1. ordinal solution to bargaining problems An ordinal solution to bargaining problems with many players Z. Safra, D. Samet Click left mouse-button to proceed, or one of the links to see  Samet’s home page (web) Samet’s home page  a short PowerPoint tutorial (in this document)a short PowerPoint tutorial Click left mouse-button to proceed, or one of the links to see  Samet’s home page (web) Samet’s home page  a short PowerPoint tutorial (in this document)a short PowerPoint tutorial

2. A solution is a function  which assigns to each problem (a,S) a point  (a,S) in R N. Bargaining problems 3 a 2 1 S A bargaining problem for a set of players N is a pair (a,S) where a - a status quo point in R N, That is, if x,y  S, and x  y then x = y. This means that for each i the projection of S on R N\i is all of R N\i. (The picture shows only part of the infinite surface.) S - a Pareto surface in R N.

3. If U i is a utility function of player i, and μ i is an order-preserving transformation of R, then μ i (U i ) represents the same preferences of i. If U i is a utility function of player i, and μ i is an order-preserving transformation of R, then μ i (U i ) represents the same preferences of i. Ordinal transformations An ordinal transformation of the utility space R N is obtained by applying to each player’s utility a continuous, order-preserving transformation. Formally: An ordinal transformation of the utility space R N is obtained by applying to each player’s utility a continuous, order-preserving transformation. Formally:

4. Ordinal transformations A vector (μ i ) i  N of continuous, order-preserving functions from R onto R, defines an ordinal transformation μ : R N → R N, by μ(x) = (μ i (x i )) i  N. A vector (μ i ) i  N of continuous, order-preserving functions from R onto R, defines an ordinal transformation μ : R N → R N, by μ(x) = (μ i (x i )) i  N. The ordinal transformation μ transforms a bargaining problem (a,S) to the problem (μ(a), μ(S)), where μ(S) = { μ(x) | x  S }. The ordinal transformation μ transforms a bargaining problem (a,S) to the problem (μ(a), μ(S)), where μ(S) = { μ(x) | x  S }.

5. An ordinal solution S 3 a 2 1 1 μ(S)μ(S) 3 μ(a) 2. (a, S). (a, S).  (μ(a), μ(S)) A solution  is ordinal if it depends only on the preferences of the players and not on their representations by utility functions. Thus, the two problems, (a, S) and (μ(a), μ(S)) … …which represent the same problem in terms of preferences, must have solutions … that represent the same preferences. I.e., μ((a, S)) = (μ(a), μ(S))μ((a, S)) = (μ(a), μ(S)) μ((a, S)) = (μ(a), μ(S))μ((a, S)) = (μ(a), μ(S)) μ

6. ShapleyShapley constructed a solution , for problems with three players, which is,  ordinal,  efficient and feasible (that is, for each problem (a, S),  (a, S)  S),  symmetric (that is, covariant with permutation of players). ShapleyShapley constructed a solution , for problems with three players, which is,  ordinal,  efficient and feasible (that is, for each problem (a, S),  (a, S)  S),  symmetric (that is, covariant with permutation of players). He has also shownHe has also shown that for problems with two players there is no solution with these properties. He has also shownHe has also shown that for problems with two players there is no solution with these properties. We construct an ordinal, efficient, feasible, and symmetric solution to problems with any number of players greater than two, which extends Shapley’s solution. The basic building blocks for the construction are … We construct an ordinal, efficient, feasible, and symmetric solution to problems with any number of players greater than two, which extends Shapley’s solution. The basic building blocks for the construction are …

7. Pareto functions The function π i : R N → R i is i’s Pareto function. it is:  independent of x i x. S 3 a 2 1 (x 1, x 2, π 3 (x)). Starting from any point x… change i’s payoff until S is reached. i’s payoff at this point is denoted by π i (x). S is reached at one point at least because it is a surface, and at one point at most because it is Pareto. x’. = (x 1, x 2, π 3 (x’))

8. Pareto functions The function π i : R N → R i is i’s Pareto function. it is:  independent of x i  decreasing in x j  continuous x. S 3 a 2 1 (x 1, x 2, π 3 (x)). Starting from any point x… change i’s payoff until S is reached. i’s payoff at this point is denoted by π i (x).

9. Ideal points The ideal point of x is π(x) = ( π 1 (x), π 2 (x), π 3 (x) ) The ideal point of x is π(x) = ( π 1 (x), π 2 (x), π 3 (x) ) S.. 3 2 1... (π 1 (x), x 2, x 3 ) π(x) a x (x 1, x 2, π 3 (x)) (x 1, π 2 (x), x 3 ) Food for thought: Which point is closer to S, x or π(x)? The relation between x and π(x) is one of three, and it defines direction with respect to S. x < π(x), x is below S x = π(x), x is on S x > π(x), x is above S x < π(x), x is below S x = π(x), x is on S x > π(x), x is above S

10. Ideal points π(x). S. 3 2 1. a x.. Here is a point above S. The ideal point of x is π(x) = ( π 1 (x), π 2 (x), π 3 (x) ) The ideal point of x is π(x) = ( π 1 (x), π 2 (x), π 3 (x) ) x < π(x), x is below S x = π(x), x is on S x > π(x), x is above S x < π(x), x is below S x = π(x), x is on S x > π(x), x is above S

11. Ordinality of ideal points. S 3 a 2 1 y = (x 1, x 2, π 3 (x)).. μ(S) 3 μ(a) 2 1. μ(x) = (μ 1 (x 1 ), μ 2 (x 2 ), μ 3 (x 3 )) x = (x 1, x 2, x 3 ) Consider the problems (a, S) and (μ(a), μ(S))… Start with x and change i’s payoff until S is reached at y. Similarly, start with μ(x) and change i’s payoff until μ(S) is reached at z. z = (μ 1 (x 1 ), μ 2 (x 2 ), π 3 (μ(x)))

12. Ordinality of ideal points. S 3 a 2 1.. μ(S) 3 μ(a) 2 1. μ(x) = (μ 1 (x 1 ), μ 2 (x 2 ), μ 3 (x 3 )) μ(y) y = (x 1, x 2, π 3 (x)) z = (μ 1 (x 1 ), μ 2 (x 2 ), π 3 (μ(x))) The payoffs of all players other than i are the same in μ(y) and in z. By definition both μ(y) and z are on the Pareto surface μ(S). μ(y) and z coincide also for i, that is, μ i (π i (x)) = π i (μ(x))) x = (x 1, x 2, x 3 ) μ(y) = (μ 1 (x 1 ), μ 2 (x 2 ), μ 3 (π 3 (x))) z = (μ 1 (x 1 ), μ 2 (x 2 ), π 3 (μ(x)))

13. Ordinality of ideal points For each player i, μ i (π i (x)) = π i (μ(x)), and hence, μ(π(x)) = π(μ(x)). For each player i, μ i (π i (x)) = π i (μ(x)), and hence, μ(π(x)) = π(μ(x)).

14. π(x) = a means that for each i, π i (x) = a i. That is, for each i, (x -i, a i )  S. A source for an ideal point 3 2 1 a.... x = (x 1, x 2, a 3 ) = ( a 1, x 2, x 3 ) = (x 1, a 2, x 3 ) Is the point a the ideal point of some x? That is, is there x for which π(x) = a? (x 1, x 2, π 3 (x)) (π 1 (x), x 2, x 3 ) (x 1, π 2 (x), x 3 ) Such a point x is illustrated now. π(x) = Claim (Shapley): For three players, there exists a unique point x such that π(x) = a. Claim (Shapley): For three players, there exists a unique point x such that π(x) = a.

15. A source for an ideal point: n=3 3 2 1 a... (x 1, x 2, a 3 ) (a 1, x 2, x 3 ) (x 1, a 2, x 3 ). (a 1, x 2, a 3 ) Consider the three points: (x 1, x 2, a 3 ), (x 1, a 2, x 3 ), (a 1, x 2, x 3 ) projected on S by the required point x. Starting from (x 1, x 2, a 3 ) we reach (a 1, x 2, x 3 ) as follows. There are several ways to prove the claim. The “trick” is to find a proof that can be extended to more than three players. First, reduce 1’s payoff to status quo a 1 … Then, increase 3’s payoff to π 3 (a 1, x 2, a 3 ) = x 3. We call π 3 (a 1, x 2, a 3 ), 3’s gains over 1 at (x 1, x 2, a 3 ).

16. A source for an ideal point: n=3 3 2 1 a... (x 1, x 2, a 3 ) (a 1, x 2, x 3 ) (x 1, a 2, x 3 ) (x 1, a 2, a 3 ).. (a 1, x 2, a 3 ) Consider the three points: (x 1, x 2, a 3 ), (x 1, a 2, x 3 ), (a 1, x 2, x 3 ) projected on S by the required point x. Similarly, from (x 1, x 2, a 3 ) we reach (x 1, a 2, x 3 ) as follows. First, reduce 2’s payoff to status quo a 2 … Then, increase 3’s payoff to π 3 (x 1, a 2, a 3 ) = x 3. Similarly, π 3 (x 1, a 2, a 3 ) is 3’s gains over 2 at (x 1, x 2, a 3 ).

17. A source for an ideal point: n=3 3 2 1 a... (x 1, x 2, a 3 ) (a 1, x 2, x 3 ) (x 1, a 2, x 3 ) (x 1, a 2, a 3 ).. (a 1, x 2, a 3 ) Consider the three points: (x 1, x 2, a 3 ), (x 1, a 2, x 3 ), (a 1, x 2, x 3 ) projected on S by the required point x.. π 3 (x 1, a 2, a 3 ) = π 3 (a 1, x 2, a 3 ) = x 3 Note that 3’s gains over 1 and 2 are the same. That is π 3 (x 1, a 2, a 3 ) = π 3 (a 1, x 2, a 3 ) Note that 3’s gains over 1 and 2 are the same. That is π 3 (x 1, a 2, a 3 ) = π 3 (a 1, x 2, a 3 ) This gives rise to the following two conditions which are equivalent to π(x) = a.

18. A source for an ideal point: n=3 3 2 1 a..... (x 1, x 2, a 3 )  S (x 1, a 2, x 3 )  S (a 1, x 2, x 3 )  S (x 1, x 2, a 3 )  S π 3 (x 1, a 2, a 3 ) = x 3 π 3 (a 1, x 2, a 3 ) = x 3 (x 1, a 2, a 3 ) (a 1, x 2, a 3 ) π(x) = a Thus, to prove the claim we need to show the existence of a unique (x 1, x 2, a 3 )  S such that π 3 (x 1, a 2, a 3 ) = π 3 (a 1, x 2, a 3 ) Thus, to prove the claim we need to show the existence of a unique (x 1, x 2, a 3 )  S such that π 3 (x 1, a 2, a 3 ) = π 3 (a 1, x 2, a 3 ) The last equivalence follows immediately from the definition of the Pareto functions π i. π 3 (x 1, a 2, a 3 ) = π 3 (a 1, x 2, a 3 ) = x 3 (x 1, x 2, a 3 ) (a 1, x 2, x 3 ) (x 1, a 2, x 3 ). This condition says that there exists a point (x 1, x 2, a 3 ) in S at which 3’s gains over 1 and 2 are the same.

19. π(x) = a - existence: n=3 3 2 1 a Reminder: Looking for (x 1, x 2, a 3 )  S such that π 3 (x 1, a 2, a 3 ) = π 3 (a 1, x 2, a 3 ) Consider the set D = {(x 1, x 2, a 3 )  S} and the functions on D, f 1 (x 1, x 2, a 3 ) = π 3 (a 1, x 2, a 3 ) f 2 (x 1, x 2, a 3 ) = π 3 (x 1, a 2, a 3 ) D (x 1, x 2, a 3 ).... (a 1, x 2, a 3 )... (x 1, a 2, a 3 ) Computing f 1 at (x 1, x 2, a 3 ) … Computing f 2 at the same point … f 1 (x 1, x 2, a 3 ) = π 3 (a 1, x 2, a 3 ) f 2 (x 1, x 2, a 3 ) = π 3 (x 1, a 2, a 3 )

20. π(x) = a - existence: n=3 3 2 1 a Reminder: Looking for (x 1, x 2, a 3 )  S such that π 3 (x 1, a 2, a 3 ) = π 3 (a 1, x 2, a 3 ) Consider the set D = {(x 1, x 2, a 3 )  S} and the functions on D, f 1 (x 1, x 2, a 3 ) = π 3 (a 1, x 2, a 3 ) f 2 (x 1, x 2, a 3 ) = π 3 (x 1, a 2, a 3 ) D (x 1, x 2, a 3 ).... (a 1, x 2, a 3 )... (x 1, a 2, a 3 ) f 1 (x 1, x 2, a 3 ) = π 3 (a 1, x 2, a 3 ) f 2 (x 1, x 2, a 3 ) = π 3 (x 1, a 2, a 3 ) First edge… Compare f 1 and f 2 at the edges of D.. (a 1, x 2, a 3 ) f 1 < f 2 f 2 (a 1, x 2, a 3 ) = π 3 (a 1, a 2, a 3 ) = π 3 (a) f 1 (a 1, x 2, a 3 ) = π 3 (a 1, x 2, a 3 ) = a 3 Because, (a 1, x 2, a 3 ) is on S π 3 (a )

21. Second edge… π(x) = a - existence: n=3 3 2 1 a Reminder: Looking for (x 1, x 2, a 3 )  S such that π 3 (x 1, a 2, a 3 ) = π 3 (a 1, x 2, a 3 ) Consider the set D = {(x 1, x 2, a 3 )  S} and the functions on D, f 1 (x 1, x 2, a 3 ) = π 3 (a 1, x 2, a 3 ) f 2 (x 1, x 2, a 3 ) = π 3 (x 1, a 2, a 3 ) D (x 1, x 2, a 3 ).... (a 1, x 2, a 3 )... (x 1, a 2, a 3 ) f 1 (x 1, x 2, a 3 ) = π 3 (a 1, x 2, a 3 ) f 2 (x 1, x 2, a 3 ) = π 3 (x 1, a 2, a 3 ) Compare f 1 and f 2 at the edges of D.. (a 1, x 2, a 3 ) f 1 < f 2 π 3 (a ). f 2 (x 1, a 2, a 3 ) = π 3 (x 1, a 2, a 3 ) = a 3 (x 1, a 2, a 3 ) f1 > f2f1 > f2 f 1 (x 1, a 2, a 3 ) = π 3 (a 1, a 2, a 3 ) = π 3 (a) Because, (x 1, a 2, a 3 ) is on S

22. π(x) = a - existence: n=3 3 2 1 a Reminder: Looking for (x 1, x 2, a 3 )  S such that π 3 (x 1, a 2, a 3 ) = π 3 (a 1, x 2, a 3 ) Consider the set D = {(x 1, x 2, a 3 )  S} and the functions on D, f 1 (x 1, x 2, a 3 ) = π 3 (a 1, x 2, a 3 ) f 2 (x 1, x 2, a 3 ) = π 3 (x 1, a 2, a 3 ) D (x 1, x 2, a 3 ).... (a 1, x 2, a 3 )... (x 1, a 2, a 3 ) f 1 (x 1, x 2, a 3 ) = π 3 (a 1, x 2, a 3 ) f 2 (x 1, x 2, a 3 ) = π 3 (x 1, a 2, a 3 ) Compare f 1 and f 2 at the edges of D.. (a 1, x 2, a 3 ) f 1 < f 2 π 3 (a ). (x 1, a 2, a 3 ) f1 > f2f1 > f2 The order of f 1 and f 2 is different at the two edges of the line segment D. Therefore they coincide at some point. This proves the existence of x that solves π(x) = a, for three players. Later, we generalize this proof for more than three players. We turn now to prove the uniqueness of x which is special for the case n = 3. This proves the existence of x that solves π(x) = a, for three players. Later, we generalize this proof for more than three players. We turn now to prove the uniqueness of x which is special for the case n = 3.

23. π(x) = a - uniqueness: n=3 3 2 1 a.. (x 1, x 2, a 3 ) f 1 (x 1, x 2, a 3 ) =π 3 (a 1, x 2, a 3 ). (a 1, x 2, a 3 ). Reminder: Looking for (x 1, x 2, a 3 )  S such that π 3 (x 1, a 2, a 3 ) = π 3 (a 1, x 2, a 3 ) f 1 is increasing with x 1 Consider the set D = {(x 1, x 2, a 3 )  S} and the functions on D, f 1 (x 1, x 2, a 3 ) = π 3 (a 1, x 2, a 3 ) f 2 (x 1, x 2, a 3 ) = π 3 (x 1, a 2, a 3 ) D

24. Consider the set D = {(x 1, x 2, a 3 )  S} and the functions on D, f 1 (x 1, x 2, a 3 ) = π 3 (a 1, x 2, a 3 ) f 2 (x 1, x 2, a 3 ) = π 3 (x 1, a 2, a 3 ) π(x) = a - uniqueness: n=3 3 2 1 a. (x 1, x 2, a 3 ) f 1 (x 1, x 2, a 3 ) =π 3 (a 1, x 2, a 3 ). (a 1, x 2, a 3 ).... Reminder: Looking for (x 1, x 2, a 3 )  S such that π 3 (x 1, a 2, a 3 ) = π 3 (a 1, x 2, a 3 ) f 1 is increasing with x 1. D.

25. π(x) = a - uniqueness: n=3 3 2 1 a. (x 1, a 2, a 3 ) (x 1, x 2, a 3 ).. f 2 (x 1, x 2, a 3 ) = π 3 (x 1, a 2, a 3 ) D Reminder: Looking for (x 1, x 2, a 3 )  S such that π 3 (x 1, a 2, a 3 ) = π 3 (a 1, x 2, a 3 ) f 2 is decreasing with x 1. Consider the set D = {(x 1, x 2, a 3 )  S} and the functions on D, f 1 (x 1, x 2, a 3 ) = π 3 (a 1, x 2, a 3 ) f 2 (x 1, x 2, a 3 ) = π 3 (x 1, a 2, a 3 ) f 1 is increasing with x 1

26. π(x) = a - uniqueness: n=3 3 2 1 a. (x 1, a 2, a 3 ) (x 1, x 2, a 3 ). f 2 (x 1, x 2, a 3 ) = π 3 (x 1, a 2, a 3 ).. Reminder: Looking for (x 1, x 2, a 3 )  S such that π 3 (x 1, a 2, a 3 ) = π 3 (a 1, x 2, a 3 ).. D.. Consider the set D = {(x 1, x 2, a 3 )  S} and the functions on D, f 1 (x 1, x 2, a 3 ) = π 3 (a 1, x 2, a 3 ) f 2 (x 1, x 2, a 3 ) = π 3 (x 1, a 2, a 3 ) Therefore, f 1 and f 2 coincide in D only once. f 2 is decreasing with x 1 f 1 is increasing with x 1

27. Constructing the solution We define an auxiliary solution  : 3 2 1 a.... x =  (a, S)  (a, S) is the unique point x for which π(x) = a.

28. Constructing the solution We define an auxiliary solution  : 3 2 1 a. x =  (a, S) Obviously  is symmetric. It is ordinal because π(μ(x)) = μ(π(x)) = μ(a), and therefore μ(x) =  (μ(a), μ(S)). But  is not the desired solution because  (a, S) is not on S.  (a, S) is the unique point x for which π(x) = a.

29. . 3 2 1 a. x =  (a, S)... y =  (x, S) Constructing the solution We define an auxiliary solution  :  (a, S) is the unique point x for which π(x) = a. Now, change the status quo point to x, and let y =  (x, S) (a, S)  y is an ordinal, symmetric solution of the original problem (a, S). Although y is not on S, it is closer than x to S.

30. a 0 = a Constructing the solution This suggests a recursive construction. Starting with the problem (a, S) generate the sequence a 1 =  (a 0, S) a 2 =  (a 1, S) a 3 =  (a 2, S) a 4 =  (a 3, S) a 5 =  (a 4, S) a 6 =  (a 5, S) a 7 =  (a 6, S) a 8 =  (a 7, S)... The sequence (a k ) converges to a point x on S. The solution defined by  (a,S) = x is an ordinal, efficient, feasible and symmetric solution. The sequence (a k ) converges to a point x on S. The solution defined by  (a,S) = x is an ordinal, efficient, feasible and symmetric solution.

31. We prove the claim for four players. The proof for more players is similar. Claim: For any number of players, there exists a point x such that π(x) = a. Claim: For any number of players, there exists a point x such that π(x) = a. Yes, even for two players. In this case x = π(a) is the unique point that satisfies π(x) = a. For more than three players there can be more than one point. (Sprumont, 2000)Sprumont The construction of a solution for n>3 starts also with points x whose ideal point is a. We claim: The construction of a solution for n>3 starts also with points x whose ideal point is a. We claim:

32. π(x) = a : four players (x 1, x 2, x 3, a 4 )  S (x 1, x 2, a 3, x 4 )  S (x 1, a 2, x 3, x 4 )  S (a 1, x 2, x 3, x 4 )  S (x 1, x 2, x 3, a 4 )  S π 4 (x 1, x 2, a 3, a 4 ) = x 4 π 4 (x 1, a 2, x 3, a 4 ) = x 4 π 4 (a 1, x 2, x 3, a 4 ) = x 4 The following equivalences hold by the definition of the Pareto functions π i. π(x) = a Like the case of three players we prove the existence of (x 1, x 2, x 3, a 4 )  S such that π 4 (x 1, x 2, a 3, a 4 ) = π 4 (x 1, a 2, x 3, a 4 ) = π 4 (a 1, x 2, x 3, a 4 ) Like the case of three players we prove the existence of (x 1, x 2, x 3, a 4 )  S such that π 4 (x 1, x 2, a 3, a 4 ) = π 4 (x 1, a 2, x 3, a 4 ) = π 4 (a 1, x 2, x 3, a 4 ) This conditions says that there exists a point (x 1, x 2, x 3, a 4 ) in S such that 4’s gains over 1, 2, and 3 coincide at this point.

33. π(x) = a : four players Reminder: Looking for (x 1, x 2, x 3, a 4 )  S such that π 4 (x 1, x 2, a 3, a 4 ) = π 4 (x 1, a 2, x 3, a 4 ) = π 4 (a 1, x 2, x 3, a 4 ). Let D = {(x 1, x 2, x 3, a 4 )  S} D 1 = {(a 1, x 2, x 3, a 4 )  S} D 2 = {(x 1, a 2, x 3, a 4 )  S} D 3 = {(x 1, x 2, a 3, a 4 )  S} D = {(x 1, x 2, x 3, a 4 )  S} D 1 = {(a 1, x 2, x 3, a 4 )  S} D 2 = {(x 1, a 2, x 3, a 4 )  S} D 3 = {(x 1, x 2, a 3, a 4 )  S} D is homeomorphic to the three dimensional simplex. D 1, D 2, and D 3 are homeomorphic to the sides of this simplex.

34. π(x) = a : four players Reminder: Looking for (x 1, x 2, x 3, a 4 )  S such that π 4 (x 1, x 2, a 3, a 4 ) = π 4 (x 1, a 2, x 3, a 4 ) = π 4 (a 1, x 2, x 3, a 4 ). We need to show that the three functions coincide at some point in D. D = {(x 1, x 2, x 3, a 4 )  S} D 1 = {(a 1, x 2, x 3, a 4 )  S} D 2 = {(x 1, a 2, x 3, a 4 )  S} D 3 = {(x 1, x 2, a 3, a 4 )  S} Define three functions on D: f 1 (x 1, x 2, x 3, a 4 ) = π 4 (a 1, x 2, x 3, a 4 ) f 2 (x 1, x 2, x 3, a 4 ) = π 4 (x 1, a 2, x 3, a 4 ) f 3 (x 1, x 2, x 3, a 4 ) = π 4 (x 1, x 2, a 3, a 4 )

35. π(x) = a : four players Reminder: Looking for (x 1, x 2, x 3, a 4 )  S such that π 4 (x 1, x 2, a 3, a 4 ) = π 4 (x 1, a 2, x 3, a 4 ) = π 4 (a 1, x 2, x 3, a 4 ). Let A 1 = { p  D | f 1 (p) ≤ f 2 (p), f 3 (p) } Define three functions on D: f 1 (x 1, x 2, x 3, a 4 ) = π 4 (a 1, x 2, x 3, a 4 ) f 2 (x 1, x 2, x 3, a 4 ) = π 4 (x 1, a 2, x 3, a 4 ) f 3 (x 1, x 2, x 3, a 4 ) = π 4 (x 1, x 2, a 3, a 4 ) Then D 1  A 1. Indeed, D = {(x 1, x 2, x 3, a 4 )  S} D 1 = {(a 1, x 2, x 3, a 4 )  S} D 2 = {(x 1, a 2, x 3, a 4 )  S} D 3 = {(x 1, x 2, a 3, a 4 )  S} f 1 (a 1, x 2, x 3, a 4 )π 4 (a 1, x 2, x 3, a 4 )= ≤π 4 (a 1, a 2, x 3, a 4 ) f 2 (a 1, x 2, x 3, a 4 )= A1A1 f 1 at a point in D 1 by definition of f 1 π 4 is decreasing by definition of f 2

36. Let A 1 = { p  D | f 1 (p) ≤ f 2 (p), f 3 (p) } π(x) = a : four players Reminder: Looking for (x 1, x 2, x 3, a 4 )  S such that π 4 (x 1, x 2, a 3, a 4 ) = π 4 (x 1, a 2, x 3, a 4 ) = π 4 (a 1, x 2, x 3, a 4 ). Define three functions on D: f 1 (x 1, x 2, x 3, a 4 ) = π 4 (a 1, x 2, x 3, a 4 ) f 2 (x 1, x 2, x 3, a 4 ) = π 4 (x 1, a 2, x 3, a 4 ) f 3 (x 1, x 2, x 3, a 4 ) = π 4 (x 1, x 2, a 3, a 4 ) D = {(x 1, x 2, x 3, a 4 )  S} D 1 = {(a 1, x 2, x 3, a 4 )  S} D 2 = {(x 1, a 2, x 3, a 4 )  S} D 3 = {(x 1, x 2, a 3, a 4 )  S} A 2 = { p  D | f 2 (p) ≤ f 1 (p), f 3 (p) } A 3 = { p  D | f 3 (p) ≤ f 1 (p), f 2 (p) } For a point p in the intersection of A 1, A 2, A 3 f 1 (p) = f 2 (p) = f 3 (p) A1A1 A2A2 A3A3 A 1, A 2, A 3 are closed (continuity of π 4 ) They cover D ( by their definition ) D i  A i for i=1, 2, 3 By the AP Lemma the intersection of the sets is not empty.AP Lemma A 1, A 2, A 3 are closed (continuity of π 4 ) They cover D ( by their definition ) D i  A i for i=1, 2, 3 By the AP Lemma the intersection of the sets is not empty.AP Lemma

37. Constructing the solution For n = 3, the auxiliary solution  (a, S) is the unique point x such that π(x) = a. For n > 3 we may have more than one such point. This problem is solved by defining for each player i and problem (a, S) a is below S  i (a, S) = a is on S a is above S a is below S  i (a, S) = a is on S a is above S min { x i | π(x) = a } max { x i | π(x) = a } aiai

38. Constructing the solution The auxiliary solution  thus defined is ordinal since π, the min, and the max functions are ordinal (i.e. commutes with order preserving transformations).  is obviously symmetric. However,  (a, S) is not necessarily on S. This problems is solved as in the case n=3 … It is clear, that in case there exists a unique x for which π(x) = a, then  (a, S) is this x. Thus for n=3,  is the auxiliary solution defined above. a is below S  i (a, S) = a is on S a is above S a is below S  i (a, S) = a is on S a is above S min { x i | π(x) = a } max { x i | π(x) = a } aiai

39. a 0 = a Constructing the solution Starting with the problem (a, S) generate the sequence a 1 =  (a 0, S) a 2 =  (a 1, S) a 3 =  (a 2, S) a 4 =  (a 3, S) a 5 =  (a 4, S) a 6 =  (a 5, S) a 7 =  (a 6, S) a 8 =  (a 7, S)... The sequence (a k ) converges to a point x on S. The solution defined by  (a,S) = x is an ordinal, efficient, feasible and symmetric solution. The sequence (a k ) converges to a point x on S. The solution defined by  (a,S) = x is an ordinal, efficient, feasible and symmetric solution.

40. More solutions … There are many solutions for bargaining problems that are ordinal, efficient, feasible, and symmetric. “Ordinal Solutions to Bargaining Problems” (Samet and Safra, 2001) presents an infinite family of such solutions for any number of players greater than two.Ordinal Solutions to Bargaining Problems Solutions in this family are based on ordinal solutions to agendas as proposed in “Bargaining with an Agenda” (O’Neill, Samet, Wiener and Winter, 2000).Bargaining with an Agenda This family includes also a solution that extends Shapley’s solution to n>3. This extension is different than the one presented here.

41. Alexandroff, P., and B. Pasynkoff (1957), Elementary Proof of the Essentiality of the Identical Mapping of a Simplex, Uspekhi Mat. Nauk 12, 175– 179 (in Russian). van der Laan, G., D. Talman, Z. Yang (1999), Intersection theorems on polytopes, Math. Program. 84, 25– 38. (The paper in pdf - Math. Program. site)The paper in pdf - Math. Program. site Alexandroff, P., and B. Pasynkoff (1957), Elementary Proof of the Essentiality of the Identical Mapping of a Simplex, Uspekhi Mat. Nauk 12, 175– 179 (in Russian). van der Laan, G., D. Talman, Z. Yang (1999), Intersection theorems on polytopes, Math. Program. 84, 25– 38. (The paper in pdf - Math. Program. site)The paper in pdf - Math. Program. site back The AP Lemma

42. Shapley’s solution Shubik, M., (1982), Game Theory in the Social Sciences: Concepts and Solutions, MIT, Press, Cambridge. Thomson, W., (1994), Cooperative models of bargaining, In Handbook of Game Theory 2, Aumann R. J., and Hart S.Eds, North Holland, p.1237-1248. Shubik, M., (1982), Game Theory in the Social Sciences: Concepts and Solutions, MIT, Press, Cambridge. Thomson, W., (1994), Cooperative models of bargaining, In Handbook of Game Theory 2, Aumann R. J., and Hart S.Eds, North Holland, p.1237-1248. back

43. No ordinal solution for n=2 Shapley, L., (1969), Utility Comparison and the Theory of Games. In La Décision: Agrégation et dynamique des orders de préférence, G. Th. Guilbaud ed., Editions du CNRS, Paris. Reprinted in, The Shapley value, Essays in honor of Lloyd S. Shapley, A. Roth ed., Cambridge University Press, 1988 Shapley, L., (1969), Utility Comparison and the Theory of Games. In La Décision: Agrégation et dynamique des orders de préférence, G. Th. Guilbaud ed., Editions du CNRS, Paris. Reprinted in, The Shapley value, Essays in honor of Lloyd S. Shapley, A. Roth ed., Cambridge University Press, 1988 back

44. Multiple solutions to π(x) = a Sprumont, Y., (2000). A note on ordinally equivalent Pareto surfaces, Journal of Mathematical Economics, 34, 27-38. (paper in pdf - JME site)paper in pdf - JME site Sprumont, Y., (2000). A note on ordinally equivalent Pareto surfaces, Journal of Mathematical Economics, 34, 27-38. (paper in pdf - JME site)paper in pdf - JME site back

45. How to run a PowerPoint presentation back Advance to next slide Go to previous slide Go to one of the slides Exit show Underlined words are links to the web or to other slides. When the presentation is entered through a browser, it is recommended to choose “full screen” from the right click menu. - left click, space bar, Enter - P, Backspace - right click and choose from the menu “go” and then “slide navigator” (or “by title”) - Esc