International Macroeconomics and Finance: Theory and Empirical Methods Phần 5 docx

5.1. CALIBRATING THE ONE-SECTOR GROWTH MODEL 145

where a0 = Ucg1 + βUcg2 = 0,

a1 = βUccg1g2,

a2 = Uccg2

1 + βUccg2

2 + βUcg22,

a3 = Uccg1g2,

a4 = βUcg32 + βUccg2g3,

a5 = Uccg1g3.

The derivatives are evaluated at steady state values.

A second but equivalent option is to take a secondóorder Taylor

approximation to the objective function around the steady state and to

solve the resulting quadratic optimization problem. The second option

is equivalent to the first because it yields linear firstóorder conditions

around the steady state. To pursue the second option, recall that λt =

(kt+1, kt, At)0

. Write the period utility function in the unconstrained

optimization problem as

R(λt) = U[g(λt)]. (5.20)

Let Rj = ∂R(λt)/∂λjt be the partial derivative of R(λt) with respect

to the j−th element of λt and Rij = ∂2R(λt)/(∂λit∂λjt) be the second

cross-partial derivative. Since Rij = Rji the relevant derivatives are,

R1 = Ucg1,

R2 = Ucg2,

R3 = Ucg3,

R11 = Uccg2

1,

R22 = Uccg2

2, +Ucg22

R33 = Uccg2

3,

R12 = Uccg1g2,

R13 = Uccg1g3,

R23 = Uccg2g3 + Ucg23.

The second-order Taylor expansion of the period utility function is

R(λt) = R(λ) + R1(kt+1 − k) + R2(kt − k) + R3(At − A) + 1

2

R11(kt+1 − k)

2

146 CHAPTER 5. INTERNATIONAL REAL BUSINESS CYCLES

+

1

2

R22(kt − k)

2 +

1

2

R33(At − A)

2 + R12(kt+1 − k)(kt − k)

+R13(kt+1 − k)(At − A) + R23(kt − k)(At − A).

Suppose we let q = (R1, R2, R3)0 be the 3 × 1 row vector of partial

derivatives (the gradient) of R, and Q be the 3 × 3 matrix of second

partial derivatives (the Hessian) multiplied by 1/2 where Qij = Rij/2.

Then the approximate period utility function can be compactly written

in matrix form as

R(λt) = R(λ)+[q + (λt − λ)

0

Q](λt − λ). (5.21)

The problem is now to maximize

Et

X∞

j=0

βj

R(λt+j). (5.22)

The first order conditions are for all t

0=(βR2 + R1) + βR12(kt+2 − k)+(R11 + βR22)(kt+1 − k) + R12(kt − k)

+βR23(At+1 − 1) + R13(At − 1). (5.23)

If you compare (5.23) to (5.19), youíll see that a0 = βR2 + R1,

a1 = βR12, a2 = R11 + βR22, a3 = R12, a4 = βR23, a5 = R13. This

verifies that the two approaches are indeed equivalent.

Now to solve the linearized first-order conditions, work with (5.19).

Since the data that we want to explain are in logarithms, you can con￾vert the first-order conditions into near logarithmic form. Let

aòi = kai for i = 1, 2, 3, and let a ìhatî denote the approximate log

difference from the steady state so that à

kt = (kt − k)/k ' ln(kt/k)

and Aàt = At − 1 (since the steady state value of A = 1). Now let

b1 = −aò2/aò1, b2 = −aò3/aò1, b3 = −a4/aò1, and b4 = −a4/aò1.

The secondóorder stochastic difference equation (5.19) can be writ￾ten as

(1 − b1L − b2L2

)à

kt+1 = Wt, (5.24)

where

Wt = b3Aàt+1 + b4Aàt.

5.1. CALIBRATING THE ONE-SECTOR GROWTH MODEL 147

The roots of the polynomial (1 − b1z − b2z2) = (1 − ω1L)(1 − ω2L)

satisfy b1 = ω1 + ω2 and b2 = −ω1ω2. Using the quadratic formula

and evaluating at the parameter values that we used to calibrate the

model, the roots are, z1 = (1/ω1)=[−b1−

q

b2

1 + 4b2]/(2b2) ' 1.23, and

z2 = (1/ω2)=[−b1 +

q

b2

1 + 4b2]/(2b2) ' 0.81. There is a stable root,

|z1| > 1 which lies outside the unit circle, and an unstable root, |z2| < 1

which lies inside the unit circle. The presence of an unstable root means

that the solution is a saddle path. If you try to simulate (5.24) directly,

the capital stock will diverge.

To solve the difference equation, exploit the certainty equivalence

property of quadratic optimization problems. That is, you first get

the perfect foresight solution to the problem by solving the stable root

backwards and the unstable root forwards. Then, replace future ran￾dom variables with their expected values conditional upon the time-t

information set. Begin by rewriting (5.24) as

Wt = (1 − ω1L)(1 − ω2L)à

kt+1

= (−ω2L)(−ω−1

2 L−1

)(1 − ω2L)(1 − ω1L)à

kt+1

= (−ω2L)(1 − ω−1

2 L−1

)(1 − ω1L)à

kt+1,

and rearrange to get

(1 − ω1L)à

kt+1 = −ω−1

2 L−1

1 − ω−1

2 L−1Wt

= −

µ 1

ω2

L−1

¶X∞

j=0

µ 1

ω2

¶j

Wt+j

= −X∞

j=1

µ 1

ω2

¶j

Wt+j . (5.25)

The autoregressive specification (5.18) implies the prediction formulae

EtWt+j = b3EtAàt+j+1 + b4EtAàt+j = [b3ρ + b4]ρj

Aàt.

Use this forecasting rule in (5.25) to get

X∞

j=1

µ 1

ω2

¶j

EtWt+j = [b3ρ + b4]Aàt

X∞

j=1

µ ρ

ω2

¶j

=

" ρ

ω2 − ρ

#

(b3ρ + b4)Aàt.