kids (this case is for j = 1; j = 2 case is similar). For the interior solution case, we have 1 = c (x 2 t) + p 2

Transcription

1 Problem 1 There are two subgames, or stages. At stage 1, eah ie ream parlor i (I all it firm i from now on) selets loation x i simultaneously. At stage 2, eah firm i hooses pries p i. To find SPE, we start from stage 2. At stage 2, (x 1, x 2 ) are given. If x 1 = x 2, whoever harges less gets the whole onsumers and if their pries are same, eah get half of the onsumers. Thus, only possible Nash equilibrium is p 1 = p 2 = 0, as setting any positive prie would make the other firm to underut slighly (your prie - ɛ) and take all the kids; and you want to underut the other firm slightly to take all the kids bak. Consider the ase x 1 = x 2. Without the loss of generality, I assume x 1 < x 2. Note that NE is only possible when we have an interior solution: there is a mid-point t suh that x 1 < t < x 2 and kid at t is indifferent. If we do not have interior solution, it means that kids at [0, x 1 ] or [x 2, 1]are indifferent and others prefer one firm, or all kids go to one firm, say firm i. For the latter ase, firm j would deviate to a slightly lower prie so that it an make positive profit. For the former ase, firm j would deviate to a slightly lower prie so that it an sell to [0, x 1 ] and some more, instead of selling only to x1 2 kids (this ase is for j = 1; j = 2 ase is similar). For the interior solution ase, we have Solving, we get (x 1 t) 2 + p 2 1 = (x 2 t) + p 2 p 2 p 1 x 1 + x 2 t = + 2 (x 2 x 1 ) 2 Firm 1 solves the following profit maximization problem: { } p 2 p 1 x 1 + x 2 Max p1 tp 1 = + p1 2 (x 2 x 1 ) 2 Taking FOC (first order ondition), we have ( p 2 x 22 ) p BR x (p 2 ) = Similarly, firm 2 solves { } p 2 p1 x 1 + x 2 Max p2 (1 t)p 2 = 1 + p 2 2 (x 2 x 1 ) 2 Taking FOC, we have ( ) p p BR 1 x 2 2 x (p 1 ) = (x 2 x 1 ) Solving two equations, we get NE: p 1 = (x 2 x 1 ) (x 1 + x 2 + 2), p 2 = (x 2 x 1 ) (4 x 1 x 2 ) 3 3 1

2 Profits are π 1 = (x 2 x 1 ) (x 1 + x 2 + 2), π 2 = (x 2 x 1 ) (4 x 1 x 2 ) At stage 1, hoosing x 1 = x 2 annot be an equilibrium, as in this ase, they will make zero profit at the stage 2 sine p 1 = p 2 = 0 is NE in the stage 2. Firm 1 knows that his profit will be π 1 = 2 18 (x 2 x 1 ) (x 1 + x 2 + 2) given x 2. Taking FOC, we have π 1 = (x 1 + x 2 + 2) (x 2 3x 1 2) < 0 x 1 18 as 0 x 1 < x 2 1. Thus, hoosing x 1 = 0 is optimal. Similiarly, for firm 2, π 2 = (4 + x 1 3x 2 ) (4 x 1 x 2 ) > 0 x 2 18 Therefore, hoosing x 2 = 1 is optimal. In summary, SPE is that at stage 1, firm 1 hooses x 1 = 0 and firm 2 hooses x 2 = 1. For stage 2, if x 1 = x 2, p 1 = p 2 = 0. If x 1 = x 2, p 1 = 3 (x2 x 1 ) (x 1 + x 2 + 2), p 2 = 3 ( x 2 x 1 ) (4 x 1 x 2 ). Problem 2 I denote hoies as P1 hooses L or R, P2 hooses l or r, P1 hooses A or B, P2 hooses a or b ( (from top to bottom ) order). The bottom subgame has three NE: (A, a), (B, b), 1 A + 3 B, 1 a b. At the next subgame, If the first NE is played, payoff is 3 for both players so ( P2 will hoose r. ) Otherwise, payoff is less than 2 (1 for (B, b) and 3 4 for 1 A + 3 B, 1 a b ), so P2 will hoose l. For the next (and last) subgame, if (A, a) is played and P2 played r, P1 will hoose R. Otherwise, he is indifferent between L and R. Therefore, SPE are (RA, ra), ({pl + (1 p)r} B, lb), ( {pl + (1 p)r} { 1 A + 3 B 4 4 Problem 3 The proposed strategy is for player i to offer δp i to other players and 1 δ(1 p i ) for himself, and to aept an offer if it is at least p i. To hek that this strategy is a SPE, we onsider all single deviations. Suppose i is the proposer after any history. His equilibrium strategy has payoff 1 δ(1 p i ). He an deviate to any strategy whih offers all other players k j δp j, whih would be aepted, yielding a payoff of 1 δ k j 1 δ(1 p i ). This is not a profitable deviation. He an deviate to offer some player less, k j < p j for some j. Then the offer is rejeted, and in the next round he has expeted payoff δp i < 1 δ(1 p i ). This is not a profitable deviation. Next, onsider i as a non-proposer after any history. Suppose he is offered k i δp i. If he aepts the offer, he gets either k i if all other non proposers aept, or expeted δp i in the next round. He an deviate to rejet the offer, } { }), l, 1 a + 1 b

3 and he would get an expeted payoff δp i in the next round. Thus, deviating is not profitable. If a non-proposer is offered k i < δp i, he should rejet in equilibrium, for an expeted payoff of δp i in the next round. If he deviates to aept, he gets either k i if all other players aept, or δp i if someone rejet. Both ases are weakly worse than rejeting, so deviating is not profitable. We have heked all possible single deviations and none are profitable. Problem 4 For this problem, let us define π t (1) to be the proposer in period t and π t (2), π t (3) to be the first and seond responders in period t. Let e i be the proposal that gives 1 to player i and 0 to everyone else. Define x t to be the proposal at period t and x t i be i s share of that. The states are: {k 0, p(1), p(2), p(3)}. For any division a = (p 1, p 2, 1 p 1 p 2 ) where eah player has stritly positive payoff, there is some δ where a SPE gives division a. Consider the following strategy: At t = 0, we start in state k 0. In this state, player π 0 (1) proposes x 0 = a. π 0 (2) aepts iff x 0 = a. π 0 (3) aepts if x 0 = a or x 0 π 0 (3) δ. After k 0, if x 0 = a we go to stage p(π 0 (1)). Otherwise, if π 0 (2) rejeted, we go to stage p(π 0 (2)). If π 0 (2) aepts and π 0 (3) rejeted, we go to stage p(π 0 (3)). In stage p(i), the proposer offers x t = e i. In these states, π t (2) aepts if the proposer offers x t = e i. π t (3) aepts if x t π t (3) δ OR xt = e i. After rejetion in state p(i), we stay in p(i) if x t = e i. Otherwise, if π t (2) rejets, we go to p(π t (2)). If π t (2) aepts and π t (3) rejets, we go to p(π t (2)). To hek that this is an equilibrium, we look at all single deviations after any valid history. We first hek that there are no profitable deviations at t = 0. Proposer: At t = 0, π 0 (1) offers x t = a. If he deviates and offers x 0 = a, it is rejeted by π 0 (2). Then in the next period we are in stage p(π 0 (2)), where e π0 (2) is offered and aepted. This gives payoff 0 to player π 0 (1), whih is not a profitable deviation. Next, we look at deviations by the responder at t = 0. Case: x 0 = a. In equilibrium both responders aept. If either responder hooses to rejet, we go to state p ( π 0 (1)), where both responders get payoff 0. This is learly not a profitable deviation. Case: x 0 = a and x 0 π 0 (3) δ. Consider π0 (3). Assume π 0 (2) aepts. In equilibrium, π 0 (3) aepts and gets x 0 π 0 (3) δ. If he deviates and rejets, we go to state p(π 0 (3)), whih leads to payoff δ, so deviation is not profitable. After the history where π 0 (2) does not aept, π 0 (3) is indifferent between his ations. Consider π 0 (2). In equilibrium, π 0 (2) rejets the offer, we go to state p(π 0 (2)) whih yields payoff δ. If player π 0 (2) instead deviates and aepts, 3

4 we know that π 0 (3) also aepts, whih yields payoff of x 0 π 0 (2) for π0 (2). Sine we assume that δ is large and x t π 0 0 (3) δ, x π 0 (2) must be small, and thus less than δ. This requires δ 0.5. Case: x 0 = a and x 0 π 0 (3) < δ. Consider π0 (3). Assume π 0 (2) aepts. Player π 0 (3) rejets in equilibrium and gets δ in state p(π 0 (3)). If π 0 (3) aepts, he instead gets xπ 0 < δ, whih is not profitable. if π 0 (2) rejets, π 0 0 (3) (3) is indifferent again. Consider π 0 (2). In equilibrium, π 0 (2) rejets and gets δ. If he deviates and aepts, π 0 (3) rejets and we go to p(π 0 (3)), where π 0 (2) gets 0. Next, we onsider t 1, when we are in state p(i). To see that the proposer has no profitable deviation, onsider two ases. If we are not in state p(π t (1)): In equilibrium the proposer offers e i and is aepted, so the proposer has payoff 0. If he offers any other x t = e i, the offer is rejeted and we go to p(π t (2)), where the proposer has payoff 0. The proposer is indifferent between any offer. If we are in the state p(π t (1)): In equilibrium, π t (1) offers x t = e πt (1), it is aepted and he has payoff 1. This is his maximum payoff, so there is no profitable deviation. Finally, we have to hek that responders have no profitable deviation, in three ases. 1. x t = e i 2. x t = e i and x t π t (3) δ 3. x t = e i and x t π t (3) < δ Case 1: In equilibrium all responders aept x t. By deviating to rejetion (for either player), we return to this state again in the next period, and all players are weakly worse off. Case 2: First onsider π t (3). Suppose that π t (2) aepts. In equilibrium, π t (3) aepts and gets a payoff of δ t x t π t (3). If he deviates and rejets, we go to p(π t (3)), where he gets a payoff δ t+1, whih is not a profitable deviation. If π t (2) rejeted, π t (3) is indifferent. Now onsider π t (2). In equilibrium, he rejets, and we go to p(π t (2)), where he gets δ t+1. If he deviates and aepts, he instead gets δ t x t π t (2). Beause δ is large and x t π(3) δ, we an assume that δt x t π t (2) δt+1. Case 3: First onsider π t (3). Suppose that π t (2) aepts. In equilibrium, π t (3) rejets and gets payoff δ t+1. If he deviates and aepts, he gets δ t x t π t (3) stritly less. This is not a profitable deviation. If π t (2) rejeted, π t (3) is indifferent. Now onsider π t (2). In equilibrium he rejets, and we go to p(π t (2)), where the payoff is δ t+1. If instead he deviates and aepts, π t (3) rejets, we go to p(π t (3)) and π t (2) gets payoff 0. This is not profitable. 4

5 Thus, we see that there is no single profitable deviation, and we have a subgame perfet equilibrium. 5

6 MIT OpenCourseWare Eonomi Appliations of Game Theory Fall 2012 For information about iting these materials or our Terms of Use, visit: