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Equilibrium pricing of currency options under a discontinuous model in a two-country economy

  • Yu Xing EMAIL logo and Xiaoping Yang
Published/Copyright: November 6, 2015

Abstract

(Bakshi, G., and Z. Chen. 1997. “Equilibrium Valuation of Foreign Exchange Claims.” Journal of Finance 52: 799–826) studied equilibrium valuation for foreign exchange claims in the setting of the two-country Lucas-type economy. In Bakshi and Chen (1997), they assumed the money supplies follow two-factor stochastic volatility processes. Based on their model, we add two independent Poisson-type jumps, respectively into the process of money supply in each country. By solving a partial integro-differential equation (PIDE) for currency options, we get closed-form solutions of call currency option prices. Our model is a generalization of Bakshi and Chen (1997), and can contain a class of stochastic-volatility jump-diffusion (SVJD) models as special cases.


Corresponding author: Yu Xing, School of Finance, Nanjing Audit University, Nanjing 211815, PR China

Appendix A. Proofs of Theorems

Proof of Theorem 1. As pointed out in formula (A3) in (Bakshi and Chen 1997), the risk premium on any contingent claim is determined by the Euler equation by solving the agent’s first-order condition. If the contingent claim is a bond, similarly we have

Et[eρΔtm(t)m(t+Δt)(ΔB(t)B(t)R(t)Δt)]=0.

Where

m(t)m(t+Δt)=1(Δm(t)m(t))c+(Δm(t)m(t))c2+(11+km1)ΔQm(t)+o(Δt).

Notice that B(t, τ; Y, Z) is a function related with variable t and Y, Z, there is no jump in the process of B(t, τ; Y, Z). We write

dB=Btdt+BYdY+BZdZ+122BY2(dY)2+122BZ2(dZ)2+2BYZdYdZ,

then we can get the discredited form ΔB. Combine the above equations, let Δt→0 and we obtain the PDE of the price of B(t, τ) as

Bτ+BY(θykyYσ1σyY)+BZ(θZkZZσ2σZZ)+122BY2σy2Y+122BZ2σZ2Z=RB,

Solve the above PDE with the terminal condition B(t, 0; Y, Z)=1, we can get a closed form solution of B(t, τ; Y, Z) expressed in Formula (10).■

Proof of Theorem 2. Applying Ito’s lemma with respect to C(t), we get:

dC(t)=Ctdt+Cee(dee)c+CYY(dYY)c+CZZ(dZZ)c+122Ce2e2(dee)c2+122CY2Y2(dYY)c2+122CZ2Z2(dZZ)c2+2CYeYe(dYY)c(dee)c+2CZeZe(dZZ)c(dee)c+[C(t,τ;Y,Z,e(1+km))C(t,τ;Y,Z,e)]dQm+[C(t,τ;Y,Z,e(11+km))C(t,τ;Y,Z,e)]dQm.

We write

dC(t)C(t)=(dC(t)C(t))c+kCdQm(t)+kCdQm(t),

kC=C(t,τ;Y,Z,e(1+km))C(t,τ;Y,Z,e)C(t,τ;Y,Z,e).

Substitute the discredited form ΔC into the following equation

Et[eρΔtm(t)m(t+Δt)(ΔC(t)C(t)R(t)Δt)]=0,

then let Δt→0 and we will obtain the PIDE of (13).■

Proof of Theorem 3. First, we let L(t)=ln[e(t)], then the variable e in (13) is replaced by the variable L as follows:

(A.1)CτRC+(μeσ1σeYσ2σeZ12σe2Y12σe2Z)CL+(θykyYσ1σyY)CY+(θzkzZσ2σzZ)CZ+12(σe2Y+σe2Z)2CL2+12σy2Y2CY2+12σz2Z2CZ2+σyσeY2CYL+σzσeZ2CZL+λmE[11+km(C(t,τ;Y,Z,L+ln(1+km))C(t,τ;Y,Z,L))]+λmE[C(t,τ;Y,Z,Lln(1+km))C(t,τ;Y,Z,L)]=0. (A.1)

Similar to (Heston 1993) and (Bakshi and Madan 2000), by analogy with the Black-Scholes formula, we choose a solution as the following form

(A.2)C(t,τ;Y,Z,L)=eLB(t,τ;Y,Z)Π1(t,τ;Y,Z,L)KB(t,τ;Y,Z)Π2(t,τ;Y,Z,L), (A.2)

Denote Π1^=BΠ1 and Π2^=BΠ2, then insert (A.2) into (A.1), we obtain the following two PIDEs:

(A.3)Π1^τRΠ1^+(μeσ1σeYσ2σeZ12σe2Y12σe2Z)(Π1^+Π1^L)+(θykyYσ1σyY)Π1^Y+(θzkzZσ2σzZ)Π1^Z+12σy2Y2Π1^Y2+12σz2Z2Π1^Z2+12(σe2Y+σe2Z)(Π1^+2Π1^L+2Π1^L2)+σyσeY(Π1^Y+2Π1^YL)+σzσeZ(Π1^Z+2Π1^ZL)+λmE[Π1^(t,τ;Y,Z,L+ln(1+km))11+kmΠ1^(t,τ;Y,Z,L)]+λmE[11+kmΠ1^(t,τ;Y,Z,Lln(1+km))Π1^(t,τ;Y,Z,L)]=0, (A.3)
(A.4)Π2^τRΠ2^+(μeσ1σeYσ2σeZ12σe2Y12σe2Z)Π2^L+(θykyYσ1σyY)Π2^Y+(θzkzZσ2σzZ)Π2^Z+12σy2Y2Π2^Y2+12σz2Z2Π2^Z2+12(σe2Y+σe2Z)2Π2^L2+σyσeY2Π2^YL+σzσeZ2Π2^ZL+λmE[11+km[Π2^(t,τ;Y,Z,L+ln(1+km))Π2^(t,τ;Y,Z,L)]]+λmE[Π2^(t,τ;Y,Z,Lln(1+km))Π2^(t,τ;Y,Z,L)]=0. (A.4)

According to the definition of an option, Π1 and Π2 are subject to the terminal condition:

(A.5)Πj(t,τ;Y,Z,L)=Prob(LlnK),j=1,2. (A.5)

Although it is difficult to compute Πj directly, we can compute the characteristic function fj of Πj. Let f1^=Bf1 and f2^=Bf2, with the F-K theorem, we get the PIDE of fj^ as follows:

(A.6)f1^τRf1^+(μeσ1σeYσ2σeZ12σe2Y12σe2Z)(f1^+f1^L)+(θykyYσ1σyY)f1^Y+(θzkzZσ2σzZ)f1^Z+12σy2Y2f1^Y2+12σz2Z2f1^Z2+12(σe2Y+σe2Z)(f1^+2f1^L+2f1^L2)+σyσeY(f1^Y+2f1^YL)+σzσeZ(f1^Z+2f1^ZL)+λmE[f1^(t,τ;Y,Z,L+ln(1+km))11+kmf1^(t,τ;Y,Z,L)]+λmE[11+kmf1^(t,τ;Y,Z,Lln(1+km))f1^(t,τ;Y,Z,L)]=0, (A.6)

with the terminal condition f1^(t,0,Y,Z,L,ϕ)=E(eiϕL),

and,

(A.7)f2^τRf2^+(μeσ1σeYσ2σeZ12σe2Y12σe2Z)f2^L+(θykyYσ1σyY)f2^Y+(θzkzZσ2σzZ)f2^Z+12σy2Y2f2^Y2+12σz2Z2f2^Z2+12(σe2Y+σe2Z)2f2^L2+σyσeY2f2^YL+σzσeZ2f2^ZL+λmE[11+km[f2^(t,τ;Y,Z,L+ln(1+km))f2^(t,τ;Y,Z,L)]]+λmE[f2^(t,τ;Y,Z,Lln(1+km))f2^(t,τ;Y,Z,L)]=0. (A.7)

with the terminal condition f2^(t,0,Y,Z,L,ϕ)=E(eiϕL).

Since the coefficients are linear with respect to V and V*, we follow the idea of (Heston 1993; Bakshi and Chen 1997), combine the terminal condition, and choose the form of f1^ is:

(A.8)f1^(t,τ;Y,Z,L,ϕ)=exp[u(τ)+X(τ)Y+X(τ)Z+iϕL]. (A.8)

Inserting (A.8) into (A.6), we can get three ODEs:

X(τ)τ+(iϕσyσekyσyσ1)X(τ)+12σy2X2(τ)+(1+iϕ)(η1η1+σ12σ12+12iϕσe2)η1+σ12=0,X(τ)τ+(iϕσzσekzσzσ2)X(τ)+12σz2X2(τ)+(1+iϕ)(η2η2+σ22σ22+12iϕσe2)η2+σ22=0,u(τ)τ+θyX(τ)+θzX(τ)+[(μmμm)(λmkm¯λmkm¯)]iϕρμm+λmkm¯+λmE[eiϕln(1+km)1]+λmE[e(1+iϕ)ln(1+km)1]=0, with the initial conditions X(0)=X*(0)=u(0)=0.

We solve the above ODEs to obtain (16). Using the same approach, we choose:

(A.9)f2^(t,τ;Y,Z,L,ϕ)=exp[V(τ)+W(τ)Y+W(τ)Z+iϕL]. (A.9)

Inserting (A.9) into (A.7), we can get three ODEs:

W(τ)τ+iϕσyσekyσ1σy)W(τ)+12σy2W2(τ)+iϕ(η1η1+σ12σ1212σe2)η1+σ1212ϕ2σe2=0,W(τ)τ+(iϕσzσekzσ2σz)W(τ)+12σz2W2(τ)+iϕ(η2η2+σ22σ2212σe2)η2+σ2212ϕ2σe2=0,V(τ)τ+θyW(τ)+θzW(τ)+[(μmμm)(λmkm¯λmkm¯)]iϕρμm+λmkm+λmE[e(1iϕ)ln(1+km)1]+λmE[eiϕln(1+km)1]=0,

with the initial conditions W(0)=W*(0)=V(0)=0.

We solve the above ODEs to arrive at (17).     ■

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The online version of this article (DOI: 10.1515/snde-2015-0001) offers supplementary material, available to authorized users.


Published Online: 2015-11-6
Published in Print: 2016-4-1

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