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.M. Kaminski Mathematical Social Sciences 40 2000 131 –155
a if max t 5 1 ` then Bt ,g ,t 5 [ or inf Bt ,g ,t [ Bt ,g ,t ;
j i
i j
i i
j i
i j
b if max t , 1 ` then Bt ,g ,t is closed and non-empty.
j i
i j
Condition CB specifies topological requirements on R with no reference to any
structure on T. This condition is closely related to one of Debreu’s 1954 classical conditions applied to the space T 3
R and Young’s 1994, p. 191 axiom see Section 3 for a more detailed discussion of the relationship.
Standards and rules can be linked via a concept of equity Young, 1994. Let g , . . . , g be an allocation for
t;g and 0 , ´ g . A transfer of ´ from t to t is
1 n
j j
i
R-justified if t ,g Pt ,g 2 ´.
i i
j j
]]] EQUITY: A rule F is R-equitable if for all
t;g [ P, no transfer is R-justified for ]]]
F t;g.
Equity means that all allocations generated by a rule conform closely to the standard. In the hydraulic language, a transfer from
t to t is R-justified if the water level in the
j i
vessel t is higher than in t . The reader is requested to go back to Fig. 1, fill the vessels
j i
with virtual water, and check how the metaphor works. Consider a simple example of
E
the unconstrained equal division rule E and the egalitarian standard R . Since E t;g 5
i E
1 utug for all i, for 0 , ´ 1 utug we can check that t ,g 2 ´P t ,g for all
i i
j i
E E
t ,t [ T and no transfer is R -justified. The rule E is R -equitable.
i j
Now, the ‘hydraulic’ and ‘semi-hydraulic’ properties will be defined formally for any rule F. Both names denote concisely two classes of rules constructed on the basis of the
concept of equity. SEMI-HYDRAULIC PROPERTY: F is semi-hydraulic if it is R-equitable for some
]]]] standard R.
HYDRAULIC PROPERTY: F is hydraulic if it is R-equitable for some numerical ]]]
standard R. There exist semi-hydraulic rules that are not hydraulic. An example is the lex-
icographic priority rule X, defined for T 5 R and no restriction, which divides the whole
X X
good equally among agents with the lowest type. X is R -equitable for the standard R
X
defined as ; t ,t [ T, ;g [ domt , ;g [ domt , [t , t
→ t ,g P t ,g ; t 5
i j
i i
j j
i j
i i
j j
i X
X
t →
t ,g P t ,g iff g , g ]. R is not numerical and it is clear that X cannot be
j i
i j
j i
j
R-equitable for any numerical standard R.
3. Results
In this section, T is any set of types and maxT any restriction. Theorems 1 and 2 generalize Young’s 1994 results which connect standards with rules. Theorem 1 will
1
be used as an intermediary step leading to Theorem 2 and Corollary 1.
Theorem 1. A rule satisfies B.SYM, B.CONS and B.MON if and only if it is semi- hydraulic, i.e. if there exists a standard R such that F is R-equitable. For any standard R
1
Proofs omitted in the main text are in Appendix A.
M .M. Kaminski Mathematical Social Sciences 40 2000 131 –155
137
that satisfies CB there exists a unique rule that satisfies B.SYM, B.CONS and B.MON and is R-equitable, and vice versa.
Theorem 11 of Young 1994, p. 192 connects rules with standards for claims, i.e. with T 5
R 2 h0j and maxt 5 t . It says that if a standard R is representable and
1 i
i
closed below then for every problem t;g there exists a unique allocation Ft;g that is
R-equitable. The present result asserts that the relation of equity is two-sided, i.e. equity is a one-to-one relation between all standards that satisfy CB and all rules that satisfy
B.SYM, B.CONS and B.MON, and that it holds universally for all types. The present axiom of closure below differs somewhat from Young’s formulation.
Young’s axiom imposes a stronger topological condition than the present axiom but it does not require that, when max
t , 1 `, the set Bt ,g ,t be non-empty. However,
i i
i j
without such a requirement, Young’s Theorem 11 is false since the relationship does not hold for some standards that satisfy his axioms. A counterexample to Young’s Theorem
CEA
11 is provided by a constrained equal award standard R with a following numerical
CEA
representation: r t ;g 5 2 g for all g [ domt . For t 5 1, t 5 2, g 5 3, and
i i
i i
i 1
2
t 5 t ,t a transfer of ´ is justified from t to t for all 0 , ´ , 1 and no equitable
1 2
2 1
CEA
allocation exists since 1,2 is the only allocation for t;g. The R
standard satisfies Young’s condition of closure below but not the present condition. The difficulty
CEA
disappears when we add a provision that r t ;g 5 2 g only for all 0 g , maxt
i i
i i
i CEA
while for g 5 max t , t ;g R
t ;g for all g [ domt .
i i
j j
i i
j j
In Theorem 1, no structure on the set of types was postulated while in Theorem 2, some topological conditions are assumed.
Theorem 2. Let T be a separable topological space and let F be a continuous rule. F satisfies B.SYM and B.CONS if and only if it is hydraulic, i.e. if it is R-equitable for
some numerical standard R. In addition, F satisfies B.MON.
Theorem 2 says that, for continuous rules defined in a separable topology, the hydraulic property is equivalent to obeying bilateral symmetry and bilateral consistency.
The assumption of continuity of F with respect to a separable topology on T cannot be dropped. Consider the lexicographic rule X introduced in Section 2. X is not hydraulic
but it satisfies B.SYM and B.CONS. However, X is not continuous in the usual topology on
R. If we change the topology, e.g. if we take the discrete topology on R, where any subset of reals is an open set, X becomes trivially continuous. However, such a
topological space is not separable. The main result of Young 1994, p. 193, first part of Theorem 12 asserts that for
claims, every impartial, bilaterally consistent, and continuous rule is equitable relative to a numerical standard and it is monotone. Since symmetry is a weaker condition than
impartiality, the ‘only if’ part of Theorem 2 provides a generalization of Young’s result by showing that the conclusion holds for any space of types with a separable topology.
Types in Theorem 2 can have many non-isomorphic interpretations that are not allowed in Young’s result. For instance, since creditors in real-world bankruptcy problems may
have both secured and unsecured claims, we need to represent their types as vectors
]] ]]]
rather than single numbers. Another example is the case of types interpreted as agents’
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.M. Kaminski Mathematical Social Sciences 40 2000 131 –155
utility functions, when the rationing problem becomes a variant of the social choice problem. These cases are discussed in more detail in the next section.
The third general result concerns consistent extensions of bilateral rules. In the context of bankruptcy, Aumann and Maschler 1985 proved that the Contested Garment rule
from the Talmud has a unique consistent extension. Dagan and Volij 1997 formulated necessary and sufficient conditions for deciding when any bilateral bankruptcy rule can
be consistently extended. The following criterion is closely related to Dagan–Volij’s result see discussion in Section 5.1.
2 2
Let us define P as the set of all two-type problems: P 5
ht;g [ P : utu 5 2j. A
2 2
2
bilateral rule H: P
→ R assigns to every bilateral problem t ,t ;g [ P an allocation.
1 i
j
]]]]
2
H has a consistent extension F if F is a consistent rule and F uP ; H. If H is
]]]]]]
H H
anonymous, it defines a binary relation R in the following way:
t ,g R t ,g iff
i i
j j
h t ,g h t ,g , where h
assumes values in R h`j and h t ,g 5 suphx [
ij i
i ij
j j
ij 1
ij i
i
[0,max t 1 maxt ]: Ht ,t ;x g
j. In words, the function h t ,g denotes the
i j
i j
i i
ij i
i
maximal total amount that types t and t must get jointly in order to give g to t .
i j
i i
H
t ,g R t ,g means that this total amount is for t ,g not greater than for t ,g .
i i
j j
i i
j j
H
Anonymity of H guarantees that the relation R is well-defined. Bilateral symmetry
and bilateral monotonicity alone do not imply anonymity of a bilateral rule. An example of a non-anonymous bilateral rule H that satisfies B.SYM and B.MON is as follows:
T 5 R , and for t , t , H is proportional; for t t , H is equal division.
1 1
2 1
2
Corollary 1. A bilateral rule H that satisfies B.AN and B.MON has a consistent
H
extension if and only if the relation R is transitive. This extension is unique.
The proof uses some concepts and results from Appendix A.
Proof. Only if : See Lemma 1e,f.
H
If : R is a standard and satisfies CB [see the proof of Theorem 5a]. By Theorem 1,
H
there exists a rule F that is R -equitable. We will show that F is a consistent extension of H. Let us assume, to the contrary, that for some
t;g [ P, Ft;g 5 g and
i i
F t;g 5 g but Ht ,t ;g 1 g 5 g 1 ´, where ´.0. Thus, Ht ,t ;g 1 g 5 g 2 ´.
j j
i j
i j i
i i
j i
j j j
By B.MON and definition of h , we have that h t ,g , g 1 g and h t ,g 2 ´ g 1
ij ij
i i
i j
ij j
j i
H H
g . This, by definition of R , implies t ,g P t ,g 2 ´. Thus, in the allocation Ft;g,
j i
i j
j H
a transfer of ´ from t to t is justified and F cannot be R -equitable.
j i
2 2
Uniqueness: Assume that for two consistent rules G ± F, G uP ; H and FuP ; H.
Let t;g [ P be such a problem that for some t [ t, Gt;g ± Ft;g , for instance,
i i
i
G t;g . Ft;g . There must exist t [ t such that Gt;g , Ft;g . Let us assume
i i
j j
j 2
2
that, for instance, G t;g 1 Gt;g Ft;g 1 Ft;g . But since F
uP ; H, GuP ;
i j
i j
H, and F, G are consistent, it follows that H t ,t ;Gt;g 1 Gt;g 5 Gt;g .
i j
i j i
i
F t;g 5 Ht ,t ;Ft;g 1 Ft;g and H violates B.MON. h
i i
j i
j i
In the next section, Corollary 1 is applied to Aumann–Maschler’s analysis of the bankruptcy problem from the Talmud, and social choice problems.
M .M. Kaminski Mathematical Social Sciences 40 2000 131 –155
139
4. Examples
