Price-level instability and international monetary policy coordination

Solving the optimization problems of a young lender and a young borrower

a. For a young lender: A young lender in country 1 at t seeks to maximize lnc1t1+βlnc2t+11 subject to (5) and (6). The Lagrangian is

χ=lnc1t1+βlnc2t+11+κ1[w−(c1t1+Lt11−θ1+bt1+Lt21−θ2+bt2)]+κ2{[(1−θ1)Rt+11+θ1pt1pt+11]Lt11−θ1+bt1rt+11+[(1−θ2)Rt+12+θ2pt2pt+12]Lt21−θ1+bt2rt+12−c2t+11},

where κ₁ and κ₂ are Lagrangian multipliers. I use (7) to rewrite the Lagrangian as follows

χ=lnc1t1+βlnc2t+11+κ1[w−(c1t1+Lt11−θ1+bt1+Lt21−θ2+bt2)]+κ2[rt+1(Lt11−θ1+bt1+Lt21−θ1+bt2)−c2t+11].

Taking derivatives of χ with respect to c1t1,c2t+11,(Lt11−θ1+bt1+Lt21−θ1+bt2),κ₁ and κ₂ gives

∂χ/∂c1t1=1/c1t1−κ1=0,∂χ/∂c2t+11=β/c2t+11−κ2=0,∂χ/∂(Lt11−θ1+bt1+Lt21−θ1+bt2)=−κ1+κ2rt+1=0,∂χ/∂κ1=w−(c1t1+Lt11−θ1+bt1+Lt21−θ2+bt2)=0,∂χ/∂κ2=rt+1(Lt11−θ1+bt1+Lt21−θ1+bt2)−c2t+11=0.

These then lead to

κ1/κ2=c2t+11/(βc1t1)=rt+1, c2t+11=(Lt11−θ1+bt1+Lt21−θ1+bt2)c2t+11/(βc1t1),c1t1=w−[(Lt11−θ1+bt1)+(Lt21−θ1+bt2)]c2t+11/(βc1t1).

Hence,

βc1t1=Lt11−θ1+bt1+Lt21−θ1+bt2,(1+β)c1t1=w.

As a result, the savings demand function σ=βw/(1+β) is obtained, where ς≡Lt1/(1−θ1)+bt1+Lt2/(1−θ1)+bt2.

b. For a young borrower: A young borrower in country 1 at t seeks to maximize lnC1t1+βlnC2t+12 subject to (11) and (12). The Lagrangian is

Γ=lnC1t1+βlnC2t+12+ν1(lt1−C1t1)+ν2(y−lt1Rt+11−C2t+11),

where ν₁ and ν₂ are Lagrangian multipliers. Taking derivatives of Γ with respect to C1t1,C2t+11,lt1,ν₁ and ν₁ gives

∂Γ/∂C1t1=1/C1t1−ν1=0,∂Γ/∂C2t+11=β/C2t+11−ν2=0,∂Γ/∂lt1=ν1−ν2Rt+11=0,∂Γ/∂ν1=lt1−C1t1=0,∂Γ/∂ν2=y−lt1Rt+11−C2t+11=0.

It can be deduced from the first three equations that

C1t1/C2t+11=1/(βRt+11),

while the last two equations give

C1t1/C2t+11=lt1/(y−lt1Rt+11).

As a result, the loan demand function lt1=y/(1+β)Rt+11 is obtained. A similar reasoning for country 2 gives lt2=y/(1+β)Rt+12.

Proof of Proposition 3

Let κ₁, κ₂ be the eigenvalues of the Jacobian matrix J, then κ₁, κ₂ are two roots of the characteristic polynomial p(κ)=κ²–Tκ+D=0. The trace, T, and the determinant, D, of matrix J are evaluated as follows

T=ΩR2[N1α1y/(1+β)R1+N2α2y/(1+β)R2+ϕ(θ1/σ1+θ2/σ2)]>0,D=ΩR2ϕ[(1−θ1)R1+θ1/σ1]>0.

It is immediate that p(–1)=1+T+D>0, while

p(1)=1+D−T=Ω[N1α1y/(1+β)(R1)2θ2/σ2R2+N2α2y/(1+β)(R2)2θ1/σ1R1+ϕθ1/σ1R1θ2/σ2R21−θ1+θ1/σ1R1+ϕ(1−θ1)R1−θ2/σ2R2−N1α1y/(1+β)R1R2−N2α2y/(1+β)(R2)2]=Ω[ϕ(θ1/σ1R1θ2/σ2R21−θ1+θ1/σ1R1+(1−θ1)R1−θ2/σ2R2)−N1α1y/(1+β)(R1)2(R1R2−θ2/σ2R21−θ1+θ1/σ1R1)−N2α2y/(1+β)(R2)2(1−θ1/σ1R11−θ1+θ1/σ1R1)]=Ω[ϕ(1−θ1)(1−θ2)1−θ1+θ1/σ1R1−N1α1y/(1+β)(R1)21−θ21−θ1+θ1/σ1R1−N2α2y/(1+β)(R2)21−θ11−θ1+θ1/σ1R1]=Ω(1−θ1)(1−θ2)1−θ1+θ1/σ1R1[ϕ−N1α1y/(1+β)(1−θ1)(R1)2−N2α2y/(1+β)(1−θ2)(R2)2].

Recall that f′(RH1)>0 and f′(RL1)<0. It is immediate that 1+D>T, or p(1)>0 at R1=RH1 and 1+D<T, or p(1)<0 at R1=RL1. It is easy to verify that –1, κ₁<1<κ₂ at RL1, so the stationary equilibrium at RL1 is a saddle. In order to determine whether the stationary equilibrium at RH1 is unstable, I consider

(44)D−1=ΩR2{ϕ[(1−θ1)R1+θ1/σ1]−N1α1y/(1+β)R1(θ2/σ2)+N2α2y/(1+β)R2(θ1/σ1)+ϕ(θ1/σ1)(θ2/σ2)(1−θ1)R1+θ1/σ1}=ΩR2{ϕ[(1−θ1)R1+θ1/σ1−(θ1/σ1)(θ2/σ2)(1−θ1)R1+θ1/σ1]−N1α1y/(1+β)R1(θ2/σ2)+N2α2y/(1+β)R2(θ1/σ1)(1−θ1)R1+θ1/σ1}=ΩR2{ϕ[(1−θ1)R1+(1−θ2)R2(θ1/σ1)(1−θ1)R1+θ1/σ1]−[N1α1y/(1+β)R1(θ2/σ2)+N2α2y/(1+β)R2(θ1/σ1)][(1−θ1)R1+θ1/σ1]−1}. (44)

It is then deduced from (34) that

D−1>ΩR2{[N1α1y/(1+β)(1−θ1)(R1)2+N2α2y/(1+β)(1−θ2)(R2)2][(1−θ1)R1+(1−θ2)R2θ1/σ1(1−θ1)R1+θ1/σ1]−[N1α1y/(1+β)R1(θ2/σ2)+N2α2y/(1+β)R2(θ1/σ1)][(1−θ1)R1+θ1/σ1]−1}=ΩR2{N1α1y/(1+β)R1[1+(1−θ2)R2θ1/σ1(1−θ1)R1−θ2/σ2(1−θ2)R2+θ2/σ2]+N2α2y/(1+β)R2[1+(1−θ1)R1θ2/σ2(1−θ2)R2−θ1/σ1(1−θ1)R1+θ1/σ1]}=ΩR2[N1α1y/(1+β)(1−θ1)(R1)2(1−θ2)R2+N2α2y/(1+β)(1−θ2)(R2)2(1−θ1)R1]>0

or D>1. Recall that T>0, it is easy to verify that 1<κ₁<κ₂, or that the steady state at RH1 is unstable. Note that the variables under consideration, Rtj, are jump variables. Hence, the steady state at RH1 is determinate.

Proof of Proposition 5

Let κ₁, κ₂ be the eigenvalues of the Jacobian matrix J, then κ₁, κ₂ are two roots of the characteristic polynomial p(κ)=κ²–Tκ+D=0. The trace, T, and the determinant, D, of matrix J are evaluated as follows:

T=Ψ[(1−θ1+(θ1−θ2)/σ1R1)N2α2y/(1+β)(R2)2+(1−θ2)(N1α1y/(1+β)(R1)2+ϕθ1/σ1R1)],D=Ψϕ(1−θ1+θ1/σ1R1)(1−θ2).

Now, I consider two cases:

(i) If θ₁>θ₂, then D>0 and T>0. It is immediate that p(–1)=1+T+D>0. I then consider

p(1)=1+D−T=Ψ{ϕ(1−θ1+θ1/σ1R1)(1−θ2)+(θ1−θ2)/σ1R1N2α2y/(1+β)(R2)2−(1−θ2)(N1α1y/(1+β)(R1)2+ϕθ1/σ1R1)−(1−θ1+(θ1−θ2)/σ1R1)N2α2y/(1+β)(R2)2}=Ψ[ϕ(1−θ1)(1−θ2)−(1−θ1)N2α2y/(1+β)(R2)2−(1−θ2)N1α1y/(1+β)(R1)2]=Ψ(1−θ1)(1−θ2)[ϕ−N1α1y/(1+β)(1−θ1)(R1)2−N2α2y/(1+β)(1−θ2)(R2)2].

It is easy to verify from (34) and (35) that 1+D>T, or p(1)>0 at R1=RH1 and 1+D<T, or p(1)<0 at R1=RL1. Thus, –1<κ₁<1<κ₂ at RL1, or the stationary equilibrium at RL1 is a saddle. In addition,

D−1=Ψ[ϕ(1−θ1+θ1/σ1R1)(1−θ2)−(θ1−θ2)/σ1R1N2α2y/(1+β)(R2)2]>Ψ[N1α1y/(1+β)(1−θ1)(R1)2+N2α2y/(1+β)(1−θ2)(R2)2](1−θ1+θ1/σ1R1)(1−θ2)−(θ1−θ2)/σ1R1N2α2y/(1+β)(R2)2]=Ψ{N2α2y/(1+β)R1(R2)2[(1−θ1)R1+θ2/σ1]+1−θ21−θ1N1α1y/(1+β)(R1)2(1−θ1+θ1/σ1R1)}>0

or D>1 at R1=RH1. Thus, 1<κ₁<κ₂, or the steady state at RH1 is unstable. Note that the variables under consideration, Rtj, are jump variables. Hence, the steady state at RH1 is determinate.

(ii) If θ₁<θ₂, then D<0 and T<0. Besides,

|1+D|−|T|=−Ψ(1−θ1)(1−θ2)[ϕ−N1α1y/(1+β)(1−θ1)(R1)2−N2α2y/(1+β)(1−θ2)(R2)2]

or |1+D|>|T| holds at R1=RH1 and |1+D|<|T| holds at R1=RL1. Thus, the stationary equilibrium at RL1 is a saddle. In addition,

|D|−1=−Ψ[ϕ(1−θ1+θ1/σ1R1)(1−θ2)−(θ2−θ1)/σ1R1N2α2y/(1+β)(R2)2]>−Ψ{[N1α1y/(1+β)(1−θ1)(R1)2+N2α2y/(1+β)(1−θ2)(R2)2](1−θ1+θ1/σ1R1)(1−θ2)−(θ2−θ1)/σ1R1N2α2y/(1+β)(R2)2}=−Ψ{N2α2y/(1+β)R1(R2)2[(1−θ2)R2+θ1/σ1]+1−θ21−θ1N1α1y/(1+β)(R1)2(1−θ1+θ1/σ1R1)}>0

or |D|>1 also holds at R1=RH1. Thus, the steady state at RH1 is a source.^[9] Note that the variables under consideration, Rtj, are jump variables. Hence, the steady state at RH1 is determinate.

Proof of Proposition 7

Let κ₁, κ₂ be the eigenvalues of the Jacobian matrix J, then κ₁, κ₂ are two roots of the characteristic polynomial p(κ)=κ²–Tκ+D=0. The trace, T, and the determinant, D, of matrix J are evaluated as follows

T=φ[(1−θ1+θ1/σ1R1)N2α2y/(1+β)(R2)2+(1−θ2)(N1α1y/(1+β)(R1)2+ϕθ1/σ1R1)]>0D=φϕ(1−θ1+θ1/σ1R1)(1−θ2)>0.

It is immediate that p(–1)=1+T+D>0, while

p(1)=1+D−T=φ[ϕ(1−θ1+θ1/σ1R1)(1−θ2)+θ1/σ1R1N2α2y/(1+β)(R2)2−(1−θ1+θ1/σ1R1)N2α2y/(1+β)(R2)2−(1−θ2)(N1α1y/(1+β)(R1)2+ϕθ1/σ1R1)]=φ[ϕ(1−θ1)(1−θ2)−(1−θ1)N2α2y/(1+β)(R2)2−(1−θ2)N1α1y/(1+β)(R1)2]=φ(1−θ1)(1−θ2)[ϕ−N1α1y/(1+β)(1−θ1)(R1)2−N2α2y/(1+β)(1−θ2)(R2)2].

Once again, 1+D>T, or p(1)>0 at R1=RH1, and 1+D<T, or p(1)<0 at R1=RL1. Thus, –1<κ₁<1<κ₂ at RL1, or the stationary equilibrium at RL1 is a saddle. Additionally,

D−1=φ[ϕ(1−θ1+θ1/σ1R1)(1−θ2)−θ1/σ1R1N2α2y/(1+β)(R2)2]>φ{[N1α1y/(1+β)(1−θ1)(R1)2+N2α2y/(1+β)(1−θ2)(R2)2](1−θ1+θ1/σ1R1)(1−θ2)−θ1/σ1R1N2α2y/(1+β)(R2)2}=φ[N2α2y/(1+β)(R2)2(1−θ1)+1−θ21−θ1N1α1y/(1+β)(R1)2(1−θ1+θ1/σ1R1)]>0,

or D>1 at R1=RH1. Thus, 1<κ₁<κ₂, or the steady state at RH1 is a source. Note that the variables under consideration, Rtj, are jump variables. Hence, the steady state at RH1 is determinate.