DP2014-31

Multiple Interior Steady Statesin the Ramsey M odel
with Elastic Labor Su pply
Takashi KAMIHIGASHI
July 31, 2014

Multiple Interior Steady States in the Ramsey
Model with Elastic Labor Supply∗
Takashi Kamihigashi†‡
July 31, 2014

Abstract
In this paper, we show that multiple interior steady states are possible in the Ramsey model with elastic labor supply. In particular, weestablish the following three results: (i) for any discount factor andproduction function, there is a utility function such that a continuumof interior steady states exist; (ii) the number of interior steady states
can also be any finite number; and (iii) for any discount factor andproduction function, there is a utility function such that there is nointerior steady state. Some numerical examples are provided.
Keywords: multiple steady states, Ramsey model, elastic labor supply,
neoclassical growth
JEL Classification: C61, C62, E13, O41∗

This paper is dedicated to Makoto Yano for his 60th birthday. Financial support fromthe Japan Society for the Promotion of Science is gratefully acknowledged. I would liketo thank an anonymous referee for helpful comments and suggestions.
†
RIEB, Kobe University, Rokkodai, Nada, Kobe 657-8501 Japan.
Email: tkamihig@rieb.kobe-u.ac.jp‡
IPAG Business School, 184 Bd Saint Germain, 75006 Paris, France.

1

Introduction

One-sector neoclassical optimal growth models have been used extensively toaddress two major issues in macroeconomics: long-run growth and short-run
fluctuations. A special case of this class of models with inelastic labor supply
is known as the Ramsey model, and one of its important properties is that all
optimal paths converge to a unique steady state. This convergence property
has various cross-country implications, on which there is a large empiricalliterature (e.g., Barro and Sala-i-Martin, 2004). Stochastic versions of theRamsey model with elastic labor supply are known as RBC (real business cycle) models (e.g., King et al., 1988; Christiano and Eichenbaum, 1992), which
are in turn considered to be prototype DSGE (dynamic stochastic generalequilibrium) models. These models are often solved by linear approximation
around a steady state, which is typically assumed to be unique.
In this paper, we study the possibility of multiple steady states in the
Ramsey model with elastic labor supply. Despite the importance of thismodel, to our knowledge, there has been no systematic analysis of this problem in the literature though there are various parametric models whose steadystates are solved explicitly.1
The literature on multiple steady states in one-sector neoclassical growthmodels dates back to Diamond (1965), who considered an overlapping generations model; see Galor and Ryder (1989) for subsequent results. Moreclosely related to this paper is the analysis of multiple steady states in thecontext of optimal growth, dating back to Kurz (1968). He showed thatmultiple steady states are possible if utility is a function of consumption andcapital.
In this paper, we show that multiple interior steady states are possible
if utility is a function of consumption and leisure, and if leisure is not a
normal good. In particular, we establish the following three results: (i) for
any discount rate and production function (satisfying standard conditions),
there is a utility function such that a continuum of interior steady states
exist; (ii) the number of interior steady states can also be any finite number;
1

In the case of inelastic labor supply, the aforementioned convergence property (which
implies the existence of a unique steady state) was established by Cass (1965) and Koopmans (1965) for the continuous time case, and by Brock and Mirman (1972, Section 2) for
the discrete time case. Le Van et al. (2007) showed a convergence property for the caseof elastic labor supply with heterogenous agents. Various convergence results on general
optimal growth models are also available; see Yano (1999, 2012) and the references therein.

1and (iii) for any discount factor and production function, there is a utility
function such that there is no interior steady state.
In a wider context, the connection between multiplicity of steady states
and non-normality of goods is already well recognized; see, e.g., De Hek(1998) and Bond et al. (2012, 2014). In particular, De Hek’s work is closest
to ours in the literature; he showed through numerical examples that multiple
steady states and non-monotone dynamics are possible in a closely related
model. His model differs from ours in that current output is assumed to bea function of the levels of capital and labor chosen in the previous period.2
However, the difference is insignificant as far as steady states are concerned;
thus our results apply to his model with minor modifications. Although it
may seem reasonable to rule out multiple steady states by assuming that
leisure is a normal good (e.g., Bosi et al., 2005), there is some empiricalevidence that suggests backward bending labor supply (e.g., Keane, 2011),
which implies that leisure is not entirely a normal good.
The rest of the paper is organized as follows. In the next section, we
present the model. In Section 3, we show some preliminary results. In
Section 4, we establish our main results. In Section 5, we offer some numericalexamples. All proofs appear in the Appendix unless otherwise indicated.

2

The Model

Consider the following one-sector optimal growth model:
max{ct ,nt ,kt+1 }∞
t=0

∞
X

β t u(ct , nt )

(1)

t=0

s.t. ∀t ∈ Z+ , ct + kt+1 = f (kt , nt ),0 ≤ nt ≤ 1,
ct , kt+1 ≥ 0,
k0 > 0 given,

(2)
(3)
(4)
(5)

where ct is consumption, nt is labor (1−nt is leisure), kt is capital, β ∈ (0, 1) isthe discount factor, u is the utility function, and f is the production function.
We state our assumptions after introducing some definitions.
2

In our notation, this means that the right-hand side of (2) is f (kt , nt−1 ). See Cai etal. (2014) for a stochastic version of this specification.

2

A feasible plan (from k 0 ) is a set of paths {ct , nt , kt }∞t=0 satisfying (2)–(4)(with k0 = k 0 ). An interior plan is a feasible plan such that none of theinequality constraints in (3) and (4) is binding for any t ∈ Z+ . An optimalplan is a feasible plan that solves the maximization problem (1)–(5). An
interior optimal plan is an optimal plan that is also an interior plan.
Throughout the paper, we maintain the following assumptions.
Assumption 1. (f1) f : R2+ → R+ is C 1 on R2++ , concave, strictly concavein the first argument, and linearly homogeneous. (f2) For all (k, n) ∈ R2++ ,we have
f1 (k, n) > 0, f2 (k, n) > 0.
(6)(f3) There exists k > 0 such thatf (k, 1) = k.

(7)(f4) We have k0 ∈ (0, k]. (f5) We have
lim f1 (k, 1) > 1/β > lim f1 (k, 1).
k↓0

k↑∞

(8)

Since f (k, 1) is strictly concave in k by (f1), condition (f4) implies thatfor any feasible plan {ct , nt , kt } from k0 ≤ k, we have∀t ∈ Z+ ,ct , kt ≤ k.

(9)

We restrict the domain of u accordingly:
Assumption 2. (u1) There exists c > k such that u : [0, c] × [0, 1] →R ∪ {−∞} is C 1 on (0, c) × (0, 1) and strictly concave. (u2) For all (c, n) ∈(0, c) × (0, 1), we haveu1 (c, n) > 0,

u2 (c, n) < 0.

(10)

It is easy to see that any interior optimal plan satisfies the following
conditions for all t ∈ Z+ :−u1 (f (kt , nt ) − kt+1 , nt ) + βu1 (f (kt+1 , nt+1 ) − kt+2 , nt+1 )f1 (kt+1 , nt+1 ) = 0,
(11)
u1 (f (kt , nt ) − kt+1 , nt )f2 (kt , nt ) + u2 (f (kt , nt ) − kt+1 , nt ) = 0.
(12)3Equation (11) is the Euler equation for kt+1 , and equation (12) is the firstorder condition for nt . The transversality condition islim β t u1 (f (kt , nt ) − kt+1 , nt )kt+1 = 0.
t↑∞

(13)

Since u and f are concave, (11)–(13) are sufficient for optimality.3
We define an interior steady state as a pair (k, n) ∈ (0, k) × (0, 1) suchthat
βf1 (k, n) = 1,
u1 (f (k, n) − k, n)f2 (k, n) + u2 (f (k, n) − k, n) = 0,f (k, n) − k > 0.

(14)
(15)
(16)

For any interior steady state (k, n), a feasible plan {ct , nt , kt } with kt = k andnt = n for all t ∈ Z+ is an interior optimal plan since it satisfies (11)–(13)with ct = f (k, n) − k > 0.

3

Preliminary Results

In this section, we define some constants, and provide simple characterizations of interior steady states.
Lemma 1. There exits a unique constant γ > 0 such that for any (k, n) ∈R2++ , we haveβf1 (k, n) = 1 ⇐⇒ k = γn.
(17)Lemma 1 implies that any interior steady state (k, n) satisfies k = γn; i.e.,the capital-labor ratio is constant across all possible interior steady states.
This allows us to focus on levels of labor in characterizing interior steadystates. We say that n ∈ (0, 1) is a steady state level of labor if the pair (k, n)with k = γn is an interior steady state.
We define the following constants:µ = f2 (γ, 1) > 0,
λ = f (γ, 1) − γ,1
ρ = − 1 > 0.
β3

See Kamihigashi (2005) for discussion on the transversality condition.

4

(18)
(19)
(20)

The inequality in (18) is immediate from (6). By linear homogeneity, for alln > 0, we haveµ = f2 (γn, n),λn = f (γn, n) − γn.

(21)
(22)

Thus µ and λ are the marginal product of labor and the consumption-laborratio, respectively, in any interior steady state. Differentiating (22) withrespect to n and using (17) and (22), we obtain4
λ = µ + ργ > 0.

(23)

The following result shows that the steady state levels of labor are characterized by a single equation.
Lemma 2. Let n ∈ (0, 1). Then n is a steady state level of labor if and only
ifg(n) ≡ u1 (λn, n)µ + u2 (λn, n) = 0.
(24)
The proofs of the following two corollaries are straightforward and thus
omitted.
Corollary 1. There exists an interior steady state if (25) or (26) below holds:
lim g(n) < 0 < lim g(n),(25)

lim g(n) > 0 > lim g(n).

(26)

n↓0

n↑1

n↓0

n↑1

Corollary 2. Suppose that there exist C 1 functions v : (0, c) → R ∪ {∞}and w : (0, 1) → R ∪ {−∞} such thatu(c, n) = v(c) + w(n).

(27)

Then there can be at most one interior steady state.
To obtain another useful characterization of interior steady states, consider the following maximization problem:maxs.t. c − µn = y,u(c, n)c≥0,n∈[0,1]4

λ = f1 (γn, n)γ + f2 (γn, n) − γ = γ/β + µ − γ = ργ + µ.

5

(28)

where y ∈ (0, k). Since u is strictly concave, this problem has a uniquesolution, which we denote by (c∗ (y), n∗ (y)). Provided that n ∈ (0, 1), we
have n = n∗ (y) if and only if
u1 (y + µn, n)µ + u2 (y + µn, n) = 0.

(29)

Exploring the similarity between (24) and (29), we obtain the following:
Lemma 3. Let n ∈ (0, 1). Then n is a steady state level of labor if and only
ifn = n∗ (ργn).

(30)

Since µ is the marginal product of labor in any interior steady state, µcan be interpreted as a steady state wage rate. Hence the constraint in (28)can be viewed as a budget constraint under this wage rate given non-labor
income y. We say that consumption is a normal (inferior) good if c∗ (y) is
strictly increasing (decreasing) in y whenever n∗ (y) ∈ (0, 1). We say that
leisure is a normal (inferior) good if n∗ (y) is strictly decreasing (increasing)in y whenever n∗ (y) ∈ (0, 1).

4Main Results

The properties of consumption and leisure defined above are closely related
to the possibility of multiple interior steady states, as we now see:
Proposition 1. If leisure is a normal good, then there can be at most one
interior steady state.
If consumption is an inferior good, then leisure must be a normal good.
Thus the following result is immediate from Proposition 1.
Corollary 3. If consumption is an inferior good, then there can be at mostone interior steady state.
Proposition 1 implies that there can be multiple interior steady states
only if leisure is not a normal good. The following result shows that there
can even be a continuum of interior steady states.
Proposition 2. For any discount factor β ∈ (0, 1) and production function f satisfying Assumption 1, there exists a utility function u satisfying
Assumption 2 such that any n ∈ (0, 1) is a steady state level of labor.
6

The number of interior steady states can also be any finite number:
Proposition 3. Let m ∈ N and 0 < η1 < η2 < · · · < ηm < 1. Then for any
discount factor β ∈ (0, 1) and production function f satisfying Assumption 1,
there exists a utility function u satisfying Assumption 2 such that n ∈ (0, 1)
is a steady state level of labor if and only if n ∈ {η1 , . . . , ηm }.
Finally, it is even possible that no interior steady state exists:
Proposition 4. For any discount factor β ∈ (0, 1) and production function f satisfying Assumptions 1, there exists a utility function u satisfying
Assumptions 2 such that there exists no interior steady state.

5

Numerical Examples

In this section, we provide some numerical examples to illustrate our results.
We assume that u is bounded below, so that we can rely on dynamic programming without technical concerns. The Bellman equation for the maximization
problem (1)–(5) can be written asv(k) = max{u(c, n) + βv(x) : c + x = f (k, n), n ∈ [0, 1], c, x ≥ 0},
c,n,x

(31)

where k ∈ [0, k] and v : [0, k] → R is the value function. We fix f and β asfollows:
f (k, n) = k α n1−α ,α = 0.4,

β = 0.95.

(32)

Under this specification, we have
γ = (αβ)1/(1−α) ≈ 0.245, µ = (1 − α)γ α ≈ 0.172, λ ≈ 0.185, k = 1.

(33)

In what follows, the value function and the associated policy functions
are computed for three different utility functions.5 Although we can use g(n)in (24) to compute interior steady states, we wish to confirm our results byusing a method that does not use g(n). In addition, dynamic programminghelps us understand other aspects of the model.
5
All of these functions are computed by modified policy iteration (e.g., Puterman, 2005)
with 10,000 equally spaced grid points.

7

The first utility function that we consider isu(c, n) = c0.4 − 0.2n1.2 .

(34)

Figure 1(a) depicts this utility function as a function of consumption andleisure (1 − n). By Corollary 2, there can be at most one interior steadystate under (34). This is confirmed by Figure 1(c), which shows that there
is a unique interior steady state.
The second utility function that we consider is(c − σ)2
σµ 
τ2 2
+ν c−
n− n −ζu(c, n) = −
2
ν
2


1/2
,

(35)

where
µ+λ
τ=
≈ 0.179,2τ 2 + λµν=
≈ 0.178,
λ+µ

σ=

1.2kν
≈ 1.244,µ(36)

and ζ is chosen so that u(0, 1) = 0. This function satisfies Assumption 2, as
can be seen from Figure 2(a). The utility function constructed in the proofof Proposition 2 is the expression inside the square brackets in (35). As in
the proof, the parameter values are chosen so that any n ∈ (0, 1) solves (24);i.e., any n ∈ (0, 1) is a steady state level of labor. This is confirmed by Figure2(c),(e).6
The last utility function that we consider is

σµ ∗
τ2
(c − σ)2+ν c−n − n2 − ζu(c, n) = −
2
ν
2


1/2
− (n − 1),(37)

where  = 0.01. This is a perturbed version of (35); all parameter values areas above. Since (24) holds for all n ∈ (0, 1) under (35), there is no n ∈ (0, 1)
satisfying (24) under (37); i.e., there is no interior steady state. This isconfirmed by Figure 3(c).
6

Numerical results based on (35) and the proof of Proposition 3 are not reported here
since they are visually almost indistinguishable from those in Figure 2.

8

(a)

(b)

(c)

(d)

(e)

Figure 1: Functions computed under (34): (a) utility function u(c, 1 − l); (b)value function v(k); and optimal policies for (c) capital (kt+1 ), (d) consumption (ct ), and (e) labor (nt ) as functions of current capital kt
9

(a)

(b)

(c)

(d)

(e)

Figure 2: Functions computed under (35): (a) utility function u(c, 1 − l); (b)value function v(k); and optimal policies for (c) capital (kt+1 ), (d) consumption (ct ), and (e) labor (nt ) as functions of current captial kt
10

(a)

(b)

(c)

(d)

(e)

Figure 3: Functions computed under (37): (a) utility function u(c, 1 − l); (b)value function v(k); and optimal policies for (c) capital (kt+1 ), (d) consumption (ct ), and (e) labor (nt ) as functions of current captial kt
11A

Appendix: Proofs

A.1

Proof of Lemma 1

By Assumption 1(f5), there exists γ > 0 such that βf1 (γ, 1) = 1. By linear
homogeneity, we have∀n > 0,βf1 (γn, n) = 1.

(38)

We have shown the “⇐” implication in (17).
To verify the reverse implication, let (k 0 , n0 )  0 be such that βf1 (k 0 , n0 ) =1. Let γ 0 = k 0 /n0 . To show that γ 0 = γ, define m(γ̃) = f1 (γ̃n0 , n0 ) for γ̃ > 0.
Then m(γ 0 ) = f1 (k 0 , n0 ) = 1/β = m(γ), where the last equality holds by (38).
Since f is strictly concave in the first argument, m(·) is strictly decreasing.
Hence γ 0 = γ.

A.2

Proof of Lemma 2

Let n ∈ (0, 1) satisfy (24). Let k = γn > 0. Then (14) holds by Lemma 1.
By (22), we havef (k, n) − k = λn > 0.
(39)
Thus (16) holds. Since f (k, 1) > f (k, n) > k by (39), we have k < k. We
obtain (15) from (39) and (21). Hence n is a steady state level of labor.
Conversely, let (k, n) ∈ (0, k) × (0, 1) satisfy (14)–(16). Then k = γn by
Lemma 1. Thus (39) holds. Substituting (39) and (21) into (15), we obtain
(24).

A.3

Proof of Lemma 3

Let n ∈ (0, 1). Suppose that n = n∗ (ργn). It follows from (29) thatu1 (ργn + µn, n)µ + u2 (ργn + µn, n) = 0.

(40)

Since ργ + µ = λ by (23), we obtain (24). Conversely, assume (24). Substituting (23) into (24), we obtain (40), which implies that n = n∗ (ργn).

A.4

Proof of Proposition 1

Suppose that leisure is a normal good. Then n∗ (·) is strictly decreasing.
Hence there can be at most one solution n to (30). Thus by Lemma 3, there
can be at most one interior steady state.
12

A.5

Proof of Proposition 2

Define φ : R2 → R by

1
σµ 
τ2
φ(c, n) = − (c − σ)2 + ν c −n − n2 ,
2
ν
2

(41)

where σ, τ > 0 are parameters and ν is defined byν=τ 2 + λµ
.
λ+µ

(42)


σµ 
φ2 (c, n) = ν c −− τ 2 n.
ν

(43)

For all c, n ∈ R, we have
φ1 (c, n) = (σ − c) + νn,

Direct computation shows that for all n ∈ R, we have7
φ1 (λn, n)µ + φ2 (λn, n) = 0.

(44)

It thus follows by Lemma 2 that if u = φ on [0, c] × [0, 1] for some c > k,then any n ∈ (0, 1) is a steady state level of labor. Hence it suffices to showthat φ satisfies Assumption 2 on [0, c] × [0, 1] for some σ, τ > 0 and c > k.
To this end, we need two lemmas:
Lemma 4. Let
ĉ =

σµ
.
ν

(45)

Then
∀(c, n) ∈ [0, ĉ) × [0, 1],

φ1 (c, n) > 0,

φ2 (c, n) < 0.

(46)

Proof. Let (c, n) ∈ [0, ĉ) × [0, 1]. Since ĉ < σ, we have φ1 (c, n) > 0 by (43).
Note thatσµ − τ 2 n = −τ 2 n < 0.
(47)
φ2 (c, n) < ν ĉ −νThis completes the proof.
Lemma 5. The function φ is strictly concave if and only ifµ < τ < λ.
7(48)

φ1 (λn, n)µ + φ2 (λn, n) = [(σ − λn) + νn]µ + ν(λn − σµ/ν) − τ 2 n = [ν(µ + λ) − (λµ +τ )]n + σµ − σµ = 0.
2

13

Proof. Note from (43) that the Hessian matrix of φ at any (c, n) is 
φ11 φ12−1 ν.
=
φ21 φ22ν −τ 2(49)

Thus φ is strictly concave if and only if ν 2 < τ 2 . Since ν, τ > 0, it follows
that φ is strictly concave if and only if ν < τ . Recalling (42), we see that
ν < τ ⇔ τ 2 + λµ < (λ + µ)τ ⇔ τ 2 − (λ + µ)τ + λµ < 0 ⇔ (τ − λ)(τ − µ) < 0.
The last inequality is equivalent to (48) since µ < λ by (23).
To complete the proof of Proposition 2, fix τ ∈ (µ, λ) and c > k. Let σ > 0
be such that ĉ > c, where ĉ and ν are given by (45) and (42), respectively.
Suppose that u(c, n) = φ(c, n) for all (c, n) ∈ [0, c] × [0, 1]. Then Lemmas 4
and 5 imply Assumption 2. Thus by (44) and Lemma 2, any n ∈ (0, 1) is a
steady state level of labor.

A.6

Proof of Proposition 3

Let m ∈ N and 0 < η1 < η2 < · · · < ηm < 1. It suffices to show that
there exists a utility function u satisfying Assumption 2 such that n ∈ (0, 1)
satisfies (24) if and only if n ∈ {η1 , . . . , ηm }. To this end, let J = {1, . . . , m},and define h : R+ → R by
Z nY(x − ηj )dx.
(50)h(n) =0j∈J

Q
Since h0 (n) = j∈J (n−ηj ), we have h0 (n) = 0 if and only if n ∈ {η1 , . . . , ηm }.
Let c and φ be as in the proof of Proposition 2. Let  ∈ R. Defineu(c, n) = φ(c, n) + h(n) for (c, n) ∈ [0, c] × [0, 1]. Then for any n ∈ (0, 1),
we have g(n) = h0 (n) by (44) and (24); thus (24) holds if and only if n ∈{η1 , . . . , ηm }. It remains to show that u satisfies Assumption 2 for some ∈ R.
Recalling (46) and (49), we see that if  = 0, then u1 (c, n) > 0, u2 (c, n) >0, and the Hessian matrix of u at (c, n) is negative definite for all (c, n) ∈[0, c] × [0, 1]. Since these properties are preserved for  sufficiently close to 0,
it follows that u satisfies Assumption 2 for such , as desired.

14

A.7

Proof of Proposition 4

Let c and φ be as in the proof of Proposition 2. Let  > 0. Suppose thatu(c, n) = φ(c, n) − n for all (c, n) ∈ [0, c] × [0, 1]. It is easy to see that u
satisfies Assumption 2. By (44) and (24), we have g(n) = − for all n ∈ (0, 1).
Hence there is no interior steady state by Lemma 2.

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16