Proposed Answers to Assignment #3 and #4
Takeki Sunakawa
December 9, 2015
Assignment #3 (1)
The planner’s problem is
{cmaxt,kt}
∞
X
t=0
βtln ct
subject to ct+ kt ≤kθt−1, given k−1.
[Note that A in the production function is set as A = 1.] The value function is defined as
V (kt−1) ≡ max
kt
∞
X
t=0
βtln(kt−1θ −kt). The Bellman equation is written as
V (kt−1) = max
kt
ln(kt−1θ −kt) + βV (kt) .
Assignment #3 (2)
Conjecture V (k) = A + B ln k. Then we have V (k) = max
k′ ln(k
θ−k′) + β (A + B ln k′) . 1
The FOC is
1 kθ−k′ =
βB k′ . It can be solved for
k′ = βB 1 + βBk
θ.
Pluging it into the Bellman equation, we have ln(kθ−k′) + β (A + B ln k′) ,
= ln
kθ− βB 1 + βBk
θ
+ βA + βB ln
βB 1 + βBk
θ
,
= ln
1
1 + βBk
θ
+ βA + βB ln
βB 1 + βBk
θ
,
= βA + ln
1
1 + βB
+ βB ln
βB 1 + βB
+ θ ln k + βBθ ln k. Thus we have
A = βA + ln
1
1 + βB
+ βB ln
βB 1 + βB
, B = θ(1 + βB).
Finally, we have
B = θ/(1 − βθ), and
A = (1 − β)−1ln
1
1 + βB
βB 1 + βB
βB
.
Assignment #4 (1)
The equilibrium conditions are
1 = β1 + θy k −δ
, (1 − θ)y
h = Bc, y = kθh1−θ, c + δk = y.
2
These are solved for the ratio values y k = θ
−1(β−1−1 + δ),
k h =
y k
θ−11 , c
k = y k −δ. Then we have
h = (1 − θ)
y k
Bkc ,
= (1 − θ)
y k
B(yk−δ),
= (1 − θ)(β
−1−1 + δ)
B(β−1−1 + (1 − θ)δ),
= (1 − θ)[1 − β(1 − δ)] B(1 − β[1 − (1 − θ)δ]). Also, k = (k/h)h, y = (y/k)k and c = (c/k)k are obtained.
As for the log-linearization, only two equations are new.
(1 − θ)yt ht
= Bct,
⇒ (1 − θ)(1 + ˆyt− ˆht) = Bc(1 + ˆct),
∴ yˆt− ˆht = ˆct. Note that we have used (1 − θ) = Bc.
λt = (1 − γ) + γλt−1+ εt,
⇒ λ(1 + ˆλt) = (1 − γ) + γλ(1 + ˆλt−1) + εt,
∴ λˆt = γˆλt−1+ εt.
Note that we have used λ = (1−γ)+γλ therefore λ = 1. Also, net investment is given by
xt = kt−(1 − δ)kt−1
3
⇒ x(1 + ˆxt) = k(1 + ˆkt) − (1 − δ)k(1 + ˆkt−1),
⇒ xˆxt = kˆkt−(1 − δ)kˆkt−1,
∴ δˆxt= kˆkt−(1 − δ)kˆkt−1. Note that we have used x = δk.
4
