1 Introduction
The growth model developed by Ramsey (1928) and subsequently adapted by
Cass (1965) and Koopmans (1968) is arguably the single most important
theoretical construct in modern macroeconomics. It is the foundation of both
growth theory and business cycle theory. It takes pride of place in every graduate
text in macroeconomics and growth theory.1
The geometry of the dynamics of the Ramsey model is well understood;
every graduate student ritually ponders its phase diagram. There is a paucity of
analytical solutions, however. Long and Plosser (1983) developed a discretetime
example with log utility, CobbDouglas technology, and 100 percent depreciation
which is often used to exposit RBC models.2 It is also possible to derive an
explicit solution when there is a constant gross saving rate [Kurz (1968), Barro,
Mankiw, and SalaiMartin (1995), Barro and SalaiMartin (2004, Chapter 2),
and Wälde (2005)]. Recently, Mehlum (2005) has provided a closed form solution
for the case of a Leontieff technology. While these are illuminating examples,
each involves some unappealing features. The assumption of 100 depreciation in
Long and Plosser is highly unrealistic, as is the restriction to log preferences. At
any rate, this example only yields a solution in discrete time; it doesn’t work in
continuous time [Chang (1988)] . The constant saving rate formulation requires
the elasticity of intertemporal substitution to be tied to the rates of depreciation
and time preference in an unintuitive way. The Leontieff technology has long
been known to be a razoredge case. It would be nice to have an example with a
reasonable rate of depreciation, a nonunitary elasticity of intertemporal
substitution, and diminishing returns. Absent such a solution, empirical
implementation of the model has required either a linear approximation around
the steady state [Barro and SalaiMartin (2004)] or numerical simulation
[Mulligan and SalaiMartin (1991)].
This paper will derive an explicit solution to the canonical version of the
Ramsey model with constant elasticity preferences and CobbDouglas
technology. The solution can only be obtained when capital’s share of GDP is
equal to the reciprocal of the intertemporal elasticity of substitution. The
distinguishing feature of this constellation of parameters is that it generates a
1 Some examples are Blanchard and Fischer (1989), Romer (2005), Sargent (1989, 2002),
Turnovsky (1999, 2000), Obstfeld and Rogoff (1999), and Barro and SalaiMartin (2004).
2 In this case both the utility function and production function have a unitary elasticity of
substitution. Benhabib and Rustichini (1994) show that solutions can be found for a more general
class of models, where (given CRRA preferences and CES technology) the elasticities of
substitution of the utility function and production function are equal, but not necessarily equal to
one.
linear saddle path. This permits an explicit solution for the dynamics of the
capital stock, along with an exact expression for the speed of convergence.
Although this is admittedly a special case, it provides a concrete example
that makes the dynamic workings of the model transparent. It is not intended to
supplant graphical and numerical analysis, but to serve as a complementary tool
that helps reveal the economic properties of the model. As Boucekkine and RuizTamarit (2004b) put it when motivating the same parametric device in studying
the Lucas (1988) model, “. . . there is an obvious need to complement this
computational and qualitative literature with fullfledged theoretical proofs . . .”
Two elements of this solution have been anticipated in the literature. First,
what makes the model tractable is the fact that the consumption/capital ratio is
constant when labor’s share equals the intertemporal elasticity. Barro and SalaiMartin (2004) are aware of this fact in the Ramsey model. Xie (1991) exploits this
property in the Romer (1986) model, while Xie (1994) and Boucekkine and RuizTamarit (2004a) use it to shed light on the Lucas (1988) model. The tractability
of this case has also been put to good use in some stochastic growth models. For
example, Chang (1998) solved the closely related “inverse optimal” problem for
this case, deriving the utility function and rate of time preference that generate a
constant equilibrium consumption/capital ratio. More recently, Wälde (2005) has
used this combination of parameters to solve a Poisson model of endogenous
growth cycles. The innovation here is to use the constancy of the
consumption/capital ratio to arrive, via a Bernoulli transformation, at closed form
solutions for the capital stock and the speed of convergence in the nonstochastic
Ramsey model.
Second, the derivation centers around the hypergeometric function.
Boucekkine and RuizTamarit (2004b) have recently, and independently, applied
the hypergeometric function to the Lucas (1988) model. The recursive structure of
the Lucas model allows them to derive analytical results without restricting the
intertemporal elasticity. The Ramsey model, however, is generally not recursive;
the only case where it is recursive seems to be the example studied here, where
the intertemporal elasticity is equal to the reciprocal of capital’s share.
The outline of the paper is as follows. Section 2 recapitulates the Ramsey
model. Section 3 sketches the solution, leaving the details to an appendix.
Section 4 discusses the properties of the solution. It shows vividly how the
speed of convergence varies with time, a feature of the model that is masked by
estimates of the speed of convergence calculated from a Taylorseries
approximation around the steady state. It also suggests a pedagogical application
of the model: students are often asked to solve the CobbDouglas version of the
Solow (1956) model by using a Bernoulli transformation. This model could be
adapted to serve as a complementary example of a tractable solution to the
Ramsey model.
Section 5 extends the model to study two problems that have heretofore
lacked analytical solutions. The first is the dynamic effects of a temporary change
in government expenditures; the second is to the transition dynamics of the Jones
and Manuelli (1990) model.
Section 6 calibrates the model. This is obviously a very special case, so I
do not wish to press the issue. However, I suggest that this calibration may not be
terribly unrealistic. If we adopt a broad notion of capital and embrace a degree of
agnosticism about the elasticity of intertemporal substitution it may be a
reasonable first approximation to the real world.
Section 7 offers some concluding thoughts.
2 The Model
Since the model is so familiar, my description of it will be succinct. Imagine an
economy inhabited by a large number of identical, infinitelylived consumers.
For simplicity there is neither population growth nor technological growth. I
adopt the functional forms commonly used in empirical, and in many theoretical,
applications: technology is CobbDouglas, with a capital share ofα ∈[0,1]; the
rate of depreciationδ ∈[0,1]is constant; there is a constant rate of time preference
ρ ∈[0,∞]and preferences are timeseparable, with a constant elasticity of
intertemporal substitution, σ ∈[0,∞].
It is well known that the dynamics of per capita consumption c and capital
k are determined by the following pair of differential equations:
k = k α –δk – c = σ (αk α –1 – δ – ρ) 
(1) (2) 
cc
Equation (1) is the resource constraint; Equation (2) is the Euler equation. There
are two boundary conditions: the initial capital stock ko and the transversality
condition
.
lim 0
1
→ ∞
=
–
–
t
e ρtc σ k (3)
For future reference, the steady state is
1
1 –
+
=
α
ρ α
δ
k (4)
( )
.
1
c k k k
α
ρ δ α
= α – δ = + – (5)
The system is saddlepath stable. Given the initial capital stock k0 there is a
unique value of initial consumption c0 that will satisfy the transversality
condition.
3 A Solution
The system of equations (1) – (2) is dauntingly nonlinear. To approach a solution,
define the capitaloutput ratio (a Bernoulli transformation) by z = k 1–α and the
consumptioncapital ratio by x = c k . Using these transformations the system can
be rewritten as
z = (1–α )[1– (δ + x)z]  (6)  
( ) + – – 1 
x + 
 1  (7) 
x z
x
–
= σα δ σ ρσ The Bernoulli transformation converts the capital accumulation equation
[Equation (1)] into the linear, albeit nonautonomous, differential equation in
Equation (6). Despite this simplification, the system still does not allow an
analytical solution. Equation (7) is still nonlinear.
Suppose, however, that σ =1 α . In that case, z disappears from equation
(7), and the system becomes recursive: Equation (7) reduces to
( )
,
1
x
xx
+
– +
= –
α
δ α ρ
(8)
while z evolves according to Equation (6), given the forcing process x determined
in Equation (8). This is a simple, autonomous logistic equation.
The solution to Equations (6) and (8), subject to the initial condition
k0 and the transversality condition [Equation (3)] is given by the following result.
PROPOSITION 1: If σ = 1 α the transitional paths of consumption and capital
are
( )
,
1
c k
α
ρ + δ –α
= (9)
–α ( –α –α ) –( –α )ρα+δ –α
= + –
1
1
1
1 1
0
1 t
k k k k e . (10)
Since to some extent the method is the message of this result, I will sketch its
derivation before discussing its properties.
PROOF:
Consider the logistic equation in Equation (8). This has the general
solution
xAe xt
x
x
+
=
1
(11)
where A is an arbitrary constant of integration and ( )
α
ρ +δ –α
=
1
x . Substitute
this back into Equation (6) to arrive at
( ) = – – + z 1 α 1 δ 
+ z xAe x 1 xt 
(12) 
In the appendix I demonstrate that the general solution to Equation (12) is:  
1 1(1 α,1, ;ω) α 1 (1 α )( δ ) (1 )1 α , – – – + – – + + = F d Bx e x t xAe xt z 
(13) 
2 δ
+
x
where B is another arbitrary constant of integration, d = 1+ (1–α)(δ + x) x ,
,
1 xt
xt
xAe
xAe
+
ω = and
( ) ( )( )
( )
( )( )( )
(1 )(2 ) . . .
1 2 3
1
1
1 
2 d 
d 
2 1 1 ,1; ; 1 2 3 +
+ +
– – –
+
+
– –
+
–
– = + ω
α α α
ω
α α
ω
α
α ω
d d d d
F d
(14)
is a hypergeometric function.3
The following lemma pins down the constants A and B .
LEMMA. Given the initial condition
k0 the transversality condition [Equation
(3)] will be satisfied if and only if A = 0 and
. 1 0 1 = – – δ B x z 
(15) 
+
α
x
The proof of this lemma is sufficiently involved to be relegated to the appendix.
The intuition, however, is straightforward. Notice that the transversality
condition (3) can be written as
. lim 0 → ∞ = – – – t e ρt x σ k σ 
(16) 
1
1 1
Consider the solution for x in Equation (11). If A > 0 then x would fall to zero
over time, driving marginal utility and k to infinity. If A < 0 then x would
become negative in finite time. If A = 0 x is constant and, given Equation (10),
Equation (16) is satisfied. Therefore, the transversality condition is satisfied if and
only if A = 0.
Applying this Lemma to Equations (11) and (13) completes the proof:
Since A = 0, Equation (11) implies that x = x,∀t. This yields Equation (9). Next,
evaluating Equation (13) at the values of A and B described in the Lemma yields
( )(x )t e x α δ δ – – + + 1 0 1 . 
z = 1 
δ  (17) 
z
x
+ –
+
Using the definition of z we can then back out the solution for the capital stock in
Equation (10).
4 Discussion
Note several interesting features of the equilibrium. First, Equation (9) asserts
that in every period the ratio of consumption to capital must be constant,
3 The function 2 F1(a, b, d;ω)is convergent for arbitrary a,b, and d for real –1 < ω < 1. See
Weisstein (2004).
c k = x = x. In other words, the saddlepath is linear, with a slope equal to x.
This is depicted in Figure 1. This property is now fairly well known, for a range
of models [Chang (1988), Xie (1991, 1994), Barro and SalaiMartin (2004),
Boucekkine and RuizTamarit (2004a, b), Wälde (2005)].
The Phase Diagram
Figure 1
The Phase Diagram
Second, Equation (10) provides an exact solution for the dynamics of the
capital stock. The growth rate of the capital stock is
( )
–
+
=
+
– –
+ –
= = –
–
+
–
– –
– –
1
.
1
1
1 1
0
1 1
0
α
δ α
ρ
α
α α
α α
δ α
ρ
ρ δ
α α
ρ δ
γ
k k
k k e
k k
kk
t
k
(18)
On inspection, the growth rate decreases over time: The growth rate depends upon
the deviation between net output and consumption. However, consumption grows
linearly with the capital stock, while gross output increases at a decreasing rate,
due to diminishing returns.
We can also obtain an exact, analytical calculation of the speed of
convergence to the steady state:
(1 ) .
ln
–1
+
= –
∂ ∂
= –
α
δ α
ρ
α
γ
β
k k
k
k
k (19)
Barro and SalaiMartin (2004, chapter 2), in contrast, resort to a calibration of a
linear approximation around the steady state. The accuracy of their estimate
diminishes rapidly the further away capital is from its steady state. The
numerical calculations of Mulligan and SalaiMartin (1991) are exact, but lack
the transparency of the solution in Equation (19).
Third, the model yields a clean expression for the dynamics of the gross
saving rate, s =1– c k α . From Equations (9) and (10) it follows that
( ) ( ) ( )
( ) ( ) ,
1
1
1
0
1
1 1
0
1
t
t
s s s e
s k k k e
δ α
ρ
α
δ α
ρ
α
α α α
α
ρ δ α
+
– –
+
– –
– – –
= + –
+ –
+ –
= –
(20)
where s =1– xα (δ + ρ) is the steady state saving rate. The model predicts that, if
k0 < k , the saving rate will decrease monotonically over the transition to the
steady state. The intuition is simple. In general, the increase in the capital stock
exerts offsetting income and substitution effects on saving [Barro and SalaiMartin (2004, Chapter 2, Section 6.3)]. As interest rates fall the substitution effect
tends to reduce saving. However, as the gap between current income and
permanent income diminishes, there is less incentive to save. Because people like
intertemporal substitution (σ > 1) in this calibration of the model the substitution
effect is strong, and dominates the income effect: Since the consumptioncapital
ratio is constant it follows that s =1– xk 1–α =1–αx R, where R =αk α –1 is the
gross interest rate. The saving rate therefore decreases as the interest rate falls.
This is not consistent with the empirical evidence [Barro and SalaiMartin
(2004)], since the saving rate actually increases over the transition.
Finally, the solution suggests a useful pedagogical exercise. It is well
known that the Solow (1956) model with CobbDouglas technology is an
autonomous Bernoulli equation [Chiang (1967, p. 454), Obstfeld and Rogoff
(1999), Barro and SalaiMartin (2004, p. 44)].4 Students are often asked to solve
4 Barro and SalaiMartin (2004, appendix 2.C) attribute the use of this transformation in the
Solow model to Jaume Ventura. However, it was known at least as early as Chiang (1967).
the Solow model explicitly using this “trick.” It is nontrivial to show – using
either the methods here, or those in Barro and SalaiMartin (2004, Appendix 2b)
— that the saddlepath is linear when σ =1 α . However, students may as an
exercise be told to assume this result – or tell an intuitive story to motivate it —
and then be invited to solve the rest of the model. Once they have determined the
steadystate capitaloutput ratio, they can write Equation (6) as an autonomous
Bernoulli equation, which can be solved just like the Solow model.
5  Extensions 
The model can be adapted to provide explicit solutions to a range of problems in growth theory. I will consider two here. 

5.a  A Temporary Increase in Government Expenditures 
A traditional thought exercise [Romer (2005), Turnovsky (1999)] is to trace out
the dynamic effects of a temporary increase in government expenditures in the
Ramsey model. This problem has been addressed graphically [Blanchard and
Fischer (1989), Romer (2005)] and solved using linear approximations
[Turnovsky (1999)]. I provide an explicit solution for the case where σ = 1 α. 5
First we need to incorporate the government into a decentralized version
of the model. Suppose that that government issues debt b , collects lumpsum
taxes τ, and engages in expenditures G. Given an initial debt of b0 and the
interest rate r = αk α –1 – δ , its flow budget constraint is then
b = rb + G –τ. (21)
To obtain an analytical solution I assume that government expenditures are
proportional to the capital stock, G = gk. While this assumption is just a
convenience here, it is standard in endogenous growth models [see Turnovsky
(2000), for example].6
Imagine a temporary increase in g from time zero to time T :
5 It is also possible to consider expected future changes in government expenditure, as well as
temporary and expected future changes in distortionary taxes on capital.
6
The traditional exercise involves an increase in government expenditures that is independent of
the capital stock or output [Blanchard and Fischer (1989), Romer (2005)]. This makes no
substantive difference. Alternatively, the model could be solved by making government
expenditures proportional to output.
g g t T
g t g t T
= < ≥
= ≤ <
2 1
1, 0
(22)
The dynamics of capital and consumption are now given by
, .
, 0
1 2
k k c g k t T
k k k c g k t T
= – – – >
= – – – ≤ ≤
δ
δ
α
α
(23)
= σ (αk α –1 –δ – ρ)
cc
(24)
Define ( )
1 1
1
x – g
+ –
=
α
ρ α δ
,
( )
2 2
1
x – g
+ –
=
α
ρ α δ
and .
∆g = g1 – g 2 In the
appendix I prove
PROPOSITION 2: Assuming that σ =1 α the equilibrium paths of consumption
and capital in response to a temporary increase in government expenditures from
time 0 to time T are:
( )
, .
, 0 ,
1
2
2
1
1
x k t T
k t T
e
x
xg
c
x t T
= >
≤ ≤
∆
–
=
–
(25)
( )
( )
( ) ( )( )
( )
, .
0 ,
1 ,
1
1 ,1, ;
1
1
1
1
1
1
1
1
2
2
2
2 1 1
1
1
1
k e t T
t T
e e
x
g
e
x
e g
x
g
k F d
t
T
x t T x t
x t T
x t T
>
= + + – +
≤ ≤
+  Ω –  ∆ 
∆
∆ –
– –
+
=
–
+
– –
–
–
– – +
–
–
–
–
α
δ α
ρ
α
α
α
α δ
α
ρ δ
α
ρ δ
α
α
ρ δ
α
(26)
where
1 .
1
1 1 ,1, ; 1
2
2
2
0 2 1 1
+ ∆
∆
∆ –
– –
+
Ω = – –
–
–
x T
xT
xT
e
x
g
e
x
e g
x
g
F d
x
z α
δ
(27)
Equation (25) makes the dynamics of consumption per unit capital
transparent: As shown in Figure 2, consumption initially falls, given the initial
capital stock, so that c k drops.7 Then, as the capital stock falls, c k increases
until the economy regains the saddlepath at timeT ; thereafter it remains constant.
Intuitively, people respond to the increase in government expenditures by
reducing consumption. Since the change in government expenditures is
temporary, however, they reduce consumption by less than the increase in
government expenditures. This causes the capital stock to decrease over time, so
that consumption per unit capital has to accelerate as it approaches the saddlepath.
Effects of a Temporary Increase in Government Expenditures Figure 2
Effects of a Temporary Increase in Government Expenditures
7 Figure 2 is identical to Figure 2.9 in Romer (2005) except that the k = 0 locus rotates around the
origin rather than making a parallel shift.
5.b Transition Dynamics in an Endogenous Growth Model
Another illuminating application of the model is to the transitional dynamics of
endogenous growth. This too has been analyzed graphically [Barro and SalaiMartin (2004)], but never solved explicitly.
I adopt a CobbDouglas version of the ak technology suggested by Jones
and Manuelli (1990): y = k α + ak. Since the marginal product of capital is
bounded away from zero, this is sufficient to induce long run growth. The
dynamics of k and c are now
k = k α + ak –δk – c = σ (αk α –1 + a –δ – ρ). 
(28) (29) 
cc
In order to generate an asymptotically balanced growth path I assume that
a > δ + ρ. In order to satisfy the transversality condition I also assume that
ρ > (1–α )(a –δ ).
In the appendix I derive the closed form solution to this model:
PROPOSITION 3: If σ = 1 α the transitional paths of consumption and capital in
the endogenous growth model are
( )( )
,
1
c a k
α
ρ – –δ –α
= (30)
( )
( δ ρ ) α
α α
α
δ ρ
α
δ ρ
α – – –
–
–
= – – – + + – – 1
1
1
10
a t
e
a
k
a
k . (31)
Equation (30) asserts that the consumption/capital ratio is again constant over
time. Equation (31) implies that the growth rate of capital is
( ) (a )t δ ρ – – – – 1 
a 
k α – + 
a ρ – – 
k 10 
a α δ ρ – – = 
e ρ 

a 
kk
α α
α
δ α
δ ρ
α
δ α
–
– –
–
– –
+
10
. (32)
This makes it clear that the growth rate falls over time, tending asymptotically
to(a – δ – ρ) α .
6 Calibration
We can only solve the model in the special case where the intertemporal elasticity
of substitution σ is equal to the reciprocal of capital’s share of income α. Is this
likely to be a reasonable description of the real world?
At first glance, the answer to this question would seem to be no. The
traditional estimate of physical capital’s share of GDP is about one third [Simon
(1990)]. This would require the intertemporal elasticity of substitution to be as
large as three. However, the intertemporal elasticity of substitution is usually
thought be quite small, certainly less than one [Hall (1988), Attanasio and Weber
(1993), Epstein and Zin (1991), Campbell, Lo, and MacKinlay (1997)].
Recent empirical work casts this conventional wisdom in doubt. Barro and
SaliMartin (2004) point out that if α = .3the Ramsey model does a poor job at
explaining a range of empirical features of growth. However, “all of these
shortcomings are eliminated” [Barro and SaliMartin (2004); see also Mankiw,
Romer, and Weil (1992) and Obstfeld and Rogoff (1994)] if we adopt a broader
notion of capital (so that it includes human as well a physical capital), and
increase capital’s share: Barro and SalaiMartin suggest α = .75, Obstfeld and
Rogoff (1994) α = .66. If we set α = .75, we are only restricting the elasticity of
substitution to be about σ =1.33. The last decade has seen more and more
empirical evidence that σ is in fact greater than one, with estimates ranging from
1.5to as much as 3[Bufman and Leiderman (1990), Attanasio and Weber (1989),
Koskievic (1999), VissingJorgensen (2003)]. There is also the interesting
empirical fact that, in a cross section of countries, growth rates are negatively
correlated with production risk [Ramey and Ramey (1995)]. In the canonical
stochastic Ak growth model [Smith (1996)] this can only occur if the
intertemporal elasticity is greater than one.
What of the model’s prediction that the consumptioncapital ratio should
be constant? It is a traditional “stylized fact” of growth, enunciated by Kaldor
(1963) and later supported by Maddison (1982), that the ratio of consumption to
physical capital is roughly constant over time. What can we say about the
consumptioncapital ratio if we adopt a broader notion of capital, one that that
encompasses human as well as physical capital? Empirical estimates [Jorgenson
and Fraumeni (1989), Lettau and Ludvigson (2004)] suggest that human capital
represents between two thirds and 90 percent of total wealth. In a recent paper
Zhang (2004) establishes that the ratio of consumption to total wealth is roughly
constant. He estimates the steadystate consumptioncapital ratio to be .05739, in
annual terms. Now compare this with the predicted consumptioncapital ratio of
the model: Following Barro and SalaiMartin (2004), set the rate of depreciation
at .05 and the rate of time preference at .02 per year. Let capital’s share be .66 [as
suggested by Obstfeld and Rogoff (1994)], which restricts the intertemporal
elasticity of substitution to the fairly low value of 1.5. The consumptioncapital
ratio predicted by Equation (9) is then .055. This matches the data remarkably
well.
In conclusion, if we are willing to embrace a broad notion of capital and
entertain the possibility that the elasticity of substitution may be even a bit larger
than one, this calibration may work fairly well.
7 Conclusion
This paper provides an explicit solution to the Ramsey model when capital’s share
is equal to the reciprocal of the intertemporal elasticity of substitution. This is
admittedly a special case. However, it illuminates the dynamics of the model in a
way that approximations and numerical methods cannot: it helps us to “inspect
the mechanism” [Campbell (1994)]. In principle it could be used as a benchmark
from which to undertake perturbation analyses of less restrictive cases [Judd
(1997)]. In addition, it offers a useful pedagogical exercise, providing a closedform solution to the Ramsey corresponding to the Bernoulli solution of the Solow
model.
In another paper [Smith (2005)] I derive an explicit solution to a stochastic
version of the Ramsey model. Applying the methods of Merton’s (1975) classic
paper on stochastic growth I can then derive an explicit solution for the ergodic
distribution of the capital stock. The nonlinearities of the model (in both the utility
and production functions) interact with uncertainty — via Jensen’s inequality — to
affect the steady state in complicated and even surprising ways. The presence of
uncertainty makes the “mechanism” much more complicated than generally
appreciated.
Appendix
DERIVATION OF THE GENERAL SOLUTION FOR z
Consider first the homogeneous equation
( ) 1 z = – –α δ + 
. 1 z xAe x xt + 
(A.1) 
This can be integrated to find the complementary solution z c = x α –1e –(1–α )(x+δ )t (1 + xAe xt )1–α . 
(A.2)  
To find the particular solution to Equation (12), I will use the method of  
variation of parameters.  Conjecture that the particular solution is 
z p = zcΨ, where Ψ is an unknown function of time. Substituting this conjecture
into Equation (12), it follows that
Ψ = 1 – α = (1 – α ) 1–α (1–α )(x+δ )t (1 + xt )α –1.
c
x e xAe
z
(A.3)
Integrating Equation (A.3) leads to1
( )( ) ( )( ) ( )( )
; .
1
1 1 2 1 1 , 1 ,1
–
– +
+
– +
–
+
Ψ = – +
–
x t xAe xt
x
x
x
x
e F
xx
α δ α δ
α
δ
α δ
α
(A.4)
It follows that
( ) ( )( ) ( )( )
; .
1
1 1 2 1 1 , 1 ,1
–
– +
+
– +
–
+
+
=
–
xt
xt
p xAe
x
x
x
x
F
x
xAe
z
α δ α δ
α
δ
α
(A.5)
However, a useful property of the hypergeometric function [Weisstein (2004)] is
that ( ) ( )
–
= – –
–
1
2 1 , , ; 1 2 1 , , ,
z
z
F a b c z z a F a c b c . Therefore Equation (A.4) can
be written as
.
1
1 2 1 1 ,1, ;
z p = x +δ F –α d +xAe xAe xt xt (A.6)
where d = 1+ (1–α)(δ + x) x . To verify that this is a solution; differentiate
Equation (A.6) with respect to time, using the definition of the hypergeometric
function given by Equation (14) of the text:
( )
( )( ) – – + – 1 2 1 2 2 α α α x Ae xt xt 
= 1 z p 
( )
( )( )( )
( )( )
( ) ( )( )
( )
( )( )( )
( )( )
+
+ +
– – –
+
+
– –
+
–
–
+
=
+
+ +
– – –
+
+
+ +
. . .
1 2
1 2 3
3
1
1 2
1 1 2
. . .
1 2
1 2 3
3
1
1
2
2
2
ω
α α α
ω
α α α
ω ω
δ
ω
α α α
ω
δ
x d d d d d d
x
xAe d d d d d d
x
(A.7)
Substitute this back into Equation (12) in the text to find
( ) ( )( )
( )
( )( )( )
( )( )
( ) [ ( )] ( )( )
( )
( )( )( )
( 1)( 2) . . .
1 2 3
1
1 1 2
1 1 1 1
. . .
1 2
1 2 3
3
1
1 2
1 1 2
2 3
2
+
+ +
– – –
+
+
– –
+
–
+
+
+ –
– –
=
+
+ +
– – –
+
+
– –
+
–
–
+
ω
α α α
ω
α α
ω
α
ω δ
δ
α
ω
α α α
ω
α α α
ω ω
δ
x d d d d d d
x
x d d d d d d
x
(A.8)
Collect the terms inω i (i = 0,1,2,…). Some tedious algebra reveals that the
coefficient attached to each ω i in the left hand side of Equation (A.8) equals the
coefficient attached to the corresponding ω i on the right hand side. Therefore
Equation (A.6) solves Equation (14) in the text.
Using Equations (A.2) and (A.6) leads to the general solution
α ( α )( δ ) α ( α )( δ ) ( ) α
δ x + 
x  1+xAe 
Bx e  xAe 
– – – + –
+ +
z = 1 2 F1 1– ,1,1+ 1– x + ;xAe xt xt 11 x t 1 xt 1
(A.9)
where B is an arbitrary constant.
PROOF OF THE LEMMA
Using the initial condition k0 and Equations (13) and (14), it follows that
( )
+
–
+
= –
–
xA
xA
F d
x
B A x z
1
1 0 1 2 1 1 α,,1, ;
δ
α
. (A.10)
Now consider the transversality condition in Equation (3) of the text. Recalling
that σ =1 α , use Equations (11) and (13) to write this limit as
( ) ( )( ) 1 1 1 – – – + x t α α δ 
lim 
1 1 ,1; ; – xAe F a d xt α 
1 + xAe 
2 + x δ 
1 + xAe xt 
(1 ) . + – Ae xt α 
+  xt 
→ ∞
– t
xx e
B A
x
e
t
ρ
(A.11)
We need to show that this limit is zero if and only if A = 0.
First, suppose that A = 0. Note that, from Equation (14),
(0) 1 1 . = – – B x z 
(A.12) 
0 
+ δ 
α
x
Use this fact to evaluate the limit in Equation (A.11) when A = 0 :
( )( )
0
1 1
lim 1 1
0
=
→ ∞
+
+ –
+
– – – – +
t
e
x
k
x
e
ρt α α δ x t
δ δ (A.13)
This establishes sufficiency.
To establish necessity first rewrite Equation (A.11) as
t ( xt ) xt xt ( ) (
xAe e – + – + 
x  xAe  δ  → ∞ 
α 
ρ – 
xt )[ ( )(x )]t
AB
t
xAe
e xAe F a d
x
x
ρ α δ
+ +
lim + 1+ 2 1 ,1; ;1+ 1 1
(A.14)
Now consider the limit of the second term:
( )( ) [ ( )( )]
.
lim 1 1
→ ∞
+ – + – +
t
xAe e
Ax
B
xt ρ α x δ t
(A.15)
Since ρ + (1–α)(δ + x) > 0 and – ρ + x –(1–α)(δ + x) = 0 ,
( )( ) [ ( )( )] ( )
.
lim 1 1
→ ∞
+ – + – + =
t
xAe e AB A
Ax
B
xt ρ α x δ t
(A.16)
If B(A) ≠ 0 this will be zero if and only if A = 0.
Next consider the first expression in Equation (A.14):
( ) 1 . 1 2 1 ,1; ; ++ + xAe xAe e xAe F a d 
lim 
– ρt 
xt xt α 
→ ∞
–
t
xx
xt
α
δ (A.17)
First consider the limit of the hypergeometric function. Note that
lim 1 =1.
→ ∞
+
t
xAe
xAe
xt
xt
(A.18)
[Lebedev (1972, p. 244, Equation (9.3.1))] provides a limit result about
hypergeometric functions which then establishes
( ) ( )
( ) ( )
→ ∞
= Θ >
Γ – + Γ –
Γ Γ – +
=
+
t
d d
d d
xAe
xAe
F a d
xt
xt
0.
1 1
2
1
lim 2 1 ,1; ;
α
α
(A.19)
Since the hypergeometric function converges to this finite limit as t → ∞ it follows
that
( ) = ++ + xAe xAe e xAe F a d xt t xt lim 1 2 1 ,1; ; 

( ) + Θ + – – e xAe t xt 1 . ρ α α 
–  xt  lim  α 
→ ∞
→ ∞
– t
xx
t
xx
1
ρ
α
δ δ
(A.20)
Now rewrite this as
( ) ( ) ( ) ( )
→ ∞ t 
→ ∞ lim 
Θ
+
+ Θ =
+
–
–
– –
–
e xA
xx
t
e e xA
xx
x t xt x t
lim α ρ α .
α
α
α ρ
α
δ δ (A.21)
This will go to zero if and only if A = 0. This establishes necessity.
Given A = 0, B is given by Equation (A.12). This proves the lemma.
SOLUTION FOR THE MODEL WITH A TEMPORARY INCREASE IN
GOVERNMENT EXPENDITURES
The dynamics of x and z are given by
( )
( )
, ;
1
1 , 0 ,
1 2
g x t T
g x t T
xx
+ + >
– +
= –
+ + ≤ ≤
– +
= –
α
δ α ρ
α
δ α ρ
(A.22)
( )[ ( ) ]
(1 )[1 ( ) ], .
1 1 , 0 ,
2
1
g x z t T
z g x z t T
= – – + + >
= – – + + ≤ ≤
α δ
α δ
(A.23)
First, define xi = +( – ) – g i
α
ρ 1 α δ
, and =1+ (1– )( + ), i =1,2.
x
x
d
i
i
α δ
Let
0. ∆g = g1 – g 2 > Note in passing that x + g = x + g = x = 

1  α  
1 2  2  1  . 
()ρ +δ –α
Following the derivation of the basic model without government, the general
solutions for x and z are
, ;
1
, 0 ,
1
1 2
2 2
2
1 1
1
t T
x A e
x
t T
x A e
x
x
x t
x t
>
+
=
≤ ≤
+
=
(A.24)
( )( ) ( )
( )( ) (1 ) , .
1
1 1 ,1, ;
1 , 0 ,
1
1 1 ,1, ;
1
2 2
1 1
2 2
2 2
2 2
2 1 2
1
1 1
1 1
1 1
1 1
1 1
2 1 1
21
2
1
1
1
B x e x A e t T
x A e
x A e
F d
x
B x e x A e t T
x A e
x A e
F d
x
z
x t x t
xt
x t
x t x t
x t
x t
+ + >
+
–
+
=
+ + ≤ ≤
+
–
+
=
–
– – – +
– – – + –
α α δ α
α α δ α
α
δ
α
δ
(A.25)
There are four arbitrary constants, A1, A2 , B1, and B2. To pin down these
constants, we have four boundary conditions:
(a) (b) (c) (d) 
the initial capital stock, k0,  
the transversality condition [Equation (3) in the text],  
continuity of c at T, continuity of k at T. 

Consider the A‘ s first. Using the argument developed in the previous  
appendix, the transversality condition (b) requires A2 = 0. Using the continuity condition (c) along with Equation (A.24) and the fact that A2 = 0, we find 

. 1 1 2 x T e x x g ∆ – (A.26) 
1 A 
= – 
To find
B1, substitute Equation (A.26) into Equation (A.25) evaluated at t = 0 and
using the initial condition (a), z0 = k01–α :
∆
∆ –
– – – F d z 0 2 1 1 1 1 α,1, ; 
= – ∆ – – x T e x g B x 2 11 1 1 1 α 
– xT 
(A.27) 
+ x δ 
1 
–
–
xT
e
x
e g
x
g
2
2
1
α
This implies, from Equation (A.25), that for the interval 0 ≤ t ≤ T
( )
( )
( )
( )( )
.
1
1
1
1 1 ,1, ;
1
1 1 ,1, ;
1
1
2
2
2
2
0 2 1 1
2
2
2 1 1
1
1
1
1
x t
x T
x t T
xT
xT
x t T
x t T
e
e
x
e g
x
g
e
x
e g
x
g
F d
x
z
e
x
e g
x
g
F d
x
z
α δ
α
α
δ
α
δ
– – +
–
–
–
–
–
–
–
∆
∆ –
–
∆
∆ –
– –
+
–
+
∆
∆ –
– –
+
=
(A.28)
Note that
( )( )
.
1
1
1
1 1 ,1, ;
1
1 1 ,1, ;
1
1
2
2
2
2
0 2 1 1
2
2
2 1 1
1
x T
xT x T
xT
T
e
e
x
g
x
g
e
x
e g
x
g
F d
x
z
x
g
x
g
F d
x
z
α δ
α
α
α δ
δ
– – +
–
– –
–
∆
–
∆
–
∆
∆ –
– –
+
–
+
∆
∆ –
– –
+
=
(A.29)
Finally, to find B2 substitute Equation (A.29) to the second interval (for t > T ) of
Equation (A.26), evaluated at t = T. Recalling that z =1/(x +δ )this yields
B2 = x21–α (zT – z)e(1–α )(x+δ )T Substituting this back into Equation (A.26) yields z for t > T : z z (zT z)e (1)(x)(tT ). 
(A.30)  
= +  –  —α +δ –  (A.31) 
SOLUTION FOR THE ENDOGENOUS GROWTH MODEL
Assuming thatσ = 1 α , Equations (28) and (29) in the text imply that the
dynamics of z and x are
z = (1–α )[1+ (a –δ – x)z]
(a )(1 ) x. = – –δ –σ – ρσ + 
(A.32) (A.33) 
xx
I assume that a > δ + ρ in order to generate longterm growth.
Begin with the logistic equation (A.33). This has the general solution
,
1 ˆ
ˆ x 
x 
=  (A.34) 
xAe xˆt
+
where ( )( )
α
ρ – –δ –α
=
1
xˆ a . The transversality condition will require that xˆ > 0.
Substitute Equation (A.34) back into the linear equation (A.32):
( )
z 
+
= – + – – xAe
x
z a
1 ˆ xˆt
ˆ
1 α 1 δ . (A.35)
Following the derivation of Equation (A.9), the general solution to Equation
(A.35) is
α ( α )( δ ) α ( α )( δ ) ( ) α
δ xˆ + – a 
x  1+xˆAe xt 
2  Bxˆ e  1 xˆAe 
–
– – – + –
+ +
z = 1 F1 1– ,1,1+ 1– xˆˆ+ – a ;xˆAe xˆt ˆ11 xˆ a t xt 1
.
(A.36)
The initial condition
z0 requires
( ) 1 ˆ 1 = – – B A x z α 
, ˆ 1 ,,1, ˆ; – xA F d α 
1 ˆ + xA 
ˆ 0 2 1 + – x a δ 
(A.37)
where ( )( ).
ˆ
1 ˆ
ˆ 1
x
x a
d = + –α +δ – The transversality condition requires
( ) ( )( ) ˆ ˆ ,1; ˆ; 1 ˆ 1 1 1 ˆ ˆ ˆ 2 1 – – – + – x e B A xt x a t xt xt 

(1 ˆ ) . + – xAe 
1 ˆ + + xAe 
ˆ + – x a δ 
1 ˆ ˆ + Ae 

ˆ 
lim e 
– xt α 
xAe F a d 
α α 
→ ∞
– t
xx
t
α
δ
ρ
(A.38)
Following the reasoning in the previous lemma, the transversality condition will
be satisfied if and only if (1) A = 0 and (1) ρ > (1–α)(a –δ ). The latter condition
implies that xˆ > 0, which yields Equation (30) in the text.
Setting A = 0 , Equation (A.37) yields
.
ˆ
1
ˆ
0
1
+ –
= –
–
x a
B x z
δ
α (A.39)
Substituting this into Equation (A.36) yields
( )( )
( )
( )
.
ˆ
1
ˆ
1
1
0
1 ˆ
0
a t
x a t
e
a
z
a
e
x a
z
x a
z
δ ρ
δ α α
α
δ ρ
α
δ ρ
α
δ δ
– –
–
– – + –
= – – – + + – –
+ –
+ –
+ –
=
(A.40)
Substituting z = k 1–α into Equation (A.40) yields Equation (31) in the text.
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