Axiomatization of the Choquet integral for 2-dimensional heterogeneous product sets
aa r X i v : . [ q -f i n . E C ] M a r Axiomatization of the Choquet integral for 2-dimensionalheterogeneous product sets
Mikhail Timonin a a Queen Mary University of LondonMile End Road London, UK, E1 4NS+442078826139
Abstract
We prove a representation theorem for the Choquet integral model. The preference re-lation is defined on a two-dimensional heterogeneous product set X = X × X whereelements of X and X are not necessarily comparable with each other. However, makingsuch comparisons in a meaningful way is necessary for the construction of the Choquetintegral (and any rank-dependent model). We construct the representation, study itsuniqueness properties, and look at applications in multicriteria decision analysis, state-dependent utility theory, and social choice. Previous axiomatizations of this model,developed for decision making under uncertainty, relied heavily on the notion of comono-tocity and that of a “constant act”. However, that requires X to have a special structure,namely, all factors of this set must be identical. Our characterization does not assumecommensurateness of criteria a priori, so defining comonotonicity becomes impossible. Keywords:
Choquet integral, Decision theory, MCDA
1. Introduction
Rank-dependent models appeared in axiomatic decision theory in reply to the criticismof Savage’s postulates of rationality (Savage, 1954). The renowned Ellsberg paradox(Ellsberg, 1961) has shown that people can violate Savage’s axioms and still considertheir behaviour rational. First models accounting for the so-called uncertainty aversionobserved in this paradox appeared in the 1980s, in the works of Quiggin (1982) andothers (see (Wakker, 1991a) for a review). One particular generalization of the expectedutility model (EU) characterized by Schmeidler (1989) is the Choquet expected utility(CEU), where probability is replaced by a non-additive set function (called capacity) andintegration is performed using the Choquet integral.Since Schmeidler’s paper, various versions of the same model have been characterizedin the literature (e.g. (Gilboa, 1987; Wakker, 1991b)). CEU has gained some momentumin both theoretical and applied economic literature, being used mainly for analysis ofproblems involving Knightian uncertainty. At the same time, rank-dependent models, inparticular the Choquet integral, were adopted in multiattribute utility theory (MAUT)(Keeney and Raiffa, 1976). Here the integral gained popularity due to the tractability ofnon-additive measures in this context (see (Grabisch and Labreuche, 2008) for a review).
URL: [email protected] (Mikhail Timonin)
Preprint submitted to Elsevier he model permitted various preferential phenomena, such as criteria interaction, whichwere impossible to reflect in the traditional additive models.The connection between MAUT and decision making under uncertainty has beenknown for a long time. In the case when the number of states is finite, which is assumedhereafter, states can be associated with criteria. Accordingly, acts correspond to multicri-teria alternatives. Finally, the sets of outcomes at each state can be associated with thesets of criteria values. However, this last transition is not quite trivial. It is commonlyassumed that the set of outcomes is the same in each state of the world (Savage, 1954;Schmeidler, 1989). In multicriteria decision making the opposite is true. Indeed, considerpreferences of consumers choosing cars. Each car is characterized by a number of features(criteria), such as colour, maximal speed, fuel consumption, comfort, etc. Apparently,sets of values taken by each criterion can be completely different from those of the oth-ers. In such context the ranking stage of rank-dependent models, which in decision underuncertainty involves comparing outcomes attained at various states, would amount tocomparing colours to the level of fuel consumption, and maximal speed to comfort.Indeed, the traditional additive model (Debreu, 1959; Krantz et al., 1971) only im-plies meaningful comparability of units between goods in the bundle, but not of theirabsolute levels. However, in rank-dependent models such comparability seems to be anecessary condition. This paper develops a characterization of the Choquet integral fortwo-dimensional sets with comparability (commensurateness) of the criteria not assumeda priori.Let X = X × X be a (heterogeneous) product set and < a binary relation definedon this set. In multiattribute utility theory, elements of the set X are interpreted asalternatives characterized by two criteria taking values from sets X and X . In decisionmaking under uncertainty, the factors of the set X usually correspond to outcomes invarious states of the world, and an additional restriction X = X = Y is being made.Thus in CEU, the set X is homogeneous, i.e. X = Y n .Previous axiomatizations of the Choquet integral have been given for this specialcase of X = Y n (see (K¨obberling and Wakker, 2003) for a review of approaches) and itsvariant X = R n (see (Grabisch and Labreuche, 2008) for a review). Another approachusing conditions on the utility functions was proposed in (Labreuche, 2012). A conjointaxiomatization of the Choquet integral for the case of a general X was an open problemin the literature. One related result that should be mentioned is the recent conjointaxiomatization of another non-additive integral, the Sugeno integral ((Greco et al., 2004;Bouyssou et al., 2009)).The crucial difference between our result and previous axiomatizations is that the no-tions of “comonotonicity” and “constant act” are no longer available in the heterogeneouscase. Recall that two acts are called comonotonic in CEU if their outcomes have the sameordering. A constant act is plainly an act having the same outcome in every state of theworld. Apparently, since criteria sets X and X in our model can be completely disjoint,neither of the notions can be used anymore due to the fact that there does not exist ameaningful built-in order between elements of sets X and X . New axioms and prooftechniques must be introduced to deal with this complication.The paper is organized as follows. Section 2 defines the Choquet integral and looks atits properties. Section 3 states and explains the axioms. Section 4 gives the representationtheorem. Section 5 discusses the main result and its economic interpretations. The proof2f the theorem is presented in the Appendix; in particular, necessity of axioms is discussedin section A.11.
2. Choquet integral in MAUT
Let N = { , } be a set (of criteria) and 2 N its power set. Definition 1.
Capacity (non-additive measure, fuzzy measure) is a set function ν : 2 N → R + such that:1. ν ( ∅ ) = 0 ;2. A ⊆ B ⇒ ν ( A ) ≤ ν ( B ) , ∀ A, B ∈ N .In this paper, it is also assumed that capacities are normalized, i.e. ν ( N ) = 1 . Definition 2.
The
Choquet integral of a function f : N → R with respect to a capacity ν is defined as C ( ν, f ) = ∞ Z ν ( { i ∈ N : f ( i ) ≥ r } ) dr + Z −∞ [ ν ( { i ∈ N : f ( i ) ≥ r } ) − dr Denoting the range of f : N → R as ( f , . . . , f n ), the definition can be expressed as: C ( ν, ( f , . . . , f n )) = n X i =1 ( f ( i ) − f ( i − ) ν ( { j ∈ N : f j ≥ f ( i ) } )where f (1) , . . . , f ( n ) is a permutation of f , . . . , f n such that f (1) ≤ f (2) ≤ · · · ≤ f ( n ) , and f (0) = 0. Let < be a binary relation on the set X = X × X . ≻ , ≺ , , ∼ , are defined in theusual way. We say that < can be represented by a Choquet integral, if there exists acapacity ν and functions f : X → R and f : X → R , called value functions, such that: x < y ⇐⇒ C ( ν, ( f ( x ) , f ( x )) ≥ C ( ν, ( f ( y ) , f ( y )) . As seen in the definition of the Choquet integral, its calculation involves comparisonof the f i ’s to each other. It is not immediately obvious how this operation can have anymeaning in the MAUT context. It is well-known that comparing levels of value functionsfor various attributes is meaningless in the additive model (Krantz et al., 1971) (recallthat the origin of each value function can be changed independently). In the homogeneouscase X = Y n this problem is readily solved, as we have a single set of outcomes Y (inthe context of decision making under uncertainty). The required order is either assumedas given (Wakker, 1991a) or is readily derived from the ordering of the constant acts( α, . . . , α ) (Wakker, 1991b). Since there is a single outcome set, we also have a singlevalue (utility) function U : Y → R , and thus comparing U ( y ) to U ( y ) is perfectlysensible, since U represents the order on the set Y . None of these methods can be readilyapplied in the heterogeneous case. 3 .2. Properties of the Choquet integral Below are given some important properties of the Choquet integral:1. Functions f : N → R and g : N → R are comonotonic if for no i, j ∈ N we have f ( i ) > f ( j ) and g ( i ) < g ( j ). For all comonotonic f the Choquet integral reducesto the Lebesgue integral. In the finite case, the integral is accordingly reduced to aweighted sum.2. Particular cases of the Choquet integral (e.g. (Grabisch and Labreuche, 2008)). • If ν ( { } ) = ν ( { } ) = 1, then C ( ν, ( f , f )) = max( f , f ). • If ν ( { } ) = ν ( { } ) = 0, then C ( ν, ( f , f )) = min( f , f ). • If ν ( { } ) + ν ( { } ) = 1, then C ( ν, ( f , f )) = ν ( { } ) f + ν ( { } ) f Property 1 states that the set X can be separated into subsets corresponding toparticular orderings of the value functions. In the case of two criteria there are onlytwo such sets: { x ∈ X : f ( x ) ≥ f ( x ) } and { x ∈ X : f ( x ) ≥ f ( x ) } . Since theintegral on each of the sets is reduced to a weighted sum, i.e. an additive representation,we should expect many of the axioms of the additive conjoint model to be valid on thissubsets. This is the intuition behind several of the axioms given in the following section.
3. AxiomsDefinition 3.
A relation < on X × X satisfies triple cancellation , provided that forevery a, b, c, d ∈ X and p, q, r, s ∈ X , if ap bq, ar < bs , and cp < dq , then cr < ds . Definition 4.
A relation < on X × X is independent , iff for a, b ∈ X , ap < bp forsome p ∈ X implies that aq < bq for every q ∈ X ; and, for p, q ∈ X , ap < aq for some a ∈ X implies that bp < bq for every b ∈ X . A1. < is a weak order. A2.
Weak separability - for any a i p j , b i p j ∈ X such that a i p j ≻ b i p j , we have a i q j < b i q j for all q ∈ X j , for i, j ∈ { , } .The separability condition is weaker than the one normally used. In fact, it only rulesout a reversal of strict preference. Note, that the condition implies that for any a, b ∈ X either ap < bp or bp < ap for all p ∈ X (symmetrically for the second coordinate).Apparently, transitivity also holds: if ap < bp for all p ∈ X and bp < cp for all p ∈ X ,then ap < cp for all p ∈ X . This allows to introduce the following weak orders: Definition 5.
For all a, b ∈ X define < as a < b ⇐⇒ ap < bp for all p ∈ X . Define < symmetrically. Definition 6.
We call a ∈ X minimal if b < a for all b ∈ X , and maximal if a < b for all b ∈ X . Symmetric definitions hold for X . The condition first appeared in (Bliss, 1975), and in this form in (Mak, 1984) efinition 7. For any z ∈ X define SE z = { x : x ∈ X, x < z and z < x } , and NW z = { x : x ∈ X, x < z and z < x } . The “rectangular”cones SE z and NW z play a significant role in the sequel. A3.
For any z ∈ X , triple cancellation holds either on SE z or on NW z .The axiom says that the set X can be covered by “rectangular” cones, such that triplecancellation holds within each cone. We will call such cones “3C-cones”. The axiomeffectively divides X into subsets, defined as follows. Definition 8.
We say that • x ∈ SE if: – There exists z ∈ X such that z is not maximal and z is not minimal, triplecancellation holds on SE z , and x ∈ SE z , or – x is maximal or x is minimal and for no y ∈ SE x \ x triple cancellation holdson NW x ; • x ∈ NW if: – There exists z ∈ X such that z is not maximal and z is not minimal, triplecancellation holds on SE z , and x ∈ SE z , or – x is minimal or x is maximal and for no y ∈ NW x \ x triple cancellationholds on SE x .Define also Θ = NW ∩ SE . Definition 9.
We say that i ∈ N is essential on A ⊂ X if there exist x i x j , y i x j ∈ A , i, j ∈ N , such that x i x j ≻ y i x j . Essentiality of coordinates is discussed in detail in Section A.3.
A4.
Whenever ap bq, ar < bs, cp < dq , we have that cr < ds , provided that either:a) ap, bq, ar, bs, cp, dq, cr, ds ∈ NW ( SE ), or;b) ap, bq, ar, bs ∈ NW and i = 2 is essential on NW and cp, dq, cr, ds ∈ SE orvice versa, or;c) ap, bq, cp, dq ∈ NW and i = 1 is essential on NW and cp, dq, cr, ds ∈ SE orvice versa.Informally, the meaning of the axiom is that ordering between preference differences(“intervals”) is preserved irrespective of the “measuring rods” used to measure them.However, contrary to the additive case this does not hold on all X , but only when eitherpoints involved in all four relations lie in a single 3C-cone, or points involved in tworelations lie in one 3C-cone and those involved in the other two in another.5 Whenever ap bq, cp < dq and ay ∼ x π ( a ) , by ∼ x π ( b ) , cy ∼ x π ( c ) , dy ∼ x π ( d ), and also eπ ( a ) < f π ( b ), we have eπ ( c ) < f π ( d ), for all ap, bq, cp, dq ∈ NW or SE provided coordinate i = 1 is essential on the subset which contains thesepoints, ay , by , cy , dy ∈ NW or SE , x π ( a ) , x π ( b ) , x π ( c ) , x π ( d ) ∈ NW or SE provided coordinate i = 2 is essential on the subset which contains these points, eπ ( a ) , f π ( b ) , eπ ( c ) , f π ( d ) ∈ NW or SE . Same condition holds for the other dimen-sion symmetrically.The formal statement of A5 is rather complicated, but it simply means that the orderingof the intervals is preserved across dimensions. Together with A4 the conditions aresimilar to Wakker’s trade-off consistency condition (Wakker, 1991a). The axiom bearseven stronger similarity to Axiom 5 (compatibility) from section 8.2.6 of (Krantz et al.,1971). Roughly speaking, it says that if the interval between c and d is larger thanthat between a and b , then projecting these intervals onto another dimension by meansof the equivalence relations must leave this order unchanged. We additionally requirethe comparison of intervals and projection to be consistent - meaning that quadruplesof points in each part of the statement lie in the same 3C-cone. Another version ofthis axiom, which is going to be used frequently in the proofs, is formulated in terms ofstandard sequences in Lemma 3. A6.
Bi-independence : Let ap, bp, cp, dp ∈ SE ( NW ) and ap ≻ bp . If for some q ∈ X alsoexist cq ≻ dq , then cp ≻ dp . Symmetric condition holds for the second coordinate.This axiom is similar to “strong monotonicity” in (Wakker, 1991a). We analyze itsnecessity and the intuition behind it in section A.3. A7.
Both coordinates are essential on X . A8.
Restricted solvability : if x i a j < y < x i c j , then there exists b : x i b j ∼ y , for i, j ∈{ , } . A9.
Archimedean axiom: for every z ∈ NW ( SE ) every bounded standard sequencecontained in NW z ( SE z ) is finite. Structural assumption..
For no a, b ∈ X we have ap ∼ bp for all p ∈ X . Similarly, forno p, q ∈ X we have ap ∼ aq for all a ∈ X . If such points exist, say ap ∼ bp for all p ∈ X , then we can build the representation for a set X ′ × X where X ′ = X \ a , andlater extend it to X by setting f ( a ) = f ( b ). X is order dense.. Whenever x ≻ y there exists z such that x ≻ z ≻ y . From this andrestricted solvability immediately follows that < i is order dense as well, in other words,whenever a i p j ≻ b i p j there exists c ∈ X i such that a i p j ≻ c i p j ≻ b i p j , for i, j ∈ N . ”Closedness” of SE and NW .. Finally, we extend the set X as follows. Whenever exist ap NW and bp SE , there exist also cp ∈ Θ. Similarly, whenever exist ap NW and aq SE , there exist also ar ∈ Θ. 6 .1. Discussion of axioms
Roughly speaking, for two dimensional sets the Choquet integral can be characterizedby saying that X is divided into two subsets such that < on each of them has an additiverepresentation, while the intersection of these subsets (in the representation) is the line { x : f ( x ) = f ( x ) } . In the previous characterizations locating these subsets wasstraightforward, as they are nothing else but the comonotonic subsets of X . In thispaper we take a different approach. Instead, we state that X can be separated intotwo subsets without imposing any additional constraints on their location and then useadditional axioms to characterize the intersection of these subsets and to show that it ismapped to the line { x : f ( x ) = f ( x ) } .Our axioms aim to reflect the main properties of the Choquet integral. The first oneis that the set X can be divided into subsets, such that within every such subset the pref-erence relation can be represented by an additive function. The axiom ( A3 ) we introduceis similar to the “2-graded” condition previously used for characterizing of MIN/MAXand the Sugeno integral ((Greco et al., 2004; Bouyssou et al., 2009)). At every point z ∈ X it is possible to build two “rectangular cones”: { x : x < z and z < x } , and { x : x < z and z < x } . The axiom states that triple cancellation must then hold onat least one of these cones.The second property is that the additive representations on different subsets are in-terrelated, in particular trade-offs between criteria values are consistent across subsetsboth within the same dimension and for different ones. This is reflected by two axioms( A4, A5 ), similar to the ones used in (Wakker, 1991b) and (Krantz et al., 1971) (section8.2). One, roughly speaking, states that triple cancellation holds across cones, while theother says that the ordering of intervals on any dimension must be preserved when theyare projected onto another dimension by means of equivalence relations.These axioms are complemented by a new condition called bi-independence ( A6 ) andweak separability ( A2 ) - which together reflect the monotonicity property of the integral.Standard essentiality,“comonotonic” Archimedean axiom and restricted solvability( A7,A8,A9 ) complete the list. Finally, < is supposed to be a weak order, and X isorder dense ( A1 ).Our most important axioms - A3,A4,A5,A6 , are not only sufficient, but also nec-essary. Necessity and detailed analysis of A6 is given in Section A.3, necessity of A4 and A5 is proved in Section A.11, whereas necessity of A3 is immediate (in the rep-resentation one of the regions NW z and SE z is necessarily contained in a comonotonicsubset of R ). Necessity of some of the remaining axioms is well-known Wakker (1991a);Bouyssou and Pirlot (2004).
4. Representation theoremTheorem 1.
Let < be an order on X and let X be order dense and the structuralassumption hold. Then, if axioms A1 - A9 are satisfied, there exists a capacity ν andvalue functions f : X → R , f : X → R , such that < can be represented by the Choquetintegral: x < y ⇐⇒ C ( ν, ( f ( x ) , f ( x ))) ≥ C ( ν, ( f ( y ) , f ( y ))) , (1) for all x, y ∈ X . Moreover, ν is determined uniquely and value functions have the fol-lowing uniqueness properties: . If ν ( { } ) + ν ( { } ) = 1 , then for any functions g : X → R , g : X → R such that(1) holds with f i substituted by g i , we have f i ( x i ) = αg i ( x i ) + β i for some α > .2. If ν ( { } ) ∈ (0 , and ν ( { } ) ∈ (0 , and ν ( { } ) + ν ( { } ) = 1 , then for anyfunctions g : X → R , g : X → R such that (1) holds with f i substituted by g i , wehave f i ( x i ) = αg i ( x i ) + β for some α > .3. If ν ( { } ) ∈ (0 , , ν ( { } ) ∈ { , } , then for any functions g : X → R , g : X → R such that (1) holds with f i substituted by g i , we have : • f i ( x i ) = αg i ( x i ) + β , for all x such that f ( x ) < max f ( x ) and f ( x ) > min f ( x ) ; • f i ( x i ) = ψ i ( g i ( x i )) where ψ i is an increasing function, otherwise.4. If ν ( { } ) ∈ { , } , ν ( { } ) ∈ (0 , , then for any functions g : X → R , g : X → R such that (1) holds with f i substituted by g i , we have : • f i ( x i ) = αg i ( x i ) + β , for all x such that f ( x ) < max f ( x ) and f ( x ) > min f ( x ) ; • f i ( x i ) = ψ i ( g i ( x i )) where ψ i is an increasing function, otherwise.5. If ν ( { } ) = ν ( { } ) = 0 or ν ( { } ) = ν ( { } ) = 1 , then for any functions g : X → R , g : X → R such that (1) holds with f i substituted by g i , we have : f i ( x i ) = ψ i ( g i ( x i )) where ψ i are increasing functions such that f ( x ) = f ( x ) ⇐⇒ g ( x ) = g ( x ) .4.1. Uniqueness properties imply commensurateness As the uniqueness part of Theorem 1 states, unless < can be represented by anadditive functional on all of X (Case 1), the representation implies commensurateness of levels of utility functions defined on different factors of the product set. Indeed, we havethat if f ( x ) ≥ f ( x ) in one representation, then necessarily g ( x ) ≥ g ( x ) in anotherone. This is a much stronger uniqueness result in comparison to the traditional additivemodels. In Section 5 we discuss some economic implications of this. This paper provides a characterization of the Choquet integral for two-dimensionalsets, which allows to have simpler proofs. We believe that an extension to n dimensionsis mostly a technical task. Axiom A3 would be separated into two conditions. Oneis similar to the current A3 , and holds for any pair of dimensions with the remainingcoordinates fixed, and the other is acyclicity of the absence of additivity on the n-criteriacounterparts of regions NW z and SE z in between pairs of coordinates. Just as in thepresent paper, stronger uniqueness would be due to the lack of additivity. The remainingdifferences are technical. 8 . Applications In multicriteria context our result implies that the decision maker constructs a one-to-one mapping between elements of criteria sets (their subsets to be precise). Someauthors interpret this by saying that criteria elements sharing the same utility valuespresent the same level of “satisfaction” for the decision maker (Grabisch and Labreuche,2008). Technically, such statements are meaningful, in the sense that permissible scaletransformations do not render them ambiguous or incorrect, unless the representationis additive. However, the substance of statements like “ x on criterion 1 is at least asgood as x on criterion 2” (which would correspond to f ( x ) ≥ f ( x )) is not easy tograsp. Perhaps, it is possible to think about workers performing various tasks within asingle project. From the perspective of a project manager, achievements of various work-ers, serving as criteria in this example, can be level-comparable despite being physicallydifferent, if the project has global milestones (i.e. scale) which are mapped to certainpersonal milestones for every involved person. The novelty of our characterization isthat this scale is not given a priori. Instead, we only observe preferences of the projectmanager and infer all corresponding mappings from them. State-dependent utility concept is evoked when the nature of the state itself is ofsignificance to the decision-maker. One popular example is healthcare, where variousoutcomes can have major effects on the personal value of the insurance premium (Karni,1985). In the state-dependent context preferences of the decision maker are given bya binary relation on a set X = Y . However, unlike in CEU, there exist two (by thenumber of states) utility functions U : Y → R and U : Y → R , so that x < y ⇐⇒ C ( ν, ( U ( x ) , U ( x ))) ≥ C ( ν, ( U ( x ) , U ( x ))), where x, y are acts, and x , x , y , y arerespectively outcomes of x and y in each of the states (1 , { z : z ∈ X, U ( z ) ≥ U ( z ) } and { z : z ∈ X, U ( z ) ≤ U ( z ) } do not necessarilycorrespond to comonotonic regions of X anymore. Constant acts also do not have anyspecial value in such context, since U ( x ) and U ( x ) are not necessarily equal. However,our characterization allows to construct this model. Moreover, as a further generalization,in our framework sets of outcomes in every state can be completely disjoint as well. If we think of the set N as of a society with two agents, then X is the set of allpossible welfare distributions. Moreover, contrary to the classical scenario, agents couldbe receiving completely different goods, for example X might correspond to healthcareoptions, whereas X to various educational possibilities. In this case it is not a trivial taskto build a correspondence between different options across agents. Our result basicallystates that, provided the preferences of the social planner abide by the axioms given inSection 3, the decisions are made as if the social planner has associated cardinal utilitieswith the outcomes of each agent which are unit and level comparable (cardinal fullycomparable or CFC in terms of Roberts (1980)). Such approach is not conventional insocial choice problems, where the global (social) ordering is usually not considered asgiven. Instead, the conditions are normally given on individual utility functions and9he “aggregating” functional that is used to derive the global ordering. However, oneof the important questions in social choice literature is that of the interpersonal utilitycomparability and whether it is justifiable to assume it or not. Our results show that incase the global ordering of alternatives made by the society (or the social planner) satisfycertain conditions, it is in principle possible to have individual preferences representedby utility functions that are not only unit but also level comparable between each other. Appendix
Subsequent sections are organized as follows. Section A.1 contains a brief sketchof the proof. Section A.3 investigates monotonicity properties, Sections A.2, A.4- A.9contain the main body of the proof: construction of capacity and value functions, SectionA.10 analyses uniqueness of the obtained representation, finally, necessity of the axiomsis shown in Section A.11.
A.1. Proof sketch
1. Define extreme points of SE and NW and temporarily remove them from X (Sec-tion A.4).2. Take any point z , show that there exists an additive representation for < on NW z if z ∈ NW or SE z if otherwise (Section A.4).3. Having built additive representations for < on SE z and SE z , show that thereexists an additive representation on SE z ∪ SE z (Section A.4).4. Cover all SE with 3C-cones, and show that the joint representation, call it V SE ,can also be extended to cover all SE (Section A.4).5. Perform steps 2 and 3 for NW and obtain V NW (Section A.4).6. Align and scale V SE and V NW such that V SE = V NW on the common domain, and V SE = λV NW on their common domain (Section A.5).7. Pick two points r , r from Θ and set r as a common zero. Set V SE ( r ) = 1 anddefine φ ( x ) = V SE ( x ) , φ ( x ) = V SE ( x ) /V SE ( r ) (Section A.5).8. Show that for all x ∈ X we have φ ( x ) = φ ( x ) iff x ∈ Θ unless < can berepresented by an additive functional on all X (Section A.6).9. Representations now are φ + kφ on SE and φ + λkφ on NW (Section A.6).10. Rescale so that “weights” sum up to one : k φ + k k φ , λk φ + λk λk φ (SectionA.6).11. Extend the representation to the extreme points (Section A.7).12. Show that < can be represented on X by these two representations (Section A.8).13. Show that < can be represented by the Choquet integral (Section A.9).10 .2. Technical lemmasLemma 1. If < satisfies triple cancellation then it is independent.Proof. ap ap, aq < aq, ap < bp ⇒ aq < bq. Lemma 2. X = SE ∪ NW .Proof. Assume x SE , x NW . First assume that x is such that its coordinates arenot maximal or minimal. Then, there does not exist z such that x ∈ SE z and triplecancellation holds on SE z . At the same time, there does not exist z such that x ∈ NW z and triple cancellation holds on NW z . This implies that triple cancellation does not holdon SE x or NW x (otherwise we could have taken z = x ). This violates A3 .Now assume x is maximal. x SE implies that there exists y ∈ SE x such thattriple cancellation holds on NW y . But then x ∈ NW , a contradiction. Other cases aresymmetrical. Lemma 3.
Axiom A5 implies the following condition. Let { g ( i )1 : g ( i )1 y ∼ g ( i +1)1 y , g ( i )1 ∈ X , i ∈ N } and { h ( i )2 : x h ( i )2 ∼ x h ( i +1)2 , h ( i )2 ∈ X x , i ∈ N } be two standard sequences,each entirely contained in NW or SE . Assume also, that there exist z , z ∈ X , p, q ∈ X , a, b ∈ X such that g ( i )1 p, g ( i )1 q ∈ NW or SE , and ah ( i )2 , bh ( i )2 ∈ NW or SE for all i ,and g ( i )1 p ∼ bh ( i )2 and g ( i +1)1 p ∼ bh ( i +1)2 , then for all i ∈ N , g ( i )1 p ∼ bh ( i )2 .Proof. The proof is very similar to the one from Krantz et al. (1971) (Lemma 5 in section8.3.1). Assume wlog that { g ( i )1 : g ( i )1 y ∼ g ( i +1)1 y } is an increasing standard sequence on X , which is entirely in SE , whereas { h ( i )1 : x h ( i )1 ∼ x h ( i +1)1 } is an increasing standardsequence on X , and lies entirely in NW . Assume also for some k it holds g ( k )1 y ∼ x h ( k )2 , g ( k +1)1 y ∼ x h ( k +1)2 . We need to show that g ( i )1 y ∼ x h ( i )2 for all i . We will showthat g ( k +2)1 y ∼ x h ( k +2)2 from which everything holds by induction.Assume x h ( k +2)2 ≻ g ( k +2)1 y . Since the sequences are increasing, by restricted solv-ability exists g ∈ X such that g ( k +2)1 y ∼ x g . By A5’ , g ( k )1 y ∼ g ( k +1)1 y , g ( k +1)1 y ∼ g ( k +2)1 y , x h ( k )2 ∼ x h ( k +1)2 imply x h ( k +1)2 ∼ x g . By definition of { h ( i )2 } , x h ( k +1)2 ∼ x h ( k +2)2 . Thus, x h ( k +2)2 ∼ x g and by independence x h ( k +2)2 ∼ x g , hence also g ( k +2)1 y ∼ x h ( k +2) , a contradiction. The case x h ( k +2)2 ≺ g ( k +2)1 y is symmetrical. Showingthat g ( k − y ∼ x h ( k − can be done in a similar fashion. Lemma 4.
The following statements hold: • If NW ( SE ) has only X essential, then for all x ∈ NW ( SE ) there exists y ∈ X such that x y ∈ Θ . • If NW ( SE ) has only X essential, then for all x ∈ NW ( SE ) there exists y ∈ X such that y x ∈ Θ .Proof. Immediately follows from the structural assumption and closedness of NW ( SE ).11 .3. Essentiality and monotonicity In what follows the essentiality of coordinates within various SE z ( NW z ) is criti-cal. The central mechanism to guarantee consistency in number of essential coordinateswithin various 3C-cones is bi-independence which is closely related to comonotonic strongmonotonicity of Wakker (1989).In the Choquet integral representation problem for a heterogeneous product set X = X × X , strong monotonicity is actually a necessary condition because of the following.Assume ap, bp, cp, dp ∈ SE and ap ≻ bp, cp ∼ dp . Assume also there exist cq, dq ∈ NW such that cq ≻ dq . Then, provided the representation exists, we get α f ( a ) + α f ( p ) > α f ( b ) + α f ( p ) α f ( c ) + α f ( p ) = α f ( d ) + α f ( p ) β f ( c ) + β f ( q ) > β f ( d ) + β f ( q ) . The first inequality entails α = 0. From this and the following equality follows f ( c ) = f ( d ), which contradicts with the last inequality. Thus cq ≻ dq implies cp ≻ dp but onlyin the presence of ap ≻ bp in the same “region” ( SE or NW ). This is also the reasonbehind the name we gave to this condition - “bi-independence”. Lemma 5.
Pointwise monotonicity.If for all i, j ∈ N we have a i x j < a i y j for all a i ∈ X i , then x < y .Proof. x = x x < x y < y y = y .Bi-independence, together with the structural assumption also implies some sort of“strong monotonicity”. Lemma 6. If X is essential on SE ( NW ) , a < b iff ap ≻ bp for all ap, bp ∈ NW .Proof. Let X be essential on SE . By the structural assumption, a < b implies existenceof some q ∈ X such that aq ≻ bq . If aq, bq ∈ SE the result follows by independence.If aq, bq ∈ NW the result follows by bi-independence. If bq ∈ NW , aq ∈ SE , then byclosedness assumption there exists cq ∈ Θ, and either bq ≻ cq , or cq ≻ aq . The resultfollows by independence or bi-independence.Conceptually, Lemma 6 implies that if a coordinate is essential on some 3C-cone NW z ( SE z ), then it is also essential on NW x ( SE x ) for all x ∈ NW ( SE ). This allows usto make statements like “coordinate i is essential on NW ”. A.4. Building additive value functions on NW and SE
In this section we assume that SE ( NW ) has two essential coordinates.12 .4.1. Covering SE and NW with maximal SE z and NW z In the sequel we could have covered areas SE and NW by sets SE z ( NW z ) for all z ∈ SE ( NW ), but it is convenient to introduce the following lemma. Lemma 7.
For every x ∈ SE there exists z ∈ Θ such that SE x ⊂ SE z . Accordingly, forevery y ∈ NW there exists z ∈ Θ such that NW y ⊂ NW z .Proof. Take x ∈ SE such that x NW . If there exists y ∈ NW x such that y ∈ NW ,then by closedness assumption there must exist either ay ∈ Θ and x ∈ SE ay or x p ∈ Θand x ∈ SE x p . If such y does not exist, X = Θ. Other cases are symmetrical.It follows from Lemma 7 that SE = S z ∈ Θ SE z , while NW = S z ∈ Θ NW z . Comparingthis to definitions of SE and NW we are able to define also the following notions: Definition 10.
We write x ∈ SE ext and say that x ∈ X is extreme in SE if x ∈ Θ and [ x is minimal or x is maximal ] . We write x ∈ NW ext and say that x ∈ X is extreme in NW if x ∈ Θ and [ x is minimal or x is maximal ] . x ∈ X is extreme if itis extreme in SE or in NW . Note that contrary to the homogeneous case X = Y n , extreme points for SE and NW can be asymmetric, i.e. if a point z is extreme in SE it is not necessarily extremein NW . A.4.2. Representations within SE z In the following we will build an additive representation on SE . The case of NW issymmetric. We proceed by building representations on sets SE z for all z ∈ Θ \ SE ext (i.e.for all non-extreme points of Θ). Essential coordinates..
For now we assume that both coordinates are essential on NW and SE . Theorem 2.
For any z ∈ Θ \ SE ext there exists an additive representation of < on SE z : x < y ⇔ V z ( x ) + V z ( x ) ≥ V z ( y ) + V z ( y ) . Proof. SE z is a Cartesian product, < is a weak order on SE z , < satisfies triple cancellationon SE z , < satisfies Archimedean axiom on SE z , both coordinates are essential. It remainsto show that < satisfies restricted solvability on SE z .Assume that for some xa, y, xc ∈ SE z , we have xa < y < xc , hence exists b ∈ X : xb ∼ y . We need to show that xb ∈ SE z . If xb ∼ xa or xb ∼ xc , then the result isimmediate. Hence, assume xa ≻ xb ≻ xc . By definition, x ∈ SE z , if x < z , and z < b .For xb we need to check only the latter condition. It holds, since xa ≻ xb ≻ xc , and byweak separability a < b .Therefore all conditions for the existence of an additive representation are met(Krantz et al., 1971). 13 .4.3. Joining representations for different SE z (or NW z ) This section closely follows (Wakker, 1991a).
Theorem 3.
There exists an additive interval scale V SE on S SE z , with z ∈ Θ \ SE ext ,which represents < on every SE z .Proof. Choose the “reference” points - pick any r ∈ SE and any r , r ∈ SE r such that r p < r p for every p ∈ X . Set V r ( r ) = 0 , V r ( r ) = 0 , V r ( r ) = 1. Now, we alignrepresentations on the other sets SE z with the reference one. Assume that for some z ∈ Θ we have already obtained an additive representation V z on SE z . Observe that V z and V r are additive value functions for < on SE z ∩ SE r . Morevover SE z ∩ SE r = SE q ,where q = r if r < z and q = z if the opposite is true. Similarly, q = r if az < ar for all a ∈ X and q = z in the opposite case. Hence, uniqueness results fromKrantz et al. (1971) can be applied. In particular, this means that on SE z ∩ SE r we have V ri = αV zi + β i , so the functions are defined up to a common unit and location.We choose the unit and location of V z so that V z ( x ) = V r ( x ) for all x ∈ SE z ∩ SE r .Next, we choose the location of V z so that it coincides with V r on SE z ∩ SE r .Finally, we show that V si ( x i ) = V ti ( x i ) for any s, t ∈ Θ and x ∈ SE s ∩ SE t . Thisimmediately follows, since V s and V t coincide (with V r ) on SE s ∩ SE t ∩ SE r . Thisdefines their unit and locations, hence they also coincide on SE s ∩ SE t . Now define V SE as a function which coincides with V z i on the respective domains SE z i . By the aboveargument, this function is well-defined. Theorem 4.
Representation V SE obtained in Theorem 3 is globally representing on SE \ SE ext = S z ∈ Θ \ SE ext SE z .Proof. Let x < y . There can be two cases. First, assume that x < y , but y < x (orvice versa). In this case, x and y belong to the same SE z (e.g. SE x ) and therefore V SE is a valid representation.Next, assume that x j < j y j for all i, j ∈ N . Assume, that x ∈ SE s , y ∈ SE t . Observethat x y ∈ SE s ∩ SE t because by the made assumptions, x y ∈ SE x , x y ∈ SE y . Bypointwise monotonicity (Lemma 5) x x < x y < y y , hence V ( x ) + V ( x ) ≥ V ( x ) + V ( y ) ≥ V ( y ) + V ( y ), with first inequalities lying in SE s , and second in SE t . Thereverse implication is also true. A.5. Aligning V SE and V NW First we will show that it is not possible for the common domain of V SEi and V NWi for some i to contain a single point. A.5.1. Analysis of the common domain of V SE and V NW Lemma 8.
Let a < b , and for some p ∈ X we have a p, b p ∈ Θ . Define X a b = { x : x ∈ X , b < x < a } . Then, triple cancellation holds everywhere on X a b × X .Proof. All points in the below proof are from X a b × X . Let ax by, aw < bz, cx < dy .We will show that together with the assumptions of the Lemma, this implies cw < dz .14 a p fxywzq a b dc Figure 1: Lemma 8
The case when all points belong to SE or NW , or two pairs belong to SE and twoto NW is covered by A4 . Thus, assume wlog x < p , so that ax, cx ∈ NW and the re-maining points are in SE (Fig. 1). Assume also dp < cp and b < a . Assume also ax ≻ ap (hence by independence also cx ≻ cp ), bp ≻ by (hence also dp ≻ dy ), otherwise the resultimmediately follows by A4 (e.g. if ax ∼ ap , we can replace ax by ap and cx by cp in theassumptions of the lemma, which brings all points to SE ).1. ax by . bp ≻ by , hence ax ≺ bp . ax ≻ ap , hence bp ≻ by < ax ≻ ap , bp ≻ ap , therefore, by re-stricted solvability exists f p ∼ ax . Also, f p ≻ ap, bp ≻ f p .2. cx < dy . There can be two cases:a) If cx dp , then dp < cx ≻ cp , hence exists gp ∼ cx .b) cx ≻ dp .3. aw < bz .Solve for q : aw ∼ f q . By the results in point 1 and independence we have f w ≻ aw < bz ≻ f z , therefore by restricted solvability exists q : f q ∼ aw .4. Cases correspond to those in point 2 above:a) f p ∼ ax, gp ∼ cx, aw ∼ f q , hence by A4 cw ∼ gqf p by, gp < dy, f q < bz , hence by A4 gq < dz and cw < dz .b) ax ≺ bp, cx ≻ dp, aw < bz , hence by A4 cw < dz .From this it follows that it is impossible that for some i the common domain of V SEi and V NWi includes a single point. Let (wlog) i = 1 and a ∈ X be such a point.Apparently ap ∈ Θ for all a ∈ X . Then, from Lemma 8 it follows that SE = NW = X .15 .5.2. Aligning representations on SE and NW There can be four cases, depending on the number of essential coordinates on NW and SE :1. Both areas have two essential coordinates;2. One area has two essential coordinate, another has one essential coordinate;3. Both areas have one essential coordinates;4. An area does not have any essential coordinates.We start with the case where both coordinates are essential on NW and SE . A.5.2.1. Both coordinates are essential
Lemma 9.
Choose r ∈ X and r ∈ X from the common domain of V SE and V NW such that r < r , and set V SE ( r ) = V NW ( r ) = 0 and V SE ( r ) = V NW ( r ) = 1 . Then, V SE ( x ) = V NW ( x ) on all x from their common domain.Proof. This follows directly from A4 . Assume, we want to show that V SE ( y ) = V NW ( y )for some y ∈ X . Starting from r build any standard sequence on X in SE , say { α ( i )1 : α ( i )1 y s ∼ α ( i +1)1 y s } . Then, all α ( i )1 y n , α ( i )1 y n which are in NW also form a sequence in NW : if α ( i )1 y s ∼ α ( i +1)1 y s , α ( i +1)1 y s ∼ α ( i +2)1 y s and α ( i )1 y n ∼ α ( i +1)1 y n , for some y n , y n ∈ X ,then by A4 , necessarily α ( i +1)1 y n ∼ α ( i +2)1 y n .Now let 1 = V SE ( r ) ≈ n [ V SE ( y s ) − V SE ( y s )] V SE ( y ) ≈ m [ V SE ( y s ) − V SE ( y s )] ≈ mn . Such n and m exist by the Archimedean axiom. By the argument above we get1 = V NW ( r ) ≈ n [ V NW ( y n ) − V NW ( y n )] V NW ( y ) ≈ m [ V NW ( y n ) − V NW ( y n )] ≈ mn By denserangedness, approximations become exact in the limit, so we obtain V SE ( y ) = V NW ( y ) on all y ∈ X from their common domain. Lemma 10.
Assume, V SE is an additive representation of < on SE \ SE ext , and V NW is a representation on NW \ NW ext , with V SE and V NW scaled so that they have acommon zero and unit (as in Lemma 9). Then, V SE = λV NW on the common domain.Proof. By Lemma 9, V SE = V NW on the common domain. Assume V SE ( r ) = λ, V NW ( r ) = 1. We will now show that V SE ( x ) = λV NW ( x ) for all x ∈ X fromthe common domain of these functions. Construct a standard sequence within SE z , thistime on X . By A4 , it is also a sequence in NW . We obtain λ = V SE ( r ) ≈ n [ V SE ( x s ) − V SE ( x s )] V SE ( x ) ≈ m [ V SE ( x s ) − V SE ( x s )] ≈ λmn
16y the argument above we get1 = V NW ( r ) ≈ n [ V NW ( x n ) − V NW ( x n )] V NW ( x ) ≈ m [ V NW ( x n ) − V NW ( x n )] ≈ mn From this in the limit we obtain V SE ( x ) = λV NW ( x ) on all x ∈ X from the commondomain of V SE and V NW .At this point we can drop superscripts and say that we have representations V + V on SE and V + λV on NW . Fix two non-extreme points in Θ : r and r , such that r < r and r < r . If such points do not exist, then by Lemma 8 triple cancellationholds everywhere and < can be represented by an additive function (i.e. λ = 1). Rescale V and V so that V ( r ) = 0 , V ( r ) = 0 , V ( r ) = 1. Assume that after rescaling we get V ( r ) = k . Define φ ( x ) = V ( x ) /k , i.e. φ ( r ) = 1. Define φ ( x ) = V ( x ). Thuswe get representations φ + kφ on SE and φ + λkφ on NW . Finally rescale in thefollowing way: k φ + k k φ on SE and λk φ + λk λk φ on NW . We have thus definedthe following representations: φ SE ( x ) = 11 + k φ ( x ) + k k φ ( x ) φ NW ( x ) = 11 + λk φ ( x ) + λk λk φ ( x ) . (2)Note, that it follows that φ SE ( r ) = φ NW ( r ) = 1. A.5.2.2. One area has a single essential coordinate
Assume SE has two essential coordinates and NW only has X essential. After anadditive representation V SE has been built on SE , and re-scaled as in (2) we have values φ and φ for all points in SE , in particular those in Θ. Let φ NW ( x ) = φ ( x ) + 0 φ ( x )(in other words, set λ = 0 in (2)) for those x i where φ i are defined. By structuralassumption, bi-independence and additivity φ NW represents < on those points for whichit is defined. For example, let ap, bp ∈ NW be such that ap ≻ bp . Since both coordinatesare essential on SE by bi-independence we get also aq ≻ bq for all q ∈ X such that aq, bq ∈ SE . Additivity implies φ ( a ) > φ ( b ). For the remaining x ∈ X , i.e. for x ∈ X such that there are no points in Θ first coordinate of which is x , build a simpleordinal representation. Lemma 4 shows that values for all x ∈ X have already beendefined at this point. Other cases are similar. A.5.2.3. Both areas have a single essential coordinate
An interesting result is that A3 is sufficient for characterization of cases where both SE and NW have one essential coordinate. There are two cases in total:1. X is essential on NW , X is essential on SE ;2. X is essential on NW , X is essential on SE .We will need the following lemma. 17 emma 11. Let X be essential on NW and X be essential on SE or X be essentialon NW and X be essential on SE . Then, either for all x ∈ SE exists z ∈ Θ such that z ∼ x , or for all y ∈ NW exists z ∈ Θ such that z ∼ y .Proof. We only consider one case, others being symmetrical. Assume X is essential on SE and X is essential on NW . Assume also there exists x ∈ SE such that x ≻ z for all z ∈ Θ, in particular some maximal z max . We will show that this implies that there doesnot exist y ∈ NW such that y ≻ z or z ≻ y for all z ∈ Θ.Assume such y exists. Take x y . By A3 it belongs either to SE or NW . If itbelongs to NW , by closedness assumption exists x t ∈ Θ. We get x t ∼ x ≻ z max - acontradiction. If x y ∈ SE , exists ay ∈ Θ. We have y ∼ ay which contradicts both y ≻ z and z ≻ y for all z ∈ Θ.Finally, we need to show that Θ does not have gaps. Assume there exists y ∈ NW and z , z ∈ Θ such that z ≻ y ≻ z but there is no z ∈ Θ such that z ∼ y . Since only X is essential on NW , we get z y ∈ SE . By closedness assumption exists x ∈ X suchthat x y ∈ Θ. Since only X is essential we conclude x y ∼ y , which is a contradiction.Therefore, for every y ∈ NW there exists z ∈ Θ such that y ∼ z . Defining value functions..
Lemma 11 guarantees that for all points in SE or all pointsin NW exists an equivalent point in Θ. Assume for example that Θ is such that for all x ∈ SE exists z ∈ Θ. Assume also that X is essential on SE and X is essential on NW .Now define value functions φ : X → R and φ : X → R as follows. Choose φ to beany real-valued function such that φ ( x ) > φ ( y ) iff x < y . Now for all z from Θ set φ ( z ) = φ ( z ). Finally, extend φ to the whole X by choosing any function such that φ ( x ) > φ ( y ) iff x < y . Lemma 4 guarantees that the functions have been definedfor all x ∈ X and all x ∈ X . A.5.2.4. Areas without essential coordinates
Lemma 12. If A1 - A9 and the structural assumption hold, there can not be NW z ( SE z ) with no essential coordinates.Proof. Assume for some z ∈ Θ the set NW z has no essential coordinates. By bi-independence and the structural assumption it follows that there are no essential co-ordinates on any NW z . This implies (by A7 ) that both coordinates are essential on SE .Take ap, bp ∈ NW z . Apparently, ap ∼ bp . By structural assumption there must exist q ∈ X such that aq ≻ bq . It can’t be that aq, bq ∈ NW , hence aq, bq ∈ SE .By closedness assumption there exist w, z ∈ X such that aw, bz ∈ Θ. Also, since nocoordinate is essential in NW we have aw ∼ bz . Since aq ≻ bq it must be bz ≻ bw , sinceotherwise by strict monotonicity (Lemma 6) it can’t be that aw ∼ bz .By independence we have aw ≻ bw . By definition of NW and SE we have aw ∈ SE (since by weak separability a < b ) and aw ∈ NW (since by weak separability z < w ).Hence, by independence it must be az ≻ aw (since az ∈ SE ) and az ∼ aw (since az ∈ NW ). We have arrived at a contradiction. A.6. Properties of the intersection of SE and NWLemma 13.
For any non-extreme x ∈ X we have: x ∈ Θ ⇒ φ ( x ) = φ ( x ) , nless < can be represented by an additive function (i.e λ = 1 in (2)). For the case when both NW and SE have a single essential coordinate the resultholds by definition of φ i , so for the remainder of this section we assume that SE or NW has two essential coordinates. r α (1)1 β (1)1 X X x π ( x ) π ( x ) r π ( r ) π ( r ) Figure 2: Lemma 13
Proof.
We start with a case where both coordinates are essential on SE and NW . Assumealso x < r (without loss of generality, other cases are symmetrical and can be proved bythe same technique). We are going to show that x ∈ Θ ⇒ φ ( x ) = φ ( x ) or λ = 1.Assume for now that we can find the following solutions: • Solve for π ( r ) : π ( r ) r ∼ r . • Solve for π ( r ) : r π ( r ) ∼ r . • Solve for π ( x ) : π ( x ) r ∼ x . • Solve for π ( x ) : r π ( x ) ∼ x .Pick α (1)1 ∈ X such that r < α (1)1 , it exists by denserangedness. Solve for β (1)2 : α (1)1 r ∼ r β (1)2 , exists by restricted solvability.Now build an increasing standard sequence α ( i )1 : α (0)1 = r on X which lies in SE and an increasing standard sequence β ( i )2 : β (0)2 = r on X which lies in NW (see Fig.2). Since π ( r ) r ∼ r π ( r ), by A5 (Lemma 3) we have for some m : α ( m − r < π ( r ) r < α ( m )1 r ⇐⇒ r β ( m − < r π ( r ) < r β ( m )2 . φ ( r ) = φ ( r ) = 0) it follows that φ ( π ( r )) ≈ mφ ( α (1)1 ) , φ ( π ( r )) ≈ mφ ( β (1)2 ) and, since φ SE ( π ( r ) r ) = φ SE ( r ) = φ NW ( r ) = φ NW ( r π ( r )), we obtain: 11 + k mφ ( α (1)1 ) = λk λk mφ ( β (1)2 ) . (3)Similarly, π ( x ) r ∼ r π ( x ), so by A5 (Lemma 3) we have α ( n − r < π ( x ) r < α ( n )1 r ⇐⇒ r β ( n − < r π ( x ) < r β ( n )2 . From this follows that φ ( π ( x )) ≈ nφ ( α (1)1 ) , φ ( π ( x )) ≈ nφ ( β (1)2 ) and by (3) itfollows that φ SE ( π ( x ) r ) = φ NW ( r π ( x )). Hence11 + k φ ( x ) + k k φ ( x ) = 11 + λk φ ( x ) + λk λk φ ( x ) , and so φ ( x ) = φ ( x ) or λ = 1 (i.e. the structure is additive).We need to revisit the case where solutions mentioned in the beginning do not ex-ist. Consider Figure 3. Assume this time that there does not exist π ( r ) such that r ∼ π ( r ) r . If we choose the step in the standard sequence α ( i )1 small enough sothat there exist α ( k +1)1 , α ( k +2)1 such that α ( k +1)1 r < r r and α ( k +2)2 r < r r (which wecan do by non-maximality of r and denserangedness of < ), then we can “switch”from the standard sequence α ( i )1 on X to the standard sequence γ ( i )2 on X keepingthe same increment in value between subsequent members of the sequence. Indeed, γ ( k )2 : r γ ( k )2 ∼ α ( k )1 r and γ ( k +1)2 : r γ ( k +1)2 ∼ α ( k +1)1 r exist by monotonicity and restrictedsolvability, so does x : x γ ( k )2 ∼ r γ ( k +1)2 ∼ α ( k +1)1 r , and by A5 (Lemma 3) we get[ r γ ( k )2 ∼ r β ( k )2 , r γ ( k +1)2 ∼ r β ( k +1)2 ] ⇒ γ ( i )2 ∼ β ( i )2 for all i . Note that γ ( i )2 r are in SE for all i such that r < γ ( i )2 since r and r are in SE . By monotonicity and restrictedsolvability there exists i such that x r < x γ ( i +1)2 < r < x γ ( i )2 . Finally, the increment invalue is the same between members of α i and γ ( i )2 since α k ∼ γ ( k )2 and α k +1 ∼ γ ( k +1)2 . Theresult then follows as above.Finally, we look at the case where only one coordinate is essential on either NW or SE . First assume that X is essential on NW . We defined φ NW ( x ) = 0 φ ( x ) + φ ( x ).Definition implies φ NW ( r ) = 0 , φ NW ( r ) = 1. Build a standard sequence { α ( i )1 } on X from r to r (in case there exists a solution for r ∼ π ( r ) r , otherwise use theapproach detailed in the previous paragraph), setting α (0)1 = r . Take α (1)1 r and α (2)1 r .By restricted solvability there must exist β (1)2 and β (2)2 , such that α (1)1 r ∼ r β (1)2 and α (2)1 r ∼ r β (2)2 . By closedness assumption for β (1)2 , β (2)2 there must exist x , x such that x β (1)2 ∈ Θ, x β (2)2 ∈ Θ. Also, since only X is essential, we get x β (1)2 ∼ r β (1)2 , x β (2)2 ∼ r β (2)2 . By weak monotonicity and definition of SE , α (2)1 β (1)2 < α (2)1 r ∼ x β (2)2 < x β (1)2 ,hence by restricted solvability exists z : x β (2)2 ∼ z β (1)2 . By A5 then x β (1)2 ∼ z r . Byadditivity x β (2)2 ∼ z β (1)2 and x β (1)2 ∼ z r entail φ ( b ) − φ ( b ) = φ ( b ) − φ ( r ). Fromthis the result follows as in the proof above. If now X is essential on NW repeat theproof as above this time starting the sequence from r “towards” r .20 X r α (1)1 r γ ( k )2 α ( k )1 β ( k )2 α ( k +1)1 γ ( k +1)2 x β ( k +1)2 γ ( k +2)2 α ( k +2)1 β ( k +2)2 r y γ ( k +3)2 Figure 3: Lemma 13 - changing direction
Lemma 14.
The following statements hold or < has an additive representation:1. If ap ∈ Θ then for no b ∈ X holds bp ∈ Θ and also for no q ∈ X holds aq ∈ Θ .2. x ∈ SE ⇒ φ ( x ) ≥ φ ( x ) , y ∈ NW ⇒ φ ( y ) ≥ φ ( y ) .Proof.
1. Assume ap, bp ∈ Θ. Assume first, that SE or NW has two essential coordi-nates. If ap ∼ bp , then aq ∼ bq for all q ∈ X , which violates the structural assump-tion. Let ap ≻ bp . By denserangedness there exists also cp such that ap ≻ cp ≻ bp .If p is minimal, then by definition of SE , bp ∈ SE if for no c ∈ X such that c < b holds cp ∈ NW . This is obviously violated by ap . If p is maximal, then bydefinition of NW , ap ∈ NW if for no c ∈ X such that a < c holds cp ∈ SE , whichis violated by bp . Hence, p is not maximal or minimal. If a is maximal, then cp, bp are non-extreme, hence by Lemma 13, φ ( c ) = φ ( p ) = φ ( b ), hence bq ∼ cq for all q ∈ X . If b is minimal, ap, bp are non-extreme, hence φ ( a ) = φ ( p ) = φ ( c ), hence aq ∼ cq for all q ∈ X . In both cases structural assumption is violated.2. Pick any bq ∈ SE . By Lemma 7 there exists ap ∈ Θ such that bq ∈ SE ap , hence b < a and p < q . By Lemma 13 φ ( a ) = φ ( p ). We also have φ ( b ) ≥ φ ( a ) , φ ( p ) ≥ φ ( q ). The result follows. NW case is symmetric. A.7. Extending value functions to extreme points
Value functions for the case when both SE and NW have a single essential coordinatewere fully defined in Section A.5.2.3. Thus in what follows we will consider cases where SE or NW have two essential coordinates.21s indicated in (Wakker, 1991a), value functions might be driven to infinite values atthe maximal/minimal points of rank-ordered subsets, nevertheless not implying existenceof infinite standard sequences residing entirely within comonotonic cones. Put it anotherway, it might be not possible to “reach” a maximal/minimal point with a sequence lyingentirely in NW or SE . Yet another way to say it is that for some maximal/minimalpoint z , the set NW z ( SE z ) contains no standard sequences (see also (Wakker, 1991a)Remark 24).The cornerstone of this section is Lemma 13. It plays the same role as proportionalityof value functions plays in (Wakker, 1991a), effectively guaranteeing that both valuefunctions φ and φ are limited if maximal/minimal elements exist. Lemma 15.
Assume that SE has two essential coordinates. The following statementshold: • If there exist a maximal M ∈ X , φ is bounded from above. • If there exist a minimal m ∈ X , φ is bounded from below.Assume that NW has two essential coordinates. The following statements hold: • If there exist a minimal M ∈ X , φ is bounded from below. • If there exist a maximal m ∈ X , φ is bounded from above.Proof. We shall only prove the first one. First, notice that there must exist p ∈ X suchthat M p ∈ SE . Take x ∈ X and v , w ∈ X such that v < w , and x v ∈ SE .If such points cannot be found, X has an additive representation (all x ∈ NW ), andthe result follows. So we assume such points exist. By definition of SE x v it followsthat M w , x v , M v ∈ SE . Hence, we can evoke the argument from Wakker (1991a)Lemma 20.If M w x v then we have an upper bound: V ( M ) ≤ V ( x ) + V ( v ) − V ( w ). If M w ≻ x v then by monotonicity M v < M w ≻ x v and hence exists z ∈ X suchthat M w ∼ z v , hence z v < βw for all β ∈ X . Lemma 16. If x x ∈ Θ and x x is extreme, then lim z ∈ Θ ,z → x φ ( z ) = lim z ∈ Θ ,z → x φ ( z ) . Proof.
For the case when SE or NW have two essential variables the result follows fromLemma 13, otherwise it is by definition of φ i (see Section A.5.2.3). A.7.1. Extending value functions to extreme elements of Θ Extreme elements of Θ are the only representatives of maximal/minimal equivalenceclasses of SE ( NW ).. Lemma 17.
Let X be essential on SE . If there exists z ∈ Θ such that z is minimal,then x ≻ z for all x ∈ SE . If X is essential on SE and there exists z ∈ Θ such that z is maximal, then z ≻ x for all x ∈ SE . Similarly, if X is essential on NW and thereexists z ∈ Θ such that z is maximal, then z ≻ x for all x ∈ NW . If X is essential on NW and there exists z ∈ Θ such that z is minimal, then x ≻ z for all x ∈ NW . roof. We provide the proof just for one of the cases. Let NW have two essentialvariables. Assume z is maximal. Since z ∈ Θ, for all x ∈ NW holds z < x and bymaximality z < x . Hence, by Lemma 6, z ≻ x for all x ∈ NW . The case with theminimal z is symmetric. Uniqueness of definition of φ i at the extreme elements of Θ . . Lemma 18.
If both coordinates are essential on SE and NW the values of φ i for extreme x ∈ Θ are uniquely defined. Moreover, φ ( x ) = φ ( x ) .Proof. Assume, for example x x ∈ Θ and x is minimal. Then any z x such that z < x , belongs to SE , and any equivalence relation within SE involving z x uniquelydefines φ ( x ) (see (2)). Similarly, any x z such that z < x , belongs to NW , and anyequivalence relation within NW involving x z uniquely defines φ ( x ). By Lemma 16these values are equal. Lemma 19.
If both coordinates are essential on SE but only one on NW (or vice versa)the values of φ i for extreme x ∈ Θ can be set as follows: • If x x ∈ Θ , NW has two essential coordinates and x is maximal, then φ ( x ) isuniquely defined, φ ( x ) can be set to any value greater or equal to φ ( x ) . • If x x ∈ Θ , NW has two essential coordinates and x is minimal, then φ ( x ) isuniquely defined, φ ( x ) can be set to any value less or equal to φ ( x ) . • If x x ∈ Θ , SE has two essential coordinates and x is minimal, then φ ( x ) isuniquely defined, φ ( x ) can be set to any value less or equal to φ ( x ) . • If x x ∈ Θ , SE has two essential coordinates and x is maximal, then φ ( x ) isuniquely defined, φ ( x ) can be set to any value greater or equal to φ ( x ) .Proof. Consider the first case. φ ( x ) is defined uniquely as in the proof of Lemma 18.However, this is not possible for φ ( x ). This is because x is the only point in NW having x as the first coordinate, and, by Lemma 17 there is no equivalence relation within NW which involves x . If X is essential on SE then all points from the equivalence class whichincludes x also have x as their first coordinate, which does not allow to elicit φ ( x ). Ifonly X is essential on SE , then the representations of equivalences involving x do notinclude φ ( x ).In the case where only one coordinate is essential on both SE and NW no specialtreatment is required for the extreme elements of Θ. Lemma 20. If x ∈ SE ext then for any y ∈ N W such that x ∼ y , we have: φ NW ( x ) = φ NW ( y ) . If further, x is maximal, then φ SE ( x ) > φ SE ( y ) for all y ∈ SE . If x is minimal, then φ SE ( x ) < φ SE ( y ) for all y ∈ SE .If x ∈ NW ext then for any y ∈ SE such that x ∼ y , we have: φ SE ( x ) = φ SE ( y ) . f further, x is minimal, then φ NW ( x ) < φ NW ( y ) for all y ∈ NW . If x is maximal,then φ NW ( x ) > φ NW ( y ) for all y ∈ NW .Proof. In case when SE and NW have the same number of essential variables, valuefunctions at the extreme points are defined uniquely (Lemma 18 for case when bothvariables are essential and by definition otherwise) and the second parts of each statementfollow immediately by Lemma 17. If only SE or NW have two essential variables, theresult follows by Lemma 19, Lemma 17, and definition of φ . Lemma 21.
For any x ∈ X we have: φ ( x ) = φ ( x ) ⇒ x ∈ Θ . Proof.
Assume φ ( x ) = φ ( x ) and x NW . By Lemma 7 exists z ∈ Θ such that x ∈ SE z . By structural assumption we have φ ( z ) ≥ φ ( x ) = φ ( x ) ≥ φ ( x ), with atleast one inequality being strict (otherwise x = z ).If z is non-extreme then by Lemma 13 we have φ ( z ) = φ ( z ) - a contradiction. If z is extreme, the only cases when φ ( z ) > φ ( z ) can hold is when either z is minimalor z is maximal. But it is easy to see that in this case the only points for which it is notpossible to find a non-extreme z , are the extreme points themselves. Lemma 22.
For all x ∈ X such that φ ( x ) ≥ φ ( x ) we have x ∈ SE . If x ∈ X is suchthat φ ( x ) ≥ φ ( x ) then x ∈ NW .Proof. For non-extreme points this follows from Lemma 14 and Lemma 21. Assume φ ( x ) ≥ φ ( x ). If φ ( x ) = φ ( x ) then by Lemma 21 x ∈ Θ, so we are done. Therefore,assume φ ( x ) > φ ( x ). If x ∈ NW , then by Lemma 14 it must be φ ( x ) ≥ φ ( x ),a contradiction. Therefore, x ∈ SE . For extreme points the result follows from Lemma19. Finally, we can formulate: Theorem 5.
The following statements hold: • If both NW and SE have two essential variables, then for all x ∈ X : x ∈ Θ ⇐⇒ φ ( x ) = φ ( x ) , unless < can be represented by an additive function (i.e λ = 1 in (2)). • If only NW or only SE have two essential variables, the for all non-extreme x ∈ X : x ∈ Θ ⇒ φ ( x ) = φ ( x ) , while at extreme x ∈ X , φ ( x ) and φ ( x ) are related as in Lemma 19. Finally,for all x ∈ X : φ ( x ) = φ ( x ) ⇒ x ∈ Θ , • If both NW and SE have only one essential variable, then for all x ∈ X : x ∈ Θ ⇐⇒ φ ( x ) = φ ( x ) . Proof.
Follows from Lemmas 13, 21, 18, 19.24 .8. Constructing a global representation on X Lemma 23.
Assume z ≻ z for some z , z ∈ Θ . There exists z such that z ≻ z ≻ z .Proof. By order density and closedness assumption.For all x ∈ X let φ x ( x ) be equal to φ SE ( x ) if x ∈ SE or φ NW ( x ) if x ∈ NW . Forpoints in Θ values of two latter functions coincide, so φ x ( x ) is well-defined. Lemma 24.
Let φ x ( x ) > φ y ( y ) . Then, x ≻ y .Proof. If x and y belong to SE or NW the conclusion is immediate, so we only need tolook at the remaining case. Assume x ∈ SE , y ∈ NW .First we will show that it can’t hold that x ≻ z, y ≻ z or z ≻ x, z ≻ y for all z ∈ Θ.Assume x ≻ z, y ≻ z for all z ∈ Θ. Let x < y . If x y ∈ SE , then exists z y ∈ Θ suchthat x y < z y < y y , a contradiction. If x y ∈ NW , then exists x z ∈ Θ, such that x y < x z < x x , again a contradiction. Other cases are symmetrical.Hence, assume there exists z ∈ Θ, such that x < z , y < z and z ∈ Θ such that z < x or z < y . The only non-trivial case is z ≻ x ≻ z , z ≻ y ≻ z (in other cases oneof the points z or z immediately leads to the conclusion). We have φ ( z ) > φ ( x ) > φ ( y ) > φ ( z ) , hence also φ ( z ) = 0 . φ ( z ) > . φ ( x ) > . φ ( y ) > . φ ( z ) = φ ( z ). By denseranged-ness of φ (see (Wakker, 1991a) equation (16)), we can find a point c such that0 . φ ( x ) > φ ( c ) > . φ ( y ). We have c z ∈ SE , c z ∈ NW , hence there exists c such that c c ∈ Θ. Since c c is not extreme, we have φ ( c c ) = 2 φ ( c ), and hence φ ( x ) > φ ( c ) > φ ( y ). The first inequality is in SE , while the second is in NW , hence weconclude that x ≻ y . Lemma 25.
Let x ≻ y . Then, φ ( x ) > φ ( y ) .Proof. By Lemma 24 we have x ≻ y ⇒ φ ( x ) ≥ φ ( y ). Hence, we need to show that φ ( x ) = φ ( y ). Assume, x ∈ SE , y ∈ NW . If SE or NW have only one essentialcoordinate, then by Lemma 4 exists z ∈ Θ, equivalent either to x or y , from which theconclusion is immediate. Hence, assume both areas have two essential coordinates. x and x can’t be both minimal, because otherwise x ≻ y cannot hold by pointwisemonotonicity, so assume x is not minimal. We will find a point z in SE such that x ≻ z < y . Take some z such that x < z and z x ∈ SE (it can be found by closednessand order density). If z x < y , we have φ ( x ) > φ ( z x ) ≥ φ ( y ), otherwise by restrictedsolvability we can find w such that w x ∼ y , w ∈ SE , and hence φ ( x ) > φ ( w x ) = φ ( y )(equality follows from Lemma 24). The case when x is not maximal is identical. Theorem 6.
For any x, y ∈ X we have x < y ⇐⇒ φ ( x ) ≥ φ ( y ) . Proof.
Immediate by Lemmas 24 and 25. 25 .9. The representation is a Choquet integral
Representations φ SE and φ NW uniquely define a capacity ν . For the case when SE or NW has two essential coordinates, set (using (2)): ν ( { } ) = 11 + k (from φ SE ) ν ( { } ) = λ λk (from φ NW ) ν ( { , } ) = 1 . Thus, we obtain C ( ν, φ ( x )) = φ SE ( x ) = 11 + k φ ( x ) + k k φ ( x ) , for all x ∈ SE ,C ( ν, φ ( x )) = φ NW ( x ) = 11 + λk φ ( x ) + λk λk φ ( x ) , for all x ∈ NW . Assume now, that SE and NW has only one essential coordinate. If X is essential on SE set ν (1) = 1, otherwise zero. If X is essential on NW set ν (2) = 1, otherwise zero.As above, set ν ( { , } ) = 1 We obtain: C ( ν, φ ( x )) = φ ( x ) , if X is essential on the area containing x,C ( ν, φ ( x )) = φ ( x ) , if X is essential on the area containing x, in particular C ( ν, φ ( x )) = max( φ ( x ) , φ ( x )) if X is essential on SE , X is essential on NW , C ( ν, φ ( x )) = min( φ ( x ) , φ ( x )) if X is essential on SE , X is essential on NW . A.10. Uniqueness
Uniqueness properties are similar to those obtained in the homogeneous case X = Y n ,but are modified to accommodate for the heterogeneous structure of the set X in thispaper. Lemma 26.
Representation (1) has the following uniqueness properties:1. If NW = SE = X then for any functions g : X → R , g : X → R such that (1)holds with f i substituted by g i , we have f i ( x i ) = αg i ( x i ) + β i .2. If both coordinates are essential on NW and SE , then for any functions g : X → R , g : X → R such that (1) holds with f i substituted by g i , we have f i ( x i ) = αg i ( x i ) + β .3. If both coordinates are essential on NW , but only one coordinate is essential on SE , then for any functions g : X → R , g : X → R such that (1) holds with f i substituted by g i , we have : • f i ( x i ) = αg i ( x i ) + β , for all x such that f ( x ) < max f ( x ) and f ( x ) > min f ( x ) ; • f i ( x i ) = φ i ( g i ( x i )) where φ i is an increasing function, otherwise. . If both coordinates are essential on SE , but only one coordinate is essential on NW , then for any functions g : X → R , g : X → R such that (1) holds with f i substituted by g i , we have : • f i ( x i ) = αg i ( x i ) + β , for all x such that f ( x ) < max f ( x ) and f ( x ) > min f ( x ) ; • f i ( x i ) = ψ i ( g i ( x i )) where ψ i is an increasing function, otherwise.5. If one coordinate is essential on NW and another one on SE , then for any functions g : X → R , g : X → R such that (1) holds with f i substituted by g i , we have : f i ( x i ) = ψ i ( g i ( x i )) where ψ i are increasing functions such that f ( x ) = f ( x ) ⇐⇒ g ( x ) = g ( x ) .Proof.
1. Direct by uniqueness properties of additive representations.2. Direct by uniqueness properties of additive representations, Lemma 18, Theorem6.3. Assume NW has two essential coordinates. If there exists an element in Θ suchthat x is minimal or x is maximal, then, by Lemma 15, there exist respectivelya minimal φ ( x ) and maximal φ ( x ). Points x such that f ( x ) < max f ( x )and f ( x ) > min f ( x ) are precisely the elements for which there exist a ∈ X or p ∈ X such that either ax ∈ NW or x p ∈ NW . From this follows thatuniqueness of φ i for these points is defined by the uniqueness properties of NW and definition of φ i on SE , i.e. f i ( x i ) = αg i ( x i ) + β . For the remaining points(including extreme elements of Θ), uniqueness is derived from the uniqueness ofordinal representations and Lemma 4.4. The proof is identical to the one in the previous point.5. Uniqueness properties are derived from the uniqueness of ordinal representations,Lemma 4 and definition of φ i (Section A.5.2.3).Uniqueness part of Theorem 1 directly follows from Lemma 26. A.11. Necessity of axioms
A4..
Let ap bq, ar < bs, cp < dq and assume cr ≺ ds . Let also ap, bq, cp, dq ∈ NW , ar, bs, cr, ds ∈ SE and X to be symmetric in NW (the other cases are symmetric).We obtain: α f ( a ) + α f ( p ) ≤ α f ( b ) + α f ( q ) α f ( c ) + α f ( p ) ≥ α f ( d ) + α f ( q ) β f ( a ) + β f ( r ) ≥ β f ( b ) + β f ( s ) β f ( c ) + β f ( r ) < β f ( d ) + β f ( s )From the first two inequalities and essentiality of X ( α = 0) follows f ( a ) + f ( d ) ≤ f ( b ) + f ( c ). Last two inequalities imply f ( a ) + f ( d ) > f ( b ) + f ( c ), a contradiction.We also give the “necessity” proof of the condition in Lemma 3, since comparing itwith the necessity proof of A5 allows to elicit some interesting implications of essentiality.27 emma 3. Let { g ( i )1 : g ( i )1 y ∼ g ( i +1)1 y , g ( i )1 ∈ X , i ∈ N } and { h ( i )2 : x h ( i )2 ∼ x h ( i +1)2 , h ( i )2 ∈ X x , i ∈ N } be two standard sequences, the first entirely contained in NW and thesecond in SE . Assume also, that there exist z , z ∈ X , p, q ∈ X , a, b ∈ X such that g ( i )1 p, g ( i )1 q ∈ NW , and ah ( i )2 , bh ( i )2 ∈ SE for all i , and g ( i )1 p ∼ bh ( i )2 and g ( i +1)1 p ∼ bh ( i +1)2 .Finally, assume, g ( i +2)1 p ≻ bh ( i +2)2 . Other cases are symmetric. α f ( g ( i )1 ) + α f ( y ) = α f ( g ( i +1)1 ) + α f ( y ) α f ( g ( i )1 ) + α f ( y ) = α f ( g ( i +1)1 ) + α f ( y ) β f ( x ) + β f ( h ( i )1 ) = β f ( x ) + β f ( h ( i +1)1 ) β f ( x ) + β f ( h ( i +1)1 ) = β f ( x ) + β f f ( h ( i +2)1 ) α f ( g ( i )1 ) + α f ( p ) = β f ( b ) + β f ( h ( i )1 ) α f ( g ( i +1)1 ) + α f ( p ) = β f ( b ) + β f ( h ( i +1)1 )First two equations imply α ( f ( g ( i )1 ) − f ( g ( i +1)1 )) = α ( f ( g ( i +1)1 ) − f ( g ( i +2)1 )). Thefollowing two imply β ( f ( h ( i )1 ) − f ( h ( i +1)1 )) = β ( f ( h ( i +1)1 ) − f ( h ( i +2)1 )). Finally, thelast two equations imply α ( f ( g ( i )1 ) − f ( g ( i +1)1 )) = β ( f ( h ( i )1 ) − f ( h ( i +1)1 )). Appar-ently α ( f ( g ( i +1)1 ) − f ( g ( i +21 )) < β ( f ( h ( i +1)1 ) − f ( h ( i +2)1 )) is then a contradiction.If we were to add an essentiality condition to Lemma 3, the statement can be madestronger as shown below. A5..
Assume ap bq, cp < dq and ay ∼ x π ( a ) , by ∼ x π ( b ) , cy ∼ x π ( c ) , dy ∼ x π ( d ),and also eπ ( a ) < gπ ( b ). Also, X is essential on the set ( NW or SE ) which includes ap, bq, cp, dq , and X is essential on the set ( NW or SE ), which includes x π ( a ) and x π ( b ). Finally, assume eπ ( c ) ≺ gπ ( d ).We get α f ( a ) + α f ( p ) ≤ α f ( b ) + α f ( q ) α f ( c ) + α f ( p ) ≥ α f ( d ) + α f ( q ) β f ( e ) + β f ( π ( a )) ≥ β f ( g ) + β f ( π ( b )) β f ( e ) + β f ( π ( b )) < β f ( g ) + β f ( π ( d )) γ f ( a ) + γ f ( y ) = δ f ( x ) + δ f ( π ( a )) γ f ( b ) + γ f ( y ) = δ f ( x ) + δ f ( π ( b )) γ f ( c ) + γ f ( y ) = δ f ( x ) + δ f ( π ( c )) γ f ( d ) + γ f ( y ) = δ f ( x ) + δ f ( π ( d ))First two inequalities and the essentiality of X ( α = 0) imply f ( a ) − f ( b ) ≤ f ( c ) − f ( d ). Second pair of inequalities yields f ( π ( c )) − f ( π ( d )) < f ( π ( a )) − f ( π ( b )),28hile the final pair of equations leads to γ ( f ( c ) − f ( d )) = δ ( f ( π ( c )) − f ( π ( d ))).Combining these results and due to essentiality of X (hence δ = 0) we get: γ ( f ( a ) − f ( b )) ≤ γ ( f ( c ) − f ( d )) = δ ( f ( π ( c )) − f ( π ( d ))) < δ ( f ( π ( a )) − f ( π ( b ))) , which contradicts the third pair of inequalities above, which yield γ ( f ( a ) − f ( b )) = δ ( f ( π ( a )) − f ( π ( b ))). References
L. Savage, The Foundations of Statistics, Wiley publications in statistics, John Wileyand Sons, 1954.D. Ellsberg, Risk, ambiguity, and the Savage axioms, The Quarterly Journal of Economics75 (4) (1961) 643–669, ISSN 0033-5533.J. Quiggin, A theory of anticipated utility, Journal of Economic Behavior & Organization3 (4) (1982) 323–343.P. Wakker, Additive representations on rank-ordered sets. I. The algebraic approach,Journal of Mathematical Psychology 35 (4) (1991a) 501–531.D. Schmeidler, Subjective probability and expected utility without additivity, Economet-rica: Journal of the Econometric Society 57 (3) (1989) 571–587, ISSN 0012-9682.I. Gilboa, Expected utility with purely subjective non-additive probabilities, Journal ofMathematical Economics 16 (1) (1987) 65–88.P. Wakker, Additive representations of preferences, a new foundation of decision analysis;the algebraic approach, in: Mathematical Psychology, Springer, 71–87, 1991b.R. L. Keeney, H. Raiffa, Decisions with multiple objectives: Preferences and value trade-offs, Cambridge Univ Pr, ISBN 0521438837, 1976.M. Grabisch, C. Labreuche, A decade of application of the Choquet and Sugeno integralsin multi-criteria decision aid, 4OR: A Quarterly Journal of Operations Research 6 (1)(2008) 1–44, ISSN 1619-4500.G. Debreu, Topological methods in cardinal utility theory, Cowles Foundation DiscussionPapers .D. H. Krantz, R. D. Luce, P. Suppers, A. Tversky, Foundation of Measurement. Vol. 1:Additive and Polynomial Representations, 1971.V. K¨obberling, P. P. Wakker, Preference foundations for nonexpected utility: A gen-eralized and simplified technique, Mathematics of Operations Research 28 (3) (2003)395–423. 29. Labreuche, An Axiomatization of the Choquet Integral and Its Utility Functions with-out Any Commensurability Assumption, in: S. Greco, B. Bouchon-Meunier, G. Coletti,M. Fedrizzi, B. Matarazzo, R. R. Yager (Eds.), Advances in Computational Intelligence,vol. 300 of