MACROECONOMICS 2018 Problem set n° 2 Prof. Nicola Dimitri

MACROECONOMICS 2018

Problem set n° 2 Prof. Nicola Dimitri

1) Consider the OLG with money seen in class, where the two periods endowments of the consumption good for the representative agent are e₁>0 _and e₂>0 , with the population growing according to the following equation

N

t

=(1+n)

^t

N

0

Suppose the good is perishable, that is the storage technology is characterized by a return rate

r=−1

.

a) Write down the budget constraints, for the two periods, and the inter-temporal budget constraint.

b) If m_t=M_t

p_t , discuss the dynamic stability of monetary stationary state, that is the stability of the real- money market equilibrium equation

m

_{t +1}

=h(m

_t

)

, when the utility function of the representative agent

W ( ^c

1 t

, c

_{2t +1}

)

is

i)

c

_{1 t}

+ log c

_{2t +1}

1+θ

ⁱⁱ⁾

c

_{1 t}

+ c

_2t

1+θ

2) Consider the OLG without money seen in class, where the two periods endowments of the consumption good for the representative agent are

e

₁

=0

_and

e

₂

=0

. Population grows as in problem 1. Suppose the utility function of the representative agent

W ( c

_{1 t}

, c

_{2t +1}

) =logc

_{1 t}

+ c

_{2t +1}

1+θ

and the production function

f ( ^k

t

) ^=k

t

a , with

0<a<1

. Find the dynamic law driving the evolution of the equilibrium per capita capital

k

_t , discuss its stationary states and its dynamics

3)

A) Consider the following one variable

x

_t

≥0

, discrete time, dynamical system

x

_t

=a x

_{t −1}

( ^b−x

t−1

) ^{+ c} ( ^b−x

t−1

) (Generalised logistic function) with a , b , c> 0

Draw a graph of the function, find the Stationary States and discuss their stability. Can the system ever exhibit cyclical dynamics?

4) Consider the two variable, first order linear difference equation, system x_t=2 x_t−1+y_{t −1}+5

y

_t

=0.3 y

_t−1

+1

Find the stationary state of the system and discuss whether or not is a saddle point.

Discussion

The system could be written in matrix term as

u_t=A u_{t −1}+b

Where

u

_t

= [ ^{x y}

^tt

]

^,

^{A= [ 0 0.3 2 1 ]}

^,

^{b= [ 5 1 ]}

. The stationary state is u_s=¿

[ ^{x y}

^s

^{=−45 /7}

^s

^=10/7 ]

Note that u_s solves the equation u_s=A u_s+b _hence

u

_s

=( I− A)

⁻¹

b

where

( I − A)

⁻¹

( I− A)

⁻¹

= [ ^−1−10/7 0 10 /7 ]

The eigenvalues are λ₁=2 _and λ₂=0,3 . Therefore, u_s is a saddle point and the unique path converging to it the saddle path. But what is the saddle path in this variation of the linear model, with the vector

b

adding up to

A u

_{t−1 ?}

This is how to proceed. The model can be written as

u

(¿¿ t−1+u

_s

−u

_s

)+b u

_t

= A

¿

. Define

z

_t¿

u

_t

−u

_{s and so}

u_t=A z_t−1+A u_s+b But

Au_s+b=A (I− A)⁻¹b+b=

(

^{I +A}

(

I −A

)

⁻¹

)

^b=(

(

I− A

)(

I− A

)

⁻¹+

(

I −A

)

⁻¹)b

hence

( I −A )+ A

¿

( I − A)

⁻¹

b=u

_s

Au

_s

+ b=¿

Therefore

u_t=A z_t−1+A u_s+b can be written as

z

_t

= A z

_{t −1}

As stationary state

z

_s the above system has the origin

z

_s

=0

. Therefore, the saddle path of z_t is the eigenvector v₂=¿

[ ^v

¹²

^{=−17 x v}

²²

^=R

²

^/10 ]

^{of A} associated to

λ

₂

= 0,3

. As a consequence, if

z

₀

=u

₀

−u

_s must be located on the vector

v

_{2 to} converge to

z

_s

=0

it follows that

u

₀

= z

₀

+u

_s , that is the

u

_t must be initially located on the vector

v

₂ to which the stationary state is added. This means that

u

₀ must be located over the parallel to

v

_{2 crossing}

u

_s , as we saw in class.