1
A1 ParetoA2 SymmetryA3 InvarianceA4 IIA
A1 ParetoA2 SymmetryA3 InvarianceA4 IIA
First, we show that there exists a function satisfying the axioms.
There exists a unique
satisfying A1- A4
2: f BTheorem:
Proof:
2d
,S,dFor any given bargaining problem
define g S,d
S
=
= Does such a point always exist ??Is it unique ??
Yes !!!
Proof:
4
(divide the $)
0
Proof:
Consider the bargaining problem ,Δ,0 f S,d g S,d
Uniqueness: If satisfies the axioms then: f S,d
(1,0)
(0,1)
1 12 2f ,0 ,By Pareto + Symmetry:
A2 (Symmetry)
1 2 2 1( , ) ( , )x x x xα
( , ) α α αf S d f S,d
( , ) α α αf 0 f ,0 f ,0
By definition: 1 12 2g ,0 ,
1 12 2f ,0 g ,0 ,
7
Proof:
For a given (nondegenerate) bargaining problem ,S,d
0 (1,0)
(0,1)
d
S
g S,d
(a,d2)
(d1,b)
(d1, d2)
Consider the bargaining problem ,Δ,0Find an affine transformation α
α
α
α
8
Proof:
0 (1,0)
(0,1)
d
S
g S,d
(a,d2)
(d1,b)
(d1, d2)
Find an affine transformation α
α
α
α
α x, y = μx + ν, μy + ν
1 2α 0,0 = ν,ν = d ,d 2α 1,0 = μ+ ν ν = a,d, 1α 0,1 = ν, μ ν = d+ ,b
,1 2
1 2
ν,ν
μ, μ
= d ,d
a - d b - d
,1 2a - d b - d > 0,0
9
Proof:
0 (1,0)
(0,1)
d
S
g S,d
(a,d2)(d1, d2)
α
α
α Δ = S
S
αf Δ,0 = f α ,α 0
f S,d 1 12 2α ,
1 12 2,
==
(d1,b)α
g S,d
g S,d f S,d S g S,d S
1 12 2α , 1 1
2 2α 1,0 + 0,1
1 1 1 12 12 2 2 2α 1,0 + α 0,1 a,d d ,b
= g S,d
120
A Generalization
Changing A2 (Symmetry)
A2 (nonsymmetric)
for some f(Δ,0) = α,1 - α 0 < α < 1.
Δ
(1,0)
(0,1)
A
B
α measures the strength of Player 1
α,1 - α
A
B
1 - α=
α
13d
,S,dFor any given bargaining problem
define g S,d
SA
B
With the new A2, define a different g S,d
A
B
1 - α=
α
14d
g S,dS
Does such a point always exist ??Is it unique ??
Yes !!!
A
B
Following the steps of the previous theorem,
g S,d is the unique function satisfying the 4 axioms.
Yes !!!Yes !!!
A
B
1 - α=
α
15
A brief mathematical InterludeA brief mathematical Interlude
Consider the (implicit) functionα 1-αx y = K
Find a tangent at a point (x0,y0) on the curve
α-1 1-α α -ααx y + 1 - α x y y = 0
0 0αy + 1 - α x y = 0
0
0
αyy = -
1 - α x
differentiating
y
x
16
A brief mathematical InterludeA brief mathematical Interlude
Find a tangent at a point (x0,y0) on the curve
0
0
αy-1 - α x
0
0
y - y=
x - x
x
y
The tangent’s equation:
The intersections with the axis (x=0, y=0)
0 0x y,0 , 0,
α β
17
A
B
A brief mathematical InterludeA brief mathematical Interlude
x
y
0 0
0 0
x yx , y = α ,0 + β 0,
α β
(x0/α,0)
(0, y0/β)
(x0 , y0)
A
B
1 - α=
α
18
A
B
A brief mathematical InterludeA brief mathematical Interlude
x
y
α 1-αx y = KAny tangent of the function
is split by the tangency point in the ratio 1 - α
α
19
A brief mathematical InterludeA brief mathematical Interlude
x
y
α 1-α
(x,y) Smax x y
For any convex set S, by maximizing
S
1-α
α
We find the unique point in S in which the tangent is split
in the ratio
end of mathematical Interludeend of mathematical Interlude
20
α 1-α
1 2(x,y) Smax x - d y - d
To find the Nash Bargaining Solution of a bargaining problem ,S,d
S
d
Nash Bargaining Solution
21
All axioms were used in the proof
But are they necessary?
All axioms were used in the proof
But are they necessary?
A1. Without Pareto,h(S,d) = dsatisfies the other axioms.
A2. Without Symmetry,
satisfies the other axioms.
α 1-α
1 2(x,y) Smax x - d y - d
22
All axioms were used in the proof
But are they necessary?
All axioms were used in the proof
But are they necessary?
A3. Without Invariance,
satisfies the other axioms.
(x, y) S
h(S,d) = max x + y
S
(0,1)
(0,0) (2,0)d
(x, y) Smax xy
h(S,d)
(1, 0.5)
Nash Bargaining Solution
23
All axioms were used in the proof
But are they necessary?
All axioms were used in the proof
But are they necessary?
S
d
A4. Without IIA, the following function satisfies the other axioms.
The Kalai Smorodinsky solution
(S,d)K
1 2max max, (S,d)K
24
All axioms were used in the proof
But are they necessary?
All axioms were used in the proof
But are they necessary?
S
d
A4. Without IIA, the following function satisfies the other axioms.
The Kalai Smorodinsky solution
(S,d)K
1 2max max,(S,d)K
is on the Pareto front of S
1
2
max
max
(S,d) = (x, y)
x=
y
K is on the Pareto front of S
1
2
max
max
(S,d) = (x, y)
x=
y
K