ordinal solution to bargaining problems an ordinal solution to bargaining problems with many players...

An ordinal solutionordinal solution to bargaining problemsto bargaining problems

with many players

Z. Safra, D. Samet

An ordinal solutionordinal solution to bargaining problemsto 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)

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)

A solution is a function which assigns to each problem (a,S) a point (a,S) in RN.

Bargaining problems3

A bargaining problem for a set of players N is a pair

a - a status quo point in RN,

That is, if x,y S, and x y then x = y.

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.)

S - a Pareto surface in RN.

• Ui is a utility function of player i, and

• μ i is an order-preserving transformation of R,

μ i(Ui) represents the same preferences of i.

• Ui is a utility function of player i, and

• μ i is an order-preserving transformation of R,

μ i(Ui) represents the same preferences of i.

Ordinal transformations

An ordinal transformation of the utility space RN is obtained by applying to each player’s utility a continuous, order-preserving transformation.

Formally:

An ordinal transformation of the utility space RN is obtained by applying to each player’s utility a continuous, order-preserving transformation.

Formally:

Ordinal transformations

A vector (μ i) i N of continuous, order-preserving functions

from R onto R, defines an ordinal transformation

μ : RN → RN,

μ(x) = (μ i(xi)) i N .

A vector (μ i) i N of continuous, order-preserving functions

from R onto R, defines an ordinal transformation

μ : RN → RN,

μ(x) = (μ i(xi)) 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 }.

An ordinal solution

2. (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))

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).

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.

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 …

Pareto functions

The function πi : R

N → Ri

is i’s Pareto function.

N → Ri

it is:

independent of xi

(x1, x2, π3(x)).

Starting from any point x…

change i’s payoff until S is reached.

i’s payoff at this point is denoted by πi(x).

S is reached at one point at least because it is a surface,

and at one point at most because it is Pareto.

= (x1, x2, π3(x’))

Pareto functions

N → Ri

it is:

independent of xi

decreasing in xj

continuous

(x1, x2, π3(x)).

Starting from any point x…

change i’s payoff until S is reached.

i’s payoff at this point is denoted by πi(x).

Ideal pointsThe ideal point of x is

π(x) = (π1(x), π2(x), π3(x))

The ideal point of x is

π(x) = (π1(x), π2(x), π3(x))S

(π1(x), x2, x3)

(x1, x2, π3(x))

(x1, π2(x), x3)

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

Ideal points

π(x).

x.. Here is a point above S.

π(x) = (π1(x), π2(x), π3(x))

Ordinality of ideal points

y = (x1, x2, π3(x)).

μ(S)3

μ(x) = (μ1(x1), μ2(x2), μ3(x3))x = (x1, x2, x3)

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 = (μ1(x1), μ2(x2), π3(μ(x)))

μ(S)3

μ(x) = (μ1(x1), μ2(x2), μ3(x3))

y = (x1, x2, π3(x))

z = (μ1(x1), μ2(x2), π3(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,

μi(πi(x)) = πi(μ(x)))

x = (x1, x2, x3)

μ(y) = (μ1(x1), μ2(x2), μ3(π3(x)))

z = (μ1(x1), μ2(x2), π3(μ(x)))

For each player i,

μi(πi(x)) = πi(μ(x)),

and hence,

μ(π(x)) = π(μ(x)).

For each player i,

μi(πi(x)) = πi(μ(x)),

and hence,

μ(π(x)) = π(μ(x)).

π(x) = a means

that for each i, πi(x) = ai .

That is, for each i, (x-i , ai ) S.

A source for an ideal point

= (x1, x2, a3 )

= ( a1 , x2, x3)

= (x1, a2, x3)

Is the point a the ideal point of some x?

That is, is there x for which π(x) = a?

(x1, x2, π3(x))

(π1(x), x2, x3)

(x1, π2(x), x3)

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.

A source for an ideal point: n=3

(x1, x2, a3)

(a1, x2, x3)

(x1, a2, x3) .(a1, x2, a3)

Consider the three points:

(x1, x2, a3), (x1, a2, x3), (a1, x2, x3)

projected on S by the required point x.

Starting from (x1, x2, a3) we reach (a1, x2, x3) 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 a1 …

Then, increase 3’s payoff to π3(a1, x2, a3) = x3.

We call π3(a1, x2, a3), 3’s gains over 1

at (x1, x2, a3).

We call π3(a1, x2, a3), 3’s gains over 1

at (x1, x2, a3).

(x1, x2, a3)

(a1, x2, x3)

(x1, a2, x3)

(x1, a2, a3).

.(a1, x2, a3)

(x1, x2, a3), (x1, a2, x3), (a1, x2, x3)

Similarly, from (x1, x2, a3) we reach (x1, a2, x3) as follows.

First, reduce 2’s payoff to status quo a2 …

Then, increase 3’s payoff to π3(x1, a2, a3) = x3.

Similarly, π3(x1, a2, a3) is3’s gains over 2

at (x1, x2, a3).

Similarly, π3(x1, a2, a3) is3’s gains over 2

at (x1, x2, a3).

(x1, x2, a3)

(a1, x2, x3)

(x1, a2, x3)

(x1, a2, a3).

.(a1, x2, a3)

(x1, x2, a3), (x1, a2, x3), (a1, x2, x3)

π 3(x

1, a 2

Note that 3’s gains over 1 and 2 are the same. That is

π3(x1, a2, a3) = π3(a1, x2, a3)

Note that 3’s gains over 1 and 2 are the same. That is

π3(x1, a2, a3) = π3(a1, x2, a3)

This gives rise to the following two conditions which are equivalent to π(x) = a .

(x1, x2, a3) S

(x1, a2, x3) S

(a1, x2, x3) S

(x1, x2, a3) S

π3(x1, a2, a3) = x3

π3(a1, x2, a3) = x3

(x1, a2, a3)

(a1, x2, a3)

π(x) = a

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)

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)

The last equivalence follows immediately from the definition of the Pareto functions πi .π 3

a 2, a

a 1, x

2, a 3

(x1, x2, a3)

(a1, x2, x3)

(x1, a2, x3).

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.

π(x) = a - existence: n=33

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}

and the functions on D,f1(x1, x2, a3) = π3(a1, x2, a3) f2(x1, x2, a3) = π3(x1, a2, a3)

D(x1, x2, a3)

(a1, x2, a3).

(x1, a2, a3)

Computing f1 at (x1, x2, a3) …

Computing f2 at the same point …

f1(x1, x2, a3) = π3(a1, x2, a3)

f2(x1, x2, a3)= π3(x1, a2, a3)

D(x1, x2, a3)

(a1, x2, a3).

(x1, a2, a3)

f1(x1, x2, a3) = π3(a1, x2, a3)

f2(x1, x2, a3)= π3(x1, a2, a3)

First edge…

Compare f1 and f2 at the edges of D. .(a1, x2, a3 )

f1 < f2

f2(a1, x2, a3) = π3(a1, a2, a3)

= π3(a)

f1(a1, x2, a3) = π3(a1, x2, a3)

Because, (a1, x2, a3) is on S

π3(a )

Second edge…

D(x1, x2, a3)

(a1, x2, a3).

(x1, a2, a3)

f1(x1, x2, a3) = π3(a1, x2, a3)

f2(x1, x2, a3)= π3(x1, a2, a3)

f1 < f2

π3(a )

. f2(x1, a2, a3 ) = π3(x1, a2, a3)

(x1, a2, a3 )

f1 > f2

f1(x1, a2, a3 ) = π3(a1, a2, a3)

= π3(a)

Because, (x1, a2, a3) is on S

D(x1, x2, a3)

(a1, x2, a3).

(x1, a2, a3)

f1(x1, x2, a3) = π3(a1, x2, a3)

f2(x1, x2, a3)= π3(x1, a2, a3)

f1 < f2

π3(a )

.(x1, a2, a3 )

f1 > f2

The order of f1 and f2 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.

π(x) = a - uniqueness: n=3

.. (x1, x2, a3)

f1(x1, x2, a3) =π3(a1, x2, a3). (a1, x2, a3)

f1 is increasing with x1

(x1, x2, a3)

f1(x1, x2, a3) =π3(a1, x2, a3)

.(a1, x2, a3)

(x1, a2, a3)

(x1, x2, a3)

.f2(x1, x2, a3)

= π3(x1, a2, a3)

f2 is decreasing with x1

(x1, a2, a3)

(x1, x2, a3).

f2(x1, x2, a3)= π3(x1, a2, a3)

Therefore, f1 and f2

coincide in D only once.

Therefore, f1 and f2

coincide in D only once.

f2 is decreasing with x1

Constructing the solution

We define an auxiliary solution :

.x = (a, S)

(a, S) is the unique point x for which π(x) = a.(a, S) is the unique point x for which π(x) = 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.

.x = (a, S) .

y = (x, S)

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.

a0 = a

This suggests a recursive construction.Starting with the problem (a, S) generate the sequence

a1 = (a0, S)a2 = (a1, S)

a3 = (a2, S)a4 = (a3, S)

a5 = (a4, S)

a6 = (a5, S)

a7 = (a6, S)

a8 = (a7, S)...

The sequence (ak) converges to a point x on S.

The solution defined by

(a,S) = x

is an ordinal, efficient, feasible and symmetric solution.

(a,S) = x

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)

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:

π(x) = a : four players

(x1, x2, x3, a4) S

(x1, x2, a3, x4) S

(x1, a2, x3, x4) S

(a1, x2, x3, x4) S

(x1, x2, x3, a4) S

π4(x1, x2, a3, a4) = x4

π4(x1, a2, x3, a4) = x4

π4(a1, x2, x3, a4) = x4

The following equivalences hold by the definition of the Pareto functions πi.

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)

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.

π(x) = a : four playersReminder: Looking for (x1, x2, x3, a4) S such that π4(x1, x2, a3, a4) = π4(x1, a2, x3, a4) = π4(a1, x2, x3, a4).

D = {(x1, x2, x3, a4) S} D1 = {(a1, x2, x3, a4) S} D2 = {(x1, a2, x3, a4) S} D3 = {(x1, x2, a3, a4) S}

D = {(x1, x2, x3, a4) S}

D2 = {(x1, a2, x3, a4) S}D

3 = {(x1 , x

2 , a3 , a

4 ) S}

D is homeomorphic to the three dimensional simplex.

D1, D2, and D3 are homeomorphic to the sides of

this simplex.

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}D

3 = {(x1 , x

2 , a3 , a

4 ) S}

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) }

Then D1 A1. Indeed,

D = {(x1, x2, x3, a4) S}

D2 = {(x1, a2, x3, a4) S}D

3 = {(x1 , x

2 , a3 , a

4 ) S}

f1(a1, x2, x3, a4) π4(a1, x2, x3, a4)=

≤ π4(a1, a2, x3, a4)

f2(a1, x2, x3, a4)=

f1 at a point in D1

by definition of f1

π4 is decreasing

by definition of f2

Let A1 = { p D | f1(p) ≤ f2(p), f3(p) }

D = {(x1, x2, x3, a4) S}

D2 = {(x1, a2, x3, a4) S}D

3 = {(x1 , x

2 , a3 , a

4 ) S}

A2 = { p D | f2(p) ≤ f1(p), f3(p) } A3 = { p D | f3(p) ≤ f1(p), f2(p) }

For a point p in the intersection of A1, A2 , A3

f1(p) = f2(p) = f3(p)

• A1, A2, A3 are closed (continuity of π4 )

• They cover D (by their definition)

• Di Ai for i=1, 2, 3

By the AP Lemma the intersection of

the sets is not empty.

• A1, A2, A3 are closed (continuity of π4 )

• They cover D (by their definition)

• Di Ai for i=1, 2, 3

By the AP Lemma the intersection of

the sets is not empty.

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 { xi | π(x) = a }

max { xi | π(x) = a }

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 { xi | π(x) = a }

max { xi | π(x) = a }

a0 = a

Starting with the problem (a, S) generate the sequence

a1 = (a0, S)a2 = (a1, S)

a3 = (a2, S)a4 = (a3, S)

a5 = (a4, S)

a6 = (a5, S)

a7 = (a6, S)

a8 = (a7, S)...

(a,S) = x

ordinal solution to bargaining problems an ordinal solution to bargaining problems with many players...

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