Advanced Macroeconomics Problem Set 3
Exercise 14.4 First, g∗is given by
ληβL − (λ − 1)−1g∗= θg∗+ ρ
Left hand side of the equation is decreasing in g∗and right hand side is increasing in g∗, so g∗ which satisfies above equation is only one. (Of course the condition of
ληβL > ρ > (1 − θ)(λ − 1)ηβL (1) must be hold.)
Therefore
g∗= ληβL − ρ θ + (λ − 1)−1 is unique.
Next, we will prove the transversality condition is satisfied. The transversality condition is
t→∞lim exp − Zt
0
r(s)ds Z t
0
V(ν, t|q)dν = 0
Recall that (14.17), I mean
V(ν, t|q) =q(ν, t) λη Then we can calculate
Z t 0
V(ν, t|q)dν = 1 λη
Zt 0
q(ν, t)dν
= 1
ληQ(t)
Therefore, from this equation the transvesality condition is given by
1
t→∞lim exp(−r
∗t) 1
ληQ(t)
= lim
t→∞ exp(−r
∗t) 1
ληexp(g
∗t)
= 1
ληt→∞limexp −(r
∗− g∗)t
So, it is almost the same as the previous exercises
r∗− g∗ = θg∗+ ρ − g∗
= ρ − (1 − θ) ληβL − ρ θ + (λ − 1)−1
= 1 + (λ − 1)−1ρ − (1 − θ)ληβL θ + (λ − 1)−1
= λ
(λ − 1)θ + 1 ρ − (1 − θ)(λ − 1)ηβL > 0
We use condition (1) in last inequality.
After all, as long as r∗− g∗> 0 has hold, the transversality condi- tion is satisfied; and this is the proof of the Execercis 14.4.
Exercise 14.17 (a)
ln Y(t) = ln C + n(t) ln λ Therefore
ln Y(t + Δ) − ln Y(t) = n(t + Δ) − n(t) ln λ
g∗ = Eln Y(t + Δ) − ln Y(t) Δ
= En(t + Δ) − n(t)
Δ ln λ
= 1
ΔE ln λ
= ηL∗
Rln λ
2
(Expectation of Poisson Process)
E =
∞
X
=0
(zΔ)
e−zΔ
!
= (zΔ)e−zΔ
1 +(zΔ)
1
1! + (zΔ)2
2! + . . .
= (zΔ)e−zΔezΔ
= zΔ
Next, the transversality condition in steady state
t→∞limexp(−r
∗t)V(q) = 0
V(q) = βq(t)LE r∗+ z∗
= β(L − L
∗ R)
ρ∗+ z∗ q(0) exp(g
∗t)
t→∞limexp(−r
∗t)β(L − L
∗ R)
ρ∗+ z∗ q(0) exp(g
∗t)
=β(L − L
∗ R)q(0)
ρ∗+ z∗ t→∞limexp −(ρ − g∗t)
0 < g∗< ρ
⇔ 0 < λ(1 − β)ηL − ρ <1 + λ(1 − β)ρ ln λ
⇔ ρ < λ(1 − β)ηL <ln λ + 1 + λ(1 − β)
ln λ ρ
Because
g∗= ηλ(1 − β)ηL − ρ η 1 + (1 − β)λ ln λ
= λ(1 − β)ηL − ρ 1 + (1 − β)λ ln λ g∗> 0 ⇔ λ(1 − β)ηL − ρ > 0
3
g∗< ρ ⇔λ(1 − β)ηL − ρ
1 + (1 − β)λ ln λ < ρ
(b)
In previous model, we use linear approximation like Q(t + Δ) ≈ zΔλQ(t) + (1 − z(t)Δ)Q(t)
Q(t + Δ) − Q(t)
Δ ≈ z(λ − 1)Q(t) And we obtain
Q(t)˙
Q(t) = (λ − 1)z
However, we does not use such a approximation in this model. This is different point.
As you see
ln λ ≈ λ − 1
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