Proof that the outcome set satisfies the axioms

338 S .J. Willson Mathematical Social Sciences 39 2000 323 –348

6. Proof that the outcome set satisfies the axioms

The purpose of this section is to prove the following result. Main Theorem 6.1. There exists a unique function X which assigns to each strict game G a nonempty set XG of payoff vectors from G and which satisfies the axioms in Section 2. Moreover, if rational players negotiate under the rules of play of Section 3, then the set XG consists precisely of all the possible final outcomes under these rules of play. These rules of play thus give a formal procedure that realizes the set XG. Simulation of the play via these rules by rational players will give a computation of XG. The proof of Theorem 6.1 is lengthy and will be broken into several pieces. Note that we assume throughout that the game G is strict. In particular, this implies that if two states v and w satisfy that pv pw and pv ± pw, then pv . pw for all all i players i. Theorem 6.2. There is at most one function X satisfying the axioms. Proof. Suppose that both X and Y satisfy the axioms. We know that XG consists of all the vectors Xs ; G and YG consists of all the vectors Ys ; G by the floor axiom. Hence it suffices to prove that for every permutation s and every game G we have Xs ; G 5 Ys ; G. The proof will be by induction on the number k of distinct payoff vectors for the game G. If k 5 1, then the result is trivial, since each Xs ; G and each Ys ; G must be a payoff vector of G by the form axiom; since k 5 1 there is only one such vector, so they are equal. We now assume by induction the truth of the result for any game with fewer than k distinct payoff vectors, and we prove the result for a game G with exactly k distinct payoff vectors, where k . 1. By the default reduction axiom, Xs ; G 5 Xs ; Gs and Ys ; G 5 Ys ; Gs. We show that Gs has fewer than k distinct payoff vectors; then Xs ; Gs 5 Ys ; Gs by induction, and the result follows. To see that Gs has fewer than k distinct payoff vectors, note that every payoff vector for Gs is also a payoff vector from G, so Gs has at most k distinct payoff vectors. If any payoff vector pv for G fails to satisfy pv all Fs ; G, then by the definition of Gs it follows that pv is replaced by Fs ; G, already one of the payoff vectors, wherever it occurs; hence the payoff vector pv does not occur in Gs, Gs has strictly fewer than k distinct payoff vectors, and by the inductive hypothesis, Xs ; Gs 5 Ys ; Gs. There remains only the case where every payoff vector pv for G satisfies pv all Fs ; G. In this case, the definition of Gs yields that the payoff matrix for Gs is the same as the payoff matrix for G. Hence Fs ; Gs 5 Fs ; G. The minimal possible payoff is Fs ; G for all players simultaneously. All other payoffs are strictly better simultaneously for all the players, by strictness of G. Since k 5 2, not all payoff vectors can equal Fs ; G, and there exists a state v for which pv ± Fs ; G. We now utilize the calculation in Lemma 5.5 to obtain a contradiction. Since Fs ; G is the minimal payoff vector for Gs and Fs 5 max hK s; s : s lies in S j, s 1 1 s 1 s 1 s 1 S .J. Willson Mathematical Social Sciences 39 2000 323 –348 339 it follows that for all s in S K s ; s 5 Fs ; G. Similarly, since K s ; s 1 s 1 , 1 s 1 1 s 5 max hK s; s , s : s lies in S j and K s; s is the minimal s 1 s 2 2 s 1 s 2 s 2 s 2 1 s 1 vector Fs ; G, it follows that K s ; s , s 5 Fs ; G for all s in S and all 2 s 1 s 2 s 1 s 1 s in S . Repeating this argument inductively, we see that K s ; s , . . . , s 2 s 2 n 21 s 1 s 5 Fs ; G for all s in S , all s in S , . . . , and all s in S . In s n 21 s 1 s 1 s 2 s 2 s n 21 s n 21 particular, if v is the state found above such that pv ± Fs ; G, then K s ; v , . . . , n 21 s 1 v 5 Fs ; G. But K s ; v , . . . , v 5 max h pt: t 5 v , . . . , s n 21 n 21 s 1 s n 21 s n s 1 s 1 t 5 v j pv since v is one of the candidates for t. This proves Fs ; s n 21 s n 21 all G pv yet pv Fs ; G by hypothesis, whence Fs ; G 5 pv, a contradic- all all tion, proving Theorem 6.2. h In the remainder of this section we demonstrate that the sets XG defined in Section 5 satisfy the axioms given in Section 2. This will complete the proof of Theorem 6.1. The form and nontriviality axioms are immediately seen to be satisfied by the outcomes obtained by our rules of play. The rules of play clearly do not favor any particular strategies or any particular players, and it follows that the outcome sets XG obey the symmetry axiom. We now prove the other properties. 6.1. Pareto-optimality Theorem 6.3. Pareto-optimality axiom. Each Os is Pareto-optimal. Proof. Suppose that the transiency is t, so Os 5 Os ; t, s 5 Os ; t 1 1, s for any state s. If Os is not Pareto-optimal, then there exists a state w such that Os pw all and for some player i, Os , pw. Hence Us ; t,w 5 pw. Moreover, since i pw ± Os, it must follow that Os , pw for all j since the game is strict. j For simplicity of notation, we give the proof only in the case where n 53 and the order is s 51, 2, 3. A state will be described by three strategies abc, in which a is the strategy by player 1, b is the strategy by player 2, and c is the pure strategy by player 3. We will sometimes omit the s from the notation for M, O, and U. Suppose that state w 5 abc. Hence Ut, abc 5 pw . Os for all i. i ˜ ˜ Note Mt 1 1, abc, 3 5 max hUt, u: abc [3] uj 5 max hUt, abc : c in S j Ut, 3 3 3 3 abc . Os. Hence Mt 1 1, abc, 3 ± Os. If Mt 1 1, abc, 3 5 Ut, u, then because 3 Ut, u ± Os 5 Ot,u, it follows that Ut, u 5 pu whence Ut, u . Ot, u 5 Os i for all i. We thus obtain that Mt 1 1, abc, 3 ± Os and Mt 1 1, abc, 3 . Os for i all i. ˜ ˜ Similarly Mt 1 1, abc, 2 5 max hMt 1 1, abc, 3: b in S j Mt 1 1, abc, 3. 2 2 2 Since Mt 1 1, abc, 3 . Os by the previous paragraph, we see that Mt 1 1, abc, 2 ˜ ˜ 2 ± Os. But Mt 1 1, abc, 2 5 Mt 1 1, abc, 3 [for some b ] 5 Ut, u [for some u] Os. Hence Mt 1 1, abc, 2 . Os for all i. all i ˜ ˜ Now Mt 1 1, abc, 1 5 max hMt 1 1, abc, 2: a in S j Mt 1 1, abc, 2 . Os 1 1 1 1 by the preceding paragraph. Hence Mt 1 1, abc, 1 ± Os. But Mt 1 1, abc, 1 5 340 S .J. Willson Mathematical Social Sciences 39 2000 323 –348 ˜ ˜ ˜ ˜ ˜ Mt 1 1, abc, 2 [for some a ] 5 Mt 1 1, abc, 3 [for some b ] 5 Ut, u [for some u] Os. Hence Mt 1 1, abc, 1 . Os for all i. all i Finally, Os ; t 1 1 5 Mt 1 1, abc, 1. Hence we see that Os ; t 1 1 . Os 5 i Os ; t for all i. This contradicts Theorem 5.4, which showed that Os ; t 1 1 5 Os ; t. h The agreement axiom follows immediately from the Pareto-optimality axiom since in the perhaps rare situation where there is a single state s that is simultaneously best for all players, it follows that ps is the only Pareto-optimal payoff vector. Since XG is non-empty by nontriviality, XG must be exactly h psj. 6.2. Lower bound axiom Recall that a lower bound for player i is a number L for which there exists a pure strategy a [ S such that for all s in S for which s 5 a it is true that p s L. i i i Theorem 6.4. Lower bound axiom Suppose that L is a lower bound for player i arising from the pure strategy a in S . Then for each permutation s, Os L. i i Proof. Since Os Os ; 1 by Lemma 5.2, it suffices to prove Os ; 1 5 L for all i. all i 1 1 Suppose that i 5 s j . Then Os ; 1 5 Ms ; 1, s , s1 [for any state s ] 5 max s 1 1 2 2 3 hMs; 1, t, s2: s [s1] t j 5 Ms; 1, s , s2 [for some state s ] 5 Ms; 1, s , 3 j j s3 [for some state s ] 5 . . . 5 Ms ; 1, s , s j [for some state s ]. j There are two cases to consider. If j , n then Ms ; 1, s , s j 5 max hMs; 1, t, i j ˆ ˆ s j 1 1: s [i] t j [since s j 5 i]. Among the possible t is some t for which t 5 a. i ˆ ˆ ˆ For this t it follows that Ms ; 1, t, s j 1 1 L since Ms ; 1, t, s j 1 1 will be the i payoff from a state with ith component a [because player i never thereafter changes the strategy again], and by the definition of L all the resulting payoffs to player i must be at least L. Hence the best t from the viewpoint of s j 5 i must also give payoff to i at least L. j j On the other hand, if j 5 n then Ms ; 1, s , s j 5 max h pt: s [i] t j [since i j ˆ ˆ s j 5 i]. Among the candidates t is some t for which t 5 a. Hence Ms; 1, s , i ˆ ˆ s j pt . By the definition of L, we know pt 5 L. The result follows. h i i 6.3. Floor axiom Theorem 6.5. For any permutation s, Fs 5 Os ; 1. Proof. Let s 5 s , s , . . . , s be the state obtained in the computation of Fs, so that 1 2 n Fs 5 ps. When player sn makes the choice of strategy, s is already determined s k for all k ± n. Player sn thus chooses s such that ps 5 max h pu: t [sn] u and s n s n for all k ± n, t 5 s j. s k s k Hence ps 5 Ms ; 1, u, sn for any u agreeing with s in all positions except possibly in position sn. In particular, Ms ; 1, s, sn 5 ps 5 Fs. S .J. Willson Mathematical Social Sciences 39 2000 323 –348 341 Thus, if player sn 2 1 chooses strategy s , then the assumption that sn is s n 21 rational means that the payoff will be Ms ; 1, s, sn. Hence player sn 2 1 chooses s so as to maximize Ms ; 1, s, sn, hence to obtain the payoff max hMs; s n 21 s n 21 1, u, sn: s [sn 2 1] u j. Each such Ms; 1, u, sn equals ps by the previous paragraph. This same quantity defines Ms ; 1, s, sn 2 1, whence Ms ; 1, s, sn 2 1 5 ps 5 Fs. Repeating this argument, we find that Ms ; 1, s, s j 5 ps for each j. In particular, Os ; 1 5 Os ; 1, s 5 Ms ; 1, s, s1 5 ps 5 Fs. h Corollary 6.6. For any permutation s, Fs Os. all Proof. This follows from Lemma 5.2 and the definition of Os. h Corollary 6.7. Floor axiom. For each permutation s of the players there is a uniquely determined outcome Xs in XG such that Xs Fs. Moreover, every element of all XG arises in this manner for some permutation s. Proof. Given any permutation s, let Xs 5 Os. Then Xs satisfies Xs Fs. all Conversely, any pv in XG has the form pv 5 Os for some permutation s. h 6.4. Strong dominance Theorem 6.8. Strong dominance axiom: suppose that a player say player j has a strategy s that is strongly dominated. Let G9 denote the matrix game obtained by j 9 9 eliminating the strategy s from S . Thus, we let S9 5 P S , where S 5 S for i ± j, j j i i i 9 S 5 S 2 hs j. Define p9: S9 → R by p9 5 p uS9, and let G9 5 hS9, p9j. Then XG 5 XG9. j j j Proof. For simplicity of notation, we prove the result in the case where j 5 1, S 5 h1,2, . . . ,mj, and strategy m is strongly dominated by strategy 1 for player 1. 1 Claim 1. For each permutation s, Fs, G equals Fs, G9 and also equals pv for some state v in S9. Proof of Claim 1. Suppose that the order in which players announce their irrevocable strategies is s. If player 1 is player sk, we compute Fs, G 5 Os ; 1, s; G [for any s, 2 by Theorem 6.4, so we may select s [ S9] 5 Ms; 1, s, s1; G 5 Ms; 1, s , s2; 2 k k G [for some s ] 5 . . . 5 Ms ; 1, s , sk; G [for some s ]. n In the case where k 5 n, then Fs ; G 5 max hUs; 0, t; G: s [sn]t j 5 max s n 1 n h pt: s [1] t j. Suppose that the last quantity is maximized when t 5 u. If u ± m, then 1 u ± S9 and clearly Fs ; G9 also equals pu since the only change in the computation n for Fs ; G9 would be that in finding max h pt: s [1] t j we would need to restrict t to 1 being in S9, and t 5 u would still be obtained. If instead u 5 m, consider the state v 1 n such that v 5 u for all j ± 1, v 5 1. Then s [1] v. Moreover, by strong dominance j j 1 n pv pu so max h pt: s [1] t j is also maximized when t 5 v. Thus we may 1 1 342 S .J. Willson Mathematical Social Sciences 39 2000 323 –348 utilize v rather than u and Fs ; G 5 pv. Since v [ S9 again Fs; G9 5 pv 5 Fs; G. This proves the result if k 5 n. k k If k , n then Ms ; 1, s , sk; G 5 max hMs; 1, u, sk 1 1; G: s s k k [sk]u j 5 Ms; 1, u, sk 1 1; G for some u such that s [sk] u. If u ± m, then u 1 k 12 lies in S9. Moreover, Ms ; 1, u, sk 1 1; G 5 Ms ; 1, u , sk 1 2; G [for some k 12 k 12 k 12 n u such that u [sk 1 1] u , hence satisfying u [ S9] 5 . . . 5 Ms; 1, u , n n n sn; G [for some u such that u [sn] u , hence satisfying u [ S9] 5 pv [for n some v such that u [sn] v, hence satisfying v [ S9]. Thus if u ± m, then Fs; 1 G 5 pv for some v [ S9, and again a consideration of each step in the parallel computation of Fs ; G9 shows that Fs ; G9 5 pv for this same v. This proves the result if u ± m. 1 k Otherwise, if u 5 m, define the state v such that v 5 1 and s [sk] v. We will 1 1 show that Ms ; 1, v, sk 1 1; G Ms ; 1, u, sk 1 1; G. To see this, note that 1 k 12 k 12 Ms ; 1, u, sk 1 1; G 5 Ms ; 1, u , sk 1 2; G [for some u such that u k 12 k 12 n [sk 1 1] u , hence satisfying u 5 m] 5 . . . 5 Ms ; 1, u , sn; G [for some 1 n n n n u such that u [sn] u , hence satisfying u 5 m] 5 pw [for some w such that u 1 [sn] w, hence satisfying w 5 m]. Similarly Ms ; 1, v, sk 1 1; G 5 px for some 1 state x satisfying x 5 1. But px pw by strong dominance, showing that Ms ; 1, 1 1 v , sk 1 1; G Ms ; 1, u, sk 1 1; G. On the other hand, since u was chosen to 1 maximize the expression, we also have Ms ; 1, v, sk 1 1; G Ms ; 1, u, sk 1 1; 1 G. Hence Ms ; 1, v, sk 1 1; G 5 Ms ; 1, u, sk 1 1; G 5 px. We see that if u 5 m, we may replace u by v in the computation because of a tie. Since v ± m, the 1 1 argument in the preceding paragraph then yields the result. Claim 2. If u is a state for G such that u 5 m, then pu Fs, G. 1 1 Proof. We modify the argument of Theorem 6.4. Suppose 1 5 sk. Then Fs ; G 5 1 1 Os ; 1; G [by Theorem 6.3]5Ms ; 1, s , s1 [for any state s ] 5 max hMs; 1, s 1 1 2 2 3 t, s2: s [s1] t j 5 Ms; 1, s , s2 [for some state s ] 5 Ms; 1, s , s3 3 k k [for some state s ] 5 . . . 5 Ms ; 1, s , sk [for some state s ]. We consider two k k cases. If k , n then Ms ; 1, s , sk 5 max hMs; 1, t, s j 1 1: s [1] t j [since 1 ˆ ˆ ˆ sk 5 1]. Among the possible t is some t for which t 5 1. For this t it follows that 1 ˆ ˆ Ms ; 1, t, sk 1 1 pv since Ms ; 1, t, sk 1 1 will be the payoff pw from a 1 state w with first component 1 [because player 1 never thereafter changes the strategy again], and by strong dominance pw pu. Hence the best t from the viewpoint of 1 player 1 must also yield that Ms ; 1, t, sk 1 1 pu, so that Fs ; G pu as 1 1 well. k k On the other hand, if k 5 n then Ms ; 1, s , sk 5 max h pt: s [1] t j [since 1 k ˆ ˆ sk 5 1]. Among the candidates t is some t for which t 5 1. Hence Ms; 1, s , 1 ˆ ˆ sk pt . By strong dominance we know pt pu, so that again Fs ; G 1 1 1 pu. This proves Claim 2. h Claim 3. Os ; G 5 Os ; G9 and is the payoff from a state in S9. S .J. Willson Mathematical Social Sciences 39 2000 323 –348 343 Proof. Suppose that Os ; G 5 pu for some state u not in S9; hence u 5 m. Suppose 1 that the transiency is t. Then pu 5 Os ; G 5 Os, t; G Os, t 2 1; G . . . all all Os, 1; G 5 Fs ; G pu by Lemma 5.2 and Claim 2. It follows that Fs ; all 1 G 5 pu. By Claim 1, however, Fs ; G 5 pv for some v in S9. Hence pu 5 pv, so it is also true that Os ; G is the payoff from a state v in S9. Hence in the play of the game G, no ultimate outcome needs to arise from a state s with s 5 m; at best the use of such a state merely ties the payoffs using a state from S9. 1 Clearly then player 1 can always make the same selection of strategies in the play of the game G9, avoiding strategy m, without compromising any payoff, so Os ; G 5 Os ; G9. The theorem follows from Claim 3. h Remark. The conclusion of the strong dominance axiom is false under the assumption of mere dominance, as may be seen as follows using Game 32. Suppose G is 2, 2 4, 1 1, 4 3, 3 Here the first row dominates the second but does not strongly dominate it. If we could ignore any dominated row, then we would have XG 5 XG9 for G9 with matrix 2, 2 4, 1. Clearly XG9 5 h2, 2j, while one easily sees that XG 5 h3, 3j. Remark. Much of the length and complexity of the proof of Theorem 6.8 is to deal with the possibility that some state u with u 5 m could have the same payoff as a state v 1 with v ± m. In the common circumstance when distinct states have distinct payoff 1 vectors, such ties cannot occur and the argument can be simplified dramatically. For example, Claim 2 could be then strengthened to assert that pu , Fs ; G when 1 u 5 m. 1 6.5. Default reduction n Theorem 6.9. Let G be the game with strategies S and payoff vectors p: S → R . For n each permutation s let Gs be the game with strategies S and payoff vectors q: S → R given by qs 5 ps if ps Fs ; G; all qs 5 Fs ; G otherwise. Then Os ; r 1 1, s; G 5 Os ; r, s; Gs for all s and all r 1; Ms ; r 1 1, s, s j ; G 5 Ms ; r, s, s j ; Gs for all s, all r 1, and all j; and Us ; r 1 1, s; G 5 Us ; r, s; Gs for all r 1 and all s. In particular, Os ; G 5 Os ; Gs. Proof. For all states s, we know that Os ; 1, s; G 5 Fs ; G by Theorem 6.5. Then Us ; 1, s; G 5 qs from the definition of U. Hence Ms ; 2, s, sn; G 5 max s n 344 S .J. Willson Mathematical Social Sciences 39 2000 323 –348 hUs; 1, t; G: s [sn] tj 5 max hqt: s [sn] tj 5 Ms; 1, s, sn; Gs. From this s n it follows that Ms ; 2, s, sn 2 1; G 5 max hMs; 2, t, sn; G: s [sn 2 1] s n 21 t j 5 max hMs; 1, t, sn; Gs: s [sn 2 1] tj 5 Ms; 1, s, sn 2 1; Gs. s n 21 Repeating the argument we find that for all j and for all s we have Ms ; 2, s, s j ; G 5 Ms ; 1, s, s j ; Gs. Hence Os ; 2, s; G 5 Os ; 1, s; Gs for all s. Now Us ; 2, s; G 5 ps if ps Os ; 2, s; G; otherwise Us ; 2, s; G 5 Os ; 2, all s; G. Similarly, Us ; 1, s; Gs 5 qs if qs Os ; 1, s; Gs; otherwise Us ; 1, s; all Gs 5 Os ; 1, s; Gs. Hence Us ; 2, s; G 5 Us ; 1, s; Gs unless either a ps Os ; 2, s; G, qs Os ; 1, s; Gs, but ps ± qs; or all all b ps Os ; 2, s; G; for some i, qs , Os ; 1, s; Gs, and ps ± Os ; 1, s; all i Gs; or c for some i, ps , Os ; 2, s; G; for some j, qs , Os ; 1, s; Gs, and Os ; 2, i j s; G ± Os ; 1, s; Gs; or d for some i, ps , Os ; 2, s; G; qs Os ; 1, s; Gs, and Os ; 2, s; i all G ± qs. Case c cannot occur because of the calculation above. From case a it would follow that ps Os ; 1, s; G 5 Fs ; G, so qs 5 ps by the definition of q, a all contradiction showing that case a cannot occur. From case b it would follow that ps Os ; 1, s; G 5 Fs ; G, so qs 5 ps; hence qs Os ; 2, s; G 5 Os ; all all 1, s; Gs, contradicting that for some i, qs , Os ; 1, s; Gs. From case d it would i follow that qs Os ; 1, s; Gs 5 Os ; 2, s; G, but Os ; 2, s; G ± qs; hence for all some j we have qs . Os ; 2, s; G Os ; 1, s; G 5 Fs ; G; hence qs ± Fs ; j all G, whence qs 5 ps and ps Fs ; G by the definition of q; now ps Os ; all all 1, s; Gs 5 Os ; 2, s; G contradicting that for some i, ps , Os ; 2, s; G. Hence i none of cases a, b, c, or d can occur, whence Us ; 2, s; G 5 Us ; 1, s; Gs for all s. We now repeat the paragraphs above with obvious modifications in order to yield an inductive proof of the result. h Corollary 6.10. Default reduction axiom. For each permutation s, Xs ; G 5 Xs ; Gs. Proof. Xs ; G 5 Os ; G while Xs ; Gs 5 Os ; Gs. h Corollary 6.10 completes the argument that all the axioms are satisfied by XG given by the rules of play, hence completes the proof of Theorem 6.1. h Proposition 6.11. The axioms other than default reduction completely determine the outcomes on all strict 2 by 2 games with no repeated payoff vectors. Proof. If the game G has a single strategy that is best for both players, then this outcome S .J. Willson Mathematical Social Sciences 39 2000 323 –348 345 is the only member of XG since it is the only Pareto-optimal outcome. Otherwise, the game is equivalent to one of the 57 games of strategy given in Brams 1994. For each one, we can readily verify that these axioms determine the outcome, as computed in the manner at the end of Section 2. At no stage is the default reduction axiom required. h

7. Example of the ‘play’ of a game