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Non-interference and social welfare orderings satisfying strong Pareto and anonymity. Tsuyoshi Adachi Waseda University. A social welfare ordering : a reflexive, complete and transitive binary relation on. : utility vectors. Axioms Efficiency Strong Pareto ( SP ): - PowerPoint PPT Presentation
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Non-interference and social welfare orderings satisfying strong Pareto
and anonymityTsuyoshi Adachi
Waseda University
A social welfare ordering : a reflexive, complete and transitive binary relation on
: utility vectors
Axioms•Efficiency
Strong Pareto (SP): •Impartiality
Anonymity (AN):If for some permutation ,
then .•Noninterference (Mariotti & Veneziani, 2009a)
0
2
4
6
1 2 3 4 5
0
2
4
6
1 2 3 4 50
2
4
6
1 2 3 4 5
Non-interferenceLet be as follows: There exists such thati. andii. andiii. Then, .
’s utility is only changed.
x (x’) is better than y (y’) for in both of the pairs.
The changes have the same sign
0
2
4
6
1 2 3 4 5
The left vector is better than the right vectorDecreasing“Non-interference” requires
0
2
4
6
1 2 3 4 5
0
2
4
6
1 2 3 4 50
2
4
6
1 2 3 4 50
2
4
6
1 2 3 4 5
0
2
4
6
1 2 3 4 50
2
4
6
1 2 3 4 5
Previous results (difficulty)Complete Non-Interference (CNI):
For all such that , , , and , .
SP and CNI is dictatorial. (MV, 2009a)
( SP, AN, and CNI are not compatible.)
DecreasingIncreasing
Previous results (characterization)Individual Damege Principle (IDP):
0
2
4
6
1 2 3 4 5
0
2
4
6
1 2 3 4 5
For all such that , , , , and , .
SP,AN,IDP (MV, 2009ab)
: the leximin ordering
Decreasing
Previous results (characterization)Individual Benefit Principle (IBP):
0
2
4
6
1 2 3 4 5
0
2
4
6
1 2 3 4 5
For all such that , , , , and , .
SP,AN,IBP (MV, 2009ab)
: the leximax ordering
Increasing
Previous results (characterization)Uniform Non-Interference (UNI):
1. SP,AN,UNI implies .2. is SP,AN,UNI (MV, 2009a)
For all such that , , , and , .
0
2
4
6
1 2 3 4 5
0
2
4
6
1 2 3 4 5
Let
The same value
MV (2009ab)– SP, AN and CNI are not compatible.
• IDP, IBP, and UNI (as restrictions of CNI)– Characterization of (resp. ) without equity axioms
(resp. inequity axioms)
Hammond (1976), D’Aspremont and Gevers (1977)
• Our questions:– Restrictions of CNI other than IDP, IBP, and UNI. – SWOs characterized by the combination of such axioms
Introduction of a general class of non-interference axioms
Non-interference on D
For all such that , , , and , .
Non-interference on (NI on ):
i. ii. and
is the set of possible :
• CNI NI on • IDP NI on • IBP NI on • UNI NI on
The existing axiomsThe first quadrant
The third quadrant
• Lower-Half Noninterference (LNI) denotes NI on
Let
Lower-Half Noninterference
Remark : For all SP , LNI UNI Proof
i. [LNI UNI] is clearii. UNI LNI
• By Remark,
– SP, AN, LNI implies
0
2
4
6
1 2 3 4 50
2
4
6
1 2 3 4 5
0
2
4
6
1 2 3 4 50
2
4
6
1 2 3 4 5
LNI and UNI
0
2
4
6
1 2 3 4 5 is smaller then
The same value
By UNIBy SP
NI on NI on
Existing results (recosideration)• SP, AN, UNI implies .
NI on
• SP and CNI is dictarorial SP, AN, and CNI are not compatible• SP, AN, IDP
NI on
• SP, AN, IBP
NI on
Other examples
• SP, AN, and NI on ?
The key lemma
Lemma: Let be SP, AN. Then, i.[ is NI on with ]ii.[ is NI on with ]
Let
Leximin and Leximax orderings
Proposition 1: i. is NI on ii.( )
Theorem 1: i.[SP, AN, NI on ]
[ and ] ii.( )
○ ×
Compatibility with SP and AN
By Proposition 1, ×
By Lemma,
or non-existence
Theorem 2: NI on is compatible with SP and AN , , or .
Incompatible
NI on is implied by IBP, IDP, or LNI
I. [SP, AN, NI on ]
II. [SP, AN, NI on ]
III. [ satisfies SP, AN, NI on ]
Characterization of SWOs
Theorem 3: The combination of SP, AN, and NI oni. characterizes , ii. characterizes , iii. is satisfied by , oriv. is incompatible.
Conclusions
• Generalized class of NI axioms• Characterization of NI axioms compatible with SP
and AN.– A NI axiom is compatible iff it is implied by IBP,
IDP, or LNI• SWOs characterized by SP, AN, and a NI axiom.
– The leximin, leximax, and utilitarian orderings have an important role.
Individual 2
Individual 1
NI on implies [ ]
Proof of the Lemma •The two person case : Individual 1 and 2.•Let :
i.e., there exists such that . (Note that for all )
•Let be SP, AN, NI on
Individual 2
Individual 1
Step 1: For all in ,
By SP and AN,
By NI on ,
The same length
1
By AN,
Individual 2
Individual 11
1
By Step 1, for all in ,
By Step 1, For all in ,