(2a)-P.1

The Solow Model: Agenda and Objectives

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Problems sets for practice. –

(2a)-P.2

Context: Growth Theory

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- => worth studying.

(2a)-P.3

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Y =F(K,AL)

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F(xK ,xAL) = x ⋅F(K ,AL)

x = 1AL

F(K,AL)= AL ⋅F(k,1)

k =K /(AL)

f (k) = F(k,1)

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F(0,1) = 0,FK (k,1)k→0⎯ → ⎯ ∞,FK (k,1)

k→∞⎯ → ⎯ ⎯ 0

KIKdtdK ⋅−== δ/

I = S = sY

LnLdtdL ⋅==/ AgAdtdA ⋅==/

(2a)-P.4

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Yt = F(Kt ,AtLt )

Kt+1 = Kt −δ ⋅Kt + It

It = s ⋅Yt

Yt − It

Δ

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K(t +Δt) = K(t)−δK(t)Δt + I(t)Δt - Δ

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K (t+Δt)−K (t)Δt = I(t)−δK(t)

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dK (t)dt = I(t)−δK(t)

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- stocks flow

- – – Δ

(2a)-P.5

A(0) = A0 A(t) = A0 ⋅egt

AL (

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k =K /(AL) -

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- - Variables on a balanced growth path = Steady state values * Exogenous growth trend.

(2a)-P.6

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Y = F(K ,AL) = AL ⋅F(k,1)

y = F(k,1) = f (k) -

k = K /AL

ngKK

LL

AA

KK

kk −−=−−=

- δδδ −=−=⇒−= ksyKsYKKKsYK ///

- )()( )( ngsngs kkf

ky

kk ++−=++−= δδ

dk /dt = sf (k)− (δ + g+n)k

sf (k*) = (δ + g+n)k* Equation for dk/dt and steady state condition are worth memorizing!

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sf (k)↔ (δ + g+n)k -

dk /dt = sf (k)− (δ + g+n)k

(2a)-P.7

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(2a)-P.8

(2a)-P.9

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(2a)-P.10

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sf (k*) = (δ + g+n)k*

f (k*)ds+ sf '(k*)dk* = (δ + g+n)dk*

dk*ds =

f (k*)(δ + g+n)− sf '(k*)

> 0

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y* = f (k*)

dy*ds = f '(k*)dk

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ds =f (k*) ⋅ f '(k*)

(δ + g+n)− sf '(k*)> 0

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c* = f (k*)− (δ + g+n)k*

dc*ds = [ f '(k*)− (δ + g+n)]dk

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ds

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(2a)-P.11

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αk (k) ≡ f '(k)k / f (k) -

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sf (k*) = (δ + g+n)k*

δ + g+n = s ⋅ f (k* )

k*

δ + g+n− s ⋅ f '(k*) = s ⋅ f (k* )

k*− s ⋅αk (k)

f (k)k = s f (k)

k ⋅(1−αk (k)) > 0

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dy*ds = f (k* ) f '(k* )

(δ+g+n)−sf '(k* )= f (k* )

sαk f (k

* ) /k*

f (k* ) /k*[1−αk (k* )]

= y*s

αk (k* )

1−αk (k* )

dy*y* = d ln(y*)

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d ln(y* )ds = 1

s ⋅αk (k

* )1−αk (k

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αk

d ln(y* )ds = 3 (

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dy* / y* = αk (k* )dk* / k* ]

(2a)-P.12

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dc*ds = [ f '(k*)− (δ + g+n)] dk*ds = [αk (k

*)− s] f (k* )

k*dk*ds

( f ' )−1(δ + g+n)

αk (k* )−s

s < αk (k* )

s > αk (k* )

s = αk (k* )

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s = α k (k*) [Fact to remember]

k* = ( f ' )−1(δ + g+n)

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s ≤ αk (k* )

αk

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(2a)-P.13

(2a)-P.14

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Y =F(K,AL)= Kα(AL)1−α

y = f (k) = kα

- α -

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MPK = dY / dK = α ⋅Kα −1 ⋅ (AL)1−α = α ⋅Y / K MPK ⋅K /Y =α

MPL = dY / dL = (1−α) ⋅Y / L

MPL ⋅ L /Y = 1−α -----------------------------------------------------

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- κ

ln(y) = ln( f (eκ )) = z(κ)

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z ' (κ) = 1f (eκ )

f ' (eκ )eκ =kf ' (k)f (k)

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z ' (κ) = α

z(κ) = z0 + α ⋅κ

z0

ln(y) = z0 + α ⋅ ln(k)

y = ez0 kα

z0

(2a)-P.15

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0=k

0 = s ⋅ kα −1− (δ + g+ n)

k = k* = sδ + g+ n⎛

⎝ ⎜

⎞

⎠ ⎟

11−α

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y* = k*⎛ ⎝ ⎜ ⎞

⎠ ⎟ α

=s

δ + g + n

⎛

⎝ ⎜

⎞

⎠ ⎟

α1−α

c* = (1− s) ⋅ y* = (1− s) ⋅s

δ + g + n

⎛

⎝ ⎜

⎞

⎠ ⎟

α1−α

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- α - α

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k(t) = [(k*)1−α +{(k(0))1−α − (k*)1−α} ⋅e−λt ]11−α

λ = (δ + g+n)(1−α) λ

- λ

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(2a)-P.16

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dk /dt = sf (k)− (δ + g+n)k = skα − (δ + g+n)k

z = k1−α = k / y -

dz /dt = (1−α)k−α ⋅dk /dt = [(1−α)k−α ] ⋅[skα − (δ + g+n)k]

= (1−α)s− (1−α)(δ + g+n)k1−α = (1−α)s− λz

λ = (δ + g+n)(1−α) > 0 linear

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dz /dt = −a ⋅ z+b

z(t) = ba + (z(0)− b

a) ⋅e−at

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z(t) = z* + (z(0)− z*) ⋅e−λt

z* = (1−α)sλ

= sδ+g+n

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k* = ( sδ+g+n)

1/(1−α)

k(t) = z(t)11−α = [(k*)1−α +{(k(0))1−α − (k*)1−α} ⋅e−λt ]

11−α

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k(0) = y(0)1α = Y (0)

L(0)A(0)( )1α

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y(t) = z(t)α1−α

(2a)-P.17

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λ(k) = (dk /dt)/(k* − k)

k ≠ k* [More general def. than in Romer]

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dk /dt = sf (k)− (δ + g+n)k

- λ

dk /dt ≈ [sf '(k*)− (δ + g+n)](k − k*)

λ = δ + g+n− sf '(k*)

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sf '(k*) = αk (k*)(δ + g+n)

λ = (1−αk (k*))(δ + g+n)

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λ = (1−α)(δ + g+n) is [Fact to remember]

δ + g+n ≈ 6%

α =1/3

λ ≈ 4%

λ

dk /dt ≈ −λ(k − k*)+ sf "(k* )2 (k − k*)2

λ(k) ≈ λ + (−sf "(k* )

2 )(k − k*) - λ λ λ λ

(2a)-P.18

Yt = F(Kt ,AtLt )

Kt+1 = Kt −δ ⋅Kt + It

It = s ⋅Yt

Yt − It

kt = Kt /(AtLt )

yt = f (kt ) = F(kt ,1)

Kt+1 = (1−δ) ⋅Kt + s ⋅Yt

Kt+1AtLt = s ⋅Yt AtLt + (1−δ) ⋅

KtAtLt

Kt+1AtLt = Kt+1

At+1Lt+1 ⋅At+1Lt+1

AtLt = kt+1(1+ g)(1+n)

kt+1(1+ g)(1+n) = s ⋅ f (kt )+ (1−δ) ⋅ kt

kt+1 = (1−δ )⋅kt+s⋅ f (kt )(1+g)(1+n)

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Δkt+1 = s⋅ f (kt )−{(1+g)(1+n)−(1−δ )}⋅kt(1+g)(1+n)

Δkt+1 = 0⇔ s ⋅ f (kt ) = {(1+ g)(1+n)− (1−δ)} ⋅ kt

s ⋅ f (k*) = (δ + g+n+ gn) ⋅ k*

gn --

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(2a)-P.19

Learning Objectives

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Problem sets for practice.

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