Optimization of Consumption and Savings with Uncertainty and Borrowing Constraints - Prof., Assignments of Economics

Download Optimization of Consumption and Savings with Uncertainty and Borrowing Constraints - Prof. and more Assignments Economics in PDF only on Docsity! Econ 601 Problem Set 6 (Sketch of Solutions) Exercise 1 We solve this problem by backward induction. The dynamic budget con- straint for the last period plus the solvency condition implies that 0 = (1 + r)(aT + yT cT ) cT = aT + yT For cT to be non-negative with probability one we need that aT  min(yT ) (1) Now, move back one period. From the dynamic constraint we have that aT = (1 + r)(aT1 + yT1 cT1) cT1 = aT1 + yT1 aT(1+r) For cT1 and cT to be nonnegative with probability one, assets in aT1 must satisfy the following inequality aT1  min(yT1) + min(aT ) 1 + r (2) Replacing equation (1) into equation (2) aT1  min(yT1) min(yT ) 1 + r (3) Move back another period. The dynamic budget constraint implies that aT1 = (1 + r)(aT2 + yT2 cT2) cT2 = aT2 + yT2 aT1 (1 + r) (4) For cT2; cT1 and cT to be nonnegative with probability one, assets in aT1 must satisfy the following inequality 1 aT2  min(yT2) + min(aT1) 1 + r (5) Replacing equation (4) into equation (5) aT2  min(yT2) min(yT1) 1 + r min(yT ) (1 + r) 2 (6) If one keeps iterating backward one gets that for consumption to be nonneg- ative with probability 1 in period t and all the following period it has to be true that at  TX s=t (1 + r)(st)min(ys) Exercise 2 Part a Given the assumption of no discounting and zero interest rate, the La- grangian is given by max c1;c2 L = 1ae ac1 E1 1ae ac2 +  [y1 + y2 c1 c2] First order conditions c1c eac1  = 0 (7) c2c E1e ac2  = 0 (8) c y1 + y2 = c1 + c2 (9) From equation (7) and equation (8) we get that eac1 = E1e ac2 (10) Replacing equation (9) into equation (10) eac1 = E1e a(y1+y2c1) eac1 = ea(y1c1)E1e ay2 (11) 2 Let’s …rst solve the certainty equivalence case. In this case d = 0 and then y1 = y2 = y3 = 1. Given that by assumption R = 1, the intertemporal budget constraint can be written as 3X t=1 ct = 3X t=1 yt (20) Write the Lagrangian max fctg3t=1 L = 3X t=1 u(ct) +  " 3X t=1 yt 3X t=1 ct # FOC c1  = 0 (21) c2  = 0 (22) c3  = 0 (23) Equations (21), (22) and (23) imply c1 = c2 = c3 (24) Replacing equation (24) into equation (20) we get that c1 = c2 = c3 = 1. Therefore, s1 = s2 = 0. Now, let’s solve the case with uncertainty. To do that, we have to work backwards. In period 3, s2 is given and the agent is going to consume c3 = 1+s2. This means that value function for period 3 is V3(s2) = u(1 + s2) (25) Having period 3 value function we can now solve for period 2 problem. The agent problem at period 2 is V2(s1; y2) = max s2 fu(y2 + s1 s2) + V3(s2)g (26) Replacing equation (25) into equation (26) we get V2(s1; y2) = max s2 fu(y2 + s1 s2) + u(1 + s2)g (27) 5 FOC s2c u0(y2 + s1 s2) + u0(1 + s2) = 0 Rearranging and taking inverse y2 + s1 s2 = 1 + s2 s2 = y2 + s1 1 2 (28) The actual savings in period 2 will depend on the shock we get that period. If we get a positive shock, saving will be spos2 = 1 + d+ s1 1 2 = s1 + d 2 (29) Now, the value function in period 2 with a good shock V pos2 (s1; 1 + d) = u(1 + d+ s1 s pos 2 ) + u(1 + s pos 2 ) V pos2 (s1; 1 + d) = u  1 + d+ s1 s1 + d 2  + u  1 + s1 + d 2  V pos2 (s1; 1 + d) = u  2 + d+ s1 2  + u  2 + s1 + d 2  V pos2 (s1; 1 + d) = 2u  2 + d+ s1 2  (30) Now, let’s consider what would happen if the household gets a bad shock in the second period. Equation (28) implies that if s1 < d, s2 < 0 which violates the no-borrowing constraint. Then, if s1 < d, s2 should be 0. Note that in a setting with quadratic utility and = R = 1, s1 will always be less than d. If the household were to save more than d, it would be shifting too many resources to the future. That is, the marginal utility of consumption today will be less that the expected value of the marginal utility of consumption tomorrow which would violate the Euler Equation. To see this more clearly, let’s take, for example, the case of s1 = d. In this case, u0(c1) = a bc1 = a b(1 d+ s0) = a b+ bd In period 2, after a bad shock, 6 u0neg(c2) = a bc2 = a b(1 d+ s1 sneg2 ) = a b(1 s neg 2 ) To …gure out what sneg2 is, remember that there is no more uncertainty after period 2. Euler equation between period 2 and 3 is simple c2 = c3, where is the lagrange multiplier on the no-borrowing constraint. Given that income is 1 in period 3 and they currently have 1 unit of resources (y2 = 1 d and s1 = d) sneg2 = 0 for the agent to smooth consumption across periods. Then, u0neg(c2) = a bc2 = a b(1 d+ s1 sneg2 ) = a b After a good shock, u0pos(c2) = a bc2 = a b(1 + d+ s1 spos2 ) = a b(1 + 2d s pos 2 ) Note that the resources available after a positive shock are 1+2d and period 3 resources is 1: The di¤erence is 2d so they should set spos2 = d to smooth consumption across periods. Then, u0pos(c2) = a bc2 = a b(1 + d+ s1 spos2 ) = a b bd Now, let’s calculate the expected value in period 1 of marginal utility of period 2 consumption. E1(u 0(c2)) = 1 2u 0pos(c2) + 1 2u 0neg(c2) E1(u 0(c2)) = 1 2 (a b bd) + 1 2 (a b) = a b 1 2bd Therefore, for savings s1  d, u0(c1) > E1(u0(c2)) which implies that the Euler Equation is being violated. Therefore savings are not being set at the optimal level. The household will always choose s1 < d which implies s2 will optimally be set equal to 0 after a bad shock. Then, V neg2 (s1; 1 d) = u(y neg 2 + s1 s neg 2 ) + u(1 + s neg 2 ) V neg2 (s1; 1 d) = u(1 d+ s1) + u(1) (31) Then, the value function in period one is V1(1) = max s1  u(1 s1) + 1 2 V neg2 (s1; 1 d) + 1 2 V pos2 (s1; 1 + d)  (32) Replacing equation (30) and (31) into equation (32) we get 7 2(1 s1)  2 + s1 + d 2   2 + s1 d 2  = 0 2(1 s1) = 4 + 2s1 2 s1 = 0 As expected, we get the same solution as the certainty equivalence case, which implies no precautionary savings. Note that in this case in the second period after a bad shock the household will borrow sneg2 = d2 , and it will lend in the case of a good shock s POS 2 = d 2 . Then, with no borrowing constraints, the consumption history in this case of the negative and positive shock will be  1; 1 d2 ; 1 d 2 and  1; 1 + d2 ; 1 + d 2 respectively. Part c Let’s solve …rst the problem of the lucky households. We have already solve for the value of getting a positive shock in period 2. V pos2 (s1; 1 + d) = 2u 2+d+s1 2
. We need to solve for optimal period 1 saving. The value function will be given by V1(1) = max s1  u(1 s1) + 2u  2 + d+ s1 2  FOC s1c u(1 s1) + u0  2 + d+ s1 2  = 0 Using functional forms a+ b(1 s1) + a b  2 + d+ s1 2  = 0 1 s1  2 + d+ s1 2  = 0 s1 = d 3 (35) Replacing equation (35) into equation (29) we get 10 s2 = d 3 Then, the consumption history for the lucky household will be  1 + d3 ; 1 + d 3 ; 1 + d 3 . For the unlucky households, the period 2 value function will be V neg2 (s1; 1 d) = u(1 d+ s1) + u(1) Their period 1 value function will be given by V1(1) = max s1 fu(1 s1) + u(1 d+ s1) + u(1)g FOC s1c u0(1 s1) + u0(1 d+ s1) = 0 Using functional forms a+ b(1 s1) + a b(1 d+ s1) = 0 (1 s1) (1 d+ s1) = 0 s1 = d 2 Given that the Euler Equation holds between period one and two, s2 = 0 Then, the consumption history for the unlucky household will be  1 d2 ; 1 d 2 ; 1 . Then, the economy-wide savings in period one will be s1 = 1 2s Lucky 1 + 1 2s Unlucky 1 = 1 2 d3
+ 12 d 2
= d12 Part d The economy-wide savings: - Homogeneous agents & borrowing constraints: s1 = d7 - Homogeneous agents & no borrowing constraints: s1 = 0 - Heterogeneous agents & borrowing constraints: s1 = d12 11