EC9012 - Economic Analysis: Macro
Irfan Qureshi
University of Warwick
October 27, 2015
Irfan Qureshi (University of Warwick)
EC9012 - Economic Analysis: Macro
October 27, 2015
1 / 18
Question - 1(a)
Constant Returns to scale
Remember for CRS: F (λK , λL) = λF (K , L)
For kt ≤ 2,
λ2 K 2
K2
= λ 6K −
= λF (K , L)
F (λK , λL) = 6λK −
λL
L
For kt > 2,
F (λK , λL) = 2λK + 4λL = λ (2K + 4L) = λF (K , L)
Irfan Qureshi (University of Warwick)
EC9012 - Economic Analysis: Macro
October 27, 2015
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Question - 1(a) - continued
Output per worker
F (Kt , Lt )
yt =
=F
Lt
yt =
Irfan Qureshi (University of Warwick)
Kt
,1
Lt
6k − k 2 for
2k + 4 for
= f (kt )
kt ≤ 2
kt > 2
EC9012 - Economic Analysis: Macro
October 27, 2015
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Question - 1(b)
Share of labour income: Take derivative of the
production function w.r.t to Lt
wt =
Irfan Qureshi (University of Warwick)
kt2 for
4 for
kt ≤ 2
kt > 2
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October 27, 2015
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Question - 1(b) - continued
In a competitive economy, share of labor = sL =
wL
Y
I know w from the previous part, Y and L are given.
Then:
k
6−k for kt ≤ 2
sL =
2
kt > 2
k+2 if
Irfan Qureshi (University of Warwick)
EC9012 - Economic Analysis: Macro
October 27, 2015
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Question - (c)
Solow Model: If there is no depreciation and no population growth, then
the capital accumulation equation is just:
kt+1 = sf (kt )
(1)
I know s = 1/3 (given) and f (kt ) from the previous part.
6k − k 2 /3 for kt ≤ 2
kt+1 =
≡ φ(kt )
(2k + 4) /3
for kt > 2
This is the dynamical system.
Irfan Qureshi (University of Warwick)
EC9012 - Economic Analysis: Macro
October 27, 2015
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Question - (c) - continued
Show: Unique, globally stable, non-trivial steady state
kt+1 =
6k − k 2 /3 for
(2k + 4) /3
for
kt ≤ 2
≡ φ(kt )
kt > 2
Set kt+1 = k̄ & kt = k̄. Then I can solve to obtain the steady states.
Steady states are k = 0, 4
For stability |φ0 (k̄)| < 1
There exists a unique globally stable steady state k̄ = 4.
Irfan Qureshi (University of Warwick)
EC9012 - Economic Analysis: Macro
October 27, 2015
7 / 18
Question - (d)
OLG Model:
Since individuals only consume in the second year of life - when old:
kt+1 = st = wt
⇒
kt+1 =
kt2 for
4 for
kt ≤ 2
≡ φ(kt )
kt > 2
Set kt+1 = k̄ & kt = k̄. Then I can solve to obtain the steady states.
Steady states are 0, 1 and 4.
I
I
For stability |φ0 (k̄)| < 1
Stable steady states are are 0 and 4. 0 is trivial. So the answer is
k̄ = 4.
Irfan Qureshi (University of Warwick)
EC9012 - Economic Analysis: Macro
October 27, 2015
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Question - (e)
In the Solow model savings are a constant fraction of income which is
concave with respect to capital.
In the OLG model savings are a constant fraction of the wage rate,
which is not concave with respect to capital.
The difference is not related to the endogenous saving rate. Here it is
always 1/3 in the Solow model and 1 in the OLG.
Irfan Qureshi (University of Warwick)
EC9012 - Economic Analysis: Macro
October 27, 2015
9 / 18
Question - 2
From the class notes (page 8):
wt = (1 − α)Aktα
Rt = αAktα−1
The maximization problem:
1
max ct + ct+1
ρ
s.t.
ct+1 = Rt+1 st = Rt+1 (wt − ct )
where, in a competitive economy, the individual takes prices, Rt+1 , as
given, and can be calculated precisely, due to rational expectations. Notice
that the budget constraint is simple the savings in the first period times
the return. Obviously next periods consumption cannot be greater than
this (since there is no borrowing)
Irfan Qureshi (University of Warwick)
EC9012 - Economic Analysis: Macro
October 27, 2015
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Intuition
What is ρ. Time preference. Or how much the individual cares about
the next period.
In principle there will three cases regarding the relationship between
Rt+1 and ρ
Actual capital greater than steady state capital, excess capital
implying low return
Actual capital equal to steady state capital, high return
Actual capital equal to steady state capital = equilibrium
So savings would depend on this.
Irfan Qureshi (University of Warwick)
EC9012 - Economic Analysis: Macro
October 27, 2015
11 / 18
Part - (a) (ii)
→

if
 = wt
∈ [0, wt ] if
st

=0
if
Rt+1 > ρ(ct = 0&ct+1 > 0)
Rt+1 = ρ(ct > 0&ct+1 > 0)
Rt+1 < ρ(ct > 0&ct+1 = 0)
where,
Rt+1 > ρ if
Rt+1 = ρ if
Rt+1 < ρ if
kt+1 < k̄
kt+1 = k̄
kt+1 > k̄
α−1
Recall tha Rt+1 = αAkt+1
. Then if, Rt+1 = ρ:
R
=
αAk̄ α−1 = ρ
↔
k̄
=
αA
ρ
1
1−α
This implies a particular and constant level of kt+1 (lecture notes page 10).
Irfan Qureshi (University of Warwick)
EC9012 - Economic Analysis: Macro
October 27, 2015
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Part - (a) (ii)
Clearly k̄ is an important landmark to understand the law of motion that
governs the evolution of kt . Two important points:
The individuals born in period t saving, investment decision affects
the return to this investment. Since the return to capital tomorrow
will be R = αAk̄ α−1 we can see this return as a function of kt+1
Because the individuals budget constraint in the first period is
ct + st ≤ wt then it is clear that the wage at time t will depend on
the stock of capital in the economy when the individual from the t
generation was young wt = (1 − α)Aktα , and will be an upper bound
for the stock of capital st = kt+1 ≤ wt
Irfan Qureshi (University of Warwick)
EC9012 - Economic Analysis: Macro
October 27, 2015
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Part - (a) (iii)
If st = kt+1 ≥ k̄ then Rt+1 < ρ and the economy will consume everything
in the first period and therefore savings in the second period will be zero.
So the second corner ct = wt > 0, c − t + 1 = 0 is not a feasible solution.
Then there is a critical value of kt defined by:
wt = (1 − α)Ak̂ α = k̄
→
k̂ =
k̄
(1 − α)A
Irfan Qureshi (University of Warwick)
1/α
=
1
(1 − α)A
EC9012 - Economic Analysis: Macro
αA
ρ
1
1−α
!1/α
October 27, 2015
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Part - (a) (iii)
If the individual is born in an economy with starting capital k0 < K̂
then she will save all her wage wt = (1 − α)Aktα = k1 < k̂. Clearly
k1 > k0 . If k1 < k̂ then we repaet this process.
From the opposite side if k0 is high enough such that k1 > k̂ then the
the upper bound wt will be above k̄: w1 = (1 − α)Ak1α > k̄. Then
the individual will have three alternatives for her saving decision and
here is where the fact that the individuals saving decision determines
the return to her savings (capital) and plays a crucial role:
I
I
I
If s1 = w1 > k̄ (save all, consume nothing at t = 1), then s1 = k1 > k̄
and therefore the return to savings R2 will be less than ρ. This implies
that s1 = 0 was the optimal strategy.
If s1 < k̄ (save less than k̄), then the return will be R1 > ρ which
implies that saving the whole wage w1 was the optimal strategy.
Finally the individual could choose to save s1 = k̄. In this case R1 = ρ
which implies that k1 = k̄
Irfan Qureshi (University of Warwick)
EC9012 - Economic Analysis: Macro
October 27, 2015
15 / 18
Part - (a) (iii)
One can think of this decision making process as a trial and error
process.
First the individual considers saving everything.
Then realizing that it is not optimal, it considers saving less than the
whole wage w1 and will do this until it reaches k̄ in which he satisfies
the above constraints.
If the individual starts by saving less than k̄ then she will later realize
that it reaches an equilibrium if and only if k1 = k̄.
Irfan Qureshi (University of Warwick)
EC9012 - Economic Analysis: Macro
October 27, 2015
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Part - (a) (iii)
Finally, we can define the dynamical system as:
(1 − α)Aktα if kt < k̂
kt+1 =
k̄
if kt ≥ k̂
Main message: If depreciation = 1, then savings can only occur till a
certain point due to the return on capital.
Irfan Qureshi (University of Warwick)
EC9012 - Economic Analysis: Macro
October 27, 2015
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Part - (b)
In order to ensure that the steady state has positive consumption for
agents when young we need k̂ < k̄. Otherwise steady state will be
characterized by ct = 0 ∀ t
k̂
<
k̄
↔
k̄
(1 − α)A
1/α
[(1 − α)A]
Irfan Qureshi (University of Warwick)
<
k̄
1
1−α
>
k̄ =
ρ
>
α
1−α
αA
ρ
EC9012 - Economic Analysis: Macro
1
1−α
October 27, 2015
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