ejercicio ramsey bueno

7/28/2019 Ejercicio Ramsey Bueno

1/10

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

) .


2/10

2

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


3/10

3

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

=

+

.


4/10

4

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

.


5/10


6/10

6

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

,


7/10

7

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


8/10

8

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

.


9/10

9

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.


10/10

10

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

.

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