• In the last lecture we linearized, and then solved the differential equation underlying the Solow model to derive the time paths of the endogenous variables away from steady state

Spring Semester ’12-’13 Akila Weerapana

Lecture 9: Dynamics of the Solow Model

I. OVERVIEW

• In the last lecture we linearized, and then solved the differential equation underlying the Solow model to derive the time paths of the endogenous variables away from steady state

• In today’s class, we will use these solutions to look at how policy changes and other economic changes impact the economy using the Solow model with technology. Through these exercises we will be able to gain a better understanding of what can lead to sustained economic growth and rapid increases in economic growth in countries.

II. THE DYNAMICS OF THE SOLOW MODEL An Increase in the Saving Rate s ↑

• Just to recap, the solution for the path of capital per-effective worker is

ln ˜ k _t = ω _t ln ˜ k ₀ + (1 − ω t ) ln ˜ k ^∗ where ω _t = e ^−λt and ln ˜ k ^∗ =

( 1

1 − α )

ln

( s

n + δ + g )

• According to the equation, an increase in s to s ^′ raises the steady state capital stock to ln ˜ k ^∗

=

( 1 1 −α

) ( s

n+g+δ

)

. As we have seen, the level of capital per effective worker will then be a weighted average of the starting value (the original steady state) and the new steady state. The weight on the starting value will decline rapidly from 1 initially, and gradually slow down as it approaches zero.

• So in other words, the economy starts out at ˜k 0 , will start growing towards the new steady state ˜ k ^∗ gradually and eventually reach the new steady state.

• The resulting time-path for ln ˜k is shown below. I have assumed that the economy was in steady state to begin with.

- 6

ln ˜ k

t ln ˜ k ₀ ^∗

ln ˜ k ₁ ^∗

t ₀ t ₁

q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q

- 6

ln ˜ y

t ln ˜ y ^∗ ₀

ln ˜ y ^∗ ₁

t ₀ t ₁

q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q

• You can also capture this fairly elegantly using MATLAB, following the model we did for the AD-IA model. Choose some arbitrary values such as s = 0.3, n = 0.02, g = 0.03, δ = 0.05, α = 1/3 as global parameters and calculate a baseline path (you can choose ˜ k ₀ to be the steady state or not) for ˜ k ^∗ and ˜ y ^∗ . Now change one or more of the basic parameters (an increase in s in this case) and calculate the new steady state. Then use the weights to construct the time-paths for varying values of t.

• If I simulate what happens when s rises from 0.3 to 0.4, the resulting diagram is shown below. As you can see it mirrors the diagram we drew above. The red dashed line shows the counterfactual, and the green solid lines show the two points that the weights get applied to (this convention applies to all the graphs shown from now on in this lecture)

...

..

0 .

20 .

40 .

60 .

80 .

100 .

1.7 . 1.8

. 1.9

. 2

. 2.1

.

t .

ln

˜ k ^t

...

..

0 .

20 .

40 .

60 .

80 .

100 0.55 . .

0.6

. 0.65

. 0.7

.

t .

ln ˜y t

• Once we have the graphs for ˜k and ˜y we can then deduce the behavior of k and y. Similarly, once we have the graphs for k and y we can deduce the behavior of K and Y .

• Since ˜k = _A ^k ⇒ ln ˜k + ln A = ln k. SInce A t = A ₀ e ^gt , ln A _t = ln A ₀ + gt. The solution for ln k _t becomes (as we showed in the previous class)

ln k _t = ω _t (ln ˜ k ₀ + ln A ₀ + gt) + (1 − ω t )(ln ˜ k ^∗ ₁ + ln A ₀ + gt)

• The timepath of k t is a weighted average of the time-path of k in an economy with a steady-

state based on ˜ k 0 and the time-path of k in an economy with a steady-state based on ˜ k ^∗ ₁ , with

k _t growing rapidly from the lower path to the higher path. Since the steady state path for

k is growing at a rate g and g did not change, the diagram for the timepath of capital per

worker looks like the following (a similar argument can be made for the time-path of y t ).

...

..

0 .

10 .

20 .

30 .

1.5 . 2

. 2.5

. 3

.

t .

ln k t

...

..

0 .

10 .

20 .

30 0.5 . .

1

. 1.5

.

t .

ln y t

• Since k = ^K _L ⇒ ln k + ln L = ln K. SInce L t = L ₀ e ^nt , ln L _t = ln L ₀ + nt. The solution for ln K t becomes (as we showed in the previous class)

ln K _t = ω _t (ln ˜ k ₀ + ln A ₀ + ln L ₀ + (n + g)t) + (1 − ω t )(ln ˜ k ^∗ ₁ + ln A ₀ + ln L ₀ + (g + n)t)

• The timepath of K t was a weighted average of the time-path of K in an economy with a steady-state at ˜ k ₀ and the time-path of K in an economy with a steady-state at ˜ k ₁ ^∗ , with K _t growing rapidly from the lower path to the higher path. Since the steady state path for K is growing at a rate g + n and neither g nor n changed, the diagram for the timepaths for K and Y look like the following

...

..

0 .

10 .

20 .

30 2.5 .

. 3

. 3.5

. 4

. 4.5

.

t .

ln K t

...

..

0 .

10 .

20 .

30 .

1.5 . 2

. 2.5

. 3

.

t .

ln Y t

An Decrease in the Population Growth Rate n ↓

• Suppose there is a sudden decrease in the population growth rate from n to n ^′ . This pushes the break even line down, and raises the steady state capital stock to ln ˜ k ^∗

=

( 1 1 −α

) ( s n

+g+δ

) .

• Since the initial position did not change, the economy starts out at ˜k 0 , and will start growing

towards the new steady state ˜ k ^∗ gradually and eventually reach the new steady state.

...

..

0 .

20 .

40 .

60 .

80 .

100 .

1.7 . 1.8

. 1.9

. 2

.

t .

ln

˜ k ^t

...

..

0 .

20 .

40 .

60 .

80 .

100 .

0.54 0.56 .

. 0.58

. 0.6

. 0.62

. 0.64

. 0.66

.

t .

ln ˜y t

• The solution for ln k t is

ln k _t = ω _t (ln ˜ k ₀ + ln A ₀ + gt) + (1 − ω t )(ln ˜ k ^∗ ₁ + ln A ₀ + gt)

• The timepath of k t is a weighted average of the time-path of k in an economy with a steady- state based on ˜ k 0 and the time-path of k in an economy with a steady-state based on ˜ k ^∗ ₁ , with k t growing rapidly from the lower path to the higher path. Since the steady state path for k is growing at a rate g and g did not change, the diagram for the timepath of capital per worker looks like the following (a similar argument can be made for the time-path of y _t ).

...

..

0 .

10 .

20 .

30 .

2

. 2.5

. 3

.

t .

ln k t

...

..

0 .

10 .

20 .

30 0.5 . .

1

. 1.5

.

t .

ln y t

• The solution for ln K t is

ln K t = ω t (ln ˜ k 0 + ln A 0 + ln L 0 + (n ^′ + g)t) + (1 − ω t )(ln ˜ k ^∗ + ln A 0 + ln L 0 + (n ^′ + g)t)

• The timepath of K t was a weighted average of the time-path of K in an economy with a

steady-state at ˜ k ₀ and the time-path of K in an economy with a steady-state at ˜ k ₁ ^∗ , with a

growth path of g + n ^′ .

...

..

0 .

10 .

20 .

30 2.5 .

. 3

. 3.5

. 4

.

t .

ln K t

...

..

0 .

10 .

20 .

30 .

1.5 . 2

. 2.5

. 3

.

t .

ln Y t

• As you can see the economy is better off in per-capita terms but worse off in aggregate terms.

An Decrease in the Capital Stock K 0 ↓

• Suppose there is a sudden destruction of the capital stock. This lowers the initial level of capital per effective worker. The economy sees its starting position at ˜ k ₀ drop to ˜ k ^′ ₀ but nothing else changes.

• Using our intuitive weighted average solution for the time-path, we can see that since the original steady state never changed, which means that the economy will return back to the original steady state. A simulation for a destruction of 1/3rd of the capital stock is shown below.

...

..

0 .

20 .

40 .

60 .

80 .

100 .

1 . 1.2

. 1.4

. 1.6

.

t .

ln

˜ k ^t

• The timepath of k t was a weighted average of the time-path of an economy with a steady-state

at ˜ k ^′ ₀ and an economy with a steady-state at ˜ k ^∗ , with k t growing rapidly from the lower path

to the higher path. Since the steady state path for k is growing at a rate g and g did not

change, the diagram for the timepath of capital per worker (and output per worker) looks

like the following:

...

..

0 .

10 .

20 .

30 1 . .

1.5

. 2

. .

t .

ln k t

...

..

0 .

10 .

20 .

30 .

0.5 . 1

. .

t .

ln y t

• The solution for ln K t is

ln K _t = ω _t (ln ˜ k ₀ + ln A ₀ + ln L ₀ + (n ^′ + g)t) + (1 − ω t )(ln ˜ k ^∗ + ln A ₀ + ln L ₀ + (n ^′ + g)t)

• The timepath of K t was a weighted average of the time-path of K in an economy with a steady-state at ˜ k ^′ ₀ and the time-path of K in an economy with a steady-state at ˜ k ^∗ , with a growth path of g + n ^′ .

...

..

0 .

10 .

20 .

30 .

1.5 . 2

. 2.5

. 3

. 3.5

. 4

.

t .

ln K t

...

..

0 .

10 .

20 .

30 1 . .

1.5 . 2

. 2.5

. 3

.

t .

ln Y t