Optimal growth with investment enhancing labor

Accepted Manuscript Optimal growth with investment enhancing labor Bertrand Crettez, Naila Hayek, Lisa Morhaim PII: DOI: Reference:

S0165-4896(16)30213-X http://dx.doi.org/10.1016/j.mathsocsci.2016.12.002 MATSOC 1914

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Mathematical Social Sciences

Received date: 19 December 2015 Revised date: 16 October 2016 Accepted date: 17 December 2016 Please cite this article as: Crettez, B., Hayek, N., Morhaim, L., Optimal growth with investment enhancing labor. Mathematical Social Sciences (2016), http://dx.doi.org/10.1016/j.mathsocsci.2016.12.002 This is a PDF file of an unedited manuscript that has been accepted for publication. As a service to our customers we are providing this early version of the manuscript. The manuscript will undergo copyediting, typesetting, and review of the resulting proof before it is published in its final form. Please note that during the production process errors may be discovered which could affect the content, and all legal disclaimers that apply to the journal pertain.

*Highlights (for review)

Highlights • We study a non-convex optimal growth problem with investment enhancing labor. • We prove that there exists an optimal growth path, that all optimal paths are interior and that at least one of them is monotonic. • We also study the existence and uniqueness of the steady state. We show in particular that a rise in the efficiency of the investment enhancing labor does not necessarily lead to an increase in the steady state value of this labor. • Furthermore we provide a complete study of the dynamics of the optimal solution in the special case of a logarithmic utility function and a Cobb-Douglas production function.

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Optimal Growth with Investment Enhancing Labor Second Revised version Bertrand Crettez, Naila Hayek and Lisa Morhaim Universit´e Panth´eon-Assas, Paris II, CRED⇤ October 5, 2016

Abstract We study a non-convex optimal growth problem with investment enhancing labor. We prove that there exists an optimal growth path, that all optimal paths are interior and we provide a condition under which at least one of them is monotonic. We also study the existence and uniqueness of the steady state. We show in particular that a rise in the efficiency of the investment enhancing labor does not necessarily lead to an increase in the steady state value of this labor. Furthermore we provide a complete study of the dynamics of the optimal solution in the special case of a logarithmic utility function and a Cobb-Douglas production function. Keywords: optimal growth; non-convex model, value function; Bellman equation; investment enhancing labor JEL Classification: C61, 041

1

Introduction

A well-functioning financial sector eases information processing, risk taking and the allocation of capital across firms. Therefore the long-run rise in the financial sector in modern economies has certainly been beneficial to them. Yet, there is a growing concern that there is “too much finance” (Arcand, Berkes, and Panizza, 2012, Zingales, 2015) and that the benefits of finance are inflated. While assessing the net benefits of finance is a difficult task, two facts support this concern. First, the finance’s share of national This research was completed thanks to the Novo Tempus research grant, ANR-12-BSH1-0007, Program BSH1-2012. We thank two anonymous referees for stimulating comments on two previous versions of this paper. ⇤

1

income has become higher than it has been during the last 150 years (Philippon and Reshef, 2007, Philippon and Reshef, 2013).1 Second, it seems that the percentage of graduates with positions in finance has sharply increased (see Goldin and Katz, 2008, for a study of the Harvard graduates ; Philippon and Reshef, 2007, for a study of the U.S. economy). These figures underline the need to further the study of the e↵ect of finance on an economy. This paper tackles this issue along an optimal growth approach. For sure, standard optimal growth models often assume that savings can easily be transformed into capital and that an extra unit of savings produces an extra unit of capital. In real life economies, however, this transformation rate is less than one. The value of this rate results from various transaction costs that investors must bear. Examples of these costs include, but are not limited to, the costs to define and protect property rights and the search costs to find interesting opportunities. In this paper we study an optimal growth model with investment enhancing labor. To wit, we assume that labor can be allocated between production and investment enhancing.2 The higher the share of labor devoted to investment enhancing, the higher the rate of transformation of savings to capital. Because of this assumption, our optimal growth model is non-convex. The present paper is related to Pagano (1993), where the fraction of savings which is actually invested is exogenous, and to Varoudakis and Berth´elemy (1994), and Eggoh and Villieu (2013), who assume that this fraction increases with the quantity of labor allocated to the financial sector. Eggoh and Villieu study a version of our model with a CCRA utility function and a Cobb-Douglas production function which comprises a Romer-like externality. They show that multiple endogenous growth paths can exist and that there might exist a non-linear relationship between financial development and economic growth.3 There are two di↵erences between these papers and the present one. First, these papers do not address the optimal growth problem. They rather study the properties of intertemporal competitive equilibria. Second, these papers rely on specific functional forms for the production and the utility functions. By contrast, this paper focuses on optimal growth strategies in a general setting (but without Romer like external e↵ects). These optimal strategies are solutions to a nonconvex optimal growth problem. The source of the non-convexity is di↵erent from those considered in the literature (see e.g., Azariadis and Stachurski, 2005; Majumdar, 2006; Kamihigashi and Roy, 2007; Hung, Le Van, and Michel, 2009; Akao, Kamihigashi, 1 Philippon and Reshef (2007) indicate that in the last 60 years, the U.S. financial sector has grown from 2.3% to 7.7% of GDP. 2 Findlay and Wilson (1987) consider a model where a part of labor a↵ects the net value of production. This share of labor is interpreted as being the resource devoted by society to the organization of the market system itself. 3 Of course, there are other approaches to study the e↵ect of finance on growth from a normative view point (i.e., an optimal growth problem). See, e.g., Philippon (2007) for a study using an overlapping generations model.

2

and Nishimura, 2011; Le Van, Morhaim, and Dimaria, 2002; Le Van and Dana, 2003; Le Van and H. C. Saglam, 2004). We study both the cases where the utility function is bounded from below and the case where it is not. We first study the existence of an optimal path. Next, we show that optimal paths are interior (this is especially non-trivial when the utility function is lower bounded) and we provide a condition under which at least one of them is monotonic. We study the value function associated to the optimal growth problem, as well as the existence and uniqueness of a steady state. We devote a section to the special case with a logarithmic utility function and a Cobb-Douglas production function. While our optimal growth problem is non-convex, we can, however, address the non-convexity issue in the special case mentioned above (notably by using a result of the value function). Finally, we study the comparative statics of the steady state. We show that an increase in the efficiency of the investment enhancing labor does not always rise the steady state value of this labor. We believe that this last result is useful in regard of two recent trends. Firstly, there has been a move toward financial deregulation (at least before the financial crisis of 2007-2008).4 We consider that deregulation can be interpreted as an increase in the efficiency of investment enhancing labor. In e↵ect tight financial regulation limits by definition the range of possible activities. Secondly, information and communication technologies that increase the efficiency of labor in finance have developed at a fast pace.5 Philippon and Reshef (2007) and Philippon and Reshef (2013) argue that these two trends explain at least in part the evolution of the shares of finance in national income and in the skilled labor force. Our findings suggest that this could not be a desirable outcome. The paper unfolds as follows. In the next section we set out the model and our assumptions. Section 3 studies the existence, the properties of optimal paths and the value function associated with our problem. We consider the existence and the uniqueness of the steady-state in section 4, while section 5 is devoted to the comparative statics of this steady-state. In section 6 we provide a complete study of an economy with a logarithmic utility function and a Cobb-Douglas production function. We conclude the paper in section 7. Some of the proofs are given in the Appendix as well as a study of local stability when the steady state is not unique. 4 5

See, e.g., Philippon and Reshef (2013). These technologies ease the gathering and processing of information.

3

2

Model and Assumptions

2.1

The model

Let us consider the following problem. 8 1 X > > t > max u(ct ) > > < {(ct )t ,(kt+1 )t ,(zt )t }+1 t=0 t=0 (P) = s.t. 0  kt+1 = (zt ) f (kt , 1 > > > z 0, > t 2 [0, 1], ct > : k0 > 0 being given.

zt )

ct 8t

0,

In this problem, ct denotes consumption at date t, kt is the stock the capital (for the same date), and zt is the amount of labor devoted to transform savings into capital for the next period (kt+1 ). We call zt investment enhancing labor. The term 1 zt represents the part of labor which is used as an input in the production function.6 We denote infinite sequences from t = 0 by (c, k, z): (c, k, z) = (ct )t , (kt )t , (zt )t

+1 . t=0

For any k0 2 IR+ , let ⇧(k0 ) denote the set of feasible sequences starting from k0 : ⇧(k0 ) = {(c, k, z) | zt 2 [0, 1], 0  ct  f (kt , 1

zt ),

0  kt+1 = (zt ) f (kt , 1

2.2

zt )

ct 8t

0}. (1)

Assumptions

We shall need the following assumptions. (H1a) u is defined on IR+ , strictly increasing, strictly concave, twice continuously di↵erentiable on IR⇤+ , limc!0+ u0 (c) = +1 and u(0) = 0. (H1b) u is defined on IR+ with values in R[{ 1}, strictly increasing, strictly concave, twice continuously di↵erentiable on IR⇤+ , limc!0+ u0 (c) = +1 and u(0) = 1. (H2) The function f : (k, y) 7! f (k, y) is defined on IR+ ⇥ [0, 1] with values in IR+ , is concave and has constant returns to scale. Moreover f is twice continuously di↵eren00 00 tiable on IR⇤+ ⇥]0, 1[ and satisfies f11 < 0 and f22 < 0. For any k 2 IR+ and y 2 [0, 1], f (·, y) and f (k, ·) are strictly increasing. For any y 2 [0, 1], f (0, y) = 0 and for any k 2 IR+ , f (k, 0) = 0. Capital depreciation does not appear in the present specification of the law of motion of capital. In fact, if we assume that the production function is g(kt , 1 zt ) and the depreciation rate is , we posit that the law of motion of capital can be written as: kt+1 = (zt )(g(kt , 1 zt ) + (1 )kt ct ). Then we may define the function f by f (kt , 1 zt ) = g(kt , 1 zt ) + (1 )kt . Notice that we do not posit: kt+1 = (zt )(g(kt , 1 zt ) ct ) + (1 )kt . This admittedly standard assumption would have two drawbacks. First, some capital could be used even if no labor were devoted to enhance investment (and/or organize the allocation of capital). Second, the analysis would be more complicated. In particular, it would no longer be possible to obtain a closed-form solution for the example studied in section 6. We thank two referees for drawing our attention to this issue. 6

4

(H3) : [0, 1] ! [0, 1] is strictly increasing and strictly concave and twice continuously di↵erentiable on ]0, 1[, (0) = 0. (H4) For any z in [0, 1[, lim+ x !0 +1. (H5) 0 <

@f ( (x)x, 1 @x

z)

= lim+ f10 ( (x)x, 1 x !0

z)( (x)x)0 =

< 1.

Lemma 1. Assume (H2). Then there exists non-negative real numbers , 6= 1, and 0 such that: for any (k, y) 2 R+ ⇥ [0, 1], f (k, y)  k + 0 , i.e. for any (k, y) 2 R+ ⇥ [0, 1], [0, f (k, y)] ⇢ [0, k + 0 ]. Proof. Let (k, y) 2 IR+ ⇥ [0, 1]. Since f is strictly increasing in its second variable, one has f (k, y)  f (k, 1). Now let kˆ 2 IR+ be given. Since f is strictly concave and di↵erentiable, f (k, 1)

ˆ 1)  f 0 (k, ˆ 1)(k f (k, 1

ˆ k),

(2)

ˆ 0 (k, ˆ 1)). kf 1

(3)

or: ˆ 1)k + (f (k, ˆ 1) f (k, 1)  f10 (k,

ˆ 1) if f 0 (k, ˆ 1) 6= 1 (else take = f 0 (k, ˆ 1) + ", " > 0), and We then define: = f10 (k, 1 1 0 0 ˆ ˆ ˆ = f (k, 1) kf1 (k, 1) (since f has constant returns to scale, by Euler’s Theorem is positive). Then we are done. (H6) 2.2.1

< 1 where

is given in Lemma 1.

Remarks on the assumptions

1) Note that the assumptions on the utility (H1a) and (H1b) only di↵er by the value at origin. 2) The assumption (H4) is crucial to have an interior solution in the cases where u(0) = 0. The following example illustrates this assumption. Let f (k, 1

z) = k ↵ (1

z)1

↵

, and (z) = z ✓ , 0 < ↵ < 1 and 0 < ✓ < 1.

We have: f10

x (x), 1

✓

z =↵ x

1+✓

◆↵

1

(1

z)1

↵

,

(4)

and 0

x (x) = (1 + ✓)x✓ .

(5)

Thus: f10 x (x), 1

0

z x (x) = ↵(1 5

z)1

↵

(1 + ✓)x↵(1+✓)

1

(6)

The condition limx!0+ f10 x (x), 1 ✓ < 1 ↵↵ .

z x (x)

0

= +1 is satisfied for all z in [0, 1[ i↵

3) Note that the existence of a k 2 R⇤+ such that f10 (k, 1) < 1 is a sufficient condition for (H6) ( < 1).

3

Optimal Paths

In this section we first study the existence of optimal paths. Then we prove that all optimal paths are interior. Further, we study the value function associated to the optimal growth problem. Finally, we provide a condition under which at least one optimal path is monotonic.

3.1

Definition of the objective function

Lemma 2. Assume (H1a) or P (H1b), and (H2), (H5) and (H6). Let k0 2 R+ be given. Then the limit limT !+1 Tt=0 t u(ct ) is well defined in R [ { 1}.

Proof. Let t be a given integer. Since u is concave we can always find (using the above reasoning with f ) two non-negative real numbers A and B such that for all ct : u(ct )  Act + B

(7)

Thus along a feasible path: u(ct )+  Af (kt , 1 zt ) + B,  Af (kt , 1) + B,  A ( kt + 0 ) + B, ✓ 0  A t+1 (k0 )+ 1 1

0

+

0

◆

(8a) (8b) (8c) + B,

(8d)

where the last inequality comes from the fact that since (c, k, z) 2 ⇧(k0 ), by (H2) and Lemma 1, one has 0  kt+1  kt + 0 for any t 0, which implies by induction that t 0  kt  t k0 + 0 11 . Then: u(ct )+  with a1 (k0 ) = A(k0

0

1

t+1

) and a2 = A 0 ( 1

a1 (k0 ) + a2 , + 1) + B = A 1

0

+ B.

Thus, whether u(ct ) is negative or not, we have: T X t=0

t

u(ct )+  a1 (k0 )

The conclusion follows by (H5) and (H6). 6

T X t=0

(

) t + a2

T X t=0

t

.

(9)

Let us then define the function U from ⇧(k0 ) to R [ { 1} by U (c, k, z) := lim

T !+1

T X

t

u(ct ).

(10)

t=0

We now study the problem: 8 max U (c, k, z) > < (c,k,z) (P) = s.t. (c, k, z) 2 ⇧(k0 ), > : k0 > 0 being given.

Let ⇧0 (k0 ) = {(c, k, z) 2 ⇧(k0 ) | U(c, k, z) >

3.2

1}.

Existence of a solution

Let us consider the following assumption ˆ zˆ) 2 ⇧(k0 ) such that U (ˆ ˆ zˆ) > (H7) There exists (ˆ c, k, c, k,

1, i.e. ⇧0 (k0 ) 6= ;.

Proposition 1. Assume that either [(H1a)] or [(H1b) and (H7)] hold, and that (H2)-(H3)-(H5)-(H6) hold. Then there exists a solution to the problem (P). Proof. 1. We first remark that the set ⇧(k0 ) is a closed set for the product topology since and f are continuous. 2. We now show that ⇧(k0 ) is compact in the product topology. Define: k¯0 = k0 . Define also recursively: k¯t+1 = sup

(z)f (k¯t , 1

z),

(11)

z2[0,1]

c¯t = f (k¯t , 1).

(12)

Now, we prove by induction that for any (c, k, z) 2 ⇧(k0 ), we have kt  k¯t and ct  c¯t for all t. This is true k0 . Now assume that kt  k¯t . Then: kt+1  (zt )f (kt , 1

zt )  (zt )f (k¯t , 1

zt )  k¯t+1

(13)

Moreover for all t we have ct  f (kt , 1 zt )  f (k¯t , 1 zt )  c¯t . Therefore, we have: ⇧(k0 ) ✓ ⇧+1 ¯t ] ⇥ [0, k¯t ] ⇥ [0, 1]. Since ⇧+1 ¯t ] ⇥ [0, k¯t ] ⇥ [0, 1] is compact in the t=0 [0, c t=0 [0, c product topology, and ⇧(k0 ) is closed in this topology, then ⇧(k0 ) is compact in the product topology. 3. Now let us show that U is upper semicontinuous on ⇧(k0 ). Consider a sequence (cn , k n , z n )n in ⇧(k0 ) that converges to (c, k, z) 2 ⇧(k0 ). We 7

will check that lim supn!+1 U (cn , k n , z n )  U (c, k, z). We set u(ct )+ = max{u(ct ), 0}. Since for all (c, k, z) 2 ⇧(k0 ), u(ct )+  t+1 a1 (k0 ) + a2 , we have that for all " > 0, there exists T0 such that for all T T0 : +1 X

t=T +1

t

u(ct )+  ✏.

(14)

It follows that: n

n

n

8n, U (c , k , z ) =  

1 X

t=0 T X

t

u(cnt )

t

u(cnt )

(15a) +

t=0

T X

+1 X

t

u(cnt )+

(15b)

t=T +1 t

u(cnt ) + ✏.

(15c)

t=0

Therefore: lim sup U (cn , k n , z n )  n!+1

T X

t

u(ct ) + ✏.

(16)

t=0

Since the above inequality is true for all T higher than T0 and ✏ > 0, we conclude that: lim sup U (cn , k n , z n )  n!+1

+1 X

t

u(ct ).

(17)

t=0

This proves that U is upper semicontinuous in the product topology on ⇧(k0 )(Lemma 2.39, Aliprantis and Border, 2006). 4. Since U is upper semicontinuous, and ⇧(k0 ) is a compact set, by Weierstrass Theorem, there exists a solution to (P) (Theorem 5.3.1, Aubin, 1996).

3.3

Interiority

Proposition 2. Assume either (i) (H1b)-(H2)-(H3) and (H7) or (ii) (H1a)(H2)-(H3)-(H4)-(H5)-(H6). Then any optimal path is interior, i.e. if (c, k, z) is an optimal solution, then for all t 2 N, ct > 0, kt > 0 and zt 2]0, 1[. ˆ zˆ) 2 Proof. For case (i) the proof is straightforward. Indeed, as long as there exists (ˆ c, k, ˆ ⇧(k0 ) such that U (ˆ c, k, zˆ) > 1, then any sequence (c, k, z) 2 ⇧(k0 ) such that there exists t with ct = 0 cannot be optimal. Moreover, if a feasible sequence is such that there exists t with kt = 0 then by (H2) ct = 0 and this sequence cannot be optimal. 8

On the other hand, if a feasible sequence is such that zt = 0, then kt+1 = 0 which implies by (H2) that ct+1 = 0. Again this sequence cannot be optimal. Finally, if a feasible sequence is such that there exists t with zt = 1, then by (H2) ct = 0 and this sequence cannot be optimal either. For case (ii), the proof is done through the four following lemmas and consists in constructing in each case where zt or ct could violate the interiority, a feasible sequence that would contradict the optimality. Moreover, if a feasible sequence is such that there exists t with kt = 0 then by (H2) ct = 0 which is a contradiction. Lemma 3. Let (c, k, z) be an optimal solution. Then for all t 2 N, zt > 0. Proof. Assume that there exists t 2 N such that zt = 0 and let t be the smallest integer such that zt = 0. Then (zt ) = 0 which implies that kt+1 = 0, which implies that f (kt+1 , 1 zt+1 ) = 0. Therefore ct+1 = 0 and kt+2 = 0, and so on. So 8j > t, kj = 0 and cj = 0. We also have ct = f (kt , 1). • Case kt > 0. So f (kt , 1) > 0 and there exists ✏0 2]0, 1[ such that, for all ✏ such that 0 < ✏  ✏0 , f (kt , 1 ✏) ✏ > 0. ˜ z˜) in the following way: For any such 0 < ✏  ✏0 , we now construct a process (˜c, k,

c˜t = f (kt , 1

9 c˜j = cj > = k˜j = kj 8j < t, > ; z˜j = zj ✏)

✏

h k˜t = (˜ zt 1 ) f (k˜t 1 , 1 z˜t = ✏

Therefore:

z˜t 1 )

c˜t

1

i

= kt

9 > > = > > ;

,

9 c˜t+1 = f (k˜t+1 , 1 z˜t+1 ) = f ( (✏)✏, 1 z˜t+1 )> > = h i ˜ ˜ , kt+1 = (˜ zt ) f (kt , 1 z˜t ) c˜t = (✏)✏ > > ; z˜t+1 2]0, 1[ 9 c˜j = 0 > = 8j > t + 1. k˜j = 0 > ; z˜j 2]0, 1[

9

✏

=

+1 X

t

u(˜ ct )

t=0 t

+1 X

t

u(ct ),

t=0

t t+1 u(˜ ct ) + t+1 u(˜ ct+1 ) u(ct ) u(ct+1 ), t 0 t+1 0 u (˜ ct )(˜ ct ct ) + u (˜ ct+1 )(˜ ct+1 ct+1 ) (since u is concave),  (˜ ct ct ) = t u0 (˜ ct ) + t+1 u0 (˜ ct+1 ) c˜t+1 . c˜t+1

=

So one can find 0 < ✏0  ✏0 such that

✏0

> 0 since:

lim u0 (˜ ct+1 ) = lim u0 f ( (✏)✏, 1

✏ !0

✏ !0 0

z˜t+1 ) = +1, by (H1a),

lim u0 (˜ ct ) = u (f (kt , 1)) > 0,

✏ !0

(˜ ct ct ) f (kt , 1 ✏) ✏ f (kt , 1) = lim , !0 ✏ !0 c˜t+1 f ( (✏)✏, 1 z˜t+1 ) f20 (kt , 1 ✏) 1 (by L’Hopital’s rule), = lim 0 ✏ !0 f1 ( (✏)✏, 1 z˜t+1 )( (✏)✏)0 = 0 by (H4).

lim

✏

˜ z˜) is feasible and But the existence of an ✏0 such that the process (˜c, k, tradicts the optimality of (c, k, z).

✏0

> 0 con-

• Case kt = 0, t > 0. In this case ct = 0. So let t0 < t be the greatest integer such that kt0 > 0 (t0 may be equal to 0.) So 8j > t0 , kj = 0 and cj = 0 and ct0 = f (kt0 , 1 zt0 ). – If zt0 = 1 then ct0 = 0. Then there exists ✏ 2]0, 1[ such that f (kt0 , ✏) > 0. We ˜ z˜) such as: construct a process (˜c, k,

c˜j = cj k˜j = kj z˜j = zj Thus:

9 > = > ; ✏

9 c˜t0 = f (k˜t0 , ✏) > = ˜ , 0 0 k t = kt > ; z˜t0 = 1 ✏

8j < t0 ,

=

+1 X t=0

t

u(˜ ct )

+1 X

t

u(ct ) =

c˜j = 0 k˜j = 0 z˜j 2]0, 1[

t0

9 > = > ;

, 8j > t0 .

u(˜ ct0 ) > 0,

t=0

which contradicts the optimality of (c, k, z). – If zt0 < 1 then ct0 = f (kt0 , 1 zt0 ) > 0 and there exists ✏ 2]0, 1[ such that 0 < zt0 ✏ < 1 and f (kt , 1 (zt0 ✏)) > 0.

10

˜ z˜) in the following way: We construct a process (˜c, k, 9 c˜j = cj = 8j < t0 , k˜j = kj ; z˜j = zj

Therefore:

c˜t0 = hf (kt0 , 1 (zt0 ✏)) > ct0 i k˜t0 = (˜ zt0 1 ) f k˜t0 1 , 1 z˜t0 1 c˜t0 1 = kt0 z˜t0 = zt0 ✏ 9 c˜j = 0 = 8j > t0 . k˜j = 0 ; z˜j 2]0, 1[ ✏

= =

+1 X

t

+1 X

u(˜ ct )

t=0 t0

t

9 > = > ;

,

u(ct ),

t=0

[u(˜ ct0 )

u(ct0 )] > 0,

which contradicts the optimality of (c, k, z). Thus we have for all t 2 N, zt > 0. Lemma 4. Let (c, k, z) be an optimal solution. Then for all t 2 N, zt < 1. Proof. Assume that there exists t 2 N such that zt = 1 and let t be the smallest integer such that zt = 1. So f (kt , 1 zt ) = f (kt , 0) = 0 which implies that ct = 0 and kt+1 = 0. So 8j > t, cj = 0 and 8j > t + 1, kj = 0. • Case kt > 0. So f (kt , 1) > 0 and there exists ✏ 2]0, 1[ such that f (kt , 1 ✏) > 0. ˜ z˜) in the following way: We construct a process (˜c, k, 9 9 9 c˜j = cj > c˜t = f (kt , 1 ✏) > c˜j = 0 > = = = 8j < t, , 8j k˜j = kj k˜t = kt k˜j = 0 > > > ; ; ; z˜j = zj z˜t = ✏ z˜j 2]0, 1[ Thus we have:

✏

= =

+1 X t=0 t

t

u(˜ ct )

u(˜ ct ) =

+1 X

t

u(ct ),

t=0

t

u f (kt , 1

11

✏ > 0,

t + 1.

which contradicts the optimality of (c, k, z). • Case kt = 0, t > 0. Let t0 < t be the greatest integer such that kt0 > 0 (t0 may be equal to 0.) So 8j > t0 , kj = 0 and cj = 0. And ct0 = f (kt0 , 1 zt0 ) > 0. Since 0 < zt0 < 1 by Lemma 3, there exists ✏ > 0 such that 0 < zt0 ✏ < 1. We ˜ z˜) in the following way: construct a process (˜c, k,

c˜t0 = f kt0 , 1 (zt0 k˜t0 = kt0 z˜t0 = zt0 ✏

9 c˜j = cj = 8j < t0 , k˜j = kj ; z˜j = zj 9 ✏) > ct0 = , where 0 < zt0 ;

✏ < 1 since zt0 > 0,

9 c˜j = 0 = 8j > t0 . k˜j = 0 ; z˜j 2]0, 1[

We have: ✏

= =

+1 X

t

u(˜ ct )

t=0 t0

+1 X

t

u(ct ),

t=0

[u(˜ ct0 )

u(ct0 )] > 0,

which contradicts the optimality of (c, k, z). Lemma 5. Let (c, k, z) be an optimal solution. Then for all t 2 N, ct > 0. Proof. Assume that there exists t 2 N such that ct = 0 and let t be the smallest integer such that ct = 0. So cj > 0, 8j < t. • Case kt > 0 and cj = 0, 8j

t. (t can be equal to 0).

In this case kt+1 = (zt )f (kt , 1 zt ) > 0 since ct = 0 (and we have seen above that zt 2]0, 1[). Since kt+1 > 0, there exists ✏ > 0 such that kt+1 ✏ > 0. ˜ z˜) in the following way: We construct a process (˜c, k, c˜j = cj k˜j = kj z˜j = zj and,

9 > = > ;

c˜t = 8j < t,

✏ 9 = (zt ) >

k˜t = kt z˜t = zt

12

> ;

c˜t+1 = 0 ,

9 > =

k˜t+1 = kt+1 ✏ , > ; z˜t+1 = zt+1

c˜j = 0 k˜j = (˜ zj 1 )f k˜j 1 , 1 z˜j = zj

z˜j

9 > =

1

> ;

We have: ✏

=

+1 X

t

u(˜ ct )

t=0

= But

✏

t

+1 X

t

8j > t + 1.

u(ct ),

t=0

t

u(˜ ct ) =

✏ ) > 0. (zt )

u(

> 0 contradicts the optimality of (c, k, z).

• Case where kt > 0 and there exists t0 > t such that ct0 > 0. Let t0 be the smallest integer higher than t, such that ct0 > 0. zj 1 ), 8j such that t + 1  j  t0 since

In this case kj > 0, and kj = (zj 1 )f (kj 1 , 1 cj = 0 8j such that t  j  t0 1. We have both: 0 = ct0

1

= f (kt0 1 , 1

ct0 = f (kt0 , 1

z t0 1 )

kt 0 , (zt0 1 )

kt0 +1 . (zt0 )

z t0 )

Since ct0 > 0 and kt0 > 0, there exists ✏0 > 0 such that, for all ✏ such that 0 < ✏  ✏0 , kt0 +1 f (kt0 ✏, 1 zt ) > 0. (zt ) ˜ z˜) in the following For any such 0 < ✏  ✏0 , we can construct a feasible process (˜c, k, way: 9 c˜j = cj = > ˜ 8j < t0 kj = kj > ; z˜j = zj c˜t0 = f (k˜t0 , 1

z t0 )

k˜t0 = kt0 ✏ z˜t0 = zt0

c˜t0 1,

k˜t0 z˜t0

kt0 +1 9 > (zt0 ) = > ;

13

1

,

= 1 1

✏ 9 = (zt0 1 ) >

= kt 0 = z t0

c˜j = cj k˜j = kj z˜j = zj

1

1

9 > = > ;

> ;

,

8j > t0 .

We have: ✏

= =

+1 X

t=0 t0 1

t

u(˜ ct )

+1 X

t

u(ct )

t=0 t0

u(˜ ct0 1 ) + (u(˜ ct0 ) u(ct0 )), ✓ ◆ ✓ ◆ ✏ kt0 +1 kt0 +1 0 0 t0 = t 1u + t u f (k˜t0 , 1 zt0 ) u f (kt0 , 1 zt0 ) , (zt0 1 ) (zt0 ) (zt0 ) ✓ ◆ ✓ ◆✓ ◆ ✏ ✏ kt0 +1 t0 1 0 t0 0 ˜ ˜ u + u f ( kt 0 , 1 z t 0 ) f (kt0 , 1 zt0 ) f (kt0 , 1 zt0 ) , (z 0 ) (z 0 ) (zt0 )  ✓ t 1 ◆ t 1 ✏ 1 kt0 +1 f (kt0 ✏, 1 zt0 ) f (kt0 , 1 zt0 ) 0 = ✏ t 1 u0 + u0 f (kt0 ✏, 1 zt0 ) . . (zt0 1 ) (zt0 1 ) (zt0 ) ✏ So one can find 0 < ✏0  ✏0 such that

> 0 since: ◆ ✏ 0 lim u = +1, ✏ !0 (zt0 1 ) ✓ ◆ kt0 +1 0 0 0 lim u f (kt ✏, 1 zt ) , = u0 (f (ct0 )) ✏ !0 (zt0 ) f (kt0 ✏, 1 zt0 ) f (kt0 , 1 zt0 ) lim = f10 (kt0 , 1 ✏ !0 ✏

But the existence of an ✏0 such that

✏0

✓

✏0

z t0 ) >

1.

> 0 contradicts the optimality of (c, k, z).

• Case kt = 0, t > 0. Then kt+1 = 0 and we have 8j > t, cj = 0 and kj = 0. Notice that kj > 0, 8j < t (if kj = 0 for a j < t then cj = 0 for this j < t which contradicts cj > 0, 8j < t.) kt = 0 implies ct 1 = f (kt 1 , 1 zt 1 ). Since ct 1 > 0, there exists ✏0 > 0 such that, for all ✏ such that 0 < ✏  ✏0 , ct 1 ✏ > 0. (zt 1 ) ˜ z˜) in the following way: We now construct a process (˜c, k,

c˜j = cj ˜ k j = kj z˜j = zj

c˜t = f (k˜t , 1

9 > = > ;

c˜t 8j < t

= ct

1,

z˜t ) = f (✏, 1

k˜t = ✏ z˜t 2]0, 1[

1

k˜t z˜t 9 z˜t ) > = > ;

14

,

1 1

1

= kt = zt

✏ 9 = (zt 1 ) > > ;

1

1

c˜j = 0 ˜ kj = 0 z˜j 2]0, 1[

9 > = > ;

8j > t.

We have: ✏

=

+1 X

t

u(˜ ct )

t=0 t 1

+1 X

t

u(ct ),

t=0 t

t 1 t u(˜ ct 1 ) + u(˜ ct ) u(ct 1 ) u(ct ), t 1 0 t 0 u (˜ ct 1 )(˜ ct 1 ct 1 ) + u (˜ ct )(˜ ct ct ), ✏ ✏ + t u0 f (✏, 1 z˜t ) f (✏, 1 z˜t ), = t 1 u0 ct 1 (zt 1 ) (zt 1 )  ✏ 1 f (✏, 1 z˜t ) = ✏ t 1 u0 c t 1 + t u0 f (✏, 1 z˜t ) . (zt 1 ) (zt 1 ) ✏

=

So one can find 0 < ✏0  ✏0 such that

✏0

> 0 since:

lim u0 (f (✏, 1

z˜t )) = +1,

✏ !0

f (✏, 1 z˜t ) = f10 (0, 1 z˜t ) > 0, ✏ ✏ 1 u0 (ct 1 ) )( = > 1. (zt 1 ) (zt 1 ) (zt 1 ) lim

✏ !0

lim u0 ct

✏ !0

1

But ✏0 > 0 contradicts the optimality of (c, k, z). Hence for all t 2 N, ct > 0. Lemma 6. Let (c, k, z) be an optimal solution. Then for all t 2 N, kt > 0. Proof. Assume that there exists t 2 N⇤ such that kt = 0. Then ct = 0 but this contradicts the previous lemma. The proof of Proposition 2 is now complete.

3.4

First-order conditions

Proposition 3. Assume that the assumptions of Proposition 2 hold. Along an optimal path, we have 8t 0, f20 (kt , 1

1 0 u (ct ) + f10 (kt+1 , 1 (zt )

0

(zt ) kt+1 = 0, (zt )2

(18)

zt+1 )u0 (ct+1 ) = 0,

(19)

1 kt+1 . (zt )

(20)

zt ) +

and ct = f (kt , 1

zt )

Proof. From Proposition 2, optimal paths are interior. The conditions (18) and (19) then follow directly. As the problem is non-convex, these conditions are not sufficient for optimality. 15

3.5

On the Value Function

The value function V : R+ ! R [ { 1} associated to problem P is defined as follows: V (k) =

sup (c,k,z)2⇧(k0 )

U (c, k, z)

Under the assumptions of Proposition 1, we can substitute max for sup in the above definition. Following Propositions 5.3.1 and 5.3.2. in Le Van and Dana (2003) we can prove the following Propositions 4 and 5. Proposition 4. Assume that (H1b)

(H2), (H3), (H5)

(H6) hold. Then:

1. The value function V (.) is upper semicontinuous. 2. V (0) =

1.

3. There exist two non-negative real numbers A1 , B1 such that 8k0 A1 k0 + B1 . 4. 8(c, k, z) in ⇧0 (k0 ), lim

t!+1

t

0, V (k0 ) 

V (kt ) = 0.

5. V (.) is increasing. Proposition 5. The Value function satisfies the Bellman equation: 8k0 , V (k0 ) = max{u(c) + V (k1 ) : 0  c  f (k0 , 1

z), k1 = (z)(f (k0 , 1

z)

c)} (21)

Moreover, the value function is the unique solution in S to the Bellman equation, where S is the set of functions h : R+ ! R [ { 1}, which are upper semi-continuous, and such that: 1. for all k0 > 0, for all (c, k, z) 2 ⇧(k0 ), lim supt!+1 2. 8k0 , for all (c, k, z) 2 ⇧0 (k0 ), lim

t

t!+1

3.6

t

h(kt ) = 0.

A Monotonicity Property

We noticed that along an optimal path, we have for all t: f20 (kt , 1

zt ) +

This is the first-order condition (18). So we have 0 < kt+1 <

f20 (kt ,0) (1)2 . 0 (1)

16

0

(zt ) kt+1 = 0. (zt )2

h(kt )  0.

Lemma 7. Assume lim f20 (k, y) = +1. Let k, k 0 2 R⇤+ satisfying 0 < k 0 < y!0

f20 (k,0) (1)2 0 (1)

be given. Then the equation:

f20 (k, 1

z) +

0

(z) 0 k = 0, (z)2

(22)

has a unique solution z = z(k, k 0 ). Proof. Let k, k 0 2 R⇤+ satisfying 0 < k 0 < G(z) defined on [0, 1] G(z) =

f20 (k,0) (1)2 0 (1)

f20 (k, 1

z) +

be given. Consider the function 0

(z) 0 k. (z)2

G is continuously di↵erentiable and strictly decreasing on ]0, 1[ since G0 (z) = f ”22 (k, 1 2 2 (z)( 0 (z))2 k 0 < 0. Moreover lim G(z) = +1 , lim G(z) = 1 (by the z) + ”(z)( (z))( (z)) 4 z!0

assumption on f20 ). The result follows.

z!1

The next Lemma is instrumental to prove the monotonicity property of some optimal paths under some condition. To prove the Lemma we will use the following assumption: (H8) For all k, k 0 2 R⇤+ satisfying 0 < k 0 < and c = f (k, 1 u00 (c)

z)

k0

(z)

f20 (k,0) (1)2 , 0 (1)

z = z(k, k 0 ) given by Lemma 7

we have

f10 (k, 1 z) + u0 (c) (z)

00 f12 (k, 1 z) 0 (z) (z) 3 (z)f 00 (k, 1 z) + k 0 [ 00 (z) (z) 2( 0 (z))2 ] 22

Lemma 8. Assume lim f20 (k, y) = +1. Set U (kt , kt+1 ) = u(f (kt , 1 y!0

kt+1 ). (z(kt ,kt+1 ))

0.

z(kt , kt+1 ))

Then U is supermodular if and only if (H8) holds.

Proof. We have: ✓ ◆ @U (kt , kt+1 ) kt+1 0 = u f (kt , 1 z(kt , kt+1 )) @kt (z(kt , kt+1 )) ✓ @z(kt , kt+1 ) ⇥ f10 (kt , 1 z(kt , kt+1 )) f20 (kt , 1 z(kt , kt+1 )) @kt ◆ 0 kt+1 (z(kt , kt+1 )) @z(kt , kt+1 ) + 2 . (z(kt , kt+1 )) @kt Using (22), this equation reduces to: ✓ @U (kt , kt+1 ) 0 = u f (kt , 1 z(kt , kt+1 )) @kt 17

kt+1 (z(kt , kt+1 ))

◆

f10 (kt , 1

z(kt , kt+1 )).

Using (22) again we have: ✓ ◆ @ 2 U (kt , kt+1 ) kt+1 = u00 f (kt , 1 z(kt , kt+1 )) @kt @kt+1 (z(kt , kt+1 )) " # @z(kt ,kt+1 ) 0 (z(k , k )) k (z(k , k )) t t+1 t+1 t t+1 @z(k , k ) @kt+1 t t+1 ⇥ f20 (kt , 1 z(kt , kt+1 )) 2 (z(k , k @kt+1 t t+1 )) ⇥ f10 (kt , 1 z(kt , kt+1 )) ✓ ◆ kt+1 @z(kt , kt+1 ) u0 f (kt , 1 z(kt , kt+1 )) f 00 (kt , 1 z(kt , kt+1 )) , (z(kt , kt+1 )) 12 @kt+1 ✓ ◆ 0 kt+1 f1 (kt , 1 z(kt , kt+1 )) 00 = u f (kt , 1 z(kt , kt+1 )) (z(kt , kt+1 )) (z(kt , kt+1 )) ✓ ◆ k @z(kt , kt+1 ) t+1 00 u0 f (kt , 1 z(kt , kt+1 )) f12 (kt , 1 z(kt , kt+1 )) . (z(kt , kt+1 )) @kt+1 Using (22) and di↵erentiating with respect to kt+1 we obtain: @z(kt , kt+1 ) = @kt+1 0

3 (z(k

00 t , kt+1 ))f22 (kt , 1

We readily see that

(z(kt , kt+1 )) (z(kt , kt+1 )) z(kt , kt+1 )) + kt+1 [ 00 (z(kt , kt+1 )) (z(kt , kt+1 ))

@ 2 U (kt ,kt+1 ) @kt @kt+1

2( 0 (z(kt , kt+1 )))2 ]

0 which is equivalent to U supermodular.

Remarks. • A real-valued function U (., .) defined in R2+ has increasing di↵erences in its two arguments k and k 0 if for all k1 k2 , the di↵erence U (k1 , k 0 ) U (k2 , k 0 ) is non-decreasing in k 0 . The function U is supermodular if and only if it has increasing di↵erences. Further, when U is twice continuously di↵erentiable, we know that U has increasing 2 U (k ,k t t+1 ) di↵erences if and only if @ @k 0 (Becker and Boyd, 1997, p 265). t @kt+1 • Notice that the assumption (H8) is satisfied in many cases, for example when u(c) = ln(c), f (k, 1 z) = k ↵ (1 z)1 ↵ , (0 < ↵ < 1) and any function satisfying the assumptions of the problem. Other examples are when u(c) = ln(c) and f and are any functions satisfying the assumptions of the problem and the additional assumptions: f20 (x, 1) (z(k,k0 )) is non-increasing with respect to x and 1 0 (z(k,k 0 ))(1 z(k,k 0 )) > 0 xf10 (x, 1) or f20 (x, 1) (z(k,k0 )) is non-decreasing with respect to x and 1 0 (z(k,k 0 ))(1 z(k,k 0 )) < 0. 0 xf1 (x, 1)

18

This can be seen below since in this case we have: @U (kt , kt+1 ) f10 (kt , 1 z(kt , kt+1 )) = kt+1 @kt f (kt , 1 z(kt , kt+1 )) (z(kt ,kt+1 )) = =

f10 (kt , 1 f10 (kt , 1

✓

kt 1 +

kt+1 z(kt , kt+1 ))(1 z(kt , kt+1 )) (z(kt ,kt+1 )) 1 ✓ ◆◆ . 0 f2 (kt /(1 z(kt , kt+1 )), 1) (z(kt , kt+1 )) 1 0 (z(k , k z(kt , kt+1 ))f10 (kt /(1 z(kt , kt+1 )), 1) z(kt , kt+1 )) t t+1 ))(1

z(kt , kt+1 ))kt +

kt /(1

z(kt , kt+1 ))

f20 (kt , 1

Under the assumptions and noticing that z(kt , kt+1 ) is increasing with respect to kt+1 , t ,kt+1 ) we easily see that @U (k@k is increasing with respect to kt+1 . t Proposition 6. Assume (H8) holds and lim f20 (k, y) = +1. Then there is a monoy!0

tonic optimal path. Proof. The argument draws heavily on Becker and Boyd (1997) (p 265-266). Let A(k) be the set of (non-negative) real numbers k 0 such that: k0 (z(k, k 0 ))

z(k, k 0 ))

f (k, 1

0.

(23)

Now let k1 and k2 be two positive real numbers such that k1 > k2 . Then we have: A(k2 ) ✓ A(k1 ). Indeed let x be in A(k2 ) and such that: f (k2 , 1

z(k2 , x))

x (z(k2 , x))

0.

(24)

f (k1 , 1

z(k2 , x))

x > 0, (z(k2 , x))

(25)

z(k1 , x))

x > 0, (z(k1 , x))

(26)

We then have:

and then by definition of z(k1 , x): f (k1 , 1 which proves that x 2 A(k1 ). Recall that: 0

✓

U (k, k ) = u f (k, 1

0

z(k, k ))

19

k0 (z(k, k 0 ))

◆

.

(27)

For k > 0 let M (k) = {k 0 2 A(k) | k 0 2 arg max U (k, k 0 )+ V (k 0 )}. Since u(.), f (.) and z(., .) are continuous and V (.) is upper-semi continuous, M (k) is a non-empty compact set. Let m(k) be the maximal optimizer of M (k). We want to prove that for all positive k1 and k2 , k1 > k2 ) m(k1 ) m(k2 ). To do this let x be a real number such that x  m(k2 ). Then we have by definition of m(k2 ) U (k2 , x) + V (x)  U (k2 , m(k2 )) + V (m(k2 )),

(28)

or U (k2 , m(k2 )) + V (m(k2 ))

(U (k2 , x) + V (x))

We know that the function U (., .) is such that di↵erences. That is: U (k2 , m(k2 ))

@2U

@k@k0

0.

(29)

0. Then U (., .) has increasing

U (k2 , x)  U (k1 , m(k2 ))

U (k1 , x).

(30)

Using this last inequation in (29), we obtain: U (k1 , m(k2 )) + V (m(k2 ))

U (k1 , x) + V (x),

(31)

for all x  m(k2 ). Now, if m(k1 ) < m(k2 ) then the above inequation gives: U (k1 , m(k2 )) + V (m(k2 ))

U (k1 , m(k1 )) + V (m(k1 )),

but this contradicts the definition of m(k1 ). Hence m(k1 ) m(k2 ). Now consider the optimal path defined by k0 and kt+1 = m(kt ). Suppose that k0 < k1 = m(k0 ). Since m is increasing, it follows by induction that this optimal path is non-decreasing. The same argument shows that the optimal path is non-increasing whenever k1 = m(k0 ) < k0 .

4

Existence Uniqueness and Stability of the Steady State

We first address the issue of the existence and uniqueness of the steady-state. Then we consider its stability.

4.1

Existence of the steady state

Recall that, under assumptions (i) (H1b)-(H2) and (H7) or (ii) (H1a)-(H2)-(H3)(H4)-(H5)-(H6), any solution being interior7 , by Proposition 3 an optimal path +1 (c, k, z) = (ct )t , (kt )t , (zt )t t=0 satisfies conditions (18), (19) and (20), for all t. 7

Note that any solution being interior, the problem (P) is then also equivalent to 8 P1 t 1 u f (kt , 1 zt ) > t=0 (zt ) kt+1 > max (k,z) > < s.t. zt 2]0, 1[, > 0 < kt+1 < (zt )f (kt , 1 zt ), > > : 1 ct = f (kt , 1 zt ) (zt ) kt+1 .

20

Definition 1. A steady state is a triple (c, k, z) such that: (z)f10 (k, 1

z) = 1,

(32a)

0

(z) k, (z)2 c = f (k, 1 z)

f20 (k, 1

z) =

(32b) k/ (z).

(32c)

We now study the existence of a steady state in the general case. To do this let us define = 1 k z . Then, equations (32a) and (32b) can be expressed as follows: (z)f10 ( , 1) = 1, 0 f20 ( , 1) (z)(1 z) = . 0 f1 ( , 1) (z) Let

be the value of

(33) (34)

such that: f10 ( , 1) = 1.

(35)

This is the so-called modified golden rule capital. 1 Since is increasing define z1 ( ) = 1 [ 0 ] and so the function z1 ( ) is increasf1 ( , 1) ing (recall that f10 ( , 1) is decreasing with respect to ). Let ⌦(z) =

0 (z)(1

z)

(z)

. ⌦(z) is decreasing (since (z) is concave). We can then define

z2 ( ) such that z2 ( ) = ⌦ 1 [

f20 ( ,1) ]. f10 ( ,1)

Proposition 7. The equation only if there exists a steady-state.

1

[

0 1 1 f2 ( , 1) ] = ⌦ [ ] has a solution f10 ( , 1) f10 ( , 1)

8

if and

f20 ( , 1) ], , 1) f10 ( , 1) k = (1 z), c = f (k, 1 z) k/ (z). So (33) and (34) are satisfied and z in [0, 1], k 0. Let us verify that c = f (k, 1 z) k/ (z) 0. Since f is linear-homogeneous by Euler’s Theorem we have: Proof. Let

be the solution of the equation. Set z =

f (k, 1

z)

k = f10 (k, 1 (z) = (f10 (k, 1

z)k + f20 (k, 1 z)

1

[

1

] = ⌦ 1[

f10 (

z)(1

z)

1 )k + f20 (k, 1 (z)

k , (z) z)(1

z).

z)

0.

Now using equation 1 we have: f (k, 1

8

z)

k =( (z)

1 (z)

1 )k + f20 (k, 1 (z)

Note that this solution is then in ]0, ].

21

z)(1

4.2

Uniqueness of the steady state

We will use the next assumption: f 0 ( , 1) (H9) 2 0 is strictly increasing with respect to . f1 ( , 1) Proposition 8. If assumption (H9) holds, then there exists a unique steady-state. Proof. Under the assumption (H9), the function z2 ( ) is strictly decreasing, and we have 0 < lim #0 z2 ( )  1. We also observe that z2 ( ) < z1 ( ) = 1. We also have: lim #0 z1 ( ) = 0. Since the functions zi (.) (i = 1, 2) are continuous on [0, ] and z1 is strictly increasing from 0 to 1 and z2 is strictly decreasing from a positive real number to 1 f 0 ( , 1) z2 ( ) < 1, there exists a unique 2 [0, ] such that 1 [ 0 ] = ⌦ 1[ 2 0 ], f1 ( , 1) f1 ( , 1) so by the previous proposition a steady state exists. It is also the unique possible steady state.

Remark. It is interesting to study the relevance of Assumption (H9). Let us consider a C.E.S. function:9 f (k, 1

z) = [↵k ⇢ + (1

↵)(1

1

z)⇢ ] ⇢ ,

(36)

with ⇢ < 1. We can check that: f20 ( , 1) 1 ↵ = . 0 f1 ( , 1) ↵ ⇢

(37)

We see that the left-hand side of the above equation is a strictly increasing function of i↵ ⇢ < 0. It is well known that the elasticity of substitution of the CES is equal to 1/(1 ⇢). Therefore, in the C.E.S case, Assumption (H9) is satisfied if and only if the elasticity of substitution is less than 1. This means that when the capital-labor ratio increases by 1% the ratio of the marginal productivity of labor to the marginal productivity of capital increases by more than one percent. Thus increasing the capital-labor ratio makes labor relatively more productive at the margin. The above condition on the elasticity of substitution is in fact a general result. One 9 We notice that this function does not satisfy all the assumptions of (H2). For instance, f (0, y) and f (z, 0) are non-nil if y or z are non-nil.

22

way to see this in the general case is as follows. We have: d f 0 ( , 1) f0( ( 01 ) = 10 d f2 ( , 1) f2 ( f0( = 10 f2 ( =

, 1) d f10 ( , 1) + , , 1) d f20 ( , 1) , 1) d f10 ( , 1) [1 + f 0 ( ,1) ], 1 , 1) d f20 ( , 1) 0

(38a) (38b)

f2 ( ,1)

f10 ( , 1) [1 f20 ( , 1)

1

],

(38c)

where is by definition the elasticity of substitution. We see that Assumption (H9) is equivalent to < 1. What is the empirical validity of this inequality? Figure 3, page 9, in Raval, 2015, depicts estimates of the elasticity of substitution found in the literature.10 This author finds that “[t]he median estimate in the literature is 0.54, only a few estimates are above one.” A notable exception is Piketty, 2013’s estimates which are higher than one (the di↵erence, however, is related to a di↵erent way to measure capital).

4.3

Stability of the steady state

There are two cases. Firstly, assume that there is a unique steady-state. Then, when there is at least one monotonic optimal path, this optimal path converges to this steady-state. This result provides an underpinning of the comparative statics analysis presented in the next section. Secondly, assume that there are multiple steady-states. Then, assuming again that there is a monotonic optimal path, a comparative statics analysis is meaningful when the steady-state considered is locally stable. In appendix we provide a general study of the local stability of steady-states. We also propose a study of the case where the utility function is CRRA and the production function is Cobb-Douglas.

5

Comparative Statics

We now assume that the function (.) depends on a parameter a which measures the efficiency of the investment enhancing labor. Specifically, we suppose that there is a function : [0, 1] ⇥ R+ ! [0, 1]. We assume that the function (., a) has the same properties as the function (z) when the second argument a is constant. Furthermore, we assume that (z, a) is increasing with respect to a when z is positive (and nondecreasing if z is nil). We also assume that (., .) is twice continuously di↵erentiable). Finally, we assume that lima!+1 (z, a)  1. In plain English, this means that when a increases, the fraction of the gross savings which is actually invested increases. Thus a can be thought of as a measure of the efficiency of the investment enhancing labor. These estimates vary across time, countries, level of aggregation, and also depend on whether technical progress is taken into account, as well as on the econometric techniques. 10

23

We have given two instances of an increase in this efficiency. This increase can first be caused by the move towards financial deregulation (in so far as financial regulation limits the range of possible banking and financial activities). Second, the increase in the efficiency of investment labor can be due to the surge in information and communication technologies which considerably ease the gathering and processing of information. Should these trends result in a lower or a greater share of labor devoted to transform savings into productive capital? Recall that the steady-state of the economy is described by the following equations:11 (z, a)f10 ( f20 ( f10 ( where

=

k = (z, a) f (k, 1 z) , 1) = 1, 0 , 1) z) 1 (z, a)(1 = , , 1) (z, a)

c ,

(39a) (39b) (39c)

k . 1 z

It is interesting to see how the steady-state consumption c changes with a. Using the first of the above equation, we can check that: dc = (1 da So when

dk da

> 0,

dc da

)f10 (k, 1

z)

dk k(a) + da (z, a)2 dc da

> 0. Otherwise, the sign of

0 2 (z, a).

(40)

is indeterminate.

Di↵erentiating equations (39b) and (39c) with respect to a yields:

h

0 0 1 f1

0 1

Solving for ✓ dz ◆ da d da

(1 z)

dz da

00 11

and

+

d da

0 1 @ = det(M )

(

0 )2 (1 1 2

z)

i

00

f11 f20 f10

@ @

!✓

dz da d da

◆

=

(1

z)

⇥

0 0 2 f1 00 12

0 0 1 2 2

!

⇤ .

(41)

we obtain: h

0 1

@ @ (1 z)

f20 f10

00 11

+

00

(

0 )2 (1 1 2

z)

f11

i

0 0 1 f1

1 A

(1

z)

⇥

0 0 2 f1 00 12

0 0 1 2 2

!

⇤ ,

(42)

where M is defined as M= 11

h

0 0 1 f1

0 1

(1 z)

00 11

+

(

0 )2 (1 1 2

z)

i

00

f11 @ @

f20 f10

!

,

In this section, we assume that a steady state exists, at least for some a.

24

(43)

f20

@ @

>0

f10 f20 f10 f20 f10 f20 f10 f20 f10 f20 f10 f20 f10 f20 f10 f20 f10

@ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @

>0 >0 =0 =0 =0 <0 <0 <0

⇥ ⇥ ⇥ ⇥ ⇥ ⇥ ⇥ ⇥ ⇥

00 12

0 0 1 2 2

00 12

0 0 1 2 2

00 12

0 0 1 2 2

00 12

0 0 1 2 2

00 12

0 0 1 2 2

00 12

0 0 1 2 2

00 12

0 0 1 2 2

00 12

0 0 1 2 2

00 12

0 0 1 2 2

⇤ ⇤ ⇤ ⇤ ⇤ ⇤ ⇤ ⇤ ⇤

det(M )

dz da

d da

>0

+

?

+

=0

+

-

+

<0

+

-

?

>0

+

+

+

=0

+

=0

+

<0

+

-

?

>0

?

?

?

=0

?

?

?

<0

?

?

?

Table 1: Comparative statics: e↵ect of an increase in a and det(M) =

0 0 1 f1

f20 f10

@ @

More specifically, ⇢ dz 1 0 0 @ = 2 f1 da det(M) @ ⇢  0 d 1 0 0 = 2 f1 da det(M)

00

f11

f20 f10 (1 z)



0 1

00

f11 (1 00

11

+

(1

z)

⇥

( 01 )2 (1 2

z)

00

12

z)

00

11

+

0 0 ⇤ 1 2 2

+ (1

( 01 )2 (1

z)

2

, z) 01 f10

.

(44)

(45) ⇥

00

12

0 0 ⇤ 1 2 2

.

(46)

Does an increase in the efficiency of investment enhancing labor (i.e., a rise in a) leads to a decrease in the size of the productive labor (i.e., a decrease in z)? To answer dz by that question we rely on the table 1. This table indicates notably the sign of da 00

f0

0

0

1 2 12 relying on the signs of @@ f20 and 12 In particular, table 1 gives a sufficient 2 . 1 dz condition for the steady state value of z to decrease with a. We indeed have da <0

whenever

@ @

f20 f10

> 0 and

00 12

0 0 1 2 2

= 0. To understand these conditions let us look at 00

0

0

1 2 equations (39a)-(39c). The condition 12 = 0 means that the right-hand side of 2 equation (39c) does not depend on a. To understand this right-hand side, recall that to actually invest an amount k of capital, one needs to save k/ (z). A marginal increase in investment enhancing labor z leads to a decrease in the gross amount of capital needed to actually invest k. This decrease is equal to 0 (z)k/ (z)2 . Along an optimal parth,

12

The signs of det(M ),

dz da

and

d da

are obtained from the equations (44)-(46).

25

however, this marginal savings in resources is equal to the marginal product of labor (using an additional unit of z brings about a savings in resource which is compensated 00

0

0

1 2 by a decrease in the production of good). Under the condition 12 = 0 a rise in a 2 does not a↵ect the marginal savings in resources that results from increasing z by one unit. Therefore, a marginal increase in the efficiency of investment enhancing labor only increases the net marginal product of capital (see equation (39b)). In a steadystate, this increase must me matched by a decrease in z (as can be seen from the same equation). This is conducive to a decrease in the marginal productivity of labor. But this decrease is compensated by an increase in the steady-state value of capital. There is then a kind of virtuous case. An increase in the efficiency of investment enhancing labor allows to decrease the amount of this labor while increasing actual savings. In turn, this increase in savings leads to an increase in capital and consequently to a rise in the production level (both inputs being increased). As a result, consumption increases too.

Table 1 also shows that an increase in the efficiency of investment enhancing labor can lead to an increase in the size of this labor.13 This happens for instance when ⇥ 00 0 0 ⇤ f20 @ 1 2 = 0, and 12 > 0. We have seen that the optimal value of z must be 2 @ f10 such that a marginal increase in z would lead to a marginal decrease in the amount of gross savings which is compensated by the marginal decrease in the production of good. Under the above condition a rise in a further decreases the amount of gross savings. Raising z is then welfare improving. This rise in z, in turn, causes an increase in the marginal product of capital. Staying at steady-state then requires a capital increase. The net e↵ect on the production of good is indeterminate (because the inputs do not change in the same way). Except in the two above cases, the comparative statics analysis yields indeterminate results. Therefore we cannot conclude in general that a rise in the efficiency of investment enhancing labor leads to an increase in this kind of labor.

6

An Example

In this section we present an example for which we are to compute an optimal path. Assume that: u(c) = ln c, f (k, 1 z) = k ↵ (1

z)

✓

1 ↵

, 0 < ↵ < 1,

(z) = z , 0 < ✓ < 1.

(47a) (47b) (47c)

This would correspond to what has been observed in the banking and financial sectors of developed countries, see e.g. Philippon and Reshef (2013). 13

26

The first-order conditions (18), (19) and (20) give: 1

↵ 1+✓ kt ↵ zt ( ) , ✓ 1 zt 1 zt+1 (1 + 1 ✓ ↵ ) ↵ ✓ 1 zt+1 = , 1 ↵ 1 ↵ zt 1 zt (1 + ✓ ) kt ↵ 1 ↵ ct = ( ) (1 zt (1 + )). 1 zt ✓ kt+1 =

(48a) (48b) (48c)

We can check that the following expressions satisfy equations (48a)-(48c): 1 1 + 1↵

zt =

↵ ✓

,

(49a)

z(1 + 1 ✓ ↵ ) , (1 z)↵ 1 ↵ z 1+✓ = ⇢kt↵ , where ⇢ = . ✓ (1 z)↵

ct = kt↵ , where kt+1

=

1

(49b) (49c)

From the last equation, we see that: ln kt = (1

↵t ) ln k1 + ↵t ln k0 , where k1 = ⇢ 1

1 ↵

.

(50)

With the above expressions of ct , kt and z we can compute the corresponding value of U (c, z, k) and we obtain: ✓ ◆ ln k0 ln (1 ↵) Vˆ (k0 ) = ↵ + +↵ ln k1 . (51) 1 ↵ 1 (1 )(1 ↵ ) Our aim is to show that Vˆ (k0 ) is the value function. We first check that Vˆ (k0 ) satisfies the Bellman equation. Proposition 9. The function Vˆ (.) satisfies the Bellman equation. Proof. Given the specifications given to the functions u, f , and , the Bellman equation is as follows: n o v(k) = max ln(k ↵ (1 z)1 ↵ k 0 z ✓ ) + v(k 0 ) . (52) (k0 ,z)2]0,k↵ (1 z)1

↵ z ✓ [⇥]0,1[

Let k be fixed and let us then study the following problem: max G(k 0 , z)

(k0 ,z)2⌦

27

(53)

where 0

⇢

z)1 ↵ k 0 z ✓ ✓ ◆ ⇣ ln (1 ↵) + +↵ ln k1 1 (1 )(1 ↵ ) ⌘ ↵ + ln k 0 , 1 ↵ ⌦ = {(k 0 , z) s.t. z 2]0, 1[ and 0 < k 0 < k ↵ (1 z)1

G(k , z) ⌘

ln k ↵ (1

(54) ↵ ✓

z }.

(55)

We need the following lemma to complete the proof. Lemma 9. The solution to the problem (53) is: ↵) 1+✓ k ↵ z ( ) , ✓ 1 z 1 z= = z1 . 1 + 1↵ ↵✓

k0 =

(1

(56a) (56b)

Proof. Notice that ⌦ is open and bounded and that G is continuous on ⌦. Moreover lim G(k 0 , z) = 1. Hence G attains its maximum on ⌦. To see this, we 0 0 (k ,z)!(k0 ,z0 )2@⌦

first remark that G is upper bounded on ⌦. Indeed: ✓ ◆ ⇣ ln ⌘ (1 ↵) ↵ G(k 0 , z)  sup {ln(k ↵ k 0 ) + +↵ ln k1 + ln k 0 }, 0 ↵ 1 (1 )(1 ↵ ) 1 ↵ k 1, there is a pair (kn0 , zn ) 2 ⌦ such that: V

1  G(kn0 , zn ) n

(58)

Since ⌦ is bounded the sequence (kn0 , zn )n has a convergent subsequence (kn0 s , zns )ns . 0 0 0 Let (k , z) be its limit. We necessarily have (k , z) 2 ⌦. If not (k , z) 2 @⌦ will imply lim 0 G(k 0 , z) = 1 which contradicts V n1s  G(kn0 s , zns ). 0 ,z (kn )!(k ,z)2@⌦ s ns

0

Since G(., .) is continuous on ⌦, lims!+1 G(kn0 s , zns ) = G(k , z) = V. Hence G attains 0 its maximum at (k , z). 0

From the preceding discussion, we can see that (k , z) is necessarily an interior point. Therefore it satisfies the first-order conditions: z ✓ ↵ + = 0, ↵ 1 ↵ 0 ✓ k (1 z) kz (1 ↵ )k 0 (1 ↵)k ↵ (1 z) ↵ + ✓k 0 z (1+✓) = 0. 28

(59a) (59b)

Solving the two above equations, we find that: 1 1 + 1↵

z=

(1

k0 =

↵) ✓

= z1 ,

↵ ✓

z 1+✓ (

(60a) k

1

z

)↵ .

(60b)

Since there is a unique solution to the first-order condition, the above expressions are a solution to our problem. Inserting (56a) and (56b) in equation (54), we find after some algebra that Vˆ (k) solves the Bellman equation (52). Now let us show that the function Vˆ is the value function. This follows directly from the following Proposition which shows that Vˆ belongs to the set S (see Proposition 5). Proposition 10. The function Vˆ defined above is in S. Proof. 1. We first prove that: for all k0 > 0, for all (c, z, k) 2 ⇧(k0 ), lim sup will be convenient to define: ✓ ◆ ln (1 ↵) = +↵ ln k1 . 1 (1 )(1 ↵ ) t

Since for all t, kt  tˆ

V (kt ) 

t

=

t

✓

✓

k0 +

0

1

tˆ

V (kt )  0. It (61)

and Vˆ is increasing we have:

✓ ◆◆ 0 ↵ t + ln k0 + , 1 ↵ 1 ✓ ✓ 0 ◆◆ ✓ ↵ + ⇥ ln + ln 1 + 1 ↵ 1

(62) t

(1

)k0 0

◆◆

.

(63)

We then see that: lim sup t!1

tˆ

V (kt )  0.

(64) tˆ

2. We now prove that 8k0 , for all (c, z, k) 2 ⇧(k0 ), limt

V (kt ) = 0.

We have: U (c, z, k) = =

1 X t=0 T X

t

ln ct ,

t

ln ct +

t=0

(65) T +1

1 X s=0

29

s

ln cT +1+s .

(66)

Since ct  kt↵ and kt+1  kt↵  k0↵ U (c, z, k) 

T X

t=0 T X

t

t+1

, we have: 1 X

T +1

ln ct +

(↵ )s ↵ ln kT +1 ,

(67a)

s=0

↵ T +1 ln kT +1 , 1 ↵ t=0  T X ↵ t T +1  ln ct + ln kT +1 + 1 ↵ t=0 

=

T X

t

t

ln ct +

⇣

T +1

ln ct +

t=0

U (c, z, k)  lim

T !1

T X

t

ln ct + lim inf T !1

t=0

T +1 ˆ

= U (c, z, k) + lim inf

⇣

Vˆ (kT +1 )

V (kT +1 ).

T !1

(67c)

⌘

Vˆ (kT +1 )

T +1

(67b)

(67d)

⌘

,

(68) (69)

from which it follows that: T +1 ˆ

0  lim inf

V (kT +1 ).

T !1

(70)

Combining this result with the result obtained in part 1 of the proof, we have: tˆ

lim

V (kt ) = 0,

t!1

(71)

as was to be shown. According to the above result, the values given by equations (49a), (49b) and (49c) yield the optimal solution. We observe that the sequences converge (globally) to a steady state. This steady state is as follows: z= c= k=

1 1 + 1↵

,

(72a)

z(1 + 1 ✓ ↵ ) ↵ k , (1 z)↵

1 

↵ ✓

1

↵ ✓

z 1+✓ (1 z)↵

(72b)

1 1 ↵

.

(72c)

We notice that the function f (., .) used in the example does not satisfy assumption (H9). The later is thus only sufficient, but not necessary to get a unique steady-state. To apply the comparative statics results of section 5, we can assume that (z, a) = ⇥ 00 0 0 ⇤ f0 1 2 ⌘(a)(z) where (z) = z ✓ . Then we have both @@ f20 = 0 and 12 = 0. We 2 1

30

0

f0

dz 2 1 can then check that da = 0, and that dda = . In this case, as k = (1 z) , we f100 dk have da > 0. The capital stock increases with a. From equation (40) we see that a rise in the efficiency of the investment enhancing labor leads to a rise in the steady state value of consumption.

The example studied in this section is close to the optimal growth model with embodied technical progress considered in section 4 of Krusell (1998). When there is only one capital service, the Krusell’s model comprises the following equations: zt ) = kt↵ (1 zt )1 ↵ = ct + it , kt+1 = (1 )kt + Tt+1 it , Tt+1 = Tt H(nt ),

f (kt , 1

(73) (74) (75)

where Tt denotes embodied technical progress. Our model is a variant of the above specification, with: Tt+1 = H(nt ), for all t, with H(nt ) 2 [0, 1] and

(76)

= 1.14

Krusell writes only the necessary conditions for an optimal solution.15 By contrast, we give a general study of the optimal growth problem. We also study the steady state (and perform a comparative statics analysis of what amount of a change in H(.)). In addition to a study of the monotonicity property of optimal path, we give a complete study of an example (we not only write the necessary conditions, but also prove that they are sufficient). Our model is also close to the paper by Boucekkine, Martinez, and C. Saglam (2006).16 Their model can be described as follows: f (kt , 1

zt ) = At kt↵ (1 zt )1 ↵ = ct + it , kt+1 = qt it + (1 )kt ,

(77) (78)

+ dt zt✓ (qto

(79)

qt = q t

1

qt ),

where At corresponds to the disembodied technological progress, qt to the embodied technological progress and qto to the foreign embodied technical progress. Our model is a variant of the above specification, with: qt = dt u✓t , dt = 1, At = 1. Boucekkine, Martinez, and C. Saglam (2006) present simulation results, assuming that u(ct ) = log ct . As was mentioned above, we have some results in the general case and we propose a complete study of an example. Boucekkine, Martinez, and C. Saglam (2006) As was seen before, we can handle capital depreciation in another way. His paper concentrates on the study of intertemporal equilibria. 16 The two previous models generalize the model studied in Greenwood, Hercowitz, and Hu↵man, 1988, in the sense that they endogenize embodied technical progress. On the other hand, Greenwood, Hercowitz, and Hu↵man (1988) endogenize the depreciation rate of capital. The present paper does not tackle this issue. These authors endogenize the embodied technical progress but labor is left aside of the mechanism considered. 14 15

31

also make an interesting study of the e↵ect of a change in d on the endogenous variables, and notably u (which would correspond to z in our model). They find that u decreases with d. We have shown that this is not generally the case (for instance in the example of this section, a change in a has no e↵ect).

7

Conclusion

In this paper we have considered an optimal growth problem with investment enhancing labor. Taking into account this kind of labor leads to a non trivial non convex problem. We have proved that this problem has an interior solution both in the cases where the utility function is bounded from below and the case where it is not. We have also provided some conditions under which there exists a unique steady-state and conditions under which there is a monotonic optimal path. Relying on a result obtained in the study of the value function, we have characterized the solution of our problem in the special case of a logarithmic utility function and a Cobb-Douglas production function. The comparative statics of the steady state shows that an increase in the efficiency of the investment enhancing labor is not always conducive to an increase in the steady state value of this labor. A natural extension of this work consists of studying the existence and optimality properties of intertemporal (or recursive) equilibria, as well as the policies which can decentralize the solutions of our optimal growth problem. Another issue that would be worth studying is the case where labor supply is endogenous.

References Akao, Ken-Ichi, Takashi Kamihigashi, and Kazuo Nishimura (2011). “Monotonicity and continuity of the critical capital stock in the Dechert–Nishimura model”. Journal of Mathematical Economics 47.6, pp. 677–682. Aliprantis, Charalambos D and Kim Border (2006). Infinite dimensional analysis: a hitchhiker’s guide. Springer Science & Business Media. Arcand, Jean-Louis, Enrico Berkes, and Ugo Panizza (2012). “Too much finance?” Aubin, Jean-Pierre (1996). Initiation `a l’Analyse Appliqu´ee. Azariadis, Costas and John Stachurski (2005). “Poverty traps”. Handbook of economic growth 1, pp. 295–384. Becker, Robert A and John Harvey Boyd (1997). Capital theory, equilibrium analysis, and recursive utility. Blackwell. Boucekkine, Raouf, Blanca Martinez, and Cagri Saglam (2006). “The development problem under embodiment”. Review of development economics 10.1, pp. 42–58. Eggoh, Jude C and Patrick Villieu (2013). “Un r´eexamen de la non-lin´earit´e entre le d´eveloppement financier et la croissance ´economique”. Revue d’´economie politique 123.2, pp. 211–236. Findlay, Ronald and John D Wilson (1987). The political economy of Leviathan. Springer. 32

Goldin, Claudia and Lawrence F Katz (2008). “Transitions: Career and family life cycles of the educational elite”. The American Economic Review 98.2, pp. 363–369. Greenwood, Jeremy, Zvi Hercowitz, and Gregory W Hu↵man (1988). “Investment, capacity utilization, and the real business cycle”. The American Economic Review, pp. 402–417. Hung, Nguyen Manh, Cuong Le Van, and Philippe Michel (2009). “Non-convex aggregate technology and optimal economic growth”. Economic Theory 40.3, pp. 457– 471. Kamihigashi, Takashi and Santanu Roy (2007). “A nonsmooth, nonconvex model of optimal growth”. Journal of Economic Theory 132.1, pp. 435–460. Krusell, Per (1998). “Investment-specific R and D and the decline in the relative price of capital”. Journal of Economic Growth 3.2, pp. 131–141. Le Van, Cuong and Rose-Anne Dana (2003). Dynamic programming in economics. Vol. 5. Springer Science & Business Media. Le Van, Cuong, Lisa Morhaim, and Charles-Henri Dimaria (2002). “The discrete time version of the Romer model”. Economic theory 20.1, pp. 133–158. Le Van, Cuong and H. Cagri Saglam (2004). “Quality Of Knowledge Technology, Returns To Production Technology, And Economic Development”. Macroeconomic Dynamics 8 (02), pp. 147–161. Majumdar, Mukul (2006). “Intertemporal allocation with a non-convex technology”. In: Handbook on Optimal Growth 1. Springer, pp. 171–201. Pagano, Marco (1993). “Financial markets and growth: an overview”. European economic review 37.2, pp. 613–622. Philippon, Thomas (2007). Financiers vs. engineers: should the financial sector be taxed or subsidized? Tech. rep. National Bureau of Economic Research. Philippon, Thomas and Ariell Reshef (2007). Skill biased financial development: education, wages and occupations in the US financial sector. Tech. rep. National Bureau of Economic Research. – (2013). “An international look at the growth of modern finance”. The Journal of Economic Perspectives 27.2, pp. 73–96. Piketty, Thomas (2013). Le capital au XXIe si`ecle. Seuil. Raval, Devesh (2015). “What’s Wrong with Capital in the Twenty-First Century’s Model?” Varoudakis, Aristom`ene and Jean-Claude Berth´elemy (1994). “Interm´ediation financi`ere et croissance endog`ene”. Revue ´economique 3, pp. 737–750. Zingales, Luigi (2015). “Presidential Address: Does Finance Benefit Society?” The Journal of Finance 70.4, pp. 1327–1363.

33

Appendix

A

Local Stability of a Steady State

We address in this Appendix a local stability study of a steady state especially when it is not unique. We give conditions for local stability using the linearized system around a steady-state. This is first done in a general case and then applied to our problem. We also illustrate this by an example.

A.1

The dynamic problem

We investigate the stability property of the following dynamic problem:

kt+1 f10 (kt+1 , 1

(zt )(f (kt , 1

zt ) ct ) = 0, u0 (ct ) zt+1 ) (zt )u0 (ct+1 ) = 0,

f20 (kt , 1

zt ) (zt )2 +

0

(80) (81)

(zt )kt+1 = 0.

(82)

Substituting the first equation in the third one (for kt+1 ) and linearizing around a steady-state, we obtain: 0

kt+1 00 u0 (f11 kt+1 00 (f21 kt

00 f22 zt ) + f20

(f

00 f12 zt+1 ) + f10 u0 0

zt

00

(f

[f10 kt

c) zt 0

c) zt

f20 zt

zt + f10 u00 ct+1 0

(f10 kt

u00

f20 zt

ct ] = 0, (83) ct = 0, (84) ct ) = 0. (85)

From the last equation, we find that: ct =

( 0 f10

00 f21 ) 0

kt

(2f20

0

00 f22

00 0

(f

c))

zt .

(86)

Substituting this equation in the first two above equations, we obtain the following result: !✓ ◆ 1 0 k t+1 00 0 = 00 00 00 00 00 k u f1 zt+1 u0 f11 + f10 u00 0 ( 0 f10 f21 ) u0 f12 (2f20 0 f22 ) 0 !✓ 00 00 k ◆ 0 2f 0 0 f22 0 00 kt f10 f1 f21 ) k f20 + ( 2 ) 0( 0 . (87) 0 0 00 ) 00 f21 u00 ( f1 00 00 k 1 zt f10 u0 0 u (2f20 0 f22 ) 0 0

A.2

A More General Problem

We consider the following dynamic problem: 34

✓

b11 b12 b21 b22

◆✓ 1 ◆ ✓ ◆ ✓ 1◆ yt+1 a11 a12 yt = . 2 yt+1 a21 a22 yt2

(88)

We let B be the matrix in the left-hand side of the equation, and A the matrix in the right-hand side. Assume B is invertible. We therefore have: ✓ 1 ◆ ✓ 1◆ yt+1 y 1 = B A t2 . (89) 2 yt+1 yt Set C = B 1 A. We consider two cases. • Real eigen values ((tr(C))2

4 det(C))

We have a saddle i↵: (tr(C) + 1) < det(C) < tr(C)

1,

(90)

(1 + tr(C)).

(91)

or tr(C)

1 < det(C) <

• Complex eigen values (tr(C)2 < 4 det(C)) We have instability i↵ det(C) > 1.

A.3

Example

. Assume that: cbt 0 < b < 1, b (zt ) = zt✓ , 0 < ✓ < 1, f (kt , 1 zt ) = kt↵ (1 zt )1 ↵ , 0 < ↵ < 1. u(ct ) =

(92) (93) (94)

We have the following steady state (c, k, z) where: z=

↵ ✓ ↵ ✓+1

↵

, c=

1

↵ k , k↵ ↵ z✓

1

=

1 (1 z)↵ ↵ z✓

1

.

Lemma 10. We have a saddle if (✓ + 1 + ✓

↵2 ✓ ) 1 ↵

[

(b 1) (↵ + 1 (1 ↵ )

2 (↵b 1) 1 p )+ ]+↵+ ↵ 35

2 p > 0,

or (✓ + 1 + ✓

↵2 ✓ ) 1 ↵

[

(b 1) 2 (↵b 1) 1 2 (↵ + 1 + p ) + ] + ↵ + + p < 0. (1 ↵ ) ↵

and 1+↵

↵ ((3 + ↵)✓ + 2) > 0

Lemma 11. We have complex eigenvalues and instability if (✓ + 1 + ✓

↵2 ✓ ) 1 ↵

(✓ + 1 + ✓

↵2 ✓ ) 1 ↵

[

(b 1) (↵ + 1 (1 ↵ )

2 (↵b 1) 1 p )+ ]+↵+ ↵

2 p < 0,

[

(b 1) 2 (↵b 1) 1 2 (↵ + 1 + p ) + ] + ↵ + + p > 0. (1 ↵ ) ↵

and

36