Suggested Solutions to Problem Set 1

Econ6012: Macro Theory
Instructor: Dr. Yulei Luo
SEF, HKU
October 2014
Suggested Solutions to Problem Set 1
1. [10 points] Consider the following lifetime optimal consumption-saving problem:
T
1
X
1
t ct
v (a0 ) = max
(1)
1
fct ;at+1 g
t=0
subject to
at+1 = R (at
ct ) ; t = 0;
a0 = a (0) ; aT +1
;T
(2)
0;
(3)
where is the consumer’s rate of time preference (
1), is the inverse of
the elasticity of intertemporal substitution, R = 1 + r is the gross interest rate,
and given initial level of asset holdings (a0 ).
(a) Use the ‡ow budget constraints, (2), and the terminal condition to derive
the intertemporal (lifetime) budget constraint.
Solution: (2) implies that
a1 = R (a0
c0 ) ;
a2 = R (a1
c1 ) ;
aT +1 = R (aT
cT ) :
Combining all equations together and eliminating a1 ; a2 ;
aT +1
cT
+
+
T
+1
R
RT
+
c1
+ c0
R
T
X
ct
t=0
Rt
; aT gives
= a0 =)
= a0
where we use the fact that aT +1 = 0 (and then limT !1
(4)
aT +1
RT +1
= 0):
(b) Use optimal control (the Lagrange multiplier method) to derive the consumption Euler equation that links consumption in two consecutive periods,
t and t + 1; and then combine it with the intertemporal budget constraint
to …nd optimal consumption (ct ) as a function of asset holding at and model
parameters (R; ; T ). What is the consumption function when R = 1?
1
Solution: The Lagrangian is
L=
T
X
t
u(ct ) +
T
X
ct
Rt
a0
t=0
t=0
!
where is the constant Lagrangian multiplier for the lifetime budget constraint
(4). The FOCs for an optimum are then
1
; where t = 0;
Rt
t 0
u (ct ) =
Since
; T:
(5)
is a constant, the above FOCs implies that the Euler equations are
u0 (ct ) = Ru0 (ct+1 ); where t = 0;
Since u(ct ) =
c1t
1
1
;T
1:
, we have
1
ct
= R
1
; where t = 0;
ct+1
;T
1;
(6)
which means that
ct+1 = ( R)1= ct ; where t = 0;
;T
1
Combining it with the lifetime budget constraint (4) gives
T
( R)1=
X
Rt
t=0
t
c0 = a0 =)
t=0
c0 =
where ,
follows
( R)1=
R
T
X
1
!t !
(7)
, and consumption in the periods t
ct = ( R)1=
t
c0 = ( R)1=
c0 = a0 =)
a0 ;
T +1
1
( R)1=
R
1
t
T +1
1
1 can be recovered as
a0 ; 8t
1:
(8)
Similarly, we can determine the optimal time path of asset holdings:
at+1 = Rat
( R)1=
1
t
1
T +1
a0 :
(9)
Note that since a1 = R (a0 c0 ) and c0 = 1 1 T +1 a0 , a0 can be rewritten
as a function of a1 and thus c1 can be expressed as a function of a1 . Following
2
the same logic, we can always write ct as a function of at which is just the
consumption function:
! 1
T t
X
1
ct =
a =
at :
(10)
T t+1 t
1
=0
In the special case in which R = 1, we obtain
ct =
1
T +1
1
1
at+1 = Rat
ct =
a0 ; 8t
1
T t
X
R
=0
(11)
a0 ;
T +1
!
1;
(12)
1
at :
(13)
(c) Assume that T = 1. Given the same restriction R = 1, use dynamic
programming (the Bellman equation) to solve for the consumption function for
the same optimization problem, (1).
Solution: The Bellman equation for this special case
(
)
c1t
1
+ J (at+1 ) ;
J (at ) = max
ct
1
ct ). Substituting the constraint into (14) gives
(
)
c1t
1
J (at ) = max
+ J (R (at ct )) :
ct
1
(14)
where at+1 = R (at
(15)
The FOC is thus
RJ 0 (R (at
ct
ct )) = 0;
(16)
which gives optimal consumption as c = c (a). Substituting it into (15):
J (a) =
c (a)1
1
1
+ J (R (a
c (a))) :
(17)
Taking di¤erentiation w.r.t. a gives
J 0 (a) = c (a)
= c (a)
c0 (a) + RJ 0 (R (a
c0 (a) + J 0 (R (a
c (a))) 1
c (a))) 1
c0 (a) ;
c0 (a) if R = 1; (18)
which means that
J 0 (a) = u0 (c) = c (a)
3
:
(19)
Combining (18) with (19), we have
u0 (ct ) = u0 (ct+1 ) =) ct = ct+1 :
Substituting ct = ct+1 into the intertemporal budget constraint,
we have
!
1
1
X
X
1
ct
= c0
= a0 =)
Rt
Rt
P1
ct
t=0 Rt
= a0 ,
t=0
t=0
c0 =
Given that ct = c0 ; 8t
(20)
R
1
R
a0 .
(21)
1, we have
ct =
R
1
R
a0 :
(22)
Similarly, the optimal time path of asset holdings:
R
at+1 = R at
1
R
a0 ; 8t
0;
(23)
which means that
at = a0 ; 8t
1:
(24)
Therefore, the consumption function at time t is
ct =
R
1
R
at :
(25)
Note that we can also solve for the consumption function by guessing and
verifying the value function, J (a).
2. [8 points] Consider the following continuous-time optimal consumption-saving
problem:
!
Z T
ct1
1
dt;
(26)
max
exp ( t)
ct
1
0
subject to
a0t = r (at
ct ) ;
(27)
where is the discount rate, r is the instantaneous interest rate, given a0 , and
aT 0.
(a) Find the second-order di¤erential equation (SODE) governing the evolution
of at .
Solution: The Hamiltonian for this problem is:
4
ct1
1
H(t; ct ; at ; t ) = exp ( t)
+ t r(at ct );
1
where t is the co-state variable. The FOCs w.r.t. ct ,
at , and
t
are:
Hc = 0 =) exp (
0
t
Ha +
H
= 0 =) r
t
t) ct
+
= 0 =) r(at
0
t
tr
= 0;
(28)
= 0;
(29)
ct ) = a0t ;
(30)
respectively. The TVC are: T = 0; aT = 0; T aT = 0. Note that T must be
positive; otherwise, the Inada condition is not satis…ed because cT goes to 1 when
T = 0.
Following the procedure discussed in the lecture, we can easily derive the consumption Euler equation:
r
c0t
=
:
(31)
ct
Taking derivatives w.r.t. t on both sides of the budget constraint yields
a00t = r a0t
c0t ;
(32)
and combining it with the Euler equation and the budget constraint leads to the
second-order di¤erential equation in terms of at as follows:
a00t +
r
r(
r a0t
r)
at = 0:
(33)
The characteristic equation corresponding to this equation is simply:
2
+
r
r(
r
and the two eigenvalues are are 1 =
of this SODE is
at = A exp (
r
1 t)
and
r)
2
= 0;
= r. Hence, the general solution
+ B exp (
2 t) ;
(34)
where the two undetermined coe¢ cients, A and B, can be pinned down using two
boundary conditions a(0) = a0 and aT = 0:
A + B = a0 , A exp (
1T )
+ B exp (
2T )
= 0.
Solving them yields
A=
exp ( 2 T )
exp ( 1 T ) exp (
2T )
a0 ; B =
5
exp ( 1 T )
exp ( 1 T ) exp (
2T )
a0 :
1. (b) Find optimal consumption as a function of t and model parameters (T , r,
).
Solution: Using the expression for at , (34), and the budget constraint, ct =
rat a0t
, we can easily solve for the consumption function:
r
ct =
where A, B,
a0 .
1,
1
1
and
r
2
A exp (
1 t)
+ 1
2
r
B exp (
2 t) :
(35)
are determined by model parameters: , , r, T , and
2. [12 points] Consider the following Ramsey-Cass-Koopmans optimal growth
model. First, we assume that the growth rate of population in the model
economy is exogenously given and is n > 0, i.e.,
Lt+1
= 1 + n;
Lt
(36)
where L0 is given,. Second, we assume that the resource constraint in the
economy is
Kt+1 = (1
) Kt + Kt (ALt )1
Ct ;
(37)
where K0 is given, A > 0;
2 (0; 1) ; and
2 (0; 1) is the depreciation
rate. Third, we assume that the benevolent planner chooses sequences of
consumption and capital in per capita terms to solves the following dynamic
optimization problem:
!
1
1
X
c
t
t
;
(38)
max
1
fct ;kt+1 g
t=0
subject to the resource constraint in per capita terms, where ct = Ct =Lt and
kt+1 = Kt+1 =Lt+1 , where > 0 and 2 (0; 1).
(a) Derive the resource constraint in per capita terms. Set up the Lagrangian
function and …nd the consumption Euler equation for this model.
Solution: The resource constraint in per capital terms:
(1 + n) kt+1 = (1
) k t + A1
kt
ct :
(39)
(b) Find the intertemporal equilibrium (i.e., the steady state) for this model,
linearize the model around the steady state, and use the resulting two di¤erence equations system to show that the model economy is saddle-point stable
in the neighborhood of the steady state.
6
ct+1
ct
Solution: In the steady state,
k; c , satis…es
e 1
kt+1
kt
= 1 and
+ A1
k
k + A1
k
1
= 1, and the steady state,
= 1;
(40)
c = 0:
(41)
From which, we obtain that
where
gives
c
1
k =
1
A1
c =
k + A1
1
e
1=(
1
+
e
1)
1
1=(1
A1
+
=
)
> 0;
k > 0;
= 1+ . Linearizing the dynamic system, (??) and (??), around k; c
c) = e 1
(ct+1
kt+1 = k
+ A1
(ct
1
k
1
c
+ A1
c) + 1
c) + c e (
(ct
1
k
kt
k ;
kt+1
k ;
1) A1
ct+1
c = (ct
kt+1 = k
c) +
(ct
by using the fact that e1 = 1
x
et+1 = xt+1 x; x = k; c)
"
1e
1
(
1) A1 k
|
0
{z
2
c
1
k
u
Iu
K =
=
J
"
1
M=
1e
(
1
0
(43)
1e
1) A1
v
M
M
"
k ;
kt
# "
# " #
#"
1 0
e
ct+1
e
ct
0
+
=
:
1
e
e
1
kt+1
kt
0
{z
}| {z } | {z }
}| {z } |
u
1
1
k
and which can be written (denote
Multiplying J 1 on both sides gives
"
#"
"
#"
#
1
0
1 0
e
ct+1
+J 1
1
e
1
0 1
kt+1
| {z }| {z }
|
{z
}|
where
c
#"
J
I
2
k
+ A1
c) + 1
+A1
1
1) A1
1) A1
(
k
d
Kv = 0;
2
c
1
k
2
c
1
(
7
#
1) A1
1
1
d
#
" #
e
ct
0
=
e
kt
0
{z }
| {z }
v
1"
k
1
1
2
c
(44)
0
1
#
2
kt+1
k ;
(42)
which can be reduced to
e (
k
;
#
To determine the stability properties of the model, we need to know the eigenvalues (b1 ; b2 ) of the 2-by-2 coe¢ cient matrix K. Two important results in
linear algebra are that the trace and determinant of K are the sum and product of the eigenvalues (b1 ; b2 ), respectively. So
T race (K) = b1 + b2 = 1
Det (K) = b1 b2 =
1
> 1:
e
1e
1) A1
(
k
2
c+
1
> 2;
e
Hence, the discriminant should be positive because
= T race (K)2
>
1
1e
(
4Det (K) = 1
1) A1
k
2
c
1e
1
e
1) A1
(
k
2
c+
2
> 0;
1
e
2
4
1
e
2
c > 0. Therefore, both roots are real. Also,
1) A1 k
because 1 e (
1
> 1 and T race > 2; the two roots must individually be
because Det =
positive. We can also judge the magnitudes of the two roots as follows:
jbI
Kj = 0 () p (b) = (b
b1 ) (b
p (1) = (1 b1 ) (1 b2 ) = 1
1e
(
1) A1 k
=
b2 ) = 0 =)
T race + Det
2
c<0
This can only be true if one root (say b1 ) is less than 1 and the other root is
greater than 1: We can then conclude (and con…rm the predictions of the PD)
that the equilibrium is saddle-point.
After obtaining 0 < b1 < 1 and b2 > 1; we can have the general solution for
this system:
"
# "
#
e
ct
k1 A1 bt1 + k2 A2 bt2
=
:
(45)
e
kt
A1 bt1 + A2 bt2
where k1 and k2 are two constants determined by the roots (here we ignore
their detailed values). Given k0 ;
e
k0 = k0
k = A1 + A2
e
c0 = c0
c = k 1 A1 + k 2 A2
Note that bt2 ! 1 as t ! 1 because b2 > 1: To guarantee a convergent time
path to the I.E. (i.e., to kill the explosive path), the endogenous e
c0 need to be
8
set in the right way and make A2 be 0:
c0
c = k1 k0
k :
(46)
Hence, given any k0 ; we can …nd c0 s.t. the economy “jump” to the pair of
stable branches and then move to the saddle point equilibrium.
[Optional ] The following are about the details about the transitional dynamics
of e
ct ; e
kt we obtained above. Given that the economy is characterized by the
following …rst-order di¤erence equation system:
#
#
"
"
e
ct
e
ct+1
;
(47)
=K
e
e
kt
kt+1
where
K=
"
1
1e
(
1) A1
k
2
1
c
1) A1
(
1
e
1
k
2
c
#
;
and as shown above 0 < b1 < 1 and b2 > 1 are two corresponding eigenvalues. Note that given the model parameters and steady state values, we can
decompose the 2-by-2 matrix K as follows
K=Q Q
1
;
(48)
where Q has the eigenvectors of K as its columns, and
K down its diagonal:
"
#
b2 0
=
:
0 b1
The solution to system (47) can be written
"
#
"
#
e
ct+1
bt2 0
=Q
Q
e
kt+1
0 bt1
1
"
e
c0
e
k0
#
has eigenvectors of
(49)
;
(50)
where e
k0 is given. Since b2 > 1, we have one explosive eigenvalue and the
system will blow-up as t ! 1 unless the endogenous variable e
c0 “jumps” in
the right way so as to kill the explosive dynamics. To see how this works, we
can write the system as follows
"
#"
#"
#"
#
"
#
q11 q12
bt2 0
q22
q12
e
c0
e
ct+1
1
;
=
e
e
det (Q) q21 q22
0 bt1
q21 q11
k0
kt+1
"
#"
#
q11 bt2 q12 bt1
q22 e
c0 q12 e
k0
1
=
;
det (Q) q21 bt2 q22 bt1
q21 e
c0 + q11 e
k0
2
3
t q e
t
e
e
q
b
c
q
k
+
q
b
q
e
c
+
q
k
11 2
22 0
12 0
12 1
21 0
11 0
1
4
5;
=
t
t
det (Q) q21 b q22 e
c0 q12 e
k0 + q22 b
q21 e
c0 + q11 e
k0
2
1
9
#
q11 q12
. Since bt2 ! 1 as t ! 1, we have to
where we assume that Q =
q21 q22
shut these explosive paths down by setting e
c0 just right (this is our one degree
of freedom, it is the only endogenous variable not pinned down yet). In other
words, we choose consumption so that we are not on the explosive path that
violates the transversality condition (TVC). Apparently, we should set
"
e
c0 =
q12 e
k0
q22
to rule out the unstable dynamics. Hence our complete solution is
2
3
#
"
q12 e
t
e
q
b
q
k
+
q
k
12 1
21 q22 0
11 0
e
ct+1
1
4
5;
=
q
e
t
12
det (Q) q22 b1
kt+1
q21 q22 e
k0 + q11 e
k0
2
3
q12
+
q
q
q
11
12
21 q22
1
4
5 bt1 e
=
k0 :
q12
det (Q) q22
q21 q22 + q11
(51)
(52)
In practice we can …rst compute the matrix K, then compute its eigenvalues
and eigenvectors to get the matrix Q and pick the stable eigenvalue so that we
have a bounded solution.
(c) Use the phase diagram to show that the steady state is saddle-point stable.
Solution: For the phase diagram analysis, refer to lecture notes.
10