Valuing contingent exotic options: a discounted density approach

Σχετικά έγγραφα
Part III - Pricing A Down-And-Out Call Option

P AND P. P : actual probability. P : risk neutral probability. Realtionship: mutual absolute continuity P P. For example:

EE512: Error Control Coding

2 Composition. Invertible Mappings

Probability and Random Processes (Part II)

Partial Differential Equations in Biology The boundary element method. March 26, 2013

Other Test Constructions: Likelihood Ratio & Bayes Tests

Statistical Inference I Locally most powerful tests

Solutions to Exercise Sheet 5

Congruence Classes of Invertible Matrices of Order 3 over F 2

Jesse Maassen and Mark Lundstrom Purdue University November 25, 2013

k A = [k, k]( )[a 1, a 2 ] = [ka 1,ka 2 ] 4For the division of two intervals of confidence in R +

Example Sheet 3 Solutions

SOLUTIONS TO MATH38181 EXTREME VALUES AND FINANCIAL RISK EXAM

CHAPTER 101 FOURIER SERIES FOR PERIODIC FUNCTIONS OF PERIOD

6.1. Dirac Equation. Hamiltonian. Dirac Eq.

ST5224: Advanced Statistical Theory II

Phys460.nb Solution for the t-dependent Schrodinger s equation How did we find the solution? (not required)

Chapter 6: Systems of Linear Differential. be continuous functions on the interval

w o = R 1 p. (1) R = p =. = 1

Areas and Lengths in Polar Coordinates

Mean-Variance Analysis

Lecture 2: Dirac notation and a review of linear algebra Read Sakurai chapter 1, Baym chatper 3

Inverse trigonometric functions & General Solution of Trigonometric Equations

Section 8.3 Trigonometric Equations

Math221: HW# 1 solutions

C.S. 430 Assignment 6, Sample Solutions

Areas and Lengths in Polar Coordinates

6.3 Forecasting ARMA processes

The challenges of non-stable predicates

Approximation of distance between locations on earth given by latitude and longitude

DiracDelta. Notations. Primary definition. Specific values. General characteristics. Traditional name. Traditional notation

3.4 SUM AND DIFFERENCE FORMULAS. NOTE: cos(α+β) cos α + cos β cos(α-β) cos α -cos β

HOMEWORK 4 = G. In order to plot the stress versus the stretch we define a normalized stretch:

Quadratic Expressions

MathCity.org Merging man and maths

Tridiagonal matrices. Gérard MEURANT. October, 2008

ENGR 691/692 Section 66 (Fall 06): Machine Learning Assigned: August 30 Homework 1: Bayesian Decision Theory (solutions) Due: September 13

derivation of the Laplacian from rectangular to spherical coordinates

Homework 3 Solutions

Homework 8 Model Solution Section

Solution Series 9. i=1 x i and i=1 x i.

Forced Pendulum Numerical approach

Reminders: linear functions

CRASH COURSE IN PRECALCULUS

Second Order Partial Differential Equations

Srednicki Chapter 55

SOLUTIONS TO MATH38181 EXTREME VALUES AND FINANCIAL RISK EXAM

SCHOOL OF MATHEMATICAL SCIENCES G11LMA Linear Mathematics Examination Solutions

Arithmetical applications of lagrangian interpolation. Tanguy Rivoal. Institut Fourier CNRS and Université de Grenoble 1

Fractional Colorings and Zykov Products of graphs

Matrices and Determinants

Math 6 SL Probability Distributions Practice Test Mark Scheme

SPECIAL FUNCTIONS and POLYNOMIALS

Finite Field Problems: Solutions

Concrete Mathematics Exercises from 30 September 2016

If we restrict the domain of y = sin x to [ π, π ], the restrict function. y = sin x, π 2 x π 2

Mock Exam 7. 1 Hong Kong Educational Publishing Company. Section A 1. Reference: HKDSE Math M Q2 (a) (1 + kx) n 1M + 1A = (1) =

ANSWERSHEET (TOPIC = DIFFERENTIAL CALCULUS) COLLECTION #2. h 0 h h 0 h h 0 ( ) g k = g 0 + g 1 + g g 2009 =?

Ordinal Arithmetic: Addition, Multiplication, Exponentiation and Limit

MATH423 String Theory Solutions 4. = 0 τ = f(s). (1) dτ ds = dxµ dτ f (s) (2) dτ 2 [f (s)] 2 + dxµ. dτ f (s) (3)

D Alembert s Solution to the Wave Equation

ω ω ω ω ω ω+2 ω ω+2 + ω ω ω ω+2 + ω ω+1 ω ω+2 2 ω ω ω ω ω ω ω ω+1 ω ω2 ω ω2 + ω ω ω2 + ω ω ω ω2 + ω ω+1 ω ω2 + ω ω+1 + ω ω ω ω2 + ω

Commutative Monoids in Intuitionistic Fuzzy Sets

If we restrict the domain of y = sin x to [ π 2, π 2

Aquinas College. Edexcel Mathematical formulae and statistics tables DO NOT WRITE ON THIS BOOKLET

Lecture 21: Properties and robustness of LSE

Calculating the propagation delay of coaxial cable

Chapter 6: Systems of Linear Differential. be continuous functions on the interval

Second Order RLC Filters

The Simply Typed Lambda Calculus

5.4 The Poisson Distribution.

Credit Risk. Finance and Insurance - Stochastic Analysis and Practical Methods Spring School Jena, March 2009

forms This gives Remark 1. How to remember the above formulas: Substituting these into the equation we obtain with

Bayesian statistics. DS GA 1002 Probability and Statistics for Data Science.

CHAPTER 25 SOLVING EQUATIONS BY ITERATIVE METHODS

( y) Partial Differential Equations

Practice Exam 2. Conceptual Questions. 1. State a Basic identity and then verify it. (a) Identity: Solution: One identity is csc(θ) = 1

ΚΥΠΡΙΑΚΗ ΕΤΑΙΡΕΙΑ ΠΛΗΡΟΦΟΡΙΚΗΣ CYPRUS COMPUTER SOCIETY ΠΑΓΚΥΠΡΙΟΣ ΜΑΘΗΤΙΚΟΣ ΔΙΑΓΩΝΙΣΜΟΣ ΠΛΗΡΟΦΟΡΙΚΗΣ 19/5/2007

Section 7.6 Double and Half Angle Formulas

Partial Trace and Partial Transpose

On a four-dimensional hyperbolic manifold with finite volume

Η ΨΥΧΙΑΤΡΙΚΗ - ΨΥΧΟΛΟΓΙΚΗ ΠΡΑΓΜΑΤΟΓΝΩΜΟΣΥΝΗ ΣΤΗΝ ΠΟΙΝΙΚΗ ΔΙΚΗ

2. THEORY OF EQUATIONS. PREVIOUS EAMCET Bits.

Appendix to On the stability of a compressible axisymmetric rotating flow in a pipe. By Z. Rusak & J. H. Lee

Απόκριση σε Μοναδιαία Ωστική Δύναμη (Unit Impulse) Απόκριση σε Δυνάμεις Αυθαίρετα Μεταβαλλόμενες με το Χρόνο. Απόστολος Σ.

Statistics 104: Quantitative Methods for Economics Formula and Theorem Review

DESIGN OF MACHINERY SOLUTION MANUAL h in h 4 0.

A Two-Sided Laplace Inversion Algorithm with Computable Error Bounds and Its Applications in Financial Engineering

ΚΥΠΡΙΑΚΗ ΕΤΑΙΡΕΙΑ ΠΛΗΡΟΦΟΡΙΚΗΣ CYPRUS COMPUTER SOCIETY ΠΑΓΚΥΠΡΙΟΣ ΜΑΘΗΤΙΚΟΣ ΔΙΑΓΩΝΙΣΜΟΣ ΠΛΗΡΟΦΟΡΙΚΗΣ 24/3/2007

Uniform Convergence of Fourier Series Michael Taylor

ECE Spring Prof. David R. Jackson ECE Dept. Notes 2

CHAPTER 48 APPLICATIONS OF MATRICES AND DETERMINANTS

: Ω F F 0 t T P F 0 t T F 0 P Q. Merton 1974 XT T X T XT. T t. V t t X d T = XT [V t/t ]. τ 0 < τ < X d T = XT I {V τ T } δt XT I {V τ<t } I A

Instruction Execution Times

Problem Set 9 Solutions. θ + 1. θ 2 + cotθ ( ) sinθ e iφ is an eigenfunction of the ˆ L 2 operator. / θ 2. φ 2. sin 2 θ φ 2. ( ) = e iφ. = e iφ cosθ.

Trigonometric Formula Sheet

= λ 1 1 e. = λ 1 =12. has the properties e 1. e 3,V(Y

Dynamic types, Lambda calculus machines Section and Practice Problems Apr 21 22, 2016

1 String with massive end-points

Transcript:

Title Valuing contingent exotic options: a discounted density approach Author(s) Yang, H; Gerber, HU; Shiu, ESW Citation The 2012 International Conference on Actuarial Science and Risk Management (ASRM 2012), Xiamen, China, 24-26 June 2012. Issued Date 2012 URL http://hdl.handle.net/10722/165735 Rights This work is licensed under a Creative Commons Attribution- NonCommercial-NoDerivatives 4.0 International License.

Valuing contingent exotic options: a discounted density approach Department of Statistics and Actuarial Science The University of Hong Kong Hong Kong Based on a paper with Hans Gerber and Elias Shiu

Brownian motion (Wiener process) X (t) = µt + σw (t) {W (t)}: notation: standard Wiener process D = σ2 2 running minimum: running maximum: m(t) = min 0 s t X (s) M(t) = max 0 s t X (s)

2 Exponential stopping of Brownian motion τ: exponential random variable independent of {X (t)} f τ (t) = λe λt, t > 0 We are interested in X (τ), M(τ),... δ: force of interest used for discounting

Three discounted density functions f δ X (τ) (x) = 0 e δt f X (t) (x)f τ (t)dt f δ M(τ) (m) = 0 e δt f M(t) (m)f τ (t)dt f δ X (τ),m(τ) (x, m) = 0 e δt f X (t),m(t) (x, m)f τ (t)dt

α < 0 and β > 0 solutions of the quadratic equation Dξ 2 + µξ (λ + δ) = 0 1). f δ X (τ),m(τ) (x, m) = λ D e αx (β α)m = λ D eα(m x) βm, < x m, m 0 2). f δ M(τ) (m) = λ λ+δ βe βm, m 0 3). f δ X (τ) (x) = { λ D(β α) e αx, if x < 0, λ D(β α) e βx, if x > 0.

f δ X (τ),m(τ) X (τ) (x, z) = λ D e βx (β α)z, z max( x, 0). f δ M(τ),M(τ) X (τ) (y, z) = λ D e βy+αz, y 0, z 0. f δ M(τ) X (τ) (z) = λ βd eαz = λ λ + δ ( α)eαz, z 0.

m(t) = max{ X (s); 0 s t} If E[e δτ g(x (τ), M(τ))] = h(α, β), then we can translate it to E[e δτ g( X (τ), m(τ))] = h( β, α).

f δ X (τ),m(τ) (x, y) = λ D e βx+(β α)y, y min(x, 0), f δ X (τ),x (τ) m(τ) (x, z) = λ D e αx (β α)z, z max(x, 0), f δ m(τ),x (τ) m(τ) (y, z) = λ D e αy βz, y 0, z 0, fm(τ) δ (y) = λ βd e αy = f δ X (τ) m(τ) (z) = λ αd e βz = λ λ + δ ( α)e αy, y 0, λ λ + δ βe βz, z 0.

Factorization formula If τ is exponential with mean 1/λ, then the following factorization formula holds, E[e δτ g τ (X )] = E[e δτ ] E[g τ (X )], where τ is an exponential random variable with mean 1/(λ + δ) and independent of X. Remarks (i) E[e δτ ] = λ λ+δ. (ii) The condition δ > 0 can be replaced by the condition δ > λ.

3. Financial applications S(t): stock price S(t) = S(0)e X (t) = S(0)e µt+σw (t), t 0 a contingent option provides a payoff at time τ Example: τ: time of death GMDB (Guaranteed Minimum Death Benefits)

3. Lookback options Many equity-indexed annuities credit interest using a high water mark method or a low water mark method

Out-of-the-money fixed strike lookback call option Payoff: Time-0 value k [S(0)e M(τ) K] + [S(0)e y K]fM(τ) δ (y)dy = λ [S(0)βe (β 1)k λ + δ β 1 [ ] S(0) β =. λ λ + δ Another expression for the option value [ ] λ K S(0) β. D αβ(β 1) K K β 1 K Ke βk ]

In-the-money fixed strike lookback call option Payoff Rewriting as max(h, S(0)e M(τ) ) K. H K + [S(0)e M(τ) H] + Time-0 value { λ H K + H λ + δ β 1 [ S(0) H ] β }.

Floating strike lookback put option Payoff where H S(0). Time-0 value { λ H + H λ + δ β 1 max(h, max S(t)) S(τ), (1) 0 t τ [ S(0) H ] β } E[e δτ S(τ)].

Floating strike lookback put option Special case: H = S(0), the time-0 value λ β λ + δ β 1 S(0) E[e δτ S(τ)] = 1 α α E[e δτ S(τ)] E[e δτ S(τ)] = 1 α E[e δτ S(τ)]. (2) This result can be reformulated as E[e δτ max S(t)] = 0 t τ ( 1 α + 1 ) E[e δτ S(τ)].

Floating strike lookback put option Milevsky and Posner (2001) have evaluated (1) with a risk-neutral stock price process and H = S(0). They also assume that the stock pays dividends continuously at a rate proportional to its price. With l denoting the dividend yield rate, δ = r, and µ = r D l, the RHS of (2) is 2D (r D l) + (r D l) 2 + 4D(λ + r) S(0) λ λ + l. Although it seems rather different from formula (38) on page 117 of Milevsky and Posner (2001), they are the same.

Fractional floating strike lookback put option Payoff Notice [γ max 0 t τ S(t) S(τ)] + = S(0)[γe M(τ) e X (τ) ] +. [γe M(τ) e X (τ) ] + = e M(τ) [γ e X (τ) M(τ) ] +

Fractional floating strike lookback put option Hence E[e δτ e M(τ) [γ e X (τ) M(τ) ] + ] = e y [γ e z ] + fm(τ),m(τ) X δ (τ)(y, z)dydz 0 0 = λ [ ][ ] e y e βy dy [γ e z ] + e αz dz D 0 = λ 1 D β 1 α(1 α) = γ 1 α λ β λ + δ (1 α)(β 1) = γ 1 α 1 α E[e δτ e X (τ) ]. γ 1 α 0

Fractional floating strike lookback put option This can be rewritten as E[e δτ [γe M(τ) e X (τ) ] + ] = γ 1 α E[e δτ (e M(τ) e X (τ) )]. Time-0 value E[e δτ [γ max 0 t τ S(t) S(τ)] +] = γ1 α α E[e δτ S(τ)],

Out-of-the-money fixed strike lookback put option Payoff [K S(0)e m(τ) ] +, Time-0 value k [K S(0)e y ]fm(τ) δ (y)dy = λ [ ] K K α. λ + δ 1 α S(0)

In-the-money fixed strike lookback put option Payoff K min(h, S(0)e m(τ) ) = K H + [H S(0)e m(τ) ] +, Time-0 value { λ K H + H [ ] H α }. λ + δ 1 α S(0)

Floating strike lookback call option Payoff where 0 < H S(0). Time-0 value E[e δτ S(τ)] + S(τ) min(h, min 0 t τ S(t)), λ { H + λ + δ H [ ] H α }. 1 α S(0) In the special case where H = S(0), the time-0 value E[e δτ S(τ)] λ λ + δ α 1 α S(0) = E[e δτ S(τ)] β 1 β E[e δτ S(τ)] = 1 β E[e δτ S(τ)]. This result can be reformulated as ( )

Fractional floating strike lookback call option Payoff [S(τ) γ min 0 t τ S(t)] + = S(0)[e X (τ) γe m(τ) ] +. = S(0)e m(τ) [e X (τ) m(τ) γ] +

Fractional floating strike lookback call option Its expected discounted value is S(0) times the following expectation E[e δτ e m(τ) [e X (τ) m(τ) γ] + ] = λ [ 0 ][ ] e y e αy dy [e z γ] + e βz dz D 0 = λ 1 γ 1 β D 1 α β(β 1) 1 λ α = γ β 1 λ + δ (1 α)(β 1) 1 1 = γ β 1 β E[e δτ e X (τ) ].

Fractional floating strike lookback call option This can be rewritten as E[e δτ [e X (τ) γe m(τ) ] + ] = γ (β 1) E[e δτ (e X (τ) e m(τ) )]. We have E[e δτ [S(τ) γ min 0 t τ S(t)] +] = 1 βγ β 1 E[e δτ S(τ)].

High-low option Payoff max(h, max S(t)) min(h, min S(t)), 0 t τ 0 t τ where 0 < H S(0) H. Time-0 value { λ H + H [ ] S(0) β H + H [ ] H α }. λ + δ β 1 H 1 α S(0) In the special case where H = S(0) = H, time-0 value λ β α S(0) λ + δ (β 1)(1 α) = β α αβ E[e δτ S(τ)]. This can be rewritten as ( 1 α + 1 ) E[e δτ S(τ)], β

Barrier options A barrier option is an option whose payoff depends on whether or not the price of the underlying asset has reached a predetermined level or barrier. Knock-out options are those which go out of existence if the asset price reaches the barrier, and knock-in options are those which come into existence if the barrier is reached.

Parity relation Knock-out option + Knock-in option = Ordinary option. Notation: L denotes the barrier and l = ln[l/s(0)]

Up-and-out and up-and-in options (L > S(0) (l > 0)) Payoffs I ([max0 t τ S(t)]<L)b(S(τ)) = I (M(τ)<l) b(s(0)e X (τ) ) I ([max0 t τ S(t)] L)b(S(τ)) = I (M(τ) l) b(s(0)e X (τ) )

The expected discounted values Up-and-out Up-and-in λ D 0 = λ D l [ y I (y<l) b(s(0)e x )fx δ (τ),m(τ) ]dy (x, y)dx [ y l 0 b(s(0)e x )e αx dx ] e (β α)y dy [ y ] b(s(0)e x )e αx dx e (β α)y dy;

Down-and-out and down-and-in options (0 < L < S(0) (l < 0)) Payoffs I ([min0 t τ S(t)]>L)b(S(τ)) = I (m(τ)>l) b(s(0)e X (τ) ) I ([min0 t τ S(t)] L)b(S(τ)) = I (m(τ) l) b(s(0)e X (τ) )

The expected discounted values λ D 0 l [ y ] b(s(0)e x )e βx dx e (β α)y dy λ l [ ] b(s(0)e x )e βx dx e (β α)y dy, D y

Notation A 1 (n) = λ S(0) n D (n α)(β n), A 2 (n) = λ L n [ ] S(0) β, D (n α)(β n) L A 3 (n) = λ L n [ ] L α, D (n α)(β n) S(0) A 4 = λ K n [ ] K α = κk n [ ] K α, D (n α)(β α) S(0) n α S(0)

Notation A 5 = λ K n α L α [ ] S(0) β = κk n α L α [ ] S(0) β, D (n α)(β α) L n α L A 6 = λ K n [ ] S(0) β = κk n [ ] S(0) β, D (β n)(β α) K β n K A 7 = λ K (β n) L β [ ] L α = κk (β n) L β [ ] L α, D (β n)(β α) S(0) β n S(0) A 8 = λ [ ] K K α [ ] κk K α =, D α(1 α)(β α) S(0) α(1 α) S(0)

Notation A 9 = λ K 1 α L α [ ] S(0) β D α(1 α)(β α) L = κk 1 α L α [ ] S(0) β, α(1 α) L A 10 = λ [ ] K S(0) β = D β(β 1)(β α) K A 11 = λ K (β 1) L β [ ] L α D β(β 1)(β α) S(0) = κk (β 1) L β [ ] L α. β(β 1) S(0) κk β(β 1) [ ] S(0) β, K

Up-and-out all-or-nothing call option The option value is 0, if L < K, λ l D 0 [ y k S(0)n e nx e αx dx]e (β α)y dy, if L K and S(0) > K, λ l D k [ y k S(0)n e nx e αx dx]e (β α)y dy, if L K and S(0) K 0, if L < K, = A 1 (n) A 2 (n) A 4 + A 5, if L K and S(0) > K, A 6 A 2 (n) + A 5, if L K and S(0) K.

Up-and-out all-or-nothing put option The option value is = λ l D 0 [ y λ D S(0)n e nx e αx dx]e (β α)y dy, if L < K, l 0 [ k S(0)n e nx e αx dx]e (β α)y dy, if L K&S(0) > K λ D { k 0 [ y S(0)n e nx e αx dx]e (β α)y dy + l k [ k S(0)n e nx e αx dx]e (β α)y dy}, A 1 (n) A 2 (n), if L < K, A 4 A 5, if L K and S(0) > K, A 1 (n) A 5 A 6, if L K and S(0) K. if L K&S(0) K

up-and-out option with payoff S(τ) n λ D l 0 [ y ] S(0) n e nx e αx dx e (β α)y dy = A 1 (n) A 2 (n). This is the sum of the value of the up-and-out all-or-nothing put option and the value of the up-and-out all-or-nothing call option.

Up-and-out call option The value is 0, if L < K, A 1 (1) A 2 (1) A 1 (0)K +A 2 (0)K + A 8 A 9, if L K and S(0) > K, A 2 (0)K + A 10 A 2 (1) A 9, if L K and S(0) K.

Up-and-out put option The value is A 1 (0)K A 2 (0)K A 1 (1) + A 2 (1), if L < K, A 8 A 9, if L K and S(0) > K, A 1 (0)K A 1 (1) + A 10 A 9, if L K and S(0) K.

Double barrier option Payoff: π(s(τ))i {a < m(τ), M(τ) < b}

Double barrier option Using a formula from B & S (2002), we have = E[b(S(0)e X (τ) )I {a < m(τ), M(τ) < b}] [ κ 0 0 Ξ Ξ 1 Ξ b(s(0)e z )e αz dz + Ξ 1 b(s(0)e z )e βz dz b a b +Ξ b(s(0)e z )e βz dz + Ξ 1 b(s(0)e z )e αz dz 0 0 b b ] Υ 1 b(s(0)e z )e αz dz Υ b(s(0)e z )e βz dz, a a a where Ξ = e 1 2 (b a)(β α) and Υ = e 1 2 (a+b)(β α).

Double barrier option When δ > 0, we have E[e δτ b(s(0)e X (τ) )I {a < m(τ), M(τ) < b}] = λ e (λ+δ)t b(s(0)e X (t) )I {a < m(t), M(t) < b }dt 0 [ κ 0 0 = Ξ Ξ 1 Ξ b(s(0)e z )e αz dz + Ξ 1 b(s(0)e z )e βz dz b a b +Ξ b(s(0)e z )e βz dz + Ξ 1 b(s(0)e z )e αz dz 0 0 b b ] Υ 1 b(s(0)e z )e αz dz Υ b(s(0)e z )e βz dz. a a a

Several stocks µ X(t) = (X 1 (t), X 2 (t),, X n (t)) n-dimensional Brownian motion. the mean vector C the covariance matrix of X(1) h g t (X) a real-valued functional of the process up to time t. an n-dimensional vector of real numbers

E[e δτ e h X(τ) g τ (X)] = E[e δ(h)τ g τ (X); h], (3) where δ(h) = δ ln[m X(1) (h)] = δ h µ 1 2 h Ch.

Proof of (3) Conditioning on τ = t, the LHS (3) is 0 e δt E[e h X(t) g t (X)]f τ (t)dt. By the factorization formula in the method of Esscher transforms, the expectation inside the integrand can be written as the product of two expectations, Hence 0 E[e h X(t) ] E[g t (X); h] = [M X(1) (h)] t E[g t (X); h]. e δt E[e h X(t) g t (X)]f τ (t)dt = 0 e δ(h)t E[g t (X); h]f τ (t)dt.

Application of (3) k q t (k X) n-dimensional vector of real numbers real-valued functional of the process up to time t E[e δτ e h X(τ) q τ (k X)] = E[e δ(h)τ q τ (k X); h]. The quadratic equation becomes 1 2 Var[k X(1); h]ξ 2 + E[k X(1); h]ξ [λ + δ(h)] = 1 2 k Ckξ 2 + k (µ + Ch)ξ (λ + δ h µ 1 2 h Ch)

Special case: n = 2 S 1 (t) = S 1 (0)e X 1(t) and S 2 (t) = S 2 (0)e X 2(t) µ = (µ 1, µ 2 ) ( ) σ 2 C = 1 ρσ 1 σ 2 ρσ 1 σ 2 σ2 2

Margrabe option Payoff: [S 1 (τ) S 2 (τ)] +. (4) If we rewrite (4) as e X 2(τ) [S 1 (0)e X 1(τ) X 2 (τ) S 2 (0)] +,

[ ] E[e δτ [S 1 (τ) S 2 (τ)] + S 1 (0) < S 2 (0)] = κ S 2 (0) S1 (0) β β (β. 1) S 2 (0) Here, κ = λ D (β α ), D = 1 2 Var[X 1(1) X 2 (1)] = 1 2 (σ2 1 + σ 2 2 2ρσ 1 σ 2 ), and α < 0 and β > 0 are the zeros of D ξ 2 + (µ 1 µ 2 + ρσ 1 σ 2 σ 2 2)ξ (λ + δ µ 2 1 2 σ2 2) = ln[m X(1) ((ξ, 1 ξ) )] (λ + δ).

If we write (4) as Here, e X 1(τ) [S 1 (0) S 2 (0)e X 2(τ) X 1 (τ) ] +, E[e δτ [S 1 (τ) S 2 (τ)] + S 1 (0) < S 2 (0)] κ [ ] S 1 (0) S1 (0) α = α (1 α. ) S 2 (0) κ = λ D (β α ), D = 1 2 Var[X 2(1) X 1 (1)] = D, and α < 0 and β > 0 are the zeros of ln[m X(1) ((1 ξ, ξ) )] (λ + δ).

Hence α = 1 β and β = 1 α. Thus, κ = κ