Notes for ECON 970 and ECON 973 Loris Rubini University of New Hampshire

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1 Notes for ECON 970 and ECON 973 Loris Rubini University of New Hampshire 1 Introduction Economics studies resource allocation problems. In macroeconomics, we study economywide resource allocation problems. To solve a resource allocation problem, we need to know (i) what the feasibility set is; and (ii) what the payoff (utility) for each individual is under each feasible set. Given this information, we want to know what the best possible allocation (an efficient allocation) of resources is, and next we want to know whether there is a decentralised process that will deliver this allocation. 1.1 Big Issues in Macroeconomics Growth. Why is GDP growth in South Korea roughly 6 times larger than in Cameroon? Development. Why is average consumption in the U.S. roughly 20 times larger than average consumption in Africa? Business Cycles. Why do allocations fluctuate over time within a given country? Labor Market. Why is time allocated to work 25 percent larger in the US than in Europe? 1.2 Static Endowment Economy Let the economy E be described by the list {(X i, u i, w i ), i = 1,..., I}, where for each i = 1,..., I X i R K + (the consumption set with K goods) u i : X i R (utility functions) w i X i (endowment) This economy describes a resource allocation problem, where the feasible choices are the allocations of goods across individuals such that the total allocation of good k does not not exceed the total amount available of such good and each allocation is associated with a payoff given by the utility function. These notes are extracts of the courses Macro I and II taught by R. Rogerson at ASU. 1

2 1.3 Some Useful Definitions An allocation for this economy is a list (x i, i = 1,..., I) such that x i X i. (rules out, for example, negative amounts). An allocation for this economy is feasible if I x ik i=1 I w ik for all k = 1,..., K i=1 An allocation for this economy is Pareto efficient if (x i, i = 1,..., I) is feasible; and There is no other feasible allocation ( x i, i = 1,..., I) such that 1.4 Competitive Equilibrium u i ( x i ) u i (x i ) for all i = 1,..., I and u i ( x i ) >u i (x i ) for at least one i A Competitive Equilibrium for this economy is an allocation (x i, i = 1,..., I) and a price vector p R K + such that (Consumer Maximization) For each i, i = 1,..., I, taking p, x i is a solution to subject to x i u i (x i ) K p kx ik k=1 K p kw ik k=1 x i X i Market Clearing I x ik i=1 I i=1 w ik for all k = 1,..., K There are three classic questions regarding a competitive equilibrium: Does a competitive equilibrium exist? In such case, is it unique? If (p, x ) is a competitive equilibrium, is x Pareto efficient? (First welfare theorem) If x is a Pareto Efficient allocation, is there a price p such that (p, x ) is a competitive equilibrium? (Second welfare theorem) 2

3 1.5 Adding Production Let Y R K be a production set. If y k > 0, then the good k is an output of production If y k < 0, then the good k is an input of production If y k = 0, then the good k is neither an input nor an output of production Example: let y = f(n) be a production function, where n is labor services. What is the production set? There are two goods: y and n, Thus, Y R 2. Y = { (y, n) R 2 : y f( n), n 0, y 0 } 1.6 Static Production Economy Let the economy E be described by the list {(X i, u i, w i ), i = 1,..., I, Y j where for each i = 1,..., I = 1,..., J}, X i R K + (the consumption set with K goods) u i : X i R (utility functions) w i X i (endowment) An allocation for this economy is a list (x i, i = 1,..., I, y j, j = 1,..., J) such that x i X i for all i = 1,..., I and y j Y j for all j = 1,..., J. An allocation for this economy is feasible if I I I x ik w ik + y jk for all k = 1,..., K i=1 i=1 i=1 An allocation for this economy is Pareto efficient if (x i, i = 1,..., I, y j, j = 1,..., J) is feasible; and There is no other feasible allocation ( x i, i = 1,..., I, ỹ j, j = 1,..., J) such that 1.7 Competitive Equilibrium u i ( x i ) u i (x i ) for all i = 1,..., I and u i ( x i ) >u i (x i ) for at least one i A Competitive Equilibrium for this economy is an allocation (x i, i = 1,..., I, y j, j = 1,..., J), a price vector p R K + and profits π j R such that 3

4 (Consumer Maximization) For each i = 1,..., I, taking p, x i is a solution to subject to x i u i (x i ) K p kx ik k=1 K J p kw ik + θ ij πj k=1 j=1 x i X i where θ ij is the fraction of firm j profits going to consumer i. (Profit Maximization) For each j = 1,..., J, taking p as given, y j solves and π j = K k=1 p k y jk. Market Clearing subject to y j K p ky jk k=1 y j Y j I x ik i=1 I w ik + J i=1 j=1 y jk for all k = 1,..., K 2 The Solow Growth Model 2 key trade-offs in macro are: Consumption vs. saving Work vs. leisure Growth models usually includes both tradeoffs. The simplest of the modern growth models is the Solow growth model. To simplify things, this model assumes C t = (1 s)y t, 0 s 1 L t = 1 That is, consumers save a constant fraction of their income and they use all the time they have available to work. This implies that we do not need to model the preference side, and can concentrate on the technology side of things. Technology is Y t = F (K t, L t ) 4

5 and the law of motion for capital is Feasibility requires K t+1 = (1 δ)k t + I t C t + I t = Y t where Y is output, L is labor services, K is the stock of capital, I is investment and δ is the depreciation rate. The economy starts with K 0 units of capital today. 2.1 Assumptions Let F is C 2, is strictly increasing in both arguments when (k, h) >> 0, has constant returns to scale, with F (0, h) = F (k, 0) = 0 for all h, k F is strictly concave in k and h individually lim k 0 F 1 (k, h) = for all h > 0 lim h 0 F 2 (k, h) = for all k > 0 lim k F 1 (k, h) = 0 for all h > 0 lim k F 2 (k, h) = 0 for all k > 0 δ (0, 1) f(k t ) = F (K t, 1) where k t = K t /L t. f is output per capita. To see this, divide F by L and use the constant returns to scale assumption: ( F (K, L) K = F L L, L ) = f(k) L What allocations do these assumptions generate? Start with k = k 0. y 0 = f(k 0 ) i 0 = sf(k 0 ) and c 0 = (1 s)f(k 0 ) k 1 = (1 δ)k 0 + i 0 y 1 = f(k 1 ) i 1 = sf(k 1 ) and c 1 = (1 s)f(k 1 ) k 2 = (1 δ)k 1 + i 1 we can generate the entire allocation this way. We can summarize this by one equation: Notice g(0) = 0 k t+1 = (1 δ)k t + sf(k t ) g(k t ) 5

6 g (k) > 0 g (k) < 0 lim k 0 g (k) = lim k g (k) = 1 δ The function g crosses the 45 degree line once, and from below. This guarantees that there is one steady state, and that it is stable. Proposition 1 Suppose s > 0 and 0 < k 0 < k ss. Then {k t } is a strictly increasing sequence and lim t k t = k ss Proof: We know that if k < k ss k < g(k) and g(k) < k ss, so k < k ss, that is, the sequence is bounded. Also, k t < k t+1 < k ss, that is, the sequence is also increasing. Any sequence that is increasing and bounded converges to a number. We need to show that that number is k ss. Suppose that number is ˆk k ss. Take limits as t of k t+1 = g(k t ) ˆk = g(ˆk), but this can only happen if ˆk = k ss. 6

7 We can show convergence similarly if k 0 > k ss. So we have convergence. What about the speed of convergence? k t+1 = (1 δ)k t + sf(k t ) γ k = k t+1 k t = sf(k t) δ k t k t The Euler theorem says γ k k = (k)k f(k) sf k 2 f(k) = F (k, 1) = F 1 (k, 1)k + F 2 (k, 1) = f (k)k + F 2 (k, 1) Since F 2 > 0, we have that f(k) > f (k)k and therefore γ k. Thus, the rate of growth is k decreasing in k. Also lim γ k = lim s f (k) k 0 k 0 k lim γ k = lim s f (k) k k k = s lim k 0 f (k) lim k 0 k = s lim k f (k) lim k k = = 0 In Class. Matlab example. 2.2 The Golden Rule The golden rule is the level of capital that imizes consumption in steady state. This is Maximizing, c(s) = (1 s)f(k(s)), k(s) = (s/δ) 1/(1 α) c(s) = (1 s)(s/δ) α/(1 α) s g = α Question: suppose s < s g. Is it good to increase s? The answer is not, because while it increases consumption in the long run, it reduces it in the present. What if s > s g? Then it is optimal to reduce s, since it will increase consumption in the present and in the future. 2.3 Long Run Growth The fact that the model reaches a steady state implies there is no long run growth. But this is inconsistent with the data, since the US has been growing at roughly 2% per year since the mid 1850s. A solution is to introduce exogenous growth. Change the production function so that Y t = F (K t, A t L t ) 7

8 where A t+1 = (1 + γ)a t. This is called labor enhancing productivity change. Turns out this type of exogenous growth will deliver a balanced growth path, that is, in the long run, all variables will grow at a constant rate. Intuitively, the economy does not grow because one factor does not (labor), and this means the returns to the other factor (capital) are lower and lower. By increasing the productivity of labor, we increase the returns to capital, and it makes sense to keep investing more and more in the long run, generating long run growth. Dividing through by L t K t+1 = (1 δ)k t + sf (K t, A t L t ) k t+1 = (1 δ)k t + sf (k t, A t ) Thus, this time the trick does not work. Instead, divide variables in period t by L t (1+γ) t, and let ˆk t = Finally Kt L t(1+γ). t Let 1 ˆδ = 1 δ 1+γ and ŝ = s 1+γ, then K t+1 L t (1 + γ) t = (1 δ)ˆk t + sf(ˆk t ) ˆk t+1 (1 + γ) = (1 δ)ˆk t + sf(ˆk t ) ˆk t+1 = (1 ˆδ)ˆk t + ŝf(ˆk t ) This is the same as before, so we already know that ˆk converges to a constant. This means that under balanced path growth, k grows at the rate γ. Homework. Solve the model when the population grows at rate n > 0. 3 The Standard Growth Model The Standard Growth Model is the one described by Ramsey and Cass Koopmans. It delivers the optimal consumption and growth path of an economy with no uncertainty. Time is discrete and runs t = 0, 1,...,. There is an inter temporal discount factor β (0, 1) (why?). There is a single representative consumer with preferences given by the following utility function U = β t u(c t ) The feasible consumption set is simply c t 0 for all t. The technology is described by a production function y t = F (k t, h t ) 8

9 a feasibililty constraint c t + i t = y t and a law of motion k t+1 = (1 δ)k t + i t where y is output, h is labor services, k is the stock of capital, i is investment and δ is the depreciation rate. The representative consumer is endowed with k 0 units of capital today and 1 unit of time each period. 3.1 Assumptions β (0, 1) u is C 2 and strictly concave, with lim c 0 u (c) = 0 F is C 2, is strictly increasing in both arguments when (k, h) >> 0, has constant returns to scale, with F (0, h) = F (k, 0) = 0 for all h, k F is strictly concave in k and h individually lim k 0 F 1 (k, h) = for all h > 0 lim h 0 F 2 (k, h) = for all k > 0 lim k F 1 (k, h) = 0 for all h > 0 lim k F 2 (k, h) = 0 for all k > 0 δ (0, 1) 3.2 The Pareto Efficient Allocation subject to {k t,h t,c t,i t,y t} β t u(c t ) c t 0, k t 0, 0 h t 1 y t = F (k t, h t ) c t + i t = y t k t+1 = (1 δ)k t + i t k 0 given 9

10 Reducing this problem. If k 0 > 0, then h t = 1 for all t. From the restrictions, we can further reduce to subject to {k t,c t} β t u(c t ) c t 0, k t 0 Let f(k) = F (k, 1). Reduce once more to subject to {k t} k t+1 = (1 δ)k t + F (k t, 1) c t k 0 given β t u((1 δ)k t + f(k t ) k t+1 ) (1 δ)k t + f(k t ) k t+1 0, k t 0, k 0 given Remark: we can use the Weierstrass theorem to prove existence of a solution. Weierstrass says that a continuous function over a compact set will always achieve a imum. u is continuous. The restriction is not compact, but we can assume that the solution to this problem is the same as the solution to a problem where k k, which is a compact set. We just need to choose a big enough k. So existence is OK. The convexity of u gives us uniqueness. The first order condition to this problem is Simplifying, we get the Euler equation β t u (c t ) + β t+1 u (c t+1 )(1 δ + f (k t+1 )) = 0 or u(c t ) βu (c t+1 ) = 1 δ + f (k t+1 ) u((1 δ)k t + f(k t ) k t+1 ) βu ((1 δ)k t+1 + f(k t+1 ) k t+2 ) = 1 δ + f (k t+1 ) (1) Here we have choice variables. To solve, we must impose a transversality condition: or lim t βt u (f(k t + (1 δ)k t k t+1 )(f(k t + (1 δ)k t k t+1 )k t+1 = 0 lim t βt 1 u (f(k t + (1 δ)k t k t+1 )k t+1 = 0 10

11 To motivate, think of the problem with finite time. subject to {k t} T T β t u((1 δ)k t + f(k t ) k t+1 ) Taking first order conditions with respect to k T, (1 δ)k t + f(k t ) k t+1 0, k t 0, k 0 given β T u ((1 δ)k T + f(k T ) k T +1 )((1 δ) + f (k T )) 0 we know the world ends in T + 1. Complementary slackness says that if β T u ((1 δ)k T + f(k T ) k T +1 )((1 δ) + f (k T )) > 0, then k T +1 = 0, and if β T u ((1 δ)k T + f(k T ) k T +1 )((1 δ) + f (k T )) = 0, then k T +1 0, therefore, β T u ((1 δ)k T + f(k T ) k T +1 )((1 δ) + f (k T ))k T +t = 0 As T, we recover the transversality condition. Equation (??) is a difference equation of second degree. We need two border conditions to solve. One is k 0. The other one is the transversality condition. However, we generally know that the system converges to a steady state, so we can replace the transversality condition with the steady state equation. Steady state is when k t = k t+1 = k. In this case, Thus, u((1 δ)k + f(k) k) βu ((1 δ)k + f(k) k) = 1 δ + f (k) f (k) = 1 β 1 + δ The assumption that f is strictly concave, together with f (k) k = 0 and f (k) k 0 = imply that this equation has a unique solution. It is easy to show that this satisfies the transversality condition The Golden Rule The Golden Rule is the level of capital that imizes steady state consumption. We can compute this as f(k) δk f (k) = δ k The social planner chooses a lower level of capital in steady state because of the impatience of the consumer. 11

12 3.3 Descentralized Equilibrium Three types of solutions to the decentralised equilibrium Arrow Debreu Competitive Equilibrium Sequence of Markets Competitive Equilibrium Recursive Competitive Equilibrium All three give the same solution, they are simply different ways of computing the equilibrium. Before defining the equilibrium, need to answer the following questions: (i) who owns the capital? Consumers or firms?; and (ii) how many firms are there? Many or one? Assume capital is owned by the consumer and there is a single stand in firm. We can show that the solution to this problem is the same as if the firms held the capital, or there were many firms Arrow-Debreu An Arrow Debreu Competitive Equilibrium is a list of allocations for the consumer {c c t, k c t, i c t, h c t}, allocations for the firm {y f t, k f t, h f t }, and prices {p t, w t, r t } such that 1. Taking prices {p t, w t, r t } as given for all t, {c c t, k c t, i c t, h c t} solve, for all t, subject to {c t,k t,i t,h t} β t u(c t ) [p t c t + p t i t ] = [w t h t + r t k t ] k t+1 = (1 δ)k t + i t c t 0, 0 h t 1, k 0 given 2. Taking prices {p t, w t, r t } as given for all t, {y f t, k f t, h f t } solve, for all t, subject to {y t,k t,h t} [p t y t w t h t r t k t ] y t 0, h t 0, k t 0 3. Markets clear h c t =h f t k c t =k f t c c t + i c t =y f t 12

13 Arrow Debreu solves everything in period 0, and then just follows its decisions. Notice that we can simplify this. For once, we can drop the superscripts c and f. Also, if consumers choose the path for k t, then they implicitly choose i t. We can also drop y t from the list. Finally, since consumers do not value leisure, h t = 1 for all t. Given all this, an Arrow Debreu Competitive Equilibrium is a list of allocations {c t, k t, h t = 1} and prices {p t, w t, r t } such that 1. Taking prices {p t, w t, r t } as given for all t, {c t, k t } solve, for all t, subject to {c t,k t} β t u(c t ) [p t c t + p t (k t+1 (1 δ)k t )] = c t 0, k 0 given [w t + r t k t ] 2. Taking prices {p t, w t, r t } as given for all t, {k t, h t = 1} solve, for all t, subject to {k t,h t} [p t F (k t, h t ) w t h t r t k t ] h t 0, k t 0 3. Markets clear c t + k t+1 (1 δ)k t =F (k t, 1) Solution. Consumers β t u (c t ) = λp t p t = p t+1 ((1 δ) + r t+1 ) From these, u (c t ) βu (c t+1 ) = p t p t+1 = (1 δ) + r t+1 Firms F 1 (k t, 1) = r t F 2 (k t, 1) = w t Market clearing, c t = (1 δ)k t + F (k t, 1) k t+1 13

14 Combining all together, u ((1 δ)k t + F (k t, 1) k t+1) βu ((1 δ)k t+1 + F (k t+1, 1) k t+2) = (1 δ) + F 1(k t+1, 1) (2) Notice that equation (??) is the same as equation (??). We will not show it, but the transversality condition is the same as before. Thus, the two problems have the same solution, and therefore the AD competitive equilibrium is Pareto efficient. Notice though that while pt p t+1 is well defined in terms of the allocations, p t is not. We need to choose a numeraire (p 0 = 1) Sequence of Markets A Sequence of Markets competitive equilibrium is a list {c t, k t, h t = 1, b t } and prices {p t, q t, w t, r t } such that 1. Taking prices {p t, q t, w t, r t } as given for all t, {c t, k t } solve, for all t, subject to {c t,k t} β t u(c t ) p t c t + p t (k t+1 (1 δ)k t ) + q t p t+1 b t+1 = w t + r t k t + p t b t c t 0, k 0 given, b 0 given b t > B > The last expression rules out Ponzi schemes. 2. Taking prices {p t, w t, r t } as given for all t, {k t, h t = 1} solve, for all t, subject to {k t,h t} [p t F (k t, h t ) w t h t r t k t ] h t 0, k t 0 3. Markets clear c t + k t+1 (1 δ)k t =F (k t, 1) b t =0 Solution. Consumers β t u (c t ) = λ t p t λ t p t = λ t+1 p t+1 ((1 δ) + r t+1 ) λ t p t q t = λ t+1 p t+1 14

15 Remark: Notice that λ t and p t always come together. Thus, the consumer cares only about λ t p t. As such, we need to choose now a numeraire for every period. Thus, set p t = 1 for all t. From the First Order Conditions, u (c t ) βu (c t+1 ) = 1 q t = (1 δ) + r t+1 These are the same than under Arrow Debreu. Notice that in equilibrium, bond market clearing imposes b = 0, just as we would have in Arrow Debreu. Since the firm first order conditions are the same too, it is easy to show that the solution is the same. 3.4 Elastic Labor So far, leisure did not enter the utility function. Thus, as long as the wage rate in equilibrium is positive, labor supply will be inelastice and equal to 1. What if people actually like leisure? Change the utility function to β t u(c t, l t ) where l t is leisure. Now, since l t = 1 h t, where h t is hours worked, we can define the utility function as β t u(c t, 1 h t ) We can write the problem of the social planner as β t u(c t, 1 h t ) {c t,h t,k t} s.t. c t + k t+1 (1 δ)k t = F (k t, h t ) c t 0, 0 h t 1, k t 0, k 0 given. Again, we can reduce this to β t u(f (k t, h t ) k t+1 + (1 δ)k t, 1 h t ) {h t,k t} s.t. F (k t, h t ) k t+1 + (1 δ)k t 0, 0 h t 1, k t 0, k 0 given. The solution to this problem is given by the first order conditions: u 1 (F (k t, h t ) k t+1 + (1 δ)k t, 1 h t ) k t+1 : βu 1 (F (k t+1, h t+1 ) k t+2 + (1 δ)k t+1, 1 h t+1 ) = F 1(k t+1, h t+1 ) h t+1 : u 2(F (k t, h t ) k t+1 + (1 δ)k t, 1 h t ) u 1 (F (k t, h t ) k t+1 + (1 δ)k t, 1 h t ) = F 2(k t, h t ) 15

16 The first equation is the Euler condition, and as before, it is a second degree difference equation. The second equation is not intertemporal, and can usually be solved period by period as a function of k t+1, so that the sequence {k t } is the only unknown, as before. As before, we need also a transversality condition. The steady state is easy to compute as a system of 2 equations and 2 unknowns: 1 = βf 1 (k ss, h ss ) u 2 (F (k ss, h ss ) δk ss, 1 h ss ) u 1 (F (k ss, h ss ) δk ss, 1 h ss ) = F 2(k ss, h ss ) Homework: Define an Arrow Debreu Competitive Equilibrium with elastic labor and show that its solution is the same as the social planner s, both in the steady state and everywhere else. 4 Endogenous Growth The model above does not feature long run growth. However, it is simple to modify it as we did in the Solow case to deliver growth if we exogenously make a technology parameter change. Do this as a Homework. A more interesting case arises when the model endogenously delivers growth. There is a vast literature on this, and we will see the three most salient models: the AK model, Lucas model of human capital, and Romer s model of spillovers, or Kk model. 4.1 AK Model There is a representative consumer with preferences β t u(c t ), u(c) = c1 σ 1 1 σ, σ > 0 Representative firm with technology: Law of motion: Y t = AK t Endowment: K t+1 = (1 δ)k t + I t K 0 given. Solve for the social planner solution: β t u(c t ) s.t. {k t} c t + K t+1 (1 δ)k t = AK t c t 0, K t 0, K 0 given. 16

17 FOC: Look for a steady state: u (c t ) βu (c t+1 ) = A + 1 δ 1 = (A + 1 δ)β This can happen, but just if the parameters are right. Otherwise, this will not hold, and there is no steady state. The alternative is to look for a balanced growth path, where capital grows at rate g. Recalling the FOC, Replacing by the functional form for u: u (AK t K t+1 + (1 δ)k t ) βu (AK t+1 K t+2 + (1 δ)k t+1 ) = A + 1 δ u (AK t gk t + (1 δ)k t ) βu (AgK t g 2 K t + (1 δ)gk t ) = A + 1 δ u (K(A g + (1 δ))) βu (gk(a g + (1 δ))) = A + 1 δ (K(A g + (1 δ))) σ (gk(a g + (1 δ))) σ = gσ = A + 1 δ Thus, there is balanced growth path, where capital grows at rate g = (A + 1 δ) 1/σ. Notice that the only thing determined in the long run is g, so there are a continuum of balanced growth paths depending on K 0. This is an unrealistic example, but it is simple, and illustrates the essential elements of a model with endogenous growth. The first important element is a utility function that allows for balanced growth, as the one we used. But the most important element is that there are no decreasing returns to scale in the accumulable inputs (K in this case). We will see two more examples, but the logic is always the same. 4.2 Lucas Human Capital Model There is a representative consumer with preferences Representative firm with technology: β t u(c t ), u(c) = c1 σ 1 1 σ, σ > 0 Y t = F (k t, e t ), k t 0, e t 0 where F features constant returns to scale, and E is efficiency units of labor. specifically, More e t = h t s t 17

18 where h is hours spent working and s is the level of skills, and this can be accumulated. Endowments: 1 unit of labor k 0 units of capital s 0 units of skills Solve the social planner s problem: FOC: {k t},{s t} c t + i kt + i st = Y t k t+1 = (1 δ k )k t + i kt s t+1 = (1 δ s )s t + i st β t u(f (k t, s t ) [k t+1 (1 δ k )k t ] [s t+1 (1 δ s )s t ]) u (c t ) k t : βu (c t+1 ) = F 1(k t+1, s t+1 ) + 1 δ k u (c t ) s t : βu (c t+1 ) = F 2(k t+1, s t+1 ) + 1 δ s Together with two transversality conditions and two initial conditions: s 0 and k 0 given. In steady state, 1 β = F 1(k ss, s ss ) + 1 δ k = F 2 (k ss, s ss ) + 1 δ s This looks like 2 equations and two unknowns, but it is not really. The reason is that F 1 and F 2 are homogeneous of degree 0, and as such, do not depend on k and s, but on the ratio of k/s. Thus, it is 2 equations and 1 unknown. That is, ( ) kss F 1 (k ss, s ss ) = F 1, 1 s ss = F 2 ( kss s ss, 1 As in the AK model, only by chance will these equations hold all together. To solve, we need to assume growth of k and s. Look for a path where s and k grow at a common rate γ, and so does c. γ σ β = F 1 ( ) ( ) kss kss, δ k = F 2, δ s s ss s ss Now we have two equations and 2 unknowns. Note again that k ss and s ss are not determined. Rather, their ratio is, so there is a continuum of balanced growth paths. 18 )

19 4.3 Comparing AK to Lucas Recall that in the AK model, starting from whichever level of K 0, the economy was in a balanced growth path. This is not the case in Lucas, but it is similar. The economy might not start with the balanced growth path level of k/s. But it will only accumulate the factor that is lacking, and let the other one depreciate, until the proper ratio is reached. From then on, the economy will grow at a steady rate. 4.4 Romer s Kk model So the problem of traditional models in generating sustained growth is that the accumulable factors exhibit decreasing returns to scale. This would imply that to generate growth, we can simply use a function with increasing returns to scale. But the problem is that then it is hard to define a competitive equilibrium, since only one firm would produce everything and become a monopolist. A solution devised by Romer is to use increasing returns to scale, but have firm owners behave as if they had decreasing returns. This is done by assuming that there are spillovers to knowledge. Basically, each time a firm increases its stock of capital, it discovers something new that everyone else can use, increasing everyone s productivity. The idea hinges on the technology, which is Y t = F (k t, K t ) where k t is the level of capital of one particular firm and K t is a measure of the general level of capital, and cannot be affected by a single firm. In this model there is no representative firm. Rather, there are N firms, all producing the same good. There is a representative consumer with preferences There are N firms, with technologies given by β t u(c t ), u(c) = c1 σ 1 1 σ, σ > 0 y t = F (K i,t, k it ) k it is knowledge of firm i in period t, K i,t = j i k j,t. Let K t = i k it. F has CRS in both arguments, and DRS in each argument. INADA conditions hold for k it given a positive level of K it. F 12 > 0 and F is C 2 and strictly concave on each argument. The law of motion is Feasibility is k i,t+1 = (1 δ)k it + i it c t + N i it = i=1 N i=1 y it Endowments are k i0 given. Assume k i0 = k 0 for all i to guarantee an interior solution for all i when i it 0. 19

20 The social planner problem can be written as ( N ) N β t u F (k it, K it ) (k it+1 (1 δ)k it ) {k it } i=1 i=1 Proposition 2 k it = k t for all i = 1,..., N Proof: Homework. Given this proposition, the problem of the social planner can be simplified to {k t} β t u (N(F (k t, (N 1)k t ) (k t+1 (1 δ)k t ))) The first order condition is u (c t ) βu (c t+1 ) = [F 1(k t+1, (N 1)k t+1 ) + (N 1)F 2 (k t+1, (N 1)k t+1 ) + (1 δ)] As before, a steady state might not exist. This is because the right hand side of the equation above does not depend on k t, given that the functions F 1 and F 2 are homogeneous of degree zero. Thus, we look for a balanced growth path where The equilibrium γ solves k t+1 = (1 + γ)k t, c t+1 = (1 + γ)c t (1 + γ) σ β = [F 1 (k, (N 1)k) + (N 1)F 2 (k, (N 1)k) + (1 δ)] Once again, there are multiple balanced growth paths, and starting from any k 0, an economy is immediately in a balanced growth path. What about a decentralized equilibrium? Suppose the firms accumulate capital and are owned by the consumers to define an Arrow Debreu Decentralized Equilibrium: An ADCE is a list of sequences {c t, π it, k it, p t } such that Consumer. Taking {π it, p t } as given, {c t } solves s.t. {c t} β t u(c t ) p t c t = c t 0 N i=1 π it 20

21 Firms. Taking {p t } as given, {π it, k it} solve Market clearing. s.t. k it p t [F (k it, K it ) k i,t+1 + (1 δ)k it ] k it 0, k i,t+1 (1 δ)k it, k i0 given. π it = p t [ F (k it, K it ) k i,t+1 + (1 δ)k it ] c t + Characterizing the ADCE: Consumer s FOC: N (k i,t+1 ) (1 δ)k it = i=1 N F (kit, K it ) i=1 u (c t ) βu (c t+1 ) = p t p t+1 Firms. p t = p t+1 [F 1 (k i,t+1, K 1,t+1 ) + 1 δ] p t p t+1 = F 1 (k i,t+1, K i,t+1 ) + 1 δ Proposition 3 k it = k t for all i = 1,..., N Proof: Homework. Combining the FOC of the consumer with the firm s: Look for a balanced growth path: u (c t ) βu (c t+1 ) = F 1(k t+1, (N 1)k t+1 ) + 1 δ (1 + γ) σ β = F 1 (k, (N 1)k) + 1 δ Notice that the solution is not the same as the social planner s. The reason is that knowledge creates a spillover that the central planner internalizes, but the competitive equilibrium does not. Clearly, the rate of growth chosen by the social planner is larger than the equilibrium rate of growth. A subsidy to knowledge could improve welfare in this situation. 21

22 5 Recursive Problems Recall the central planner problem of optimal growth. The central planner solved, at t = 0, the entire sequence of {c t, k t }. Another option would rely on contingencies. Suppose we know that {k t } is the optimal path for capital. What happens if we were to reoptimize at t = 14? That is, if someone let us change our original decisions? Naturally, we would look at k 14 and then choose k 15 = k 15, that is, we choose not to change. This suggests that a different approach would be to choose a policy function (or contingent rule) as our solution. That is, a function that says how much k should be in the future based on what it is now. Consider the following finite time problem. subject to {s t,a t} T β t u(s t, a t ) g(s t, a t ) 0 t = 0, 1,..., T s t+1 = h(s t, a t ) t = 0, 1,..., T 1 s 0 given In order to solve this problem, we need to know the value of s 0. This way s t is the value of the state at time t: it provides all the information we need to know at time t to solve the problem in the future. a t is the value of the action at time t. Example: finite horizon standard growth model. s t k t a t c t g(s t, a t ) c t 0, k t 0 h(s t, a t ) f(k t ) c t + (1 δ)k t The way to solve this is by backward induction. Given s T, the problem is subject to a T u(s T, a t ) g(s T, a T ) 0 Let a T = π 0 (s T ) be the solution to this problem and let V 0 (s T ) = u(s T, π 0 (s T )). In period T 1, the problem is, given s T 1, subject to u(s T 1, a T 1 ) + βv 0 (s T ) s T,a T 1 g(s T 1, a T 1 ) 0 s T = h(s T 1, a T 1 ) 22

23 But given the function h, this is the same as solving subject to a T 1 u(s T 1, a T 1 ) + βv 0 (s T ) g(s T 1, a T 1 ) 0 s T = h(s T 1, a T 1 ) since this determines also s T. Again, let a T 1 = π 1 (s T 1 ) and V 1 (s T 1 ) = u(s T 1, π 1 (s T 1 ))+ βv 0 (h(s T 1, π 1 (s T 1 ))). In T 2, the problem is More generally subject to a T 2 u(s T 2, a T 2 ) + βv 1 (s T 1 ) g(s T 2, a T 2 ) 0 s T 1 = h(s T 2, a T 2 ) V t (s t ) = a t u(s t, a t ) + βv T t 1 (s t+1 ) subject to g(s t, a t ) 0 s t+1 = h(s t, a t ) and the solution is a t = π t (s t ). In this way, we solve T problems with only one variable each. If we had done Arrow Debreu, we would have 1 problem with t variables. Many times, especially with uncertainty, it is easier to do the T problems with 1 variable each. This is called the curse of dimensionality. What happens as t? Then V T 1 V T. The problem becomes V (s t ) = a t u(s t, a t ) + βv (s t+1 ) subject to g(s t, a t ) 0 s t+1 = h(s t, a t ) This is a functional equation. The unknowns are entire functions, not variables. In the standard growth model this is V (k t ) = c t u(c t ) + βv (k t+1 ) subject to c t 0, k t 0 k t+1 = f(k t ) c t + (1 δ)k t We can solve a problem this way when it is recursive. That is, when we solve the same problem every period, and the only thing changing is the state variable. 23

24 6 Bellman Equations V (s) = U(s, s ) + βv (s ) s s.t. s Γ(s) How do we find the function V (s)? Consider the right hand side of this equation. For an arbitrary function f(s), we can easily compute s U(s, s ) + βf(s ). In particular, for each s, this operator will deliver some value. This RHS can be seen as mapping a real valued function into a real valued function. We represent this as T : F F In particular, applying this operator to the function f, f = T f. Thus, the Bellman equation can be written as V = T V That is, V is a fixed point of the operator T. To know something about the existence and uniqueness of a solution, we need to appeal to the contraction mapping theorem. 6.1 Contraction Mapping Theorem Definition. Contraction mapping. Let (M, d) be a complete metric space. Then T : M M is a contraction mapping if there exists some β (0, 1) such that d(t x, T y) βd(x, y) for all (x, y) M. If T is a contraction mapping and ˆβ is the minimum value that fits with the data, then we say that T is a contraction of modulus ˆβ. In words, applying the contraction shrinks the distance between x and y. Theorem. Contraction Mapping. If (M, d) is a complete metric space and T : M M is a contraction mapping of modulus β, then: 1. T has a unique fixed point m M 2. Let m 0 be an arbitrary point in M. Let m 1 = T m 0, m 2 = T m 1,..., then m = lim t m t This theorem provides existence, uniqueness, and how to find the fixed point. Is the Bellman equation a contraction mapping? Let B(s) be the set of continuous, bounded, real valued functions on S R m. For any f, g B(s), define d(f, g) as the supreme norm, that is, d(f, g) = sup s S f(s) g(s). We can show that (B(s), d) is a complete metric space. Our Bellman equation is f (s) = u(s, s ) + βf(s ) subject to s Γ(s) s 24

25 For the Bellman operator to be a contraction, we need to show that f(s), f (s) B(s). Since we choose f, choose f B, so we need to show that f is continuous, bounded, and real valued. The Theorem of the Maximum guarantees that if u is bounded and continuous, and Γ is compact, then a solution exists and the value function is continuous and compact, and therefore f B. 6.2 Blackwell Sufficient Conditions The Blackwell Sufficient Conditions are sufficient conditions for a contraction. Thus, to show that T is a contraction, one might only show that it satisfies these conditions. Let B(s) be the set of bounded, real valued functions on S. Let T : B(s) B(s). T is a contraction of modulus β if it satisfies the following conditions: Monotonicity. For all f, g B(s), f(s) > g(s) s S T F (s) T g(s) s S. Discounting. There exists some β (0, 1) : f B(s), a R +, T (f + a)(s) T f(s) + βa s S, where (f + a)(s) = f(s) + a. For a proof, see Chapter 3 of Stokey Lucas Prescott. To show that the Blackwell sufficient conditions are satisfied in our example, recall T f(s) = U(s, s ) + βf(s ) s s.t. s Γ(s) where T : C(s) C(s) (set of continuous functions). Monotonicity. some s, Let f(s), g(s) C(s), and assume that f(s) g(s) s S. For T g(s) = U(s, s ) + βg(s ) s s.t. s Γ(s) Let s be the imizer. Then T g(s) = U(s, s ) + βg(s ). Since s Γ(s), T f(s) u(s, s ) + βf(s ) u(s, s ) + βg(s ) = T g(s) Since this holds for all s S, this proves monotonicity. Discounting. Let f(s) C(s), and let a R +. For an arbitrary s S, T (f + a)(s) = s Γ(s) U(s, s ) + βf(s ) + βa = T f(s) + βa 25

26 6.3 Properties of the Bellman Equation We often want to show that V is increasing, or concave. This section establishes a method to show different properties of V. The idea is to start with an initial guess f that has the property we are interested in, and showing that T f also has that property. Then T (T (f)) = T 2 f also has it. Since V = lim n T n f, if the property is preserved in the limit, V has the property. Let P be a closed subset of C(s). Let P P, P not necessarily closed. If T maps P into itself, we know V P. This is because conditions on a closed set are preserved in the limit. But then if T maps P into itself, this does not imply that V P. However, if T maps P into P, then V P. To see this, we can easily show V = lim n T n f P. Apply the operator once more. Since V P, then T V P. Since V = T V, V P Monotonicity of the Value Function Assume: U(s, s ) is weakly increasing in s. V (s) = s Γ(s) U(s, s ) + βv (s ) Γ(s) is weakly increasing in s, i.e., for any s 1 s 2 Γ(s 2 ) Γ(s 1 ) Proposition 4 T maps the set of increasing functions into itself. Proof: Let f C(s) be an increasing function. For any (s 1, s 2 ) S : s 1 s 2, T f(s 2 ) = s Γ(s 2 ) U(s 2, s ) + βv (s ) Let s be the imizer. Since s Γ(s 2 ) Γ(s 1 ) T f(s 1 ) U(s 1, s ) + βv (s ) U(s 2, s ) + βv (s ) = T f(s 2 ) Corollary 1 To show strict monotonicity, assume that U(s, s ) is strictly increasing in s and replace the weak inequalities by strict ones, then T f(s 1 ) U(s 1, s ) + βv (s ) > U(s 2, s ) + βv (s ) = T f(s 2 ) Concavity of the Value Function Assume: U(s, s ) is jointly concave in s and s. For all (s 1, s 2 ), θ [0, 1], s 1 Γ(s 1 ), s 2 Γ(s 2 ) θs 1 +(1 θ)s 2 Γ(θs 1 +(1 θ)s 2 ) 26

27 Proposition 5 T maps the set of concave functions into itself. Proof: Let (s 1, s 2 ) S, θ [0, 1] and f be weakly concave. Let s 1 and s 2 be the imizers. θt f(s 1 ) + (1 θ)t f(s 2 ) = θ(u(s 1, s 1) + βf(s 1)) + (1 θ)(u(s 2, s 2) + βf(s 2)) = θu(s 1, s 1) + (1 θ)u(s 2, s 2) + β(θf(s 1) + (1 θ)f(s 2)) U(θs 1 + (1 θ)s 2, θs 1 + (1 θ)s 2) + βf(θs 1 + (1 θ)s 2) T f(θs 1 + (1 θ)s 2 ) since θs 1 + (1 θ)s 2 Γ(θs 1 + (1 θ)s 2 ). Corollary 2 To show strict concavity, assume that U(s, s ) is strictly jointly concave in s and s and replace the weak inequalities by strict ones, then θt f(s 1 ) + (1 θ)t f(s 2 ) = θ(u(s 1, s 1) + βf(s 1)) + (1 θ)(u(s 2, s 2) + βf(s 2)) = θu(s 1, s 1) + (1 θ)u(s 2, s 2) + β(θf(s 1) + (1 θ)f(s 2)) < U(θs 1 + (1 θ)s 2, θs 1 + (1 θ)s 2) + βf(θs 1 + (1 θ)s 2) T f(θs 1 + (1 θ)s 2 ) The Benveniste Scheinkman Theorem What is the derivative of the value function? The Benveniste Scheinkman theorem is an application of the envelope theorem. Let s = h(s) be the policy function. When the value function is differentiable, V (s) =U 1 (s, h(s)) + U 2 (s, h(s))h (s) + βv (h(s))h (s) =U 1 (s, h(s)) + h (s)[u 2 (s, h(s)) + βv (h(s))] The term between brackets is zero, given by the first order conditions. Thus, the theorem says V (s) = U 1 (s, h(s)) When is the value function differentiable? We know that a strictly concave function is differentiable almost everywhere, so we can work with strictly concave functions. 6.4 The Standard Growth Model Setting up the planner s problem recursively Assumptions: V (k) = k u(f(k) k + (1 δ)k) + βv (k ) u is strictly increasing, strictly concave and differentiable f is strictly increasing, strictly concave and differentiable 27

28 We can prove that V is strictly increasing and strictly concave. Then by the imum theorem the policy rule is single valued (a function) and V is differentiable. Let k = g(k) be the policy function. Proposition 6 g is strictly increasing. Proof: The proof works by contradiction. Suppose for some k 1 > k 2, g(k 1 ) g(k 2 ). The first order conditions are Since k 1 > k 2 and g(k 1 ) g(k 2 ), u (f(k 1 ) g(k 1 ) + (1 δ)k 1 ) = βv (g(k 1 )) u (f(k 2 ) g(k 2 ) + (1 δ)k 2 ) = βv (g(k 2 )) f(k 1 ) g(k 1 ) + (1 δ)k 1 > f(k 2 ) g(k 2 ) + (1 δ)k 2 u (f(k 1 ) g(k 1 ) + (1 δ)k 1 ) < u (f(k 2 ) g(k 2 ) + (1 δ)k 2 ) V (g(k 1 )) < V (g(k 2 )) Since V is strictly concave, this implies g(k 2 ) < g(k 1 ), which is a contradiction. Proposition 7 The sequence {k 0, k 1, k 2,... } is either strictly increasing, strictly decreasing, or constant. Proof: Homework. Proposition 8 c(k) is a strictly increasing function. Proof: Homework The Steady State The first order conditions are u (f(k) g(k) + (1 δ)k) = βv (g(k)) Using Benveniste Scheinkman, V (g(k)) = u (f(g(k)) g(g(k)) + (1 δ)g(k))(f (g(k)) + (1 δ)) Thus, u (f(k) g(k) + (1 δ)k) = βu (f(g(k)) g(g(k)) + (1 δ)g(k))(f (g(k)) + (1 δ)) In steady state, g(k ss ) = k ss, so 1 = β(f (k ss ) + (1 δ)) Thus, the solution is the same as we had before. 28

29 6.5 Stochastic Dynamic Programming The deterministic Dynamic Programming Problem is β t u(s t, s t+1 ) s.t. s t+1 Γ(s t ) s 0 given. The stochastic Dynamic Programming Problem is E β t u(s t, s t+1, ɛ t ) s.t. s t+1 Γ(s t, ɛ t ) s 0, ɛ 0 given. We also need to specify the process for ɛ t. Let Q(ɛ t, ɛ t+1 ) be the probability of the shock being ɛ t+1 tomorrow given that today it is ɛ t. In recursive form, V (s, ɛ) = u(s, s Γ(s,ɛ) s, ɛ) + βev (s, ɛ ) = u(s, s Γ(s,ɛ) s, ɛ) + β V (s, ɛ )Q(ɛ t, ɛ t+1 )dɛ t+1 s is the endogenous state variable and ɛ is the exogenous state variable. In theory, the way to find the value function is the same as before, by making some initial guess and iterating. Except that now we are not interested in monotonicity or concavity properties of V with respect to ɛ. 7 Search Theory Consider the problem of a worker who is either unemployed receiving unemployment benefits or employed at a given wage. For now assume that if he is employed, he does not choose anything. Works at a given wage, and never gets fired. If he is unemployed, every period he receives a wage offer and he has to choose whether to accept it or not. His preferences are given by βt c t. When unemployed, every period he receives benefits b and draws a wage w from the distribution F (w), with associated density f(w). If he accepts the offer, he will become employed next period. Otherwise, he will enter the following period again as an unemployed. Formulate this problem recursively. Consider the unemployed. Let V (w) be the future discounted value of having an offer w. { } w V (w) = 1 β, b + βev (w ) (3) Let U be the value of being unemployed before getting an offer. { } w U = E 1 β, b + βu (4) 29

30 Consider the Bellman equation defined in equation (??). Show this is a contraction mapping. Proceed in two steps: Step 1. Show it maps the set of bounded, continuous functions into itself. Pick some bounded, continuous function g : [0, w] R. { } { w w w } T g(w) = 1 β, b + βeg(w ) = 1 β, b + β g(w )f(w )dw 0 T g is bounded and continuous. Step 2. Show the operator satisfies the Blackwell sufficient conditions. 1. Monotonicity. Take two bounded, continuous functions g(w), h(w) such that g(w) h(w) w [0, w]. { } { } w w T h(w) = 1 β, b + βeh(w ) 1 β, b + βeg(w ) = T g(w)! Tg(w)&! +!"(!! )! Th(w)&! 1!!! +!h(!! )! 2. Discounting. Take a bounded, continuous function g(w) and a number a R +, { } w T (g + a)(w) = 1 β, b + β (g(w ) + a)f(w )dw { } w = 1 β, b + β g(w )f(w )dw + βa { } w 1 β, b + β g(w )f(w )dw + βa = T g(w) + βa Thus, T is a contraction mapping. Homework. Prove that U as defined in equation (??) is a contraction mapping. Next, let s characterize the optimal policy rule. To do this, first we need to find the value function by iteration. Once we have this, { } w V (w) = 1 β, b + βev (w ) 30

31 w 1 β V(w) b + EV(w! ) w* The policy function is to accept any wage offer with w > w, where w = b + βev (w ) Graphically, This implies that the time spent in unemployment is a random variable. The probability that a worker who is unemployed today will be employed tomorrow is 1 F (w ). What is the expected duration of unemployment? Let p = 1 F (w ) be the probability that search results in an offer being accepted. Probability of being unemployed for one period p Probability of being unemployed for two periods (1 p)p Probability of being unemployed for three periods (1 p) 2 p. Probability of being unemployed for n periods t=1 t=1. (1 p) n 1 p Expected duration = t(1 p) t 1 p = p t(1 p) t 1 = p t=1 (1 p)t p 7.1 Comparative Statics = p (1 p) (1 p)t p = p 1 p = 1 2 p = p 1 p 1 (1 p) p = p 1 p p p The parameters in the model are β, b and F (w). How do changes in these affect the reservation wage? The reservation wage is determined by w 1 β = b + βev (w ) = b + β 31 w 0 V (w)f(w)dw (5)

32 Since V (w) depends on the parameters, we cannot simply differentiate this expression. We first need to write V as a function of the parameters. From equation (??), we can write the value function as Thus, EV (w ) = V (w) = w Replacing this in equation (??), Subtract βw from both sides (1 β)w = b(1 β) + β w = b + β 1 β 0 { } w 1 β, w 1 β = 1 1 β {w, w } V (w )f(w )dw = 1 w {w, w } f(w )dw 1 β 0 w 1 β = b + β w 1 β w = b(1 β) + β w 0 w Next, integrate by parts. Recall 0 w {w, w } f(w )dw 0 {w, w } f(w )dw {w w, 0} f(w )dw = b(1 β) + β w w (w w )f(w )dw w (w w )f(w )dw (6) b a udv = uv b a Let u = 1 F (w ) and v = w w. Then, b a vdu Notice w w (1 F (w ))dw = (1 F (w ))(w w ) w w + (w w )f(w )dw w w (1 F (w ))(w w ) w w = (1 F ( w))( w w ) (1 F (w ))(w w ) = 0 so Therefore w w (w w )f(w )dw = w = b + w w (1 F (w ))dw β w (1 F (w ))dw (7) 1 β w 32

33 This equation determines w as a function of the parameters. A change in b. Rewrite equation (??) as w = b + β [ w w ] w f(w )dw w f(w )dw 1 β w w Differentiating this expression, and noticing that the derivative of the limit of integration is multiplied by zero, w (b) b w (b) b w = 1 w (b) f(w )dw b w (b) [ w ] 1 + f(w )dw = 1 w (b) So w (b) > 0. Intuitively, if the benefits go up, I don t mind waiting a bit more in b unemployment, so I wait for better offers. A change in β. Homework. A change in F (w). Model a mean preserving increase in spread. That is, the mean wage rate stays constant, but the variance increases. Let F (w, θ) be a family of distributions on [0, w] for θ [0, 1] and f(w, θ) the associated density. Assumptions: Mean preserving. For all θ 1, θ 2 [0, 1], w 0 wf(w, θ 1 )dw = w 0 wf(w, θ 2 )dw Single crossing property. Take any θ 1, θ 2 [0, 1], θ 1 > θ 2. Then there exists some ŵ [0, w] such that F (w, θ 1 ) >F (w, θ 2 ) w [0, ŵ) F (w, θ 1 ) <F (w, θ 2 ) F (ŵ, θ 1 ) =F (ŵ, θ 2 ) Intuition: the larger the θ, the larger the variance. These properties imply that w 0 w Homework. Prove these implications. First notice 0 w (ŵ, w] (1 F (w, θ))dw = constant for all θ (8) (F (w, θ 1 ) F (w, θ 2 ))dw 0 w and θ 1 θ 2 (9) w 0 (1 F (w, θ 1 ))dw = w 0 (1 F (w, θ 2 ))dw 33

34 Next, w From (??) we know w Rewriting equation (??) 0 (1 F (w, θ 1 ))dw w 0 w w 0 (1 F (w, θ 2 ))dw = (1 F (w, θ 1 ))dw w (1 F (w, θ 1 ))dw 0 (1 F (w, θ 1 ))dw w w = b + w w (1 F (w, θ))dw θ w 0 w (1 F (w, θ 2 ))dw + w (1 F (w, θ 2 ))dw 0 (1 F (w, θ 2 ))dw 0, which implies > 0 β w (1 F (w, θ))dw 1 β w An increase in θ increases the RHS. Thus, w > 0. Intuitively, an increase in variance θ (risk) increases the reservation wage. An increase in variance is good for the worker. This is because the worker only cares about the top of the distribution. And with more variance, higher wages will come more often. 7.2 Endogenous Job Destruction Suppose we allow for quits. The value of having a wage offer w is V (w) = {w + βv (w), b + βev (w)} Is the solution a reservation wage w? In the case without quits we knew this. The value function was { } w V (w) =, b + βev (w) 1 β We know that the first term is strictly increasing and the second term a constant. With quits, the second term is a constant, but is the first term strictly increasing? If V (w) is a contraction, a sufficient condition is that V (w) is increasing. To show this, make an initial guess V 0 (w) increasing. Then V 1 (w) is increasing. Iterating, V (w) is increasing. Thus, the solution is again a reservation wage. Suppose we want to find V (w) numerically. Start by guessing V 0 (w) = 0. Iterating, V 0 (w) = 0 V 1 (w) = {w + βv 0 (w), b + βev 0 (w )} = {w, b} V 2 (w) = {w + βv 1 (w), b + βev 1 (w )}. V (w) = {w + βv (w), b + βev (w )} 34

35 Now let s change the model slightly. Assume that the representative consumer lives for only T periods. We can find the value function by backward iteration. V 0 (w) = {w, b} V 1 (w) = {w + βv 0 (w), b + βev 0 (w )} V 2 (w) = {w + βv 1 (w), b + βev 1 (w )}. V (w) = {w + βv (w), b + βev (w )} Thus, we get the same as if time would run to infinity and the initial guess is V 0 (w) = 0, except that the number of iterations in this case is finite. 8 Deterministic Recursive Competitive Equilibrium A problem with solving for the competitive equilibrium recursively is that we need a way to represent prices recursively. To represent prices, we use the knowledge that prices depend on allocations (for example, the capital stock). Thus, we need to keep track of those variables that matter to determine prices. These are the state variables. There are two types of such variables: individual state variables, and aggregate state variables. The individual are the ones that the agent controls directly. The aggregate state variables are those that determine prices, and are the distribution of the individual state variables. In the standard growth model, the individual state variable is the level of capital k. We denote the aggregate state variable by K. In equilibrium, since there is a representative household, k = K. Take for example the Arrow Debreu price for capital r t. Representing this recursively, we write it as a function r(k). The aggregate state variable follows a law of motion K (K), determined in equilibrium. Thus, with the functions r(k) and K (K) we can determine the entire sequence {r t }. Notice that the individual s decisions depend on their capital stock and the prices. Thus, they depend both on k and on K. A Recursive Competitive Equilibrium for the standard growth model is a list of functions V (k, K), k (k, K), K (K), w(k) and r(k) such that 1. (Consumer imization.) Taking the functions K (K), w(k) and r(k) as given, V (k, K) solves V (k, K) = u(w(k) + r(k)k (k (1 δ)k)) + βv (k, K (K)) k subject to w(k) + r(k)k (k (1 δ)k) 0 and k (k, K) is an optimal decision rule. 2. (Firm imization.) Taking w(k) and r(k) as given, k = K and h = 1 are a solution to F (k, h) r(k)k w(k)h s.t. k 0, h 0 k,h 35

36 3. (Consistency.) For all K, k (k, K) = K (K) 4. Markets clear. For all K, w(k) + r(k)k = F (K, 1) 8.1 A Second State Variable: Bond Market Suppose we add bonds to this economy. This is a second state variable, since we need to know the bond holdings of the consumer to determine their optimal choices. Let b, B be the individual and aggregate state variables. A Recursive Competitive Equilibrium for the standard growth model with bonds is a list of functions V (k, b, K, B), k (k, b, K, B), b (k, b, K, B), K (K, B), B (K, B), w(k, B), r(k, B) and q(b, K) such that 1. (Consumer imization.) Taking the functions K (K), w(k) and r(k) as given, V (k, K) solves V (k, b, K, B) = k,b u(w(k, B) + r(k, B)k + b (k (1 δ)k) q(k, B)b )+ subject to βv (k, b, K (K, B), B (K, B)) w(k, B) + r(k, B)k + b (k (1 δ)k) q(k, B)b 0 and k (k, b, K, B) and b (k, b, K, B) are optimal decision rules. 2. (Firm imization.) Taking w(k, B) and r(k, B) as given, k = K and h = 1 are a solution to F (k, h) r(k, B)k w(k, B)h s.t. k 0, h 0 k,h 3. (Consistency.) For all K, B, k (k, b, K, B) = K (K, B) B (k, b, K, B) = B (K, B) 4. Markets clear. For all K, B, w(k, B) + r(k, B)k = F (K, 1) B (K, B) = B 36

37 9 Stochastic Recursive Competitive Equilibrium 9.1 Building Intuition To establish ideas, start with a deterministic endowment economy with many goods. There are J goods, denoted by j, j = 1,..., J and I consumers, i = 1,..., I. An Arrow Debreu competitive equilibrium for this economy is an allocation {c i R J +} and a price vector p R J + such that 1. Taking p as given, {c i } solve 2. Markets clear. For all j, u i (c i ) s.t. p c i = p ω i I c ij = i=1 I i=1 ω ij Now think of this economy dynamic, with t = 0, 1,..., T. The way to solve this is by defining a commodity as a good in a point in time, so the problem is the same as before, except that now there are J T commodities. We can reduce the commodity space by assuming that only present time markets are open at any point in time, and adding a market for borrowing and lending. Now go back to the static problem, but assume there are S states of nature, denoted by s = 1,..., S. Let I s = {1,..., S}. Consumers make their decisions before the state is realized. Each individual is described by a pair u i (c i, s), ω i (s), where u i : R J + I s and ω i (s) R J + for each s I s. Let π s be the probability that state s occurs. Thus, S s=1 π s = 1, π s 0 s I s. Also, assume each individual can purchase an insurance i(s) in terms of good 1. By convention, i(s) > 0 is a net payment that the individual must make in state s and i(s) 0 is a net income that the individual receives in state s. An Arrow Debreu competitive equilibrium for this economy is an allocation {c i (s), i = 1,..., I} and a price vector p R J S + such that 1. Taking p as given, {c i } solve S π s u i (c i (s), s) s=1 s.t. J p j(s)c ij (s) + p 1 (s)i(s) = j=1 J p j(s)ω ij (s) s = {1,..., S} j=1 2. Markets clear. For all j, s, I c ij(s) = i=1 I ω ij (s) i=1 37

38 Example: Suppose I = 2, J = 1, S = 2, u i (c, s) = log(c) and 1 if i = 1, s = 1 0 if i = 1, s = 2 ω is = 0 if i = 2, s = 1 1 if i = 2, s = 2 Normalize the price is state 1 as 1, so that p is the price in state 2. Consumer decisions are Adding this to market clearing, p = 1, so c 1 (1) = 1 2, c 1(2) = 1 2p, c 2(1) = 1 2p, c 2(2) = 1 2 c 1 (1) = 1 2, c 1(2) = 1 2, c 2(1) = 1 2, c 2(2) = 1 2 Consumption is the same regardless of the state. Consumers insure themselves. This implies that we can think of an alternative specification of this problem with insurance. Let i 1 be the amount of insurance purchased by consumer 1. That is, the amount of consumption if state 2 should occur. Let q be its price. Notice that q cannot be indexed by state because it is purchased before the state is realised. Consumer 1 solves 1 i 1 2 log(c 1(1)) log(c 1(2)) subject to c 1 (1) = 1 qi 1 c 1 (2) = i 1 Consumer 2 solves 1 i 2 2 log(c 2(1)) log(c 2(2)) subject to c 2 (1) = qi 2 c 2 (2) = 1 i 2 From the budget constraints, we can eliminate i i to obtain c 1 (1) = 1 c 1 (2) i 1 = i 2 Thus, if we think of c 1 (1) as a different good than c 1 (2), then q is the relative price. In equilibrium, q = 1 and c 1 (1) = 1/2 = c 1 (2), as before. 9.2 Deterministic Dynamic vs. Stochastic Static Assume there are I individuals, J goods, and T periods. A sequence of markets competitive equilibrium for this economy is an allocation {c ijt, b t } and a price vector {p jt, q t } such that 1. Taking p as given, {c ijt, b t } solve T β t u i (c it ) s.t. J p jtc ijt + qt b t+1 b t = j=1 J j=1 p jtω ijt 38

39 2. Markets clear. For all j, t, I c ijt = i=1 I i=1 ω ijt Now assume that there is only one period, but S possible states of nature. An Arrow Debreu competitive equilibrium for this economy is an allocation {c ijs} and a price vector {p js} such that 1. Taking p as given, {c ijs} solve T π(s)u i (c i (s)) s.t. J p jsc ijs + p 1 (s)i(s) = j=1 J p jsω ijs s = {1,..., S} j=1 2. Markets clear. For all j, s, I c ijs = i=1 I i=1 ω ijs These two economies are very similar: one need only to replace the t by s. Next, we want to combine these two to set up a dynamic, stochastic general equilibrium framework. 9.3 Dynamic Stochastic General Equilibrium One possibility is to define an Arrow Debreu equilibrium. One only needs to define one commodity as the good in per period and per state of nature. The problem is that the system becomes unmanageable very quickly. Consider a simple economy with one good and 2 states of nature. In period 1 there are 2 commodities. In period 2, 4 commodities. In period 3, 8 commodities. In period 4, 16 commodities. In summary, the number of commodities in period t is equal to 2 t. This can explode very easily. This is why we might want to use recursive techniques. Recursively, one chooses only between two possible states each period. Recall the curse of dimensionality: in this case we might be better off solving many small problems than one huge one. Modify the standard growth model by adding technology shocks. The technology of the representative firm is y t = z t F (k t, h t ), and z t follows a Markov process with evolution given by Q(z t, z t+1 ). The social planner problem is V (k, z) = k u(zf (k, 1) k + (1 δ)k) + βe[v (k ) z] V (k )Q(z, z )dz V (k, z) = u(zf (k, 1) k + (1 δ)k) + β k 10 Real Business Cycles A business cycle is an increase or decrease in economic activity over time. But not all changes are business cycles. We need to distinguish between the trend and the cycle. Hodrick and Prescott developed the HP filter to do this. 39

40 10.1 The Hodrick Prescott Filter Let Y t be the data on GDP. Let y t = log(y t ). We can decompose y t into a trend component and a business cycle component. The HP filter helps determine what fraction of y t is trend (y T t ) and what is cycle (y Ct ) by solving the following problem min y Ct,y T t N yct 2 + λ t=1 N [ (yt,t+1 y T,t ) 2 (y T,t y T,t 1 ) 2] t=1 λ is the smoothing parameter. For λ, the solution is y T t = y T t 1. For λ = 0, the solution is y T t = y t. λ determines how smooth the trend should be. With quarterly data, the usual value used is λ = 1, 600. For annual data, λ = 400. Figure?? shows the business cycle estimation in the US since 1900 for different λ s. A small λ attributes most of the fluctuation to changes in trends, while larger λ s assume the trend changes less and all movement is due to high frequency changes. Figure?? works with logs. The advantage is that we can interpret the business cycles as percent deviations. 3 Figure 1: US GDP, Trend, and Business Cycles US Real GDP per Capita and HP Trends (Source: Maddison) x Data = 1 1 = 400 = 10, US Real Business Cycles (Source: Maddison) Table 14.2 shows some key statistics for business cycles. Especially noteworthy is that consumption fluctuates less than output, investment fluctuates more, and productivity fluctuates about the same. This has motivated plenty of research that assume that cycles 40

41 Figure 2: Log US GDP, Trend, and Business Cycles 11 Log US Real GDP per Capita and HP Trends (Source: Maddison) 10 Data = 1 9 = 400 = 10, US Real Business Cycles (Source: Maddison) originate from shocks to technology (productivity). Consumption fluctuates less than output because individuals smooth consumption, which implies that investment (savings) must fluctuate more than output RBC in the Standard Growth Model Preferences Technology Endowments 1 unit of time each period k 0 units of capital at time 0. β t u(c t, 1 h t ) y t = z t F (k t, A t, h t ) y t = c t + i t k t+1 = (1 δ)k t + i t A t+1 = (1 + γ)a t 41

42 10.3 Calibration Step 1: Pick functional forms u(c, 1 h) = a log(c t ) + (1 a) log(1 h) F (k, Ah) = k θ (Ah) 1 θ log z t+1 = ρ 0 + ρ 1 log(z t ) + ɛ t, ɛ t N(µ, σ 2 ) Step 2: Pick parameter values. Choose β, θ, δ, γ, a to match the real interest rate, the ratios k, i, hours worked, and the y y growth rate along the steady state. For the productivity process, we estimate the AR process. First we need to identify something in the data that resembles productivity. This is the Solow residual. To see how this works, assume continuous time and a production function Then Divide by y(t), y(t) = A(t)F (k(t), h(t)) ẏ(t) = Ȧ(t)F (k(t), h(t)) + A(t)F 1(k(t), h(t)) k(t) + A(t)F 2 (k(t), h(t))ḣ(t) ẏ(t) y(t) = A(t) A(t) + F 1(k(t), h(t))a(t)k(t) k(t) A(t)F (k(t), h(t)) A(t) = A(t) + r(t)k(t) k(t) y(t) k(t) + w(t)h(t) ḣ(t) y(t) h(t) }{{}}{{} capital share labor share k(t) + F 2(k(t), h(t))a(t)h(t) A(t)F (k(t), h(t)) ḣ(t) h(t) Consider ρ 1 = 0.95, ρ 0 = µ = 0 (so that unconditional mean of z is 1), and σ 2 = Table?? model shows the performance of the model. Why is consumption less volatile than output, while investment is more volatile? Consumption smoothing. 42

43 Table 1: RBC Simulation: Model vs. Data Std. Dev. Correlation with Output Cyclical Comp. Cyclical Comp. Model Data Model Data Output Consumption Investment Hours Labor Productivity

44 11 Models with Heterogenity Until now we have only seen models with a representative agent, or representative firm. Here we will change that assumption. We will first introduce heterogeneous individuals, and then heterogeneous firms Simple Dynamic Model with Heterogeneous Agents Recall the static model with 2 agents, but extend now to a 2 period model. Individuals are indexed by i = 1, 2, with preferences U i = E log(c i1 ) + β log(c i2 ) with 0 < β < 1, c 0. The endowment is stochastic. Assume there is no aggregate uncertainty or growth, so that 2 i=1 ω it = Ω for t = 1, 2. With probability π, the endowments are ω 11 (1) = 1, ω 12 (1) = 0 ω 21 (1) = 0, ω 22 (1) = 1 With probability 1 π, the endowments are ω 11 (2) = 0, ω 12 (2) = 1 ω 21 (2) = 1, ω 22 (2) = 0 An insurance equilibrium for this economy is an allocation {c it(s)} and prices {p(s), q} such that 1. Taking prices as given, the allocation solves, for all i, 2 π(s) [log(c i1 (s)) + β log(c i2 (s))] s=1 s.t. c i1 (1) + p(1)c i2 (1) = ω i1 (1) + p(1)ω i2 (1) i i q c i1 (2) + p(2)c i2 (2) = ω i1 (2) + p(2)ω i2 (2) + i i 2. Goods and insurance markets clear 2 2 c it (s) = ω it (s) i=1 i=1 for all t, s i 1 = i 2 Working through the first order conditions, we get c i1 (2) c i1 (1) = 1/q c i2 (s) βc i1 (s) = p(s) 44 for all s

45 Guess and verify the solution q = 1, p(1) = p(2) = 1/β. This implies c it (s) = 1/2 i 1 = 1/2 i 2 = 1/2 for all i, t, s It is straightforward to check that this satisfies all the conditions in the problem. Now assume an alternative with no insurance, but with borrowing and lending. A sequence of markets with no insurance equilibrium for this economy is an allocation {c it(s), b i (s)} and prices {p(s), r(s)} such that 1. Taking prices as given, the allocation solves, for all i, 2 π(s) [log(c i1 (s)) + β log(c i2 (s))] s=1 s.t. c i1 (s) = ω i1 (s) b 1 (s) c i2 (s) = ω i2 (s) + (1 + r(s))b 1 (s) for all s for all s 2. Goods and borrowing and lending markets clear 2 2 c it (s) = ω it (s) i=1 i=1 b 1 (s) = b 2 (s) for all t, s The first order conditions imply b i (s) = β(1 + r s)ω i1 (s) ω i2 (s) (1 + r(s))(1 + β) Guess and verify β(1 + r s ) = 1. This implies For consumer 2 b 1 (1) = β 1 + β, b 1(2) = β 1 + β c 11 (1) = β > 1 2, c 12(1) = β > 1 2 c 11 (2) = β 1 + β < 1 2, c 12(2) = β 1 + β < 1 2 b 2 (1) = β 1 + β, b 2(2) = β 1 + β c 21 (1) = β 1 + β < 1 2, c 12(1) = β 1 + β < 1 2 c 21 (2) = β > 1 2, c 22(2) = β <

46 That is, without insurance consumers no longer smooth consumption across states (still smooth across periods). They face uncertainty in consumption. What about welfare? With insurance, U ins = 1/2 [log(1/2) + β log(1/2)] + 1/2 [log(1/2) + β log(1/2)] = (1 + β) log(1/2) Without insurance [ ( ) ( )] β β U b l =1/2 log + β log 1 + β 1 + β [ ( ) β (1 + β) 1/2 log + 1/2 log 1 + β Jensen s inequality says U ins > U b l. [ + 1/2 ( β log )] ( ) ( )] β log = 1 + β 1 + β 11.2 More Complicated Model of Heterogeneous Agents There is a continuum of ex ante identical agents uniformly distributed over the unit interval. Their preferences are given by β t u(c t ) Each period, each individual receives an endowment shock e it. Assume e it { e 1, e 2,..., e N}, where e 1 < e 2 < < e N. Each period, the probability that e it = e j is π j. These realizations are iid across time and across individuals. We can also think of this process as following a Markov chain. That is, the probability of having income e t+1 tomorrow depends on income today e t. Define π ji = P rob(e t+1 = e j e t = e i ) Then there is a transition matrix Π that determines the law of motion of the income realizations. This matrix is π 11 π π N1 π 12 π π N2 Π =.... π 1N π 2N... π NN Notice that each row should add up to 1. For example, row 1 gives the probabilities of going to any realization starting from e 1. Under certain assumptions that we will always make, Π implies an invariant distribution of earnings Π, where Π = lim T ΠT We typically assume that in period 0 agents draw their earnings from a distribution with these probabilities. 46

47 Social Planner Solution There are many solutions to the social planner problem because there are many different objective functions. Consider one where all individuals receive equal weights. The problem is 1 0 β t u(c it )di s.t. 1 0 c it di = N π j e j t, c it 0 i, t. j=1 FOC β t u (c it ) = λ t Thus, consumption should be the same across individuals and across time Competitive Equilibrium If we solve an Arrow Debreu equilibrium, we would be assume there exists a full set of markets for borrowing and lending and insurance, and the solution would be the same. What if we look for an equilibrium where not all of these markets exist? On the extreme, if we eliminate all borrowing and lending and insurance markets, then c it = e it. Less extremely, suppose we only close borrowing and lending. Then we would recover the social planner solution. The reason is that people insure themselves. What if we close the insurance market, but allow for borrowing and lending? We would not recover the social planner solution, but we would not have c it = e it either, because people will save and borrow and use this as an imperfect insurance. Individual state variables: endowment shock e and asset position a. Aggregate state variables: joint distribution of endowment shocks e and asset positions a, µ(e, a), where 1 µ(e 0 i, a i )di = a e µ(e, a)da = 1, since there is a total of 1 individuals. A Recursive Competitive Equilibrium for this economy is a list of functions V (e, a, µ), g(e, a, µ), G(µ), r(µ), and Λ(µ) such that 1. (Consumer imisation.) Taking r(µ) and Λ(µ) as given, V (e, a, µ) solves ( ) a V (e, a, µ) = u a + e + βe e V (e, a, Λ(µ)) a 1 + r(µ) a s.t. a + e 0, a B 1 + r(µ) and g(e, a, µ) is an optimal decision rule. 2. Consistency. B e N e=e 1 g(e, a, µ)µ(e, a)da = G(µ) 47

48 For convenience, write this as g(e, a, µ)dµ = G(µ) and µ + g(e, a, µ)+ next period s distribution of e s determine Λ(µ) 3. Market Clearing ( G(µ) = 0 s.t. ) a dµ = 0 Special case. Suppose a { a 1, a 2,..., a N}. We can represent µ as an N M matrix, where the ij th element is the mass of individuals with e = e i and a = a j. Since the total mass of individuals is 1, the sum of all elements of µ must be 1, i.e., a e µ(e, a) = 1. The consistency condition for Λ(µ) is, Define the operator A as i,j s.t. g(i,j,µ)=j µ ij π ii = Λ(µ) i j µ = Aµ where A is an N N matrix. Solving for an equilibrium is difficult. Markets are incomplete, so we cannot find the social planner s solution. The equilibrium is hard because prices depend on the aggregate state, which is a high dimensional object. Solving for the steady state is somewhat easier, and this is what we will do. An issue that comes up is what is a steady state. It is a state where the aggregate state variable does not change. This does not mean that nothing changes: even when the distribution µ(e, a) is constant, individuals will be moving around in steady state. This is why we usually call this an invariant distribution Solution Algorithm What makes things easier in this case is that there is only one price to consider: r(µ). The idea is to make an initial guess on r, and then check whether the equilibrium conditions for r being constant are met. If they are not, update the guess. Given this guess, solve the consumer problem. V (e, a, r) = u a s.t. a + e a 1 + r ) (a + e a + βe e V (e, a, r) 1 + r 0, a B This determines a policy rule g(e, a, r). Given g(e, a, r) and π we know the transition Λ(µ). In steady state, µ = Λ(µ). 48

49 Recall the operator A, µ = Aµ There are two ways to find the steady state. The first solves for µ ss, where µ ss = Aµ ss µ ss (I A) = 0 The second possibility computes µ 1 = Aµ 0 µ 2 = Aµ 1 µ 3 = Aµ 2. If this process converges to a fixed point, this is the steady state. Note that so far we have not imposed market clearing. Only under the steady state r will markets clear. Thus, compute g(e, a, r)µ(e, a) = X If X > 0, there is too much saving r is too high If X < 0, there is too much saving r is too low a e 11.3 The Standard Growth Model with Heterogeneous Agents There is a continuum of ex ante agents of mass 1. The preferences are There is a stand in firm with technology β t u(c t ) y t = F (k t, e t ) (e t : efficiency units of labor services) The law of motion for capital is k t+1 = (1 δ)k t + i t Output can be used for consumption and investment y t = c t + i t Individual i has k i0 units of capital at time 0. Each individual has 1 unit of time each period, but random efficiency units of this time use. We assume e it follows a Markov process, with transition matrix Q(e it, e i,t+1 ). There is no insurance markets, but individuals can save and borrow. 49

50 Defining the Equilibrium Individual State Variables. e, k, b Aggregate State Variables. µ(e, k, b) A Recursive Competitive Equilibrium for this economy is a list of functions V (e, k, b, µ), k (e, k, b, µ), b (e, k, b, µ), K (µ), B (µ), w(µ), r(µ), and Λ(µ) such that 1. (Consumer imisation.) Taking w(µ), r(µ) and Λ(µ) as given, V (e, k, b, µ) solves V (e, k, b, µ) = u (w(µ)e + r(µ)k + (1 δ)k k ) + βev (e, k, b, Λ(µ)) a s.t. w(µ)e + r(µ)k + (1 δ)k k 0, k 0, b B and k (e, a, µ) is an optimal decision rule. 2. (Firm Maximization. ) Taking w(µ), r(µ) as given, for all µ, K, E solve F (K, E) r(µ)k w(µ)e st K 0, E 0 K,E 3. Consistency. K (µ) = B (µ) = k (e, k, b, µ)dµ b (e, k, b, µ)dµ Given µ, the policy rule k (e, k, b, µ) and Q leads to Λ(µ). 4. Market Clearing K = E = kdµ edµ The goods market clears due to Walras law The Standard Growth Model with Heterogeneous Firms There is one representative agent with preferences given by β t u(c t ) There is a continuum of firms of mass N. Each firm has technology y it = A it f(k it, h it ) f(k, h) is strictly jointly concave in k, h. A it is an idiosyncratic shock that evolves according to the transition law Q(A it, A i,t+1 ). Output can be used for consumption or investment. Capital can also be turned into output. The representative individual is endowed with 1 unit of time each period and k 0 units of capital. He owns the firms. {A i0 } and k 0 are given. 50

51 Defining the Equilibrium Equilibrium for this economy is a list of functions V (k, K, µ), k (k, K, µ), K (K, µ),k f (A, K, µ), h f (A, K, µ), w Individual State variables. k for the individual, A for the firm. Aggregate State variables. K and the distribution µ(a). A Recursive Competitive and Λ(K, µ) such that 1. (Consumer imisation.) Taking w(k, µ), r(k, µ) and Λ(K, µ) as given, V (k, µ) solves V (k, K, µ) = u (w(k, µ) + r(k, µ)k + (1 δ)k k ) + βev (k, K, Λ(µ)) k s.t. w(k, µ)e + r(k, µ)k + (1 δ)k k 0, k 0 and k (k, K, µ) is an optimal decision rule. 2. (Firm Maximization. ) Taking w(k, µ), r(k, µ) as given, for all K, µ, k f (A, K, µ), h f (A, K, µ), solve 3. Consistency. AF (k, h) r(k, µ)k w(k, µ)h st k 0, h 0 k,h Λ(K, µ) is consistent with Q and K. K (K, µ) = k (k, K, µ) 4. Market Clearing K = 1 = k f (A, K, µ)dµ h f (A, K, µ)dµ The goods market clears due to Walras law. Remark We have assumed the measure of firms is fixed. We can easily endogenize this measure. Adding a per period, fixed production cost implies that firms that receive a very low shock A will exit endogenously. On the other hand, if we assume that there is a fixed cost to draw the initial productivity A, there is endogenous entry. These fixed costs introduce new conditions that the firms must satisfy and a new decision rule. Assume there is a fixed per period cost κ units of output. Then the problem of the firm is I (AF (k, h) r(k, µ)k w(k, µ)h κ) k 0,h 0,I {0,1} where, assuming that the reservation value is 0, { 1 if the firm decides to produce I(A, K, µ) = 0 otherwise 51

52 On the other hand, if there is a fixed cost to draw the initial productivity κ e units of output, firms will enter until the expected value of operating equals the entry cost. The problem of the firm becomes dynamic. Assume that A gets drawn from a distribution G(A) v f (A, K, µ) = k,h AF (k, h) r(k, µ)k w(k, µ)h + 1 q(k, µ) E A Av f (A, K, µ) where q(k, µ) = r(k, µ) + 1 δ. The free entry condition is κ e + v f (A, K, µ)dg(a) 0 12 Indivisible Labor Suppose individuals choose whether to work or not, but there is no adjustment on the hours worked. That is, h {0, 1}. The economy is static, with a continuum of individuals of measure 1 and a representative firm. Preferences are u(c) + v(1 h) where u(c) is increasing and strictly concave and v is a number multiplying the 1 h. The technology uses only labor y = f(h), f strictly concave and strictly increasing. Assume f (1) = 0 (to stay away from corners). Individual own the firm in equal shares Competitive Equilibrium A competitive equilibrium for this economy is a list c (i), h (i), H, π, w such that 1. (Consumer imization.) Taking w and π as given, for each i, c (i), h (i) is a solution to u(c) + v(1 h) s.t. c = c 0,h {0,1} w h + π 2. (Firm imization). Taking w as given, H solves H f(h) w H and π = f(h ) w H 3. Markets clear c (i)di = f(h ) h (i)di = H The optimal decision on h (i) is 1 if u(w + π ) > u(π ) + v h (i) = 0 if u(w + π ) < u(π ) + v h [0, 1] if u(w + π ) = u(π ) + v 52

53 From this, if and only if w = w, people are indifferent between working and not working, where w is such that u(w + π ) = u(π ) + v. When workers are indifferent between working and not working, the equilibrium number of workers will be given by the demand for labor at that wage. Thus, that wage determines the profits of the firm, which itself affects the cutoff wage w. The equilibrium w is such that the expected profits are equal to the actual profits. Assume only λ (0, 1) individuals work. The following must hold: h h 1 Labor supply 1 Labor supply Higher profits, higher w Labor demand Labor demand w w w w w = f (λ) π = f(λ) f (λ)λ c w = w + π = f(λ) + (1 λ)f (λ) c n = π = f(λ) f (λ)λ Let U be the sum of utilities across individuals, that is, the expected utility. U = λu(c w ) + (1 λ)(u(c n ) + v) = u(c w ) The last equality follows from u(c w ) = u(c n ) + v in equilibrium. Claim. This equilibrium is NOT Pareto efficient. To show this, consider the alternative feasible allocation. Give everyone consumption c = λc w + (1 λ)c n, and anyone can work with probability λ. The expected utility is U l = u(c ) + (1 λ)v Notice that this is higher than the expected utility under the equilibrium: U l = u(c ) + (1 λ)v = u(λc w + (1 λ)c n ) + (1 λ)v > λu(c w ) + (1 λ)(u(c n ) + v) = U 12.2 The Optimal Allocation What would the social planner choose? λu(c w) + (1 λ)(u(c n ) + v) s.t. λc w + (1 λ)c n = f(λ) c w 0,c n 0,λ [0,1] 53