14.04 Problem Set 2 Solutions

Problem Set 2 Solutions 14.04, Fall 2020 1 Uncertainty and Risk Aversion A farmer believes that there is a 50-50 chance that the next growing season will be abnormally rainy. His expected utility function has the form U (YN R , YR ) = 1 YN1−θ 1 YR1−θ R + 21−θ 21−θ where YN R and YR represent the farmer’s income in the states of ”normal rain” and ”rainy” respectively. a) Let θ = 1/2. Suppose the farmer must choose between two crops that promise the following income prospects Crop Wheat Corn YN R $28,000 19,000 YR 10,000 15,000 Which of the crops will he plant? Solution: If the farmer plants wheat, his expected utility will be 28, 0001/2 + 10, 0001/2 ≈ 267.3 Notice that the 12 probabilities cancel with 1 − θ. If the farmer plants corn, his expected utility will be 19, 0001/2 + 15, 0001/2 ≈ 260.3 Therefore the farmer chooses to plant wheat because while wheat is more risky, his risk aversion is not that high. He is willing to take on more risk for higher expected utility from a successful wheat crop when it isn’t abnormally rain. b) Suppose the farmer can plant half his field with each crop. This means that in the YN R state he will earn $14,000+9,500=23,500 and in the YR state he will earn $5,000+7,500=12,500. Would he choose to do this? Solution: Expected utility from planting half of his field with each crop would be 23, 5001/2 + 12, 5001/2 ≈ 265.1 He would not choose to do this since this is lower expected utility than planting wheat alone. This is because the farmer has low risk aversion, and wheat maximizes expected output 12 (YN R + YR ) 1 c) What mix of wheat and corn would provide maximum expected utility to this farmer? Solution: The farmer is solving the following optimization problem max (x28, 000 + (1 − x)19, 000)1/2 + (x10, 000 + (1 − x)15, 000)1/2 x∈[0,1] The optimum is achieved at x = 1 (a corner solution), which we might have guessed at from the earlier answers since the farmer has low risk aversion and therefore would prefer to plant only wheat. d) Let θ = 3/4. How would your answer to (c) change in this case? Give an intuitive reason why a higher risk aversion parameter changes his decisions. Solution: Now the farmer has higher risk aversion, which would suggest that he would want to plant less wheat, which is a very risky crop. The optimization problem is now max (x28, 000 + (1 − x)19, 000)1/4 + (x10, 000 + (1 − x)15, 000)1/4 x∈[0,1] Note the scalar multiple does not affect the optimal solution of x. The optimal x is ≈ .694 2 Pareto Optimality Suppose individuals denoted by i have happiness functions that only depend on one’s own consumption, called ui (xi ) where xi is a vector of one’s own consumption. However, individuals are altruists in that they care about other people’s happiness also. Let I be the total number of people. Each person’s overall utility function is Ui (x1 , . . . , xI ) where Ui has the form Ui (x1 , . . . , xI ) = Ui (u1 (x1 ), u2 (x2 ), . . . , uI (xI )) Assume that Ui is strictly increasing in all the arguments. Here, individual i’s utility depends on all the other people’s happiness functions u1 (x1 ), u2 (x2 ), . . . , uI (xI ). a) Show that if consumption allocation x = (x1 , . . . , xI ) is Pareto optimal under altruistic utility function Ui (·), then allocation x is also Pareto optimal under the individualistic utility function ui (·) (where individuals only care about their own happiness). Solution: We can prove the contrapositive that if x is not Pareto optimal under ui (·), then x is not Pareto optimal under Ui (·). If x is not Pareto optimal under ui (·), then there exists y that Pareto dominates x under ui . Then ui (yi ) ≥ ui (xi ) for all i and ui (yi ) > ui (xi ) for some i. Thus Ui (y) > Ui (x) for every i since Ui is strictly increasing in the u1 , . . . , uI values. This proves that y Pareto dominates x under Ui and therefore x is not Pareto optimal under Ui . b) Does this mean that a community of altruists can use standard competitive markets to attain Pareto optimality? (Optional) 2 3 Dynamic Programming Consider a farmer who lives for T periods. In each period t, the farmer chooses how much to consume (ct ) and how many seeds to plant to be available in the next period (Kt+1 ). In particular, consumption at date t is given by ct = αKt − Kt+1 where α > 0 is the yield to seed ratio. The farmer then solves the problem T X max {Kt+1 ,ct }T t=1 β t−1 u(ct ) (1) t=1 s.t. ct = αKt − Kt+1 ct ≥ 0 (2) where u(ct ) is the utility the farmer gets from consumption in period t, β ∈ (0, 1) and K1 > 0 is given exogenously (e.g. seeds the farmer inherited when he was born). Let {Kt∗+1 , c∗t }Tt=1 solve the above problem. That is, the farmer picks a sequence of consumption and seed-planting decisions in each period to solve the above problem. a) Argue that, regardless of the value of T , we will always have that KT∗ +1 = 0. Solution: If the farmer’s solution involved KT∗ +1 > 0, then by decreasing KT +1 they could increase cT (without affecting consumption in any previous period). So it cannot be an optimum. Intuitively, planting is only useful to the farmer because it enables them to consume more next period, and this benefit is irrelevant in the final period. b) Now let’s look at the case where T = 2. Use the fact that K3∗ = 0 and the constraints to eliminate ct and write the farmer’s problem as an optimization problem in one variable only, K2 . What constraint must K2 satisfy? Solution: Because K3∗ = 0 we have immediately c2 = αK2 . Furthermore c1 = αK1 − K2 . So we can substitute these into the utility function to get that the farmer maximizes u(αK1 − K2 ) + βu(αK2 ) Because consumption must be non-negative in each period, K2 must satisfy 0 ≤ K2 ≤ αK1 . c) Suppose the farmer’s per-period utility takes the form u(ct ) = ct . Solve for the farmer’s optimal planting decision, K2∗ . (Hint: there will be two cases, depending on the values of α and β). What is consumption in each period? 3 Solution: Now the farmer’s problem is max K2 :0≤K2 ≤αK1 αK1 − K2 + αβK2 If αβ > 1, they maximize this by setting K2 = αK1 , i.e. planting everything in the first period. In this case consumption in each period is c1 = 0, c2 = α2 K1 . If αβ < 1, they maximize this by setting K2 = 0, i.e. by eating everything in the first period. Consumption is c1 = αK1 and c2 = 0. 4