Journal of Mathematical Economics 35 (2001) 547–562
Rank-dependent preferences without
ranking axioms
Zvi Safra a,∗ , Uzi Segal b,c
b
a Faculty of Management, Tel Aviv University, Tel Aviv 69978, Israel
Department of Economics, University of Western Ontario, London ON, Canada N6A 5C2
c Department of Economics, Boston College, Chestnut Hill, MA 02467-3806, USA
Received 14 March 2000; received in revised form 16 November 2000; accepted 28 March 2001
Abstract
A key feature of the rank-dependent model for decision making under risk is that the weighting of
an outcome depends on its relative rank. This theory received numerous axiomatizations, however,
all these sets of axioms need to make an explicit reference to the ranking of the outcomes. This
situation is unsatisfactory, as it seems to be desirable to get the ranking property of this model as
a consequence of the model, rather than as an assumption. [Yaari, M.E., 1987. The dual theory of
choice under risk. Econometrica 55, 95–115] offered a special version of this model (called dual
theory), where the utility function is linear. This paper offers a set of axioms implying Yaari’s dual
theory without making any reference to the order of the outcomes. The main axiom is called semi
betweenness, which, unlike the usual case, is made on random variables rather than on distribution
functions. © 2001 Elsevier Science B.V. All rights reserved.
Keywords: Rank-dependent model; Axioms; Dual theory
1. Introduction
The rank-dependent model for decision making under risk established itself during the
past two decades as a major alternative to expected utility theory. Similarly to the latter
model, it assumes that outcomes are evaluated by the utility level, the decision maker
receives from them. But, whereas the traditional theory evaluates lotteries with respect to
the expected value of these utilities, the rank-dependent model takes the expected value
of these utilities with respect to a transformation of the distribution function. Formally, it
∗ Corresponding author.
E-mail addresses: safraz@post.tau.ac.il (Z. Safra), segalu@bc.edu (U. Segal).
0304-4068/01/$ – see front matter © 2001 Elsevier Science B.V. All rights reserved.
PII: S 0 3 0 4 - 4 0 6 8 ( 0 1 ) 0 0 0 6 0 - X
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Z. Safra, U. Segal / Journal of Mathematical Economics 35 (2001) 547–562
assumes the existence of a continuous, increasing, and onto transformation function g :
[0, 1] → [0, 1] such that
V (F ) = u(x) dg(F (x))
where F is a cumulative distribution function.
A key feature of this theory is that the weighting of an outcome depends on its relative
rank. This theory received numerous axiomatizations (see Weymark, 1981; Quiggin, 1982;
Chew and Epstein, 1989; Segal, 1990; Quiggin and Wakker, 1994; Wakker, 1994). However,
all these sets of axioms need to make an explicit reference to the ranking of the outcomes.
This situation is unsatisfactory, as it seems to be desirable to get the ranking property of
this model as a consequence of the model, rather than as an assumption.
Yaari (1987) offered a special version of this model (called dual theory), where the utility
function u is linear. In this paper, we offer a different set of axioms, implying the general
form of Yaari’s dual theory without making any reference to the order of the outcomes. 1
We present our axioms and results for finite dimensional spaces in the next section. In
Section 4, we apply the results to the analysis of preferences over finite random variables
and extend them to continuous random variables in Section 5. We explain our claim that
the axioms of this paper do not depend on the order of the outcomes in Section 3. In this
section, we also explain why our results need some topological structure on the outcomes
set. The main results are proved in Sections 6 and 7.
2. Axioms and results
Let e = (1, . . . , 1) ∈ Rn and let D be the main diagonal of Rn , i.e. D = {λe : λ ∈ R}.
For x, y ∈ Rn , let [x, y] = {αx + (1 − α)y : α ∈ [0, 1]} be the chord connecting x and
y. Let Π n be the set of permutations of {1,. . . ,n}. For every permutation π ∈ Π n , let
π(x) = (xπ(1) , . . . , xπ(n) ). For any permutation π ∈ Π n , let Kπ = {x ∈ Rn : xπ(1) ≤
· · · ≤ xπ(n) }. That is, Kπ is the set of points in Rn that is ordered (from the lowest to the
highest co-ordinate) by π. Let π id be the identity permutation. Then Kπ id is the set {x ∈
that orders its
Rn : x1 ≤ · · · ≤ xn }. Let xord be the vector obtained from x by a permutation
n = x ∈ Rn :
let
Δ
xi = 1 be
co-ordinates. 2 That is, x ord = π(x) iff x ∈ Kπ . Finally,
+
the (n−1)-dimensional simplex in Rn and let H n = x ∈ Rn : xi = 1 be the hyperplane
containing Δn .
Consider a complete and transitive preference relation on Rn . For x ∈ Rn , let I (x) =
(y : y ∼ x) be the indifference set through x. On we make the following assumptions.
S (Symmetry). For every permutation π ∈ Π n , π(x) ∼ x.
C (Continuity). For every x, the sets {y : y x} and {y : x y} are closed.
DE (Diagonal Equivalence). For every x there is y ∈ D such that x ∼ y.
1 In Safra and Segal (1998), we offered, among other things, a set of axioms that imply a very restrictive version
of Yaari’s model, where the transformation functions g is quadratic.
2 The permutation needs not be unique (e.g. when x in on D), but xord is nevertheless well defined.
Z. Safra, U. Segal / Journal of Mathematical Economics 35 (2001) 547–562
549
LN (Local Nonatiation). For every x there are x k → x and y k → x such that for every
k, x k x y k .
SB (Semi Betweenness). x ∼ y implies the existence of permutations π and π such that
[π(x), π (y)] ⊂ I (x).
For x ∈ Rn , let C(x) = {λe ∈ D : λe ∼ x}. By DE, C(x) is not empty.
Fact 1. For every c ∈ C(x), [x, c] ⊂ I (X).
Proof. For every c ∈ D and for every π, π(c) = c. By SB, there is a permutation π such
that [π(x), c] ⊂ I (x). By Symmetry, [x, c] ⊂ I (x).
Trivially, the stronger assumption “for x ∼ y, the chord [x, y] is contained in I(x)” implies
(together with continuity) that can be represented by a linear function V (x) = a · x (see
also the discussion of the SB axiom in Section 3 below). The first main result of this paper
is the following theorem.
Theorem 1. The following two conditions are eqvivalent.
1. The preference relation satisfies S, C, DE, LN, and SB
2. (a) The preference relation is strictly monotonic along the main diagonal D. That
is, either for all λ > λ , λe λ e, or for all λ > λ , λ e λe; and
(b) There is a unique continuous function s : R → H n , where s(λ) = s λ = (s1λ , . . . , snλ ),
satisfying for every k = 1, . . . , n and λ < λ,
k
k
siλ ≤
siλ
i=1
(1)
i=1
such that
x ∼ λe ⇔ s λ · x ord = λ
(2)
The proof of this theorem appears in Section 6. Note that the second condition of Theorem
1 is equivalent to the following statement:
The preferences can be represented implicitly by one of the following functions.
1. V (x) = λ for the unique λ solving s λ · x ord = λ; or
2. V (x) = −λ for the unique λ solving s λ · x ord = λ.
where the continuous function sλ satisfies the requirement of inequality (1) above.
Consider the case n = 2. Along an indifference curve through λe = (λ, λ) the preferences
satisfy (x1 , x2 ) ∼ λe iff V λ (x1 , x2 ) = V λ (λ, λ), where
s1λ x1 + s2λ x2 x1 ≤ x2
V λ (x1 , x2 ) =
s1λ x2 + s2λ x1 x2 ≤ x1
Fig. 1 depicts this case. Along indifference curve , s1λ < 0 and s2λ >1. Along indifference
curve , s1λ = 0 and s2λ = 1. Along indifference curves and , 0 < s1λ , s2λ < 1. Along
indifference curve , s1λ = 1 and s2λ = 0. Finally, along indifference curves and , s1λ >1
and s2λ < 0.
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Z. Safra, U. Segal / Journal of Mathematical Economics 35 (2001) 547–562
Fig. 1. Indifference curves.
Preferences in Fig. 1 may either always increase along the main diagonal, or always
decrease along it. Moreover, even in the first case, preferences are not necessarily monotonic
in each co-ordinate away from the main diagonal.
As is evident from Fig. 1, the slopes of the indifference curves on one side of the diagonal
change monotonically. Moreover, this is all that is needed to guarantee that on each side of
the main diagonal, indifference curves do not intersect. Thenext lemmashows that for the
general, n-dimensional case, the requirement λ < λ ⇒ ∀k ki=1 siλ ≤ ki=1 siλ is the most
general restriction on the function s. That is, the above inequality implies that indifference
sets do not intersect.
Lemma 1. Let the function s : R → H n satisfy the requirements of inequality (1). Then
for every x there is at most one value of λ such that s λ · x ord = λ.
Proof. Let λ > λ . For k = 1, . . . , n, define tk = ki=1 siλ − ki=1 siλ . By inequality (1),
tk ≥ 0 for all k. For k > 1, we obtain
k
k
k−1
k−1
skλ − skλ =
siλ −
siλ −
siλ −
siλ
i=1
i=1
i=1
i=1
i=1
i=1
i=1
i=1
k
k−1
k
k−1
=
siλ −
siλ −
siλ −
siλ = −tk + tk−1
For i = 1, . . . , n − 1, define
follows that
mi
=
ei+1
− ei ,
where ei denotes the ith unit vector of Rn . It
s λ − s λ = (−t1 , t1 − t2 , . . . , tn−2 − tn−1 , tn−1 ) =
n−1
ti mi
i=1
Z. Safra, U. Segal / Journal of Mathematical Economics 35 (2001) 547–562
551
Obviously, mi · x ord ≥ 0 for i = 1, . . . , n − 1, hence
λ
s ·x
ord
λ
−s ·x
ord
n−1
=
ti mi · x ord ≥ 0
i=1
Therefore, if s λ · x ord = λ, then s λ · x ord = λ for all λ < λ.
Theorem 1 and Lemma 1 depict indifference sets of the preferences , but the axioms
are not rich enough to put further restrictions on the functions s λ beyond those imposed by
inequality (1). More assumptions are needed for a more uniform behavior of indifference
sets.
DL (Diagonal Linearity). x ∼ y ⇒ ∀λ, x + λe ∼ y + λe.
H (Homogeneity). x ∼ y ⇒ ∀λ > 0, λx ∼ λy.
Theorem 2. The following two conditions are equivalent.
1. The preference relation satisfies S, C, DE, LN, and SB and either DL or H.
2. The preference relation can be represented by the function
V (x) = s · x ord
(3)
for some s ∈ H n .
Proof. (2) ⇒ (1): the function of Eq. (3) is homogeneous, therefore H. Also, V (x + λe) =
V (x) + λ, hence assumption DL.
(1) ⇒ (2): by Theorem 1, x ∼ λe iff s λ · x ord = λ. Assume DL, suppose s λ = s λ , and
λ
λ
choose x such that s · x = λ, but s · x = λ. Then x ∼ λe, and by DL, x + (λ − λ)e ∼ λ e.
However, s λ ·(x +(λ −λ)e) = s λ ·x +λ −λ = λ , a contradiction. A similar contradiction
is obtained if we assume H.
3. Discussion of the axioms
In Section 1, we claimed that our axioms are not based on ordering properties of the set
of outcomes. We now turn into an explanation of this claim.
Clearly, Symmetry does not depend on any properties of the set of outcomes. Continuity
and Local Nonsatiation are topological properties. They require the existence of a topology
on the set of outcomes, but they do not depend on the existence of an order on this set.
The analysis of Semi Betweenness is more difficult. Mathematically, it is weaker than the
following two axioms.
B (Betweenness). x ∼ y implies for all α ∈ [0, 1], [x, y] ⊂ I (x).
CB (Comonotonic Betweenness). If x and y are comonotonic, i.e. if for all i and j,
(xi − xj )(yi − yj ) ≥ 0, then x ∼ y implies for all α ∈ [0, 1][x, y] ⊂ I (x).
Betweenness is defined with no reference to the order of the outcomes, and is too strong
for our analysis (it implies that can be represented by V (x) = s · x). CB depends very
strongly on the order of the outcomes. And given the rest of the axioms, it is not hard to prove
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Z. Safra, U. Segal / Journal of Mathematical Economics 35 (2001) 547–562
that this axiom implies the preferences of Theorem 1. Most of the proof of this theorem is
devoted to showing that the permutations π and π of the SB axiom are transforming x and
y into the same comonotonic sector of Rn . To that extent, SB implies CB without making
references to the order of the outcomes.
Like the rest of the axioms, Diagonal Equivalence does not make any reference to the
order of the outcomes. In no way does this axiom imply monotonicity. As is demonstrated by
the two-dimensional example of Section 2, the point on the main diagonal that is indifferent
to x may be smaller or larger than x in all its co-ordinates (see indifference curves and
in Fig. 1). Note, however, that SB implies weak monotonicity along the main diagonal D
(see Claim 5 below).
The two linearity axioms, DL and H, require linearity of the set of outcomes, but they too
do not depend on any ordering of the outcomes. Note that for every x, x + λe, and λx are
comonotonic. This does not imply that these axioms use properties regarding any natural
ordering of the outcomes, because the axioms do not require that x and y are comonotonic.
Although, together all the axioms imply that the treatment of the ith co-ordinate of x depends
on its ranking in x, none of the axioms by itself implies such a dependency.
If we add monotonicity to our axioms (x y implies x y), then the vectors sλ must
be non-negative, that is, for every λ, s λ ∈ Δn . Otherwise, if siλ < 0, obtain a contradiction
by changing only the value of xiord .
Although our axioms do not depend on the order of the outcomes, they do depend on the
fact that the set of outcomes has some strict topological properties. Several axiomatizations
of the rank-dependent model do not restrict the set of outcomes and they can be of any type,
while preferences over acts induce a natural order over these outcomes. Such models usually
work within the probability simplex while holding outcomes fixed (see, e.g. Abdellaoui,
1999). We work with random variables where the probabilities are fixed, but outcomes vary.
To that extent, we cannot dispense with some topological structure on the set of outcomes.
Clearly, our axioms and proofs require linearity of that set.
4. Risk
The preferences discussed above were defined over Rn . One possible interpretation of
these preferences involves decision making under risk. Suppose that there are n equally probable states for the world, s1 , . . . , sn . A lottery X is a function from {S1 , . . . , Sn } to R, and
since these events are fixed, it can be written as the vector (X(S1 ), . . . , X(Sn )) ∈ Rn . The
cumulative distribution function obtained from X is given by FX (x) = Pr(X ≤ x). Trivially,
this probability equals 1/n times the number of the outcomes of X that are not greater than
x. Denote the set of these distribution functions F. It follows that preferences over F can be
described as preferences over Rn . As usual, we denote by δx , the constant random variable x.
Our axioms are meaningful in this context (Symmetry depends on the requirement that
Pr(S1 ) = · · · = Pr(Sn )), hence Theorems 1 and 2 apply. Let s0λ = 0, and define the functions
g λ : [0, 1] → R as follows. For p ∈ [(i − 1)/n, i/n)], let
λ
g (p) =
i−1
k=1
skλ
i−1 λ
+n p−
si
n
(4)
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553
Observe that under the monotonicity assumption, s λ ∈ Δn (see Section 3 above). In that
case, gλ is an increasing function.
Theorem 1 now implies the following representation: for every X and λ, X ∼ δλ iff
x dg λ (FX (x)) = λ
Theorem 2 implies that either for all X and Y, X Y iff
x dg(FX (x)) ≥ x dg(FY (x))
(5)
or for all X and Y, X Y iff
x dg(FX (x)) ≤ x dg(FY (x))
The function g is defined similarly to the definition of gλ above, with s replacing sλ . Observe
that the functional of Eq. (5) is the one suggested by Yaari (1987).
The restriction imposed on sλ by Lemma 1 has some implications for the analysis of
decision making under risk. The maximal insurance premium a decision maker with income
λ is willing to pay to entirely avoid the risk X (which is added to λ) is the difference between
the expected value of λ + X, λ + E(X), and his certainty equivalent of λ + X, which we
denoted C(λ + X). Denote this difference ρ(λ + X). The next lemma shows that as the
wealth level λ goes up, decision makers are willing to pay higher premium for full insurance
against a given risk X.
Lemma 2. Under the assumptions of Theorem 1, ∀X, ρ(λ+X) is a non-decreasing function
of λ.
Proof. Let a (λ) = C(λ + X) be the certainty equivalent of the random variable λ + X.
By Eq. (5)
x dg a(λ) (Fλ+X (x))
= (x + λ) dFX (x) − (x + λ) dg a(λ) (FX (x))
= x dFX (x) − x dg a(λ) (FX (x))
ρ(λ + X) =
x dFλ+X (x) −
For λ > λ we obtain
ρ(λ + X) − ρ(λ + X) =
x dg a(λ ) (FX (x)) −
x dg a(λ) (FX (x)) ≥ 0
where the last inequality follows from Lemma 1 and from Yaari (1987).
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Z. Safra, U. Segal / Journal of Mathematical Economics 35 (2001) 547–562
5. Continuous random variables
In this section, we analyze preferences over general sets of random variables, based on
the analysis of the finite case. Let Ω = (S, Σ, P ) be a probability measure space and let X
be the set of real bounded random variables on it. 3 For X ∈ X , let FX be the distribution
function of X. Denote as before by F the set of distribution functions obtained from elements
of X . With a slight abuse of notations, we denote by λ the constant random variable with
the value λ, and its distribution function by δ λ .
On X we assume the existence of a complete and transitive preference relation . On we make the six assumptions of Section 2 (S, C, DE, LN, SB, and either DL or H), modified
for the case of continuous distributions. Only the first, second, and the fifth assumptions
need to be rewritten.
S (Symmetry). If FX = FY , then X ∼ Y .
C (Continuity). is continuous in the topology of weak convergence.
SB (Semi Betweenness). For every n and x, y ∈ Rn , if Pr(S1 ) = · · · = Pr(Sn ) =
1/n, and X = (x1 , S1 ; . . . ; xn , Sn ) ∼ Y = (y1 , S1 ; . . . ; yn , Sn ), then there are permutations π, π ∈ Π n such that for all α ∈ [0, 1], (αxπ(1) + (1 − α)yπ (1) , S1 ; . . . ; αxπ(n) +
(1 − α)yπ (n) , Sn ) ∼ X ∼ Y .
Theorem 3. The following two conditions on the preference relation over X are
equivalent.
1. satisfies S, C, DE, LN, SB, and either DL or H.
2. There is a unique continuous function g : [0, 1] → R, satisfying g(0) = 0 and g(1) = 1,
such that can be represented either by
V (X) = x dg(FX (x))
or by
V (X) = −
x dg(FX (x))
The proof of this theorem appears in Section 7.
6. Proof of Theorem 1
(2) ⇒ (1): by Lemma 1, we know that condition 2(b) is well defined. That is, if for some
λ, s λ · x ord = λ, then for every λ = λ, s λ · x ord = λ . Next we show that for every x ∈ Rn
λ
ord
there is λ such that s · x = λ. Pick λ ∈ Rn , and assume, without loss of generality, that
s λ · x ord = λ > λ . By the proof of Lemma 1, s λ · x ord ≤ s λ · x ord = λ . Therefore, by
λ
ord
continuity, there is λ ∈ (λ , λ ) such that s · x = λ. We, thus, proved that satisfies
the DE axiom.
3 This does not imply that all random variables are uniformly bounded. Below, we explicitly use the fact that if
X ∈ X , then for every a, X + a ∈ X .
Z. Safra, U. Segal / Journal of Mathematical Economics 35 (2001) 547–562
555
Symmetry follows by the fact that in Eq. (2), sλ multiplies the vector xord . For continuity,
observe that by the above discussion, if s λ · x = λ, then for all y such that s λ · y ≥ λ,
there exists λ ≥ λ such that s λ · y = λ , and for all y such that s λ · y ≤ λ, there exists
λ ≤ λ such that s λ · y = λ . Therefore, the -upper and lower sets of x are closed. Local
Nonsatiation follows in a similar way. Semi Betweenness is satisfied when π orders x and
π orders y.
(1) ⇒ (2): we prove this direction through a sequence of claims.
Claim 1. Either for all λ and λ , λ ≥ λ implies λe λ e, or for all λ and λ , λ ≤ λ
implies λe λ e.
Proof. By continuity, if λ > λ > λ and either λe λ e and λ e λ e, or λ e λe
and λ e λ e, then there are γ > λ > γ such that γ e ∼ γ e ∼ λ e, a contradiction to
Fact 1.
Assume, without loss of generality, that
(∗) λ > λ implies λe λ e.
Claim 2. For all λ∗ there are λ > λ∗ > λ such that λ e λ∗ e λ e.
Proof. Suppose that for some λ∗ and for all λ > λ∗ , λ e ∼ λ∗ e. By LN, there exists
x λ∗ e. By assumption DE, there is λ such that λ e ∼ x λ∗ e, and by condition (∗),
λ > λ∗ . The existence of λ is similarly proved.
Claim 3. For x ∈
/ D and λ∗ ∈ R there exists λ such that x + λe ∼ λ∗ e.
Proof. Suppose not, for example, suppose that for every λ, λ∗ e x + λe. Let λ > λ∗ such
that λ e λ∗ e (the existence of such λ follows by Claim 2). For each k, let λk e ∈ C(x +ke),
hence λ∗ e λk e, and by (∗) it follows that λk < λ∗ for all k. There is a sequence y k =
α k (x + ke) + (1 − α k )λk e such that y k → λ e. By Fact 1, y k ∼ λk e, hence, by continuity,
λ∗ e limλk e ∼ limy k . On the other hand, y k → λ e λ∗ e, a contradiction.
Claim 4. If y = x + λe for some λ = 0, then x ∼ y.
Proof. Let x, y ∈
/ D such that y = x + λe for some λ>0. Suppose x ∼ y, and let
λ∗ e ∈ C(x). Denote I (x) by I∗ . We now construct by induction a sequence S 2 , . . . , S n
where S i = {x i,1 , . . . , x i,i , λ∗ e} such that
1. Conv(Si ) is of dimension i,
2. Conv(S i ) ⊂ I ∗ .
3. D ⊂ span (S i − λ∗ e), where S i − λ∗ e = {z − λ∗ e : z ∈ S i }. That is, all points in D can
be represented as linear combinations of points in S i − λ∗ e.
Suppose, we had constructed these sets, and consider the set Conv(Sn ). It is convex and of
full dimension. Pick a point z in its interior, and obtain a contradiction to the LN assumption.
Therefore, if y = x + λe for λ = 0, then x ∼ y. In particular, if λ > 0, then condition (∗)
implies that y x.
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Z. Safra, U. Segal / Journal of Mathematical Economics 35 (2001) 547–562
We now turn to the proof of the existence of the sequence S 2 , . . . , S n . Let x 2,1 = x and
= y. Let z ∈ Conv(x 2,1 , x 2,2 , λ∗ e). If z ∈
/ I ∗ , then ∃λ e ∈ C(z) such that λ = λ∗ .
2,1
∗
2,2
∗
∗
By Fact 1, [x , λ e], [x , λ e] ⊂ I and [z, λ e] ⊂ I (z). Now either [x 2,1 , λ∗ e] ∩
[x 2,2 , λ e] or [x 2,2 , λ∗ e] ∩ [z, λ e] is not empty and belongs to two different indifference
/ D, the dimension of Conv(S 2 ) = Conv(x 2,1 , x 2,2 , λ∗ e)
sets, a contradiction. Since, x 2,1 ∈
2,2
2,1
is 2. Since x = x + λe it follows that D ⊂ span(S 2 − λe).
Suppose we defined S 2 , . . . , S i , i < n, and define Si +1 as follows. Pick w ∈ I ∗ such
/ span(S i − λ∗ e). We show first that such a point w exists. By
that for all π ∈ Π n , π(w) ∈
induction, the dimension of Si is less than n. Consider the set
x 2,2
G = {π(x) : x ∈ span(S i − λ∗ e) and π ∈ Π n }
/ G.
Since G is the union of n! linear sets of dimension less than n, it is not Rn . Pick w ∈
As we have proved above, there exists λ such that w = w + λe ∈ I ∗ . Since D ⊂
/ π(span(S i − λ∗ e)), and hence
π(span(S i − λ∗ e)) for all π ∈ Π n , it follows that w ∈
n
i
i
w∈
/ G. For each π ∈ Π , define Q (π ) = {z ∈ Conv(S ) : [z, π(w)] ⊂ I ∗ }. By SB, each
z ∈ Conv(S i ) belongs to at least one of these sets. It is well known that under the above
structure, there exists π ∗ such that C(Qi (π ∗ )) has a non-empty interior (relative to the
i-dimensional plane containing Si ). By continuity, Qi (π ∗ ) is a closed set, therefore, Qi (π ∗ )
is of dimension i. Define x i+1,1 , . . . , x i+1,i to be i points in Qi (π ∗ ) such that the dimension
of span(x i+1,1 −λ∗ e, . . . , x i+1,i −λ∗ e) is i, and Conv(x i+1,1 , . . . , x i+1,i ) ⊂ Int(Qi (π ∗ )).
Let x i+1,i+1 = w. Define S i+1 = {x i+1,1 , . . . , x i+1,i , w, λ∗ e}.
We have to show that
1. The dimension of Conv(Si +1 ) is i + 1;
2. Conv(S i+1 ) ⊂ I ∗ ; and
3. D ⊂ span(S i+1 − λ∗ e).
The first requirement follows from the fact that the dimension of the set span(x i+1,1 −
, x i+1,i −λ∗ e) is i, but w−λ∗ e ∈
/ span(S i −λ∗ e) ⊃ span(x i+1,1 −λ∗ e, . . . , x i+1,i −
∗
i
∗
/ span(S − λ e), but D ⊂ span(S i − λ∗ e), hence w − λ∗ e ∈
/
λ e). (Recall that w ∈
i
∗
span(S − λ e)).
The second requirement follows by the following argument. By construction, the convex
combination of L = {x i+1,1 , . . . , x i+1,i , w} is in I∗ (recall that Conv(x i+1,1 , . . . , x i+1,i ) ⊂
Int(Qi (π ∗ ))). By Fact 1, the convex combination of each point in L and λ∗ e is in I∗ . Finally,
each point in Conv(Si +1 ) is a convex combination of a point in L and λ∗ e.
For the third requirement, note that by the induction hypothesis, D ⊂ span(S i − λ∗ e).
Also, the dimension of span(S i − λ∗ e) is the same as the dimension of span(x i+1,1 −
λ∗ e, . . . , x i+1,i −λ∗ e), therefore, span(S i −λ∗ e) = span(x i+1,1 −λ∗ e, . . . , x i+1,i −λ∗ e).
Finally, we show that the claim holds for points x and y in D. Suppose that for some
/ D such that z ∼ λe (the existence of such a point z was proved
λ > λ, λ e ∼ λe. Pick z ∈
in Claim 3 above). The points (1/2)z + (1/2)λe and (1/2)z + (1/2)λ e violate the above
proof for the case where x and y are not in D.
From Claims 3 and 4, we obtain:
λ∗ e, . . .
Conclusion 1. For every x and λ∗ there exists a unique λ such that x + λe ∼ λ∗ e.
Z. Safra, U. Segal / Journal of Mathematical Economics 35 (2001) 547–562
557
The analysis so far is sufficient for the case n = 2. For each point x in R2 , the chord
between x and C(x) is in I(x), and by Conclusion 1, the whole ray from C(x) in the direction
of x is in I(x). From this point on, we assume that n ≥ 3.
Claim 5. ∀x, y ∈ Rn such that x ∼ y there exists π ∈ Π n such that [x, π(y)] ⊂ I (x).
Proof. By assumption SB, there exists π and π such that [π (x), π (y)] ⊂ I (x). By
assumption S, [x, {(π )−1 oπ }(y)] ⊂ I (x).
For x, y ∈ Rn define
Φ(x, y) = {π ∈ Π n : [x, π(y)] ⊂ I (x)}
Note that if x ∼ y, then by SB, Φ(x, y) = ∅. Also, Φ(y, x) = {π −1 : π ∈ Φ(x, y)}. By
the continuity of we obtain:
Fact 2. Φ is upper-hemi continuous, i.e. if x k → x and y k → y, and if ∀k, π ∈ Φ(x k , y k ),
then π ∈ Φ(x, y).
Claim 6. Let I be an indifference set of in Rn , and let x̄, ȳ ∈ I . For every open (relative
to I) neighborhoods V (x̄) of x̄ and V (ȳ) of ȳ, there are x ∈ V (x̄) and y ∈ V (ȳ) and open
neighborhoods V (x) ⊂ V (x̄), V (y) ⊂ V (ȳ) such that Φ is constant on V (x) × V (y).
Proof. By Fact 2, for each x, y ∈ I , there are neighborhoods V(x),V (y) ⊂ I such that for
all x ∈ V (x) and y ∈ V (y), Φ(x , y ) ⊂ Φ(x, y). The claim follows from the fact that
Φ(x̄, ȳ) is finite.
Claim 7. Let V1 and V2 be the intersections of the open balls B1 and B2 with I. If Φ =
Π n∗ ⊂ Π n constantly on V1 × V2 , then there are hyperplanes H1 and H2 in Rn such that
Vi = Hi ∩ Bi , i = 1, 2. Moreover, for every π ∈ Π n∗ , Conv(V1 ∪ π(V2 )) ⊂ H1 .
Proof. We prove that if x1 , x2 ∈ V1 , then [x1 , x2 ] ⊂ V1 . Suppose not. Let α ∗ = min{α ∈
[0, 1] : [x1 , αx1 +(1 − α)x2 ] ⊂ V1 } (α ∗ may equal 1). By Claim 4, for every w ∈ [x1 , x2 ]
there exists λw such that w + λw e ∈ I . Pick w ∈ [x2 , α ∗ x1 + (1 − α ∗ )x2 ] sufficiently
/ V1 . Then by continuity, z = w + λw e ∈ B1 ,
close to α ∗ x1 + (1 − α ∗ )x2 such that w ∈
hence z ∈ V1 (see Fig. 2). Choose π ∈ Π n∗ and y ∈ V2 := π(V2 ). By definition, [x1 , y],
[x2 , y], [z, y] ⊂ I . Choose y ∈ [x2 , y] ∩ V2 , y = y, and obtain that [x1 , y ] ⊂ I . Define
w = [w, y] ∩ [x1 , y ], and let λ = 0 such that z = w + λ e ∈ [z, y]. Since w , z ∈ I ,
we obtain a violation of Claim 4, hence [x1 , x2 ] ⊂ V1 . Since V1 is convex, it follows once
more by Claim 4 that it is contained in a hyperplane.
If V1 is contained in any lower dimensional plane, then, by continuity, either for all
x ∈ B1 \V1 and y ∈ V1 , x y, or for all such x and y, y x. Both cases contradict
Claim 4 with respect to y + λe and y − λe. Therefore, V1 is the intersection of a hyperplane
with an open set in Rn . By construction, it follows that π(V2 ) ⊂ H1 . The proof for V2 is
similar.
Summarizing, let I be an indifference set of . For every x̄, ȳ ∈ I and for every open
(relative to I) neighborhoods V1 of x̄ and V2 of ȳ, there are non-empty open (relative to I)
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Z. Safra, U. Segal / Journal of Mathematical Economics 35 (2001) 547–562
sets Vi ⊂ Vi , i = 1,2, a hyperplane H, and a permutation π , such that the convex hull of
V1 ∪ π(V2 ) is the intersection of H with an open set in Rn .
Claim 8. Let H be a hyperplane in Rn that does not contain D, and let V and V be open
neighborhoods in H such that V ⊂ Int(Kπ ) and V ⊂ Int(Kπ ) for some π = π . Then
there are x ∈ V and x ∈ V such that every y = (y1 , . . . , yn )∈[x, x ] has at most one pair
of equal co-ordinates.
Proof. Let x and x be as in the statement of claim, and assume that for some y ∈ [x, x ],
yi = yj and yk = y (j may equal k, but then i = ). The linear subspace J = {z ∈ Rn :
zi = zj and zk = z } is of dimension n−2, and it contains D. Since H does not contain D, the
intersection of J and H is of dimension n−3. Therefore, there are V1 ⊂ V and V1 ⊂ V such
that for every z ∈ V1 and z ∈ V1 for all y ∈ [z, z ], if yi = yj then yk = y , and vice versa.
The claim now follows by repeating the argument a finite number of times.
Claim 9. For every π ∈ Π n and for every indifference set I, I ∩ Kπ is contained in a
hyperplane.
Proof. Let I be an indifference set, and let V ⊂ Kπ be an open neighborhood in I that
/ H . By the above
is contained in a hyperplane H. Let x ∈ I ∩ Kπ , and suppose that x ∈
discussion, there is a neighborhood V ⊂ I ∩ Kπ close to x that is contained in a hyperplane
H = H , and there is a permutation π such that π (V ) ⊂ H (see Fig. 3). We show next
that π (V ) ⊂ H .
As was argued above, we may assume, without loss of generality, that Conv(V ∪π (V )) ⊂
H ∩ I , and its dimension is n − 1. Denote this set A. By Claim 4, D ⊂
/ H . Therefore, by
Claim 8, we can assume, without loss of generality, that π = πm · · · π1 , such that
1. For every k, π k permutes only two numbers; and
2. For every y ∈ V and y ∈ π (V ), the chord [y, y ] goes sequentially through the sections
Kπ , Kπ1 π , . . . , Kπm−1 ···π1 π , Kπm ···π1 π = Kπ π .
We now show by induction that πk · · · π1 (V ) ⊂ H . We prove the case of k = 1, the other
steps are similarly proved. Let H = {y : a·y = c}. Then π1 (V ) ⊂ π1 (H ) = {y : π1 (a)·y =
Fig. 2. Proof of Claim 7.
Z. Safra, U. Segal / Journal of Mathematical Economics 35 (2001) 547–562
559
Fig. 3. Proof of Claim 9.
c}. Denote a = π1 (a) and assume, for simplicity, that π1 (1) = 2 and π1 (2) = 1. Then
a1 = a2 , a2 = a1 , and ai = ai for i = 3, . . . , n. Pick z ∈ A ∩ Int(Kπ ) close to the common
boundary of Kπ and Kπ1 π . Consider now the number λ such that π1 (z) + λe ∈ A. Since
π1 (z) ∈ I , it follows by Claim 4 that λ = 0. Hence π1 (z) ∈ H , and since z1 = z2 , it follows
that a1 = a2 . Therefore, π1 (V ) ⊂ H , and, by induction, π (V ) ⊂ H .
Since π (V ) and π (V ) are both in H, V and V are contained in the same hyperplane.
Since V ⊂ H , so is V . A contradiction, hence x ∈ H .
It now follows that for all π, I ∩Kπ is the intersection of a hyperplane H = {x : a π ·x = λ}
with Kπ , where λ satisfies λe ∈ I . Define
sλ =
aπ
id
||a π ||
id
where, as before, π id is the identity permutation. Obviously, sλ is uniquely defined.
λ
To complete the proof, weneed to show
k thatλs satisfies the requirement that for every
k
λ
k = 1, . . . , n and λ < λ, i=1 si ≤ i=1 si . Suppose to the contrary that for some k
and λ < λ, ki=1 siλ > ki=1 siλ . Let
k
λ i=1 si (λ − λ)
k λ
λ
i=1 si −
i=1 si
x 2 = λ + k
and
λ − λ
k λ < x2
λ
i=1 si −
i=1 si
x 1 = x 2 − k
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Z. Safra, U. Segal / Journal of Mathematical Economics 35 (2001) 547–562
Observe that
k
k
λ
λ
si x 1 + 1 −
si x2 = λ
i=1
and
i=1
k
k
λ
λ
si x1 + 1 −
si x2 = λ
i=1
i=1
In other words, λe ∼ λ e, a contradiction to the monotonicity of along the main diagonal
(Claim 4).
7. Proof of Theorem 3
Clearly, (2) ⇒ (1), so we prove here that (1) ⇒ (2). By Claim 1, either for all λ >
λ , λ λ or for all λ > λ , λ λ. To simplify the arguments, we assume the first
throughout these proof.
Let X n be the set of random variables (x1 , S1 ; . . . ; x2n , S2n ), where 4 for every i =
1, . . . , 2n , Pr(Si ) = 2−n . Following Eq. (5), there exists a function gn such that preferences
over X n can be represented by
V (X) = x dg n (FX (x))
(6)
where the function gn is defined as in Eq. (4).
Fact 3. For every n > n and i = 0, . . . , 2n ,
n −n
i
n
n i2
=g
g
n
2
2n
Proof. The claim follows from the fact that Xn ⊂ Xn .
Claim 10. The functions g n are uniformly bounded. That is, there exists ¯ such for all
¯
p ∈ [0, 1] and for all n, |g n (p)| ≤ .
¯
Proof. By Eq. (4), it is enough to show that for every i = 0, . . . , 2n , |g n (i/2n )| < .
Suppose that there exists a sequence {in } such that g n (in /2n ) is unbounded, say it goes to
∞. By Eq. (6),
in
in
n in
→ −∞
0, n ; 1, 1 − n ∼ 1 − g
2
2
2n
4 Observe the slight change in notations. In previous section, n stood for the number of different possible outcomes,
while here the number of different possible outcomes is 2n .
Z. Safra, U. Segal / Journal of Mathematical Economics 35 (2001) 547–562
561
Without loss of generality, in /2n → p. By DE there exists λ such that (0, p; 1, 1 − p) ∼ λ,
a violation of assumption C.
Claim 11. For every p, the limit g n (p) exists.
Proof. For every n, g n (0) = 0 and g n (1) = 1, so let p ∈ (0, 1). Since g n (p) is bounded
(see Claim 10), there is a subsequence {nj } such that g nj (p) → . Suppose that there exists
a sequence mj such that g mj (p) → . For every n, let i(n, p) be the number such that
i(n, p)/2n ≤ p < [i(n, p) + 1]/2n . The four sequences {i(nj , p)/2nj }, {[i(nj , p) +
1]/2nj }, {i(mj , p)/2mj }, and {[i(mj , p) + 1]/2mj } all converge to p. By the continuity assumption, it follows that the four sequences {g nj (i(nj , p)/2nj )}, {g nj ([i(nj , p) +
1]/2nj )}, {g mj (i(mj , p)/2mj )}, and {g mj ([i(mj , p) + 1]/2mj )} all converge to the same
limit. Since g nj (p) ∈ [g nj (i(nj , p)/2nj ), g nj ([i(nj , p) + 1]/2nj )] and g mj (p) ∈ [g mj
(i(mj , p)/2mj ), g mj ([i(mj , p) + 1]/2mj )], it follows that = .
Define a function g : [0, 1] → R by g(p) = lim g n (p). Since the functions are uniformly
bounded, the function gn is bounded. Observe that for every n and i = 0, . . . , 2n , g(i/2n ) =
g n (i/2n ). Therefore, for all X ∈ Xn ,
(7)
C(X) = x dg(FX (x))
(As before, C(X) is that number λ such that X ∼ λ). Denote by K the set of points {i/2n :
i = 0, . . . , 2n , n = 1, 2, . . . }.
Claim 12. The function g is continuous.
Proof. Let pm → p. Since g is bounded, we can assume, without loss of generality, that
g(pm ) → μ. Suppose that μ = g(p). We have seen in the proof of Claim 11 that for every
p∗ ,
∗
i(n, p∗ )
n i(n, p ) + 1
=
limg
g(p ∗ ) = limg n
2n
2n
Take a sequence nm such that |g nm (i(nm , pm )/2nm )−g(pm )| < (1/m). Clearly (i(nm , pm )/
2nm ) → p and g nm (i(nm , pm )/2nm ) → μ. On the other hand, the sequence i(n, p)/2n → p
and the sequence g n (i(n, p)/2n ) → g(p). We obtain that both i(nm , pm )/2n → p and
i(nm , p)/2nm → p. But
i(nm , pm )
g nm
→μ
2 nm
which means, by Theorem 2, that (0, p; 1, 1 − p) ∼ μ, while
n i(n, p)
→ g(p)
g
2n
which means that (0, p; 1, 1 − p) ∼ g(p), a contradiction.
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Z. Safra, U. Segal / Journal of Mathematical Economics 35 (2001) 547–562
We now show that x dg(FX (x)) represents the preferences . Pick X ∈ X , and let
Xn → X such that for every n, Xn ∈ Xn . Since X is bounded, it is possible to pick a
sequence
such that all the random variables Xn are uniformly bounded. By Eq. (7), Xn ∼
x dg(FXn (x)).
Since Xn → X, it follows by the continuity assumption that C(Xn ) →
C(X). Hence x dg(Fxn (x)) converges, and to C(X). Since the function g is continuous
on a closed segment it is uniformly continuous, and since the random variables
Xn are
(x))
→
x
dg(F
uniformly
bounded
and
converge
to
X
it
follows
that
x
dg(F
X
X (x)),
n
hence x dg(FX (x)) = C(X).
Acknowledgements
We thank the referee for his very helpful comments and SSHRCC and the Israel Institute
of Business Research for financial support.
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