University of Warwick
EC9A2 Advanced Macroeconomic Analysis
Problem Set #1: Exogenous Growth Models
Jorge F. Chavez∗
December 3, 2012
Question 1
Production is given by:
Yt ≡ F (Kt , Lt ) = AKtα L1−α
t
where Lt+1 = (1 + n)Lt and α ∈ (0, 1).
a. Show that F exhibits a constant return to scale technology
Solution. This implies showing that F : R2 → R is homogeneous of degree 1. That’s it, we need to
show that for λ > 0, F (λKt , λLt ) = λF (Kt , Lt ). Then:
α
1−α
F (λKt , λLt ) = A(λKt ) (λLt )
= λAKtα L1−α
= λF (Kt , Lt )
t
b. Express output as a function of the capital labor ratio kt = Kt /Lt .
Solution. Use the fact that F is homogeneous of degree 1:
( )α
Kt
F (Kt , Lt ) = AKtα L1−α
=
L
A
= Lt Aktα ;
t
t
Lt
c. Find the wage rate per worker and the rental rate per capital.
Solution. Profit maximization (either by a social planner or by the representative firm in the economy)
yields:
wt
=
Rt
=
∂F (Kt , Lt )
= (1 − α) AKtα L−α
= (1 − α) Aktα
t
∂Lt
∂F (Kt , Lt )
= αAKtα−1 L1−α
= αAktα−1
t
∂Kt
Note that this assumes that the good acts here as a numeraire (its price is set to 1).
∗
e-mail:j.chavez-cotrado@warwick.ac.uk
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EC9A2 (Fall 2012)
Problem Set # 1
d. Find the dynamical system (describing the evolution of kt over time) under the assumption
that the saving rate is s ∈ (0, 1) and the depreciation rate is δ ∈ (0, 1].
Solution. We need to characterize the path of {kt }∞
t=0 given s ∈ (0, 1) and δ ∈ (0, 1]. That is, we need
to find the non-linear difference equation kt+1 = ϕ(kt ).
Start from the law of motion of the aggregate stock of capital:
Kt+1 = sYt + (1 − δ)Kt
Put everything in per-worker terms by dividing both sides by Lt+1 :
kt+1
=
Yt Lt
Kt Lt
Kt+1
=s
+ (1 − δ)
Lt+1
Lt Lt+1
Lt Lt+1
=
sAktα + (1 − δ) kt
≡ ϕ (kt )
1+n
(1)
where I am using yt = f (kt ) = Aktα . Sometimes it useful to re-express this condition in terms of
∆kt+1 . To do so subtract kt from both sides of (1) to get:
∆kt+1 =
sAktα − (n + δ) kt
1+n
e. What is the growth rate of kt , γkt ≡ (kt+1 − kt ) /kt ?
Solution.
γkt =
kt+1 − kt
sAktα−1 − (δ + n)
=
kt
1+n
f. Find the steady state level of the stock of capital per worker k, income per worker y and
consumption per capita c.
ss
= ktss = k, which implies that γk = 0. Then from the expression for
Solution. In steady-state kt+1
γk t :
(
k=
sA
δ+n
)1/(1−α)
Income per worker is
(
α
y = f (k) = Ak = A
1
1−α
s
δ+n
α
) 1−α
Finally, to get consumption per worker in steady state note that in general yt = ct + it , it = syt and
ct = (1 − s)yt . Then, in steady state:
(
c = (1 − s) y = (1 − s) A
Jorge F. Chávez
1
1−α
s
δ+n
α
) 1−α
2
EC9A2 (Fall 2012)
Problem Set # 1
g. What is the Golden Rule value of k? (k in steady state s.th. the consumption in steady state
is maximized?)
Solution. The Golden Rule value of kt is the stock of capital per worker that maximizes consumption
in steady-state. Recall that in general:
ct
= (1 − s)yt
= f (kt ) − sf (kt )
Note that there is no way to maximize consumption in all states as consumption is a function of f (kt )
which is unbounded (we can only get the “corner” solution kt = 0 if we look for a stationary point).
However, the steady-state condition is:
sf (k)
| {z }
=
Investment per capita
(n + δ) k
| {z }
(2)
Part of k that is “lost” because of pop.
growth & depreciation
ss
which is true for any steady-state (any setting in which kt+1
= ktss = k). Then because ct =
f (kt ) − sf (kt ), in steady state:
c = f (k) − (n + δ)k
which now will accept an interior solution. The FONC for the maximization problem is:
∂c
= f ′ (k) − (n + δ) = 0
∂k
⇒
(
)
f ′ k GR = n + δ
For y = Ak α we can get a closed form solution for k GR :
(
)
f ′ k GR
=
k GR
=
(
)α−1
αA k GR
=n+δ
(
)1/(1−α)
αA
n+δ
(3)
The concept of the Golden Rule for the case of the Solow model is illustrated in figure 1.( There
) you can
see three alternative saving rates, and you can visualize the Golden Rule condition f ′ k GR = n + δ:
the slope of f (·) must be equal to the slope of the (n + δ)k line.
h. What saving rate is needed to yield the Golden Rule?
Solution. Comparing equation (3) with the expression for capital in steady state it is straightforward
to see that the saving rate that allows the economy to reach k GR is sGR = α.
Alternatively recall that f ′ (k GR ) = n + δ and kt f ′ (kt )/f (kt ) = α. Then:
(
)
k GR f ′ k GR
k GR (n + δ)
=
=α
GR
f (k )
f (k GR )
But then the Golden Rule level of capital also satisfies the more general condition for all steady-states.
Combining that condition (2) with the above one we can see that sGR = α.
i. Find the elasticity of y with respect to s (in steady-state). Can observed differences in saving
rates explain the observed differences in income per-capita across the world?
Jorge F. Chávez
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EC9A2 (Fall 2012)
Problem Set # 1
Figure 1: The Golden Rule
f (k ss )
c1
cGR
(n + δ)k ss
s1 f (k ss )
c2
sGR f (k GR )
s2 f (k ss )
k2 k GR
k1
k ss
Solution. Recall that:
(
)α/(1−α)
sA
y = Ak α = A
n+δ
Taking logs1 :
ln y =
α
α
ln s + . . .
ln A +
1−α
1−α
| {z }
εys
Suppose that α = 1/3 as it is usually found in empirical studies and there is a difference in saving rates
of 3 times across countries (300%). Then the elasticity will be α/(1 − α) = 0.5 which implies that the
difference in y according to this model (with Cobb-Douglas technology) should be 0.5 × 300% = 150%.
However the observed difference is nearly 20 times (this comes from the lecture notes).
j. Find the dynamical system describing the evolution of yt under the assumption of full depreciation δ = 1.
Solution. With δ = 1 the law of motion of the stock of capital per worker is just:
kt+1 =
1
sAktα
1+n
To see this:
∂ log y
=
∂ log x
Jorge F. Chávez
∂ log y
∂y
∂y
∂ log x
∂x
∂x
≈
∆y
y
∆x
x
=
%∆y
%∆x
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EC9A2 (Fall 2012)
Problem Set # 1
Note that the numerator is investment in per capita terms. Then, this expression says that the new
capital stock (which is completely renewed each period) is lower than investment per worker, due to
population growth.2 We want to characterize {yt }∞
t=0 by analyzing a non-linear difference equation
yt+1 = φ(yt ). Recall that yt = Aktα . Then:
(
)α
(
)α
sAktα
syt
α
yt+1 = Akt+1 = A
=A
≡ φ (yt )
1+n
1+n
( i
)
k. Suppose you estimate a regression in which ln yt+1
/yti is on the left hand side (i.e.
you)
(
i
i
estimate the log of γyt + 1 across countries, where i indexes countries) and ln s , ln 1 + ni
and ln yti are on the right hand side. According to the dynamical system you defined in
question j, what would be the coefficients on your explanatory variables? How would you
interpret these coefficients? Is there “β convergence”? How would you interpret the constant
term?
Solution. Consider the regression:
( i )
)
(
yt+1
ln
= ln A + γ1 ln si + γ2 ln 1 + ni + γ3 ln yti + εit
i
yt
From the model we can get an estimable form by taking logs to the law of motion of yt+1 when δ = 0:
i
ln yt+1
= ln A + α ln s − α ln (1 + n) + α ln yti
Then by subtracting lnyti from both sides of the equation above we can see that γ0 = ln A, γ1 = α,
γ2 = −α and γ3 = −(1 − α).
2
Investment occurs in t while the new capital stock will be readily available in t + 1.
Jorge F. Chávez
5
EC9A2 (Fall 2012)
Problem Set # 1
Question 2
Output per worker is an increasing and concave function of capital per worker, given by: yt =
f (kt ). Output is divided between labor income and capital income according to their marginal
productivity. Namely,
[
]
f (kt ) ≡ f (kt ) − f ′ (kt ) kt + f ′ (kt ) kt = wt + rt kt
Suppose that the rate of saving from wage income is sw ∈ [0, 1] and the rate of saving from
capital income is sr ∈ [0, 1]. Therefore, total saving are given by
[
]
st = sw f (kt ) − f ′ (kt ) kt + sr f ′ (kt ) kt
Population and technology are constant and the rate of capital depreciation δ ∈ (0, 1)
a. Derive the dynamical system governing the evolution of capital per capita: kt+1 = ϕ(kt ).
Solution. Recall that the law of motion of the stock of capital per capita comes from the law of motion
of the aggregate stock per capita:
Kt+1 = It + (1 − δ)Kt
where It denotes aggregate investment. Because there are two distinct saving rates, we can think of
aggregate savings to be determined by an “average” saving rate (s̄) which is a weighted sum of sw
and sr :
Kt+1 = s̄Yt + (1 − δ)Kt
Now, because there is no population growth, if we divide everything by L we get:
kt+1 = st + (1 − δ)kt
where st = s̄f (kt ) = sw [f (kt ) − f ′ (kt )kt ] + sr f ′ (kt )kt . Then:
kt+1
= sw [f (kt ) − f ′ (kt )kt ] + sr f ′ (kt )kt + (1 − δ)kt
= sw f (kt ) + (sr − sw ) f ′ (kt )kt ≡ ϕ(kt )
Note that if sw = sr , then the expression for φ(kt ) is the same we had before with a single saving
rate.
b. Suppose there exist a range of kt where f ′′′ (k) = 0 (the third derivative is zero). Find a
condition on the saving rates sw and sr such that the dynamical system kt+1 = ϕ(kt ) is
convex (ϕ′′ (kt ) > 0 in the range of f ′′′ (kt ) = 0.
Solution. First let’s get the derivatives:
ϕ′ (kt )
=
sw f ′ (kt ) + (sr − sw ) [f ′′ (kt ) kt + f ′ (kt )] + (1 − δ)
ϕ′′ (kt )
=
sw f ′′ (kt ) + (sr − sw ) [f ′′′ (kt ) kt + f ′′ (kt )]
Let k̄ be the maximum stock of capital per worker that can be attained by economy and assume that
Jorge F. Chávez
6
EC9A2 (Fall 2012)
Problem Set # 1
∃k1 ≤ k2 ∈ [0, k̄] such that ∀kt ∈ [k1 , k2 ], f ′′′ (kt ) = 0. Then, for those values of kt , using (4):
ϕ′′ (kt )|kt ∈[k1 ,k2 ]
=
sw f ′′ (kt ) + (sr − sw ) f ′′ (kt )
=
(2sr − sw ) f ′′ (kt )
Therefore, for ϕ(kt ) > 0 we need 2sr < sw and strict concavity in the range [k1 , k2 ] (concavity is not
enough) of f .3
Suppose now that sr = 0 and f (kt ) = ln (1 + kt )
c. Derive the dynamical system kt+1 = ϕ(kt ).
Solution. Now savings come only from wage income:
kt+1 = sw f (kt ) − sw f ′ (kt ) kt + (1 − δ) kt ≡ ϕ(kt )
Replacing the functional form for f (·) we get that f ′ (kt ) = 1/(1 + kt ). Then:
kt+1 = sw log (1 + kt ) − sw
kt
+ (1 − δ) kt ≡ ϕ (kt )
1 + kt
d. Find ϕ′ (kt ), limkt →0 ϕ′ (kt ), limkt →+∞ ϕ′ (kt ), and ϕ′′ (kt ).
Solution.
(i) First derivative
′
ϕ (kt )
=
=
sw
− sw
1 + kt
sw kt
(1 + kt )
2
(
(1 + kt ) − kt
2
(1 + kt )
)
+ (1 − δ)
+ (1 − δ)
(ii) Limit when kt → 0
lim ϕ′ (kt ) = 1 − δ
kt →0
(iii) Limit when kt → +∞
lim ϕ′ (kt )
kt →+∞
=
=
lim
kt →+∞
sw kt
2
(1 + kt )
+ (1 − δ)
1−δ
Note that the limit of the first term of the RHS is not undetermined (no need to apply L’Hospital
rule)
3
Strict concavity implies that f ′′ (kt ) < 0
Jorge F. Chávez
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EC9A2 (Fall 2012)
Problem Set # 1
(iv) Second derivative:
ϕ′′ (kt ) =
[
sw
= s
4
1 + 2kt + kt2 − 2kt − 2kt2
(1 + kt )
(
]
4
)
1 − kt2
w
= s
]
(1 + kt )
[
w
2
(1 + kt ) − 2 (1 + kt ) kt
4
(1 + kt )
e. Is the trivial steady state, k = 0, locally stable? Explain.
Solution. To analyze stability of the trivial steady-state (k = 0), we need to evaluate the limit of the
first derivative of ϕ′ (kt ):
lim |ϕ′ (kt ) | = (1 − δ) ∈ (0, 1)
kt →0
Therefore k = 0 is locally stable.
Remark 1. Recall that for a non-linear difference equation like xt+1 = g(xt ) with fixed-point or
steady state xss4 . Then:
• If |f ′ (xss )| < 1 then xss is locally asymptotically stable.
• If |f ′ (xss )| > 1 then xss is locally asymptotically instable.
Remark 2. Recall that in the standard case:
kt+1 = sf (kt ) + (1 − δ)kt
Then, to analyze the stability of k = 0 we need:
k = 0 ⇒ ϕ′ (0) = s f ′ (0) + (1 − δ) > 1
| {z }
→+∞
which implies that in that case k = 0 is unstable.
f. Find the range of kt in which the dynamical system is strictly convex.
Solution. We want to find k1 and k2 such that ∀kt ∈ [k1 , k2 ], ϕ′′ (kt ) > 0. Recall that:
ϕ′′ (kt ) = sw
1 − kt2
(1 + kt )
4
Now, setting ϕ′′ (kt ) > 0 ⇒ 1 − kt2 > 0 ⇒ kt2 < 1 ⇒ kt < 1 or kt > 1. But kt is the stock of physical
capital per worker, so in principle kt ≥ 0. Thus, because ϕ′′ (0) > 0, the range we were looking for is
[0, 1].
g. Show that for δ = 0 a non-trivial steady state level of k does not exist (that is, explain why
there exists no k̄ > 0 such that k̄ = ϕ(k̄)). Find the growth rate of kt (i.e., kt+1 /kt − 1) as
kt → ∞ for δ = 0.
4
That is xss = g(xss )
Jorge F. Chávez
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EC9A2 (Fall 2012)
Problem Set # 1
Solution. WTS if δ = 0 then @ a non-trivial steady state k > 0 for {kt }. Note that with δ = 0:
lim ϕ′ (kt ) =
kt →0
lim ϕ′ (kt ) = 1
kt →+∞
which means that this system does not have an interior steady-state.
More formally:
Claim 1. There is no non-zero fixed point for ϕ(kt ) (that is @k > 0 s.th. k = ϕ(k)).
Proof. Suppose not: ∃k > 0 s.th k = ϕ(k). Then:
ϕ (kt )|δ=0 = sw log (1 + kt ) − sw
kt
+ kt
1 + kt
Then k > 0 must satisfy:
k = ϕ (k)|δ=0 = sw log (1 + k) −
sw k + k + k 2
1+k
Manipulating this expression a little bit we get to an expression that k > 0 must satisfy:
log (1 + k) (1 + k)
=
k
1+k
=
exp
(
k
1+k
)
(4)
Think about slopes. The LHS is a linear function with a constant slope equal(to 1.) The RHS is an
k
exponential function with slope < 1 for all values of k ≥ 0.5 For 1 + k = exp 1+k
to be true, the
two functions must intersect at some point k > 0. They only intersect (in fact they are only tangent)
at k = 0. This is a contradiction, which means that our starting premise was wrong. Q.E.D
Finally, the growth rate of kt is:
γkt
5
kt+1 − kt
≈ log kt+1 − log kt
k
[t
]
1
w log (1 + kt )
= s
−
kt
1 + kt
=
(5)
(6)
To see this note that the slope is
(
)
k
(
)
∂ exp 1+k
1
k
=
exp
<1
∂k
1+k
(1 + k)2
|
{z
}
| {z }
<1
Jorge F. Chávez
<1
9
EC9A2 (Fall 2012)
Problem Set # 1
where the second equality comes from replacing kt+1 with the expression for ϕ(kt ). Taking limits:
]
1
log (1 + kt )
= lim s
−
kt →∞
kt
1 + kt
[
]
log (1 + kt )
1
w
= s
lim
− lim
kt →∞
kt →∞ 1 + kt
kt
[
]
1
1
1+kt
w
= s
lim
− lim
kt →∞ 1
kt →∞ 1 + kt
[
lim γkt
kt →∞
w
=
0
where again, the second-to-last equality applies L’Hospital rule for undetermined limits.
Jorge F. Chávez
10
EC9A2 (Fall 2012)
Problem Set # 1
Question 3
Production if given by:
Yt ≡ F (Kt , Lt ) = Ktα (Bt Lt )1−α
where Lt+1 = (1 + n)Lt , Bt+1 = (1 + g)Bt and α ∈ (0, 1).
a. Find the dynamical system describing the evolution of kt = Kt / (Bt Lt ) (that is the stock of
capital per-effective worker). Find the steady state level of kt . What is the growth rate of
output per worker yt = Yt /Lt in the steady state? What is the growth rate of capital per
worker Kt /Lt = Bt kt in the steady state?
Solution.
i. We want to find the non-linear difference equation kt+1 = φ(kt ) for kt ≡ Kt /(Bt Lt )
Start from the law of motion of the aggregate stock of capital Kt+1 = It + (1 − δ) Kt . Recall that
this is a closed economy and that there is a constant (and exogenous) saving rate s:
St = It = sYt
Putting everything in per-effective worker terms (by dividing by Bt Lt ):
Kt+1
F (Kt , Lt ) Bt Lt
Kt
Bt Lt
=s
+ (1 − δ)
Bt+1 Lt+1
Bt Lt Bt+1 Lt+1
Bt L Bt+1 Lt+1
We get the non-linear difference equation:
kt+1 =
sktα + (1 − δ) kt
≡ φ (kt )
(1 + g) (1 + n)
As before it is useful to also express the law of motion of kt in terms of ∆kt+1 ≡ kt+1 − kt . To
do this subtract kt from both sides of φ(kt ):
∆kt+1
=
=
sktα + (1 − δ) kt − (1 + n + g + gn − 1 + δ) kt
(1 + g) (1 + n)
sktα − (n + g + gn + δ) kt
(1 + g) (1 + n)
ii. Steady state for kt
ss
= ktss = k. Then:
In steady-state kt+1
∆ktss = 0 ⇒ sk α − (n + g + gn + δ) k = 0
Solving for k:
[
s
k=
(n + g + gn + δ)
]1/(1−α)
To check stability of the steady-state we need to check whether lim |φ′ (kt )| is less than unity:
kt →∞
lim |φ′ (kt )| = kt →∞
Jorge F. Chávez
1−δ
<1
(1 + g) (1 + n) 11
EC9A2 (Fall 2012)
Problem Set # 1
Therefore, the steady-state k is stable.
iii. Growth rate of output per worker yt = Yt /Lt in steady state
Note that:
( )α
(
)α
1−α
Yt
Ktα (Bt Lt )
Kt
Kt
1−α
=
=
=
Bt
Bt = ktα Bt
Lt
Lt
Lt
B t Lt
Now, once the economy reaches an steady-state
ytss = k α Bt
Then, we can see:
ss
yt+1
Bt+1
−1=
−1=g
ss
yt
Bt
Alternatively we can take logs to ytss = k α Bt :
ss
− log ytss = log Bt+1 − log Bt ≈ g
log yt+1
iv. Growth rate of capital per worker Kt /Lt = Bt kt , where kt = Kt /(Bt Lt ).
Same as above. First define the per-capita stock of capital k̃t = Kt /Lt :
Kt
≡ k̃t = kt Bt
Lt
Then get the growth rates directly or by the logarithm approximation:
log k̃t+1 − log k̃t = log Bt+1 − log Bt ≈ g
b. According to Solow’s growth accounting, the growth rate in Total Factor Productivity (TFP)
is calculated by:
∆yt
∆k̃t
∆At
=
− α̃
At
yt
k̃t
where yt ≡ Yt /Lt , k̃t ≡ Kt /Lt , α̃ is the elasticity of yt with respect to kt (which is equal to
the share of capital in a competitive economy). Find TFP growth rate ∆At /At in the steady
state according to the model in part a.
Solution. According to part (a) both yt = Yt /Lt and k̃t = Kt /Lt will grow at a rate g in steady-state.
Therefore:
∆At
At
Jorge F. Chávez
∆yt
∆k̃t
− α̃
yt
k̃t
= g − α̃g
=
12