y_τI_fτ=t 1gjF^t ¹ +E Y_τI_fτ>t 1gjF^t ¹

= E Y_t ₁I_f_τ=t 1gjF^t ¹ +E Y_τ_{_}_tI_f_τ>t 1gjF^t ¹ ^.⁽⁹⁾ First, we’re going to tackle the second term in the equality stated above.

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Lemma 1 IV

Snell’s problem

Given that τ_t, or the maximum of the two stopping times t andτ, is itself a stopping time, which besides belongs to Λt since, by its very de…nition, it takes its values in the set f^{t, ...,}^Tg, the induction hypothesis then allows us to state that

Zt E[Y_τ_{_}tjF^t]. (10)

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Lemma 1 V

Snell’s problem

On another front, f^τ>t 1g 2 F^t ¹ ^since f^τ>t 1g^c 2 F^t ¹^{. Indeed,}

f^τ ^t ¹g=

t[1

j=0

f^τ=jg

| {z }

2F^j F^t ¹

| {z }

2F^t ¹

2 F^t ¹ ⁽¹¹⁾

implies that I_f_τ>t 1g is F^t ¹ measurable.

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Lemma 1 VI

Snell’s problem

It follows that

E Y_τ_{_}_tI_f_τ>t 1gjF^t ¹

= I_fτ>t 1gE[Y_τ_{_}t jF^t ¹]

= _I_f_τ_>_t ₁_gE[E[Y_τ_{_}tjF^t]jF^t ¹] I_fτ>t 1gE[ZtjF^t ¹]

I_f_τ>t 1gmax_| f^Y^t ¹^,_{z^E[Z_tjF^t ¹]g_}

=Zt 1

= I_f_τ>t 1gZ_t ₁

where the …rst inequality comes from expression (10) Z_t E[Y_τ_{_}_tjF^t].

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Lemma 1 VII

Snell’s problem

What about the …rst term in equation (9) E[Y_τjF^t ¹] = E Yt 1I_fτ=t 1gjF^t ¹ +E Y_τ_{_}tI_fτ>t 1gjF^t ¹ ^? From the de…nition of Z_t ₁, we have that

Z_t ₁ =maxf^Y^t ¹^,^E[Z_tjF^t ¹]g ^Y^t ¹^, hence

E Y_t ₁I_fτ=t 1gjF^t ¹ ^E ^Z^t ¹^I_fτ=t 1gjF^t ¹

= Zt 1I_fτ=t 1g.