ANSWER KEY 4 GAME THEORY, ECON 395

ANSWER KEY 4 GAME THEORY, ECON 395 PROFESSOR A. JOSEPH GUSE (1) (Gibbons 2.1) Suppose a parent and child play the following game. First, the child takes an action A, that produces net income for the child, I C (A), and net income for the parent, I P (A). Second the parent observes incomes I C and I P and then chooses a bequest B to leave to the child. Payoffs to child and parent are W C and W P as follows. W C (I C,B) = U(I C +B) W P (I P,I C,B) = V(I P B)+kU(I C +B) where 0 < k < 1, where U and V are both increasing strictly concave functions and where I C (A)andI P (A)arestrictlyconcavewhichbothreachtheirimumatfiniteandpossibly different values of A. B can anything on [ I C,I P ]. Prove the Rotten Kid Theorem : in the backwards induction outcome, the child chooses the action that imizes the family s aggregate income, I C (A)+I P (A), even though only the parent s payoff exhibits altruism. ANSWER A convenient way to think about this problem is to define the total consumption flowing to the parent and to the child as ĨP(A) I P (A) B and ĨC(A) I C (A)+B. Using backward induction let us characterize the parent s problem and solution first. Note that by choosing B, the parents, in effect, choose ĨP(A) and ĨC(A) with the constraint that Ĩ P +ĨC = I P (A)+I C (A) This can be illustrated in our familiar indifference curve and budget line framework as seen in Figure 1. On the axes are the two quantities which the parents care about - ĨP and Ĩ C. We also see two hypothetical budget lines representing two possible actions taken by the child. For example, if the child takes action A 1, then the total income available will be I C (A 1 )+I P (A 1 ). Using the bequest the parent can choose to be anywhere on the budget line. Put another way, the child chooses the budget line (by choosing an action A) and the parents choose where on the budget line to be. To see how the parents choose, it is clear that if V and U are both differentiable functions, the parent will choose the pair (ĨC,ĨP) such that (1) 1 = 1

2 PROFESSOR A. JOSEPH GUSE Ĩ c I c (A 2 )+I p (A 2 ) A 2 I c (A 1 )+I p (A 1 ) A 1 Ĩ C (A 2 ) Ĩ C (A 1 ) Z 1 Z 2 H Child s Indifference Curves Parent s Indifference Curves ĨP(A1) ĨP(A2) Ic(A1) + Ip(A1) Ic(A2) + Ip(A2) Ĩ p Figure 1. Note that child s indifference curves in this space are horizontal lines, since they only care about ĨC. The parents are altruistic, caring about both their own total income ĨP and their child s ĨC. Hence their indifference curves are downward sloping. That they obey diminishing MRS is proved here. For example if the child chooses action A 1 then the parent will choose the tangency point Z 1 = (ĨP(A 2 ),ĨC(A 2 )) and according to the figure if the child chooses action A 2 the parent will choose point Z 2 = (ĨP(A 2 ),ĨC(A 2 )). The question is, does the child have incentive to choose a budget line which is as far out as possible? According to the figure, he does. The child clearly prefers the budget line generated by A 2 to that of A 1, since it results in his getting an income of ĨC(A 2 ) which (again according to the figure) is greater than ĨC(A 1 ). But how do we know this is just a result of how the figure was drawn? To see why it must be so, consider the value of the marginal rate of substitution at point Z 1. Since the parent is optimizing it must be equal to 1. What about at point H? If the law of diminishing MRS holds for the parent s preference in this setting then the (absolute value of the ) MRS must be less than 1 at H. And if that is true then it must be that figure accurately depicts the optimal choice for the parents when facing the budget line generate by A 2 in the sense that it involves more ĨC. (Put another way, if the law of diminishing MRS holds, then ĨC must be a normal good.) To verify that the law of diminishing MRS holds, take the derivative of W P/ ĨP w.r.t. Ĩ P : ĨP ( ) = ĨP = V (ĨP) ku (ĨC) ( ) V (ĨP) ku (ĨC)

ANSWER KEY 4 GAME THEORY, ECON 395 3 by assumption V is strictly concave. Therefore V < 0. Meanwhile U is increasing in its ( argument. ) Therefore U > 0. Putting these two facts together, it must be that WP / ĨP is negative. Hence diminishing MRS certainly holds and the figure is a ĨP correction depiction. QED. Aternate Answer Another (equivalent) way to think about this is to consider the parents problem and the first order condition B V(I P(A) B)+kU(I C (A)+B) The optimal choice of B must satisfy this first order condition V (I P (A) B)+kU (I C (A)+B) = 0 The optimal choice will depend on the choice of A from the child s previous move. Let s differentiate this first order condition w.r.t. to A taking the fact that a function B(A) is defined implicitly by this first order condition and re-arrange to learn something about how the child s final consumption level (I C +B) must change with A.. V (I P (A) B(A))(I P(A) B (A))+kU (I C (A)+B(A))(I C(A)+B (A)) = 0 I C(A)+B (A) = V (I P (A) B(A))(I P (A) B (A)) ku (I C (A)+B(A)) Note that since V < 0 and U < 0 this equation tells us that in any equilibrium the sign of I C (A)+B (A) must be the same as the sign of I P (A) B (A). In other words (2) I C(A)+B (A) > 0 I P(A) B (A) > 0 An important corollary is that whenever the sum of these two objects is positive each of them are positive. In other words (2) implies that I C(A)+I P(A) > 0 I C(A)+B (A) > 0 AND I P(A) B (A) > 0 But this is exactly what the problem asks us to show: when A increases total family income, it effectively increases the child consumption (through the parents FOC). In other words we want to show that QED. I C(A)+I P(A) > 0 I C(A)+B (A) > 0

4 PROFESSOR A. JOSEPH GUSE (2) (Gibbons 2.2) Samaritan s Dilemma. Another child and parent game. Let income I C and I P be fixed exogenously. In the first period the child decides how much of I C to save, S, consuming the rest (I C S) in the current period. In the second period, the parent chooses a bequest level B. The child s payoff W C and the parent s payoff W P are as follows. W C (I C,S,B) = U 1 (I C S)+U 2 (S +B) W P (I P,I C,S,B) = V(I P B)+k[U 1 (I C S)+U 2 (S +B)] Assume that the utility functions U 1, U 2 and V are increasing and strictly concave. Show that in the backwards induction outcome, the child saves too little, so as to induce the parent to leave a larger bequest (i.e., both the parent s and the child s payoffs could be increased if S were suiltably larger and B suitable smaller) ANSWER. Parents Desiction Since the parents decide on the bequest last, we examine their decision first. B V(I P B)+k[U 1 (I C S)+U 2 (S +B)] The optimal choice of B(S) as a function of the Child s savings decision must satisfy the following first order condition V (I p B(S)) = ku 2(S +B(S)) In words, this says that at the optimal bequest level B(S), the marginal cost to the parent of increasing the bequest in term of the value of their own consumption should just equal the marginal value the parents place on the child s 2nd period consumption. We learn about B (S) by totally differentiating this FOC (or applying the Implicit Function Rule)... B (S) = ku ku 2 (S +B(S)) 2 (S +B(S))+V (I p B(S)) Since by assumption U < 0 and V < 0, we conclude that B (S) < 0. That is, the more the child saves, the lower the bequest. Child s Decision Now let s examine the child s decision problem. In a SPNE, the child chooses a level of savings S looking ahead to the Parent s response, B(S). The first order condition is S U 1(I c S)+U 2 (S +B(S))

ANSWER KEY 4 GAME THEORY, ECON 395 5 Child s Second Period Consumption (S +B) I c +B(S ) I c +B(I c ) I c S +B(S ) S B(0) Orphan s Budget I c (S ) I c Child s BudgetWhen B = B(S) Slope = 1 + B (S) 1 Child s Budget When B = B(S ) Slope = 1 Child s First Period Consumption (I c S) Figure 2. Child Saving s Decision Problem. The Black Budget line represents child s range of choices when Parent reacts to his savings decision. The Green Budget line represents the child s range of choices when Parent s promise to fix bequest at B(S ) independent of S. U 1(I c S)+(1+B (S))U 2(S +B(S)) = 0 1+B (S) = U 1 (I c S) U 2 (S +B(S)) Let S be the optimal choice in this problem and B(S ) be the parents response. Now suppose instead that the parents threatened to set B(S) = B(S ) regardless of what the child chooses to save. In this case, the value of V would not change from its value in the SPNE. What about the value of U 1 anfd U 2. Let s reanalyze the childs choice. S U 1(I c S)+U 2 (S +B(S )) Let S be the solution to this problem. The FOC is 1 = U 1 (I c S ) U 2 (S +B(S )) Since the Child can still achieve the same point in this problem and but chooses not to, he must be better off. (U 1 +U 2 is bigger) and since V is the same, the parents are better off as well. (3) (Gibbons 2.3) Suppose players in Rubinstein s infinite-horizon bargaining game have different discount factors, δ 1 and δ 2. Show that in any subgame perfect equilibrium (SPNE), player 1 offers

6 PROFESSOR A. JOSEPH GUSE ( 1 δ2, δ ) 2(1 δ 1 ) 1 δ 1 δ 2 1 δ 1 δ 2 ANSWER. Let Ω = {ω 1,ω 2,...} be the set of all SPNE in this game. Suppose for the moment that this set is non-empty and either finite or infinite. Let M(ω) be the payoff going to player 1 in the game when strategies are played according to equilibrium ω. Define the following two quantities. 1 M ω Ω M(ω) M min min ω Ω M(ω) Now consider the game at t=2. Since this game looks just like the game beginning at t = 0, M must also be the most player 1 can expect if the game reaches this point. This means that at t = 1 player 2 will offer player 1 at most δ 1 M, keeping at least 1 δ 1 M for herself. This, in turns means that at t = 0, player 1 could never expect 2 to accept anything more than δ 2 (1 δ 1 M ) and therefore could never expect to keep more than 1 δ 2 (1 δ 1 M ). But this means that, by definition, M = 1 δ 2 (1 δ 1 M ) M (1 δ 2 δ 1 ) = 1 δ 2 M = 1 δ 2 (1 δ 2 δ 1 ) Return now to t = 2 and repeat a similar line of argument involving M min as follows. M min is the least player 1 can expect if the game reaches t = 2. This means that at t = 1 player 2 will offer player 1 at least δ 1 M min, keeping at most 1 δ 1 M min for herself. This, in turns means that at t = 0, player 1 can expect 2 to always anything less than δ 2 (1 δ 1 M min ) and therefore could always expect to keep at least 1 δ 2 (1 δ 1 M min ). But this means that, by definition, M min = 1 δ 2 (1 δ 1 M min ) M min (1 δ 2 δ 1 ) = 1 δ 2 M min = 1 δ 2 (1 δ 2 δ 1 ) But this is the same value that we came up with for M. Hence M = M min = M, which means that if any SPNE exist they involve player 1 offering to keep exactly M = 1 In the case where Ω is an infinite set and the and min are not guaranteed to exist, let M be the least upper bound and M min be the greatest lower bound

1 δ 2 (1 δ 2 δ 1 ) ANSWER KEY 4 GAME THEORY, ECON 395 7 at t = 0. We have only to verify that this is an equlibrium. (Remember we began by assuming the set of SPNE was non-empty.) Recall that a strategy profile must specify a complete plan of action for both players whenever it might be their turn to move. Suppose that the players use the following strategies for proposal making and responding. PROPOSING 1 s Share 2 s Share When? Player 1 M 1 M even t Player 2 δ 1 M 1 δm odd t RESPONDING Accept Values Reject Values (1 s Share) (1 s Share) When? Player 1 [δ 1 M,1] [0,δ 1 M) odd t Player 2 [0,M] (M,1] even t It is easy to verify that this is a subgame perfect equilibrium.