Semi-global solutions to DSGE models: perturbation around a deterministic path

Appendix

A Proofs for Section 5

Proof of proposition 5.1: The proof is by induction on i. Suppose that i=0. For the time T from (52) we have

ETuT+1=BTuT+Q21,TsT+Ψ2,TETηT+1.

As B_T is invertible, we have

uT,T=−KT,TsT−gT,0+LT,T−1ETuT+1,

where KT,T=BT−1Q21,T;gT,0=−BT−1Ψ2,TETηT+1 and LT,T−1=BT−1. From (53), (54) and (56) it follows that the inductive assumption is proved for i=0. Assuming that (55) holds for i>0, we will prove it for i+1. To this end, consider Equation (52) for the time t=T−i−1. As the matrix B_T−i is invertible, we obtain

uT,T−i−1=−BT−i−1−1Q21,T−i−1sT−i−1−BT−i−1−1Ψ2,T−i−1ET−i−1ηT−i+BT−i−1−1ET−i−1uT,T−i.

Substituting the induction assumption (55) for u_T,T−i yields

uT,T−i−1=−BT−i−1−1Q21,T−i−1sT−i−1−BT−i−1−1Ψ2,T−i−1ET−i−1ηT−i+BT−i−1−1ET−i−1[−KT,T−isT−i+gT,i+(∏k=1i+1LT,T−i+k−1)ET−i(uT+1)].

Substituting (51) for E_T−i−1(s_T−i ) and using the law of iterated expectations gives

uT,T−i−1=−BT−i−1Q21,T−isT−i−1−BT−i−1Ψ2,T−iET−i−1ηT−i+BT−i−1gT,i+BT−i−1(∏k=1i+1LT,T−i+k−1)ET−i−1(uT+1)+BT−i−1[−KT,T−i(AT−isT−i−1+Q12,T−iuT,T−i−1+Ψ1,T−iET−i−1ηT−i)].

Collecting the terms with u_T,T−i−1, s_T−i−1 and η_T−i , we get

(I+BT−i−1KT,T−iQ12,T−i)uT,T−i−1=−BT−i−1[(Q21,T−i+KT,T−iAT−i)sT−i−1+(Ψ2,T−i+KT,T−iΨ1,T−i)ET−i−1ηT−i+gT,i+(∏k=1i+1LT,T−i+k−1)ET−i−1(uT+1)]

Suppose for the moment that the matrix

ZT,T−i=I+BT−i−1KT,T−iQ12,T−i

is invertible. Pre-multiplying the last equation by ZT,T−i−1, we obtain

uT,T−i−1=−ZT,T−i−1BT−i−1[(Q21,T−i+KT,T−iAT−i)sT−i−1+(Ψ2,T−i+KT,T−iΨ1,T−i)ET−i−1ηT−i+gT,i+(∏k=1i+1LT,T−i+k−1)ET−i−1(uT+1)].

Note that L_T,T−i =B_T−iZ_T,T−i ; then using the definition of K_T,T−i−1 (53), we see that

Using the definition of g_T,i and L_T−i,T−i+_j [(54) and (56)], we deduce that

From (91) and (92) it follows that

uT,T−i−1=−KT,T−i−1sT−i−1+gT,i+1+(∏k=1i+2LT,T−i−1+k−1)ET−i−1(uT+1).□

Proof of proposition 5.2: We begin by rewriting (53) as

(BT−i+KT,T−iQ12,T−i)KT,T−(i+1)=(Q21,T−i+KT,T−iAT−i).

Rearranging terms, we have

Taking the norms and using the norm properties gives

∥KT,T−(i+1)∥≤∥BT−i−1∥⋅∥Q21,T−i∥+∥BT−i−1∥⋅∥KT,T−i∥⋅∥AT−i∥+∥BT−i−1∥⋅∥KT,T−i∥⋅∥Q12,T−i∥⋅∥KT,T−(i+1)∥.

Rearranging terms, we get

Inequality (94) is a difference inequality with respect to ||K_T,T−i ||, i=0, 1, …, T, and with the time-varying coefficients ||A_T−i ||, ∥BT−i−1∥, ||Q_12,T−i || and ||Q_21,T−i ||. In (94) we assume that

1−∥BT−i−1∥⋅∥KT,T−i∥⋅∥Q12,T−i∥≠0.

This is obviously true if ||K_T,T−i ||=0. We shall show that if the initial condition ||K_T,T+_i ||=0, then (1−∥BT−i−1∥⋅∥KT,T−i∥⋅∥Q12,T−i∥)>0,i=1, 2, …, T. Indeed, consider the difference equation:

Lemma A.1If inequality (57) holds, then the difference equation (95) has two fixed points

wheres1∗is a stable fixed point whereass2∗is an unstable one. Moreover, under the initial conditions₀=0 the solutions_i , i=1, 2, …, is an increasing sequence and converges tos1∗.

The lemma can be proved by direct calculation. From (48)–(47) the values a, b, c and d majorize ||A_T−i ||, ∥BT−i−1∥, ||Q_{12,T−i || and ||Q21,T−i ||, respectively. If we consider Equation (92) and inequality (95) as initial value problems with the initial conditions ||KT,T+1||=0 and s0=0, then their solutions obviously satisfy the inequality ||KT,T−i ||≤si+1}, i=1, 2, …, T. In other words, ||K_T,T−i || is majorized by s_i . From the last inequality and Lemma A.1 it may be concluded that

From (96), (97) and (48) it follows that

From (57) we see that 2b²dc<(1−ab)²/2. Substituting this inequality into (98) gives

where the last inequality follows from (50).□

Proof of proposition 5.4: The assertion of the proposition is true if there exist constants M and r such that 0<r<1 and for T∈ℕ

Note now that K_T,j (K_T+1,_j ) is a solution to the matrix difference equation (53) at i=T−j (i=T+1−j) with the initial condition K_T,T+1=0 (K_{T+1,T+2=0). Subtracting (93) for KT,T−(i+1)} from that for K_T+1,_T−(_i+1), we have

KT,T−(i+1)−KT+1,T−(i+1)=BT−i−1(KT,T−i)−KT+1,T−i)AT−i−BT−i−1KT,T−i)Q12,T−iKT,T−(i+1)+BT−i−1KT+1,T−iQ12,T−iKT+1,T−(i+1).

Adding and subtracting BT−i−1⋅KT,T−i⋅Q12,T−i⋅KT+1,T−(i+1) in the right hand side gives

KT,T−(i+1)−KT+1,T−(i+1)=BT−i−1(KT,T−i)−KT+1,T−i)AT−i−BT−i−1⋅KT,T−i⋅Q12,T−i(KT,T−(i+1)−KT+1,T−(i+1))−BT−i−1(KT,T−i−KT+1,T−i)Q12,T−i⋅KT+1,T−(i+1).

Rearranging terms yields

(I+BT−i−1KT,T−iQ12,T−i)(KT,T−(i+1)−KT+1,T−(i+1))=BT−i−1(KT,T−i−KT+1,T−i)AT−i−BT−i−1(KT,T−i−KT+1,T−i)Q12,T−iKT+1,T−(i+1).

From Proposition 5.3 it follows that the matrix

ZT,T−i=(I+BT−i−1KT,T−iQ12,T−i)

is invertible, then pre-multiplying the last equation by this matrix yields

KT,T−(i+1)−KT+1,T−(i+1)=ZT,T−i−1(BT−i−1(KT,T−i−KT+1,T−i)AT−i−BT−i−1(KT,T−i)−KT+1,T−i)Q12,T−iKT+1,T−(i+1)).

Taking the norms, using the norm property and the triangle inequality, we get

From (47) and (99) we have

From the norm property and Golub and Loan (1996, Lemma 2.3.3) we get the estimate

∥ZT,T−i−1∥=∥(I+BT−i−1KT,T−iQ12,T−i)−1∥≤11−∥BT−i−1KT,T−iQ12,T−i∥≤11−∥BT−i−1∥⋅∥KT,T−i∥⋅∥Q12,T−i∥

By (99), we have

∥ZT,T−i−1∥≤11−1−ba2=21+ba

Substituting the last inequality into (102) gives

Using (103) successively for i=−1, 0, 1, …, T−1, and taking into account K_T,T+1=0 and KT+1,T+1=BT+2−1Q21,T+2 results in

Recall that Q_{21, T} depends on the solution to the deterministic problem (10), i.e.

Q21,T=Q21(xT+1(0), xT(0), zT+1(0), zT(0)).

From Hartmann (1982, Corollary 5.1) and differentiability of Q₂₁ with respect to the state variables it follows that

where α is the largest eigenvalue modulus of the matrix A from (40), C is some constant and θ is arbitrary small positive number. In fact, α+θ determines the speed of convergence for the deterministic solution to the steady state. Inserting (105) into (104), we can conclude

∥KT,j−KT+1,j∥<bC(α+θ)T+2, j=0, 1, 2, …

Denoting M=bC(α+θ) and r=α+θ we finally obtain (100).□

B The first order system

For n=1 we have

[s0(1)u0(1)]=Z1Z−1[x0(1)y0(1)]=0,

From (43) for the time T we have

uT(1)=−BT+1−1Q21,T+1sT(1)−BT+1−1Π2,t+1zT(1)+BT+1−1ETuT+1(1).

Denoting KT,T=BT+1−1Q21,T+1 and RT=BT+1−1Π2,t+1 gives

For T−1 we have

Taking conditional expectations at the time T−1 from both side (106) and inserting (2) we get

Inserting (108) into (107) gives

Inserting now E_T−1s_T into (109) from (42) yields

uT−1(1)=−BT−1Q21,TsT−1(1)−BT−1Π2,tzT−1(1)+BT−1ET−1[−KT,T(AT−1sT−1(1)+Q12,TuT−1(1)+Π1,TzT−1(1))−RTΛzT−1(1)+BT+1−1ETuT+1(1)].

Reshuffling terms, we have

Multiplying (110) by (I+BT−1KT,TQ12,T)−1 yields

uT−1(1)=−(I+BT−1KT,TQ12,T)−1BT−1(Q21,T+BT−1KT,TAT−1)sT−1(1)−(I+BT−1KT,TQ12,T)−1BT−1(Π2,t+KT,TΠ1,T+RTΛ)zT−1(1)+(I+BT−1KT,TQ12,T)−1BT−1BT+1−1ETuT+1(1).

or

uT−1(1)=−(BT+KT,TQ12,T)−1(Q21,T+BT−1KT,TAT−1)sT−1(1)−(BT+KT,TQ12,T)−1(Π2,t+KT,TΠ1,T+RTΛ)zT−1(1)+(BT+KT,TQ12,T)−1BT+1−1ETuT+1(1).

Denoting L_T,T−1=(B_T−1+K_T,TQ_12,T−1), we obtain

uT−1(1)=−LT,T−1−1(Q21,T+BT−1KT,TAT−1)sT−1(1)−LT,T−1−1(Π2,t+KT,TΠ1,T+RTΛ)zT−1(1)+LT,T−1−1BT+1−1ETuT+1(1).

Denoting

KT,T−1=LT,T−1−1(Q21,T+BT−1KT,TAT−1)

and

RT−1=LT,T−1−1(Π2,t+KT,TΠ1,T+RTΛ),

we have

uT−1(1)=−KT,T−1sT−1(1)−RT−1zT−1(1)+LT,T−1−1LT,T−1ETuT+1(1).

Following the same derivation as in A for the proof of Proposition 5.1, we obtain the following representation:

where R_t can be computed by backward recursion

Rt=LT,t+1−1(Π2,t+KT,tΠ1,t+Rt+1Λ)

Inserting (42) into (111) gives

Etst+1(1)=Ast(1)+Q11,t+1st(1)+Q12,t+1(−KT,tst(1)−Rtzt(1))+Π1,t+1zt(1)

After reshuffling we get

Etst+1(1)=(At+1−Q12,t+1KT,t)st(1)+(−Q12,t+1Rt+Π1,t+1)zt(1).

Denoting 𝔸_t =A_t+1−Q_12,t+1K_T,t and ℙ_t =−Q_12,t+1R_t +Π_1,t+1, we have

It is easy to see that

[st+1(1)ut+1(1)]−Et[st+1(1)ut+1(1)]=[ℝ1,tℝ2,t]εt+1=ZΦt+1−1f5,t+1εt+1.

From (112) it follows that

(Etst+1−st+1)+st+1=Atst+ℙtzt,

thus, we obtain

Recall now that the initial conditions are s0(1)=0 and z0(1)=0, then for t=1 from (113) we have

s1(1)=ℝ1,0ε1;

for t=2

s2(1)=(A1ℝ1,0+ℙt)ε1+ℝ1,1ε2.

Continuing in this fashion, we get the moving-average representation of st(1)​:

where the coefficients γ_t,t−i can be obtained by forward recursion in t=1, 2, …, T and backward recursion in i=0, 1, …, t−1

γt,t=ℝ1,t−1,γt,t−1=At−1γt−1,t−1+ℙt−1,…γt,t−i=At−1γt−1,t−i+ℙt−1Λi−1,…γt,1=At−1γt−1,1+ℙt−1Λt−2

Indeed, inserting (114) into (113) and taking into account z_t =ε_t +Λε_t−1+…+Λ^t−1ε₁, we obtain

Collecting terms with ε_j gives

st+1=ℝ1,tεt+1+(Atγt,t+ℙt)εt+(Atγt,t−1+ℙtΛ)εt−1+⋯+(Atγt,1+ℙtΛt−1)ε1.

Thus, for each t we compute γ_t,i , starting with the first index t=1, then decreasing the index i=t, t−1, …, 1 and using at each step γ_t−1,_i . For the variable ut(1)we also have a moving-average representation. Inserting the moving-average representation of the process zt(1)and (114) in (111), we have

or in the shorter form

where δ_t,i =−K_T,tγ_t,i −R_t Λⁱ⁻¹.

Taking into account that xt(1)=Z11st(1)+Z12ut(1) and yt(1)=Z21st(1)+Z22ut(1), we get the moving-average representation for original variables

xt(1)=ρt,txεt+ρt,t−1xεt−1+⋯+ρt,2xε2+ρt,1xε1,

yt(1)=ρt,tyεt+ρt,t−1yεt−1+⋯+ρt,2yε2+ρt,1yε1,

where ρt,ix=Z11γt,i+Z12δt,i and ρt,iy=Z21γt,i+Z22δt,i.