Dynamic Problem Set 1 Solutions

Transcription

1 Dynamic Problem Set 1 Solutions Jonathan Kreamer July 15, 2011 Question 1 Consider the following multi-period optimal storage problem: An economic agent imizes: c t} T β t u(c t ) (1) subject to the period-by-period budget constraints k t+1 = Ak t + y t c t (2) for t = 0, 1,..., T, and the non-negativity constraint k t 0 for t = 1,..., T + 1, with k 0 given, where c t is consumption expenditure assumed to be incurred at the end of period t, k t is the capital stock at the beginning of period t, y t is labor income received at the end of period t, β (0, 1) is the discount factor, and the storage technology is linear, Ak t. Suppose that u(c) = 1 (c c)2 2 where c is the bliss level of consumption. For the time being, ignore the non-negativity constraint for t = 1, 2...T hoping that the solution you derive ignoring the non-negativity constraint is such that the capital stock is positive (except for T + 1), in whcih case the non-negativity constraint is irrelevant in the first place. (i) Set up the Lagrangian function, derive the first-order conditions, and eliminate the Lagrangian multipliers to obtain the Euler equation. The Lagrangian is L = The first-order conditions are β t 1 2 (c c t) 2 + β t λ t [Ak t + y t c t k t+1 ] λ t = c c t λ t 1 = βaλ t Ak t + y t = c t + k t+1 1

2 The first equation gives us that the shadow value each period is equal to the marginal utility of consumption. The third equation is the period budget constraint. The second equation gives us the Euler equation once we once we substitute in for the shadow value: c c t = βa (c c t+1 ) (ii) Derive the lifetime budget constraint corresponding to the period-by-period budget constraints (2). The period T budget constraint is k T +1 = Ak T + y T c T Substituting in the period T-1 budget constraint yields Rearranging a bit, this yields k T +1 = A (Ak T 1 + y T 1 c T 1 ) + y T c T If we consider in the manner, we get k T +1 = A 2 k T 1 + A (y T 1 c T 1 ) + y T c T k T +1 = A T +1 k 0 + A s (y T s c T s ) Alternatively, dividing by A T +1 and counting up instead of down, we get s=0 k T +1 A T +1 = k 0 + y t c t A t If we interpret A as the relative price of goods between periods t and t + 1 (that is, the price of period t goods in terms of period t + 1 goods), then this budget constraint is T k T +1 A T +1 + c t A t = k 0 + Which is the budget constraint in terms of period 0 goods, with the left-hand side being total expenditures, and the right-hand side being total resources. Note that 1 A is the price of period t goods in terms of period t 0 goods. (iii) Combine the Euler equation and the lifetime budget constraint to obtain the solution for c 0. Is your solution linear in the initial capital stock k 0? Letting Y = k 0 + T y t A t Writing in terms of c c t, we get y t A t be lifetime wealth, and setting k T +1 = 0, the lifetime budget constraint becomes c T A t = c t A t = Y c c t A t + Y 2

3 The Euler equation is c c t = βa (c c t+1 ). Starting with period 0 and iterating, we get c c 0 = β t A t (c c t ). Substituting this into the lifetime budget constraint yields T c A t = c c 0 (βa 2 ) t + Y Using the equation for geometric summation, T rt = [ 1 A (T +1) 1 A 1 ] c = [ 1 ( βa 2 ) (T +1) 1 (βa 2 ) rt 1 r, we get ] (c c 0 ) + Y Solving for c 0, we get c 0 = c + ( 1 ( βa 2 ) 1 1 (βa 2 (T +1) ) ) [ ( 1 A (T +1) Y 1 A 1 ) ] c Recall that we have Y = k 0 + T y t A, so c t 0 is linear in k 0. (iv) Now we wish to use dynamic programming to solve this problem. Let v t (k t ) denote the period t value function. a) Write down the Bellman equation for period t. The Bellman equation is v t (k t ) = 1 } c t,k t+1 2 (c c t) 2 + βv t+1 (k t+1 ) subject to k t+1 = Ak t + y t c t. Embedding the constraint into the Bellman equation directly yields v t (k t ) = 1 } c t,k t+1 2 (c c t) 2 + βv t+1 (k t+1 ) + λ t (Ak t + y t c t k t+1 ) b) Show that the decision rule for period T is given by c T = Ak T + y T and that the value function for period T is given by v T (k T ) = α 0T + α 1T k T 1 2 α 2T k 2 T (Determine the coefficients α 0T, α 1T, and α 2T ). Thus the value function is quadratic in k T. Now conjecture that the period t + 1 value function is quadratic in k t+1 : v t+1 (k t+1 ) = α 0,t+1 + α 1,t+1 k t α 2,t+1k 2 t+1 Show that the period t value function then is quadratic in k t, and that the decision rule is linear in k t. Observe that there is no value from periods after T, so we have v T +1 (k) = 0. So v T is defined as v T (k T ) = c T,k T +1 subject to k T First-order conditions are 1 2 (c c T ) 2 + λ T (Ak T + y T c T k T +1 ) } 3

4 c c T = λ T k T +1 λ T = 0 Ak T + y T = c T + k T +1 Assuming we can t obtain c T = c, we set k T +1 = 0. Therefore c T is derived from the budget constraint, and is Therefore the value function is c T = Ak T + y T v T (k T ) = 1 2 (c y T Ak T ) 2 which is clearly quadratic in k T. In particular, the value function is v T (k T ) = 1 2 (c y T ) 2 + A (c y T ) k T 1 2 A2 k 2 T Now suppose that the period t + 1 value function is quadratic. Then the period t value function is v t (k t ) = 1 c t,k t+1 2 (c c t) 2 + β ( } α 0,t + α 1,t k t+1 + α 2,t kt+1) 2 + λt (Ak t + y t c t k t+1 ) First-order conditions are c c t = λ t β (α 1,t + 2α 2,t k t+1 ) = λ t Ak t + y t = c t + k t+1 Combining the first two yields βα 1,t + 2βα 2,t k t+1 = c c t. Combined with the budget constraint, this yields k t+1 = c Ak t y t βα 1,t c t = Ak t + y t c Ak t y t βα 1,t Or equivalently c t = ( ) 2Aβα2,t k t c βα 1,t 2βα 2,t y t Call these decision rules k t+1 = γ 0 + γ 1 k t and c t = ψ 0 + ψ 1 k t. Note that both decision rules are linear in k t. Substituting these into the expression for v t (k t ) yields v t (k t ) = 1 2 [c (ψ 0 + ψ 1 k t )] 2 + β [α 0,t + α 1,t (γ 0 + γ 1 k t ) + α 2,t (γ 0 + γ 1 k t ) 2] which is quadratic in k t. 4

5 (v) We have implicitly assumed that the parameters of the problem (including k 0 and y 0, y 1,..., y T ) are such that the bliss point c is never reached (that is, the agent does not want to throw away capital) and that the agent never wishes the capital stock to be negative (that is, never wishes to borrow. For the two-period case (T=1), provide an example where the true solution (that is, the solution under (2) and (3)) is c 0 = c 1 = c, and either the period-by-period budget constraints, or the non-negativity constraint for k T +1 hold with strict inequality. (This is the case of the super-rich agent, who is so rich that she or he throws away resources.) Also, provide an example where the true solution calls for k 1 = 0 (that is, the non-negativity constraint is binding). Does the Euler equation hold for this case? (The agent in this situation is sometimes said to be liquidity constrained.) Consider the general problem 12 (c c 0) 2 12 } (c c 1) 2 c 0,c 1 subject to c 0 = Ak 0 + y 0 k 1 and c 1 = Ak 1 + y 1 k 2, along with non-negativity constraints on c 0, c 1, k 0, k 1, k 2. The Lagrangian is L = 1 2 (c c 0) β (c c 1) 2 +λ 0 (Ak 0 + y 0 k 1 c 0 )+βλ 1 (Ak 1 + y 1 k 2 c 1 )+µ 0 c 0 +µ 1 c 1 +γ 1 k 1 +γ 2 k 2 First-order conditions are c c 0 λ 0 β (c c 1 ) βλ 1 λ 0 βaλ 1 βλ 1 0 Ak 0 + y 0 k 1 c 0 = 0 Ak 1 + y 1 k 2 c 1 = 0 For an example of the first, consider y 0 = y 1 = c and k 0 = 1. Since the agent has income equal to the bliss level of consumption, he has no need of his initial unit of capital. Mathematically, the only solution that respects these expressions and the non-negativity constraints is λ 0 = λ 1 = 0, which implies c 0 = c 1 = c. For an example of the second, consider the case where k 0 = 0, y 0 = 0, and y 1 = (1 + A)c. Now the period 0 budget constraint is Ak 0 + y 0 k 1 c 0 = 0, which for k 0 = y 0 = 0 becomes k 1 + c 0 = 0. Combined with non-negativity constraints on c 0 and k 1, this implies c 0 = 0. Then from the second budget constraint we get Ak 1 + y 1 k 2 c 1 = 0, or k 2 + c 1 = (1 + A)c. The consumer will never choose c 1 c (this should be obvious, but mathematically it follows from λ 1 0 combined with the second equation above). This means that we have c 1 = c, which implies λ 1 = 0, which allows us to set k 2 > 0 to soak up the remainder of our wealth. The liquidity (borrowing) constraint can be seen in the equation λ 0 > βaλ 1. This implies that the marginal utility from consumption in period 0 is greater than from consumption in period 1. Therefore at the margin we would like to shift some consumption from period 1 (where is yields less utility to us) to period 0 (where it yields more). However, we are prevented from doing this because we cannot borrow in period 0 against our future income. Intuitively, if the value of the marginal utility of your income is growing faster than βa, and you can only save and not borrow (k 0), then you will be liquidity constrained. Remark: Why is this called liquidity constrained? ) 5

6 Question 2 An optimal cake-eating problem Consider a consumer who has the following preferences over the consumption of cake: c t} Y β t u(c t ) Where c t is the amount of cake consumed and β is a parameter of voracity, determining how patient the consumer is in his preferences for cake. The initial size of the cake if x 0 = 1. Assume that the cake does not spoil at all until period T (so that the storage technology is perfect), but is entirely consumed by rates between period T and T + 1 (so that the consumer can gain no utility by caving any of the cake to period T + 1). (a) What are the state and control variables in this problem (note that there is more than one possible answer to this question). Control variables are c t and x t+1 (the latter I write under protest). State variables are x t. (b) Assume that T = 1 (a two-period problem) and that u(c t ) = ln(c t ). Solve for the optimal consumption of cake in each period. The problem is ln c 0 + β ln c 1 } subject to budget constraint c 0 + c 1 1 and non-negativity constraints c 0, c 1 0. The Lagrangian is L = ln c 0 + β ln c 1 + λ (1 c 0 c 1 ). First-order conditions are 1 β c 0 = λ, c 1 = λ, and c 0 + c 1 = 1. Combining the first two yields c 1 = βc 0. Substituting in from the last yields 1 c 0 = βc 0, so we have c 0 = 1 1+β and therefore c 1 = β 1+β. We could also write this in terms of value functions. We have v 0 (x 0 ) = c0 ln c 0 + βv 1 (x 0 c 0 )} and v 1 (x 1 ) = c1 ln c 1 + βv 2 (x 1 c 1 )}. We get c 1 = x 1, which implies v 1 (x 1 ) = ln x 1. Substituting this into the expression for v 0 and setting x 0 = 1, we get v 0 (x 0 ) = c0 ln c 0 + β ln(1 c 0 )}. The first-order condition is 1 c 0 = β 1 c 0, which implies c 0 = 1 1+β, and therefore x 1 = c 1 = β 1+β. For the remaining parts of the question, assume that T = (the rates never come and the consumer lives forever). (c) Assume that u(c t ) = ln(c t ). Solve for the problem s Euler equation and for the optimal consumption of cake in each period. How does the path of cake consumption depend on β? The Euler equation is 1 c t = β 1 c t+1, or c t+1 = βc t. We don t need to worry about non-negativity constraints for c 0 and c 1 because ln( ) satisfies the Inada conditions. Is there a choice of c t } that satisfies the Euler equation? Suppose we( have c 0. Then we have c 1 = βc 0, c 2 = β 2 c 0,...,c t = β t c 0. The budget constraint is c 0 + c = 1, or c β + β ) 1 = 1. Assuming we have β < 1, this infinite sum is equal to 1 β, so we get c 0 = 1 β, and therefore c 1 = β(1 β), c 2 = β 2 (1 β) and so forth. If we have β 1 there is no solution. (d) Now assume that u(c t ) = (c t 0.1) 2. How does the problem change? What additional constraint is required to avoid unrealistic results? The Euler equation is u (c t ) βu (c t+1 ). We don t have strict equality because we now have to consider the possibility that the non-negativity constraint will be binding since the Inada conditions do not hold for this utility function. The marginal utility function is 2c t 0.2. Marginal utility is increase with c t. In other words, the utility function is convex. This reverses the usual forces that equalize marginal utility. Typically we consider concave utility functions, in which case consumers will tend to smooth consumption. Why is this? If marginal utility 6

7 is higher in one period than another, they will try to shift consumption towards the period with higher marginal utility. If we have a concave utility function, this will increase the marginal utility of the period we are shifting consumption out of, and decrease marginal utility in the period towards which we are shifting consumption. However, if we have a convex utility function, when we shift consumption towards a period with higher marginal utility, we get yet higher marginal utility in that period. The solution with a convex function is then to concentrate our consumption in one period. Which period will we choose? If we have β < 1, we should choose the first period. If we have β = 1, then we are indifferent between choosing any period (but we still should choose just one). If we have β > 1, then we always want to delay our consumption still further, and the problem blows up. (e) Assume that u(c t ) = c t. Solve the problem s Euler equation. Why does such an odd result occur? What is the optimal path of cake-earing when β = 1, β < 1, and β > 1? The Euler equation is u (c t ) βu (c t+1 ). Since marginal utility is just 1, this is 1 β. If we have c t = 0 then we have β 1; if we have c t+1 = 0 then we have β 1; and c t, c t+1 > 0 is only possible if β = 1. 7