ADVANCED MACROECONOMICS 2015 FINAL EXAMINATION FOR THE FIRST HALF OF SPRING SEMESTER

Transcription

1 ADVANCED MACROECONOMICS 2015 FINAL EXAMINATION FOR THE FIRST HALF OF SPRING SEMESTER Hiroyuki Ozaki Keio University, Faculty of Economics June 2, 2015 Important Remarks: You must write all your answers in English although I do not deduct it about the English mistake. More importantly, you need to justify your answers as much as possible. The mark depends on the correctness of your exposition. Consider the problem of maximizing U 0 c) = Uc 0, c 1, c 2,...) := t + c1 θ over the set of all consumption streams in the feasible set 0c R + given the parameters θ [0, 1) and x 0 > 0. 1 x R ) } + c x 1 = 6x 0 and t 1) c t + x t+1 = 6x t Note that the utility function is recursive with the aggregator function W : R + R R defined by W c, u) = + u Answer both of the following two problems. Important Hints: Notice that for any c R +, Uc, c, c,...) = You may get some marks by proving this.) A good strategy to tackle these problems is to begin by finding a steady-state investment level, that is, an investment level which is optimal to hold forever if it is ever attained. 1. Problem 1. Assume that θ = 0. Derive the set of optimal consumption streams. Then, determine the value function. Problem 2. Assume that θ 0, 1) and x 0 = θ /5. Derive the set of optimal consumption streams. Then, find the value of the utility function evaluated by these consumption streams. 1

2 ANSWER KEY Hiroyuki Ozaki Keio University, Faculty of Economics June 3, 2015 In this paragraph, I prove that U 0 the same utility function as in the exam) defined by U 0 0 c) = Uc 0, c 1, c 2,...) := t + c1 θ is recursive with the aggregator function W : R + R R defined by W c, u) = + u ) That is, what I said right after the exam is wrong. To see this, note that U 0 0 c) = t + c1 θ = = = t + c1 θ t + c1 θ t + = + U 0 1 c) + 1 1, t 1 1 ) t 1 1 where the first line holds by the definition of U 0 ; the second line holds by the continuity of ; the third and the fourth lines are trivial; and the last line holds again by the definition of U 0. This shows that U 0 solves Koopmans equation and hence it is recursive with the aggregator function 1). Note that U 0 0, 0, 0,...) = 1. Next, I define the utility function U 1 by U 1 0 c) := t + c1 θ t and I prove that U 1 is recursive with the aggregator function 1). To this end, I do the same thing as above: U 1 0 c) = t + c1 θ 2 t + 1 1

3 = = = Note that U 1 0, 0, 0,...) = 0. t + c1 θ t + c1 θ t + = + U 1 1 c) t t ) t Clearly, U 0 and U 1 are distinct from each other. Furthermore, both are recursive with the identical aggregator function 1). This means that the aggregator function dose not uniquely determine the utility function. This is something that I did not intend. I know some condition under which the aggregator function uniquely determines the utility function and I thought that this condition is now satisfied. I was wrong and I will talk more about this later. Now I proceed in any case. To solve this maximization problem, we need to find the value of u := Uc, c, c,...). Since u satisfies Koopmans equation, we know that it must be the case that u = + u + 1 1, from which it follows that u + 1 = + u + 1. This is simply the quadratic equation of u + 1 and it can be easily solved for u to get and u = u = 1 1 2) 1. 3) Both are increasing in c and hence are the well-defined utility values over constant consumption streams. Furthermore, it is immediate that 2) corresponds to U 0 and 3) corresponds to U 1. However, in what follows, I exploit the fact that there exists a steady-state consumption level which satisfies the Euler equation. And, to guarantee its existence, I require that the utility level corresponding to the steady-state consumption stream be given by 3). To sum up, at this very stage, the utility function need be defined by U 1 instead of U 0. While I do not stint my apology for this mistake, I conclude, after a moment s reflection, that you are given enough information to solve this problem. That is, 1), 3) and the production function that are everything you need to solve this problem. So, I have decided not to give any special treatment in grading your examination papers. 3

4 Here is the answer. First, in order to derive the Euler equation, use Bellman s equation along the line given in the lecture. This is justified because the utility function is recursive. Bellman s equation is V x t ) = max W c t, V x t+1 )) c t + x t+1 = F x t ), c t, x t+1 0 } = max W F x t ) x t+1, V x t+1 )) F x t ) x t+1 0 }. By assuming the interiority of the solution, the first-order necessary condition turns out to be W 1 c t, u t+1 ) = W 2 c t, u t+1 ) V x t+1 ), ) where W i denotes the partial derivative of W with respect to the i-th argument and u t denotes Uc t, c t+1, c t+2,...). The value function V is known to be differentiable by the Benveniste- Scheinkman Theorem and the envelop theorem shows that its derivative V is given by V x t ) = W 1 c t, u t+1 ) F x t ). 5) By combining ) and 5), we finally reach the Euler equation for a one-sector growth model with a recursive-utility-maximizing agent: W 1 c t, u t+1 ) = W 2 c t, u t+1 ) W 1 c t+1, u t+2 ) F x t+1 ). Specifying the aggregator function by 1) and the production function by F x) = 6x leads to the Euler equation for the current problem: c θ t = 3 u t+ 1) 1/2 c θ t+1. 6) I now find the steady-state consumption stream which satisfies the Euler equation 6). Since t) c t = c and u t+1 is given by 3), we have 1 = 3 1/2 which can be easily solved for the steady-state consumption level c : c = 6 1/1 θ). Notice that I would get stuck here if I used 2). Answer to Problem 1. Set θ = 0. Then, c = 6 and the steady-state capital stock x is given by x = 6/5 from the feasibility constraint: c + x = 6 x. Define ˆtx 0 ) by ˆtx 0 ) := min t 0 6 t x 0 > 6/5 }.

5 I hereafter suppress x 0 ).) Consider the consumption stream 0 c given by 0c = 0, 0,..., 0, 6ˆt x 0 6/5, 6, 6,...), 7) where 6ˆt x 0 6/5 is the ˆt-th period s consumption, and the capital stream 1 x defined by 1x = 6x 0, 6 2 x 0,..., 6ˆt 1 x 0, 6/5, 6/5,...), where 6ˆt 1 x 0 is the ˆ)-th period s capital stock. Notice that, in this particular problem, the Inada condition is not satisfied and hence a zero consumption can not be excluded. In fact, the first-order necessary condition ) should be replaced by the inequality recall the Kuhn-Tucker theory in the first lecture), W1 c t, u t+1 ) = W 2 c t, u t+1 ) V x t+1 ) if c t > 0 W 1 c t, u t+1 ) W 2 c t, u t+1 ) V x t+1 ) if c t = 0 and hence, the Euler equation is now the Euler inequality: W1 c t, u t+1 ) = W 2 c t, u t+1 ) W 1 c t+1, u t+2 ) F x t+1 ) if c t > 0 W 1 c t, u t+1 ) W 2 c t, u t+1 ) W 1 c t+1, u t+2 ) F x t+1 ) if c t = 0. This means that the Euler equation 6) now turns out to be 1 = 3 ut+ 1) 1/2 if c t > u t+ 1) 1/2 if c t = 0. 8) Consider the consumption stream 0 c defined by 7). First, note that c ˆt = 6ˆt x 0 6/5 is positive by definition and hence that the first half of 8) is met. Second, note that for any t, U t c ) U6, 6, 6,...) because 6ˆt x 0 6/5 6. Therefore, the second half of 8) is also met. I thus conclude that 7) satisfies 8). Next, I need to show that 7) is the unique consumption stream which satisfies 8). A lot of work might be necessary for this and I do not like to do that here. Instead, I simply assume it. You do not lose any mark without doing this, off course.) Finally, I show that the optimal consumption stream certainly exists by showing that the utility function together with the given production function) is upper-convergent. Here, the upper-convergence theorem in the lecture is invoked.) Since any consumption stream which violates the Euler equation can not be optimal and since we know that there exists an optimal consumption stream, 7) must be the unique optimal consumption stream I am seeking. To this end, notice that the variable discount factor for the aggregator function 1) is given by 1/2) u + 1) 1/2. This shows that the discount factor is bounded above by a half. Furthermore, the higher the future utility is, the closer to zero the discount factor becomes. While the economy can grow without bound with the constant rate of growth since the production 5

6 function is of Ak-type), the future is discounted nonlinearly. From analogy to L Hospital s rule, we see that far future will be less and less important for the consumer and the upper-convergence thus follows. This is only intuition behind the fact that the upper-convergence is actually satisfied but I think that it is enough for the purpose of the exam. I can give a formal proof that the utility function is upper-convergent given the aggregator function 1).) The value function J is J x 0 ) = U 1 0, 0,..., 0, 6ˆt x 0 6/5, 6, 6,...) = W c 0, W c 1, W cˆt 1, W cˆt, )) )) uˆt+1 = W 0, W 0, W 0, W cˆt, 8)) )) ) )) = W 0, W 0, W 0, cˆt + 3)1/2 1 = W )) 0, W 0, cˆt + 3)1/ 1 = cˆt + 3)1/2)ˆ, where cˆt = 6ˆt x 0 6/5. Since ˆt itself is the function of x 0, J is the well-defined value function. Answer to Problem 2. Since θ 0, 1), the Inada condition requires that the consumption be strictly positive. Let x = 6 1/1 θ) /5. Then, c + x = 6 1/1 θ) + 6 1/1 θ) /5 = 6 6 1/1 θ) /5 = 6 x = 6x 0. This shows that c, c, c,...) and x, x, x,...) are steady-state consumption and capital streams which are feasible from x 0. Notice that here is no transition path and the economy stays in the steady state from the beginning. By the argument around the equation not inequality) 6), this path clearly satisfies the Euler equation. The same logic shows that c, c, c,...) is the unique optimal consumption stream from x 0 and the value of the utility function is then given by ) J 6 1/1 θ) /5 = U 1 c, c, c,...) = 1 = 8. Important Announcement. As I mentioned above, I did not expect the aggregator function 1) generates multiple utility functions. As I said there, I though some condition which guarantees the uniqueness is satisfied but I missed something. What did I miss? If you figure out this problem completely, I will add some marks so as to upgrade your final score for the first half of this semester), say, from B to A. Submit your result as a term paper via an by June 20. You may like to see the book by Becker, R. E. and J. H. Boyd III, Capital Theory, Equilibrium Analysis and Recursive Utility, 1997, Blackwell. 6