Lecture Notes on Bargaining

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1 Lecture Notes on Bargaining Levent Koçkesen 1 Axiomatic Bargaining and Nash Solution 1.1 Preliminaries The axiomatic theory of bargaining originated in a fundamental paper by Nash (1950, Econometrica). The scenario is that two agents have access to any of the alternatives in some set S, called the feasible set. Their preferences over the alternatives in the feasible set di er. If they agree on a particular alternative, that is what they get. Otherwise, they end up at a prespeci ed alternative in the feasible set called the disagreement point, which is denoted by d. The feasible set is given in the utility space. More formally, a bargaining problem is de ned by the tuple (S; d) whereweassumethats is a non-empty, convex, and compact subset of R 2 ; and d 2 S We further assume that there exists an s 2 S, such that s À d 1 Denote the set of all such bargaining problems by B A solution on B is any function f B!S Example 1 Two individuals are bargaining over a cake of size ¼; some of which they may discard if they wish. So let X = x 2 R 2 + x 1 + x 2 ¼ ª be the set of outcomes that the bargaining agents may agree upon. If they fail to agree, they each receive x i =0 Also let the von Neumann-Morgenstern preferences of the agents are represented by strictly increasing and concave utility functions u 1 X! R and u 2 X! R Assume that u 1 and u 2 are normalized, so that d i = u i (0) = 0; i=1; 2 Then, the bargaining problem is (S;0) ; where S (v 1 ;v 2 ) 2 R 2 9x 2 X with u 1 (x) =v 1 ;u 2 (x) =v 2 ª Exercise 1.1 Verify that (S; 0) as de ned above is a bargaining problem. i 1 Given x; y 2 R n ;x y if x i y i for all i; x > y if x i y i for all i and x 6= y; and x À y if x i >y i for all 1

2 1.2 Nash Bargaining Solution Nash bargaining solution, denoted f N (S; d) ; is given by f N (S; d) =argmax d s2s (s 1 d 1 )(s 2 d 2 ) for all (S; d) 2B The beauty of this solution concept is that it is the unique solution which satis es the following four axioms Axiom 1 (Pareto E ciency (PAR)) If s>f(s; d) ; then s=2 S Axiom 2 (Symmetry (SYM)) If (s 1 ;s 2 ) 2 S implies (s 2 ;s 1 ) 2 S; then f 1 (S; d) =f 2 (S; d) Axiom 3 (Invariance (INV)) Let (S; d) be a bargaining problem and de ne the bargaining problem (S 0 ;d 0 ) as S 0 = f 1 s 1 + 1; 2 s s 2 Sg and where 1 ; 2 > 0 Then we have d 0 i = id i + i f i (S 0 ;d 0 )= i f i (S; d)+ i for i =1; 2 Axiom 4 (Independence of Irrelevant Alternatives (IIA)) If (S; d) and (T;d) are bargaining problems with S ½ T and f (T;d) 2 S; then f (S; d) =f (T;d) Theorem 2 Asolutionf satis es PAR, SYM, INV, and IIA if and only if f (S; d) =f N (S; d) Proof. Exercise 1.2 Verify that f N satis es the four axioms. To prove the necessity, take any bargaining problem (S; d) and suppose that f satis es all the axioms. We need to show that f (S; d) =f N (S; d) Step 1. Let f N (S; d) =z Since by assumption there exists s 2 S such that s i >d i for i =1; 2; from the de nition of f N we have z i >d i for i =1; 2 Let (S 0 ;d 0 ) be the bargaining problem de ned by S 0 = f 1 s 1 + 1; 2 s s 2 Sg and d 0 i = id i + i 2

3 where i = 1 2(z i d i ) d i i = 2(z i d i ) Notice that d 0 =(0; 0) and i f N (S; d)+ i = i z i + i =1=2 for i =1; 2 Since both f and f N satisfy INV we have f i (S 0 ; 0) = i f i (S; d)+ i f N i (S 0 ; 0) = i f N i (S; d)+ i for i =1; 2; so that f (S; d) =f N (S; d) if and only if f (S 0 ; 0) = f N (S 0 ; 0) = (1=2; 1=2) Step 2. For all (s 0 1 ;s0 2 ) 2 S0 we have s 0 1 +s0 2 1 To see this, suppose there exists (s0 1 ;s0 2 ) 2 S0 such that s s0 2 > 1 Let (t 1 ;t 2 ) = ((1 ")(1=2) + "s 0 1 ; (1 ")(1=2) + "s0 2 ) where " 2 (0; 1). Since S 0 is convex, (t 1 ;t 2 ) 2 S 0 But for small enough "; t 1 t 2 > 1=4; contradicting the fact that f N (S 0 ; 0) = 1=4 Step 3. Since S 0 is bounded, the result of Step 2 ensures that we can nd a rectangle T that is symmetric about the 45 o line and that contains S 0 ; on the boundary of which is (1=2; 1=2) (See Figure 1). Step 4. By PAR and SYM we have f (T;0) = (1=2; 1=2) Step 5. SinceS 0 ½ T and f (T;0) = (1=2; 1=2) 2 S 0 ; by IIA we have f (S 0 ; 0) = f (T;0) = (1=2; 1=2) ; completing the proof. s 2 T 1 1/2 s 1 s 2 = 1/4 0 1/2 S' 1 s 1 3

4 Example 3 Suppose individual 1 and individual 2 are trying to divide 10 dollars and for all x 2 [0; 10] Then, u 1 (x) = x u 2 (x) = ln(1+x) S = f(u 1 ;u 2 )0 u 1 10; 0 u 2 ln (11 u 1 )g Assume d =(u 1 (0) ;u 2 (0)) = (0; 0) To nd the Nash solution we have to solve max x ln (11 x) x2[0;10] Solution is obviously interior, and hence satis es the rst order condition ln (11 x) x 11 x =0 whichissolvedatx ' 657 Notice that the less risk averse individual gets a larger share. (See Figure 2) u S 6.57 u 1 Exercise 1.3 Two individuals are bargaining over a cake of size 1; some of which they may discard if they wish. So let X = x 2 R 2 + x 1 + x 2 1 ª be the set of outcomes that the bargaining agents may agree upon. If they fail to agree, they each receive x i =0 Also let the von Neumann-Morgenstern preferences of the agents are represented by strictly increasing, concave, and di erentiable utility functions u 1 X! R and u 2 X! R Assume that u 1 and u 2 are normalized, so that d i = u i (0) = 0; i=1; 2 Then, the bargaining problem is (S;0) ; where S (v 1 ;v 2 ) 2 R 2 9x 2 X with u 1 (x) =v 1 ;u 2 (x) =v 2 ª 4

5 Now, consider the bargaining problem (S 0 ; 0) which di ers from (S; 0) only in that u 2 is replaced with w 2 h±u 2 where h R! R is a strictly increasing and strictly concave function with h (0) = 0 Show that f2 N (S 0 ; 0) <f2 N (S; 0) Axiom 5 (Strong Individual Rationality (SIR)) F (S; d) À d Theorem 4 A bargaining solution f satisfy PAR, SIR, INV, and IIA if and only if there exists 2 (0; 1) such that f (S; d) =f (S; d) 2 arg max d s2s s 1 s1 2 f is called an asymmetric (if 6= 1=2) Nash bargaining solution. Proof. Exercise. The parameter re ects the relative bargaining power of agent 1. 2 Alternating O ers (Rubinstein) Bargaining 2.1 Preliminaries Two players, A and B, bargain over a cake of size 1. At time 0 player A makes an o er x A 2 [0; 1] to player B. If player B accepts the o er, agreement is reached and player A receives x A and player B receives 1 x A If player B rejects the o er, she makes a countero er x B 2 [0; 1] at time > 0 If this countero er is accepted by player A, then player B receives x B and player A receives 1 x B Otherwise, player A makes another o er at time 2 This process continues inde nitely until a player accepts an o er. If the players reach an agreement at time t on a partition that gives player i asharex i of the cake, then player i s payo is x i e rit where r i > 0 is player i s discount rate. If players never reach an agreement, then each player s payo is zero. For notational convenience de ne ± i = e ri so that payo to share x i at period t is ± t ix i 2.2 Stationary No-delay Equilibrium We will rst characterize the SPE with the following two properties, and then show that all SPE have these properties. 1. (No Delay) All equilibrium o ers are accepted. 2. (Stationarity) A player always makes the same o er in equilibrium. 5

6 Let x i denote the equilibrium o er by player i Given properties 1 and 2, the current present value of rejecting an o er x A is ± Bx B for player B. This implies that in equilibrium 1 x A = ± Bx B (1) Similarly 1 x B = ± Ax A (2) Therefore, there is a unique solution x A = 1 ± B (3) x B = 1 ± A (4) Thus, there exists at most one SPE satisfying the two properties. But we still have to verify there exists such an equilibrium. Consider the strategy pro le (s A ;s B ) de ned as Player A Always o er x A ; accept any x B with 1 x B ± A x A Player B Always o er x B ; accept any x A with 1 x A ± B x B Proposition 5 The above strategy pro le is a SPE. Proof. We will rst prove Lemma 6 One-deviation property holds for the Rubinstein bargaining game. Sketch of Proof. Suppose s is not SPE. Then there exists a history h, aplayeri; and s i such that s i is a pro table deviation from history h on. The deviation cannot lead to perpetual disagreement, as the payo to disagreement is zero. So, it must be that the deviation leads to acceptance of an o er by some player at a nite period, say t 0. We can now de ne a new strategy which is the same as s i until (and including) t 0 andthenitisidenticaltos i This, however, is a pro table deviation which di ers from s i in only nitely many histories. From then on we can nd a one-shot pro table deviation as we did in the earlier proof of one-deviation property in nite games.k Consider any period when A has to make an o er. Her payo to s A is x A If A o ers x A <x A then 1 x A >± B x B by equation (1) and hence B accepts any such o er which gives A a payo less than x A If she o ers x A >x A ; then B rejects and o ers x B ; A accepts giving her a payo of ± A (1 x B) <x A 6

7 by equation (1). Therefore, there is no pro table one-shot deviation in any subgame starting with her o er. Now, consider subgames starting with player A responding. If player A rejects o er x B with 1 x B ± A x A ; then she will o er x A herself and get ± Ax A So this is not a pro table deviation. By a symmetric argument, it follows that player B s strategy is optimal in every subgame as well. 2.3 Unique Subgame Perfect Equilibrium Theorem 7 The strategy pro le s istheuniquespe. Proof. Let i denote any subgame that starts with player i making an o er. Clearly, all such subgames are strategically equivalent (since preferences are stationary, i.e., does not depend on calendar time) and thus all have the same SPE. Let G i denote the set of SPE in any subgame i and let M i = supg i m i = infg i Lemma 8 There exists a unique SPE payo pro le of A given by (x A ; 1 x A ) and a unique SPE payo pro le of A given by (x B ; 1 x B ) Proof of Lemma 8. Claim 1. m i 1 ± j M j ;i6= j Proof of Claim 1. First note that player B accepts any o er x A such that 1 x A >± B M B Now suppose m A < 1 ± B M B for contradiction. Then, there exists an equilibrium of A yielding u A < 1 ± B M B [If, u A 1 ± B M B >m A ; for all u A 2 G A ; we would contradict the fact that m A =infg A ; i.e., or all ">0; 9u A 2 G A such that m A u A <m A +"] But, player A can pro tably deviate from such an equilibrium by o ering x A such that u A <x A < 1 ± B M B k Claim 2. m i 1 ± j M j ;i6= j ProofofClaim2. Suppose m A > 1 ± B M B for contradiction. Note that for all ">0; 9u B 2 G B such that M B "<u B M B So, there exists and equilibrium of A in which A receives 1 ± B u B < 1 ± B (M B ") =1 ± B M B + ± B " 7

8 [The following is one such equilibrium A o ers ± B u B to B, B accepts any o er which gives her ± B u B and rejects any other o er. Following B s rejection the play resumes according to the equilibrium of B which yields B the payo u B ] In other words, for all ">0 there exists u A 2 G A such that which contradicts m A > 1 ± B M B k Claim 3. M i 1 ± j M j ;i6= j u A < 1 ± B M B + ± B " ProofofClaim3.Player B rejects any o er which gives her less than ± B m B and following rejection she never o ers more than ± A M A Therefore, 8 < M A max 1 ± {z Bm B} max when B accepts ; ± 2 AM {z A } 9 = ; max when B rejects and hence M A 1 ± B m B k Claims1and2implythat m A = 1 ± B M B m B = 1 ± A M A Claim 3, then yields or Similarly, Then, by Claim 1, M B 1 ± A (1 ± B M B ) M B M A 1 ± A 1 ± B 1 ± A m A 1 ± B M B 1 ± B = 1 ± B Since M A m A we then have M A 1 ± B and hence M A = 1 ± B 8

9 Similarly, we have M B = 1 ± A So, the unique payo pro le in A is (x A ; 1 x A ) and the unique payo pro le in B is (x B ; 1 x B ).k We can now complete the proof of the theorem using Lemma 8. We rst show that all equilibrium o ers are accepted in any SPE. Suppose there exists a SPE in which player A s o er is rejected. By Lemma 8, A s equilibrium payo in this subgame is x A But by Lemma 8. A s payo in subgame following rejection is (1 x B ), and hence, the equilibrium payo of A in the subgame in which her o er is rejected must be ± A (1 x B ) But, this implies x A = ± A (1 x B )=±2 Ax A ; a contradiction. Similarly, player B s equilibrium o ers must be accepted. Second, we show that in all SPE A o ers x A and B o ers x B Suppose A o ers x A >x A in aspeofa A. This o er must be rejected by B in equilibrium, because otherwise B would get less than 1 x A in that subgame which contradicts Lemma 8. This, in turn, contradicts that no equilibrium o er is rejected. Now suppose x A <x A in a SPE of a A. This o er, too, must be rejected by B, because otherwise A would get less than x A in that subgame, contradicting Lemma 8. So, A must be o ering x A in all SPE. Similarly, B must always be o ering x B Since there is a unique SPE satisfying these properties, as proved in Proposition 5, the proof is complete. 2.4 Properties of the Equilibrium (1) Equilibrium is Unique and E cient This is the case for ± i < 1 That is, there has to be some friction. Otherwise there exists a continuum of equilibria, including ine cient ones. (2) Bargaining Power Note that the share of player A is ¼ A = x A = 1 ± B = 1 e r B 1 e r A e r B andthatofbis ¼ B =1 x A = ± B (1 ± A ) and hence the share of player i is increasing in ± i and decreasing in ± j The bargaining power comes from patience. The more patient a player is, the higher her share. 9

10 If the payers are equally patient, i.e., ± A = ± B = ±; then ¼ A = 1 1+± > ± 1+± = ¼ B In other words, there is a rst mover advantage. The rst mover advantage disappears as! 0 lim ¼ A =!0 lim ¼ B =!0 r B r A + r B r A r A + r B In particular, if r A = r B ; the division is (3) Relationship with Nash Bargaining Solution Let S = (s A ;s B ) 2 R 2 + s A + s B 1 ª and d =(0; 0) Then, as! 0; the SPE of the Rubinstein game converges to an asymmetric Nash bargaining solution, i.e., f (S; d) = arg max s2s s As 1 B with = r B r A + r B 10