ejercicio ramsey bueno
TRANSCRIPT
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Economics 205A
Fall 2012
K. Kletzer
Problem Set 2 Sample Answers
1. Ans: a) The command optimum solves
m a x
{ c
t
, k
t
}
0
e
t
c
1
1
1
d t
subject to
k = f ( k ) k c ,
k
t
0 for all t 0
given initial k equal to k0
.
b) The current-time Hamiltonian is
H =
c
1
1
1
+ q ( f ( k ) ( n + ) k c ) .
The necessary conditions are
c
= q ,
q q =
f
( k )
q ,
k = f ( k ) k c
and
l i m
t
q
t
e
t
k
t
= 0 and k0
given.
c) The system you work with is
c
c
= f
(
k
)
and
k = f ( k ) k c .
This has a steady state given by
f
( k
) = +
and
c
= f ( k
) k
.
The solution path is given by the stable saddle path in your phase diagram. This converges to the steady
state and is the optimum because it satisfies the transversality condition,
l i m
s
c
t
e
t
k
t
= 0 .
d) Linearizing, you get
c =
c
f
( k
) ( k k
)
and
k =
f
( k
)
( k k
) ( c c
) = ( k k
) ( c c
) .
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The eigenvalues are given by
=
1
2
2
4 f
( k
)
c
1 / 2
.
The root,
, is negative, and the root +
> > 0 , so that we have saddl- path stability. The stable path is
given by:
k
t
k
= ( k
0
k
) e
t and ct
c
= ( c
0
c
) e
t
,
wherec
0
= c
+ ( k
0
k
) (
).
To first-order approximation, output is
y
t
y
= ( y
0
y
) e
t
(fromy
t
y
= f
( k
) ( k
t
k
)). Thus,
y
t
y
t
=
( y
t
y
)
y
t
=
1
y
y
t
,
so that at timet = 0
, the growth rate of output is given by
y
0
y
0
=
1
y
y
0
which is increasing iny
/ y
0
. The lowery
0
relative toy
, the higher is the growth rate of output.
2. Ans: a) The command optimum solves
m a x
{ c
t
,
t
, n
t
, k
t
+ 1
}
t = 0
t
u ( c
t
,
t
)
subject to
k
t + 1
= f ( k
t
, n
t
) + ( 1 ) k
t
c
t
,
1 n
t
+
t
0 and nt
,
t
, k
t
0 for all t 0
given initial k equal to k0
.
b) The necessary conditions for an interior optimum are
u ( c
t
,
t
)
c
t
=
1 +
f ( k
t + 1
, n
t + 1
)
k
t + 1
u ( c
t + 1
,
t + 1
)
c
t + 1
f ( k
t
, n
t
)
n
t
u ( c
t
,
t
)
c
t
=
u ( c
t
,
t
)
t
for 1 > t
> 0 ,
k
t + 1
= f ( k
t
, n
t
) + ( 1 ) k
t
c
t
,
1 = n
t
+
t
and
l i m
t
t
u ( c
t
,
t
)
c
t
k
t + 1
= 0 and k0
given.
These assume thatl i m
c
t
0
u ( c
t
,
t
)
c
t
= ,
l i m
t
0
u ( c
t
,
t
)
t
=
and that the Inada conditions hold for
f ( k
t
, n
t
) . We implicitly assume that f ( kt
, n
t
) and u ( ct
,
t
) are strictly concave in their arguments and that
f ( k
t
, 0 ) = f ( 0 , n
t
) = 0 .
Corner solutions in t
and nt
are possible. The additional first-order conditions are
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f ( k
t
, n
t
)
n
t
u ( c
t
,
t
)
c
t
u ( c
t
,
t
)
t
for nt
= 1 , t
= 0 ,
and f ( k
t
, n
t
)
n
t
u ( c
t
,
t
)
c
t
u ( c
t
,
t
)
t
for nt
= 0 , t
= 1 .
Now, let f ( k , n ) = A k
n
1
for 0 < 0 constants. Let u ( c , ) = l o g c + l o g . These
will lead to solutions such that nt
> 0 and t
> 0 , so that nt
+
t
= 1 for all t . The necessary conditions
become
1
c
t
=
1 + A k
1
t + 1
n
1
t + 1
1
c
t + 1
c
t + 1
c
t
=
1 + A
k
t + 1
n
t + 1
1
,
( 1 ) A k
t
n
t
1
c
t
=
1
t
c
t
t
= ( 1 ) A
k
t
n
t
,
k
t + 1
= A k
t
n
1
t
+ ( 1 ) k
t
c
t
,
1 = n
t
+
t
and
l i m
t
t
k
t + 1
c
t
= 0 and k0
given.
c) Begin with the three conditions
c
t + 1
c
t
=
1 + A
k
t + 1
n
t + 1
1
,
c
t
1 n
t
= ( 1 ) A
k
t
n
t
and
k
t + 1
k
t
=
A
k
t
n
t
1
k
t
c
t
.
The steady state is given by
A
k
n
1
=
1
1 + = +
c
1 n
= ( 1 ) A
k
n
andc
k
= A
k
n
1
.
These simplify to
k
n
=
A
+
1
1
,
c
1 n
= A
1
1
( 1 )
+
1
andc
k
=
+
.
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You might simplify further to find thatn
is given by
1
n
= 1 +
1
1
+
which is strictly between zero and one for > 0
and1 > > 0
.
We can reduce the system of three equations to a dynamical system in c and k by eliminating n . However,
lets linearize all three and simplify our system from there. Linearizing the Euler condition about the steady
state leads to the equation
d c
t + 1
c
d c
t
c
= ( 1 ) A
k
n
1
d k
t + 1
k
d n
t + 1
n
= ( 1 ) ( + )
d k
t + 1
k
d n
t + 1
n
where the notationd x
t
x
t
x
. Next, linearize the first-order condition for consumption of goods and
leisure. This isd c
t
c
d n
t
1 n
=
d k
t
k
d n
t
n
.
Lastly, linearize the equation of motion fork
as
d k
t + 1
d k
t
= A
k
n
1
d k
t
+ ( 1 ) A
k
n
d n
t
d k
t
d c
t
= d k
t
+ A
1
1
( 1 )
+
1
d n
t
d c
t
.
Now, we need to eliminate d kt + 1
and d nt + 1
from the linearized Euler condition so that we have d kt
and
d n
t
on the right-hand side of the condition. To do this, use the first-order condition for goods-leisure choice
to write
1
1 n
+
1
n
d n
t
=
d k
t
k
d c
t
c
.
d c
t
c
+
d n
t
1 n
=
d c
t
c
+
1
1 +
1 n
n
d k
t
k
d c
t
c
Then used c
t + 1
= ( d c
t + 1
d c
t
) + d c
t
= c
t + 1
+ d c
t
andd k
t + 1
= k
t + 1
+ d k
t
to substitute into the Euler
condition to get (in steps)
c
t + 1
c
= ( 1 ) ( + )
d k
t + 1
k
d n
t + 1
n
= ( 1 ) ( + )
d k
t
k
d n
t
n
( 1 ) ( + )
k
t + 1
k
n
t + 1
n
=
1
( + )
d c
t
c
+
d n
t
1 n
1
( + )
c
t + 1
c
+
n
t + 1
1 n
=
1
( + )
1 +
1 n
n
1
1 n
n
d c
t
c
+
d k
t
k
+
1 n
n
c
t + 1
c
+
k
t + 1
k
We replace kt + 1
using the equation of motion for k and rearrange to get
c
t + 1
c
1 +
( 1 ) ( + )
1 +
1 n
n
1 n
n
=
( 1 ) ( + )
1 +
1 n
n
1 n
n
c
k
d c
t
c
+
1
d k
t
k
+
1
( + )
d n
t
n
.
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Start with the effect of the productivity shock on the equation of motion where k
t
= 0because the
economy starts in the steady state:
d ( k
t + 1
+ c
t
) = k
n
1
d A + ( 1 ) A k
n
d n
t
.
We use the linearized first-order condition for the goods-leisure choice to find d nt
,
d c
t
c
=
d k
t
k
d n
t
n
d n
t
1 n
=
d n
t
n
d n
t
1 n
where withd k
t
= k
t
k
= 0
in the steady state. This gives us
d n
t
=
1 n
c
1 n
n
+ 1
d c
t
which substitutes into the equation of motion to get
d ( k
t + 1
+ c
t
) =
A
+
k
d A ( 1 ) A
A
+
1
A
1
1
+
1
( 1 )
1 n
n
+ 1
d c
t
=
A
+
k
d A
1
1
n
n
+ 1
d c
t
.
after substituting expressions for k
n
and 1 n
c
from part c.
This gives us one of the relationships we need,
d k
t + 1
+
1 n
n
+ 2
1 n
n
+ 1
d c
t
=
A
+
k
d A .
Another is the equation of the stable eigenvector which as the positive slope,
d c
t + 1
d k
t + 1
=
m
2 2
m
2 1
.
Now, you can see we need the Euler condition to get a relationship between d ct + 1
and d ct
. Use one of the
earlier versions that has kt + 1
in it,
c
t + 1
=
( 1 ) ( + )
1 +
1 n
n
1
( 1 ) ( + )
1 +
1 n
n
1 n
n
1
1 n
n
d c
t
+
c
k
d k
t
+
c
k
k
t + 1
=
( 1 ) ( + )
1 +
1 n
n
1
( 1 ) ( + )
1 +
1 n
n
1 n
n
1
1 n
n
d c
t
+
c
k
d k
t + 1
where the last equation uses d kt
= 0 .
The three conditions solve for d kt + 1
and d ct
simultaneously from
d k
t + 1
+
1 n
n
+ 2
1
n
n
+ 1
d c
t
=
A
+
k
d A ,
and
1
( 1 ) ( + )
1 n
n
1 +
1 n
n
1
( 1 ) ( + )
1 +
1 n
n
1 n
n
1
d c
t
=
m
2 2
m
2 1
+
( 1 ) ( + )
1 +
1 n
n
1
( 1 ) ( + )
1 +
1 n
n
1 n
n
1
c
k
d k
t + 1
,
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which imply thatd c
t
d A
> 0 andd k
t + 1
d A
> 0 ,
we haved n
t
d c
t
=
1 n
c
1 n
n
+ 1
0 . The eigenvectors satisfy
=
k
1
.
The slope of the saddle path that converges to the steady state is negative, and the slope of the divergent
saddle path is positive.
Note that the locus, k
t + 1
= 0
, is has zero slope and the locus, q
t + 1
= 0
, as negative slope. The
slope of the (linearized) stable saddle path is between these. This can be seen by comparing the slope of
q
t + 1
= 0
, d q td k
t
=
f
( k
)
f
( k
) k
, to the slope of the saddle path, k
,
k
f
( k
)
f
( k
) k
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number) gives us
2
k
f
( k
)
f
( k
) k
f
( k
) k
2
=
f
( k
) k
2
4
f
( k
) k
k
f
( k
)
f
( k
) k
+ 4
k
f
( k
)
f
( k
) k
2
=
f
( k
) k
2
4
k
f
( k
) + 4
k
f
( k
)
f
( k
) k
2
>
f
( k
) k
2
4
k
f
( k
) .
d) Letk
0
be less thank
, the steady-state capital stock. How do investment andq
behave over time?
For k0
< k
, the solution for q0
is given by the requirement that the transversality condition is satisfied.
This is l i mt
t
q
t
k
t + 1
= 0 . The is given by a pair ( k0
, q
0
) on the stable saddle path. Thus,
q
0
q
=
k
( k
k
0
)
where q = 1 + and k by f ( k ) = + +
+
2
. Both kt
and qt
converge toward their steady-state
values at the proportionate rate
. As k grows, q decreases.
4. Ans: a) The problem is
V ( k
t
) = m a x
{ c
t
, k
t
+
1
}
c
1
t
1
1
+ V ( k
t + 1
)
subject to
k
t + 1
= ( 1 + A ) k
t
c
t
for ct
0 and kt + 1
0 .
The necessary conditions include the first-order conditions,
c
t
=
t
V
( k
t + 1
) =
t + 1
,
and the envelope condition,
V
( k
t
) = ( 1 + A )
t
which also can be written
V
( k
t + 1
) = ( 1 + A )
t + 1
.
The necessary conditions include the equation of motion,k
t
+ 1
= ( 1 + A
)
k
t
c
t
, the constraints,c
t
0
and kt + 1
0 and a transversality condition, l i mT
t + T
T
k
t + T
0 .
b) First, eliminate the derivative of the value function, V ( kt
) and V ( kt + 1
) , to write the Euler condition
in terms of consumption as
c
t
= ( 1 + A ) c
t + 1
.
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Next, solve the Euler condition backward and the equation of motion for the capital stock forward to get
c
t + 1
c
t
= ( ( 1 + A ) )
1
c
s
c
t
= ( ( 1 + A ) )
1
( s t )
and
( 1 + A ) k
t
=
s = t
1
1 + A
t s
c
s
+ l i m
T
1
1 + A
T
k
T + 1
.
Imposing the transversality condition and non-negativity constraint for kT + 1
, we get
( 1 + A ) k
t
=
s = t
1
1 + A
t s
c
s
=
s = t
1
1 + A
s t
( ( 1 + A ) )
1
( s t )
c
t
which leads to
( 1 + A ) k
t
=
1
1
1
( 1 + A )
1
c
t
.
The solution for ct
is
c
t
=
1
1
( 1 + A )
1
( 1 + A ) k
t
.
c) First, calculate
m a x U
t
=
s = t
s t
c
1
s
1
1
=
s = t
s t
c
1
t
( ( 1 + A ) )
1
( s t )
1
1
=
c
1
t
1
s = t
s t
( ( 1 + A ) )
1
( s t )
s = t
s t
1
1
=
c
1
t
1
1
1
1
( 1 + A )
1
1
1
1
1
.
Next, use the solution for ct
in terms of kt
to substitute for ct
in m a x Ut
:
m a x U
t
=
1
1
( 1 + A )
1
1
1
1
( 1 + A )
1
( 1 + A )
1
k
1
t
1
1
1
1
1
=
( 1 + A )
1
1
k
1
t
1
1
1
1
1
.
The value function is given by
V ( k
t
) = m a x U
t
=
( 1 + A )
1
1
k
1
t
1
1
1
1
1
.
Note that this function has the form,
V ( k
t
) = C
k
1
t
1
+ D
for constantsC
andD
.
d) Substitute your function into the dynamic programming problem,
C
k
1
t
1
+ D = m a x
{ c
t
, k
t
+
1
}
c
1
t
1
1
+
C
k
1
t + 1
1
+ D
subject to
k
t + 1
= ( 1 + A ) k
t
c
t
forc
t
0and
k
t + 1
0.
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The first-order conditions are
C k
t
=
t
( 1 + A ) , C k
t + 1
=
t
and c t
=
t
.
These lead to the equations,
k
t + 1
=
1
( 1 + A )
1
k
t
and ct + 1
=
1
( 1 + A )
1
c
t
,
which imply that ct
= Z k
t
for some constant Z . Substituting into the integrated resource identity,
( 1 + A ) k
t
=
s = t
1
1 + A
t s
c
s
,
we verify the solution,
c
t
=
1
1
( 1 + A )
1
( 1 + A ) k
t
.