歐式下出局選擇權之均勻漸近展開

國立臺灣大學理學院數學系 碩士論文

Department of Mathematics College of Science

National Taiwan University Master Thesis

歐式下出局選擇權之均勻漸近展開

Uniform Asymptotic Expansions for European Down-and-Out Barrier Options

許姍祺 Shan-Chi Hsu

指導教授: 姜祖恕博士 張志中博士

Advisor: Tzuu-Shuh Chiang, Ph.D.

Chih-Chung Chang, Ph.D.

中華民國 100 年 6 月

June, 2011

摘要

在本文中，我們計算出在隨機波動率模型下的歐式下出局選擇權的均勻漸近展 開，其中波動率的隨機性來自一具備均數復歸特性的隨機過程。並於文末針對此 一計算模型在以文內提出的方式進行修正後，是否能求得更為精確之選擇權價格 估計值一點進行討論。　

關鍵字

歐式障礙選擇權。均數復歸隨機過程。選擇權定價。隨機波動率。均勻漸近展 開。

Abstract

We calculate the uniform asymptotic expansion for European down-and-out barrier options under the stochastic volatility model, where the volatility is driven by a mean-reverting diffusion process. We also discuss whether the modified method of uniform asymptotic expansion which we used in the last chapter can approximate the price of option with more accuracy.

Key Words

European barrier option. Mean-reverting process. Option pricing. Stochastic volatility. Uniform asymptotic expansion.

Contents

摘要 . . . i

Abstract . . . ii

Chapter 1 Introduction . . . 2

Chapter 2 Asymptotic Expansion for European Derivatives . . . 4

Chapter 3 Uniform Asymptotic Expansion for European Derivatives . . . 14

Chapter 4 Down-and-Out Barrier Options . . . 24

Chapter 5 Uniform Asymptotic Expansion for Down-and-Out Barrier Options . 31 Bibliography . . . 41

Chapter 1 Introduction

When it comes to the derivative pricing, we always let the price of derivative be a martingale such that it keeps the attribute of fairness. For instance, the well- known Black-Scholes model was derived under the assumptions that the derivative price is fair and the volatility of underlying asset price is deterministic. However, the volatility in the real world is obviously more unpredictable than the one with deterministicity. To correct the dissimilarity of pricing model to the reality arised from the setting of deterministic volatility, it is more suitable to substitute for the one with randomness, that is, a stochastic volatility, to capture the unprediction of underlying asset price.

In this article, we consider the uniform asymptotic expansion for European derivatives and derive the result of that for down-and-out barrier options. The model of asymptotic expansion is mentioned in [1] and modified in [2]. The main difference in these two model is that the stochastic process which drives the volatil- ity of underlying asset price in [1] is an Ornstein-Uhlenbeck process on the real line R, which has the nature of mean reversion, whereas that in [2] is a more general diffusion process on a circle S, a compact set with nonzero circumference, such that the process can also attain it invariant distribution.

There are several flaws in the model in [1] as we consider the asymptotic expan- sion of derivative price. Due to the use of OU process onR, we cannot uniquely de- cide the second-order term of derivative price. Moreover, we cannot get the outcome of uniform asymptotic expansion. Hence, in the following chapters, it is convenient

to use this setting in [2] to present the related results in [1] and [2] for the final conclusion and to develop the discussion on down-and-out barrier options.

The rest of the article is arranged as follows. In Chapter 2, we introduce the asymptotic expansion for European derivatives mentioned in [1]. In Chapter 3, we present the uniform asymptotic expansion for European derivatives appeared in [2]. In Chapter 4, we illustrate the calculations of the zero-order term and the first-order term of European down-and-out call options. In Chapter 5, we conclude the accuracy of approximation of European down-and-out options calculated by the method of uniform asymptotic expansion.

Chapter 2

Asymptotic Expansion for European Derivatives

We shall introduce the asymptotic expansion in this chapter. The calculation method is used in [1] and then referred in [2]. However, to achieve the desirable conclusion in the following chapters, it is convenient to substitute some of the set- tings in [2] for those originally in [1].

Consider a risky asset price S_t such that

dS_t = µS_tdt + σ_tS_tdfW_t, (2.1) where fW_t is a standard Brownian motion under the subjective probability P and σ_t is a stochastic volatility driven by a diffusion process eY_t such that

d eYt= m( eYt) dt + β( eYt) dZ_t^∗, (2.2) where Z_t^∗ is a standard Brownian motion under P, dependent of fW_t.

To construct eY_t such that it has the feature of mean reversion, we shall define eY_t as a diffusion process on a circle S with nenzero circumference. Since S is a compact set, the definition implies that eY_t will finally attain its invariant distribution.

Let Y_t= eY_t/ε, bZ_t=√

εZ_t/ε^∗ , and σ_t= f (Y_t), then (2.1) and (2.2) become

dSt= µStdt + f (Yt)StdfWt (2.3) dY_t= 1

εm(Y_t) dt + 1

√εβ(Y_t) d bZ_t, (2.4)

respectively.

Since bZ_t is dependent of fW_t, there exists |ρ| < 1 such that d bZt= ρ dfWt+√

1− ρ²d eZt, (2.5)

where eZ_t is a standard Brownian motion under P, independent of fW_t. Thus, (2.4) becomes

dYt= 1

εm(Yt) dt + 1

√εβ(Yt) (

ρ dfWt+√

1− ρ²d eZt

)

. (2.6)

Introduce the risk-neutral probability P^(γ) such that dfWt= dWt− µ− r

f (Y_t)dt (2.7)

d eZt= dZt− γ(Yt) dt, (2.8) where W_t and Z_t are two independent Brownian motions under P^(γ), then (2.3) and (2.6) become

dS_t= rS_tdt + f (Y_t)S_tdW_t (2.9)

dY_t= [1

εm(Y_t)− 1

√εβ(Y_t)Λ(Y_t) ]

dt + 1

√εβ(Y_t) (

ρ dW_t+√

1− ρ²dZ_t )

, (2.10) respectively, where

Λ(y) = ρµ− r

f (y) + γ(y)√

1− ρ². Let X_t= log S_t, then

dX_t= (

r−1 2f²(Y_t)

)

dt + f (Y_t) dW_t.

Consider a European derivative on the underlying asset price S_twith nonnegative payoff functionH(s) and maturity time T. Let P^ε(t, x, y) be the price of derivative and h(x) =H(e^x), then by Feynman-Kac formula,

P^ε(t, x, y) = E^(γ)[

e^{−r(T −t)}h(X_T)X_t= x, Y_t= y] . By applying Ito’s formula to e^−rtP^ε(t, x, y),

d(

e^−rtP^ε(t, x, y))

= {1

ε [

m(Y_t)∂P^ε

∂y +1

2β²(Y_t)∂²P^ε

∂y² ]

+ 1

√ε [

ρβ(Y_t)f (Y_t)∂²P^ε

∂x∂y − β(Yt)Λ(Y_t)∂P^ε

∂y ]

+ [∂P^ε

∂t + 1

2f²(Y_t)∂²P^ε

∂x² + (

r− 1 2f²(Y_t)

)∂P^ε

∂x − rP^ε ]}

dt + dMt,

where M_t represents martingale part.

Let

L0 = m(y) ∂

∂y +1

2β²(y) ∂²

∂y² L1 = ρβ(y)f (y) ∂²

∂x∂y − β(y)Λ(y) ∂

∂y L2 =LBS(f (y)) = ∂

∂t+1

2f²(y) ∂²

∂x² + (

r−1 2f²(y)

) ∂

∂x − r ·,

whereL0 is the infinitesimal generator of Y_tandLBS(f (y)) is Black-Scholes operator calculated at the logarithm of underlying asset price and the volatility level f (y), and

L = 1

εL0+ 1

√εL1+L2,

then P^ε(t, x, y) is the solution of

LP^ε(t, x, y) = 0 (2.11)

with the terminal condition

P^ε(T, x, y) = h(x). (2.12)

Define P (t, x, y) as the outer expansion and let P^ε= P in this chapter, then by expanding P (t, x, y) in powers of√

ε, P^ε(t, x, y) = P (t, x, y)

= P₀(t, x, y) +√

εP₁(t, x, y) + εP₂(t, x, y) + ε√

εP₃(t, x, y) + ε²P₄(t, x, y) +· · · and

LP^ε = (1

εL0+ 1

√εL1+L2

) (P₀ +√

εP₁+ εP₂+ ε√

εP₃+ ε²P₄+· · ·)

= 1

εL0P₀+ 1

√ε(L0P₁+L1P₀) + (L0P₂+L1P₁+L2P₀)

+√

ε(L0P₃ +L1P₂+L2P₁) + ε(L0P₄+L1P₃+L2P₂) +· · · .

Choose P_i’s to satisfy

L0P₀ = 0

L0P1+L1P0 = 0

L0P₂+L1P₁+L2P₀ = 0 L0P₃+L1P₂+L2P₁ = 0 L0P₄+L1P₃+L2P₂ = 0

(2.13)

...

with terminal conditions

P₀(T, x, y) = h(x) (2.14)

P₁(T, x, y) = 0. (2.15)

Here we make the following assumptions for several functions mentioned above.

Assumption 2.1.

1. f (y), γ(y), β(y), and m(y) are bounded and sufficiently smooth in y ∈ S such that f²(y) > 0, γ(y) > 0, and β²(y) > 0, ∀y ∈ S.

2. h(x) is bounded and sufficiently smooth in x∈ R.

These assumptions hold throughout the article and ensure the feasibility of asym- totic expansion.

Consider the term of order 1/ε,

L0P₀ = 0. (2.16)

SinceL0 is the infinitesimal generator of Y_t, which is also an ergodic Markov process, P₀ must be independent of y, that is,

P₀ = P₀(t, x). (2.17)

Consider terms of order 1/√ ε,

L0P1+L1P0 = 0. (2.18)

Since all the terms inL1 take derivatives of y, by (2.17),

L1P₀ = 0. (2.19)

Thus, (2.18) reduces to

L0P₁ = 0.

Using the same argument as we discussed (2.16), we get that

P1 = P1(t, x). (2.20)

Consider terms of order 1,

L0P₂+L1P₁+L2P₀ = 0. (2.21)

Similarly as we discussed (2.19), we get that L1P₁ = 0.

Thus, (2.21) reduces to

L0P₂+L2P₀ = 0, (2.22)

a Poisson equation for P₂ with respect to L0.

To solve the Poisson equation, we state a lemma mentioned both in [1] and [2].

Lemma 2.1. Suppose L is an infinitesimal generator of some ergodic Markov process

Y_t and g(y) and its derivatives are bounded in y∈ S.

1. The Poisson equation

Lu(y) = g(y) (2.23)

has a solution if and only if the centering condition

⟨g⟩ =

∫

S

g(y) p(y) dy = 0 (2.24)

is satisfied, where p(y) is the stationary density associated with L.

2. Any solution of (2.23) can be written as

u(y) = u₀(y) + c,

where u0(y) is the unique solution of (2.23) with⟨u0⟩ = 0 and c is independent of y.

By Lemma 2.1, (2.22) has a solution if and only if the centering condition

⟨L2P₀⟩ =

∫

S

L2P₀(t, x, y) Φ(y) dy = 0 (2.25)

is satisfied, where Φ(y) is the invariant distribution of Y_t. Since P₀ is independent of y,

⟨L2P₀⟩ = ⟨L2⟩P0

= [∂

∂t +1

2⟨f²⟩ ∂²

∂x² + (

r− 1 2⟨f²⟩

) ∂

∂x − r · ]

P₀

=LBS(¯σ)P₀

= 0,

(2.26)

where ¯σ =√

⟨f²⟩, that is, ⟨L2⟩ is Black-Scholes operator calculated at the logarithm of underlying asset price and the volatility level ¯σ.

Hence, by (2.26) and (2.14), P₀(t, x) is the solution of

LBS(¯σ)P₀ = 0 (2.27)

with the terminal condition

P₀(T, x) = h(x). (2.28)

If (2.25) is satisfied, then by (2.22), L0P₂ =−L2P₀

=− (L2P0− ⟨L2P0⟩)

=−1 2

(f²(y)− ⟨f²⟩) (∂²P₀

∂x² (t, x)− ∂P₀

∂x (t, x) )

.

Let ψ(y) be the solution of

L0ψ = f²(y)− ⟨f²⟩ (2.29)

with

⟨ψ⟩ = 0, (2.30)

then ψ(y) is uniquely defined and P₂(t, x, y) =−1

2

(∂²P₀

∂x² (t, x)− ∂P₀

∂x (t, x) )

ψ(y) + A₁(t, x), (2.31) where A1(t, x) is to be determined.

Consider terms of order √ ε,

L0P₃+L1P₂+L2P₁ = 0. (2.32)

Similarly as we addressed (2.22), (2.32) has a solution if and only if

⟨L1P₂+L2P₁⟩ = 0,

equivalent to

⟨L2P1⟩ = −⟨L1P2⟩. (2.33)

Since P₁ does not depend on y, similarly as (2.26),

⟨L2P₁⟩ = ⟨L2⟩P1 =LBS(¯σ)P₁.

Since A₁ does not depend on y, either,

⟨L1A₁⟩ = 0.

Then,

−⟨L1P₂⟩ = 1 2

[

ρ⟨βfψ^′⟩

(∂³P₀

∂x³ (t, x)−∂²P₀

∂x² (t, x) )

−⟨βΛψ^′⟩

(∂²P₀

∂x² (t, x)− ∂P₀

∂x (t, x) )]

= U₁∂P₀

∂x + U₂∂²P₀

∂x² + U₃∂³P₀

∂x³

= H(t, x),

where

U₁ = 1

2⟨βΛψ^′⟩ U₂ =−1

2(ρ⟨βfψ^′⟩ + ⟨βΛψ^′⟩) U₃ = 1

2ρ⟨βfψ^′⟩.

According to (2.33) and (2.15), we can express P₁(t, x) as the solution of

LBS(¯σ)P1 = H(t, x) (2.34)

with the terminal condition

P₁(T, x) = 0. (2.35)

It can be showed that ∀n ∈ N, LBS(¯σ)∂ⁿP₀

∂xⁿ = ∂ⁿ

∂xⁿLBS(¯σ)P₀ = 0. (2.36) Using (2.36) to check

P₁(t, x) = −(T − t)H(t, x), (2.37) we get

LBS(¯σ)P₁ =LBS(¯σ)[−(T − t)H(t, x)]

= H− (T − t)LBS(¯σ)H

= H and

P1(T, x) = 0.

At this stage, we want to show that ∀(t, x, y) ∈ [0, T ] × R × S, P^ε(t, x, y)− (P0(t, x) +√

εP₁(t, x))=O(ε).

Let

Z_p3(t, x, y) = (P₀(t, x) +√

εP₁(t, x) + εP₂(t, x, y) + ε√

εP₃(t, x, y))

− P^ε(t, x, y),

then Z_p3(t, x, y) can be expressed as the solution of LZp3(t, x, y) =

(1

εL0+ 1

√εL1+L2

)

(P₀+√

εP₁ + εP₂+ ε√

εP₃)− LP

= 1

εL0P0+ 1

√ε(L0P1+L1P0) + (L0P2+L1P1+L2P0)

+√

ε(L0P₃+L1P₂+L2P₁) + ε(L1P₃+L2P₂+√

εL2P₃)

= ε(L1P₃+L2P₂+√

εL2P₃)

= εF_p^ε(t, x, y) with the terminal condition

Z_p3(T, x, y) = ε(P₂(T, x, y) +√

εP₃(T, x, y))

= εG^ε_p(x, y).

To solve Z_p3, applying Ito’s formula to e^−rsZ_p3(s, X_s, Y_s) to get d(

e^−rsZ_p3(s, X_s, Y_s))

= e^−rsLZp3(s, X_s, Y_s) ds + dM_s

= εe^−rsF_p^ε(s, Xs, Ys) ds + dMs

(2.38)

and integrating (2.38) over s from t to T to get

e^−rTZ_p3(T, X_T, Y_T)− e^−rtZ_p3(t, x, y) = ε

∫ T t

e^−rsF_p^ε(s, X_s, Y_s) ds + (M_T − Mt).

(2.39) Since

e^−rTZ_p3(T, X_T, Y_T) = εe^−rTG^ε_p(X_T, Y_T), (2.40) by (2.39) and (2.40), Zp3 has probabilistic representation

Z_p3(t, x, y) = εE^(γ) [

e^{−r(T −t)}G^ε_p(X_T, Y_T)−

∫ _T

t

e^−r(s−t)F_p^ε(s, X_s, Y_s) ds

X^t= x, Y_t= y ]

. By Assumption 2.1, F_p and G_p are bounded for any given (x, y), which imply that

Z_p3(t, x, y) =O(ε).

Hence,

P^ε(t, x, y)− (P0(t, x) +√

εP₁(t, x))

=Z_p3(t, x, y)− ε(P2(t, x, y) +√

εP₃(t, x, y))

=O(ε).

Chapter 3

Uniform Asymptotic Expansion for European Derivatives

In the previous chapter, we mentioned the pricing system for European options, where the stochastic volatility is driven by an ergodic process Yt on a circle S with nonzero circumference, and computed the asymptotic approximation of the zero- order term and the first correction to the price, that is, P₀+√

εP₁ to P^ε. However, this model is restricted by the essence of Y_t. The reason why we chose it to drive the volatility is that we can produce good approximation to the price at the expiration date when the distribution of Y_t is close enough to its invariant distribution. It means that if the starting time t is too close to the maturity time T, the process Y_t will not have enough time to attain its invariant distribution and we can only get bad approximation.

Moreover, under the original structure of model, we decide the zero-order term P₀ and the first-order term P₁, but there is still part of the second-order term P₂ undetermined, explicitly, A₁ in (2.31). These two problems limit the initial model we used and cause the difficulty in enhancing the accuracy of approximation. To eliminate these problems, we need to introduce the inner expansion, Q, where

Q(τ, x, y) = Q₀(τ, x, y) +√

εQ₁(τ, x, y) + εQ₂(τ, x, y) + ε√

εQ₃(τ, x, y) + ε²Q₄(τ, x, y) +· · · and

τ = T − t ε ,

contrast to the outer expansion P, which we mentioned in Chapter 2.

The definitions of parameters in this chapter which have already appeared are basicly the same as in Chapter 2 except what we specifically cite.

Since the main objective of introduction of Q is to correct the approximation problem which happens when t is too close to T, the consideration is unnecessary when t is far enough from T and Q shall disappear in this situation. According to the definition of τ, it means that

Q_i(τ, x, y)→ 0 as τ → ∞. (3.1)

Now, the option price P^ε becomes

P^ε(t, x, y) = P (t, x, y) + Q(τ, x, y)

= P₀(t, x) +√

εP₁(t, x) + εP₂(t, x, y) + ε√

εP3(t, x, y) + ε²P4(t, x, y) +· · · + Q₀(τ, x, y) +√

εQ₁(τ, x, y) + εQ₂(τ, x, y) + ε√

εQ₃(τ, x, y) + ε²Q₄(τ, x, y) +· · · .

As in Chapter 2, P^ε(t, x, y) is the solution of (2.11) with the terminal condition (2.12).

Expand LP^ε in powers of√

ε, then LP^ε(t, x, y) =LP (t, x, y) + LQ(τ, x, y)

= (1

εL0+ 1

√εL1+L2

)

P (t, x, y) + (1

εLe0+ 1

√εLe1 + eL2

)

Q(τ, x, y)

= 1

εL0P₀+ 1

√ε(L0P₁ +L1P₀) + (L0P₂+L1P₁+L2P₀)

+√

ε(L0P₃+L1P₂ +L2P₁) + ε(L0P₄+L1P₃+L2P₂) +· · · +1

εLe0Q₀+ 1

√ε( eL0Q₁+ eL1Q₀) + ( eL0Q₂ + eL1Q₁+ eL2Q₀)

+√

ε( eL0Q₃+ eL1Q₂+ eL2Q₁) + ε( eL0Q₄+ eL1Q₃+ eL2Q₂) +· · · ,

where

Le0 =− ∂

∂τ + m(y) ∂

∂y +1

2β²(y) ∂²

∂y² =− ∂

∂τ +L0

Le1 = ρβ(y)f (y) ∂²

∂x∂y − β(y)Λ(y) ∂

∂y =L1

Le2 = 1

2f²(y) ∂²

∂x² + (

r− 1 2f²(y)

) ∂

∂x − r · Choose P_i’s and Q_i’s to satisfy (2.13) and

Le0Q₀ = 0

Le0Q₁+ eL1Q₀ = 0

Le0Q2+ eL1Q1+ eL2Q0 = 0 Le0Q₃+ eL1Q₂+ eL2Q₁ = 0 Le0Q₄+ eL1Q₃+ eL2Q₂ = 0

(3.2)

...

respectively, with terminal conditions

P₀(T, x) + Q₀(0, x, y) = h(x) (3.3) P_j(T, x, y) + Q_j(0, x, y) = 0, j = 1, . . . , 4. (3.4) Since the calculations of Pi’s are the same as what we did in Chapter 2, we will use the results of P_i’s directly and concentrate on the calculations of Q_i’s here.

Consider the term of order 1/ε inLQ,

Le0Q₀ = 0. (3.5)

By (2.14) and (3.3),

Q₀(0, x, y) = 0. (3.6)

Then, Q0(τ, x, y) is the solution of (3.5) with the initial condition (3.6).

Since eL0contains only variables τ and y, we can view the variable x as a constant in the differential problem above. Then,

Q₀(τ, x, y) =

∫

S

S₀(τ, x, y− z) Q0(0, x, z) dz = 0, (3.7)

where S₀(τ, x, y) is the source function of (3.5).

Consider terms of order 1 in LQ,

Le0Q₁+ eL1Q₀ = 0. (3.8)

Because of (3.7), (3.8) reduces to

Le0Q₁ = 0. (3.9)

By (2.15) and (3.4),

Q₁(0, x, y) = 0. (3.10)

Then, Q₁(τ, x, y) is the solution of (3.9) with the initial condition (3.10).

Using the same argument as we concluded (3.7), we get that

Q₁(τ, x, y) = 0. (3.11)

Consider terms of order √

ε inLQ,

Le0Q₂+ eL1Q₁+ eL2Q₀ = 0. (3.12)

Because of (3.7) and (3.11), (3.12) reduces to

Le0Q₂ = 0. (3.13)

According to (3.4), Q2(τ, x, y) can be expressed as the solution of (3.13) with the initial condition

Q₂(0, x, y) =−P2(T, x, y)

= 1 2

(∂²P₀

∂x² (T, x)− ∂P₀

∂x (T, x) )

ψ(y)− A1(T, x)

= 1

2(h^′′(x)− h^′(x)) ψ(y)− A1(T, x)

(3.14)

By the ergodicity of Y_t,

Q2(τ, x, y) → −⟨P2(T, x,·)⟩ as τ → ∞. (3.15) Because of (3.1), (3.15) implies that

⟨P2(T, x,·)⟩ = 0. (3.16)

According to (2.30) and (2.31), (3.16) implies that

A1(T, x) = 0. (3.17)

According to (3.13) and (3.14), we can rewrite Q₂ as Q₂(τ, x, y) = 1

2(h^′′(x)− h^′(x)) φ(τ, y), (3.18) where φ(τ, y) is the solution of

Le0φ = 0

with the initial condition

φ(0, y) = ψ(y), which defines φ uniquely.

To determine A₁(t, x), first, come back to consider (2.32). Since each term inL1

has derivatives with respect to y, L1P₂ is uniquely determined.

Define eP₃ to be the solution of

L0Pe3 =−L1P2− L2P1

with

⟨ eP3⟩ = 0, then eP₃ is uniquely determined.

We can represent P₃(t, x, y) as

P3(t, x, y) = eP3(t, x, y) + A2(t, x), (3.19) then A₂(t, x) is to be determined.

In addition, consider terms of order ε in LP,

L0P4+L1P3+L2P2 = 0. (3.20) According to Lemma 2.1, suppose

⟨L2P2⟩ = −⟨L1P3⟩. (3.21)

SinceL1P₃ =L1Pe₃, L1P₃ is uniquely defined and so is ⟨L1P₃⟩.

We can rewrite (3.21) as

LBS(¯σ)A₁(t, x) + 1

4⟨f²ψ⟩G(t, x) = F (t, x), where

F (t, x) = −⟨L1P₃⟩ G(t, x) = −

(∂⁴P₀

∂x⁴ (t, x)− 2∂³P₀

∂x³ (t, x) + ∂²P₀

∂x² (t, x) )

. Then, A₁(t, x) is the solution of

LBS(¯σ)A₁(t, x) = F (t, x)− 1

4⟨f²ψ⟩G(t, x)

with the terminal condition (3.17), which determines A1 uniquely and so does P2. To determine A2(t, x), we use similar operations as we solved A1(t, x). First, consider (3.20) again and define P₄(t, x, y) to be the solution of

L0Pe4 =−L1P3− L2P2

with

⟨ eP₄⟩ = 0,

then eP₄ is uniquely determined.

Representing P₄(t, x, y) as

P4(t, x, y) = eP4(t, x, y) + A3(t, x), then A₃(t, x) is to be determined.

In addition, consider terms of order ε^3/2 inLP, L0P₅+L1P₄+L2P₃ = 0.

According to Lemma 2.1, suppose

⟨L2P3⟩ = −⟨L1P4⟩. (3.22)

SinceL1P₄ =L1Pe₄, L1P₄ is uniquely defined and so is ⟨L1P₄⟩.

Because of (3.19), we can express A₂(t, x) as the solution of LBS(¯σ)A2 =−⟨L1P4⟩ − ⟨L2Pe3⟩

by the rewrite of (3.22), with the terminal condition

A₂(T, x) =⟨P3(T, x,·)⟩. (3.23) Due to (3.4), we need to consider terms of order√

ε in LQ,

Le0Q₃+ eL1Q₂+ eL2Q₁ = 0, (3.24)

to solve A₂(t, x).

By (3.11), (3.24) reduces to

Le0Q₃+ eL1Q₂ = 0,

equivalent to

∂Q₃

∂τ =L0Q₃+L1Q₂. (3.25)

Then, Q₃(τ, x, y) is the solution of (3.25) with the initial condition

Q₃(0, x, y) =−P3(T, x, y). (3.26) Integrating (3.25) over s from 0 to τ, then

Q₃(τ, x, y) = Q₃(0, x, y) +

∫ _τ

0

L0Q₃(s, x, y) ds +

∫ _τ

0

L1Q₂(s, x, y) ds. (3.27)

In the stationary condition, (3.27) becomes

⟨Q3(τ, x,·)⟩ = ⟨Q3(0, x,·)⟩ +

∫ τ 0

∫

S

L0Q3(s, x, y) Φ(y) dy ds +

∫ τ 0

∫

S

L1Q₂(s, x, y)Φ(y) dy ds

(3.28)

and (3.26) becomes

⟨Q3(0, x,·)⟩ = −⟨P3(T, x,·)⟩. (3.29)

SinceL0 is the infinitesimal generator of Y_t and Φ(y) is the invariant distribution of that,

∫ τ 0

∫

S

L0Q₃(s, x, y) Φ(y) dy ds =

∫ τ 0

∫

S

Q₃(s, x, y)L^∗0Φ(y) dy ds = 0, whereL^∗0 is the adjoint ofL0.

Then, (3.28) reduces to

⟨Q3(τ, x,·)⟩ = ⟨Q3(0, x,·)⟩ +

∫ _τ

0

∫

SL1Q₂(s, x, y)Φ(y) dy ds. (3.30) Let τ → ∞, then by (3.1), (3.30) becomes

⟨Q3(0, x,·)⟩ +

∫ _∞

0

∫

S

L1Q₂(s, x, y)Φ(y) dy ds = 0.

According to (3.23) and (3.29), we get

A₂(T, x) =

∫ _∞

0

∫

S

L1Q₂(s, x, y)Φ(y) dy ds.

Now, A₂ is uniquely determined and so are P₃ and Q₃.

We can apply the same method to determine P₄(t, x, y) and Q₄(τ, x, y). Because we just want to approximate the derivative price up to O(

ε^3/2)

, it is sufficient to set A₃(t, x) = 0.

At this stage, we want to show that for (t, x, y)∈ [0, T ] × R × S,

sup

t,x,y

P^ε(t, x, y)− (

P₀(t, x) +√

εP₁(t, x) + εP₂(t, x, y) + εQ₂

(T − t

ε , x, y))  = O(

ε^3/2) .

Let

Z_q4(t, x, y) =[(

P₀(t, x) +√

εP₁(t, x) + εP₂(t, x, y) + ε√

εP₃(t, x, y) + ε²P₄(t, x, y)) +(

εQ₂(τ, x, y) + ε√

εQ₃(τ, x, y) + ε²Q₄(τ, x, y))]

− P^ε(t, x, y),

then Z_q4(t, x, y) can be expressed as the solution of LZq4(t, x, y) =

(1

εL0+ 1

√εL1+L2

) (P0+√

εP1+ εP2+ ε√

εP3+ ε²P4

)

+ (1

εLe0+ 1

√εLe1+ eL2

) (εQ2+ ε√

εQ3+ ε²Q4

)− LP

= 1

εL0P₀+ 1

√ε(L0P₁ +L1P₀) + (L0P₂+L1P₁+L2P₀) +√

ε(L0P₃+L1P₂ +L2P₁) + ε(L0P₄+L1P₃+L2P₂) + ε√

ε(L1P₄+L2P₃+√

εL2P₄) + eL0Q₂+√

ε( eL0Q₃+ eL1Q₂) + ε( eL0Q₄+ eL1Q₃+ eL2Q₂) + ε√

ε( eL1Q4+ eL2Q3+√

ε eL2Q4)

= ε√

ε[(L1P₄+L2P₃+√

εL2P₄) + ( eL1Q₄+ eL2Q₃+√

ε eL2Q₄)]

= ε^3/2F_q(t, x, y) with the terminal condition

Z_q4(T, x, y) = 0.

Using the same argument as in Chapter 2, Zq4 has probabilistic representation

Zq4(t, x, y) = ε^3/2E^(γ) [

−

∫ T t

e^−r(s−t)Fq(s, Xs, Ys) ds

X^t= x, Yt= y ]

. (3.31) For i = 0, . . . , 4, by Assumption 2.1, P_i and its derivatives are bounded in (x, y) and the smoothness of functions implies that

sup

t,x,y

[|L1P₄(t, x, y)| + |L2P₃(t, x, y)| + |L2P₄(t, x, y)|] < ∞ (3.32)

sup

t,x,y

[|P3(t, x, y)| + |P4(t, x, y)|] < ∞. (3.33) Furthermore, since the source function S_i(τ, x, y) with respect to Q_i(τ, x, y) must decays exponentially fast as τ → ∞, there exists C, k > 0 such that for all (x, y) and l, n = 1, . . . , 4 with l≤ n,

|Qi(τ, x, y)| ≤ Ce^−kτ ∂ⁿQ_i(τ, x, y)

∂x^l∂y^n−l

 ≤ Ce^−kτ,

that is, as ε→ 0,

sup

t,x,y

[ eL¹Q₄

(T − t

ε , x, y)  +

 eL²Q₃

(T − t

ε , x, y)  +

 eL²Q₄

(T − t

ε , x, y) ]

=O(1) (3.34)

sup

t,x,y

[Q³( T − t

ε , x, y)  +

Q⁴( T − t

ε , x, y) ]

=O(1). (3.35) Thus,

sup

t,x,y|Fq(t, x, y)| = O(1) by (3.32) and (3.34) and we get that

sup

t,x,y|Zq4(t, x, y)| = O( ε^3/2)

(3.36)

from (3.31).

Therefore, by (3.33), (3.35), and (3.36), we conclude that sup

t,x,y

P^ε(t, x, y)− (

P₀(t, x) +√

εP₁(t, x) + εP₂(t, x, y) + εQ₂

(T − t

ε , x, y)) 

= sup

t,x,y

Z^q4(t, x, y)− ε√ ε

[

(P₃(t, x, y) +√

εP₄(t, x, y)) +

( Q₃

(T − t ε , x, y

) +√

εQ₄

(T − t

ε , x, y))]

=O( ε^3/2)

.

Chapter 4

Down-and-Out Barrier Options

We now introduce the pricing for European down-and-out barrier call options. The kind of option gives the holder the right to buy the underlying asset at the maturity time T by the strike price K if the price of underlying asset never hit the barrier E before time T ; otherwise the option becomes worthless. In the general condition, we assume E < K.

Many of the following operations are essentially the same as we did in Chapter 2, but there are still some distinct results due to the definite setting on the terminal condition, the addition of boundary condition, and the use of underlying asset price in substitution for its logarithm. Here we use the price of underlying asset directly to illustrate the calculation of option pricing, where the results are mentioned in [1]

and the detailed method are referred in [3]. The stochastic differential equations we face are (2.9) and (2.10).

Let P(t, s, y) be the price of European down-and-out barrier call option in this chapter and

L₀ = m(y) ∂

∂y + 1

2β²(y) ∂²

∂y² L₁ = ρβ(y)f (y)s ∂²

∂s∂y − β(y)Λ(y) ∂

∂y L₂ = L_BS(f (y)) = ∂

∂t+1

2f²(y)s² ∂

∂s² + r (

s ∂

∂s − · )

,

where L_BS(f (y)) is Black-Scholes operator calculated at the underlying asset price

and the volatility level f (y), and L = 1

εL₀+ 1

√εL₁+ L₂, then P(t, s, y) is the solution of

LP = 0 with the terminal condition

P(T, s, y) = (s − K)⁺

and the boundary condition

P(t, E, y) = 0.

Expand P(t, s, y) in powers of √

ε, then by the results in Chapter 2, the zero- order termP0 and the first-order term P1 are both independent of y, that is,

P0 =P0(t, s) P1 =P1(t, s).

In this situation, P0(t, s) is the solution of

L_BS(¯σ)P0 = 0 (4.1)

with the terminal condition

P0(t, s) = (s− K)⁺ (4.2)

and the boundary condition

P0(t, E) = 0. (4.3)

Let

t = T − 2

¯

σ²η (4.4)

s = Ke^ζ (4.5)

P0(t, s) = Ke^aη+bζu(η, ζ), (4.6)

then (4.1) becomes

∂u

∂η = ∂²u

∂ζ² + [2b + (k− 1)]∂u

∂ζ +{[

b²+ (k− 1)b − k]

− a} u, where k = 2r/¯σ².

To eliminate ∂u/∂x and u terms, let a =−1

4(k + 1)² b =−1

2(k− 1), then (4.1) reduces to

∂u

∂η = ∂²u

∂ζ² (4.7)

and (4.2) becomes to

u(0, ζ) = (

e^12(k+1)ζ− e^12(k−1)ζ)+

. Let ζ₀ = log(E/K), then (4.3) becomes

u(η, ζ0) = 0.

Now we shall solve u(η, ζ) by the method of image. Consider two solutions of diffusion equation (4.7); one with an initial condition u(0, ζ) in ζ > ζ₀ and another with −u(0, 2ζ0− ζ) in ζ < ζ0. Since (4.7) has no source, suppose the first question has a solution u₁(η, ζ) and the second one has u₂(η, ζ), their antisymmetric initial conditions with respect to the barrier ζ₀ suggest that

u2(η, ζ) =−u1(η, 2ζ0− ζ),

which guarantees that u₁ + u₂ solves both questions if we restrict in either ζ > ζ₀ or ζ < ζ₀ and

(u₁+ u₂)(η, ζ₀) = 0.

So, we can solve u(η, ζ) by defining

u₀(ζ) = (

e^12(k+1)ζ− e^12(k−1)ζ)+

,

redefining

u(0, ζ) = u₀(ζ)− u0(2ζ₀− ζ), (4.8) and solving the reconstructed question of diffusion equation (4.7) with the initial condition (4.8). The solution restricted in ζ > ζ₀ is what we want.

We have an easy way to represent the solution of initial question. Let C_BS(t, s) be the price of European option at the volatility ¯σ with no barrier, then by (4.6),

C_BS(t, s) = Ke^aη+bζu₁(η, ζ).

Hence,

P0(t, s) = Ke^aη+bζ(u₁(η, ζ) + u₂(η, ζ))

= Ke^aη+bζ(u₁(η, ζ)− u1(η, 2ζ₀− ζ)

= C_BS(t, s)−(s E

)1−k

C_BS (

t,E² s

)

= C_BS(t, s)− MCBS(t, s), s > E,

whereM is the mirror operator such that for some given function g(t, s), Mg(t, s) =(s

E )1−k

g (

t,E² s

) .

Next, we considerP1(t, s). Due to the use of underlying asset price in substitution for its logarithm, by slightly altering the result in Chapter ch02, we can express P1(t, s) as the solution of

L_BS(¯σ)P1 =AP0, where

A = V2s² ∂²

∂s² + V₃s³ ∂³

∂s³ V₂ = ρ⟨βfψ^′⟩ −1

2⟨βΛψ^′⟩ V₃ = 1

2ρ⟨βfψ^′⟩, with the terminal condition

P1(T, s) = 0 (4.9)

and the boundary condition

P1(t, E) = 0. (4.10)

Define

F (t, s) =AP0(t, s)

= V₂s²∂²C_BS

∂s² (t, s) + V₃s³∂³C_BS

∂s³ (t, s)

−(s E

)1−k( V₂E⁴

s²

∂²C_BS

∂s² (

t,E² s

)

− V3

E⁶ s³

∂³C_BS

∂s³ (

t,E² s

)

+q₁ (

t,E² s

)) , where

q₁(t, s) = κ₀C_BS(t, s) + κ₁s∂C_BS

∂s (t, s) + κ₂s²∂²C_BS

∂s² (t, s) κ₀ = k(k− 1)[V2− (k + 1)V3]

κ1 = 2kV2− 3k (2r

¯ σ² + 1

) V3

κ₂ =−3(k + 1)V3.

We shall use the method of image again. Since we still face the differential operator L_BS(¯σ), we can apply similar definitions as in (4.4), (4.5),

P1(t, s) = Ke^aη+bζv(η, ζ),

and

F (t, s) = Ke^aη+bζG(η, ζ) to transform L_BS(¯σ) into the diffusion operator.

Then, the question under consideration becomes

∂v

∂η − ∂²v

∂ζ² = G(η, ζ) with the initial condition

v(0, ζ) = 0