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(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
- [ ]
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
*
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
*
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
* )
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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