1 / 40

An ordinal solution to bargaining problems with many players Z. Safra , D. Samet

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) a short PowerPoint tutorial (in this document). Bargaining problems. 3. S.

dash
Download Presentation

An ordinal solution to bargaining problems with many players Z. Safra , D. Samet

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. An ordinal solutionto bargaining problemswith many playersZ. Safra, D. Samet • Click left mouse-button to proceed, or one of the links to see • Samet’s home page(web) • a short PowerPoint tutorial(in this document)

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

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

  4. Ordinal transformations A vector (μi)i Nof continuous, order-preservingfunctions from R onto R, defines an ordinal transformation μ : RN →RN, by μ(x)= (μi(xi))i N. The ordinal transformation μ transformsa bargaining problem (a,S) to the problem(μ(a), μ(S)), where μ(S) = { μ(x) | x  S }.

  5. An ordinal solution 3 S 3 μ(S) μ . (μ(a),μ(S)) .(a,S) 2 2 μ(a) 1 a 1 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))

  6. We construct an ordinal, efficient, feasible, and symmetric solution to problems with any number of players greater thantwo, which extends Shapley’s solution. The basic building blocks for the construction are … • Shapley 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 shown that for problems with two players there is no solution with these properties.

  7. Pareto functions 3 S 2 a 1 Starting from any point x… changei’s payoff until S is reached. i’s payoff at this point is denoted by πi(x). . (x1, x2, π3(x)) = (x1, x2, π3(x’)) The function πi: RN → Ri is i’s Pareto function. S is reached at one pointatleast because it is a surface, and at one pointat most because it is Pareto. . . x’ itis: x • independent of xi

  8. Pareto functions 3 S 2 a 1 Starting from any point x… changei’s payoff until S is reached. i’s payoff at this point is denoted by πi(x). . (x1, x2, π3(x)) The function πi: RN → Ri is i’s Pareto function. . itis: x • independent of xi • decreasing in xj • continuous

  9. Ideal points Food for thought:Which point is closer to S, x or π(x)? . Theideal point of x is π(x) = (π1(x), π2(x), π3(x)) 3 S The relation between x and π(x) is one of three, and it defines direction with respect to S. . π(x) (x1, x2, π3(x)) . 2 (x1, π2(x), x3) . . x • x < π(x), x is belowS • x = π(x), x is onS • x > π(x), x is aboveS a (π1(x), x2, x3) 1

  10. Ideal points . . . Theideal point of x is π(x) = (π1(x), π2(x), π3(x)) 3 S . x Here is a point above S. . 2 π(x) • x < π(x), x is belowS • x = π(x), x is onS • x > π(x), x is aboveS a 1

  11. Ordinality of ideal points 3 S 3 μ(S) 2 2 a μ(a) 1 1 Start withx and changei’s payoff untilSis reached aty. Consider the problems (a,S) and (μ(a),μ(S))… Similarly, start withμ(x) and changei’s payoff untilμ(S)is reached atz. . y = (x1, x2, π3(x)) . z = (μ1(x1), μ2(x2), π3(μ(x))) . . x = (x1, x2, x3) μ(x) = (μ1(x1), μ2(x2), μ3(x3))

  12. Ordinality of ideal points 3 S μ(y) 3 μ(S) 2 2 a μ(a) 1 1 The payoffs of all players other thaniare the same inμ(y) and in z. μ(y) and z coincide also for i, that is, μi(πi(x)) = πi(μ(x))) By definition bothμ(y)andzareon the Pareto surfaceμ(S). . y = (x1, x2, π3(x)) . z = (μ1(x1), μ2(x2), π3(μ(x))) . μ(y) = (μ1(x1), μ2(x2), μ3(π3(x))) z = (μ1(x1), μ2(x2), π3(μ(x))) . x = (x1, x2, x3) μ(x) = (μ1(x1), μ2(x2), μ3(x3))

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

  14. A source for an ideal point π(x) = Is the point a the ideal point of some x? That is, is there x for which π(x) =a? 3 Claim (Shapley): For three players, there exists a unique point x such that π(x) =a. π(x) =ameans that for eachi, πi(x) =ai. That is, for eachi, (x-i , ai) S. (π1(x), x2, x3) . = ( a1, x2, x3) . x . 2 . (x1, π2(x), x3) = (x1, a2, x3) (x1, x2, π3(x)) a = (x1, x2, a3) Such a pointxis illustrated now. 1

  15. A source for an ideal point: n=3 . (a1, x2, x3) . . (x1, a2, x3) (x1, x2, a3) Consider the three points: (x1, x2, a3), (x1, a2, x3), (a1, x2, x3) projected on S by the required point x. There are several ways to prove the claim. The “trick” is to find a proof that can be extended to more than three players. 3 Starting from (x1, x2, a3) we reach (a1, x2, x3) as follows. First, reduce1’s payoff to status quo a1… 2 . (a1, x2, a3) Then, increase 3’s payoff to π3(a1, x2, a3) = x3. a We call π3(a1, x2, a3), 3’s gains over 1 at (x1, x2, a3). 1

  16. A source for an ideal point: n=3 . (a1, x2, x3) . . (x1, a2, x3) (x1, x2, a3) Consider the three points: (x1, x2, a3), (x1, a2, x3), (a1, x2, x3) projected on S by the required point x. 3 Similarly, from (x1, x2, a3) we reach (x1, a2, x3) as follows. First, reduce2’s payoff to status quo a2… 2 . (a1, x2, a3) Then, increase 3’s payoff to π3(x1, a2, a3) = x3. . a Similarly, π3(x1, a2, a3) is3’s gains over 2 at (x1, x2, a3). (x1, a2, a3) 1

  17. A source for an ideal point: n=3 . (a1, x2, x3) . . (x1, a2, x3) (x1, x2, a3) Consider the three points: (x1, x2, a3), (x1, a2, x3), (a1, x2, x3) projected on S by the required point x. 3 π3(x1, a2, a3) = π3(a1, x2, a3) = x3 Note that 3’s gains over 1 and 2 are the same. That is π3(x1, a2, a3) =π3(a1, x2, a3) . 2 . (a1, x2, a3) . a This gives rise to the following two conditions which are equivalenttoπ(x) = a . (x1, a2, a3) 1

  18. A source for an ideal point: n=3 (a1, x2, x3) (x1, a2, x3) (x1, x2, a3) (x1, x2, a3) S π3(x1, a2, a3) =x3 π3(a1, x2, a3) =x3 (x1, x2, a3) S (x1, a2, x3) S (a1, x2, x3) S 3 π(x) = a π3(x1, a2, a3) = π3(a1, x2, a3) = x3 The last equivalence follows immediately from the definition of the Pareto functionsπi . . . . 2 . Thus, to prove the claim we need to show the existence of a unique (x1, x2, a3) S such that π3(x1, a2, a3) = π3(a1, x2, a3) (a1, x2, a3) . This condition says that there exists a point (x1, x2, a3) in S at which 3’s gains over 1 and 2 are the same. . a (x1, a2, a3) 1

  19. π(x) =a- existence: n=3 Reminder: Looking for (x1, x2, a3) S such that π3(x1, a2, a3) = π3(a1, x2, a3) Consider the set D = {(x1, x2, a3) S} 3 and the functions on D,f1(x1, x2, a3) = π3(a1, x2, a3) f2(x1, x2, a3) = π3(x1, a2, a3) . f1(x1, x2, a3) = π3(a1, x2, a3) . f2(x1, x2, a3)= π3(x1, a2, a3) 2 . Computing f1 at (x1, x2, a3) … . . (a1, x2, a3) . (x1, x2, a3) a Computing f2 at the same point … . D (x1, a2, a3) 1

  20. π(x) =a- existence: n=3 Reminder: Looking for (x1, x2, a3) S such that π3(x1, a2, a3) = π3(a1, x2, a3) Consider the set D = {(x1, x2, a3) S} 3 π3(a ) and the functions on D,f1(x1, x2, a3) = π3(a1, x2, a3) f2(x1, x2, a3) = π3(x1, a2, a3) Because, (a1, x2, a3) is on S . f1(x1, x2, a3) = π3(a1, x2, a3) . f1< f2 f2(x1, x2, a3)= π3(x1, a2, a3) Compare f1 and f2 at the edges of D. 2 . (a1, x2, a3 ) . . . f1(a1, x2, a3) = π3(a1, x2, a3) (a1, x2, a3) . (x1, x2, a3) = a3 a . First edge… D f2(a1, x2, a3) = π3(a1, a2, a3) = π3(a) (x1, a2, a3) 1

  21. π(x) =a- existence: n=3 Reminder: Looking for (x1, x2, a3) S such that π3(x1, a2, a3) = π3(a1, x2, a3) Consider the set D = {(x1, x2, a3) S} 3 π3(a ) and the functions on D,f1(x1, x2, a3) = π3(a1, x2, a3) f2(x1, x2, a3) = π3(x1, a2, a3) Because, (x1, a2, a3) is on S . f1(x1, x2, a3) = π3(a1, x2, a3) . f1< f2 f2(x1, x2, a3)= π3(x1, a2, a3) Compare f1 and f2 at the edges of D. 2 . (a1, x2, a3 ) . . . f1(x1, a2, a3) = π3(a1, a2, a3) = π3(a) (a1, x2, a3) . (x1, x2, a3) a . Second edge… D . f2(x1, a2, a3) = π3(x1, a2, a3) = a3 (x1, a2, a3) (x1, a2, a3 ) 1 f1>f2

  22. π(x) =a- existence: n=3 Reminder: Looking for (x1, x2, a3) S such that π3(x1, a2, a3) = π3(a1, x2, a3) Consider the set D = {(x1, x2, a3) S} 3 π3(a ) and the functions on D,f1(x1, x2, a3) = π3(a1, x2, a3) f2(x1, x2, a3) = π3(x1, a2, a3) 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. . f1(x1, x2, a3) = π3(a1, x2, a3) . f1< f2 f2(x1, x2, a3)= π3(x1, a2, a3) Compare f1 and f2 at the edges of D. 2 . (a1, x2, a3 ) . . . The order of f1andf2is different at the two edges of the line segment D. Therefore they coincide at some point. (a1, x2, a3) . (x1, x2, a3) a . D . (x1, a2, a3) (x1, a2, a3 ) 1 f1>f2

  23. π(x) =a- uniqueness: n=3 Reminder: Looking for (x1, x2, a3) S such that π3(x1, a2, a3) = π3(a1, x2, a3) Consider the set D = {(x1, x2, a3) S} 3 and the functions on D,f1(x1, x2, a3) = π3(a1, x2, a3) f2(x1, x2, a3) = π3(x1, a2, a3) . 2 f1(x1, x2, a3) =π3(a1, x2, a3) . . . (a1, x2, a3) (x1, x2, a3) a f1 is increasing with x1 D 1

  24. π(x) =a- uniqueness: n=3 Reminder: Looking for (x1, x2, a3) S such that π3(x1, a2, a3) = π3(a1, x2, a3) Consider the set D = {(x1, x2, a3) S} 3 and the functions on D,f1(x1, x2, a3) = π3(a1, x2, a3) f2(x1, x2, a3) = π3(x1, a2, a3) . f1(x1, x2, a3) =π3(a1, x2, a3) . . 2 . . . . . (a1, x2, a3) (x1, x2, a3) a f1 is increasing with x1 D 1

  25. π(x) =a- uniqueness: n=3 Reminder: Looking for (x1, x2, a3) S such that π3(x1, a2, a3) = π3(a1, x2, a3) Consider the set D = {(x1, x2, a3) S} 3 and the functions on D,f1(x1, x2, a3) = π3(a1, x2, a3) f2(x1, x2, a3) = π3(x1, a2, a3) f2(x1, x2, a3)= π3(x1, a2, a3) . . 2 . (x1, x2, a3) . a f1 is increasing with x1 D (x1, a2, a3) f2 is decreasing with x1 1

  26. π(x) =a- uniqueness: n=3 Reminder: Looking for (x1, x2, a3) S such that π3(x1, a2, a3) = π3(a1, x2, a3) Consider the set D = {(x1, x2, a3) S} 3 Therefore, f1andf2coincide in Donly once. and the functions on D,f1(x1, x2, a3) = π3(a1, x2, a3) f2(x1, x2, a3) = π3(x1, a2, a3) . . f2(x1, x2, a3)= π3(x1, a2, a3) 2 . . . . . (x1, x2, a3) a f1 is increasing with x1 . D f2 is decreasing with x1 (x1, a2, a3) 1

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

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

  29. Constructing the solution We define an auxiliary solution : 3 (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) . . x =  (a,S) (a,S) y is an ordinal, symmetric solution of the original problem (a,S). . 2 . . y =  (x,S) Although y is not onS, it iscloserthanxtoS. a 1

  30. Constructing the solution a0 = a This suggests a recursive construction. Starting with the problem(a,S) generate the sequence The sequence (ak) converges to a point xonS. The solution defined by (a,S) = x is an ordinal, efficient, feasible and symmetric solution. a1 =(a0,S) a2 =(a1,S) a3 =(a2,S) a4 =(a3,S) a5 =(a4,S) a6 =(a5,S) a7 =(a6,S) a8 =(a7,S) . . .

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

  32. π(x) =a :four players The following equivalences hold by the definition of the Pareto functions πi. (x1, x2, x3, a4) S π4(x1, x2, a3, a4) =x4 π4(x1, a2, x3, a4) =x4 π4(a1, x2, x3, a4) =x4 (x1, x2, x3, a4) S (x1, x2, a3, x4) S (x1, a2, x3, x4) S (a1, x2, x3, x4) S π(x) = a Like the case of three players we prove the existence of (x1, x2, x3, a4) S such that π4(x1, x2, a3, a4) =π4(x1, a2, x3, a4) =π4(a1, x2, x3, a4) This conditions says that there exists a point (x1, x2, x3, a4) 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 (x1, x2, x3, a4) S such thatπ4(x1, x2, a3, a4) = π4(x1, a2, x3, a4) = π4(a1, x2, x3, a4). Let D = {(x1, x2, x3, a4) S} D1 = {(a1, x2, x3,a4) S} D2 = {(x1, a2, x3, a4) S} D3 = {(x1, x2, a3, a4) S} Dis homeomorphic to the three dimensional simplex. D3 = {(x1, x2, a3, a4) S} D1 = {(a1, x2, x3,a4) S} D = {(x1, x2, x3, a4) S} D1, D2,and D3are homeomorphic to the sides of this simplex. D2 = {(x1, a2, x3, a4) S}

  34. π(x) =a :four players Reminder: Looking for (x1, x2, x3, a4) S such thatπ4(x1, x2, a3, a4) = π4(x1, a2, x3, a4) = π4(a1, x2, x3, a4). Define three functions on D: f1(x1, x2, x3, a4) = π4(a1, x2, x3, a4) f2(x1, x2, x3, a4) = π4(x1, a2, x3, a4) f3(x1, x2, x3, a4) = π4(x1, x2, a3, a4) D3 = {(x1, x2, a3, a4) S} D1 = {(a1, x2, x3,a4) S} We need to show that the three functions coincide at some point in D. D = {(x1, x2, x3, a4) S} D2 = {(x1, a2, x3, a4) S}

  35. π(x) =a :four players = π4(a1, x2, x3, a4) ≤ π4(a1, a2, x3, a4) = f2(a1, x2, x3, a4) Reminder: Looking for (x1, x2, x3, a4) S such thatπ4(x1, x2, a3, a4) = π4(x1, a2, x3, a4) = π4(a1, x2, x3, a4). by definition of f1 Define three functions on D: f1(x1, x2, x3, a4) = π4(a1, x2, x3, a4) f2(x1, x2, x3, a4) = π4(x1, a2, x3, a4) f3(x1, x2, x3, a4) = π4(x1, x2, a3, a4) π4 is decreasing by definition of f2 Let A1= { pD | f1(p) ≤ f2(p), f3(p) } Then D1 A1. Indeed, D3 = {(x1, x2, a3, a4) S} D1 = {(a1, x2, x3,a4) S} A1 f1(a1, x2, x3, a4) D = {(x1, x2, x3, a4) S} f1at a point in D1 D2 = {(x1, a2, x3, a4) S}

  36. π(x) =a :four players • A1,A2,A3are closed(continuity of π4 ) • They cover D (by their definition) • DiAi for i=1, 2, 3 • By the AP Lemmathe intersection of the sets is not empty. Reminder: Looking for (x1, x2, x3, a4) S such thatπ4(x1, x2, a3, a4) = π4(x1, a2, x3, a4) = π4(a1, x2, x3, a4). Define three functions on D: f1(x1, x2, x3, a4) = π4(a1, x2, x3, a4) f2(x1, x2, x3, a4) = π4(x1, a2, x3, a4) f3(x1, x2, x3, a4) = π4(x1, x2, a3, a4) Let A1= { pD | f1(p) ≤ f2(p), f3(p) } A2= { pD | f2(p) ≤ f1(p), f3(p) } A3= { pD | f3(p) ≤ f1(p), f2(p) } A3 D3 = {(x1, x2, a3, a4) S} D1 = {(a1, x2, x3,a4) S} A1 D = {(x1, x2, x3, a4) S} For a point p in the intersection of A1, A2,A3 f1(p) = f2(p) = f3(p) A2 D2 = {(x1, a2, x3, a4) S}

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

  38. Constructing the solution ais below S i(a,S) = ais on S ais above S min { xi| π(x)=a } ai max { xi| π(x)=a } It is clear, that in case there exists a unique x for which π(x)=a, then(a,S)is this x.Thusfor n=3,  is the auxiliary solution defined above. The auxiliary solutionthus defined isordinalsince π, themin, andthe max functions are ordinal (i.e. commutes with order preserving transformations). is obviouslysymmetric. However, (a,S)is not necessarily on S. This problems is solved as in the casen=3…

  39. Constructing the solution a0 = a Starting with the problem(a,S) generate the sequence The sequence (ak) converges to a point xonS. The solution defined by (a,S) = x is an ordinal, efficient, feasible and symmetric solution. a1 =(a0,S) a2 =(a1,S) a3 =(a2,S) a4 =(a3,S) a5 =(a4,S) a6 =(a5,S) a7 =(a6,S) a8 =(a7,S) . . .

  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) presentsaninfinitefamily of such solutions for any number of players greater thantwo. 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). This family includes also a solution that extends Shapley’s solution to n>3. This extension is different than the one presented here.

More Related