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A noncooperative interpretation of the Aumann–Davis–Maschler bargaining set
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Mathematical Social Sciences 54 (2007) 203 – 217 www.elsevier.com/locate/econbase
A noncooperative interpretation of the Aumann–Davis–Maschler bargaining set ☆ Cheng-Cheng Hu Center for General Education, Southern Taiwan University of Technology, Tainan, 71005, Taiwan Received 13 December 2006; received in revised form 20 April 2007; accepted 2 May 2007 Available online 18 May 2007
Abstract An extensive form game is proposed in the paper to give a noncooperative interpretation of the Aumann– Davis–Maschler bargaining set for transferable utility games. Inspired by Stearns' transfer scheme procedure, a bilateral negotiation between the objector and counterobjector is devised in the extensive form game. © 2007 Elsevier B.V. All rights reserved. Keywords: Bargaining set; Nash program; Subgame perfect equilibrium JEL classification: C71; C72
1. Introduction Stability is one of the most important properties of the solution concepts for transferable utility games (TU games). How do players distribute among themselves the surplus in such a way that some stability requirements will be satisfied? There are various approaches to distribute the payoff among players in different stable senses. The core is one of the most important solution concepts for TU games. In the case of the core, an allocation is stable if no coalition can deviate profitably. The idea of the core is to select all stable allocations. Unfortunately, the core is not always nonempty. The Aumann–Davis–Maschler (ADM) bargaining set was introduced in Davis and Maschler (1963, 1967) and in Aumann and Maschler (1964) to overcome the emptiness shortcoming of the core. It is well known that the ADM bargaining set is nonempty if the game is essential. The idea behind the ADM bargaining set is to weak the stability implicated in the core by objection and ☆
The author is grateful to Chen-Ying Huang and an anonymous referee for helpful comments. E-mail address: [email protected]. 0165-4896/$ - see front matter © 2007 Elsevier B.V. All rights reserved. doi:10.1016/j.mathsocsci.2007.05.002
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counterobjection argument. The ADM bargaining set requires that each pair (i, j) of players is in a specific stable situation in such a way that each objection proposed by player i at an allocation in the ADM bargaining set against player j would be nullified by a counterobjection proposed by player j. Every player can not propose an objection at any core allocation. Hence, the core is contained in the ADM bargaining set. We can view that an allocation is in the ADM bargaining set if and only if it is in equilibrium for each pair of players in the sense of objection and counterobjection. Stearns (1968) introduced a dynamic system that converges to points in the ADM bargaining set. Stearns' dynamic system based on a two-person transfer procedure. For each TU game v, the set Mij ðvÞ consists of all allocations at which each objection proposed by player i would be nullified by a counterobjection proposed by player j. If an allocation is not in Mij ðvÞ, it implies that there exists a justified objection of player i against player j at it. Stearns' two-person transfer procedure at this allocation describes that there exists a minimal transfer amount t from player j to player i and by transferring the amount t from player j to player i a new allocation in Mij ðvÞ would be reached. Thus, an allocation is in the ADM bargaining set if and only if each pair of players does not need to make a two-person transfer procedure at it. The Stearns' approach can be viewed as a characterization of a specific type of stability for the ADM bargaining set. There are two approaches to justify whether a solution concept is game-theoretically sound. One is to provide an axiomatic foundation of the solution concept. The other is to give the solution concept a noncooperative interpretation. Since the ADM bargaining set is not axiomatized yet, it is quite interesting to justify it by providing a bargaining foundation. In the literature, several notions of bargaining sets have been investigated from the noncooperative viewpoint. Einy and Wettstein (1999) suggested two mechanisms to implement the Mas-Colell bargaining set and the ADM bargaining set. Nevertheless, their mechanisms violate feasibility out of equilibrium. Pérez-Castrillo and Wettstein (2000) provided two mechanisms to give noncooperative interpretations of two bargaining sets in super-additive environments. The first bargaining set is a close variation of the one proposed by Zhou (1994) and the second is the Mas-Colell bargaining set. They added exogenous players in the noncooperative game to avoid the problem of feasibility of the out-of-equilibrium outcomes. Serrano and Vohra (2002a,b) proposed their game models to give the bargaining foundations of the ADM bargaining set and the Mas-Colell bargaining set. In their multi-stage bargaining games, the outcome of each strategy profile is feasible, in and out of equilibrium. In the paper, we provide a noncooperative interpretation of the ADM bargaining set for transferable utility games. Our extensive form game is related to the one proposed in Serrano and Vohra (2002a). Both games are based on the definition of the ADM bargaining set. We also use one of their designs to ensure that each equilibrium would induce a unanimous agreement and to determine the protocol in further negotiations. Nevertheless, there are some differences between our extensive form game and theirs. Inspired by Stearns' two-person transfer procedure, we construct a feasible noncooperative game in which a bilateral negotiation between the objector and counterobjector is devised. Besides winning or losing the objection/counterobjection battle, the device of the bilateral negotiation provides an opportunity for both the objector and the counterobjector to arrive at an equilibrium state. Thus, the specific stability behind the ADM bargaining set is revealed through the bilateral negotiation in our game model. 2. Definitions and conventions Let N = {1, 2, ⋯, n} be a finite set of players. A coalition is a nonempty finite subset of N. A transferable utility game (TU game) with the player set N is a characteristic function v that assigns to each T ⊆ N a real number and v(∅) is assumed to be 0.
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For S ⊆ N, let |S| be the cardinality of S. We denote by ℝS the real |S|-dimensional space with coordinates named after the members of S. LetP x ¼ ðxi ÞiaN aℝN . x is called a payoff vector. We S write xS for the restriction of x on ℝ , xðSÞ ¼ iaS xi , and x(∅) is defined to P be 0. Let v be a TU game with the player set N. We say v is essential if vðN ÞN iaN vðfigÞ and 0-normalized if v({i}) = 0 for all i ∈ N. In the paper, we only consider the essential and 0normalized TU games. Let xaℝN . x is called an imputation if xðN Þ ¼ vðN Þ and xi zvðfigÞ; for all iaN : Let I(v) denote the set of all imputations of the game v. Let x be an imputation and let i, j be two distinct players in N. We denote F ij ¼ fSoN : iaS and jgSg and V ðSÞ ¼ fyaℝS : yðSÞVvðSÞg for all S ⊆ N. An objection of i against j at x is a pair (S, y) satisfying: SaF ij ; yaV ðSÞ; yi Nxi ; yk zxk for all k in Snfig: Let (S, y) be an objection of i against j at x. A counterobjection to this objection is a pair (T, z) satisfying: T aF ji ; zaV ðT Þ; zk zyk for all k a T \ S; zk zxk for all k a T nS: We say that an objection is justified if it has no counterobjection. With these definitions we arrive at the Aumann–Davis–Maschler (ADM) bargaining set MðvÞ ¼ fxaIðvÞ : there exists no justified objection at xg: It is known that the ADM bargaining set MðvÞ is nonempty. For more reference, please see Peleg and Sudhölter (2003). 3. A noncooperative game model Given a TU game v with the player set N, we shall construct an extensive form game Γ(v) associated with it. It consists of 5 stages. We denote Π to be the set of all permutations of N, in other words, one-to-one and onto functions from N to N. In Stage 1 each player proposes an imputation, a permutation in Π and a negative real number simultaneously. Assume that player i proposes πi ∈ Π, ∀i ∈ N. We denote the composition of these permutations which are proposed by players to be p ¼ p1 ∘p2 ∘ N ∘pn . π has two functions. It is used to determine who will be the current proposer and the order of moves in Stages 4 and 5 if necessary. π can be interpreted to be an endogenously determined protocol in our extensive form game. This particular design was recently used by Thomson (2005) and Serrano and Vohra (2002a,b).
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Let i, j ∈ N, i ≠ j. We denote Gij ¼ fðS; yÞ : SaF ij and yaV ðSÞg The noncooperative game Γ(v) is defined in the following: Stage 1 Let ℝ denote the set of all negative real numbers. At the beginning of the game, every player proposes simultaneously a proposal in IðvÞ P ℝ . Player i's choice will be denoted by (xi, πi, di), ∀i ∈ N. (i) If xi ≠ xj, for some i, j ∈ N, the game ends. The outcome of the game is that player π(1) gets d, where d ¼ maxfd i g. All other players get 0. iaN
(ii) If xi = xj = x, ∀i, j ∈ N, we will refer to x as the binding proposal. Player π(1) becomes the current proposer. The game proceeds to Stage 2. Stage 2 Assume that π(1) = i. In Stage 2, there are two actions “take leave” (L) or “choose a player in N \ {i}, say j, and make a proposal in Gij to j” can be taken by the current proposer. (i) If L is taken, the binding proposal x is the payoff of the game. The game ends. (ii) If (S, y) in Gij is proposed to j, the game proceeds to Stage 3. Stage 3 In Stage 3, there are three actions can be taken by player j: accept (S, y), keep quiet, or reject (S, y) by proposing a proposal in Gji. (i) If player j accepts (S, y), the game ends. Player i gets yi, player j gets xi + xj − yi and players in N \ {i, j} obtain xN \{i,j}. Note that x is the binding proposal formed in Stage 1. (ii) If player j keeps quiet, the game proceeds to Stage 5. (iii) If player j rejects (S, y) by proposing a proposal (T, z) in Gji, the game proceeds to Stage 4. Stage 4 In Stage 4, players in T \ { j} respond to the proposal (T, z) sequentially according to the order induced by π. If (T, z) is accepted by all players in T \ { j} or T = { j}, the game ends. Players in T obtain z, each player k in N \ (T ⋃ {i}) obtains xk and player i obtains v(N ) − z(T ) − x(N \ (T ⋃ {i})). If any player in T \ { j} rejects (T, z), the game proceeds to Stage 5. Stage 5 In Stage 5, players in S \ {i} respond to the proposal (S, y) sequentially according to the order induced by π.
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If (S, y) is accepted by all players in S \ {i} or S = {i}, the game ends. Players in S obtain y, each player k in N \ (S ⋃ { j}) obtains xk and player j obtains v(N ) − y(S ) − x(N \ (S ⋃ { j})). If any player in S \ {i} rejects (S, y), the game ends. The final outcome is x. Remark 1. The TU feature is important in our game model. If player j accepts player i's proposal in Stage 3, then player i receives the transfer amount from player j. If player j rejects player i's proposal, then the objection/counterobjection battle would be launched. The loser obtains the residual and all outcomes in the game are feasible, in and out of equilibrium. Remark 2. On the basis of Stearns' transfer scheme procedure, we devise a bilateral negotiation between the objector and the counterobjector in Stages 2 and 3 of our extensive form game. Suppose that player i proposes (S, y) to player j in Stage 2. If (S, y) is accepted by j, players i and j's payoffs are yi and xi + xj − yi, respectively. Therefore, it can be viewed that the proposal (yi, xi + xj − yi) is brought out to player j implicitly when the proposal (S, y) is made by player i. Since players in S \ {i} do not respond to (S, y) yet and only players i and j have the rights to make decisions in Stages 2 and 3, it can be regarded that there is a bilateral negotiation between i and j in the two stages. Besides winning or losing the objection/counterobjection battle, this bilateral negotiation provides an opportunity for both the objector and the counterobjector to reach a state in which they are in equilibrium. Like the extensive form game proposed by Serrano and Vohra (2002a), our game model is based on the definition of the ADM bargaining set. Due to this factor, both games consist of the stages for the objection/counterobjection battle. For convenience, we call their extensive form mechanism the SV game. Both games have other similarities: (i) The first two stages of our extensive form game are very similar to the ones in the SV game. We follow the design of the first stage of the SV game to ensure that each equilibrium would induce a unanimous agreement and to determine the protocol in further negotiations. (ii) In the second stages of both games, the proposer (the objector) can take the following action: choose a player (the counterobjector) and make a proposal (a potential objection). Nevertheless, besides Remark 2, there are some other differences between the SV game and ours: (1) In Stage 2 of our extensive form game, the objector not only can make a potential objection, but also can take the action L to end the game. (2) Unlike the SV game, if the potential objection (S, y) is proposed, players in S would not be asked for instant responses in our game model. They would decide to accept or reject the potential objection in the final stage, if necessary. (3) In the SV game, the counterobjector needs to name a coalition containing at least two players to make a counterobjection. We do not need the restriction here. (4) In our game, the counterobjector can not reject the agreed objection (or the agreed counterobjection) in our extensive form game. In fact, the final outcome is determined immediately if the potential objection (or the counter-objection) is accepted. (5) Players who are neither the objector nor the counterobjector can keep their payments not less than the ones assigned by the binding proposal in our game model.
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4. Properties In this section, we introduce some properties that will be used to show the main results of the paper. Before proving these properties, we need some necessary notations. Let v be a TU game with the player set N. Let x be an imputation, let i, j ∈ N, i ≠ j and let (S, y) ∈ Gij. We call (S, y) a quasi-objection of i against j at x if it satisfies yk zxk for all kaSnfig: Note that if (S, y) is a quasi-objection of i against j at x satisfying yi N xi, it is an objection of i against j at x. Let (S, y) be a quasi-objection of i against j at x and let (T, z) ∈ Gji. If (T, z) satisfies zk zyk for all kaT \ S; zk zxk for all k aT nðS [ fjgÞ; it is called a quasi-counterobjection to (S, y). Note that if (T, z) is a quasi-counterobjection to the objection (S, y) satisfying zj ≥ xj, it is a counterobjection to (S, y). Let (S, y) be a quasi-objection of i against j at x. We say (S, y) is a maximum quasi-objection of i against j at x if yi ¼ maxfyi V: ðSV; yVÞ is a quasi−objection of i against j at xg: Let (S, y) be a quasi-objection of i against j at x and let (T, z) be a quasi-counterobjection to (S, y). We say (T, z) is a maximum quasi-counterobjection of j to (S, y) if zj ¼ maxfzVj : ðT ;V z VÞ is a quasi−counterobjection to ðS; yÞg;
ð1Þ
and zj is called the maximum surplus of j against (S, y) at x. Note that zj ≥ 0. Let i, j ∈ N, i ≠ j, and let Mij ðvÞ ¼ fxaIðvÞ : for each objection of i against j at x, there exists a counterobjection to it}. It is clear that MðvÞ ¼ \ Mij ðvÞ. i; jaN ;i pj Given x ∈ I(v), i, j ∈ N, i ≠ j and t ≥ 0, define the payoff vector x′ in ℝN as follows: xVi ¼ xi þ t; xj V¼ xj t; xkV ¼ xk ; for all kaN nfi; jg: x′ is said to result from x by a transfer of size t from j to i. The following result is established by Stearns (1968). Lemma 3. Given distinct players i, j in N and x ∈ I(v), there is a minimal t ≥ 0 such that there is a point x′ in Mij ðvÞ which results from x by a transfer of size t from j to i. Denote mij(x) to be the t of Lemma 3. Note that 0 ≤ mij(x) ≤ xj. We call mij(x) the minimal transfer size from j to i at x. Remark 4. The following three statements are equivalent. (i) xaMðvÞ. (ii) xaMij ðvÞ for all i, j ∈ N, i ≠ j. (iii) mij(x) = 0 for all i, j ∈ N, i ≠ j.
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Let x ∈ I(v). Suppose that mij(x) N 0. For each positive integer t, define the payoff vector xt as follows. 1 xti ¼ xi þ 1 mij ðxÞ; t 1 ð2Þ xtj ¼ xj 1 mij ðxÞ; t xtk ¼ xk ; for all kaN nfi; jg: t Clearly, x ∈1I(v). Since 1 t mij ðxÞbmij ðxÞ for all t ≥ 1, it holds that xt gMij ðvÞ by Lemma 3. We derive that there exists a justified objection (St, yt) of i against j at xt for all t ≥ 1. By the facts that 0 ≤ yht ≤ v (St) for all h ∈ St and the cardinality of the set F ij is finite, it holds that there exists a coalition S ⁎ aF ij and a subsequence {ytk} of { yt} such that
S ⁎ ¼ S tk for all k; and y⁎ ¼ lim ytk : kYl
We call such (S⁎, y⁎) a limited objection of i against j at x. Note that (S⁎, y⁎) is also a quasiobjection of i against j at x. Lemma 5. Let mij(x) N 0 and (S⁎, y⁎) be a limited objection of i against j at x. Then zj ≤ xj − mij(x) for all quasi-counterobjection (T, z) to (S⁎, y⁎). Proof. If not, there exists a quasi-counterobjection (T, z) to (S⁎, y⁎) such that zj N xj − mij(x). Let e ¼ jT1 j ðzj ðxj mij ðxÞÞÞ, z′j = xj − mij(x) + ε and z′h = zh + ε for all h ∈ T \ {j}. Since (T, z) is a quasi-counterobjection to (S⁎, y⁎), we have zh ≥ yh⁎ for h ∈ S⁎ ∩ T and zh ≥ xh for h ∈ T \ (S⁎ ⋃ {j}). Since (S⁎, y⁎) is a limited objection of i against j at x, there exists a subsequence fytk g; ytk aℝS ⁎ , such that limkYl ytk ¼ y⁎ and each pair (S⁎, ytk) is a justified objection of i against j at xtk, where xtk is defined by Eq. (2). It holds that (T, z′) is a counterobjection to (S⁎, ytk) at xtk for some sufficiently large positive integer tk. This contradicts to the fact that (S⁎, ytk) is a justified objection of i against j at xtk. We obtain the desired contradiction. □ 5. The main results In this section, we will show that each payoff vector in MðvÞ can be supported by a subgame perfect equilibrium of the game Γ(v) and every subgame perfect equilibrium outcome is in MðvÞ. First, we shall construct a subgame perfect equilibrium f for each payoff vector ¯x in MðvÞ. For each player i and each imputation x, denote Fi ðxÞ ¼ k a N nfig : mik ðxÞN0 and k a arg max ½mil ðxÞ : laN nfig
Let ¯ xaMðvÞ. The strategy profile f is defined in the following. (1) At the beginning of the game, player i proposes (x¯, π I, − 1) for all i ∈ N, where πI is the identity mapping on N. In the following, we assume that x is the binding proposal.
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(2) In Stage 2, if player i is the current proposer, i takes the action:
L; propose ðS ⁎ ; ðxi þ mij ðxÞ; y⁎ S ⁎ nfig ÞÞ to j;
if Fi ðxÞ ¼ F; o:w:;
where (S⁎, y⁎) is a limited objection of i against j at x, and j is the smallest index among Fi(x) according to the order of πI. (3) Suppose that player i proposes (S, y) to player j in Stage 2. Player j is going to respond to the proposal (S, y) in Stage 3. There are two cases to be discussed. (i) If (S, y) is a quasi-objection of i against j at x, then player j takes the action: proposeðT V; zVÞ; if zj VNxj þ xi yi ; accept ðS; yÞ; o:w:; where (T′, z′) is a maximum quasi-counterobjection of j to (S, y). (ii) If (S, y) is not a quasi-objection of i against j at x, then player j takes the action: 8 if xj þ xi yi zmaxfzWj ; xj g; < accept ðS; yÞ; propose ðT W; zWÞ; if z Wj Nmaxfxj þ xi yi ; xj g; : keep quiet; o:w:; where (T″, z″ ) is a maximum quasi-objection of j against i at x. (4) Suppose that player j proposes (T, z) in Stage 3 to reject the proposal (S, y) proposed in Stage 2 by player i. Player k, k ∈ T \ { j}, is going to respond to (T, z) in Stage 4. (i) If k ∈ S, there are two cases to be discussed. Case 1 (S, y) is a quasi-objection of i against j at x. Player k responds: accept; if zk zyk ; reject; o:w: Case 2 (S, y) is not a quasi-objection of i against j at x. Player k responds: accept; if zk zxk ; reject; o:w: (ii) If k ∉ S, then player k responds:
accept; if zk zxk ; reject; o:w:
(5) In Stage 5, player k, k ∈ S \ {i}, is going to respond to the proposal (S, y), then player k responds: accept; if yk zxk ; reject; o:w:
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It is easy to see that the outcome of the strategy profile f is x¯. By Remark 4, mij (x¯) = 0 for all i, j ∈ N, i ≠ j. Then Fi(x¯) = 0 for all i ∈ N. By following f, each player proposes (x¯, πI, − 1). By the game rule, x¯ is the binding proposal and πI(1) = 1 is the current proposer. Since F1(x¯) = ∅, player 1 will take L in Stage 2, the game ends with the outcome x¯. To show that f is a subgame perfect equilibrium, we need the following results. Lemma 6. In Stage 5, suppose that player k is going to make a response. If all players in N \ {k} follow f−k, then it is the best for player k to follow fk. Proof. Suppose that x is the binding proposal and player k is going to respond to the proposal (S, y) proposed by i in Stage 2. There are two cases to be considered. Case 1 yk ≥ xk. By following fk, player k will accept (S, y). If yh ≥ xh for all h ∈ S \ {i, k}, then (S, y) is accepted by all players in S \ {i} and player k obtains yk. If player k deviates by rejecting (S, y), the game ends and player k's payoff is xk ≤ yk. If yh b xh for some h ∈ S \ {i, k}, then (S, y) is rejected by player h. The game ends and player k's payoff is xk whatever player k deviates or not. Case 2 yk b xk. By following fk , player k will reject (S, y). The game ends and player k's payoff is xk. If player k deviates by accepting (S, y), then either (S, y) is accepted by all players in S \ {i} or is rejected by some player in S \ {i, k}. We derive that player k obtains at most xk. By Cases 1 and 2, it is not better off for player k to deviate. □ Lemma 7. In Stage 4, suppose that player k is going to make a response. If all players in N \ {k} follow f−k, then it is the best for player k to follow fk. Proof. Suppose that x is the binding proposal. We assume that player i proposes (S, y) to player j in Stage 2 and player j rejects (S, y) by proposing (T, z) in Stage 3. In Stage 4, player k, k ∈ T \ { j}, is going to respond to the proposal (T, z). We only discuss the case that k ∈ S and (S, y) is a quasiobjection of i against j at x. The remaining cases can be derived by the similar arguments. There are two possibilities to be considered. Case 1 zk ≥ yk. Player k will accept (T, z) by following fk. (i) If (T, z) is accepted by all players in T \ { j, k}, the game ends and player k obtains zk. If player k deviates by rejecting (T, z), the game proceeds to Stage 5. Player k obtains at most yk by Lemma 6. It is not better off for player k to deviate. (ii) If (T, z) is rejected by some player h in T \ { j, k} by following fh, the game proceeds to Stage 5. Player k obtains yk by the assumptions that all players in N \ {k} follow f−k and (S, y) is a quasi-objection of i against j at x. If player k deviates, the game still proceeds to Stage 5 by the assumption that player h rejects (T, z) by following fh. Player k will obtain at most yk by Lemma 6. It is not better off for player k to deviate. Case 2 zk b yk. Player k will reject (T, z) by following fk. The game proceeds to Stage 5, (S, y) will be accepted by all players in S \ {i} and player k obtains yk. Assume that player k deviates. Then either the game proceeds to Stage 5, or (T, z) is accepted by all players in T \ { j}. In case of the game proceeding to Stage 5, player k obtains at most yk by Lemma 6. In case of all players in T \ {j} accepting (T, z), player k obtains zk b yk. It is not better off for player k to deviate. □
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Remark 8. Suppose that x is the binding proposal, players i and j are proposers in Stages 2 and 3, respectively. If players in N \ {i, j} follow fN \{i, j} in Stages 4 and 5, then each player k in N \ {i, j} obtains xk at least by Lemmas 6 and 7. In fact, each player who needs to make a response in Stages 4 or 5 can reject i or j's proposal such that the final outcome of the game is x. Assume that, at the beginning of the game Γ(v), player i proposes ðx; pi ; d i ÞaIðvÞ P ℝ for each i ∈ N. Let d ¼ maxiaN fd i g. Since each player proposes the same allocation x, x is the binding proposal. Under such assumptions, we shall show Lemmas 9 and 10. Lemma 9. In Stage 3, player j is going to respond to i's proposal (S, y). If all players in N \ { j} follow f−j, then it is the best for player j to follow fj. Proof. There are two cases to be considered. Case 1 (S, y) is a quasi-objection of i against j at x. Then yh zxh for all h a Snfig:
ð3Þ
Let (T′, z′) be a maximum quasi-counterobjection to (S, y). There are two subcases to be discussed. Subcase 1 xj + xi − yi ≥ z′j. By following fj, player j accepts (S, y). The game ends and player j obtains xj + xi − yi. (1) Suppose that player j deviates by proposing (T, z). If (T, z) is a quasi-counterobjection to (S, y), then (T, z) is accepted by all players in T \ { j} following fT \{j}. The game ends and player j obtains zj ≤ zj′ ≤ xj + xi − yi. It is not better off for player j to deviate. If (T, z) is not a quasi-counterobjection to (S, y), then (T, z) is rejected by some player in T \ { j}. The game proceeds to Stage 5. Claim A: Player j obtains at most xj + xi − yi. Since (S, y) is a quasi-objection of i against j at x, (S, y) is accepted by all players in S \ {i} following fS \{i}. The game ends and player j obtains v(N) −y(S) −x(N \ (S ⋃ {j}) ≤xj +xi −yi by Eq. (3). By Claim A, it is not better off for player j to deviate. (2) Suppose that player j deviates by keeping quiet. Then the game proceeds to Stage 5. By Claim A, player j obtains at most xj + xi − yi. It is not better off for player j to deviate. Subcase 2 xj + xi − yi b z′j. By following fj, player j rejects (S, y) by proposing (T′, z′). (T′, z′) will be accepted by all players in T ′\ {j} following fT′ \{j}, player j obtains z′j. If player j deviates by accepting (S, y), then player j obtains xj + xi − yi. It is not better off for player j to deviate. If player j deviates by proposing (T, z) to reject (S, y), then, by the same arguments of Subcase 1, it is not better off for player j to deviate. If player j deviates by keeping quiet, then player j obtains at most xj + xi − yi by Claim A. It is not better off for player i to deviate. Case 2 (S, y) is not a quasi-objection of i against j at x. Let (T″, z″ ) be a maximum quasi-objection of j against i at x. There are three subcases to be considered. Subcase (i) xj + xi − yi ≥ max{z″j, xj}. By following fj, player j will accept (S, y). The game ends and player j obtains xj + xi − yi. If player j deviates, then either he proposes (T, z) to reject (S, y) or keeps quiet.
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If j keeps quiet, the game proceeds to Stage 5. Claim 1: If the game proceeds to Stage 5, then player j obtains xj. By the assumption that (S, y) is not a quasi-objection of i against j at x, we have yh b xh for some h ∈ S \ {i}. Player h rejects (S, y) by following fh and the game ends. Player j obtains xj. By Claim 1, it is not better off for player j to deviate. If j proposes (T, z) to reject (S, y), the game proceeds to Stage 4. Claim 2: Player j's payoff is at most max{z″j, xj}. If (T, z) is a quasi-objection of j against i at x, then all players in T \ { j} accept (T, z) by following fT \ { j}. Then player j obtains zj ≤ z″j. If (T, z) is not a quasi-objection of j against i at x, then zh b xh for some h in T \ { j}. Player h rejects (T, z) by following fh in Stage 4 and the game proceeds to Stage 5. Player j obtains xj by Claim 1. We derive that player j's payoff is at most max {z″j, xj}. By Claim 2, player j obtains at most max{z″j, xj} ≤ xj + xi − yi. It is not better off for player j to deviate. Subcase (ii) z″j N max{xj + xi − yi, xj}. By following fj, player j will propose (T″, z″) to reject (S, y) and the game proceeds to Stage 4. All players in T″ \ {j} will accept (T″, z″) by following fT″ \{ j}. The game ends and player j obtains z″j. If player j deviates, then either player j accepts (S, y), or keeps quiet, or proposes (T, z) to reject (S, y). If player j accepts (S, y), the game ends and player j obtains xj + xi − yi b z″j. If player j keeps quiet, the game proceeds to Stage 5. By Claim 1, player j's payoff is xj b z″j. If player j proposes (T, z) to reject (S, y), then player j obtains at most max{z″j, xj} = z″j by Claim 2. Subcase (iii) xj + xi − yi b max{z″j, xj} and z″j ≤ max{xj + xi − yi, xj}. It holds that max{xj + xi − yi, z″j} ≤ xj. By following fj, player j keeps quiet and the game proceeds to Stage 5. By Claim 1, player j obtains xj. If player j deviates, then either player j accepts (S, y) or proposes (T, z) to reject (S, y). If player j accepts (S, y), the game ends and player j obtains xj + xi − yi ≤ xj. If player j proposes (T, z) to reject (S, y), the game proceeds to Stage 4. By Claim 2, player j obtains at most max{z″j, xj} = xj. It is not better off for player j to deviate. By Cases 1 and 2, the proof is complete. □ Lemma 10. Suppose that player i is the current proposer in Stage 2. Then, it is the best for player i to follow fi. Proof. There are two cases to be considered. Case 1 Fi(x) = ∅. By following fi, player i takes the action L, the game ends and player i obtains xi. If player i deviates by proposing (S, y) to some player, say j, there are two subcases to be discussed. Subcase 1 (S, y) is an objection of i against j at x. Since Fi(x) = ∅, we have that mij(x) = 0 and xaMij ðvÞ by Lemma 3. By the assumption that (S, y) is an objection of i against j at x, we derive that there exists a counterobjection (T, z) to (S, y) and zj ≥ xj. Let (T′, z′) be a maximum quasi-counterobjection to (S, y). Then, z′j ≥ zj ≥ xj. Since (S, y) is an objection of i against j at x, it holds that yi N xi. We derive that zj′ N xj + xi − yi.
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By the assumption that all players in N \ {i} following f−i, player j will reject (S, y) by proposing (T′, z′) and (T′, z′) will be accepted by all players in T \ {j}. The game ends and player i obtains X vðN Þ zVðT VÞ xðN nðT V [ figÞÞ ¼ xðT V [ figÞ z VðT ÞV V xi þ ðxk zkVÞV xi : kaT V
It is not better off for player i to deviate. Subcase 2 (S, y) is not an objection of i against j at x. It means that yi ≤ xi or yh b xh for some h ∈ S \ {i}. Note that each player h in N \ {i, j} obtains xh at least by Remark 8. (1) Suppose that yi ≤ xi. If player j accepts (S, y), then the game ends and player j obtains xj + xi − yi ≥ xj. By Lemma 9, player j will obtain xj at least by following fj. We derive that player i obtains at most xi. (2) Suppose that yh b xh for some h ∈ S \ {i}. If player j keeps quiet, then the game proceeds to Stage 5. Player h rejects (S, y) by following fh and the game ends. Player j obtains xj. By Lemma 9, player j will obtain xj at least by following fj. We derive that player i obtains at most xi. It is not better off for player i to deviate. Case 2 Fi(x) ≠ ∅. By following fi, player i proposes ðS ⁎ ; ðxi þ mij ðxÞ; y⁎ S ⁎ nfig ÞÞ
ð4Þ
to j, where j is the smallest index among Fi(x) according to the order of πI and (S⁎, y⁎) is a limited objection of i against j at x. Note that Eq. (4) is an objection of i against j at x. Let (T′, z′) be a maximum quasi-counterobjection to Eq. (4). Then (T′, z′) is also a quasi-counterobjection to (S⁎, y⁎). By Lemma 5, we have zj′ ≤ xj − mij (x). Player j accepts the proposal Eq. (4) by following fj and the game ends. Player i obtains xi + mij(x). If player i deviates by taking the action L, the game ends. Player i obtains xi b xi + mij(x). If player i deviates by proposing (S, y) to some player, say k, there are two subcases to be discussed. Subcase 1 (S, y) is a quasi-objection of i against k at x. There are two possibilities to be discussed. (1) yi N xi + mik(x). then (S, y) is an objection of i against k at the payoff vector x′ = (xi + mik(x), xk − mik(x), xN \ {i,k}). We have xVaMik ðvÞ by Lemma 3. Then there exists a counterobjection (T, z) to (S, y). Let (T′, z′) be a maximum quasi-counterobjection to (S, y). It holds that zk′ ≥ zk ≥ xk − mik(x) N xi +xk − yi. By following fk, player k will propose the maximum quasi-counterobjection (T′, z′) to reject (S, y). The game proceeds to Stage 4 and all players in T′ \ {k} will accept (T′, z′) by following fT′ \{k}. The game ends and player i obtains vðN Þ zVðT VÞ xðN nðT V [ figÞÞ ¼ xðT V [ figÞ zVðT VÞV xi þ xk zkV X þ ðxh zV h ÞV xi þ mik ðxÞV xi þ mij ðxÞ: haT Vnf kg
It is not better off for player i to deviate. (2) yi ≤ xi + mik(x). If player k accepts (S, y), then the game ends. Player k obtains xi + xk − yi. By Lemma 9, player k will obtain xi + xk − yi at least by following fk. Each player h in N \ {i, k} obtains xh at least by Remark 8. It holds that player i gets at most yi ≤ xi + mik(x) ≤ xi + mij(x). It is not better off for player i to deviate. Subcase 2 (S, y) is not a quasi-objection of i against k at x. Then yh b xh for some h ∈ S \ {i}. If player k keeps quiet, the game proceeds to Stage 5 and player h will reject (S, y) by following fh.
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The game ends and player k obtains xk. By Lemma 9, player k obtains xk at least by following fk. It implies that player i gets at most xi ≤ xi + mij(x). It is not better off for player i to deviate. □ In the following, we will show that f is a subgame perfect equilibrium. Theorem 11. f is a subgame perfect equilibrium of Γ(v). Proof. By the above Lemmas, it remains to show that it is the best for player i to follow fi at the beginning of the game. We have mentioned that, by following f, x¯ is the binding proposal and πI(1) = 1 is the current proposer. Since F1(x¯) = ∅, player 1 will take L in Stage 2, the game ends with the outcome x¯. If player i deviates at the beginning of the game by proposing (x, π′, d′), there are two cases to be considered. Case 1 x ≠ x¯. No binding proposal is formed. Then player i obtain max{− 1, d′} b x¯i. Case 2 x = x¯. The binding proposal is still x¯. Suppose that the current proposer is player h. If h ≠ i, then player h takes L in Stage 2 by following fh, the game ends and player i obtains x¯i. If h = i, then it is the best for player i to follow fi by Lemma 10. That is, player i takes L, the game ends and player i obtains x¯i. By Cases 1 and 2, it is not better off for player i to deviate. □ Next, we show that every subgame perfect equilibrium outcome of the game Γ(v) is in the ADM bargaining set MðvÞ. Let g = (g1, g2, ⋯, gn) be an arbitrary subgame perfect equilibrium of the game Γ(v). Lemma 12. Suppose that x is the binding proposal and the game proceeds to Stage 5. If all players in S \ {i} are going to respond to the proposal (S, y) proposed by i and yh N xh for all h ∈ S \ {i}, then (S, y) will be accepted by all players in S \ {i} following gS \{i}. Lemma 12 is easy to derive, we omit the proof. Lemma 13. Suppose that x is the binding proposal and the game proceeds to Stage 3 in which j is going to respond to (S, y) proposed by player i. If yh N xh for all h ∈ S \ {i} and the maximum surplus of j against (S, y) at x is less than xj + xi − yi, then player j will accept (S, y) by following gj. Proof. We shall show that player j will accept (S, y) and obtain xj + xi − yi by following gj. If not, there are two cases to be discussed. Case 1 By following gj, player j rejects (S, y) by proposing (T, z). There are two possibilities to be considered. Subcase 1 (T, z) is accepted by all players in T \ { j} following gT \{j}. The game ends and players in T obtain z. We shall show that (T, z) is a quasi-counterobjection to (S, y). If not, then either zh b yh for some h ∈ S ∩ T or zh b xh for some h ∈ T \ (S ⋃ {j}). In case of zh b yh for some h ∈ S ∩ T, player h can deviate by the following. Player h rejects (T, z) in Stage 4 and he follows gh in Stage 5. Then the game will move to Stage 5 and, by Lemma 12, (S, y) will be accepted by all players in S \ {i}. Player h obtains yh N zh. It violates that g is a subgame perfect equilibrium. In case of zh b xh for some h ∈ T \ (S ⋃ {j}), player h can deviate by rejecting (T, z). The game proceeds to Stage 5 and player h's payoff is xh N zh. It violates that g is a subgame perfect equilibrium. We derive that (T, z) is a quasi-counterobjection to (S, y). It holds that zj b xj + xi − yi by the assumption that the maximum surplus of j against (S, y) is less than xj + xi − yi. If player j accepts (S, y), he obtains xj + xi − yi. It is better off for player j to deviate. This violates that g is a subgame perfect equilibrium.
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Subcase 2 (T, z) is rejected by some player h in T \ {j} following gh. The game proceeds to Stage 5. In Stage 5, (S, y) will be accepted by all players in S \ {i} by Lemma 12 and the game ends. Player j obtains v (N) − y(S) − x(N \ (S ⋃ {j})) b xj + xi − yi. This violates that g is a subgame perfect equilibrium. Case 2 Player j keeps quiet by following gj. The game proceeds to Stage 5. By the same arguments of Subcase 2, we obtain the desired contradiction. By Cases 1 and 2, the proof is complete. □ Theorem 14. The subgame perfect equilibrium outcome of g is in the ADM bargaining set MðvÞ. Proof. Suppose that, at the beginning of the game, i proposes ðxi ; pi ; d i ÞaIðvÞ P ℝ by following gi, ∀i ∈ N. Step 1: The subgame perfect equilibrium g must induce a binding proposal. We claim that xi = xj for all i, j ∈ N. Denote π = π1 ∘ π2 ∘ ⋯ ∘ πn and assume that π(1) = k ∈ N. Suppose that no binding proposal is formed. It implies that xi ≠ xj for some i, j ∈ N, i ≠ j. By the game rule, player k obtains d ¼ maxiaN fd i gb0, other players obtain 0. Let k deviate by proposing (xk, πk, d / 2) at the beginning of the game. Since xi ≠ x j for some i, j ∈ N, i ≠ j, no binding proposal is formed. Hence, player k obtains d / 2 N d. Player k is better off by deviation. We obtain the desired contradiction. Step 2: Let x be the binding proposal induced by g. We will show that x is the subgame perfect equilibrium outcome. Let x′ be the outcome of the game by following g. We will show that x′i ≥ xi for all i in N. Let player i deviate by the following. Player i proposes (x, π¯i, di ) at the beginning of the game, where π¯i satisfies i iþ1 n pð1Þ ¼ p1 ∘p2 ∘ : : : ∘pi1 ∘ p ¯ ¯ ∘p ∘ : : : ∘p ð1Þ ¼ i:
ð5Þ
Then player i becomes the current proposer. Player i takes the action L in Stage 2, the game ends and player i gets xi. By subgame perfection, we derive that x′i ≥ xi. Since the binding proposal x is formed, the game would not end in Stage 1. We derive that x′ (N) = x(N) = v(N) by the game rule and x′ = x. Step 3: xaMðvÞ. If not, there exist i, j ∈ N, i ≠ j such that mij(x) N 0 by Remark 4. Let 0 b ε b mij (x) and x″ = (xi + ε, xj − ε, xN \ {i,j}). Since mij(x) is the minimal transfer size from j to i at x, there exists a justified objection (S, y) of i against j at x″ satisfying yk N xk for all k ∈ S \ {i}. We derive that zj b x″j = xj − ε for all (T, z) in Gji satisfying zðT Þ ¼ vðT Þ; zh ¼ yh for haS \ T ; zh ¼ xh for haT nðS [ fjgÞ: Let y′i = xi + ε and y′k = yk for all k ∈ S \ {i}. It holds that the maximum surplus of j against (S, y′) at x is less then xj − ε. Let player i deviate by the following. Player i makes the proposal (x, π¯i, di) at the beginning of the game, where π¯i satisfies Eq. (5). The binding proposal is still x and player i becomes the current proposer. In Stage 2, player i makes the proposal (S, y′) to player j and player j will accept (S, y′) in Stage 3 by Lemma 13. The game ends and player i obtains xi + ε. Player i is better off by deviation. We obtain a desired contradiction. By Steps 1, 2 and 3, the proof is complete. □
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References Aumann, R.J., Maschler, M., 1964. The bargaining set for cooperative games. In: Dresher, M., Shapley, L.S., Tucker, A.W. (Eds.), Advances in Game Theory. Princeton Univ. Press, Princeton, NJ, pp. 443–476. Davis, M., Maschler, M., 1963. Existence of stable payoff configurations for cooperative games: abstract. Bull. Am. Math. Soc. 69, 106–108. Davis, M., Maschler, M., 1967. Existence of stable payoff configurations for cooperative games. In: Shubik, M. (Ed.), Essays in Mathematical Economics in Honor of Oskar Morgenstern. Princeton Univ. Press, Princeton, NJ, pp. 39–52. Einy, E., Wettstein, D., 1999. A non-cooperative interpretation of bargaining sets. Rev. Econ. Des. 4, 219–230. Peleg, B., Sudhölter, P., 2003. Introduction to the Theory of Cooperative Games. Kluwer Academic Publishers, Dordrecht, The Netherlands. Pérez-Castrillo, D., Wettstein, D., 2000. Implementation of bargaining sets via simple mechanisms. Games Econ. Behav. 31, 106–120. Serrano, R., Vohra, R., 2002a. Bargaining and bargaining sets. Games Econ. Behav. 39, 292–308. Serrano, R., Vohra, R., 2002b. Implementing the Mas-Colell bargaining set. Investig. Econ. 26, 285–298. Stearns, R.E., 1968. Convergent transfer schemes for N-person games. Trans. Am. Math. Soc. 134, 449–459. Thomson, W., 2005. Divide-and-permute. Games Econ. Behav. 52, 186–200. Zhou, L., 1994. A new bargaining set of an N-person game and endogenous coalition formation. Games Econ. Behav. 6, 512–526.
Mathematical Social Sciences 54 (2007) 203 – 217 www.elsevier.com/locate/econbase
A noncooperative interpretation of the Aumann–Davis–Maschler bargaining set ☆ Cheng-Cheng Hu Center for General Education, Southern Taiwan University of Technology, Tainan, 71005, Taiwan Received 13 December 2006; received in revised form 20 April 2007; accepted 2 May 2007 Available online 18 May 2007
Abstract An extensive form game is proposed in the paper to give a noncooperative interpretation of the Aumann– Davis–Maschler bargaining set for transferable utility games. Inspired by Stearns' transfer scheme procedure, a bilateral negotiation between the objector and counterobjector is devised in the extensive form game. © 2007 Elsevier B.V. All rights reserved. Keywords: Bargaining set; Nash program; Subgame perfect equilibrium JEL classification: C71; C72
1. Introduction Stability is one of the most important properties of the solution concepts for transferable utility games (TU games). How do players distribute among themselves the surplus in such a way that some stability requirements will be satisfied? There are various approaches to distribute the payoff among players in different stable senses. The core is one of the most important solution concepts for TU games. In the case of the core, an allocation is stable if no coalition can deviate profitably. The idea of the core is to select all stable allocations. Unfortunately, the core is not always nonempty. The Aumann–Davis–Maschler (ADM) bargaining set was introduced in Davis and Maschler (1963, 1967) and in Aumann and Maschler (1964) to overcome the emptiness shortcoming of the core. It is well known that the ADM bargaining set is nonempty if the game is essential. The idea behind the ADM bargaining set is to weak the stability implicated in the core by objection and ☆
The author is grateful to Chen-Ying Huang and an anonymous referee for helpful comments. E-mail address: [email protected]. 0165-4896/$ - see front matter © 2007 Elsevier B.V. All rights reserved. doi:10.1016/j.mathsocsci.2007.05.002
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counterobjection argument. The ADM bargaining set requires that each pair (i, j) of players is in a specific stable situation in such a way that each objection proposed by player i at an allocation in the ADM bargaining set against player j would be nullified by a counterobjection proposed by player j. Every player can not propose an objection at any core allocation. Hence, the core is contained in the ADM bargaining set. We can view that an allocation is in the ADM bargaining set if and only if it is in equilibrium for each pair of players in the sense of objection and counterobjection. Stearns (1968) introduced a dynamic system that converges to points in the ADM bargaining set. Stearns' dynamic system based on a two-person transfer procedure. For each TU game v, the set Mij ðvÞ consists of all allocations at which each objection proposed by player i would be nullified by a counterobjection proposed by player j. If an allocation is not in Mij ðvÞ, it implies that there exists a justified objection of player i against player j at it. Stearns' two-person transfer procedure at this allocation describes that there exists a minimal transfer amount t from player j to player i and by transferring the amount t from player j to player i a new allocation in Mij ðvÞ would be reached. Thus, an allocation is in the ADM bargaining set if and only if each pair of players does not need to make a two-person transfer procedure at it. The Stearns' approach can be viewed as a characterization of a specific type of stability for the ADM bargaining set. There are two approaches to justify whether a solution concept is game-theoretically sound. One is to provide an axiomatic foundation of the solution concept. The other is to give the solution concept a noncooperative interpretation. Since the ADM bargaining set is not axiomatized yet, it is quite interesting to justify it by providing a bargaining foundation. In the literature, several notions of bargaining sets have been investigated from the noncooperative viewpoint. Einy and Wettstein (1999) suggested two mechanisms to implement the Mas-Colell bargaining set and the ADM bargaining set. Nevertheless, their mechanisms violate feasibility out of equilibrium. Pérez-Castrillo and Wettstein (2000) provided two mechanisms to give noncooperative interpretations of two bargaining sets in super-additive environments. The first bargaining set is a close variation of the one proposed by Zhou (1994) and the second is the Mas-Colell bargaining set. They added exogenous players in the noncooperative game to avoid the problem of feasibility of the out-of-equilibrium outcomes. Serrano and Vohra (2002a,b) proposed their game models to give the bargaining foundations of the ADM bargaining set and the Mas-Colell bargaining set. In their multi-stage bargaining games, the outcome of each strategy profile is feasible, in and out of equilibrium. In the paper, we provide a noncooperative interpretation of the ADM bargaining set for transferable utility games. Our extensive form game is related to the one proposed in Serrano and Vohra (2002a). Both games are based on the definition of the ADM bargaining set. We also use one of their designs to ensure that each equilibrium would induce a unanimous agreement and to determine the protocol in further negotiations. Nevertheless, there are some differences between our extensive form game and theirs. Inspired by Stearns' two-person transfer procedure, we construct a feasible noncooperative game in which a bilateral negotiation between the objector and counterobjector is devised. Besides winning or losing the objection/counterobjection battle, the device of the bilateral negotiation provides an opportunity for both the objector and the counterobjector to arrive at an equilibrium state. Thus, the specific stability behind the ADM bargaining set is revealed through the bilateral negotiation in our game model. 2. Definitions and conventions Let N = {1, 2, ⋯, n} be a finite set of players. A coalition is a nonempty finite subset of N. A transferable utility game (TU game) with the player set N is a characteristic function v that assigns to each T ⊆ N a real number and v(∅) is assumed to be 0.
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For S ⊆ N, let |S| be the cardinality of S. We denote by ℝS the real |S|-dimensional space with coordinates named after the members of S. LetP x ¼ ðxi ÞiaN aℝN . x is called a payoff vector. We S write xS for the restriction of x on ℝ , xðSÞ ¼ iaS xi , and x(∅) is defined to P be 0. Let v be a TU game with the player set N. We say v is essential if vðN ÞN iaN vðfigÞ and 0-normalized if v({i}) = 0 for all i ∈ N. In the paper, we only consider the essential and 0normalized TU games. Let xaℝN . x is called an imputation if xðN Þ ¼ vðN Þ and xi zvðfigÞ; for all iaN : Let I(v) denote the set of all imputations of the game v. Let x be an imputation and let i, j be two distinct players in N. We denote F ij ¼ fSoN : iaS and jgSg and V ðSÞ ¼ fyaℝS : yðSÞVvðSÞg for all S ⊆ N. An objection of i against j at x is a pair (S, y) satisfying: SaF ij ; yaV ðSÞ; yi Nxi ; yk zxk for all k in Snfig: Let (S, y) be an objection of i against j at x. A counterobjection to this objection is a pair (T, z) satisfying: T aF ji ; zaV ðT Þ; zk zyk for all k a T \ S; zk zxk for all k a T nS: We say that an objection is justified if it has no counterobjection. With these definitions we arrive at the Aumann–Davis–Maschler (ADM) bargaining set MðvÞ ¼ fxaIðvÞ : there exists no justified objection at xg: It is known that the ADM bargaining set MðvÞ is nonempty. For more reference, please see Peleg and Sudhölter (2003). 3. A noncooperative game model Given a TU game v with the player set N, we shall construct an extensive form game Γ(v) associated with it. It consists of 5 stages. We denote Π to be the set of all permutations of N, in other words, one-to-one and onto functions from N to N. In Stage 1 each player proposes an imputation, a permutation in Π and a negative real number simultaneously. Assume that player i proposes πi ∈ Π, ∀i ∈ N. We denote the composition of these permutations which are proposed by players to be p ¼ p1 ∘p2 ∘ N ∘pn . π has two functions. It is used to determine who will be the current proposer and the order of moves in Stages 4 and 5 if necessary. π can be interpreted to be an endogenously determined protocol in our extensive form game. This particular design was recently used by Thomson (2005) and Serrano and Vohra (2002a,b).
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Let i, j ∈ N, i ≠ j. We denote Gij ¼ fðS; yÞ : SaF ij and yaV ðSÞg The noncooperative game Γ(v) is defined in the following: Stage 1 Let ℝ denote the set of all negative real numbers. At the beginning of the game, every player proposes simultaneously a proposal in IðvÞ P ℝ . Player i's choice will be denoted by (xi, πi, di), ∀i ∈ N. (i) If xi ≠ xj, for some i, j ∈ N, the game ends. The outcome of the game is that player π(1) gets d, where d ¼ maxfd i g. All other players get 0. iaN
(ii) If xi = xj = x, ∀i, j ∈ N, we will refer to x as the binding proposal. Player π(1) becomes the current proposer. The game proceeds to Stage 2. Stage 2 Assume that π(1) = i. In Stage 2, there are two actions “take leave” (L) or “choose a player in N \ {i}, say j, and make a proposal in Gij to j” can be taken by the current proposer. (i) If L is taken, the binding proposal x is the payoff of the game. The game ends. (ii) If (S, y) in Gij is proposed to j, the game proceeds to Stage 3. Stage 3 In Stage 3, there are three actions can be taken by player j: accept (S, y), keep quiet, or reject (S, y) by proposing a proposal in Gji. (i) If player j accepts (S, y), the game ends. Player i gets yi, player j gets xi + xj − yi and players in N \ {i, j} obtain xN \{i,j}. Note that x is the binding proposal formed in Stage 1. (ii) If player j keeps quiet, the game proceeds to Stage 5. (iii) If player j rejects (S, y) by proposing a proposal (T, z) in Gji, the game proceeds to Stage 4. Stage 4 In Stage 4, players in T \ { j} respond to the proposal (T, z) sequentially according to the order induced by π. If (T, z) is accepted by all players in T \ { j} or T = { j}, the game ends. Players in T obtain z, each player k in N \ (T ⋃ {i}) obtains xk and player i obtains v(N ) − z(T ) − x(N \ (T ⋃ {i})). If any player in T \ { j} rejects (T, z), the game proceeds to Stage 5. Stage 5 In Stage 5, players in S \ {i} respond to the proposal (S, y) sequentially according to the order induced by π.
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If (S, y) is accepted by all players in S \ {i} or S = {i}, the game ends. Players in S obtain y, each player k in N \ (S ⋃ { j}) obtains xk and player j obtains v(N ) − y(S ) − x(N \ (S ⋃ { j})). If any player in S \ {i} rejects (S, y), the game ends. The final outcome is x. Remark 1. The TU feature is important in our game model. If player j accepts player i's proposal in Stage 3, then player i receives the transfer amount from player j. If player j rejects player i's proposal, then the objection/counterobjection battle would be launched. The loser obtains the residual and all outcomes in the game are feasible, in and out of equilibrium. Remark 2. On the basis of Stearns' transfer scheme procedure, we devise a bilateral negotiation between the objector and the counterobjector in Stages 2 and 3 of our extensive form game. Suppose that player i proposes (S, y) to player j in Stage 2. If (S, y) is accepted by j, players i and j's payoffs are yi and xi + xj − yi, respectively. Therefore, it can be viewed that the proposal (yi, xi + xj − yi) is brought out to player j implicitly when the proposal (S, y) is made by player i. Since players in S \ {i} do not respond to (S, y) yet and only players i and j have the rights to make decisions in Stages 2 and 3, it can be regarded that there is a bilateral negotiation between i and j in the two stages. Besides winning or losing the objection/counterobjection battle, this bilateral negotiation provides an opportunity for both the objector and the counterobjector to reach a state in which they are in equilibrium. Like the extensive form game proposed by Serrano and Vohra (2002a), our game model is based on the definition of the ADM bargaining set. Due to this factor, both games consist of the stages for the objection/counterobjection battle. For convenience, we call their extensive form mechanism the SV game. Both games have other similarities: (i) The first two stages of our extensive form game are very similar to the ones in the SV game. We follow the design of the first stage of the SV game to ensure that each equilibrium would induce a unanimous agreement and to determine the protocol in further negotiations. (ii) In the second stages of both games, the proposer (the objector) can take the following action: choose a player (the counterobjector) and make a proposal (a potential objection). Nevertheless, besides Remark 2, there are some other differences between the SV game and ours: (1) In Stage 2 of our extensive form game, the objector not only can make a potential objection, but also can take the action L to end the game. (2) Unlike the SV game, if the potential objection (S, y) is proposed, players in S would not be asked for instant responses in our game model. They would decide to accept or reject the potential objection in the final stage, if necessary. (3) In the SV game, the counterobjector needs to name a coalition containing at least two players to make a counterobjection. We do not need the restriction here. (4) In our game, the counterobjector can not reject the agreed objection (or the agreed counterobjection) in our extensive form game. In fact, the final outcome is determined immediately if the potential objection (or the counter-objection) is accepted. (5) Players who are neither the objector nor the counterobjector can keep their payments not less than the ones assigned by the binding proposal in our game model.
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4. Properties In this section, we introduce some properties that will be used to show the main results of the paper. Before proving these properties, we need some necessary notations. Let v be a TU game with the player set N. Let x be an imputation, let i, j ∈ N, i ≠ j and let (S, y) ∈ Gij. We call (S, y) a quasi-objection of i against j at x if it satisfies yk zxk for all kaSnfig: Note that if (S, y) is a quasi-objection of i against j at x satisfying yi N xi, it is an objection of i against j at x. Let (S, y) be a quasi-objection of i against j at x and let (T, z) ∈ Gji. If (T, z) satisfies zk zyk for all kaT \ S; zk zxk for all k aT nðS [ fjgÞ; it is called a quasi-counterobjection to (S, y). Note that if (T, z) is a quasi-counterobjection to the objection (S, y) satisfying zj ≥ xj, it is a counterobjection to (S, y). Let (S, y) be a quasi-objection of i against j at x. We say (S, y) is a maximum quasi-objection of i against j at x if yi ¼ maxfyi V: ðSV; yVÞ is a quasi−objection of i against j at xg: Let (S, y) be a quasi-objection of i against j at x and let (T, z) be a quasi-counterobjection to (S, y). We say (T, z) is a maximum quasi-counterobjection of j to (S, y) if zj ¼ maxfzVj : ðT ;V z VÞ is a quasi−counterobjection to ðS; yÞg;
ð1Þ
and zj is called the maximum surplus of j against (S, y) at x. Note that zj ≥ 0. Let i, j ∈ N, i ≠ j, and let Mij ðvÞ ¼ fxaIðvÞ : for each objection of i against j at x, there exists a counterobjection to it}. It is clear that MðvÞ ¼ \ Mij ðvÞ. i; jaN ;i pj Given x ∈ I(v), i, j ∈ N, i ≠ j and t ≥ 0, define the payoff vector x′ in ℝN as follows: xVi ¼ xi þ t; xj V¼ xj t; xkV ¼ xk ; for all kaN nfi; jg: x′ is said to result from x by a transfer of size t from j to i. The following result is established by Stearns (1968). Lemma 3. Given distinct players i, j in N and x ∈ I(v), there is a minimal t ≥ 0 such that there is a point x′ in Mij ðvÞ which results from x by a transfer of size t from j to i. Denote mij(x) to be the t of Lemma 3. Note that 0 ≤ mij(x) ≤ xj. We call mij(x) the minimal transfer size from j to i at x. Remark 4. The following three statements are equivalent. (i) xaMðvÞ. (ii) xaMij ðvÞ for all i, j ∈ N, i ≠ j. (iii) mij(x) = 0 for all i, j ∈ N, i ≠ j.
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Let x ∈ I(v). Suppose that mij(x) N 0. For each positive integer t, define the payoff vector xt as follows. 1 xti ¼ xi þ 1 mij ðxÞ; t 1 ð2Þ xtj ¼ xj 1 mij ðxÞ; t xtk ¼ xk ; for all kaN nfi; jg: t Clearly, x ∈1I(v). Since 1 t mij ðxÞbmij ðxÞ for all t ≥ 1, it holds that xt gMij ðvÞ by Lemma 3. We derive that there exists a justified objection (St, yt) of i against j at xt for all t ≥ 1. By the facts that 0 ≤ yht ≤ v (St) for all h ∈ St and the cardinality of the set F ij is finite, it holds that there exists a coalition S ⁎ aF ij and a subsequence {ytk} of { yt} such that
S ⁎ ¼ S tk for all k; and y⁎ ¼ lim ytk : kYl
We call such (S⁎, y⁎) a limited objection of i against j at x. Note that (S⁎, y⁎) is also a quasiobjection of i against j at x. Lemma 5. Let mij(x) N 0 and (S⁎, y⁎) be a limited objection of i against j at x. Then zj ≤ xj − mij(x) for all quasi-counterobjection (T, z) to (S⁎, y⁎). Proof. If not, there exists a quasi-counterobjection (T, z) to (S⁎, y⁎) such that zj N xj − mij(x). Let e ¼ jT1 j ðzj ðxj mij ðxÞÞÞ, z′j = xj − mij(x) + ε and z′h = zh + ε for all h ∈ T \ {j}. Since (T, z) is a quasi-counterobjection to (S⁎, y⁎), we have zh ≥ yh⁎ for h ∈ S⁎ ∩ T and zh ≥ xh for h ∈ T \ (S⁎ ⋃ {j}). Since (S⁎, y⁎) is a limited objection of i against j at x, there exists a subsequence fytk g; ytk aℝS ⁎ , such that limkYl ytk ¼ y⁎ and each pair (S⁎, ytk) is a justified objection of i against j at xtk, where xtk is defined by Eq. (2). It holds that (T, z′) is a counterobjection to (S⁎, ytk) at xtk for some sufficiently large positive integer tk. This contradicts to the fact that (S⁎, ytk) is a justified objection of i against j at xtk. We obtain the desired contradiction. □ 5. The main results In this section, we will show that each payoff vector in MðvÞ can be supported by a subgame perfect equilibrium of the game Γ(v) and every subgame perfect equilibrium outcome is in MðvÞ. First, we shall construct a subgame perfect equilibrium f for each payoff vector ¯x in MðvÞ. For each player i and each imputation x, denote Fi ðxÞ ¼ k a N nfig : mik ðxÞN0 and k a arg max ½mil ðxÞ : laN nfig
Let ¯ xaMðvÞ. The strategy profile f is defined in the following. (1) At the beginning of the game, player i proposes (x¯, π I, − 1) for all i ∈ N, where πI is the identity mapping on N. In the following, we assume that x is the binding proposal.
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(2) In Stage 2, if player i is the current proposer, i takes the action:
L; propose ðS ⁎ ; ðxi þ mij ðxÞ; y⁎ S ⁎ nfig ÞÞ to j;
if Fi ðxÞ ¼ F; o:w:;
where (S⁎, y⁎) is a limited objection of i against j at x, and j is the smallest index among Fi(x) according to the order of πI. (3) Suppose that player i proposes (S, y) to player j in Stage 2. Player j is going to respond to the proposal (S, y) in Stage 3. There are two cases to be discussed. (i) If (S, y) is a quasi-objection of i against j at x, then player j takes the action: proposeðT V; zVÞ; if zj VNxj þ xi yi ; accept ðS; yÞ; o:w:; where (T′, z′) is a maximum quasi-counterobjection of j to (S, y). (ii) If (S, y) is not a quasi-objection of i against j at x, then player j takes the action: 8 if xj þ xi yi zmaxfzWj ; xj g; < accept ðS; yÞ; propose ðT W; zWÞ; if z Wj Nmaxfxj þ xi yi ; xj g; : keep quiet; o:w:; where (T″, z″ ) is a maximum quasi-objection of j against i at x. (4) Suppose that player j proposes (T, z) in Stage 3 to reject the proposal (S, y) proposed in Stage 2 by player i. Player k, k ∈ T \ { j}, is going to respond to (T, z) in Stage 4. (i) If k ∈ S, there are two cases to be discussed. Case 1 (S, y) is a quasi-objection of i against j at x. Player k responds: accept; if zk zyk ; reject; o:w: Case 2 (S, y) is not a quasi-objection of i against j at x. Player k responds: accept; if zk zxk ; reject; o:w: (ii) If k ∉ S, then player k responds:
accept; if zk zxk ; reject; o:w:
(5) In Stage 5, player k, k ∈ S \ {i}, is going to respond to the proposal (S, y), then player k responds: accept; if yk zxk ; reject; o:w:
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It is easy to see that the outcome of the strategy profile f is x¯. By Remark 4, mij (x¯) = 0 for all i, j ∈ N, i ≠ j. Then Fi(x¯) = 0 for all i ∈ N. By following f, each player proposes (x¯, πI, − 1). By the game rule, x¯ is the binding proposal and πI(1) = 1 is the current proposer. Since F1(x¯) = ∅, player 1 will take L in Stage 2, the game ends with the outcome x¯. To show that f is a subgame perfect equilibrium, we need the following results. Lemma 6. In Stage 5, suppose that player k is going to make a response. If all players in N \ {k} follow f−k, then it is the best for player k to follow fk. Proof. Suppose that x is the binding proposal and player k is going to respond to the proposal (S, y) proposed by i in Stage 2. There are two cases to be considered. Case 1 yk ≥ xk. By following fk, player k will accept (S, y). If yh ≥ xh for all h ∈ S \ {i, k}, then (S, y) is accepted by all players in S \ {i} and player k obtains yk. If player k deviates by rejecting (S, y), the game ends and player k's payoff is xk ≤ yk. If yh b xh for some h ∈ S \ {i, k}, then (S, y) is rejected by player h. The game ends and player k's payoff is xk whatever player k deviates or not. Case 2 yk b xk. By following fk , player k will reject (S, y). The game ends and player k's payoff is xk. If player k deviates by accepting (S, y), then either (S, y) is accepted by all players in S \ {i} or is rejected by some player in S \ {i, k}. We derive that player k obtains at most xk. By Cases 1 and 2, it is not better off for player k to deviate. □ Lemma 7. In Stage 4, suppose that player k is going to make a response. If all players in N \ {k} follow f−k, then it is the best for player k to follow fk. Proof. Suppose that x is the binding proposal. We assume that player i proposes (S, y) to player j in Stage 2 and player j rejects (S, y) by proposing (T, z) in Stage 3. In Stage 4, player k, k ∈ T \ { j}, is going to respond to the proposal (T, z). We only discuss the case that k ∈ S and (S, y) is a quasiobjection of i against j at x. The remaining cases can be derived by the similar arguments. There are two possibilities to be considered. Case 1 zk ≥ yk. Player k will accept (T, z) by following fk. (i) If (T, z) is accepted by all players in T \ { j, k}, the game ends and player k obtains zk. If player k deviates by rejecting (T, z), the game proceeds to Stage 5. Player k obtains at most yk by Lemma 6. It is not better off for player k to deviate. (ii) If (T, z) is rejected by some player h in T \ { j, k} by following fh, the game proceeds to Stage 5. Player k obtains yk by the assumptions that all players in N \ {k} follow f−k and (S, y) is a quasi-objection of i against j at x. If player k deviates, the game still proceeds to Stage 5 by the assumption that player h rejects (T, z) by following fh. Player k will obtain at most yk by Lemma 6. It is not better off for player k to deviate. Case 2 zk b yk. Player k will reject (T, z) by following fk. The game proceeds to Stage 5, (S, y) will be accepted by all players in S \ {i} and player k obtains yk. Assume that player k deviates. Then either the game proceeds to Stage 5, or (T, z) is accepted by all players in T \ { j}. In case of the game proceeding to Stage 5, player k obtains at most yk by Lemma 6. In case of all players in T \ {j} accepting (T, z), player k obtains zk b yk. It is not better off for player k to deviate. □
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Remark 8. Suppose that x is the binding proposal, players i and j are proposers in Stages 2 and 3, respectively. If players in N \ {i, j} follow fN \{i, j} in Stages 4 and 5, then each player k in N \ {i, j} obtains xk at least by Lemmas 6 and 7. In fact, each player who needs to make a response in Stages 4 or 5 can reject i or j's proposal such that the final outcome of the game is x. Assume that, at the beginning of the game Γ(v), player i proposes ðx; pi ; d i ÞaIðvÞ P ℝ for each i ∈ N. Let d ¼ maxiaN fd i g. Since each player proposes the same allocation x, x is the binding proposal. Under such assumptions, we shall show Lemmas 9 and 10. Lemma 9. In Stage 3, player j is going to respond to i's proposal (S, y). If all players in N \ { j} follow f−j, then it is the best for player j to follow fj. Proof. There are two cases to be considered. Case 1 (S, y) is a quasi-objection of i against j at x. Then yh zxh for all h a Snfig:
ð3Þ
Let (T′, z′) be a maximum quasi-counterobjection to (S, y). There are two subcases to be discussed. Subcase 1 xj + xi − yi ≥ z′j. By following fj, player j accepts (S, y). The game ends and player j obtains xj + xi − yi. (1) Suppose that player j deviates by proposing (T, z). If (T, z) is a quasi-counterobjection to (S, y), then (T, z) is accepted by all players in T \ { j} following fT \{j}. The game ends and player j obtains zj ≤ zj′ ≤ xj + xi − yi. It is not better off for player j to deviate. If (T, z) is not a quasi-counterobjection to (S, y), then (T, z) is rejected by some player in T \ { j}. The game proceeds to Stage 5. Claim A: Player j obtains at most xj + xi − yi. Since (S, y) is a quasi-objection of i against j at x, (S, y) is accepted by all players in S \ {i} following fS \{i}. The game ends and player j obtains v(N) −y(S) −x(N \ (S ⋃ {j}) ≤xj +xi −yi by Eq. (3). By Claim A, it is not better off for player j to deviate. (2) Suppose that player j deviates by keeping quiet. Then the game proceeds to Stage 5. By Claim A, player j obtains at most xj + xi − yi. It is not better off for player j to deviate. Subcase 2 xj + xi − yi b z′j. By following fj, player j rejects (S, y) by proposing (T′, z′). (T′, z′) will be accepted by all players in T ′\ {j} following fT′ \{j}, player j obtains z′j. If player j deviates by accepting (S, y), then player j obtains xj + xi − yi. It is not better off for player j to deviate. If player j deviates by proposing (T, z) to reject (S, y), then, by the same arguments of Subcase 1, it is not better off for player j to deviate. If player j deviates by keeping quiet, then player j obtains at most xj + xi − yi by Claim A. It is not better off for player i to deviate. Case 2 (S, y) is not a quasi-objection of i against j at x. Let (T″, z″ ) be a maximum quasi-objection of j against i at x. There are three subcases to be considered. Subcase (i) xj + xi − yi ≥ max{z″j, xj}. By following fj, player j will accept (S, y). The game ends and player j obtains xj + xi − yi. If player j deviates, then either he proposes (T, z) to reject (S, y) or keeps quiet.
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If j keeps quiet, the game proceeds to Stage 5. Claim 1: If the game proceeds to Stage 5, then player j obtains xj. By the assumption that (S, y) is not a quasi-objection of i against j at x, we have yh b xh for some h ∈ S \ {i}. Player h rejects (S, y) by following fh and the game ends. Player j obtains xj. By Claim 1, it is not better off for player j to deviate. If j proposes (T, z) to reject (S, y), the game proceeds to Stage 4. Claim 2: Player j's payoff is at most max{z″j, xj}. If (T, z) is a quasi-objection of j against i at x, then all players in T \ { j} accept (T, z) by following fT \ { j}. Then player j obtains zj ≤ z″j. If (T, z) is not a quasi-objection of j against i at x, then zh b xh for some h in T \ { j}. Player h rejects (T, z) by following fh in Stage 4 and the game proceeds to Stage 5. Player j obtains xj by Claim 1. We derive that player j's payoff is at most max {z″j, xj}. By Claim 2, player j obtains at most max{z″j, xj} ≤ xj + xi − yi. It is not better off for player j to deviate. Subcase (ii) z″j N max{xj + xi − yi, xj}. By following fj, player j will propose (T″, z″) to reject (S, y) and the game proceeds to Stage 4. All players in T″ \ {j} will accept (T″, z″) by following fT″ \{ j}. The game ends and player j obtains z″j. If player j deviates, then either player j accepts (S, y), or keeps quiet, or proposes (T, z) to reject (S, y). If player j accepts (S, y), the game ends and player j obtains xj + xi − yi b z″j. If player j keeps quiet, the game proceeds to Stage 5. By Claim 1, player j's payoff is xj b z″j. If player j proposes (T, z) to reject (S, y), then player j obtains at most max{z″j, xj} = z″j by Claim 2. Subcase (iii) xj + xi − yi b max{z″j, xj} and z″j ≤ max{xj + xi − yi, xj}. It holds that max{xj + xi − yi, z″j} ≤ xj. By following fj, player j keeps quiet and the game proceeds to Stage 5. By Claim 1, player j obtains xj. If player j deviates, then either player j accepts (S, y) or proposes (T, z) to reject (S, y). If player j accepts (S, y), the game ends and player j obtains xj + xi − yi ≤ xj. If player j proposes (T, z) to reject (S, y), the game proceeds to Stage 4. By Claim 2, player j obtains at most max{z″j, xj} = xj. It is not better off for player j to deviate. By Cases 1 and 2, the proof is complete. □ Lemma 10. Suppose that player i is the current proposer in Stage 2. Then, it is the best for player i to follow fi. Proof. There are two cases to be considered. Case 1 Fi(x) = ∅. By following fi, player i takes the action L, the game ends and player i obtains xi. If player i deviates by proposing (S, y) to some player, say j, there are two subcases to be discussed. Subcase 1 (S, y) is an objection of i against j at x. Since Fi(x) = ∅, we have that mij(x) = 0 and xaMij ðvÞ by Lemma 3. By the assumption that (S, y) is an objection of i against j at x, we derive that there exists a counterobjection (T, z) to (S, y) and zj ≥ xj. Let (T′, z′) be a maximum quasi-counterobjection to (S, y). Then, z′j ≥ zj ≥ xj. Since (S, y) is an objection of i against j at x, it holds that yi N xi. We derive that zj′ N xj + xi − yi.
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By the assumption that all players in N \ {i} following f−i, player j will reject (S, y) by proposing (T′, z′) and (T′, z′) will be accepted by all players in T \ {j}. The game ends and player i obtains X vðN Þ zVðT VÞ xðN nðT V [ figÞÞ ¼ xðT V [ figÞ z VðT ÞV V xi þ ðxk zkVÞV xi : kaT V
It is not better off for player i to deviate. Subcase 2 (S, y) is not an objection of i against j at x. It means that yi ≤ xi or yh b xh for some h ∈ S \ {i}. Note that each player h in N \ {i, j} obtains xh at least by Remark 8. (1) Suppose that yi ≤ xi. If player j accepts (S, y), then the game ends and player j obtains xj + xi − yi ≥ xj. By Lemma 9, player j will obtain xj at least by following fj. We derive that player i obtains at most xi. (2) Suppose that yh b xh for some h ∈ S \ {i}. If player j keeps quiet, then the game proceeds to Stage 5. Player h rejects (S, y) by following fh and the game ends. Player j obtains xj. By Lemma 9, player j will obtain xj at least by following fj. We derive that player i obtains at most xi. It is not better off for player i to deviate. Case 2 Fi(x) ≠ ∅. By following fi, player i proposes ðS ⁎ ; ðxi þ mij ðxÞ; y⁎ S ⁎ nfig ÞÞ
ð4Þ
to j, where j is the smallest index among Fi(x) according to the order of πI and (S⁎, y⁎) is a limited objection of i against j at x. Note that Eq. (4) is an objection of i against j at x. Let (T′, z′) be a maximum quasi-counterobjection to Eq. (4). Then (T′, z′) is also a quasi-counterobjection to (S⁎, y⁎). By Lemma 5, we have zj′ ≤ xj − mij (x). Player j accepts the proposal Eq. (4) by following fj and the game ends. Player i obtains xi + mij(x). If player i deviates by taking the action L, the game ends. Player i obtains xi b xi + mij(x). If player i deviates by proposing (S, y) to some player, say k, there are two subcases to be discussed. Subcase 1 (S, y) is a quasi-objection of i against k at x. There are two possibilities to be discussed. (1) yi N xi + mik(x). then (S, y) is an objection of i against k at the payoff vector x′ = (xi + mik(x), xk − mik(x), xN \ {i,k}). We have xVaMik ðvÞ by Lemma 3. Then there exists a counterobjection (T, z) to (S, y). Let (T′, z′) be a maximum quasi-counterobjection to (S, y). It holds that zk′ ≥ zk ≥ xk − mik(x) N xi +xk − yi. By following fk, player k will propose the maximum quasi-counterobjection (T′, z′) to reject (S, y). The game proceeds to Stage 4 and all players in T′ \ {k} will accept (T′, z′) by following fT′ \{k}. The game ends and player i obtains vðN Þ zVðT VÞ xðN nðT V [ figÞÞ ¼ xðT V [ figÞ zVðT VÞV xi þ xk zkV X þ ðxh zV h ÞV xi þ mik ðxÞV xi þ mij ðxÞ: haT Vnf kg
It is not better off for player i to deviate. (2) yi ≤ xi + mik(x). If player k accepts (S, y), then the game ends. Player k obtains xi + xk − yi. By Lemma 9, player k will obtain xi + xk − yi at least by following fk. Each player h in N \ {i, k} obtains xh at least by Remark 8. It holds that player i gets at most yi ≤ xi + mik(x) ≤ xi + mij(x). It is not better off for player i to deviate. Subcase 2 (S, y) is not a quasi-objection of i against k at x. Then yh b xh for some h ∈ S \ {i}. If player k keeps quiet, the game proceeds to Stage 5 and player h will reject (S, y) by following fh.
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The game ends and player k obtains xk. By Lemma 9, player k obtains xk at least by following fk. It implies that player i gets at most xi ≤ xi + mij(x). It is not better off for player i to deviate. □ In the following, we will show that f is a subgame perfect equilibrium. Theorem 11. f is a subgame perfect equilibrium of Γ(v). Proof. By the above Lemmas, it remains to show that it is the best for player i to follow fi at the beginning of the game. We have mentioned that, by following f, x¯ is the binding proposal and πI(1) = 1 is the current proposer. Since F1(x¯) = ∅, player 1 will take L in Stage 2, the game ends with the outcome x¯. If player i deviates at the beginning of the game by proposing (x, π′, d′), there are two cases to be considered. Case 1 x ≠ x¯. No binding proposal is formed. Then player i obtain max{− 1, d′} b x¯i. Case 2 x = x¯. The binding proposal is still x¯. Suppose that the current proposer is player h. If h ≠ i, then player h takes L in Stage 2 by following fh, the game ends and player i obtains x¯i. If h = i, then it is the best for player i to follow fi by Lemma 10. That is, player i takes L, the game ends and player i obtains x¯i. By Cases 1 and 2, it is not better off for player i to deviate. □ Next, we show that every subgame perfect equilibrium outcome of the game Γ(v) is in the ADM bargaining set MðvÞ. Let g = (g1, g2, ⋯, gn) be an arbitrary subgame perfect equilibrium of the game Γ(v). Lemma 12. Suppose that x is the binding proposal and the game proceeds to Stage 5. If all players in S \ {i} are going to respond to the proposal (S, y) proposed by i and yh N xh for all h ∈ S \ {i}, then (S, y) will be accepted by all players in S \ {i} following gS \{i}. Lemma 12 is easy to derive, we omit the proof. Lemma 13. Suppose that x is the binding proposal and the game proceeds to Stage 3 in which j is going to respond to (S, y) proposed by player i. If yh N xh for all h ∈ S \ {i} and the maximum surplus of j against (S, y) at x is less than xj + xi − yi, then player j will accept (S, y) by following gj. Proof. We shall show that player j will accept (S, y) and obtain xj + xi − yi by following gj. If not, there are two cases to be discussed. Case 1 By following gj, player j rejects (S, y) by proposing (T, z). There are two possibilities to be considered. Subcase 1 (T, z) is accepted by all players in T \ { j} following gT \{j}. The game ends and players in T obtain z. We shall show that (T, z) is a quasi-counterobjection to (S, y). If not, then either zh b yh for some h ∈ S ∩ T or zh b xh for some h ∈ T \ (S ⋃ {j}). In case of zh b yh for some h ∈ S ∩ T, player h can deviate by the following. Player h rejects (T, z) in Stage 4 and he follows gh in Stage 5. Then the game will move to Stage 5 and, by Lemma 12, (S, y) will be accepted by all players in S \ {i}. Player h obtains yh N zh. It violates that g is a subgame perfect equilibrium. In case of zh b xh for some h ∈ T \ (S ⋃ {j}), player h can deviate by rejecting (T, z). The game proceeds to Stage 5 and player h's payoff is xh N zh. It violates that g is a subgame perfect equilibrium. We derive that (T, z) is a quasi-counterobjection to (S, y). It holds that zj b xj + xi − yi by the assumption that the maximum surplus of j against (S, y) is less than xj + xi − yi. If player j accepts (S, y), he obtains xj + xi − yi. It is better off for player j to deviate. This violates that g is a subgame perfect equilibrium.
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Subcase 2 (T, z) is rejected by some player h in T \ {j} following gh. The game proceeds to Stage 5. In Stage 5, (S, y) will be accepted by all players in S \ {i} by Lemma 12 and the game ends. Player j obtains v (N) − y(S) − x(N \ (S ⋃ {j})) b xj + xi − yi. This violates that g is a subgame perfect equilibrium. Case 2 Player j keeps quiet by following gj. The game proceeds to Stage 5. By the same arguments of Subcase 2, we obtain the desired contradiction. By Cases 1 and 2, the proof is complete. □ Theorem 14. The subgame perfect equilibrium outcome of g is in the ADM bargaining set MðvÞ. Proof. Suppose that, at the beginning of the game, i proposes ðxi ; pi ; d i ÞaIðvÞ P ℝ by following gi, ∀i ∈ N. Step 1: The subgame perfect equilibrium g must induce a binding proposal. We claim that xi = xj for all i, j ∈ N. Denote π = π1 ∘ π2 ∘ ⋯ ∘ πn and assume that π(1) = k ∈ N. Suppose that no binding proposal is formed. It implies that xi ≠ xj for some i, j ∈ N, i ≠ j. By the game rule, player k obtains d ¼ maxiaN fd i gb0, other players obtain 0. Let k deviate by proposing (xk, πk, d / 2) at the beginning of the game. Since xi ≠ x j for some i, j ∈ N, i ≠ j, no binding proposal is formed. Hence, player k obtains d / 2 N d. Player k is better off by deviation. We obtain the desired contradiction. Step 2: Let x be the binding proposal induced by g. We will show that x is the subgame perfect equilibrium outcome. Let x′ be the outcome of the game by following g. We will show that x′i ≥ xi for all i in N. Let player i deviate by the following. Player i proposes (x, π¯i, di ) at the beginning of the game, where π¯i satisfies i iþ1 n pð1Þ ¼ p1 ∘p2 ∘ : : : ∘pi1 ∘ p ¯ ¯ ∘p ∘ : : : ∘p ð1Þ ¼ i:
ð5Þ
Then player i becomes the current proposer. Player i takes the action L in Stage 2, the game ends and player i gets xi. By subgame perfection, we derive that x′i ≥ xi. Since the binding proposal x is formed, the game would not end in Stage 1. We derive that x′ (N) = x(N) = v(N) by the game rule and x′ = x. Step 3: xaMðvÞ. If not, there exist i, j ∈ N, i ≠ j such that mij(x) N 0 by Remark 4. Let 0 b ε b mij (x) and x″ = (xi + ε, xj − ε, xN \ {i,j}). Since mij(x) is the minimal transfer size from j to i at x, there exists a justified objection (S, y) of i against j at x″ satisfying yk N xk for all k ∈ S \ {i}. We derive that zj b x″j = xj − ε for all (T, z) in Gji satisfying zðT Þ ¼ vðT Þ; zh ¼ yh for haS \ T ; zh ¼ xh for haT nðS [ fjgÞ: Let y′i = xi + ε and y′k = yk for all k ∈ S \ {i}. It holds that the maximum surplus of j against (S, y′) at x is less then xj − ε. Let player i deviate by the following. Player i makes the proposal (x, π¯i, di) at the beginning of the game, where π¯i satisfies Eq. (5). The binding proposal is still x and player i becomes the current proposer. In Stage 2, player i makes the proposal (S, y′) to player j and player j will accept (S, y′) in Stage 3 by Lemma 13. The game ends and player i obtains xi + ε. Player i is better off by deviation. We obtain a desired contradiction. By Steps 1, 2 and 3, the proof is complete. □
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