314 J
. Naeve Mathematical Social Sciences 40 2000 313 –325
behavior stems from its application to implementation in Merlin and Naeve 1998 who are particularly concerned with the use of their results in the context of voting rules.
` The organization of the paper is quite similar to Barbera and Jackson 1988. We first
introduce some notation and define the three criteria. In listing the axioms, we take care that the definition of the axioms make sense without having to invoke other axioms,
which becomes important when we consider their independence. Then we give the characterization results. Though the propositions and their proofs are basically adapta-
` tions of the theorems of Barbera and Jackson 1988, we chose to present them in full,
both to make the exposition easier to follow and to be able to clarify some minor points that are not made explicit in the original proofs. Given the parallelism of our paper to
` Barbera and Jackson’s, our demonstration of the independence of the axioms immedi-
ately yields the same for theirs. In the concluding remarks, we discuss how our criteria `
compare to the ones presented by Barbera and Jackson 1988. As to the interpretation, we prefer to think of the vectors in 5 describing utility
consequences of an action in different states of the world. Indeed, we feel, that certain axioms are less suited for the alternative interpretation as consequences measured in
real numbers a social arrangement has for different members of a society. We defer a more detailed discussion of this point to the concluding remarks.
2. Notation and definitions
Let N denote the set of natural numbers, not including 0, let R denote the set of real
numbers and let 5 denote the set of finite dimensional vectors of real numbers. It is `
useful to be more explicit at this point. Barbera and Jackson 1988 implicitly assume that for a vector x with dimension dx, the set of indices is
h1, . . . ,dxj. We will take a slightly different approach, which avoids ambiguities later on cf. Remark 5 and
consists in explicitly defining the index sets. Let: 5 I , N: 1 uIu , `
h j
where uIu denotes the cardinality of the set I. Then we define:
I
5 5 R
I [ I
I
where R 5 hx : I
→ Rj. Following the usual notation for vectors, for x [ R , and i [ I,
we will denote the image of i under x as x 5 xi . We can then define x 5 x
, i.e. as
i 2i
I \
hi j
the restriction of x to I\
hij and write x as x 5 x ,x . Given x [ 5, we will denote the
i 2i
I
x
index set of x, by I , i.e. x [ R . The dimension of x [ 5 is dx 5 uI u. We use the
x x
following shorthand notations. Let x [ 5, then: minx 5 min x : i [ I
h j
i x
and maxx 5 max x : i [ I
h j
i x
For a [ R and x [ 5 let Ja,x 5 hi [ I : x aj
x i
J . Naeve Mathematical Social Sciences 40 2000 313 –325
315
What we are looking for is a binary relation s on 5 which ranks the vectors and has certain desirable properties we will list as axioms.
Definition 1. Given such a binary relation s on 5, we will define the following two associated binary relations, namely the indifference relation | by:
x | y : ⇔
I 5 I and neither x s y nor y s x
f g
x y
and the relation K by: xKy :
⇔ x s y or x | y
f g
Of course, this definition is intended to be applied to relations which also satisfy Axiom 0 below, thus the restriction to vectors with the same index set. We give it at this stage in
order to make sure, that Axiom 5 below is well defined also in absence of Axiom 1.
Remark 1. It is important to notice , that without further assumptions on s, the
relations | and K may fail to have many nice properties cf. also Remark 3 after
I
Axiom 1. However, | will be symmetric, and K restricted to each
R complete by construction.
We now proceed to define the three criteria we are going to consider in this paper. `
They correspond to the minimax, the leximax, and the protective criterion of Barbera and Jackson 1988. In addition we formally define the relation s
on 5, which is
P
weak Pareto dominance if we were thinking in terms of social consequences. We prefer `
the notation s over as used by Barbera and Jackson 1988 for two reasons: It
P
stresses the fact that we deal with a relation on 5 3 5 and it allows us to refer to the associated relations |
and K .
P P
Definition 2. The relation s is defined by:
P
x s y ⇔
[I 5 I ,x ± y , and x y ; i [ I ]
P x
y i
i x
Definition 3. The maximax criterion s is defined by:
M
x s y ⇔
[I 5 I and a [ R : Ja,x ± 5 and Ja,y 5 5]
M x
y
Definition 4. The leximax criterion s is defined by:
L
I 5 I and
x y
x s y ⇔
L
a [ R: [uJa, xu . uJa, yu and ;b . a, uJb, xu 5 uJb, yu]
Definition 5. The demanding criterion s is defined by:
D
I 5 I and
x y
x s y ⇔
D
a [ R: [Ja, x . Ja, y and ;b . a, Jb, x 5 Jb, y]
where . denotes proper containment.
316 J
. Naeve Mathematical Social Sciences 40 2000 313 –325
We illustrate the definitions by an example. Let I 5 h1,2,3j, x 5 5,4,3 and y 5
5,2,4. Then we have for a [ R:
5 if a . 5
5 if a . 5
h1j if a [ [5,4
h1j if a [ [5,4
Ja,x 5 and
Ja, y 5 h1,2j
if a [ [4,3 h1,3j
if a [ [4,2
5 5
I if a 3
I if a 2
Then obviously, x | y. Furthermore x | y. To see that x s y, consider a 5 3 with
P M
L
uJ3,xu 5 3 . 2 5 uJ3,yu and: if b . 5
1 if b [ [5,4
uJb,yu 5
5
2 if a [ [4,2
For the demanding criterion, however, we have J3,x 5 I . h1,3j 5 J3,y but J4,x 5
h1,2j ± h1,3j 5 J4,y, so x | y.
D
It is easy to see that for all x, y [ 5, x s y ⇒
x s y and x s y . Also x s f
g
P L
D
y ⇒
x s y and x s y , and x s y ⇒
x s y. f
g
M L
D D
L
2.1. Axioms
Axiom 0. x s y
⇒ I 5 I .
x y
This axiom has a clear interpretation in the context of choice under uncertainty: It only makes sense to compare two vectors of consequences if they are consequences in
different states in the same underlying set of possible states.
` Remark 2. Strictly speaking
, our Axiom 0 is stronger than that of Barbera and Jackson 1988. This is an artifact of our more explicit modelling of the index sets, however.
Axiom 1. s is irreflexive, transitive, and asymmetric.
This axiom states, that s should have the properties of a strict preference ordering.
Remark 3. Note that both the relations | as K defined in Definition 2 may fail to be
` transitive even if s satisfies Axiom
1 cf. the remark after Theorem 3 in Barbera and Jackson
, 1988. Consider, for example, a relation s on R which is defined by 1 s 2
and 1 s 3. This relation is irreflexive, transitive and asymmetric. Then we have 3 | 4 and 4 | 1, but 3 ¿ 1, and in the same way 3K4 and 4K1, but 3Q1. Both | and K are
reflexive, however.
Axiom 2 Symmetry. For x, y [ 5 with I 5 I , let s be a permutation on I . Define
x y
x
sx by sx 5 x . Then we have x s y
⇔ sx s s y.
21
i s
i
Symmetry states, that the labeling of states within one index set should not matter in the
J . Naeve Mathematical Social Sciences 40 2000 313 –325
317
ranking of actions. It is justified if we assume that the agents are ignorant as to the states. Symmetry would have no appeal, however, if we would assume that agents had a
priori non-uniform assessments of the probabilities of different states.
Remark 4. In our context
, one could also formulate a stronger version of Axiom 2. Again this has to do with the explicit modelling of index sets
. The stronger axiom considers changing the index set
, rather than just relabeling within one set.
Axiom 29 Neutrality. For x, y [ 5 with I 5 I , let s : I
→ I9 be a bijection to some
x y
x
other index set I9 with the same cardinality. Define sx by sx 5 x for all
21
j s
j
j [ I9 5 I . Then we have x s y
⇔ sx s s y.
s x
The characterization results will show, however, that Axiom 29 follows from Axiom 2 in conjunction with the other axioms used, which means, that always using initial segments
of the natural numbers actually does not constitute any loss of generality.
Axiom 3 Domination. x s y
⇒ x s y
P
The ranking should be monotonic with respect to the vector order. If in all states the consequences of action x are at least as good as those of action y, and in some states
strictly better, then x should be strictly preferred to y.
Axiom 4 Independence of Duplicated States. Let x, y [ 5 be such that for some i, j [
N, x 5 x and y 5 y . Then:
i j
i j
x s y ⇔
x s
y ⇔
x s
y
2i 2i
2j 2j
Again, for this axiom to make sense, it is important to assume that agents do not have any type of a priori assessment of probabilities of the states.
` Remark 5. Barbera and Jackson
1988 remark in a footnote that without symmetry, there is the need to make explicit how the operation of removing one component of a
vector is to be interpreted exactly . We have to address this point explicitly, since we have
to consider Axioms 4 and 6 without Axiom 2 in Proposition 4, where we demonstrate
the independence of the axioms . It turns out, however, that working with explicit index
sets instead of only considering the dimension of a vector , there is no ambiguity. Let
x [ 5 with index set I . Consider x . Then I
5 I \
hij, and x 5 x , for all j [ I .
x 2 i
x x
2i j j
x
2i 2i
Axiom 5 Concavity. Let x, y,z [ 5 be such that I 5 I 5 I , then: i x s y and
f
x y
z 1
1 ]
]
x s z ⇒
x s y 1 z; ii xKy and xKz ⇒
xK y 1 z. g
f g
2 2
` This is the only axiom that differs from those given by Barbera and Jackson 1988,
other than by the formal presentation. Instead of risk-aversion captured by convexity, we assume that agents are risk-loving. If an action is preferred to two others, it will also be
preferred to their average.
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. Naeve Mathematical Social Sciences 40 2000 313 –325
Remark 6. Axiom 5 actually consists of two parts which in the terminology of Debreu
1959, pp. 59f. are concavity i and weak concavity ii. Weak concavity follows from concavity only under additional assumptions
cf. Debreu, 1959, 1, p. 60. By Lemma 2 below
, restricting Axiom 5 to concavity in Debreu’s sense alone is not enough to obtain the characterization results
.
Axiom 6 Independence of Identical Consequences. For x, y [ 5, if x 5 y we have
i i
x s y ⇒
x s
y .
2i 2i
Comparison of actions which have identical consequences in some states are based on the other states in which they differ.
k k
Axiom 7 Continuity. Let x and y be two sequences of vectors in 5.
k k
k k
f
x s y , ;k
g
⇒ lim x Klim y
k →
` k
→ `
provided the limits exist. This axiom requires that small changes in the consequences should not reverse the
ranking of actions.
Axiom 8 Shuffling. Let I 5 I . Let s and · be two permutations of I . Then:
x y
x
x s y ⇒
sx s · y This is a much more demanding form of Axiom 2, which allows for different ways of
relabeling of states for two actions. Again one could formulate a stronger version in the spirit of Axiom 29.
Before considering general pairs of elements in 5, we restrict attention to a particular subset of 5 3 5. Let 6 5 x, y [ 5 3 5 : maxx ± max y . 6 is the set of pairs of
h j
vectors the maximal coordinates of which are different. On that set, all three criteria defined above coincide. For any two vectors with the same dimension, the one with the
largest maximum is preferred.
Lemma 1. The maximax , the leximax, and the demanding criterion agree on 6. They
are characterized on 6 by Axioms 0 –5 on 6 .
Proof. The agreement of the three criteria on 6 is obvious.
Let s be any binary relation on 6 satisfying Axioms 0–5. We proceed in two steps. To avoid unnecessarily complicated notation, we take an index set I 5
hi, jj and write a,b to denote the vector x with xi 5 a and x j 5 b. We shall follow this approach of
simply listing the components of a vector without explicit reference to the indices for vectors of arbitrary dimension, as long as no confusion is to be expected.
Step 1. For a,b,c,d [ R we show that:
J . Naeve Mathematical Social Sciences 40 2000 313 –325
319
[a . b, a . c, a . d] ⇒
[a, b s c, d ] 1
This is proven by contradiction. Suppose, it were otherwise. Then we would have a . b, a . c, and a . d but c,d Ka,b. Choose e [
R such that a 2 e . b, a 2 e . c,
1
and a 2 e . d. Then by Axiom 3, a 2 e,a 2 e s c,d and by Axiom 1 transitivity, a 2 e,a 2 eKa,b. By Axiom 4, a 2 e,a 2 e,a 2 eKa,a,b, and by Axiom 2, a 2
e,a 2 e,a 2 eKa,b,a. Then concavity Axiom 5 yields a 2 e,a 2 e,a 2 eK a,a 1 s
b 2,a 1 b 2 . Using Axiom 4 again, we have a 2 e,a 2 eK a,a 1 b 2 . d
s d
Repeating the same argument n times, we find for any n [ N:
n
2 2 1a 1 b ]]]]
a 2 e, a 2 eK
S
a,
D
n
2 Note that:
n
2 2 1a 1 b ]]]]
lim 5 a
n n
→ `
2 Therefore, as a 2 e , a, we can find N large enough, such that:
N
2 2 1a 1 b ]]]]
a 2 e ,
N
2 But for this N we have:
N
2 2 1a 1 b ]]]]
a 2 e, a 2 eK a,
S D
N
2 which contradicts Axiom 3. Therefore, the implication 1 must hold.
Step 2. Now consider a pair x, y [ 6, with I 5 I . Let a 5 maxx and b 5 max y.
x y
W.l.o.g., assume a . b. Let d 5 a 1 2b 3 and e 5 2a 1 b 3. Then, by Axiom 3, d,d, . . . ,d s y. Let c 5 minx. Then by Axiom 3, x s e,c, . . . ,c. Now if e 5 c, we
have e,c s d,d by Axiom 3, otherwise 1 of Step 1 yields e,c s d,d . Then by Axiom 4, e,c . . . ,c s d,d, . . . ,d , and thus by Axiom 1 transitivity, x s y.
Hence we have shown, that if I 5 I and maxx . max y, we have x s y. By Axiom
x y
1
1 Asymmetry, we know that it cannot be the case that y s x at the same time. Thus, we have an ordering ordering on 6 that coincides with maximax, leximax and the
demanding criterion. While all three criteria coincide on 6, this is not so on 5 3 5. We present the
characterizations of the three criteria in Propositions 1–3 which are analoga of Theorems `
1–3 in Barbera and Jackson 1988, and present the proofs to enhance readability. We concentrate on the characterization of the demanding criterion in Proposition 4, however,
where we demonstrate the independence of the axioms used in the characterization.
1
` This argument is not made explicit by Barbera and Jackson 1988.
320 J
. Naeve Mathematical Social Sciences 40 2000 313 –325
Proposition 1. The maximax criterion on 5 3 5 is characterized by Axioms 0 –5 on 6
and Axioms 0 and 7 on 5 3 5.
Proof. It is easy to check that s satisfies the axioms. Indeed, all axioms but Axiom 3
M
are satisfied on the whole domain 5 3 5. An example for the violation of Axiom 3 on 5 3 5 is 1,1 | 1,0.
M
By Lemma 1, it suffices to show that for x, y [ 5, with I 5 I and maxx 5 max y,
x y
the axioms imply x | y, to establish the proposition. So take two such vectors. Let
k k
k
he j be a sequence of real numbers with lim
e 5 0. Let x be defined by
k [ N
k →
` 1
k k
k k
k k
x 5 x 1 e for all i [ I , and x
by x 5 x 2 e . For all k [
N, the pairs x ,y
s d s d
1 i i
x 2
2 i i
1 k
k k
and x , y belong to 6, and by Lemma 1 we have x s y and y s x . Thus, Axiom 7
2 1
2
implies xKy and yKx. Then by Definition 1, x | y, i.e. neither x s y nor y s x.
Proposition 2. The leximax criterion on 535 is characterized by Axioms 0 and 3 –5
on 6, and Axioms 0, 6 and 8 on 535.
Remark 7. Since Axiom
8 obviously implies Axiom 2, the conditions of Lemma 1 are satisfied
.
Proof. It is easy to check that s satisfies the axioms. Indeed, all axioms but Axiom 4
L
are satisfied on the whole domain 5 3 5. An example for the violation of Axiom 4 on 5 3 5 is the following. Let x,y [ 5 3 5, with I 5 I 5
h1,2,3j. Let x 5 2,2,4 and
x y
y 5 3,3,2. Then x s y, but y s
x and y
s x
.
L 21
L 21
22 L
22
Now let x, y [ 5, with I 5 I , since otherwise Axiom 0 excludes x s y and y s x. Let
x y
uI u 5 n. By Axiom 8, we can assume I 5 I 5 h1, . . . ,nj, and moreover, x x , and
x x
y i
i 11
y y , for all i [
h1, . . . ,n 2 1j. Remove all indices i for which x 5 y and denote the
i i 11
i i
resulting vectors by x9 and y9. By construction x9, y9 [ 6. Therefore, by Lemma 1 x9 s y9
⇔ maxx9 . max y9. But maxx9 . max y9 is equivalent to:
a [ R: [uJa,xu . uJa,yu and ;b . a, uJb,xu 5 uJb,yu]
Thus, s is the leximax criterion.
Proposition 3. The demanding criterion is characterized on 5 3 5 by Axioms
0 –6.
Proof. To see that s satisfies Axioms 0–4, and 6 is straightforward. It remains to
D
demonstrate that it also satisfies concavity, i.e. Axiom 5. Suppose x s y and x s z. Let x9, y9 and z9 denote the vectors we get by elimination
D D
of all states for which x, y and z agree. By Axiom 6, x9 s y9 and x9 s z9. Let
D D
9
a 5 maxx. By the definition of s , we know that for any l with x ± a, we have
D l
9 9
9 9
a . y and a . z . For k such that x 5 a, we know a y and a z . By construction
l l
k k
k
9 9
of x9, y9 and z9, not all three can have the same component, and hence a . y or a . z .
k k
J . Naeve Mathematical Social Sciences 40 2000 313 –325
321
9 9
Therefore, for all i [ I , we have a . 1 2 y 1 z . Hence x9 s 1 2 y9 1 z9. Since
x 9 i
i D
s satisfies Axiom 6, it follows that x s 1 2 y 1 z. The argument for K is similar.
D D
D
Now we must show, that any relation s satisfying the axioms must be s . By
D
Axiom 0, only vectors with the same dimension are in the relation. We start by proving some facts for two-dimensional vectors.
Fact 1. Let a,b [ R, a . b. Then neither a,b s b,a, nor b,a s a,b.
Proof of Fact 1. Assume a,b s b,a. Then by Symmetry Axiom 2, b,a s a,b. But then Axiom 1 cannot hold, since transitivity would imply a,b s a,b contradicting
irreflexivity. The same argument can be made if b,a s a,b.
Fact 2. Let a,b,c [ R, a . b, a . c. Then neither b,a s a,c, nor a,c s b,a.
Proof of Fact 2. If b 5 c the result follows from Fact 1. So we assume b ± c. W.l.o.g., let b . c.
Assume a,c s b,a. By Axiom 3, b,a s c,a. Then by transitivity Axiom 1, a,c s c,a, which contradicts Fact 1.
So assume b,a s a,c. By Axiom 6, b,a,a s a,c,a, and by Axiom 2, b,a,a s a,a,c. Then by concavity b,a,a s a,a 1 c 2,a 1 c 2 . By Axiom 4 it follows that
s d
n
b,a s a,a 1 c 2 . This argument can be repeated n times to arrive at b,a s a,2 2
s
s d
n n
n
1a 1 c 2 . For n large enough, we have 2 2 1a 1 c 2 . b, and hence by Axiom
d
n n
3 a,2 2 1a 1 c 2 s a,b. But then Axiom 1 transitivity yields b,a s a,b
s d
which again contradicts Fact 1. We can extend Fact 2 to comparisons of vectors with arbitrary dimension.
Fact 3. Let x and y be such that I 5 I , maxx 5 max y 5 M, and such that there exists
x y
a pair i, j [ I 3 I with x 5 M . y and y 5 M . x . Then, neither x s y, nor y s x.
x x
j j
l l
Proof of Fact 3. Suppose y s x. Let minx 5 m, and let k [ I be such that x 5 m.
x k
˜ ˜
Define the vectors x and y with the same index set by: M
if i 5 j y
if i 5 j
j
˜ ˜
x 5 and
y 5
H H
i i
m if i [ I \
h jj M
if i [ I \
h jj
x x
˜ ˜
˜ ˜
By Axiom 3, we have x s x or x 5 x. Also y s y or y 5 y. Axiom 1 transitivity yields ˜
˜ y s x, which by Axiom 4 implies y ,M s M,m, contradicting Fact 2. Thus, y s x
j
cannot hold. A similar argument shows that x s y also is not possible. Now consider any two vectors x and y with I 5 I . Denote the vectors derived by
x y
elimination of all components which coincide for x and y by x9 and y9. By Axiom 6, x s y
⇔ x9 s y9. Let maxx9 5 c, and max y9 5 d.
If c 5 d, Fact 3 applies, and we have neither x9 s y9, nor y9 s x9, and hence also
322 J
. Naeve Mathematical Social Sciences 40 2000 313 –325
neither x s y nor y s x. So let c ± d. Then x9, y9 [ 6, and we know that x9 s y9 if c . d, and y9 s x9 if d . c. Since c . d iff:
a [ R : [Ja,x . Ja,y and ;b . a , Jb,x 5 Jb,y]
take a 5 c, we have for all x, y [ 5, [x s y] ⇒
[x s y]. Axiom 1 Asymmetry then
D
yields that s coincides with s .
D
Proposition 4. The Axioms 0 –6 are independent on 5 35.
Proof. We present a list of six relations, each satisfying all but one of the seven axioms. We will always just state the axiom that is not satisfied.
Axiom 0: Consider the relation s defined as follows. For x, y [ 5 :
x s y : ⇔
[x s y or dx . d y]
D
Axiom 1: Here we can actually go one step further and present three examples of relations each satisfying Axioms 0, and 2–6, one violating irreflexivity while satisfying
transitivity and asymmetry, a second violating transitivity while satisfying irreflexivity and asymmetry, and a third violating asymmetry while satisfying irreflexivity and
transitivity. Therefore, each of the three parts of Axiom 1 are independent of each other in conjunction with the rest of the axioms.
Irreflexivity: Consider the relation s defined as follows. For x, y [ 5 :
a
x s y : ⇔
[x s y or x 5 y]
a D
Transitivity: Consider the relation s which is defined as follows. For x, y [ 5, first
b
define: 9
x s y : ⇔
[I 5 I and [x s y or x, y 5 3,4,2,6]]
b x
y P
9 9
This relation clearly is not transitive, since for example 2,6 s 1,5 but not 3,4 s
b b
1,5. However, it also fails to satisfy Axioms 4 and 6. Therefore we need to consider an extension which satisfies these two axioms by making sure that all pairs x, y [ 5 3 5
which can be transformed to 3,4,2,6 by the operations considered in Axiom 4 or Axiom 6, respectively, are in the relation as well. We will formalize this by the relation
s . So this relation is defined for x, y [ 5 by:
b
x s y :
⇔ I 5 I and
b x
y
i, j [ I 3 I :
x x
x 5 3, x 5 4, y 5 2, y 5 6 and
i j
i j
;k [ I \
hi, jj,
x
x 5 y or
k k
[x 5 3 or x 5 4 and y 5 2 or y 5 6]
k k
k k
For x, y [ 5 we finally define s by:
b
J . Naeve Mathematical Social Sciences 40 2000 313 –325
323
x s : ⇔
I 5 I and [x s y or x s y]
b x
y P
b
Asymmetry: Consider the relation s defined as follows. For x, y [ 5 :
c
x s y : ⇔
[x s y or y s x]
c D
D
Axiom 2: For x [ 5, define fx by fx 5 ix . Then s is defined by:
i i
2
x s y :
⇔ fx s f y, for x, y [ 5
2 D
Axiom 3: Consider the relation s defined as follows. For x, y [ 5 :
3
x s y : ⇔
2 x s 2 y
3 D
Axiom 4: Consider the relation s defined as follows. For x, y [ 5 :
4 dx
d y
x s y : ⇔
I 5 I and
O
x .
O
y
F G
4 x
y i
i i 51
i 51
` Axiom 5: Consider the protective criterion of Barbera and Jackson 1988.
Axiom 6: Consider the relation s defined as follows. For x, y [ 5 :
6
x s y : ⇔
[x s y or x | y and x s y ] s
d
6 M
M P
Corollary 1. Axioms 0–5 are independent on 6. Proof. In view of Lemma 1, obviously Axiom 6 is implied by Axioms 0–5. The
examples in the proof of Proposition 4 show, that Axioms 0–5 are independent on 6, as well. As we noted in the introduction, Proposition 4 and Corollary 1 immediately yield
` the independence of the axioms used by Barbera and Jackson 1988 in their
characterizations of the protective criterion on 5 3 5 and on 6, respectively. We close this section by demonstrating that part 2, the weak concavity, of Axiom 5 is
essential in the characterization results of Lemma 1 and Proposition 3.
Lemma 2. The relation s satisfies Axioms
0 –4, Axiom 6, and part 1 of Axiom 5. It
P
does not satisfy part 2 of Axiom 5.
Proof. That s satisfies Axioms 0–4 and Axiom 6 is easy to see.
P
We show that s is concave, but not weakly concave. Let x s y, and x s z.
P P
P
Consider t 5 1 2 y 1 z. Clearly I 5 I . Also x t for all i [ I follows immediately,
x t
i i
x
since x y and x z for all i [ I Since x s y, there exists j [ I such that x . y .
i i
i i
x P
x j
j
But then, x . 1 2y 1 1 2z 5 t . Hence x ± t and thus x s t.
j j
j j
P
To see that s fails to satisfy weak convexity, it suffices to present an example. We
P
have 2,2 | 1,5 and 2,2 | 5,1. Hence 2,2K 1,5 and 2,2K 5,1. However,
P P
P P
1 21,5 1 5,1 5 3,3 s 2,2.
P
324 J
. Naeve Mathematical Social Sciences 40 2000 313 –325
3. Concluding remarks
