Social Welfare Functions for Sustainable Development

Transcription

1 Social Welfare Functions for Sustainable Development Thai Ha-Huy, Cuong Le Van September 9, 2015 Abstract Keywords: criterion. anonymity; sustainable development, welfare function, Rawls JEL Classification: C62, D50, D81,D84,G1. University of Evry-Val-D Essonne. IPAG Business School, CNRS, Paris School of Economics, VCREME. Cuong.Le- 1

2 1 Introduction For a sustainable development, one can distinguish two sets of criteria. The first set imposes Anonymity (i.e. Equal Treatment) and Pareto (see e.g. Diamond (1965), Basu and Mitra (2003)). The second set introduced by Chichilnisky (1996), (1997) imposes the Social Welfare Function (SWF) to exhibit No- Dictatorship of both the present and the future. Diamond (1965) shows that there exists no SWF, continuous for the l topology, which is anonymous and Pareto. Basu and Mitra (2003) prove, for sequences of utilities uniformly bounded in l, that there exists no SWF which satisfies the Dominance Axiom (a weaker version of Pareto Axiom) and the Anonymity Axiom. The SWF proposed by Rawls (1971) and by Chichilnisky (1996), (1997) are non-dictator. While the Rawls criterion is anonymous, the Chichilnisky criterion is not. All these authors consider sequences uniformly bounded in l+. This paper complements the papers by Diamond (1965) and by Basu and Mitra (2003). We introduce a stronger notion of anonymity (Strong Anonymity). When the SWF is linear, the two notions coincide. One of its purpose is to completely characterize the SWF which are strongly anonymous and continuous. We distinguish three cases: (i) The set of utility sequences is endowed with the product topology, (ii) this set is in l, (iii) this set is in l p, p < +. Cases (i) and (ii) are very usual in optimal growth models. Case (iii) is considered, in some sense, in optimal growth models without discounting (see e.g. Gale (1967), Brock (1970), Dana and Le Van (1990)). We prove that (i) For the product topology, a SWF which is continuous and anonymous will be constant. (ii) In l p, with 1 < p +, it will be purely finitely additive if it satisfies Strong Anonymity instead of Anonymity. (iii) In l 1, for any Strongly Anonymous and continuous SWF, up to a continuous, real function, it is the sum over time of the instantaneous utilities. Concerning the Pareto efficiency, we introduce a Weak Dominance Axiom which is weaker than the Pareto Axiom and Dominance Axiom. We show there exists no continuous, anonymous SWF for the product topology, or for the l - topology, which is Paretian or Dominant or Weakly Dominant. In l 1, there exist Social Welfare Functions which are continuous, strongly anonymous and dominant. Moreover, they satisfy the No-Dictatorship properties. The paper is organized as follows: In Section 2, we recall the different axioms. In Section 3, we consider the set of sequences endowed with the product topology. In Section 4, we consider sequences in the l space. In Section 5 we show that one can define a SWF in l p, p N but larger than 1, which satisfies the four axioms of sustainable development. In section 6, we consider the of l 1. Since this space is not appropriate for growth models where the consumptions 2

3 path is uniformly bounded, we study in Section 7 the existence of SWF when the state space is [0, 1]. In Section 8, we study the existence of solutions to a growth model with two kind of non-dictatorial criteria. And finally, in Section 9, we show that the Rawls model can be viewed as the limit of a sequence of Ramsey models. 2 The Welfare Axioms We will consider the set X of utility streams of real numbers. We suppose the preferences on X are represented by a SWF. Definition 1 (Anonymity) Given the space l p, p = 1, 2,..., +, a SWF ϕ : l p R satisfies the Anonymity Axiom if and only if: for all x l p, all i, j positive integers, if we define y l p as follows: y i = x j, y j = x i, y k = x k for all k i, j, then we have: ϕ(x) = ϕ(y). Remark 1: We use the definition given by of Basu and Mitra (2003). Definition 2 (Strong Anonymity) Given the space X l p, p = 1, 2,..., +, a SWF ϕ : l p R satisfies the Strong Anonymity Axiom if and only if: for all x X, all a R, all i, j positive integers, if we define y, z l p as follows: y i = x i + a, y k = x k for all k i, z j = x j + a, z k = x k for all k j, then we have: ϕ(y) = ϕ(z). One can easily check that the two definitions coincide if the SWF is linear. Proposition 1 Strong Anonymity implies Anonymity. Proof: Let ϕ be strongly anonymous. We have, for any a R, any x X This implies or equivalently ϕ(..., x i + a,..., x j,...) = ϕ(..., x i,..., x j + a,..., ) ϕ(..., x i + a a,..., x j,..., ) = ϕ(..., x i + a,..., x j a,...) ϕ(..., x i,..., x j,..., ) = ϕ(..., x i + a,..., x j a,...) Take a = x j x i we then get ϕ(..., x i,..., x j,..., ) = ϕ(..., x j,..., x i,...) Now, let X l p, and X. 3

4 Definition 3 (Pareto Axiom) For all x,y X, if x > y (i.e. x i y i, i, and x y), then ϕ(x) > ϕ(y). Definition 4 (Weak Pareto Axiom) For all x, y X, if there exists j N such that x j > y j and, for all k j, x k = y k, then ϕ(x) > ϕ(y). For all x,y X, if x >> y (i.e. x i > y i, i), then ϕ(x) > ϕ(y). Definition 5 (Dominance Axiom) (Basu and Mitra, 2003) Let ϕ : X R. ϕ satisfies the dominance property if and only if: (1) for all x X, for all y X such that y i > x i for some i, and for all j i, y j = x j, we have ϕ(y) > ϕ(x). (2) For all x, y X, if x >> y (i.e. x i > y i, i), then ϕ(x) ϕ(y). Definition 6 (Weak Dominance Axiom) Let ϕ : X R. ϕ satisfies the weak dominance property if and only if: ϕ(x, 0, 0,..., 0,...) > ϕ(y, 0, 0,..., 0,...) if (x, 0, 0,...) X, (y, 0, 0,..., ) X and x > y. The following Lemma is obvious Lemma 1 Pareto Axiom Weak Pareto Axiom Dominance Axiom Weak Dominance Axiom. This following proposition shows that Weak Dominance Axiom is strictly weaker than Dominance Axiom. Indeed, it exhibits a SWF which is anonymous and weakly dominant in l, while Basu and Mitra (2003) have proved there exists no SWF in l which is both anonymous and dominant. Proposition 2 There exists a SWF in l which is anonymous and weakly dominant. Proof: In l define the equivalence relation R by: xry iff x i y i for i in a finite set. We denote by [ x ] the equivalence class of x. Define the function f: Let f : l /R l y l, ϕ(y) = [ x ] x [ x ] (y i f([ y ]) i ) This function is well-defined since y differs from f([ y ]) for a finite number of components. It is anonymous and satisfies the Weak Dominance Axiom. i=0 4

5 Definition 7 (No Dictatorship of the Present) (Chichilnisky, 1996, 1997) Let x, y X, with ϕ(x) > ϕ(y). The SWF ϕ exhibits no dictatorship of the present if for any N, there exist k N, (z 0, z 1,...) X, (z 0, z 1,..., ) X such that ϕ(x 0, x 1,..., x k, z k+1, z k+2,...) ϕ(y 0, y 1,..., y k, z k+1, z k+2,...). Definition 8 (No Dictatorship of the Future) (Chichilnisky, 1996, 1997) Let x, y X, with ϕ(x) > ϕ(y). The SWF ϕ exhibits no dictatorship of the future if for any N, there exist k N, (z 0, z 1,..., z k ), (z 0, z 1,..., z k ) such that ϕ(z 0, z 1,..., z k, x k+1, x k+2,...) ϕ(z 0, z 1,..., z k, y k+1, y k+2,...). 3 Social Welfare Functions with the product topology We now endow X with the product topology. The following results requires neither Weak Dominance nor Pareto Axiom. Theorem 1 A SWF ϕ is continuous for the product topology and anonymous iff it is a constant function. Proof: Step 1 Let z R. For any n define z n X as follows: z n t = 0, t < n, t > n + k, z n n = z 0, z n n+1 = z 1,..., z n n+k = z k. Then z n 0, and hence ϕ(z n ) ϕ(0). Since ϕ is anonymous, we have Indeed, n, ϕ(z n ) = ϕ(z 0, z 1,..., z k, 0,...) ϕ(z 0, z 1,..., z k, 0,...) = ϕ(z 0, z 1,..., z k 1, 0,..., 0, z k, 0,...) where z k is at the (n + k + 1) th position. By induction, we have: ϕ(z 0, z 1,..., z k, 0,...) = ϕ(0, 0,..., 0, z 0,..., z k, 0...) = ϕ(z n ) where z 0 is at the n th position. Hence ϕ(z 0,..., z k, 0,..., 0,...) = ϕ(0). Step 2 Now, with any x, associate the sequences x n X defined as follows: x n t = x t, t = 0,..., n; x t = 0, t n + 1 5

6 Then x n converges to x for the product topology and ϕ(x n ) ϕ(x). But n, ϕ(x n ) = ϕ(x 0,..., x n, 0,..., 0,...), by anonymity Hence ϕ(x) = ϕ(0). The converse is obvious. = ϕ(0) from Step 1 Corollary 1 There exists no SWF ϕ, continuous for the product topology, anonymous and weakly dominant Proof: We have the contradiction ϕ(1, 0, 0, 0,...) = ϕ(0), from theorem 1 > ϕ(0) by weak dominance In view of Theorem 1, when X is endowed with the product topology, the No-Dictatorship axioms do not make sense for the SWF which are anonymous and continuous since they are constant functions. 4 Social Welfare Functions in l We now endow X with the l -topology. A function ϕ from X into R is called purely finitely additive if for any x X, any y X, we have ϕ(x) = ϕ(y) if x i = y i, i N \ I, where I is a finite set of N. Theorem 2 A SWF ϕ is strongly anonymous, continuous for the l -topology iff ϕ is purely finitely additive and continuous for the l -topology. In particular, (a) ϕ(x) = ϕ(0) if x i = 0, i N \ I, where I is a finite set (b) if x t x when t, then ϕ(x) = ϕ(x, x,..., x, x,...). Proof: Let x X and N be given. First, we will show that ϕ(x) = ϕ(0,..., 0, x N+1,..., x N+t,...). (1) Indeed, let δ = N x t. Define the sequence x(t ) as follows x(t ) 0 = x 0 δ N + T,..., x(t ) N+T 1 = x N+T 1 δ N + T x(t ) N+τ = x N+τ, τ T. 6

7 We have x x(t ) = δ N+T. This implies x(t ) x when T +. Since ϕ is continuous we have lim ϕ(x(t )) = ϕ(x). (2) T + On the other hand, since ϕ is strongly anonymous, for any T we have successively ( ) N+T 1 ϕ(x(t )) = ϕ ( x t ) δ, 0,..., 0, x N+T,..., x N+τ,... Relation (5) implies = ϕ ( ( N+T 1 t=n+1 x t ), 0, 0,..., 0, x N+T,..., x N+τ,... = ϕ(0, 0,..., 0, x N+1, x N+2,..., x N+τ,...). ϕ(x) = ϕ(0, 0,..., 0, x N+1, x N+2,..., x N+τ,...). We now show that ϕ is purely finitely additive. Let x X, and y X satisfy x i = y i, i N \ I, where I is a finite set of N. This implies the existence of N N such that x i = y i, i N. We have and hence ϕ(x) = ϕ(0,..., 0, x N, x N+1,..., x N+t,...) ϕ(y) = ϕ(0,..., 0, x N, x N+1,..., x N+t,...) ϕ(x) = ϕ(y). The converse is obvious. The proof of (a) is obvious. We now prove (b). Let x satisfy x t x when t +. Denote by x the sequence (x,..., x,...). Since ϕ is purely finitely additive, we have Define the sequence x(n) by N, ϕ(x) = ϕ(x,..., x, x N+1,..., x N+t,...). x(n) t = x, t = 0,..., N, and x(n) t = x t, t > N. Since ϕ is purely finitely additive, we have ϕ(x) = ϕ(x(n)), N (3) Obviously, x(n) x for the l -topology. Hence ϕ(x(n)) ϕ(x). From (3), ϕ(x) = ϕ(x). ) 7

8 Corollary 2 1. There is no strongly anonymous, continuous SWF in l which is Weakly Dominant (hence no SWF in l is strongly anonymous, continuous and either Paretian, or Weakly Paretian or Dominant). 2. If ϕ is a strongly anonymous, continuous SWF in l then (a) ϕ exhibits no dictatorship of the present. (b) ϕ has the dictatorship of the future. Proof: 1. We have, a R ++, ϕ(a, 0, 0,..., 0,...) = ϕ(0): the Weak Dominance Axiom cannot hold. 2.a. Let ϕ(x) > ϕ(y). Then, for all N ϕ(x 0..., x N, y N+1,..., y N+t,...) = ϕ(y) and ϕ has no dictatorship of the present. 2.b. Let ϕ(x) > ϕ(y). For any N, any (z 0,..., z N ), any (ζ 0,..., ζ N ), we have ϕ(x) = ϕ(z 0,..., z N, x N+1,..., x N+t,...) > ϕ(y) = ϕ(ζ 0,..., ζ N, y N+1,..., y N+t,...). Remark 2: In the converse part of Theorem 2, we have to assume the purely finitely additive function is continuous for the l -topology. Indeed, we now exhibit a purely finitely additive function which is not continuous for the l - topology. Define ϕ as follows: ϕ(0) = 0 ϕ(x) = ϕ(0), for any x different from 0 for a finite number of components ϕ(x) = 1, otherwise Then ϕ is purely finitely additive. But ϕ is not continuous. Indeed, let x n be defined by x n i = 1 n, n. Then xn 0 in l, but ϕ(x n ) = 1, n. Remark 3: (i) The Rawls criterion ϕ(x) = inf t (x t ) is anonymous and continuous for the l -topology. It is not strongly anonymous. The proof of its anonymity is trivial. To prove it is not strongly anonymous, consider the sequence {0, 1,..., 1,...}. Then the infimum of this sequence is 0. Let {1, 1,..., 1,...} and {0, 2, 1,..., 1,...}. Then inf{1, 1,..., 1,...} = 1 while inf{0, 2, 1,..., 1,...} = 0: strong anonymity is not satisfied. The Rawls criterion is continuous for the l -topology. For that, let x n x in this topology. Let ε be given. Then there exists N such that x n t x t ε, for all t, for all n > N. We then have inf t {x t} ε inf t {xn t } inf t {x t} + ε, n > N 8

9 In other words, ϕ(x n ) ϕ(x) ε, n > N (ii) The function ϕ(x) = lim inf t (x t ) is strongly anonymous, continuous for the l topology. It is purely finitely additive. Indeed, it is obvious that lim inf(x 0,..., x i + a, x i+1,...) = lim inf(x 0,..., x i,..., x j + a, x j+1,...) Hence strong anonymity is satisfied. We prove that this SWF is continuous for the l -topology. Let x n x in this topology. Let ε be given. Then there exists N such that x n t x t ε, for all t, for all n > N. We have, given T Hence inf {x t} ε inf t T t T {xn t } inf t T {xn t } + ε, n > N lim T {inf t T {x t}} ε lim T {inf t T {xn t }} lim T {inf t T {x t}} + ε, n > N Equivalently, ϕ(x n ) ϕ(x) ε, n > N 5 Social Welfare Functions in l p, with 1 < p < + We now endow X with the l p -topology, with 1 < p < +. Theorem 3 A SWF ϕ is strongly anonymous, continuous for the l p -topology, with 1 < p < + iff ϕ is constant function. Proof: We prove firstly that ϕ is purely finitely additive. Let x X and N be given. Like in the previous section, we show that ϕ(x) = ϕ(0,..., 0, x N+1,..., x N+t,...). (4) Indeed, let δ = N x t. Define the sequence x(t ) as follows x(t ) 0 = x 0 δ N + T,..., x(t ) N+T 1 = x N+T 1 δ N + T x(t ) N+τ = x N+τ, τ T. We have x x(t ) p p = N+T 1 δ ( N + T )p = δ p (N + T ) p 1. 9

10 This implies x(t ) x when T +. Since ϕ is continuous we have lim ϕ(x(t )) = ϕ(x). (5) T + By using the same arguments like Section 4, we have ϕ is purely finitely additive. Take any x l p. Define the sequence x(n) by x(n) t = 0, t = 0,..., N, and x(n) t = x t, t > N. Since ϕ is purely additive, we have ϕ(x(n)) = ϕ(x), N. Obviously, x(n) 0 for the l p -topology. Hence ϕ(x) = ϕ(0). 6 Social Welfare Functions in l 1 We now take X = l 1. Theorem 4 A SWF is strongly anonymous, continuous for the l 1 -topology iff it is expressed in the following form: with f : R R continuous. ϕ(x) = f( x t ) Proof: Suppose that ϕ : l 1 R is strongly anonymous, continuous. Define f : R R as: f(x) = ϕ(x, 0, 0,...) By the continuity of ϕ, we have f is continuous. For x l 1, define x(n) as follows: x 0 (n) = n 1 x t x k (n) = 0 for all 1 k n 1 x k (n) = x k for all k n and x = ( x t, 0, 0,...). Notice that ϕ(x(n)) = ϕ(x) for all n by the strong anonymity property. We have: x(n) x = x t + x t. t=n 10 t=n

11 Since x l 1, the RHS of this equality converges to 0 when n conveges to. This implies Since ϕ is continuous, we have lim x(n) x = 0. n ϕ(x) = lim ϕ(x(n)) = ϕ(x) = f( x t ). n Conversely, it is obvious that if ϕ(x) = f( x t) with f continuous, then ϕ is continuous for the l 1 topology, strongly anonymous. Let ϕ be a SWF which is strongly anonymous and continuous for l 1. From Theorem 4, there exists a continuous real function defined on R, f, such that x l p, ϕ(x) = f( x t). In this case we say that f represents ϕ. Lemma 2 ϕ is Weakly Dominant iff its representation f is strictly increasing. Proof: : It is obvious. The following corollary is immediate. Corollary 3 If the function f which represents a SWF ϕ is increasing, then ϕ is Paretian and hence Weakly Paretian, Dominant and Weakly Dominant. Remark 4: (i) The lim inf SWF is continuous for the l 1 -topology and strongly anonymous. Its representation is the function 0. Indeed, for any x l 1, the value of this sequence obtained with this SWF is 0. Actually, any purely finitely additive SWF, which is strongly anonymous, continuous for the l 1 -topology, has the function 0 as representation. (ii) The Rawls SWF is not strongly anonymous in l 1. Indeed, consider the sequence { 2, 0, 0,...} which is in l 1. Then min{ 1, 0, 0,...} = 1 > 2 min{ 2, 1, 0,...}. Theorem 5 Suppose that ϕ is a strongly anonymous, continuous, weakly dominant SWF on l 1. Then ϕ satisfies the No-dictatorship properties. Proof: (a) Let ϕ(x) > ϕ(y). From Theorem 4, it is equivalent to x t > y t. For any N N, let K satisfy Define z as follows: N x t < y t + K z t = y t, t = 0,..., N; z N+1 = K; z t = 0, t > N

12 Then Hence x t < N y t + K = z t ϕ(x) < ϕ(y 0,..., y N, K, 0, 0,...) and ϕ satisfies the No-dictatorship of the present. (b) Let ϕ(x) > ϕ(y). For any N N, let K satisfy K > x t Define z as follows: t=n z 0 = K, z t = 0, t = 1,..., N 1, z t = y t, t N Then x t < z t and ϕ(x) = f( x t ) < f( y t z t ) = ϕ(z 0,..., z N 1, y N,..., y N+t,...) and ϕ satisfies the No-dictatorship of the future. 7 Social Welfare Functions in [0, 1] The previous section deals with SWF in l p with p < +. But in growth models, usually, the state space is [0, 1]. This section focuses on the SWF defined on the unit ball of l +. The first result is quite pessimistic. Proposition 3 There exists no SWF W from X into R + : W (x) = a t x t + ϕ(x) with a t 0, t, a t = 1, ϕ purely finitely additive, which satisfies the two No-dictatorship properties. Proof: Claim 1. If ϕ(1) > 1, then W does not satisfy the No-dictatorship of the future property. Indeed, let W (x) > W (y) where x t = 1, t and y t = 0, t. Let N be the first index s.t. a N > 0. Then k N, (z 0,..., z k ) R k +, (z 0,..., z k ) Rk +, W (z 0,..., z k, x k+1,...x k+t ) = > k a t z t + a t t=k+1 k a t z t + = W (z 0,..., z k, y k+1,...). 12 a t x t + ϕ(1) ϕ(1) t=k+1 a t y t + ϕ(0)

13 Claim 2. If 0 ϕ(1) < 1, then W does not satisfy the No-dictatorship of the present property. To prove that, let ε > 0 satisfy 1 ε > ϕ(1) and let x t = 1, t and y t = 0, t. Hence W (x) > W (y). There exists N s.t. k N, k a tx t = k a t > ϕ(1) + t=k+1 a t. Then, for any k N, for any (z k+1,...) [0, 1], we have W (y 0,..., y k, z k+1,...) = < t=k+1 a t z t + ϕ(y 0,..., y k, z k+1,...) ϕ(1) + k a t x t + t=k+1 = W (x 0,..., x k, z k+1,...). a t z t + ϕ(x 0,..., x k, z k+1,...) t=k+1 Claim 3. Assume ϕ(1) = 1. Then W does not satisfy the No-dictatorship of the present property. Again let x t = 1, t and y t = 0, t. Hence W (x) > W (y). For any k, any (z 0,..., z k ) R k +, any (z 0,..., z k ) Rk +,we always have a t W (z 0,..., z k, x k+1,...) > k a t z t + ϕ(1) 1 k a t z t = k a t z t + ϕ(0) = W (z 0,..., z k, y k+1,...). The proof of the proposition is over. However, observe that the previous results are based on the assumption a t = 1. We now assume a t = + and, for simplicity, that a t > 0, t. Let Z = {x X : a tx t < + }. If (a t ) satisfies t, 0 < a a t a, then (x) Z x t 0. In this case, ϕ(x) = 0. We consider thus, for any x Z, the SWF W with the following form: W (x) = a t x t. Observe that W is upper semi-continuous for the product topology: W : Z a tx t. Proposition 4 Let (a t ) satisfy (i) t, 0 < a a t a (ii) a t = +. Let Z = {x X : a tx t < + }, and W (x) = a tx t for x Z. Then W satisfies the two No-dictatorship axioms. 13

14 Proof: (a) No-Dictatorship of the present Let W (x) > W (y). We know that y t 0. Let T be given. Define the following sequences: z t n = ε (0, 1), for t = T + 1,..., T + n, z t n = 0, t > T + n. The sequence y n = (y 0,..., y T, z T n +1,...) belongs to Z for any n. W (y n ) = a 0 y a t y T + ε T +n t=t +1 a t < + and y n Z. But W (y n ) +. (b) No-Dictatorship of the future Let W (x) > W (y). There exists N s.t. T N, T +1 a tx t < T 0 a t since a t = +. Let z t = 0, t = 0,..., T and z t = 1, t = 0,..., T. In this case, 0 W (z 0,..., z T, x T +1,...) = a t x t < t=t +1 T a t + t=t +1 T a t a t y t = W (z 0,..., z T, y T +1,...). More generally, let γ be a correspondence from R into R : x [0, 1] R γ(x) [0, 1], the graph of which has a non-empty interior. Let us assume there exists x > 0 such that (x, x) intgraphγ. We consider as before a sequence a which satisfies a t > 0, t 0 a t = +, 0 < a a t a, t. Let { } Z = x X : a t (x t x) exists in R, x t+1 γ(x t ), t 0 Assume also that x Z x t x when t tends to. Let W (x) = a t(x t x), for any x Z. Proposition 5 W satisfies the two No-dictatorship axioms. Proof: (a) No-Dictatorship of the present Let W (x) > W (y) with x Z, y Z. Since y t x and (x, x) intgraphγ, there exists x close enough to x, and T large enough, s.t. for any t T, x γ(x t+1 ), x γ( x), x γ( x) and x < x. Take N T. 14

15 Define z n as follows: z n t = x t, for t = 0,..., N + 1, z n t = x, for t = N + 2,..., N + n + 1, z n t = x, for t > N + n + 1. Then z n t+1 γ(zn t ), t > 0, and W (z n ) = = N+1 N+1 N+n+1 a t (x t x) + t=n+2 a t ( x x) N+n+1 a t (x t x) + ( x x) t=n+2 a t R. We have z n Z for any n, and W (z n ) when n + since ( x x) N+n+1 t=n+2 a t na( x x). Therefore, for any (z n N+2,...) s.t. W (y 0,..., y N+1, z n N+2,...) R, we have W (x 0,..., x N+1, z n N+2,...) < W (y 0,..., y N+1, z n N+2,...) for n large enough. (b) No-Dictatorship of the future Again, let W (x) > W (y) with x Z, y Z. As in (a), there exists x < x, close enough to x, s.t. for any t k, x t γ( x). We have lim T T a t( x x) =. Let z t = x for t = 0,..., T k. Then T a t (z t x) + a t (x t x) a t (y t x), t=t +1 for any T large enough, since T a t(z t x), t=t +1 a t(x t x) 0 when T +. 8 Solving the one sector Ramsey model with two non-dictatorial criteria 8.1 Criterion 1 Consider two problems: R : max β t u(c t ) C : max β t u(c t ) + ϕ(u(c 1 ), u(c 2 ),...) under the constraint c t + k t+1 f(k t ) for all t, with k 0 > 0 given. Let Π(k 0 ) denote the set of feasible capital stocks. We make the following assumptions. 15

16 H0 0 < β < 1. H1 The function u : R R is twice continuously differentiable and satisfies u(0) = 0. Moreover, its derivatives satisfy u > 0 (strictly increasing) and u < 0 (strictly concave). H2 Inada condition u (0) = +. H3 The function f : R R is twice continuously differentiable and satisfies f(0) = 0. Its derivatives satisfy f > 0 (strictly increasing), f < 0, strictly concave, lim x f (x) < 1. H4 ϕ is purely finitely additive, continuous for the l -topology, homogeneous of degree 1, non decreasing, ϕ(0) = 0, ϕ(1) > 0. Lemma 3 Let u l be such that lim t u t = u, then ϕ(u) = ϕ(u, u,...) Proof: Denote u = (u, u,...). Construct u n as: u n t = u for 0 t n u n t = u t for t > n Since ϕ is purely finitely additive, so for all n, ϕ(u n ) = ϕ(u). On the other hand, u n u = sup t>n u t u lim n u n u = 0, then ϕ(u) = lim n ϕ(un ) = ϕ(u) For each k Π(k 0 ), define: Φ(k) = β t u(f(k t ) k t+1 ) + ϕ(u(f(k 0 ) k 1 ), u(f(k 1 ) k 2 ),...) and W (k) = β t u(f(k t ) k t+1 ) Theorem 6 Given k 0 > 0, suppose that there exists a solution to problem C, then this solution k satisfies: i: For all t > 0, 0 < k t+1 < f(k t ). ii: (Euler equation), for all t, we have: u (f(k t ) k t+1) = βu (f(k t+1) k t+2)f (k t+1) 16

17 Proof: The proof is easy. Theorem 7 If the sequence k is solution to problem C, then it is also solution to problem R. Proof: For each k, denote f t (k) = f(f(...(k)...), t times. First case f (0) 1. In this case, for all feasible sequence k, we have lim t k t = 0, then ϕ(u(c 0 ), u(c 1 ),...) = 0 for all k Π(k 0 ), and the problem is over. Second case f (0) > 1 β. Denote k the unique positive solution to the equation f(k) = k. For all x > 0, we have lim t f t (x) = k. (a) Assume k 0 k. Then for all t > 0, kt < k since kt+1 < f(k t ), t. For each ϵ > 0, choose T sufficiently large such that for all k Π(k 0 ), we have β t u(f(k t ) k t+1 ) < ϵ 2 t=t Choose arbitrarily k Π(k 0 ). We know that lim t f t (k T ) = k. We will prove that there exists τ > 0 such that kt +τ < f τ (k T ). Suppose not, then for all t > 0, f t (k T ) kt +t < k lim t kt = k lim t u(f(kt ) kt+1 ) = 0. From Lemma 3 we have: Φ(k ) = β t u(c t ) for all k Π(k 0 ), we have: W (k ) = Φ(k ) = W (k ) Φ(k ) = β t u(c t) + ϕ(u(c 0), u(c 1),...) β t u(c t) = W (k ) Then k is the solution to problem R. We know that in this case lim t k t = k s < k: a contradiction. So there exists τ > 0 such that k T +τ < f τ (k T ). We construct the sequence k as follows: k t = k t for all 0 t T k T +t = f t (k T ) for all 1 t τ 1 k T +τ+t = k T +τ+t for all t 0 17

18 Since k T +τ < f τ (k T ), we can easily check that k is feasible. Observe that for all t T + τ, k t = k t ϕ(c ) = ϕ(c ), where c t = f(k t ) k t+1, c t = f(k t) k t+1. 0 Φ(k ) Φ(k ) = β t u(c t ) β t u(c t) because T βt u(c t ) < ϵ 2. Then we have: T 1 β t u(c t ) β t u(c t ) + ϵ 2 β t u(c t ) β t u(c t ) + ϵ β t u(c t ) β t u(c t ) ϵ That is true for all ϵ > 0, hence k is solution to problem R. (b) Now assume k 0 > k. Suppose such that for all t 0, k t k. Then for all t we have k t+1 < f(k t ) k t k t k k. We have k f(k ) k k = k. Then u(f(k t ) k t+1 ) 0 ϕ(u(c 0 ), u(c 1 ),...) = 0 W (k ) = Φ(k ). Consider k Π(k 0 ), then W (k ) = Φ(k ) = Φ(k) β t u(c t ) + ϕ(u(c 0 ), u(c 1 ),...) β t u(c t ) = W (k) So k is the solution to problem R lim t kt = k s < k: a contradiction. Hence, there exist T 0 such that kt 0 < k. We then go back to case (a). So, if f (0) > 1 β, the solutions to the two problems coincide. Third case 1 < f (0) 1 β. Then f (k) < 1 β for all k > 0. From the Euler s equation we have for all t: u (c t ) = βu (c t+1)f (k t+1) u (c t+1) c t c t+1 and c t c. If c > 0, then we have βf (k t ) 1. If f (0) < 1 β, we have obviously a contradiction, since f is decreasing. If f (0) = 1 β then k t 0: a contradiction. Thus c = 0 and ϕ(u(c 0 ), u(c 1 ),...) = 0 W (k ) = Φ(k ). 18

19 Consider k Π(k 0 ), then W (k ) = Φ(k ) = k is solution to problem R. Φ(k) β t u(c t ) + ϕ(u(c 0 ), u(c 1 ),...) β t u(c t ) = W (k) Theorem 8 Suppose that f (0) > 1, then the solution to problem R is NOT a solution to problem C. Proof: Suppose that k is solution to problem R. We know that k t k s, the steady state. Let x be the solution to equation f (x) = 1. Observe that for all x k > 0, f(k) k < f(x) x. Denote c = f(x) x. Then u(c s ) < u(c) u(c s )ϕ(1) = ϕ(u(c s ), u(c s ),...) < u(c)ϕ(1) = ϕ(u(c), u(c),...) Choose ϵ > 0 such that: 0 < ϵ < ϕ(u(c), u(c),...) ϕ(u(c s ), u(c s ),...) Choose T > 0 such that for any feasible path k Π(k 0 ) we have: t=t +1 β t u(c t ) < ϵ, where c t = f(k t ) k t+1 Denote, as above, by k the unique solution to equation f(x) = x. Notice that x < k. We know also lim t f t (kt ) = k. Then there exists τ > 0 such that f τ (kt ) > x. We construct the sequence k as follows: k t = k t for all 0 t T k T +t = f t (k T ) for all 1 t τ 1 k T +τ+t = x for all t 0 We can verify that k is feasible, and for t sufficiently large, k t = x. Then we 19

20 have: Φ(k) = = = > > β t u(c t ) + ϕ(u(c 0 ), u(c 1 ),...) β t u(c t ) + ϕ(u(c), u(c),...) T 1 β t u(c t ) + ϕ(u(c), u(c),...) + β t u(c t ) t=t T 1 β t u(c t ) + ϕ(u(c), u(c),...) β t u(c t ) + ϕ(u(c), u(c),...) ϵ β t u(c t ) + ϕ(u(c s ), u(c s ),...) = Φ(k ) Φ(k) > Φ(k ), then k is NOT a solution to problem C. Corollary 4 Assume f (0) > 1. Then problem C has no solution. Proof: It comes from Theorems 7 and 8 Theorem 9 Suppose f (0) 1. Problem C has a solution which coincides with the one to problem R. Proof: Any feasible path (k t ) converges to zero. Let k be the solution to R. We have Φ(k ) = W (k ) W (k) = W (k) + ϕ(0) = W (k) + ϕ(u(c 0 ), u(c 1 ),...) = Φ(k) with c t = f(k t ) k t+1. Thus k is a solution to Problem C. The converse comes from Theorem Criterion 2 We consider the model usually called the Rawls model: max inf t 0 c t 20

21 under the constraint c t + k t+1 f(k t ) for all t, with k 0 > 0 given. Hypothesis 1 f : R + R + is strictly concave, strictly increasing, differentiable, and f(0) = 0. Hypothesis 2 lim x f (x) < 1. Define Π(k 0 ) as the set of feasible paths. We know that Π(k 0 ) is compact in the product topology. Define also the reward function V : Π(k 0 ) R as: V (k) = inf t 0 c t with c t = f(k t ) k t+1 Observe that if f (0) 1, then for all feasible path k we have k t 0 and so V (k) = 0. The problem is over. Now we will consider only the case f (0) > 1. With this additional assumption, we have the lemma: Lemma 4 sup k Π(k0 ) V (k) > 0. Proof: Define k as the unique solution to the equation f(k) = k. We distinguish two cases: Case 1: 0 < k 0 < k. Then k 0 < f(k 0 ), so the path k = (k 0, k 0,...) is feasible, and V (k) = f(k 0 ) k 0 > 0. Case 1: k 0 k. By choosing 0 < k 1 < k f(k 0 ), we have that the path k = (k 0, k 1, k 1,...) is a feasible path, and V (k) = f(k 1 ) k 1 > 0. Theorem 10 V is upper semi-continuous for the product topology. Proof: Suppose that the sequence of feasible paths k n k for the product topology. For all ϵ > 0, there exists T such that c T < inf t 0 c t + ϵ. We have lim n c n T = c T, then for all n sufficient large c n T < c T +ϵ inf t 0 c n t c n T < c T + ϵ < inf t 0 c t + 2ϵ lim sup n V (k n ) < V (k) + 2ϵ. Since this is true for all ϵ > 0, we get the result. Remark V is not continuous. For example, let k Π(k 0 ) with V (k) > 0. Define k n as: k n t = k t for all 0 t n and k n t = 0 if t > n. Then lim n k n = k in the product topology, whereas for all n, V (k n ) = inf t 0 c n t = 0. From the upper semi-continuity of V and the compactness of Π(k 0 ), we know that there exists always k Π(k 0 ) such that V (k ) = sup Π(k0 ) V (k). The next theorem will give us the solutions to the problem. Let x satisfy f (x) = 1. Note that x is the unique solution to problem max x 0 [f(x) x]. Theorem 11 If k 0 x, then the problem has a unique solution k = (k 0, k 0,...). If x < k 0, then the problem has an infinity of solutions. 21

22 Proof: Case 1 k 0 x. Denote k as a solution to the problem. From the Lemma 4, V (k ) V (k) = f(k 0 ) k 0 with k = (k 0, k 0,...). So, we have f(kt ) kt+1 f(k 0) k 0 for all t 0. Let t = 0. We then have k1 k 0. Observe that k 0 kt+1 f(k 0 ) f(kt ) f (k 0 )(k 0 kt ) k 0 kt if kt k 0. By induction, we find that k 0 kt for all t, and furthermore, the sequence (kt ) is decreasing and then converges to ˆk k 0. From the continuity of f, we have that f(ˆk) ˆk f(k 0 ) k 0. But the function f(x) x is increasing in [0, x]. Thus, f(ˆk) ˆk f(k 0 ) k 0, then ˆk = k 0, and kt = k 0 for all t, because k 0 kt k 0. Case 2 k 0 > x. First, consider the sequence k = (k 0, x, x,...) which is feasible. So, if k is a solution then V (k ) V (k) = f(x) x. We have that for all t 0, f(kt ) kt+1 f(x) x x k t+1 f(x) f(k t ) x kt. So, kt ˆk and f(ˆk) ˆk f(x) x ˆk = x and V (k ) = f(x) x. Since k 0 > x, by induction, we can construct a sequence k which satisfies: for all t, x < k t+1 < f(k t ) f(x) + x. With this sequence, we have f(k t ) k t+1 > f(x) x, and k t+1 < k t, since f(k t ) k t < f(x) x. So k t ˆk. Using the same argument as above, we have ˆk = x and V (k) = f(x) x. We have an infinity number of sequences like that. That means the problem has an infinity of solutions. Now we consider the case more general: max inf t 0 F (k t, k t+1 ) under the constraint k t+1 Γ(k t ) t, k 0 > 0 given. Hypothesis 1 The correspondence Γ : X X is continuous, with non-empty, compact values. Moreover 0 Γ(0). Define: GraphΓ := {(x, y) y Γ(x)}. Hypothesis 2 GraphΓ is closed, convex. Hypothesis 3 F : GraphΓ R + is concave, increasing with respect to the first variable and decreasing with respect to the second variable. We also assume F (0, 0) = 0. Hypothesis 4 The set {x x Γ(x)} is compact. With this hypothesis, we know that there exists x which satisfies F (x, x) = max F (x, x). Theorem 12 If k 0 x, then there exists a unique solution which is k = (k 0, k 0,...). Proof: Since k 0 x, then there exists 0 λ 1 such that k 0 = λx. But both (0, 0), (x, x) belong to GraphΓ, and this set is convex. So, we have (k 0, k 0 ) = 22

23 λ(x, x) and belongs to GraphΓ too. Thus k 0 Γ(k 0 ) (k 0, k 0,...) is a feasible path. Denote the solution to problem as k. We have that for all t 0, F (k t, k t+1 ) F (k 0, k 0 ). We know that k 1 k 0. By induction, we will prove that k t is decreasing. First, we prove that F (x, x) is increasing in [0, x]. If it is not the case, there exist 0 x < y x which satisfy F (x, x) > F (y, y). There exists 0 < λ 1 such that y = (1 λ)x + λx. And we have: F (x, x) > F (y, y) (1 λ)f (x, x) + λf (x, x) F (x, x) which is a contradiction. Suppose that k t k 0. We have: F (k 0, k t+1) F (k t, k t+1) F (k 0, k 0 ) k t+1 k 0. Then we have: F (k t, k t+1) F (k 0, k 0 ) F (k t, k t ) for all t k t+1 k t and lim t k t =: ˆk. By the continuity of F we have F (ˆk, ˆk) F (k 0, k 0 ). So, ˆk = k 0 and then k t = k 0 for all t, because k 0 k t k 0. Theorem 13 If k 0 > x, then we have an infinity of solutions, and the value of the problem is F (x, x). Proof: We recall that k = (x, x,...) is a feasible path. Then if k is a solution, we have that for all t, F (k t, k t+1 ) F (x, x) F (k t, k t ) for all t, k t k t+1 and then k t ˆk. We have F (ˆk, ˆk) F (x, x) ˆk = x and the value of problem is F (x, x). Conversely, consider a feasible path k which satisfies k t x. Then, for all t, F (k t, k t+1 ) F (x, k t+1 ) F (x, x) and lim t F (k t, k t+1 ) = F (x, x). Then k is a solution to problem. In other words, we have an infinity of solutions. 9 Rawls model as the limit of a sequence of Ramsey models With each σ > 0, we associate the Ramsey problem: max β t u(c t) 1 1 σ 1 1 σ 23

24 under the constraint c t + k t+1 f(k t ) for all t, and k 0 > 0 given. From now on, we consider the model in the case f (0) > 1. We will prove that when σ 0, the solutions of the Ramsey problem converge to a solution of the Rawls model. Moreover, the optimal consumption paths of the Ramsey models converge to a constant consumption path. Lemma 5 For each a l+, we have: [ β t a θ t lim θ ] 1 θ = sup a t t 0 Proof: Fix any ϵ > 0. There exists T such that a T > sup t a t ϵ. Then we have: sup a θ t β t β t a θ t β T a θ T. t 0 That implies 1 (1 β) 1 θ sup a t t 0 [ β t a θ t ] 1 θ β T θ (sup a t ϵ). t 0 [ Let θ converge to infinity, we have sup t a t lim θ βt a θ ] 1 θ t for all ϵ > 0. sup t a t ϵ Let θ := 1 σ σ. Since σ 0, we have θ + and our problem becomes max k Π(k 0 ) β t u θ (c t ). θ Obviously, this problem is equivalent to min β t u θ (c t ). k Π(k 0 ) or to : For each k Π(k 0 ), define min k Π(k 0 ) [ β t u θ (c t ) ] 1 θ. Define [ V (k, θ) := β t u θ (c t ) ] 1 θ G(k 0, θ) = {k Π(k 0 ) V (k, θ) < + }.. 24

25 Lemma 6 The set G(k 0, θ) is non empty. Proof: Since f (0) > 1, by Theorem 8.6, there exists a solution k of the Rawls model which satisfies c t = c 0 for all t 2. We can verify easily that V (k, θ) < +. Lemma 7 V (, θ) is lower semi-continuous in product topology. Proof: Take any k in Π(k 0 ), and the sequence k n converges to k for the product topology. Fix any N N. We have: V (k n, θ) θ = β t u θ (c n t ) N β t u θ (c n t ). Let n converges to infinity, we have lim inf V (k n, θ) θ N βt u θ (c t ), for all N. Let N converges to infinity, we have the conclusion. Lemma 8 There exists min Π(k0 ) V (k, θ). Proof: We use the lower semi-continuity of V (, θ) and the compactness of Π(k 0 ) for the product topology. Denote by k(θ) as the solution of min Π(k0 ) V (k, θ). Lemma 9 For all t, we have 0 < k t+1 (θ) < f(k t (θ)). And we have also the Euler s equalities: u θ 1 (c t (θ))u (c t (θ)) = βu θ 1 (c t+1 (θ))u (c t+1 (θ))f (k t+1 (θ)). Proof: We prove the first claim. Indeed, V (k(θ), θ) < +. That implies c t (θ) > 0 for all t. Also we have f(k t (θ)) > c t (θ) > 0. This implies k t (θ) > 0 for all t. By the very definition of k(θ), we have for all t, k t+1 (θ) is the solution of min 0 y f(k t (θ)) u θ (f(k t (θ)) y) + βu θ (f(y) k t+2 (θ)). Since 0 < k t+1 (θ)) < f(k t (θ)), by the first order condition, we have: θu θ 1 (f(k t (θ)) k t+1 (θ))u (f(k t (θ)) k t+1 (θ)) = θβu θ 1 (f(k t+1 (θ)) k t+2 (θ))u (f(k t+1 (θ)) k t+2 (θ))f (k t+1 (θ)). That implies u θ 1 (c t (θ))u (c t (θ)) = βu θ 1 (c t+1 (θ))u (c t+1 (θ))f (k t+1 (θ)). 25

26 Theorem 14 Denote by k one solution of the Rawls model. Define c t = f(k t ) k t+1. We have that every limit point of {k(θ)} is a solution of the Rawls model and lim θ V (k(θ), θ) = sup u 1 (c t ) = t 0 1 inf t 0 u(c t ). Proof: Without loss of generality, we suppose that for the product topology, k(θ) converges to k. For each t, define c t = f(k t ) k t+1 and c t (θ) = f(k t (θ)) k t+1 (θ). Fix ϵ > 0 such that ϵ < sup t u 1 (c t ). There exists T > 0 such that u 1 (c T ) > sup t 0 u 1 (c t ) ϵ/2. For θ big enough, we have u 1 (c T (θ)) > u 1 (c T ) ϵ/2 > sup u 1 (c t ) ϵ. t 0 Observe that V (k(θ), θ) β T θ u 1 (c T (θ)) > β T θ [ ] sup u 1 (c t ) ϵ. t 0 This implies lim inf θ V (k(θ), θ) sup u 1 (c t ) ϵ. t 0 Since this inequality is true for all ϵ > 0, hence lim inf θ V (k(θ), θ) sup t 0 u 1 (c t ). Observe that for all θ, k G(k 0, θ) and we have V (k, θ) V (k(θ), θ). This implies: lim θ V (k, θ) lim sup V (k(θ), θ) lim inf θ θ V (k(θ), θ) sup u 1 (c t ). (6) By Lemma 5, we have lim θ V (k, θ) = sup t 0 u 1 (c t ). Hence sup t u 1 (c t ) sup t 0 u 1 (c t ). This implies inf t u(c t ) inf t u(c t ). Since k is a solution of the Rawls model, we have inf t u(c t ) = inf t u(c t ) and hence the inequalities in (6) are equalities. This implies that k is also a solution of the Rawls model and lim θ V (k(θ), θ) = sup u 1 (c t ) = t 0 t 0 1 inf t 0 u(c t ). Theorem 15 If k 0 x, then lim θ c t (θ) = f(k 0 ) k 0 for all t. If k 0 > x, then lim θ c t (θ) = f(x) x for all t. Proof: When k 0 x, the Rawls problem has a unique solution (k 0, k 0,... ). So, k(θ) converges to (k 0, k 0,... ). 26

27 Now consider the case k 0 x. Without loss of generality, suppose that for the product topology, k(θ) converges to k solution to the Rawls problem. From the Euler equation we have for all t: [ ] u u(ct (θ)) 1+θ (c t (θ)) = β u (c t+1(θ))f (k t+1(θ)). u(c t+1 (θ)) If c t > c t+1, then u(c t ) > u(c t+1 ), then we have: [ ] u (c t ) = lim u (c t (θ)) = u (c t+1 )f u(ct (θ)) 1+θ (k t+1 ) lim = θ θ u(c t+1 (θ)) which is a contradiction. If c t < c t+1, using the similar argument, we find that u (c t ) = 0, also a contradiction. Hence for every t 0, c t = c t+1. There exists only one solution of the Rawls model which satisfies this property. That is the solution defined by: k t+1 = f(k t ) f(x) + x for all t. The consumption path is constant, and c t = f(x) x for all t. References [1] Basu, K. and T. Mitra, Aggregating infinite utility streams with intergenerational equity: The impossibility of being paretian, Econometrica, Vol. 71, No. 5 (2003), [2] Brock, W.A., On the existence of weakly maximal programmes in a multisector economy, Review of Economic Studies, Vol. 37 (1970), [3] Chichilnisky, G., An axiomatic approach to sustainable development, Social Choice and Welfare, Vol. 13, No. 2 (1996), [4] Chichilnisky, G., What is sustainable development?, Land Economics, Vol. 7, No.4 (1997), [5] Dana, R.A. and C. Le Van, On the Bellman equation of the overtaking criterion, Journal of Optimization Theory and Applications, Vol. 78, No.3 (1990), [6] Peter A. Diamond, The evaluation of infinite utility streams, Econometrica, Vol. 33, No. 1 (1965). 27

28 [7] Gale, D., On optimal development in a multi-sector economy, Review of Economic Studies, Vol. 34, No.97 (1967), [8] Rawls, J., A Theory of Justice, Oxford, England: Clarendon (1971). 28