Lecture 8: Bargaining and Repeated Games G5212: Game Theory Mark Dean Spring 2017 Lecture 8: Bargaining and Repeated Games Bargaining We will now apply the concept of SPNE to bargaining A bit of background Bargaining is hugely interesting but complicated to model It turns out that the outcome depends a lot on the details of the game i.e. the bargaining protocol you assume Has led some to thing that non-cooperative game theory is not the way to go See Nash 1950 Lecture 8: Bargaining and Repeated Games Bargaining However, (cid:145)Sequential Bargaining(cid:146)is still a classic framework One side makes an o⁄er The other side can either accept or reject If they reject then either the game ends or they get to make a counter o⁄er We will analyze a (cid:133)nite (two stage) bargaining game Then an in(cid:133)nite game Rubinstein bargaining Lecture 8: Bargaining and Repeated Games Two Stage Bargaining Example Two Stage Bargaining There are two players bargaining over a pie of size 1: There are two rounds: Player 1 (cid:133)rst propose a way to split the pie Player 2 either accepts or rejects it. If an o⁄er is accepted, the pie is split accordingly. Otherwise, the pie shrinks by 10% and the game proceed to the next round. Player 2 proposes in the second round, and player 1 either accepts it or rejects it. If the proposal is rejected, the pie disappears. Otherwise, the pie is split accordingly. Lecture 8: Bargaining and Repeated Games In(cid:133)nite Period Bargaining Example Rubinstein Bargaining. Now suppose there are 2n periods: *In t = 2n; P2 proposes (0;1); *in t = 2n 1, P1 proposes (1 (cid:14);(cid:14)); (cid:0) (cid:0) *in t = 2n 2; ((1 (cid:14))(cid:14);1 (1 (cid:14))(cid:14)) = (cid:14) (cid:14)2;1 (cid:14)+(cid:14)2 ; (cid:0) (cid:0) (cid:0) (cid:0) (cid:0) (cid:0) *in t = 2n 3; 1 (cid:14) 1 (cid:14)+(cid:14)2 ;(cid:14) 1 (cid:14)+(cid:14)2 = (cid:0) (cid:0) (cid:0) (cid:0)(cid:0) (cid:1) 1 (cid:14)+(cid:14)2 (cid:14)3;(cid:14) (cid:14)2+(cid:14)3 : (cid:0) (cid:0) (cid:0) (cid:0) (cid:0) (cid:1) (cid:0) (cid:1)(cid:1) *Inductively, we (cid:133)nd, in period 1, (cid:0) (cid:1) 1 (cid:14)+(cid:14)2 (cid:14)3+:::+(cid:14)2n 2 (cid:14)2n 1;(cid:14) (cid:14)2+(cid:14)3:: = (cid:0) (cid:0) (cid:0) (cid:0) (cid:0) (cid:0) 2n 1( (cid:14))k;1 2n 1( (cid:14))k = 1 (cid:14)2n; (cid:14)+(cid:14)2n (cid:0) k=(cid:0)0 (cid:0) (cid:0) k=(cid:0)0 (cid:0) (cid:0)1+(cid:14) 1+(cid:14) (cid:1) (cid:16)*IPf n ; the diviPsion goes to (cid:17) 1 (cid:16); (cid:14) (cid:17) ! 1 1+(cid:14) 1+(cid:14) (cid:16) (cid:17) Lecture 8: Bargaining and Repeated Games In(cid:133)nite Period Bargaining In(cid:133)nite Horizon Bargaining (Rubinstein 1982 Ecma). There are two players bargaining over a pie of size 1: Player 1 proposes at t = 1;3;::: and player 2 proposes at t = 2;4;::: They discounts future payo⁄with a discount factor (cid:14) (0;1). 2 Theorem. There is a unique SPNE: in any period, proposer o⁄ers (cid:14) to the other and keeps 1 for himself; responder 1+(cid:14) 1+(cid:14) accepts an o⁄er i⁄it is at least (cid:14) : 1+(cid:14) Lecture 8: Bargaining and Repeated Games In(cid:133)nite Period Bargaining Proof of Uniqueness Uniqueness: Let m be the smallest amount a proposer receives over all SPNE in all subgames. Let M be the highest amount. Claim 1. m 1 (cid:14)M (cid:21) (cid:0) The reason is that the responder today will be the proposer tomorrow (cid:150)this player will get at most M tomorrow. Hence, today(cid:146)s proposer, can ensure an acceptance by o⁄ering (cid:14)M to the receiver. Hence, his payo⁄must be no smaller than 1 (cid:14)M: (cid:0) Claim 2. M 1 (cid:14)m: (cid:20) (cid:0) The responder today has a payo⁄of at least (cid:14)m (because he gets at least m tomorrow as a proposer). So the responder will not accept anything less than (cid:14)m today. If (cid:14)m is accepted today, then the proposer receives a payo⁄of 1 (cid:14)m; if it is (cid:0) rejected, the responder today (i.e., the proposer tomorrow) will not o⁄er more than (cid:14)M tomorrow (which is only (cid:14)2M discounted back today). Therefore, M max 1 (cid:14)m;(cid:14)2M : (cid:20) (cid:0) (cid:8) (cid:9) Lecture 8: Bargaining and Repeated Games In(cid:133)nite Period Bargaining Proof of Uniqueness (conti.) Putting together the two claims, we have 1 m 1 (cid:14)M 1 (cid:14)(1 (cid:14)m) = m (cid:21) (cid:0) (cid:21) (cid:0) (cid:0) ) (cid:21) 1+(cid:14) 1 M 1 (cid:14)m 1 (cid:14)(1 (cid:14)M) = M (cid:20) (cid:0) (cid:20) (cid:0) (cid:0) ) (cid:20) 1+(cid:14) Therefore, 1 m = M = : 1+(cid:14) Lecture 8: Bargaining and Repeated Games In(cid:133)nite Period Bargaining Comments on Rubinstein Bargaining Immediate agreement. First mover advantage Discounting 1 (cid:14) is a (cid:147)friction(cid:148)(cid:150)basically the pie shrinks (cid:0) over time. If the friction is larger, i.e., (cid:14) smaller, then the proposer gets a larger share. If the friction disappears, i.e., (cid:14) 1; then shares 1 (this ! ! 2 is the Nash bargaining solution). Lecture 8: Bargaining and Repeated Games Repeated Games Think back to the prisoner(cid:146)s dilemma game Is it reasonable to think that people cannot sustain the (cid:145)good(cid:146)outcome of keeping quiet? Perhaps the problem is that people play prisoner(cid:146)s dilemma more than once Can that solve the problem?