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Trichotomous preferences for gambles
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Journal of Mathematical Psychology 48 (2004) 385–398 www.elsevier.com/locate/jmp
Trichotomous preferences for gambles Yutaka Nakamura Graduate School of Systems and Information Engineering, Division of Policy and Planning Sciences, University of Tsukuba, 1-1-1 Tennoudai, Tsukuba, Ibaraki 305-8573, Japan Received 9 July 2002; revised 23 April 2004 Available online 10 November 2004
Abstract Let GX be the set of all gambles on a set X of decision outcomes. Trichotomous preferences divide GX into three disjoint subsets, viz., the set Pþ of preferable gambles, the set P0 of indifferent gambles, and the set P of unpreferable gambles. Those preference comparisons are made against the status quo, i.e., the do-nothing alternative. This paper presents and discusses dyadic representations of such trichotomous preferences and identifies necessary and sufficient axioms for the existence of those representations. r 2004 Elsevier Inc. All rights reserved. Keywords: Trichotomous preferences; Decisions under risk; Non-expected utility; Bounded rationality; Satisficing criterion
1. Introduction Let GX be the set of all gambles defined on a set X of decision outcomes. A gamble is a real-valued function p from X into the closed unitP interval ½0; 1 for which fx 2 X : pðxÞa0g is finite and x2X pðxÞ ¼ 1: Each pðxÞ is interpreted as the probability with which decision outcome x is obtained. The traditional approaches for rational choice under risk are concerned with optimality criteria that identify the most preferred gamble out of a given set of gambles. The most influential principle of optimality is maximization of expected utility (EU). However, a vast amount of evidence, accumulated in the last halfcentury, uncovered that people do not actually behave like an EU maximizer in many choice situations. As a response to this empirical fact, numerous descriptive and axiomatic alternatives to the EU maximization principle have been proposed and tested for their validity (see Fishburn, 1988; Karni and Schmeidler, 1991; Quiggin, 1993; Schmidt, 1998 and many others for Fax.: +81 29 853 5070.
E-mail address: [email protected]. URL: http://infoshako.sk.tsukuba.ac.jp/ nakamura/. 0022-2496/$ - see front matter r 2004 Elsevier Inc. All rights reserved. doi:10.1016/j.jmp.2004.08.006
surveys). Like EU theories, those are also concerned with seeking optimality criteria. In these alternatives, it is presumed that the decision maker has his/her wellstructured preferences for gambles, so that quantitative representations of preferences are axiomatically obtained. On the other hand, this paper is concerned with a satisficing criterion which is based on trichotomous preferences for gambles that identify preferable/unpreferable gambles out of a given set of gambles. Trichotomous preferences divide GX into three disjoint subsets, Pþ ; P0 ; and P ; which are interpreted as follows: p 2 Pþ 3 p is preferable; p 2 P0 3 p is indifferent; p 2 P 3 p is unpreferable: Those trichotomous judgments are made against the status quo, i.e., the do-nothing alternative. Although axiomatic analyses of satisficing criteria are important in view of modeling bounded rationality of human judgments, there seems to be no attempt to axiomatize satisficing criteria for preference judgments in the literature except Goldstein (1991) who axiomatized
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trichotomous preferences for multiattributed alternatives under certainty. The aim of the paper is to present and discuss dyadic representations of a tripartite partition P3 ¼ fPþ ; P0 ; P g of GX and to identify qualitative conditions on P3 that are necessary and sufficient for the existence of those representations. The most general dyadic representation that we shall consider yields two realvalued functions f and c on X such that, for all p 2 GX ; X X pðxÞfðxÞ40 and pðxÞcðxÞ40; p 2 Pþ 3 x2X
p 2 P 3
X x2X
x2X
pðxÞfðxÞo0
and
X
pðxÞcðxÞo0:
x2X
This representation may be interpreted as follows. There are two criteria that, respectively, divide decision outcomes into three categories: preferable, indifferent, and unpreferable. Thus, a decision outcome is preferred (respectively, unpreferred) if it is preferable (respectively, unpreferable) on both criteria. To extend those trichotomous judgments to the set of all gambles, two functions f and c on X are introduced to represent ‘‘intensity of preference’’ in each criterion, where negative values of f and c represent ‘‘intensity of undesirability.’’ Then a gamble is judged to be preferred (respectively, less preferred) to the status-quo when expected intensities of preference with respect to both criteria are positive (respectively, negative). When f and c are related by a positive multiplicative constant, the two criteria give identical trichotomous judgments for gambles. In this case, c and P pðxÞcðxÞ may be interpreted as a utility function x2X on X and expected utility of gamble p, respectively, which are conditioned P Pupon the status quo. Therefore, if pðxÞcðxÞ404 x2X x2X qðxÞcðxÞ; then the decision maker should play gamble p and should not play gamble q, since p is preferred to the status quo which is more preferable than gamble q. However, trichotomous judgments say nothing about which gamble the decision maker should take among preferable gambles. In general, the dyadic representations for trichotomous judgments can be transformed into utility representations with threshold conditional upon the status quo. It is shown that there exist two real-valued functions u and o on X such that, for all p 2 GX ; X X þ pðxÞuðxÞ4 pðxÞoðxÞ; p2P 3 x2X x2X X X p2P 3 pðxÞuðxÞo pðxÞoðxÞ; x2X x2X where u and o may be regarded as a utility function and a threshold function, respectively. This paper is organized as follows. Section 2 shows the uniqueness properties of the dyadic and threshold
representations, and explores the relationships between those representations. Section 3 identifies qualitative conditions of P3 that are necessary and sufficient for the existence of dyadic representations. We conclude the paper in Section 4. The proof of the main representation theorem is deferred to the appendix.
2. Dyadic and threshold representations Throughout the paper, we shall assume that P3 ¼ fPþ ; P0 ; P g is a tripartite partition of GX : First, we show the uniqueness of the dyadic representations. Then we examine the relationship between dyadic and threshold representations. 2.1. Dyadic representations For a probability number 0plp1; a probability mixture of two gambles p and q, denoted plq; is a gamble that yields outcome x with probability lpðxÞ þ ð1 lÞqðxÞ: Given a real-valued function f P on X, we shall extend the domain of f to GX by f ðpÞ ¼ x2X f ðxÞpðxÞ for all p 2 GX : Then it is easy to see that f on GX is linear, i.e., for all p; q 2 GX and all 0olo1; f ðplqÞ ¼ lf ðpÞ þ ð1 lÞf ðqÞ: We say that ðGX ; P3 Þ has a dyadic representation, denoted ðf; cÞ; if f and c are linear functions on GX such that, for all p 2 GX ; p 2 Pþ 3 fðpÞ40 and cðpÞ40; p 2 P 3 fðpÞo0 and cðpÞo0: Note that p 2 P0 if and only if fðpÞcðpÞp0: The uniqueness of dyadic representations is stated as follows: Theorem 2.1. Suppose that Pþ and P are not empty, and ðGX ; P3 Þ has a dyadic representation ðf; cÞ: Then ðf0 ; c0 Þ is also a dyadic representation for ðGX ; P3 Þ if and only if there are positive numbers, a and b; such that ff0 ; c0 g ¼ faf; bcg: Proof. Suppose that ðGX ; P3 Þ has a dyadic representation ðf; cÞ: If ff0 ; c0 g ¼ faf; bcg for some a40 and b40; then it easily follows that ðf0 ; c0 Þ is a dyadic representation for ðGX ; P3 Þ: Suppose that ðf0 ; c0 Þ is also a dyadic representation for ðGX ; P3 Þ: For any linear function f on GX ; let Iðf Þ ¼ fp 2 GX : f ðpÞ ¼ 0g: Then IðfÞ; IðcÞ; Iðf0 Þ; and Iðc0 Þ are not empty. Take any p 2 IðfÞ: Assume that f0 ðpÞc0 ðpÞo0: Let q 2 Pþ and r 2 P : Then ðf0 ; c0 Þ representation implies that paq; pbr 2 P0 for some 0oao1 and 0obo1: On the other hand, it follows from ðf; cÞ representation that plq 2 Pþ for all 0olo1; or plr 2 P for all 0olo1: This is a contradiction. Therefore, f0 ðpÞc0 ðpÞ ¼ 0:
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Hence, since I sets are convex, we can assume with no loss of generality that IðfÞ ¼ Iðf0 Þ and IðcÞ ¼ Iðc0 Þ: It suffices to show that f0 ¼ af for some positive number a: Let fðpÞ40 for a fixed p 2 Pþ and define a by a¼
f0 ðpÞ ; fðpÞ
which must be positive. We prove in turn that, for all q 2 GX ; (1) fðqÞo0 ) f0 ðqÞ ¼ afðqÞ; (2) fðqÞ40 ) f0 ðqÞ ¼ afðqÞ: (1) Given fðqÞo0 along with fðpÞ40; define l by lfðpÞ þ ð1 lÞfðqÞ ¼ 0: Then by linearity, fðplqÞ ¼ 0: Since IðfÞ ¼ Iðf0 Þ; we have f0 ðplqÞ ¼ lf0 ðpÞ þ ð1 lÞf0 ðqÞ ¼ 0: Hence f0 ðqÞ ¼
l f0 ðpÞfðqÞ f0 ðpÞ ¼ ¼ afðqÞ: 1l fðpÞ
(2) Take any r 2 GX for which fðrÞo0: Then fðqbrÞo0 for some 0obo1: By (1), f0 ðrÞ ¼ afðrÞ and f0 ðqbrÞ ¼ afðqbrÞ: Then linearity of f and f0 gives f0 ðqÞ ¼ afðqÞ: & 2.2. Threshold representations We say that ðGX ; P3 Þ has a threshold representation, denoted ðu; wÞ; if u and w are linear functions on GX such that, for all p 2 GX ; p 2 Pþ 3 uðpÞ4jwðpÞj;
p2P
The uniqueness of the threshold representation is given by Theorem 2.3. Suppose that Pþ and P are not empty, and ðGX ; P3 Þ has a threshold representation ðu; wÞ: Then ðu0 ; w0 Þ is also a threshold representation for ðGX ; P3 Þ if and only if there are two real numbers a and b such that a þ b40; a b40; u0 ¼ au þ bw; and w0 ¼ bu þ aw or w0 ¼ bu aw: Proof. Suppose that ðu; wÞ is a threshold representation for ðGX ; P3 Þ: Assume first that ðu0 ; w0 Þ is also a threshold representation for ðGX ; P3 Þ: Then ðu þ w; u wÞ and ðu0 þ w0 ; u0 w0 Þ are, respectively, the corresponding dyadic representations. Thus by the uniqueness of the dyadic representation, there are positive numbers a0 and b0 such that fu0 þ w0 ; u0 w0 g ¼ fa0 ðu þ wÞ; b0 ðu wÞg: Let a ¼ ða0 þ b0 Þ=2 and b ¼ ða0 b0 Þ=2; so a þ b40 and a b40: If u0 þ w0 ¼ a0 ðu þ wÞ and u0 w0 ¼ b0 ðu wÞ; then solving these with respect to u0 and w0 ; we obtain that u0 ¼ au þ bw and w0 ¼ bu þ aw: If u0 þ w0 ¼ b0 ðu wÞ and u0 w0 ¼ a0 ðu þ wÞ; then similarly we obtain that u0 ¼ au þ bw and w0 ¼ bu aw: Assume next that a þ b40; a b40; u0 ¼ au þ bw; and w0 ¼ bu þ aw or w0 ¼ bu aw: Then for all p 2 GX ; u0 ðpÞ4jw0 ðpÞj 3u0 ðpÞ4w0 ðpÞ and u0 ðpÞ4 w0 ðpÞ 3auðpÞ þ bwðpÞ4buðpÞ þ awðpÞ and auðpÞ þ bwðpÞ4 buðpÞ awðpÞ 3ða bÞuðpÞ4ða bÞwðpÞ
3 uðpÞo jwðpÞj:
Note that p 2 P0 if and only if juðpÞjpjwðpÞj: The following theorem shows the relationship between dyadic and threshold representations. þ
Theorem 2.2. Suppose that P and P are not empty. Then ðGX ; P3 Þ has a dyadic representation ðf; cÞ if and only if ðf þ c; f cÞ is a threshold representation for ðGX ; P3 Þ: Proof. Let ðf; cÞ be a dyadic representation for ðGX ; P3 Þ: Then to see that ðf þ c; f cÞ is a threshold representation for ðGX ; P3 Þ; it suffices to observe that, for all p 2 GX ; p 2 Pþ 3fðpÞ40 and cðpÞ40 3fðpÞ þ cðpÞ4cðpÞ fðpÞ and fðpÞ þ cðpÞ4fðpÞ cðpÞ 3fðpÞ þ cðpÞ4jfðpÞ cðpÞj: The converse is similar.
387
&
It follows from Theorems 2.1 and 2.2 that ðu; wÞ is a threshold representation for ðGX ; P3 Þ if and only if ðu þ w; u wÞ is a dyadic representation for ðGX ; P3 Þ:
and ða þ bÞuðpÞ4 ða þ bÞwðpÞ 3uðpÞ4jwðpÞj: Similarly, for all p 2 GX ; u0 ðpÞo jw0 ðpÞj if and only if uðpÞo jwðpÞj: Hence the desired result obtains. &
3. Axiomatizations For any positive integer n and gambles p1 ; . . . ; pn 2 GX ; we define ½p1 ; . . . ; pn ¼ fð ððp1 l1 p2 Þl2 p3 Þ . . .Þln1 pn 2 GX : 0pli p1 for i ¼ 1; . . . ; n 1g; hp1 ; . . . ; pn i; ¼ fð ððp1 l1 p2 Þl2 p3 Þ . . .Þln1 pn 2 GX : 0oli o1 for i ¼ 1; . . . ; n 1g; where ½p1 ; . . . ; pn is a convex hull generated by p1 ; . . . ; pn ; and hp1 ; . . . ; pn i is the interior of ½p1 ; . . . ; pn
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when p1 ; . . . ; pn are distinct gambles. We note that hp; pi ¼ fpg; since p ¼ plp for all 0olo1: The following axioms apply to all p; q; r 2 GX ; all 0olo1; and all 2 fþ; g; where P should be read as either Pþ or P : A1. If p 2 P ; then, for every qeP ; paq 2 P for some 0oao1: A2. If p; q 2 P and hr; pi P ; then hr; qi P : A3. If p 2 P and q 2 P0 ; then hp; qi P [ P0 : A4. If p; q 2 P ; and fr; plr; qlrg P0 ; then hplr; qlri P0 : Axiom A1 is an Archimedean axiom, which says, for example, that if gamble p is preferable (i.e., p 2 Pþ ), then, for every gamble q that is not preferable, there is a probability mixture of p and q that should be preferable. This axiom seems to be defended by the argument that if a in the axiom is very close to 1, paq should be preferred (respectively, unpreferred) to the status quo when þ is substituted for (respectively, for ). Axioms A2–A4 are concerned with convexity properties of Pþ ; P0 ; and P : Axiom A2 implies the following convexity property of Pþ and P : for all p; q 2 GX and 2 fþ; g;
p; q 2 P
) hp; qi P :
We shall use this fact by referring to axiom A2. Note that we may have reP in axiom A2. Suppose that gambles p and q are preferable and gamble r is not preferable. Then axiom A2 requires that if a probability mixture of p and r is not preferable, then there must be a probability mixture of q and r that is not preferable. This is a requirement of the threshold representation ðu; wÞ with wðrÞa0: Axiom A3 says, for example, that if p is preferable and q is indifferent, then there is no probability mixture of p and q that is unpreferable. Although P0 is not necessarily convex, axiom A4 requires that for a convex hull ½p; q; r; if p; q 2 P and r 2 P0 ; then P0 \ ½p; q; r must be convex. Observe for example that if p; q 2 Pþ and r 2 P ; then P0 \ ½p; q; r may not be convex. The main representation theorem is given by Theorem 3.1. Suppose that P and Pþ are not empty. Then ðGX ; P3 Þ has a dyadic representation if and only if axioms A1–A4 hold. Proof. See appendix
&
In what follows, we examine three special cases of dyadic representations. Suppose that ðu; wÞ is a threshold representation for ðGX ; P3 Þ: Then we say that ðGX ; P3 Þ has a linear threshold representation if wX0; a constant threshold representation if w is constant, and a threshold-free representation if w ¼ 0: It follows from Theorems 2.1 and 2.2 that, for any
number l40; ðGX ; P3 Þ has a dyadic representation ðf; cÞ 3it has a dyadic representation ðf; lcÞ 3it has a threshold representation ðu; wÞ; where u ¼ f þ lc and w ¼ f lc: Therefore, ðGX ; P3 Þ has a linear threshold representation 3fXac for some a40; a constant threshold representation 3f ¼ ac þ b for some a40 and b; a threshold-free representation 3f ¼ ac for some a40: First, we consider axioms for threshold-free representation. The following axioms apply to all p; q 2 GX ; all 0olo1; and 2 fþ; g; where P should be read as either Pþ or P : A10 : If p 2 Pþ and q 2 P ; then paq 2 P0 for some 0oao1: A20 : If p 2 P and q 2 P [ P0 ; then plq 2 P : A30 : If p; q 2 P0 ; then hp; qi P0 : Axioms A10 is a solvability axiom. Axiom A20 says, for example, that if gamble p is preferable and gamble q is preferable or indifferent, then every probability mixture between p and q is preferable. This requires that threshold vanishes. Convexity of Pþ and P follows from this axiom. Axiom A30 requires that P0 be convex. The representational implication of axioms A10 –A30 is stated by the following theorem. Theorem 3.2. Suppose that Pþ and P are not empty. Then ðGX ; P3 Þ has a threshold-free representation if and only if axioms A10 –A30 hold. Proof. Suppose that Pþ and P are not empty. The necessity of axioms A10 –A30 easily obtains. Their sufficiency follows from Lemma 3 in Fishburn (1982). To see this, we assume that axioms A10 –A30 hold. Then, by axiom A10 ; P0 a;: Fix a gamble r 2 P0 : Define two binary relations, r and r on GX as follows: for all p 2 GX ; p 2 Pþ ) pr r; p 2 P ) rr p; p 2 P0 ) p r r and r r p: Fishburn’s lemma claims that there is a linear function u on GX such that, for all p 2 GX ; pr r 3 uðpÞ40; rr p 3 uðpÞo0; whenever the following three conditions hold, understood as applying to all p; q 2 GX and all 0olo1:
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B1. If pr r and rr q; then paq r r for some 0oao1: B2. If pr r and :ðrr qÞ; then plqr r; if rr p and :ðqr rÞ; then rr plq: B3. If p r r and q r r; then plq r r: Note that B1–B3, respectively, follow from axioms A10 –A30 : This completes the proof. & It is easy to construct examples that axiom A10 holds but A1 fails to hold, and vice versa. We shall identify each outcome x 2 X with the gamble that yields x with probability one. Let X ¼ fx; yg: Assuming that x 2 Pþ ; y 2 P ; and hx; yi P0 ; we conclude that axiom A10 hold, but not axiom A1. Next we assume that fxay : a rational and 0oap1g Pþ ; fxay : a irrational and 0pao1g P : Hence it readily follows that axiom A1 hold, but not axiom A10 : When Pþ and P are convex, axiom A1 implies axiom A10 as shown by the following lemma. Lemma 3.1. Axioms A1 and A2 imply axiom A10 : Proof. Suppose that axioms A1 and A2 hold. Assume that p 2 Pþ and q 2 P : Let a ¼ inffl : plq 2 Pþ and 0olp1g: Then, by axiom A1, 0oao1: Axiom A2 implies that plq 2 Pþ
if aolo1;
plq 2 P
if 0oloa:
Hence, by axiom A1, paq 2 P0 :
&
The following implications of axioms A1–A3 will be used in the proof of Theorems 3.1 in the appendix and the proof of Theorem 3.3 below. Lemma 3.2. Suppose that axioms A1–A3 hold. (1) If 2 fþ; g; p 2 P ; and q 2 P0 ; then hp; paqi P and ½q; paq P0 for some 0pao1: (2) If p 2 Pþ and q 2 P ; then hp; pbqi Pþ ; ½pbq; paq P0 ; and hpaq; qi P for some 0oapbo1: Proof. Suppose that Axioms A1–A3 hold. (1) Suppose that 2 fþ; g; p 2 P ; and q 2 P0 : Then by axiom A3, hp; qi P [ P0 : It follows from axiom A2 that pmq 2 P0 whenever plq 2 P0 and 0pmolo1: Let a ¼ supfl : plq 2 P0 and 0plo1g: Then, by axiom A1, 0pao1 and paq 2 P0 : Hence hp; paqi P and ½q; paq P0 : (2) Suppose that p 2 Pþ and q 2 P : Then, by Lemma 3.1 and axiom A10 ; there is an 0oao1 such that paq 2 P0 : Hence the conclusion of (2) follows from (1). &
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The following axiom, which is understood as applying to all p; q; r; s 2 GX ; is devised for constant threshold representation. A40 : If p; q 2 P0 ; r; seP0 ; and, for all 0omo1; pmr 2 P0 3qmr 2 P0 ; then, for all 0omo1; pms 2 P0 3qms 2 P0 : Suppose that gambles p and q are indifferent and r is preferable. Then axiom A40 says that if the same probability mixtures between p and r and between q and r are preferable, then gamble r can be replaced by any other gamble s that is not indifferent. Although the axiom looks complicated, it seems intuitively clear that threshold must be constant. Axiomatic characterizations of linear threshold and constant threshold representations are given by Theorem 3.3. Suppose that Pþ and P are not empty. (1) ðGX ; P3 Þ has a linear threshold representation if and only if axioms A1, A2, and A30 hold; (2) ðGX ; P3 Þ has a constant threshold representation if and only if axioms A1, A2, A30 ; and A40 holds. Proof. The necessity of the axioms in (1) and (2) easily follows. We show sufficiency of those axioms for each representation. (1) Suppose that axioms A1, A2, A30 hold. Axiom A4 follows from axiom A30 : To see that axiom A3 holds, assume that p 2 Pþ ; q 2 P0 ; and paq 2 P for some 0oao1: It follows from Lemma 3.2(2) that pbðpaqÞ ¼ pða þ b abÞq 2 P0 for some 0obo1: Then by A30 ; a q ¼ paq 2 P0 ; ðpða þ b abÞqÞ a þ b ab a contradiction. Hence A3 obtains. By Theorem 3.1, there is a dyadic representation ðf; cÞ for ðGX ; P3 Þ: Then we define R ¼ fðfðpÞ; cðpÞÞ 2 R2 : p 2 GX g: Note that R is convex. For any real number l; half-spaces are defined by 2 Hþ l ¼ fða; bÞ 2 R : aXlbg; 2 H l ¼ fða; bÞ 2 R : aplbg: We are to show that R H þ l or R H l for some l40: Thus either cðpÞXlfðpÞ for all p 2 GX or cðpÞplfðpÞ for all p 2 GX : Hence, by Theorems 2.1 and 2.2, ðGX ; P3 Þ has a linear threshold representation. Since Pþ and P are not empty, there are ða; bÞ; ða0 ; b0 Þ 2 R such that a40; b40; a0 o0; and b0 o0: Assume first that R has no interior point. Then it easily follows that R is contained in a straight line L, where L ¼ fða; bÞ 2 R2 : a ¼ lb þ gg for some l40: Hence R H þ l or R H l : Assume next that R has an interior point. Let R0 be the interior of R; so it is an open convex set in R2 : By axiom A30 ; P0 is convex, so is fða; bÞ 2 R : abp0g: Therefore, axiom A30 forces R0 to exclude the origin
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ð0; 0Þ: Since R is convex, we must have that R H þ l or R H : l (2) Suppose that axioms A1, A2, A30 ; and A40 hold. Then by (1), there is a dyadic representation ðf; cÞ with, say, fXc for ðGX ; P3 Þ: Take any p; q 2 P0 : Let s 2 P and r 2 Pþ : Then ½p; plr [ ½p; pms P0 ; hr; plri Pþ ; and hs; pmsi P for some 0olp1 and 0omp1: Thus lcðpÞ þ ð1 lÞcðrÞ ¼ 0 and lfðpÞ þ ð1 lÞfðsÞ ¼ 0: Axiom A40 requires that lcðqÞ þ ð1 lÞcðrÞ ¼ 0 3 mfðqÞ þ ð1 lÞfðsÞ ¼ 0 or equivalently, cðpÞ ¼ cðqÞp03fðpÞ ¼ fðqÞX0: This implies that R ¼ fðfðpÞ; cðpÞÞ 2 R2 : p 2 GX g must be a straight line in R2 ; i.e., c ¼ af þ b for some numbers a40 and b: &
4. Conclusions We studied a satisficing criterion by axiomatizing trichotomous judgments in decision making under risk. In particular, we proposed dyadic representations of trichotomous preferences for gambles and explored their axiomatizations. Trichotomous judgments are modeled by three unary relations, Pþ ; P ; and P0 ; on GX : These sets are mutually exclusive and exhaustive, and respectively interpreted as the set of preferred gambles, the set of unpreferred gambles, and the set of indifferent gambles, where these judgments are made against the status quo. We say that a subset A of GX is a decision problem when the decision maker may choose one element of A. For such a decision problem, a satisficing procedure based on trichotomous preferences may go as follows. The decision maker sequentially examines the gambles in A according to a prespecified ordering. Once he/she confronts a gamble that is a member of the set Pþ ; he/she stops and chooses it. For the case where no element of A belongs to Pþ ; he/she will maintain the status quo. The satisficing procedure may be numerically characterized by introducing a real valued function V on GX for which Pþ ¼ fp 2 GX : V ðpÞ40g: The decision maker searches for a gamble that satisfies the condition that V-value be positive. Note that instead of having a maximization problem, maxp2A V ðpÞ; the decision maker seems to solve a simpler problem, i.e., find a p 2 A for which V ðpÞ40: Dyadic representations provide a V characterization of the satisficing procedure which consists of two linear functions on GX : That is, there are two linear functions u and o on GX such that V ðpÞ ¼ uðpÞ joðpÞj for all p 2 GX : A natural generalization of dyadic representations is to consider a set U of linear functions on GX for which,
for all p 2 GX ; p 2 Pþ 3 uðpÞ40 for all u 2 U; p 2 P 3 uðpÞo0
for all u 2 U;
where U has at most two elements for dyadic representations. U may be infinite. When U has more than two elements, it may lose threshold-interpretation as demonstrated in Theorem 2.2 for the dyadic representations. It remains to be an open problem to develop sufficient (preferably necessary) axioms for the existence of U: Note that axioms A1–A3 are necessary for U-representations. However, axioms A4 fails to hold when U contains more than two elements that are mutually independent in the sense that one is not a positive linear transformation of the other. Necessary and sufficient axioms for dyadic representations were devised by an Archimedean property and convexity properties of Pþ ; P ; and P0 : Convexity properties are empirically testable. In particular, Pþ and P must be convex. Thus, for example, for any preferred gambles p and q, there seems to be no difficulty to empirically test whether any probability mixture of p and q is a preferred gamble or not. Although P0 need not be convex, the set of indifferent gambles in any convex hull ½p1 ; . . . ; pn must be convex whenever p1 ; . . . ; pn 2 Pþ [ P0 or p1 ; . . . ; pn 2 P [ P0 : In other words, if pi 2 Pþ and pj 2 P for some distinct i and j, then the set of indifferent gambles in ½p1 ; . . . ; pn may not be convex. Finally, we comment on the intuitive relationship between trichotomous judgments and the traditional preference judgments. Traditionally, preference judgments are modeled by a subset P of GX GX : When a pair ðp; qÞ of gambles belongs to P, gamble p is judged to be preferred to gamble q. Suppose that the decision maker already holds the right to play gamble r. It is also assumed that he/she is offered to exchange r for some other gamble. Then trichotomous judgments are made against the status quo (i.e., holding the right to play the gamble r) according to his/her preferences as follows. For all p 2 GX ; ðp; rÞ 2 P ) p 2 Pþ ; ðr; pÞ 2 P ) p 2 P and if neither ðp; rÞ nor ðr; pÞ belong to P, then p is in P0 : Thus, Pþ (respectively, P ) may be interpreted as the sets of gambles for which you are willing (respectively, unwilling) to exchange the gamble r, while P0 may be the set of gambles that are indifferent to the gamble r.
Acknowledgments The author gratefully acknowledges numerous helpful suggestions and comments of two anonymous referees
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and John Miyamoto. This research is financially supported in part by the Ministry of Education, Culture, Sports, Science and Technology under Grant 15653012 in Japan.
Appendix A A.1. Proof of the main representation theorem Throughout the appendix, we shall assume that Pþ and P are not empty. First, we assume that ðGX ; P3 Þ has a dyadic representation ðf; cÞ; i.e., for all p 2 GX ; p 2 Pþ 3 fðpÞ40 and
p2P
3 fðpÞo0
and
cðpÞ40; cðpÞo0:
Necessity of A1–A3 easily follows. To show necessity of A4, we assume that p; q 2 Pþ and fr; plr; qlrg P0 for 0olo1: When p; q 2 P ; the proof is similar. Then we have fðpÞ40;
cðpÞ40;
fðqÞ40; cðqÞ40; fðrÞcðrÞp0; fðplrÞcðplrÞp0; fðqlrÞcðqlrÞp0: Since, by linearity of f and c; fððplrÞgðqlrÞÞcððplrÞgðqlrÞÞ ¼ g2 fðplrÞcðplrÞ þ ð1 gÞ2 fðqlrÞcðqlrÞ þ gð1 gÞ½fðplrÞcðqlrÞ þ fðqlrÞcðplrÞ
Fig. 1. A barycentric representation of a compound gamble ðpaqÞbr:
By axiom A2, this definition does not depend on the choice of r. Since P0 a;; Lemma 3.2 implies that T a;: However, we may have T þ \ T ¼ ;: We say that p 2 T þ is upper threshold-free, and p 2 T is lower thresholdfree. p 2 T þ \ T is simply said to be threshold-free. The aim of the step is to prove the following claim, which characterizes threshold-free gambles. Claim A.1. Let 2 fþ; g: If ½p; q [ ½q; r P0 and hp; q; ri P ; then q is threshold-free. To prove the claim, we need two lemmas.
for all 0ogo1; it suffices to show that
Lemma A.1.1. Let 2 fþ; g: If hp; qi P ; p; q 2 P0 ; and reP [ P0 ; then there is an s 2 hp; q; ri such that
fðplrÞcðqlrÞ þ fðqlrÞcðplrÞp0:
hp; q; si P ;
If fðplrÞp0 and cðqlrÞp0; then fðrÞo0 and cðrÞo0; since fðpÞ40 and cðqÞ40: This contradicts fðrÞcðrÞp0: If fðplrÞp0; then cðplrÞX0 and cðqlrÞ40; so fðqlrÞp0: If cðqlrÞp0; then fðqlrÞ40 and fðplrÞ40; so cðplrÞp0: The both cases together give the desired result. Hence A4 is necessary for dyadic representations. Next we show sufficiency of the axioms A1–A4. Suppose that axioms A1–A4 hold. Since Pþ and P are not empty, it follows from Lemma 3.1 that P0 is not empty. We show the sufficiency proof in four steps. Every gamble in the convex hull H ¼ fðpaqÞbr : 0pap1; 0pbp1 and p; q; r 2 GX g generated by gambles p, q, and r can be represented barycentrically as shown in Fig. 1. Each point in H corresponds to a point in a triangle with vertices p, q, and r. When the perpendicular distance from each side to its opposite vertex is 1, ðpaqÞbr is the point with the perpendicular distances a and b from sides rq and pq, respectively. Step 1: For 2 fþ; g; let the boundary of P0 be given by
½p; par; s [ ½q; qbr; s P0 ;
T ¼ fp 2 P0 : hp; ri P for some r 2 P g:
hr; par; qbr; si \ ðP [ P0 Þ ¼ ;; where, for some 0oao1 and 0obo1; s¼p
að1 bÞ bð1 aÞ ðqbrÞ ¼ q ðparÞ: 1 ab 1 ab
Proof (See Fig. 2 for illustration). Suppose that hp; qi Pþ ; p; q 2 P0 ; and r 2 P : When hp; qi P and r 2 Pþ ; the proof is similar. Then by Lemma 3.2(1) and A2, there are a; b 2 ð0; 1 such that hr; par; qbri P : By A1 and A2, it is impossible to have a ¼ b ¼ 1: Assume that 0oaob ¼ 1: Since p 12 q 2 Pþ ; A1 gives ðp 12 qÞgr 2 Pþ for some 0ogo1: Note that g g 1 q r : ðp2qÞgr ¼ p 2 2g g Since q 2g r 2 P and p 2 P0 ; it follows from A3 that g g q r 2 P0 [ P ; p 2 2g
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that w 2 Pþ : By A1, wgs 2 Pþ for some 0ogo1: Then there are probability numbers d and such that 0odob and wgs 2 hrds; pqi as illustrated in Fig. 3. Since rds 2 P and pq 2 P0 ; A3 gives wgs 2 P0 [ P ; a contradiction. Hence w 2 P0 ; so that hp; q; ti P0 : It follows similarly that hq; r; ti P0 : &
Fig. 2. An illustration of the proof of Lemma A.1.1.
a contradiction. When 0oboa ¼ 1; a similar contradiction obtains. Hence 0oao1 and 0obo1: Next we show that hp; qbri [ hq; pari P0 : Let s ¼ að1bÞ p 1ab t and t ¼ qbr as illustrated barycentrically in Fig. 2(a). We are to show that hqbr; si; hpar; si; hp; si; hq; si; and fsg are contained in P0 : Assume that pgt 2 P for some 0ogp að1bÞ 1ab : Then by A1, qdðpgtÞ 2 P for some 0odo1: We note that qdðpgtÞ ¼ tð1 dÞð1 gÞðpKqÞ; where K ¼ ð1 dÞ=ðd þ g dgÞ: Since pKq 2 Pþ ; A3 gives qdðpgtÞ 2 P0 [ Pþ ; a contradiction. Thus pgteP : Similarly, pgtePþ : Hence ht; si ¼ hqbr; si and fsg are contained in P0 : The other cases follow similarly. Since hr; qbr; pari P ; it remains to show that hpar; qbri [ hpar; qbr; si P and hp; q; si Pþ : We show the former case. The latter case similarly follows. Take any measure t in hpar; qbri [ hpar; qbr; si as illustrated in Fig. 2(b). By A3, t 2 P0 [ P : Assume that t 2 P0 : Let 0ogo1 be such that ht; pgqi \ hp; qbri ¼ fsg: Then take any d and for which 0oogodo1: Since hp; qi Pþ ; we have pdq; pq 2 Pþ : Let t0 and t00 be the unique measures such that ht; pdqi \ hp; qbri ¼ ft0 g and ht; pqi \ hq; pari ¼ ft00 g: Since ft; t0 ; t00 g P0 ; it follows from A4 that ht0 ; t00 i P0 : Since d and are arbitrary, we obtain that hp; q; si P0 : This contradicts A1. Hence t 2 P : &
Proof of Claim A.1. (See Fig. 4 for illustration). Let ¼ þ: When ¼ ; the proof is similar. Suppose that ½p; q [ ½q; r P0 and hp; q; ri Pþ : We have to show that q is threshold-free. Suppose on the contrary that q is not threshold-free. Take any s 2 P : By Lemma 3.2(1), hs; qgsi P and ½qgs; qi P0 for some 0ogo1: It follows from Lemma A.1.1 that there is a gamble t ¼ pKðrbsÞ 2 hp; r; si; as illustrated in Fig. 4, such that hs; pas; rbs; ti P ;
Fig. 3. An illustration of the proof of Lemma A.1.2.
Lemma A.1.2. If ½p; q [ ½q; r P0 ; hp; q; ri [ hp; ri P ; and seP [ P0 ; then there is a gamble t 2 hp; r; si such that hp; q; r; ti P and ½p; q; t [ ½q; r; t P0 ; where, for some 0oao1 and 0obo1; t¼p
að1 bÞ bð1 aÞ ðrbsÞ ¼ r ðpasÞ: 1 ab 1 ab
Proof (See Fig. 3 for illustration). We show the case for ¼ þ: When ¼ ; the proof is similar. By Lemma A.1.1, there is a gamble t 2 hp; r; si such that hp; r; ti Pþ and hr; t; rbsi [ hp; t; pasi P0 ; where t is defined in the lemma. We take any w 2 hp; q; ti and show that w 2 P0 : Since hp; q; t; ri Pþ ; A1 implies that weP : Thus assume
Fig. 4. An illustration of the proof of Claim A.1.
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½rbs; t; r [ ½pas; t; p P0 ; 0oao1; 0obo1; and K¼
and
hp; r; ti Pþ ;
where
að1 bÞ : 1 ab
Note that t 2 hs; pLri; where L ¼ að1 bÞ=ðað1 bÞ þ bð1 aÞÞ: Thus we take any d and for which 0odoLoo1: Let t0 and t00 be the unique gambles for which ft0 g ¼ hs; pdri \ ht; ri and ft00 g ¼ hs; pri \ ht; pi: Then let w0 and w00 be the unique gambles for which fw0 g ¼ hq; t0 i \ hqgs; pdri and fw00 g ¼ hq; t00 i \ hqgs; pri: Since qgs; w0 ; w00 2 P0 and pdr; pr 2 Pþ ;A4 gives hw0 ; w00 i P0 : Since hp; q; ri [ hp; r; ti Pþ ; we obtain that hq; t0 ; t00 i Pþ : By construction, hw0 ; w00 i hq; t0 ; t00 i; a contradiction. Hence q must be threshold-free. & Step 2: Recall that T þ \ T is the set of all thresholdfree gambles. Let I ¼ P0 nðT þ \ T Þ: We define a binary relation on I as follows: for all p; q 2 GX ; p q 3 ½p; q I: The aim of the step is to prove the following claim. Claim A.2. on I is an equivalence relation. To prove the claim, we need the following three lemmas. Lemma A.2.1. Let 2 fþ; g: If hp; qi P and ½p; r [ ½r; q P0 ; then there is an 0pao1 such that either 0
and
hp; q; qari P ;
0
and
hp; q; pari P :
½p; r; qar P or
½q; r; par P
Proof (See Fig. 5 for illustration). Suppose that hp; qi Pþ and ½p; r [ ½r; q P0 : When hp; qi P ; the proof is similar. By Lemma 3.2(1), there is a real-valued function f on ð0; 1Þ such that, for all 0ogo1; 0pfðgÞo1; ½r; ðpgqÞfðgÞr P0 ; and hðpgqÞfðgÞr; pgqi Pþ : By A2, either fðgÞ ¼ 0 for all 0ogo1 or fðgÞ40 for all
Fig. 5. An illustration of the proof of Lemma A.2.1.
393
0ogo1: The former case gives the desired result. Thus, we assume that fðgÞ40 for all 0ogo1: It follows from A4 that, for all g0 ; g00 2 ð0; 1Þ; ½r; ðpg0 qÞfðg0 Þr; ðpg00 qÞfðg00 Þr P0 ; hðpg0 qÞfðg0 Þr; ðpg00 qÞfðg00 Þr; pg0 q; pg00 qi Pþ : This implies that there are 0oap1 and 0obp1 such that abo1; ½r; par; qbr P0 ; and hp; q; par; qbri Pþ : It suffices to show that either a ¼ 1 or b ¼ 1: Suppose that ao1 and bo1: Then by Claim A.1, par and qbr are threshold-free. Take any s 2 P : Then hs; par; qbri P : Since pgq 2 Pþ for 0ogo1; A1 implies that t ¼ ðpgqÞds 2 Pþ for some 0odo1: As illustrated in Fig. 5, hr; ti \ hs; par; qbria;: This contradicts A3. & Lemma A.2.2. If p; q; r 2 P0 and hp; qi [ hq; ri P ; then hp; ri P0 : Proof (See Fig. 6 for illustration). Suppose that p; q; r 2 P0 and hp; qi [ hq; ri Pþ : When hp; qi [ hq; ri P the proof is similar. Take any s 2 P : By A1 and A2, ½p; par [ ½pbr; r P0 and hpar; pbri Pþ for some 0pboap1: Assume that a ¼ 1 and b ¼ 0; so that hp; ri Pþ : By Lemma A.1.1 and Claim A.1, there are threshold-free gambles t 2 hq; r; si and w 2 hp; r; si; where, for some a; b; g 2 ð0; 1Þ; gð1 bÞ ðqbsÞ; 1 bg gð1 aÞ ðpasÞ: w¼r 1 ag t¼r
Since any gamble in ½t; w is threshold-free, we take a w0 2 ht; wi: Then, as illustrated in Fig. 6, let a gamble t0 2 hqbs; pasi be such that w0 2 ht0 ; ri: Note that t0 2 P ; since, by Lemma A.1.1, hqbs; pasi P : Let 0odo1 satisfy t0 2 hpdq; si: Such a d also satisfies that w0 2 hpdq; rgsi: By A1, there is a gamble t0 ðpdqÞ; denoted t00 ;
Fig. 6. An illustration of the proof of Lemma A.2.2.
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for some 0oo1 that is contained in P : Let fw00 g ¼ ht00 ; ri \ hpdq; rgsi: Since pdq 2 Pþ and w0 is thresholdfree, we must have w00 2 Pþ :This contradicts A3. When 0ob and ao1; similar contradictions obtain. Hence hp; ri P0 : & Lemma A.2.3. If p is not threshold-free and ½p; r P0 ; then there exists at most one threshold-free gamble in ½p; r: Moreover, letting q 2 P and s 2 Pþ ; we have (1) if ½p; r contains no threshold-free gamble, then there are a; a0 ; b; b0 2 ½0; 1 such that a þ a0 o2; b þ b0 o2; ½p; paq; rbq; r [ ½p; pa0 s; rb0 s; r P0 ; hpaq; rbq; qi P
and
hpa0 s; rb0 s; ri Pþ :
(2) if ½p; r contains a threshold-free gamble pgr for some 0pgo1; then there are a; a0 ; b; b0 2 ½0; 1 such that a þ a0 o2; b þ b0 ¼ 2 if g ¼ 0; b þ b0 o2 if 0og; ½p; paq; pgr [ ½pgr; rbq; r [ ½p; pa0 s; pgr [ ½pgr; rb0 q; s P0 ; hq; paq; pgr; rbqi [ hq; paqi [ hq; rbqi P ; hs; pa0 s; pgr; rb0 si [ hs; pa0 si [ hs; rb0 si Pþ :
Proof (See Fig. 7 for illustration). Suppose that p is not threshold-free, ½p; r P0 ; q 2 P ; and s 2 Pþ : Then ½p; paq [ ½p; pa0 s P0 ; hpaq; qi P ; and hpa0 s; si Pþ for some 0oap1 and 0oa0 p1 with a þ a0 o2: Assume that ao1: When a0 o1; the proof is similar. Suppose that there are at least two threshold-free gambles in ½p; r: With no loss of generality, let r and pgr for some 0ogo1 be threshold-free gambles, so that any gamble in hr; pgri are threshold-free. It follows from A2 and Lemma 3.2(2) that hq; r; pgr; paq; sqi P and ½p; paq; pgr; sq [ ½r; pgr; paq P0 for some 0oo1: We have ðpgrÞðaKÞðsqÞ ¼ ðpaqÞðgKÞðsLrÞ;
Fig. 7. An illustration of the proof of Lemma A.2.3.
where K ¼ ð1 Þ=ðað1 Þ þ gð1 aÞÞ and L¼
gð1 aÞ : ð1 Þða gÞ þ gð1 aÞ
Let t ¼ ðpgrÞðaKÞðsqÞ: For 0odoL; let ft0 g ¼ hsdr; paqi \ ht; ri as illustrated in Fig. 7. Since ht0 ; paqi P ; paq 2 P0 ; and sdr 2 Pþ ; A3 gives a contradiction. Thus r cannot be threshold-free. Hence hp; ri contains at most one threshold-free gamble. Suppose that ½p; r contains at most one threshold-free gamble. There is a 0obp1 such that hq; rbqi P and ½rbq; r P0 : Assume first that ½p; r contains no threshold-free gamble. Then, by Claim A.1, ½paq; rbq P0 : Thus (1) obtains. Assume next that ½p; r contains one threshold-free gamble pgr for some 0pgo1: Then, by Claim A.1, ½paq; pgr [ ½pgr; rbq P0 : Hence (2) obtains. & Proof of Claim A.2. (See Fig. 8 for illustration). Since ½p; q I iff ½q; p I; is reflexive and symmetric. To show that is transitive, assume that p q and q r: We have the following two cases to examine, Case 1: either hp; ri \ Pþ a; or hp; ri \ P a;; Case 2: ½p; r P0 and ½p; rnIa;: In each case, we obtain a contradiction, so that ½p; r I; i.e., p r: Case 1: Suppose that hp; ri \ Pþ a;: When hp; ri \ P a;; the proof is similar. By Lemma 3.2(1), there are 0paobp1 such that ½r; par [ ½p; pbr P0 and hpar; pbri Pþ : First we assume that a ¼ 0 and b ¼ 1: Then by Lemma A.2.1, there is an 0ogp1 such that either (a) or (b) holds: (a) ½p; q; qgr P0 and hp; r; qgri Pþ ; (b) ½q; r; qgp P0 and hp; r; qgpi Pþ : By Claim A.1, qgr is threshold-free for (a) and qgp is threshold-free for (b). They are contradictions. Next we assume that a ¼ 0 and ba1: When If aa0 and b ¼ 1; the proof is similar. If ½q; pbr P0 ; then it follows from Lemma A.2.1 and A4 that ½q; pbr; qgr P0 and hpbr; r; qgri Pþ for some 0ogp1: Thus by Claim A.1, qgr is threshold-free, a contradiction.
Fig. 8. An illustration of the proof of Claim A.2..
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If hq; pbri \ Pþ a;; then ½q; qgðpbrÞ P0 and hqgðpbrÞ; pbri Pþ for some 0ogo1 as illustrated in Fig. 8(a), where the shaded areas are shown to be contained in P0 : Then by the similar argument of the preceding paragraph that ½qgðpbrÞ; r P0 : We note qgðpbrÞ ¼ ðqKpÞ
g r; K
where K ¼ g=ðb þ g bgÞ: By A1 and A2, ½qKp; qgðpbrÞ P0 : It follows from the similar argument of the preceding paragraph and A2 that ½qKp; pbr P0 : Thus, by Claim A.1, qKp is thresholdfree, a contradiction. Last we assume that aa0 and ba1: If ½q; pbr [ ½q; par P0 ; then by Lemma A.2.1 and A4, hq; par; pbri Pþ : Thus by Claim A.1, q is thresholdfree, a contradiction. Therefore, we assume that hq; pari \ Pþ a; or hq; pbri \ Pþ a;: Then a similar argument of the preceding paragraph gives that hr; qi (respectively, hp; qi) contains a threshold-free gamble if hq; pari \ Pþ a; (respectively, hq; pbri \ Pþ a;), a contradiction. Case 2: Suppose that ½p; r P0 and hp; ri includes a threshold-free gamble t. Let s 2 Pþ : As illustrated in Fig. 8(b), it follows from Lemma A.2.3 that there are a; b; g 2 ð0; 1Þ such that ½p; t; pgs [ ½r; t; rbs P0 ; ½p; q; qas; pgs [ ½q; r; qas; rbs P0 ; þ
hs; pgs; rbs; ti P ; hs; qas; pgsi [ hs; qas; rbsi Pþ : Let 0odog and 0oob: Then pds; rs 2 Pþ : Let ft0 g ¼ ½pgs; qas \ hq; pdsi and ft00 g ¼ ½rbs; qas \ hq; rsi: Then t0 ; t00 2 P0 : By A4, ht0 ; t00 i P0 : Since d and is arbitrary in open intervals ð0; gÞ and ð0; bÞ; respectively, we obtain that hpgs; rbs; qasi P0 : However, hpgs; rbsi Pþ which contradicts A1. & Step 3: We say that the empty set is convex. This step proves the following claim. Claim A.3. There are two disjoint convex sets, I 1 and I 2 ; such that I ¼ I 1 [ I 2 :
where gð1 b dÞ ðqdrÞ gð1 a dÞ þ að1 bÞ að1 b dÞ ¼ ðpgrÞ ðrbqÞ: gð1 a dÞ þ að1 bÞ
s ¼ ðpaqÞ
Proof (See Fig. 9 for illustration). Suppose that p 2 P0 ; q 2 P ; and r 2 Pþ : Then by Lemma 3.2, there are 0oap1; 0obo1; 0ogp1; and 0odo1 such that b þ dp1; hq; paqi [ hq; rbqi P ; hr; pgri [ hr; qdri Pþ ; and hpaq; pi [ hp; pgri [ hrbq; qdri P0 : We show that hpaq; qdri [ hpgr; rbqi P0 : To show that hpaq; qdri P0 : Assume that there is a gamble t 2 hpaq; qdri such that t 2 Pþ : A similar proof applies to obtain that hpgr; rbqi P0 : By A1, there is a gamble w 2 hq; ti such that w 2 Pþ : As illustrated in Fig. 9, w 2 hq; pri for some 0ooa: Since pr 2 P ; Lemma 3.2(1) gives wePþ ; a contradiction. Therefore, tePþ : Similarly, teP : Hence hpaq; qdri P0 : Let s be a gamble defined in the lemma. Then s 2 ½p; qKr; where K¼
ð1 bÞðgð1 aÞð1 dÞ þ ð1 gÞadÞ : að1 bÞð1 gÞ þ gð1 aÞð1 dÞ
Since ½p; qKr P0 ; it follows from Lemma A.2.3 and A3 that either (1) or (2) holds. & Lemma A.3.2. If p 2 P ; q 2 P0 ; and paq 2 P0 for 0oao1; then ½q; paq I and hq; paqi InðT [ T þ Þ: Proof. Suppose that p 2 Pþ ; q 2 P0 ; and paq 2 P0 for 0oao1: When p 2 P ; the proof is similar. Then by Lemma 3.2, ½q; pbp 2 P0 and hpbq; pi Pþ for some 0obo1: Thus apb and q is not threshold-free. Suppose on the contrary that notðq paqÞ: Then paq must be threshold-free. Take any s 2 P ; so that hpaq; si P : By A1, pgs 2 Pþ for some 0ogo1: We note that a g q ¼ ðpaqÞ s: ðpgsÞ að1 gÞ þ g bð1 gÞ þ g By Lemma 3.2(1), this measure does not belong to P ; a contradiction. Therefore, paq is not lower threshold-free.
We need the following two lemmas to prove the claim. Lemma A.3.1. If p 2 P0 ; q 2 P ; and r 2 Pþ ; then there are 0oap1; 0obo1; 0ogp1; and 0odo1 such that b þ dp1; hq; paqi [ hq; rbqi P ; hr; pgri [ hr; qdri Pþ ; and either one of (1) and (2) holds. (1) ½p; paq; rbq; qdr; pgr P0 ; hq; paq; rbqi P ; and hr; pgr; qdri Pþ ; (2) ½p; paq; s; pgr [ ½s; rbq; qdr P0 ; hq; paq; s; rbqi P ; and hr; pgr; s; qdri Pþ ;
395
Fig. 9. An illustration of the proof of Lemma A.3.1.
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Fig. 10. An illustration of the proof of Claim A.3.
Let t 2 Pþ : Then, by Lemma 3.2(1), ½q; pgt P0 and hpgt; ti Pþ for some 0pgo1: By A2 and A4, ½q; pgt; pbq P0 : Since b40; A2 implies that g40: Thus paq is not upper threshold-free. Hence ½q; paq I and hq; paqi InðT [ T þ Þ: &
Proof of Claim A.3. (See Fig. 10 for illustration). We are to show that if not ðp qÞ and not ðq rÞ for p; q; r 2 I; then p r: We have the following four cases to examine: Case 1: ½p; q \ P a; and ½q; r \ P a; for 2 fþ; g; Case 2: ½p; q \ Pþ a; and ½q; r \ P a;; Case 3: ½p; q \ P a; for 2 fþ; g and ½q; r P0 ; ½q; rnIa;; Case 4: ½p; q [ ½q; r P0 ; ½p; qnIa;; and ½q; rnIa;: Case 1: Suppose that ½p; q \ P a; and ½q; r \ P a; for 2 fþ; g: Assume the case of ¼ þ: A proof is similar for ¼ : By Lemma 3.2(1), there are 0paogp1 and 0pbodp1 such that ½q; paq [ ½p; pgq [ ½q; rbq [ ½r; rdq P0 and hpaq; pgqi [ hrbq; rdqi Pþ : By A1 and A2, hpaq; rbqi P0 [ Pþ : If hpaq; rbqi \ þ P a;; then by A1 and A2, hpaq; rbqi Pþ : Thus by Claim A.1, q is threshold-free, a contradiction. Hence, hpaq; rbqi P0 :
By A2 and A3, hp; ri P0 [ Pþ : If hp; ri \ Pþ a;; then a similar argument of the preceding paragraph gives that p and r are threshold-free, a contradiction. Hence hp; ri P0 : By A1 and A2, hpgq; rdqi P0 [ Pþ : If hpgq; rdqi \ þ P a;; then a similar argument of Lemma A.2.3 gives that ½p; pgq; pr [ ½r; rdq; pr P0 and hpgq; rdq; pri Pþ for some 0oo1 as illustrated in Fig. 10(a). Take any s 2 hpaq; rbqi; t 2 hpgq; pri; and w 2 hrdq; pri; so that s; t; w 2 P0 and hs; ti [ ht; wi [ hw; si Pþ : This is prohibited by Lemma A.2.2. Hence ½pgq; rdq P0 : If g þ da2; then any gamble in hp; ri is not thresholdfree. Therefore, p r: Assume that g ¼ d ¼ 1; and pr is threshold-free for some 0oo1: Take any gamble t 2 hq; pri for which t 2 Pþ : Let s 2 P : Then by A1, there is a gamble t0 2 ht; si for which t0 2 Pþ : Let w be a gamble such that hw; qi contains t0 as illustrated in Fig. 10(b). By A3, w 2 Pþ [ P0 : Since pr is threshold-free, we must have w 2 P : This is a contradiction. Hence hp; ri contains no threshold-free gamble, so that p r: Case 2: Suppose that ½p; q \ Pþ a; and ½r; q \ P a;: By Lemma 3.2(1), there are 0paogp1 and 0pbodp1 such that ½q; paq [ ½p; pgq [ ½q; rbq [ ½r; rdq P0 ; hpaq; pgqi Pþ ; and hrbq; rdqi P : By Case 1, it is prohibited that hp; ri \ P a; for 2 fþ; g: Thus hp; ri P0 : Suppose that pr is thresholdfree for some 0oo1: Then hpr; paq; pgqi Pþ and
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hpr; rbq; rdqi P : Take any s 2 hrbq; rdqi: By A3, hp; si P0 [ P : However, as illustrated in Fig. 10(c), there is a gamble t 2 hp; si for which t 2 hpr; paq; pgqi; a contradiction. Hence hp; ri contains no threshold-free gamble, so p r: Case 3: Suppose that ½q; r P0 ; ½q; rnIa;; and ½p; q \ P a; for 2 fþ; g: Assume the case of ¼ þ: A proof is similar for ¼ : By Lemma 3.2(1), there are 0paogp1 such that ½q; paq [ ½p; pgq P0 and hpaq; pgqi Pþ : Let rbq be a threshold-free gamble in hq; ri: By Cases 1 and 2, hp; ri P0 : Suppose that hp; ri contains a threshold-free gamble, say pr for some 0oo1: Then ½p; pgq; pr [ ½q; paq; rbq [ ½r; rbq; pr P0 and hpaq; rbq; pgq; pri Pþ : As illustrated in Fig. 10(d), take any gambles s 2 hpaq; rbqi; t 2 hpgq; pri; and w 2 hrbq; pri; so that hs; ti [ ht; wi [ hw; si Pþ : This is prohibited by Lemma A.2.2. Hence hp; ri contains no threshold-free gamble, so p r: Case 4: Suppose that ½p; q [ ½q; r P0 ; ½p; qnIa;; and ½q; rnIa;: By Case 3, hp; ri \ P ¼ ; for 2 fþ; g: Suppose that ½p; rnIa;: Let paq; qbr; and pgr be, respectively, threshold-free gambles in hp; qi; hq; ri; and hp; ri: Then every gamble in hpaq; qbri is threshold-free. Since hpaq; qbri \ hq; pgria;; hq; pgri contains more than two threshold-free gambles. This contradicts Lemma A.2.3 which says that hq; pgri contains at most one threshold-free gamble, since q is not threshold-free. Hence hp; rinI ¼ ;; so p r: & Step 4. This step completes the sufficiency proof of Theorem 3.1. For fi; jg ¼ f1; 2g we define þ Qþ i ¼ P [ ðI j nT Þ; þ Q i ¼ P [ ðI i nT Þ;
Q0i ¼ ðT þ \ I i Þ [ ðT \ T þ Þ [ ðT \ I j Þ: We need the following lemma Lemma A.4.1. For i ¼ 1; 2 and 2 f; þg; we have (1) Qþ i and Qi are not empty. þ 0 (2) if p 2 Qi and q 2 Q i ; then paq 2 Qi for some 0oao1: (3) if p 2 Qi and q 2 Qi [ Q0i ; then hp; qi Qi : (4) if p; q 2 Q0i ; then hp; qi Q0i :
Proof. (1) This follows from P a; and Pþ a;: 0 (2) We show that if p 2 Qþ 1 and q 2 Q1 ; then paq 2 Q1 for some 0oao1: When i ¼ 2; the proof is similar. We have four cases to examine Case 1: p 2 Pþ ; q 2 P ; Case 2: p 2 Pþ ; q 2 I 1 nT þ ; Case 3: p 2 I 2 nT ; q 2 P ; Case 4: p 2 I 2 nT ; q 2 I 1 nT þ : Case 1: By Lemma 3.2(2), there are 0oapbo1 such that hq; paqi P ; ½paq; pbq P0 ; and hpbq; qi Pþ : We note that paq 2 T ; pbq 2 T þ ; and either
397
½paq; pbq I 1 or ½paq; pbq I 2 or ½paq; pbq ðT \ T þ Þ: Thus we have paq 2 ðT \ I 2 Þ Q01 when ½paq; pbq I 2 ; pbq 2 ðT þ \ I 1 Þ Q01 when ½paq; pbq I 1 ; paq 2 ðT þ \ T þ Þ Q01
when ½paq; pbq ðT \ T þ Þ:
Case 2: By Lemma 3.2(1), there is an 0oao1 such that hp; paqi Pþ and ½q; paq I 1 : Then we have paq 2 T þ \ I 1 Q01 : Case 3: Similar to Case 2. Case 4: Suppose that p 2 I 2 nT and q 2 I 1 nT þ : Assume first that ½p; q \ Pþ a;: When ½p; q \ P a;; the proof is similar. Then, by Lemma 3.2(1), there are 0oaobp1 such that ½q; paq I 1 ; hpaq; pbqi Pþ ; and ½pbq; p I 2 : Note that paq 2 I 1 \ T þ and pbq 2 I 2 \ T þ : Thus paq 2 Q01 : Assume next that ½p; qnIa;; paq 2 ½p; qnI for some 0oao1: Thus paq 2 T þ \ T Q01 : (3) We show that if p 2 Q1 and q 2 Q1 [ Q01 ; then hp; qi Q1 : When i ¼ 2; the proof is similar. We prove the case of ¼ þ: The proof for ¼ is similar. We note that þ p 2 Qþ 1 3 p 2 P [ ðI 2 nT Þ; 0 þ þ q 2 Qþ 1 [ Q1 3 q 2 P [ I 2 [ ðT nI 2 Þ:
We are to show that hp; qi Pþ [ ðI 2 nT Þ: We have two cases to examine: p 2 Pþ ; p 2 I 2 nT : Case 1: ðp 2 Pþ Þ If q 2 Pþ ; then by A2, hp; qi Pþ þ Q1 : If q 2 I 2 ; then by Lemma 3.2(1), ½q; paq P0 and hpaq; qi Pþ for some 0pao1: By Lemma A.3.2, hq; paqi I 2 nðT [ T þ Þ: Since paq 2 T þ ; we obtain that hp; qi Pþ [ ðI 2 nT Þ: If q 2 T þ nI 2 ; then it follows from the definition of T þ that hp; qi Pþ : Case 2: ðp 2 I 2 nT Þ Take any s 2 P ; so that hs; pasi P and ½pas; p P0 for some 0oao1: If q 2 Pþ ; then the desired result follows from Case 1. If q 2 I 2 ; then hp; qi I 2 by the convexity of I 2 : It follows from Lemma A.2.3 that plq is threshold-free when plq 2 T : Since there is no threshold-free measure in hp; qi; we have hp; qi I 2 nT : Suppose that q 2 T þ nI 2 : Let t 2 Pþ : By definition of þ T ; hq; ti Pþ ; since q 2 T þ : It follows from A1 and A3 that hp; qi \ P ¼ ;: Assume first that hp; qi \ Pþ a;: Then hpaq; qi Pþ and ½paq; p P0 for some 0oap1: By Lemma A.3.2, ½paq; p I 2 nT : Hence, hp; qi Pþ [ ðI 2 nT Þ: Assume next that hp; qi P0 : Thus, by Lemma A.2.3, q 2 T þ \ T : Therefore, Lemma A.2.3 implies that any measure in hp; qi is not lower thresholdfree. Hence hp; qi I 2 nT : (4) We show that if p; q 2 Q01 ; then hp; qi Q01 : When i ¼ 2; the proof is similar. Thus, we assume that p; q 2 ðT þ \ I 1 Þ [ ðT \ T þ Þ [ ðT \ I 2 Þ: Take any s 2 P and any t 2 Pþ :
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Suppose that p 2 T þ \ I 1 : When p 2 T \ I 2 ; the proof is similar. If q 2 T þ \ I 1 ; then hp; qi I 1 ; since I 1 is convex. By A2, hp; qi T þ : Thus hp; qi Q01 : Assume next that q 2 T \ I 2 : By A1 and A3, hp; qi 0 P : Since p 2 I 1 and q 2 I 2 ; there must be a thresholdfree measure in hp; qi: It follows from Lemma A.2.3 that, for some a; b; l 2 ð0; 1Þ; ½p; plq; pas [ ½q; plq; qbt P0 ; hs; pas; plq; qi P ; and ht; p; plq; qbti Pþ : Thus hp; plqi T \ I 1 and hq; plqi T \ I 2 : Hence hp; qi Q01 : Assume last that q 2 T \ T þ : By A2, hp; q; ti þ T \ I 1 ; so that hp; qi Q01 : Suppose that p; q 2 T \ T þ : Then hp:qi T \ T þ : Hence hp; qi Q01 : & Proof of Theorem 3.1. It follows from Lemma A.4.1 and Theorem 3.2 that there are linear functionals, u1 and u2 ; on GX such that, for all p 2 GX and i ¼ 1; 2; p 2 Qþ i 3 ui ðpÞ40; p 2 Q0i 3 ui ðpÞ ¼ 0; p 2 Q i 3 ui ðpÞo0: Furthermore, u1 and u2 are unique up to multiplicative transformations by positive constants. Since I 1 and I 2 are disjoint, it follows from the definitions that P ¼
Q1 \ Q2 for 2 fþ; g: Thus þ p 2 Pþ 3p 2 Qþ 1 \ Q2 3u1 ðpÞ40 and u2 ðpÞ40; p 2 P 3p 2 Q 1 \ Q2 3u1 ðpÞo0 and u2 ðpÞo0:
This completes the proof.
&
References Fishburn, P. C. (1982). Nontransitive measurable utility. Journal of Mathematical Psychology, 26, 31–67. Fishburn, P. C. (1988). Nonlinear preference and utility theory. Baltimore: Johns Hopkins University Press. Goldstein, W. M. (1991). Decomposable threshold models. Journal of Mathematical Psychology, 35, 64–79. Karni, E., & Schmeidler, D. (1991). Utility theory with uncertainty. In Hildenbrand, W., & Sonnenschein, H. (Eds.), Handbook of mathematical economics, Vol. 4. Amsterdam: North-Holland. Quiggin, J. (1993). Generalized expected utility theory: the rank dependent model. Dordrecht: Kluwer Academic. Schmidt, U., 1998. Axiomatic utility theory under risk. Lecture notes in economics and mathematical systems (Vol. 461). Berlin: Springer.
Journal of Mathematical Psychology 48 (2004) 385–398 www.elsevier.com/locate/jmp
Trichotomous preferences for gambles Yutaka Nakamura Graduate School of Systems and Information Engineering, Division of Policy and Planning Sciences, University of Tsukuba, 1-1-1 Tennoudai, Tsukuba, Ibaraki 305-8573, Japan Received 9 July 2002; revised 23 April 2004 Available online 10 November 2004
Abstract Let GX be the set of all gambles on a set X of decision outcomes. Trichotomous preferences divide GX into three disjoint subsets, viz., the set Pþ of preferable gambles, the set P0 of indifferent gambles, and the set P of unpreferable gambles. Those preference comparisons are made against the status quo, i.e., the do-nothing alternative. This paper presents and discusses dyadic representations of such trichotomous preferences and identifies necessary and sufficient axioms for the existence of those representations. r 2004 Elsevier Inc. All rights reserved. Keywords: Trichotomous preferences; Decisions under risk; Non-expected utility; Bounded rationality; Satisficing criterion
1. Introduction Let GX be the set of all gambles defined on a set X of decision outcomes. A gamble is a real-valued function p from X into the closed unitP interval ½0; 1 for which fx 2 X : pðxÞa0g is finite and x2X pðxÞ ¼ 1: Each pðxÞ is interpreted as the probability with which decision outcome x is obtained. The traditional approaches for rational choice under risk are concerned with optimality criteria that identify the most preferred gamble out of a given set of gambles. The most influential principle of optimality is maximization of expected utility (EU). However, a vast amount of evidence, accumulated in the last halfcentury, uncovered that people do not actually behave like an EU maximizer in many choice situations. As a response to this empirical fact, numerous descriptive and axiomatic alternatives to the EU maximization principle have been proposed and tested for their validity (see Fishburn, 1988; Karni and Schmeidler, 1991; Quiggin, 1993; Schmidt, 1998 and many others for Fax.: +81 29 853 5070.
E-mail address: [email protected]. URL: http://infoshako.sk.tsukuba.ac.jp/ nakamura/. 0022-2496/$ - see front matter r 2004 Elsevier Inc. All rights reserved. doi:10.1016/j.jmp.2004.08.006
surveys). Like EU theories, those are also concerned with seeking optimality criteria. In these alternatives, it is presumed that the decision maker has his/her wellstructured preferences for gambles, so that quantitative representations of preferences are axiomatically obtained. On the other hand, this paper is concerned with a satisficing criterion which is based on trichotomous preferences for gambles that identify preferable/unpreferable gambles out of a given set of gambles. Trichotomous preferences divide GX into three disjoint subsets, Pþ ; P0 ; and P ; which are interpreted as follows: p 2 Pþ 3 p is preferable; p 2 P0 3 p is indifferent; p 2 P 3 p is unpreferable: Those trichotomous judgments are made against the status quo, i.e., the do-nothing alternative. Although axiomatic analyses of satisficing criteria are important in view of modeling bounded rationality of human judgments, there seems to be no attempt to axiomatize satisficing criteria for preference judgments in the literature except Goldstein (1991) who axiomatized
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trichotomous preferences for multiattributed alternatives under certainty. The aim of the paper is to present and discuss dyadic representations of a tripartite partition P3 ¼ fPþ ; P0 ; P g of GX and to identify qualitative conditions on P3 that are necessary and sufficient for the existence of those representations. The most general dyadic representation that we shall consider yields two realvalued functions f and c on X such that, for all p 2 GX ; X X pðxÞfðxÞ40 and pðxÞcðxÞ40; p 2 Pþ 3 x2X
p 2 P 3
X x2X
x2X
pðxÞfðxÞo0
and
X
pðxÞcðxÞo0:
x2X
This representation may be interpreted as follows. There are two criteria that, respectively, divide decision outcomes into three categories: preferable, indifferent, and unpreferable. Thus, a decision outcome is preferred (respectively, unpreferred) if it is preferable (respectively, unpreferable) on both criteria. To extend those trichotomous judgments to the set of all gambles, two functions f and c on X are introduced to represent ‘‘intensity of preference’’ in each criterion, where negative values of f and c represent ‘‘intensity of undesirability.’’ Then a gamble is judged to be preferred (respectively, less preferred) to the status-quo when expected intensities of preference with respect to both criteria are positive (respectively, negative). When f and c are related by a positive multiplicative constant, the two criteria give identical trichotomous judgments for gambles. In this case, c and P pðxÞcðxÞ may be interpreted as a utility function x2X on X and expected utility of gamble p, respectively, which are conditioned P Pupon the status quo. Therefore, if pðxÞcðxÞ404 x2X x2X qðxÞcðxÞ; then the decision maker should play gamble p and should not play gamble q, since p is preferred to the status quo which is more preferable than gamble q. However, trichotomous judgments say nothing about which gamble the decision maker should take among preferable gambles. In general, the dyadic representations for trichotomous judgments can be transformed into utility representations with threshold conditional upon the status quo. It is shown that there exist two real-valued functions u and o on X such that, for all p 2 GX ; X X þ pðxÞuðxÞ4 pðxÞoðxÞ; p2P 3 x2X x2X X X p2P 3 pðxÞuðxÞo pðxÞoðxÞ; x2X x2X where u and o may be regarded as a utility function and a threshold function, respectively. This paper is organized as follows. Section 2 shows the uniqueness properties of the dyadic and threshold
representations, and explores the relationships between those representations. Section 3 identifies qualitative conditions of P3 that are necessary and sufficient for the existence of dyadic representations. We conclude the paper in Section 4. The proof of the main representation theorem is deferred to the appendix.
2. Dyadic and threshold representations Throughout the paper, we shall assume that P3 ¼ fPþ ; P0 ; P g is a tripartite partition of GX : First, we show the uniqueness of the dyadic representations. Then we examine the relationship between dyadic and threshold representations. 2.1. Dyadic representations For a probability number 0plp1; a probability mixture of two gambles p and q, denoted plq; is a gamble that yields outcome x with probability lpðxÞ þ ð1 lÞqðxÞ: Given a real-valued function f P on X, we shall extend the domain of f to GX by f ðpÞ ¼ x2X f ðxÞpðxÞ for all p 2 GX : Then it is easy to see that f on GX is linear, i.e., for all p; q 2 GX and all 0olo1; f ðplqÞ ¼ lf ðpÞ þ ð1 lÞf ðqÞ: We say that ðGX ; P3 Þ has a dyadic representation, denoted ðf; cÞ; if f and c are linear functions on GX such that, for all p 2 GX ; p 2 Pþ 3 fðpÞ40 and cðpÞ40; p 2 P 3 fðpÞo0 and cðpÞo0: Note that p 2 P0 if and only if fðpÞcðpÞp0: The uniqueness of dyadic representations is stated as follows: Theorem 2.1. Suppose that Pþ and P are not empty, and ðGX ; P3 Þ has a dyadic representation ðf; cÞ: Then ðf0 ; c0 Þ is also a dyadic representation for ðGX ; P3 Þ if and only if there are positive numbers, a and b; such that ff0 ; c0 g ¼ faf; bcg: Proof. Suppose that ðGX ; P3 Þ has a dyadic representation ðf; cÞ: If ff0 ; c0 g ¼ faf; bcg for some a40 and b40; then it easily follows that ðf0 ; c0 Þ is a dyadic representation for ðGX ; P3 Þ: Suppose that ðf0 ; c0 Þ is also a dyadic representation for ðGX ; P3 Þ: For any linear function f on GX ; let Iðf Þ ¼ fp 2 GX : f ðpÞ ¼ 0g: Then IðfÞ; IðcÞ; Iðf0 Þ; and Iðc0 Þ are not empty. Take any p 2 IðfÞ: Assume that f0 ðpÞc0 ðpÞo0: Let q 2 Pþ and r 2 P : Then ðf0 ; c0 Þ representation implies that paq; pbr 2 P0 for some 0oao1 and 0obo1: On the other hand, it follows from ðf; cÞ representation that plq 2 Pþ for all 0olo1; or plr 2 P for all 0olo1: This is a contradiction. Therefore, f0 ðpÞc0 ðpÞ ¼ 0:
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Hence, since I sets are convex, we can assume with no loss of generality that IðfÞ ¼ Iðf0 Þ and IðcÞ ¼ Iðc0 Þ: It suffices to show that f0 ¼ af for some positive number a: Let fðpÞ40 for a fixed p 2 Pþ and define a by a¼
f0 ðpÞ ; fðpÞ
which must be positive. We prove in turn that, for all q 2 GX ; (1) fðqÞo0 ) f0 ðqÞ ¼ afðqÞ; (2) fðqÞ40 ) f0 ðqÞ ¼ afðqÞ: (1) Given fðqÞo0 along with fðpÞ40; define l by lfðpÞ þ ð1 lÞfðqÞ ¼ 0: Then by linearity, fðplqÞ ¼ 0: Since IðfÞ ¼ Iðf0 Þ; we have f0 ðplqÞ ¼ lf0 ðpÞ þ ð1 lÞf0 ðqÞ ¼ 0: Hence f0 ðqÞ ¼
l f0 ðpÞfðqÞ f0 ðpÞ ¼ ¼ afðqÞ: 1l fðpÞ
(2) Take any r 2 GX for which fðrÞo0: Then fðqbrÞo0 for some 0obo1: By (1), f0 ðrÞ ¼ afðrÞ and f0 ðqbrÞ ¼ afðqbrÞ: Then linearity of f and f0 gives f0 ðqÞ ¼ afðqÞ: & 2.2. Threshold representations We say that ðGX ; P3 Þ has a threshold representation, denoted ðu; wÞ; if u and w are linear functions on GX such that, for all p 2 GX ; p 2 Pþ 3 uðpÞ4jwðpÞj;
p2P
The uniqueness of the threshold representation is given by Theorem 2.3. Suppose that Pþ and P are not empty, and ðGX ; P3 Þ has a threshold representation ðu; wÞ: Then ðu0 ; w0 Þ is also a threshold representation for ðGX ; P3 Þ if and only if there are two real numbers a and b such that a þ b40; a b40; u0 ¼ au þ bw; and w0 ¼ bu þ aw or w0 ¼ bu aw: Proof. Suppose that ðu; wÞ is a threshold representation for ðGX ; P3 Þ: Assume first that ðu0 ; w0 Þ is also a threshold representation for ðGX ; P3 Þ: Then ðu þ w; u wÞ and ðu0 þ w0 ; u0 w0 Þ are, respectively, the corresponding dyadic representations. Thus by the uniqueness of the dyadic representation, there are positive numbers a0 and b0 such that fu0 þ w0 ; u0 w0 g ¼ fa0 ðu þ wÞ; b0 ðu wÞg: Let a ¼ ða0 þ b0 Þ=2 and b ¼ ða0 b0 Þ=2; so a þ b40 and a b40: If u0 þ w0 ¼ a0 ðu þ wÞ and u0 w0 ¼ b0 ðu wÞ; then solving these with respect to u0 and w0 ; we obtain that u0 ¼ au þ bw and w0 ¼ bu þ aw: If u0 þ w0 ¼ b0 ðu wÞ and u0 w0 ¼ a0 ðu þ wÞ; then similarly we obtain that u0 ¼ au þ bw and w0 ¼ bu aw: Assume next that a þ b40; a b40; u0 ¼ au þ bw; and w0 ¼ bu þ aw or w0 ¼ bu aw: Then for all p 2 GX ; u0 ðpÞ4jw0 ðpÞj 3u0 ðpÞ4w0 ðpÞ and u0 ðpÞ4 w0 ðpÞ 3auðpÞ þ bwðpÞ4buðpÞ þ awðpÞ and auðpÞ þ bwðpÞ4 buðpÞ awðpÞ 3ða bÞuðpÞ4ða bÞwðpÞ
3 uðpÞo jwðpÞj:
Note that p 2 P0 if and only if juðpÞjpjwðpÞj: The following theorem shows the relationship between dyadic and threshold representations. þ
Theorem 2.2. Suppose that P and P are not empty. Then ðGX ; P3 Þ has a dyadic representation ðf; cÞ if and only if ðf þ c; f cÞ is a threshold representation for ðGX ; P3 Þ: Proof. Let ðf; cÞ be a dyadic representation for ðGX ; P3 Þ: Then to see that ðf þ c; f cÞ is a threshold representation for ðGX ; P3 Þ; it suffices to observe that, for all p 2 GX ; p 2 Pþ 3fðpÞ40 and cðpÞ40 3fðpÞ þ cðpÞ4cðpÞ fðpÞ and fðpÞ þ cðpÞ4fðpÞ cðpÞ 3fðpÞ þ cðpÞ4jfðpÞ cðpÞj: The converse is similar.
387
&
It follows from Theorems 2.1 and 2.2 that ðu; wÞ is a threshold representation for ðGX ; P3 Þ if and only if ðu þ w; u wÞ is a dyadic representation for ðGX ; P3 Þ:
and ða þ bÞuðpÞ4 ða þ bÞwðpÞ 3uðpÞ4jwðpÞj: Similarly, for all p 2 GX ; u0 ðpÞo jw0 ðpÞj if and only if uðpÞo jwðpÞj: Hence the desired result obtains. &
3. Axiomatizations For any positive integer n and gambles p1 ; . . . ; pn 2 GX ; we define ½p1 ; . . . ; pn ¼ fð ððp1 l1 p2 Þl2 p3 Þ . . .Þln1 pn 2 GX : 0pli p1 for i ¼ 1; . . . ; n 1g; hp1 ; . . . ; pn i; ¼ fð ððp1 l1 p2 Þl2 p3 Þ . . .Þln1 pn 2 GX : 0oli o1 for i ¼ 1; . . . ; n 1g; where ½p1 ; . . . ; pn is a convex hull generated by p1 ; . . . ; pn ; and hp1 ; . . . ; pn i is the interior of ½p1 ; . . . ; pn
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when p1 ; . . . ; pn are distinct gambles. We note that hp; pi ¼ fpg; since p ¼ plp for all 0olo1: The following axioms apply to all p; q; r 2 GX ; all 0olo1; and all 2 fþ; g; where P should be read as either Pþ or P : A1. If p 2 P ; then, for every qeP ; paq 2 P for some 0oao1: A2. If p; q 2 P and hr; pi P ; then hr; qi P : A3. If p 2 P and q 2 P0 ; then hp; qi P [ P0 : A4. If p; q 2 P ; and fr; plr; qlrg P0 ; then hplr; qlri P0 : Axiom A1 is an Archimedean axiom, which says, for example, that if gamble p is preferable (i.e., p 2 Pþ ), then, for every gamble q that is not preferable, there is a probability mixture of p and q that should be preferable. This axiom seems to be defended by the argument that if a in the axiom is very close to 1, paq should be preferred (respectively, unpreferred) to the status quo when þ is substituted for (respectively, for ). Axioms A2–A4 are concerned with convexity properties of Pþ ; P0 ; and P : Axiom A2 implies the following convexity property of Pþ and P : for all p; q 2 GX and 2 fþ; g;
p; q 2 P
) hp; qi P :
We shall use this fact by referring to axiom A2. Note that we may have reP in axiom A2. Suppose that gambles p and q are preferable and gamble r is not preferable. Then axiom A2 requires that if a probability mixture of p and r is not preferable, then there must be a probability mixture of q and r that is not preferable. This is a requirement of the threshold representation ðu; wÞ with wðrÞa0: Axiom A3 says, for example, that if p is preferable and q is indifferent, then there is no probability mixture of p and q that is unpreferable. Although P0 is not necessarily convex, axiom A4 requires that for a convex hull ½p; q; r; if p; q 2 P and r 2 P0 ; then P0 \ ½p; q; r must be convex. Observe for example that if p; q 2 Pþ and r 2 P ; then P0 \ ½p; q; r may not be convex. The main representation theorem is given by Theorem 3.1. Suppose that P and Pþ are not empty. Then ðGX ; P3 Þ has a dyadic representation if and only if axioms A1–A4 hold. Proof. See appendix
&
In what follows, we examine three special cases of dyadic representations. Suppose that ðu; wÞ is a threshold representation for ðGX ; P3 Þ: Then we say that ðGX ; P3 Þ has a linear threshold representation if wX0; a constant threshold representation if w is constant, and a threshold-free representation if w ¼ 0: It follows from Theorems 2.1 and 2.2 that, for any
number l40; ðGX ; P3 Þ has a dyadic representation ðf; cÞ 3it has a dyadic representation ðf; lcÞ 3it has a threshold representation ðu; wÞ; where u ¼ f þ lc and w ¼ f lc: Therefore, ðGX ; P3 Þ has a linear threshold representation 3fXac for some a40; a constant threshold representation 3f ¼ ac þ b for some a40 and b; a threshold-free representation 3f ¼ ac for some a40: First, we consider axioms for threshold-free representation. The following axioms apply to all p; q 2 GX ; all 0olo1; and 2 fþ; g; where P should be read as either Pþ or P : A10 : If p 2 Pþ and q 2 P ; then paq 2 P0 for some 0oao1: A20 : If p 2 P and q 2 P [ P0 ; then plq 2 P : A30 : If p; q 2 P0 ; then hp; qi P0 : Axioms A10 is a solvability axiom. Axiom A20 says, for example, that if gamble p is preferable and gamble q is preferable or indifferent, then every probability mixture between p and q is preferable. This requires that threshold vanishes. Convexity of Pþ and P follows from this axiom. Axiom A30 requires that P0 be convex. The representational implication of axioms A10 –A30 is stated by the following theorem. Theorem 3.2. Suppose that Pþ and P are not empty. Then ðGX ; P3 Þ has a threshold-free representation if and only if axioms A10 –A30 hold. Proof. Suppose that Pþ and P are not empty. The necessity of axioms A10 –A30 easily obtains. Their sufficiency follows from Lemma 3 in Fishburn (1982). To see this, we assume that axioms A10 –A30 hold. Then, by axiom A10 ; P0 a;: Fix a gamble r 2 P0 : Define two binary relations, r and r on GX as follows: for all p 2 GX ; p 2 Pþ ) pr r; p 2 P ) rr p; p 2 P0 ) p r r and r r p: Fishburn’s lemma claims that there is a linear function u on GX such that, for all p 2 GX ; pr r 3 uðpÞ40; rr p 3 uðpÞo0; whenever the following three conditions hold, understood as applying to all p; q 2 GX and all 0olo1:
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B1. If pr r and rr q; then paq r r for some 0oao1: B2. If pr r and :ðrr qÞ; then plqr r; if rr p and :ðqr rÞ; then rr plq: B3. If p r r and q r r; then plq r r: Note that B1–B3, respectively, follow from axioms A10 –A30 : This completes the proof. & It is easy to construct examples that axiom A10 holds but A1 fails to hold, and vice versa. We shall identify each outcome x 2 X with the gamble that yields x with probability one. Let X ¼ fx; yg: Assuming that x 2 Pþ ; y 2 P ; and hx; yi P0 ; we conclude that axiom A10 hold, but not axiom A1. Next we assume that fxay : a rational and 0oap1g Pþ ; fxay : a irrational and 0pao1g P : Hence it readily follows that axiom A1 hold, but not axiom A10 : When Pþ and P are convex, axiom A1 implies axiom A10 as shown by the following lemma. Lemma 3.1. Axioms A1 and A2 imply axiom A10 : Proof. Suppose that axioms A1 and A2 hold. Assume that p 2 Pþ and q 2 P : Let a ¼ inffl : plq 2 Pþ and 0olp1g: Then, by axiom A1, 0oao1: Axiom A2 implies that plq 2 Pþ
if aolo1;
plq 2 P
if 0oloa:
Hence, by axiom A1, paq 2 P0 :
&
The following implications of axioms A1–A3 will be used in the proof of Theorems 3.1 in the appendix and the proof of Theorem 3.3 below. Lemma 3.2. Suppose that axioms A1–A3 hold. (1) If 2 fþ; g; p 2 P ; and q 2 P0 ; then hp; paqi P and ½q; paq P0 for some 0pao1: (2) If p 2 Pþ and q 2 P ; then hp; pbqi Pþ ; ½pbq; paq P0 ; and hpaq; qi P for some 0oapbo1: Proof. Suppose that Axioms A1–A3 hold. (1) Suppose that 2 fþ; g; p 2 P ; and q 2 P0 : Then by axiom A3, hp; qi P [ P0 : It follows from axiom A2 that pmq 2 P0 whenever plq 2 P0 and 0pmolo1: Let a ¼ supfl : plq 2 P0 and 0plo1g: Then, by axiom A1, 0pao1 and paq 2 P0 : Hence hp; paqi P and ½q; paq P0 : (2) Suppose that p 2 Pþ and q 2 P : Then, by Lemma 3.1 and axiom A10 ; there is an 0oao1 such that paq 2 P0 : Hence the conclusion of (2) follows from (1). &
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The following axiom, which is understood as applying to all p; q; r; s 2 GX ; is devised for constant threshold representation. A40 : If p; q 2 P0 ; r; seP0 ; and, for all 0omo1; pmr 2 P0 3qmr 2 P0 ; then, for all 0omo1; pms 2 P0 3qms 2 P0 : Suppose that gambles p and q are indifferent and r is preferable. Then axiom A40 says that if the same probability mixtures between p and r and between q and r are preferable, then gamble r can be replaced by any other gamble s that is not indifferent. Although the axiom looks complicated, it seems intuitively clear that threshold must be constant. Axiomatic characterizations of linear threshold and constant threshold representations are given by Theorem 3.3. Suppose that Pþ and P are not empty. (1) ðGX ; P3 Þ has a linear threshold representation if and only if axioms A1, A2, and A30 hold; (2) ðGX ; P3 Þ has a constant threshold representation if and only if axioms A1, A2, A30 ; and A40 holds. Proof. The necessity of the axioms in (1) and (2) easily follows. We show sufficiency of those axioms for each representation. (1) Suppose that axioms A1, A2, A30 hold. Axiom A4 follows from axiom A30 : To see that axiom A3 holds, assume that p 2 Pþ ; q 2 P0 ; and paq 2 P for some 0oao1: It follows from Lemma 3.2(2) that pbðpaqÞ ¼ pða þ b abÞq 2 P0 for some 0obo1: Then by A30 ; a q ¼ paq 2 P0 ; ðpða þ b abÞqÞ a þ b ab a contradiction. Hence A3 obtains. By Theorem 3.1, there is a dyadic representation ðf; cÞ for ðGX ; P3 Þ: Then we define R ¼ fðfðpÞ; cðpÞÞ 2 R2 : p 2 GX g: Note that R is convex. For any real number l; half-spaces are defined by 2 Hþ l ¼ fða; bÞ 2 R : aXlbg; 2 H l ¼ fða; bÞ 2 R : aplbg: We are to show that R H þ l or R H l for some l40: Thus either cðpÞXlfðpÞ for all p 2 GX or cðpÞplfðpÞ for all p 2 GX : Hence, by Theorems 2.1 and 2.2, ðGX ; P3 Þ has a linear threshold representation. Since Pþ and P are not empty, there are ða; bÞ; ða0 ; b0 Þ 2 R such that a40; b40; a0 o0; and b0 o0: Assume first that R has no interior point. Then it easily follows that R is contained in a straight line L, where L ¼ fða; bÞ 2 R2 : a ¼ lb þ gg for some l40: Hence R H þ l or R H l : Assume next that R has an interior point. Let R0 be the interior of R; so it is an open convex set in R2 : By axiom A30 ; P0 is convex, so is fða; bÞ 2 R : abp0g: Therefore, axiom A30 forces R0 to exclude the origin
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ð0; 0Þ: Since R is convex, we must have that R H þ l or R H : l (2) Suppose that axioms A1, A2, A30 ; and A40 hold. Then by (1), there is a dyadic representation ðf; cÞ with, say, fXc for ðGX ; P3 Þ: Take any p; q 2 P0 : Let s 2 P and r 2 Pþ : Then ½p; plr [ ½p; pms P0 ; hr; plri Pþ ; and hs; pmsi P for some 0olp1 and 0omp1: Thus lcðpÞ þ ð1 lÞcðrÞ ¼ 0 and lfðpÞ þ ð1 lÞfðsÞ ¼ 0: Axiom A40 requires that lcðqÞ þ ð1 lÞcðrÞ ¼ 0 3 mfðqÞ þ ð1 lÞfðsÞ ¼ 0 or equivalently, cðpÞ ¼ cðqÞp03fðpÞ ¼ fðqÞX0: This implies that R ¼ fðfðpÞ; cðpÞÞ 2 R2 : p 2 GX g must be a straight line in R2 ; i.e., c ¼ af þ b for some numbers a40 and b: &
4. Conclusions We studied a satisficing criterion by axiomatizing trichotomous judgments in decision making under risk. In particular, we proposed dyadic representations of trichotomous preferences for gambles and explored their axiomatizations. Trichotomous judgments are modeled by three unary relations, Pþ ; P ; and P0 ; on GX : These sets are mutually exclusive and exhaustive, and respectively interpreted as the set of preferred gambles, the set of unpreferred gambles, and the set of indifferent gambles, where these judgments are made against the status quo. We say that a subset A of GX is a decision problem when the decision maker may choose one element of A. For such a decision problem, a satisficing procedure based on trichotomous preferences may go as follows. The decision maker sequentially examines the gambles in A according to a prespecified ordering. Once he/she confronts a gamble that is a member of the set Pþ ; he/she stops and chooses it. For the case where no element of A belongs to Pþ ; he/she will maintain the status quo. The satisficing procedure may be numerically characterized by introducing a real valued function V on GX for which Pþ ¼ fp 2 GX : V ðpÞ40g: The decision maker searches for a gamble that satisfies the condition that V-value be positive. Note that instead of having a maximization problem, maxp2A V ðpÞ; the decision maker seems to solve a simpler problem, i.e., find a p 2 A for which V ðpÞ40: Dyadic representations provide a V characterization of the satisficing procedure which consists of two linear functions on GX : That is, there are two linear functions u and o on GX such that V ðpÞ ¼ uðpÞ joðpÞj for all p 2 GX : A natural generalization of dyadic representations is to consider a set U of linear functions on GX for which,
for all p 2 GX ; p 2 Pþ 3 uðpÞ40 for all u 2 U; p 2 P 3 uðpÞo0
for all u 2 U;
where U has at most two elements for dyadic representations. U may be infinite. When U has more than two elements, it may lose threshold-interpretation as demonstrated in Theorem 2.2 for the dyadic representations. It remains to be an open problem to develop sufficient (preferably necessary) axioms for the existence of U: Note that axioms A1–A3 are necessary for U-representations. However, axioms A4 fails to hold when U contains more than two elements that are mutually independent in the sense that one is not a positive linear transformation of the other. Necessary and sufficient axioms for dyadic representations were devised by an Archimedean property and convexity properties of Pþ ; P ; and P0 : Convexity properties are empirically testable. In particular, Pþ and P must be convex. Thus, for example, for any preferred gambles p and q, there seems to be no difficulty to empirically test whether any probability mixture of p and q is a preferred gamble or not. Although P0 need not be convex, the set of indifferent gambles in any convex hull ½p1 ; . . . ; pn must be convex whenever p1 ; . . . ; pn 2 Pþ [ P0 or p1 ; . . . ; pn 2 P [ P0 : In other words, if pi 2 Pþ and pj 2 P for some distinct i and j, then the set of indifferent gambles in ½p1 ; . . . ; pn may not be convex. Finally, we comment on the intuitive relationship between trichotomous judgments and the traditional preference judgments. Traditionally, preference judgments are modeled by a subset P of GX GX : When a pair ðp; qÞ of gambles belongs to P, gamble p is judged to be preferred to gamble q. Suppose that the decision maker already holds the right to play gamble r. It is also assumed that he/she is offered to exchange r for some other gamble. Then trichotomous judgments are made against the status quo (i.e., holding the right to play the gamble r) according to his/her preferences as follows. For all p 2 GX ; ðp; rÞ 2 P ) p 2 Pþ ; ðr; pÞ 2 P ) p 2 P and if neither ðp; rÞ nor ðr; pÞ belong to P, then p is in P0 : Thus, Pþ (respectively, P ) may be interpreted as the sets of gambles for which you are willing (respectively, unwilling) to exchange the gamble r, while P0 may be the set of gambles that are indifferent to the gamble r.
Acknowledgments The author gratefully acknowledges numerous helpful suggestions and comments of two anonymous referees
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and John Miyamoto. This research is financially supported in part by the Ministry of Education, Culture, Sports, Science and Technology under Grant 15653012 in Japan.
Appendix A A.1. Proof of the main representation theorem Throughout the appendix, we shall assume that Pþ and P are not empty. First, we assume that ðGX ; P3 Þ has a dyadic representation ðf; cÞ; i.e., for all p 2 GX ; p 2 Pþ 3 fðpÞ40 and
p2P
3 fðpÞo0
and
cðpÞ40; cðpÞo0:
Necessity of A1–A3 easily follows. To show necessity of A4, we assume that p; q 2 Pþ and fr; plr; qlrg P0 for 0olo1: When p; q 2 P ; the proof is similar. Then we have fðpÞ40;
cðpÞ40;
fðqÞ40; cðqÞ40; fðrÞcðrÞp0; fðplrÞcðplrÞp0; fðqlrÞcðqlrÞp0: Since, by linearity of f and c; fððplrÞgðqlrÞÞcððplrÞgðqlrÞÞ ¼ g2 fðplrÞcðplrÞ þ ð1 gÞ2 fðqlrÞcðqlrÞ þ gð1 gÞ½fðplrÞcðqlrÞ þ fðqlrÞcðplrÞ
Fig. 1. A barycentric representation of a compound gamble ðpaqÞbr:
By axiom A2, this definition does not depend on the choice of r. Since P0 a;; Lemma 3.2 implies that T a;: However, we may have T þ \ T ¼ ;: We say that p 2 T þ is upper threshold-free, and p 2 T is lower thresholdfree. p 2 T þ \ T is simply said to be threshold-free. The aim of the step is to prove the following claim, which characterizes threshold-free gambles. Claim A.1. Let 2 fþ; g: If ½p; q [ ½q; r P0 and hp; q; ri P ; then q is threshold-free. To prove the claim, we need two lemmas.
for all 0ogo1; it suffices to show that
Lemma A.1.1. Let 2 fþ; g: If hp; qi P ; p; q 2 P0 ; and reP [ P0 ; then there is an s 2 hp; q; ri such that
fðplrÞcðqlrÞ þ fðqlrÞcðplrÞp0:
hp; q; si P ;
If fðplrÞp0 and cðqlrÞp0; then fðrÞo0 and cðrÞo0; since fðpÞ40 and cðqÞ40: This contradicts fðrÞcðrÞp0: If fðplrÞp0; then cðplrÞX0 and cðqlrÞ40; so fðqlrÞp0: If cðqlrÞp0; then fðqlrÞ40 and fðplrÞ40; so cðplrÞp0: The both cases together give the desired result. Hence A4 is necessary for dyadic representations. Next we show sufficiency of the axioms A1–A4. Suppose that axioms A1–A4 hold. Since Pþ and P are not empty, it follows from Lemma 3.1 that P0 is not empty. We show the sufficiency proof in four steps. Every gamble in the convex hull H ¼ fðpaqÞbr : 0pap1; 0pbp1 and p; q; r 2 GX g generated by gambles p, q, and r can be represented barycentrically as shown in Fig. 1. Each point in H corresponds to a point in a triangle with vertices p, q, and r. When the perpendicular distance from each side to its opposite vertex is 1, ðpaqÞbr is the point with the perpendicular distances a and b from sides rq and pq, respectively. Step 1: For 2 fþ; g; let the boundary of P0 be given by
½p; par; s [ ½q; qbr; s P0 ;
T ¼ fp 2 P0 : hp; ri P for some r 2 P g:
hr; par; qbr; si \ ðP [ P0 Þ ¼ ;; where, for some 0oao1 and 0obo1; s¼p
að1 bÞ bð1 aÞ ðqbrÞ ¼ q ðparÞ: 1 ab 1 ab
Proof (See Fig. 2 for illustration). Suppose that hp; qi Pþ ; p; q 2 P0 ; and r 2 P : When hp; qi P and r 2 Pþ ; the proof is similar. Then by Lemma 3.2(1) and A2, there are a; b 2 ð0; 1 such that hr; par; qbri P : By A1 and A2, it is impossible to have a ¼ b ¼ 1: Assume that 0oaob ¼ 1: Since p 12 q 2 Pþ ; A1 gives ðp 12 qÞgr 2 Pþ for some 0ogo1: Note that g g 1 q r : ðp2qÞgr ¼ p 2 2g g Since q 2g r 2 P and p 2 P0 ; it follows from A3 that g g q r 2 P0 [ P ; p 2 2g
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that w 2 Pþ : By A1, wgs 2 Pþ for some 0ogo1: Then there are probability numbers d and such that 0odob and wgs 2 hrds; pqi as illustrated in Fig. 3. Since rds 2 P and pq 2 P0 ; A3 gives wgs 2 P0 [ P ; a contradiction. Hence w 2 P0 ; so that hp; q; ti P0 : It follows similarly that hq; r; ti P0 : &
Fig. 2. An illustration of the proof of Lemma A.1.1.
a contradiction. When 0oboa ¼ 1; a similar contradiction obtains. Hence 0oao1 and 0obo1: Next we show that hp; qbri [ hq; pari P0 : Let s ¼ að1bÞ p 1ab t and t ¼ qbr as illustrated barycentrically in Fig. 2(a). We are to show that hqbr; si; hpar; si; hp; si; hq; si; and fsg are contained in P0 : Assume that pgt 2 P for some 0ogp að1bÞ 1ab : Then by A1, qdðpgtÞ 2 P for some 0odo1: We note that qdðpgtÞ ¼ tð1 dÞð1 gÞðpKqÞ; where K ¼ ð1 dÞ=ðd þ g dgÞ: Since pKq 2 Pþ ; A3 gives qdðpgtÞ 2 P0 [ Pþ ; a contradiction. Thus pgteP : Similarly, pgtePþ : Hence ht; si ¼ hqbr; si and fsg are contained in P0 : The other cases follow similarly. Since hr; qbr; pari P ; it remains to show that hpar; qbri [ hpar; qbr; si P and hp; q; si Pþ : We show the former case. The latter case similarly follows. Take any measure t in hpar; qbri [ hpar; qbr; si as illustrated in Fig. 2(b). By A3, t 2 P0 [ P : Assume that t 2 P0 : Let 0ogo1 be such that ht; pgqi \ hp; qbri ¼ fsg: Then take any d and for which 0oogodo1: Since hp; qi Pþ ; we have pdq; pq 2 Pþ : Let t0 and t00 be the unique measures such that ht; pdqi \ hp; qbri ¼ ft0 g and ht; pqi \ hq; pari ¼ ft00 g: Since ft; t0 ; t00 g P0 ; it follows from A4 that ht0 ; t00 i P0 : Since d and are arbitrary, we obtain that hp; q; si P0 : This contradicts A1. Hence t 2 P : &
Proof of Claim A.1. (See Fig. 4 for illustration). Let ¼ þ: When ¼ ; the proof is similar. Suppose that ½p; q [ ½q; r P0 and hp; q; ri Pþ : We have to show that q is threshold-free. Suppose on the contrary that q is not threshold-free. Take any s 2 P : By Lemma 3.2(1), hs; qgsi P and ½qgs; qi P0 for some 0ogo1: It follows from Lemma A.1.1 that there is a gamble t ¼ pKðrbsÞ 2 hp; r; si; as illustrated in Fig. 4, such that hs; pas; rbs; ti P ;
Fig. 3. An illustration of the proof of Lemma A.1.2.
Lemma A.1.2. If ½p; q [ ½q; r P0 ; hp; q; ri [ hp; ri P ; and seP [ P0 ; then there is a gamble t 2 hp; r; si such that hp; q; r; ti P and ½p; q; t [ ½q; r; t P0 ; where, for some 0oao1 and 0obo1; t¼p
að1 bÞ bð1 aÞ ðrbsÞ ¼ r ðpasÞ: 1 ab 1 ab
Proof (See Fig. 3 for illustration). We show the case for ¼ þ: When ¼ ; the proof is similar. By Lemma A.1.1, there is a gamble t 2 hp; r; si such that hp; r; ti Pþ and hr; t; rbsi [ hp; t; pasi P0 ; where t is defined in the lemma. We take any w 2 hp; q; ti and show that w 2 P0 : Since hp; q; t; ri Pþ ; A1 implies that weP : Thus assume
Fig. 4. An illustration of the proof of Claim A.1.
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½rbs; t; r [ ½pas; t; p P0 ; 0oao1; 0obo1; and K¼
and
hp; r; ti Pþ ;
where
að1 bÞ : 1 ab
Note that t 2 hs; pLri; where L ¼ að1 bÞ=ðað1 bÞ þ bð1 aÞÞ: Thus we take any d and for which 0odoLoo1: Let t0 and t00 be the unique gambles for which ft0 g ¼ hs; pdri \ ht; ri and ft00 g ¼ hs; pri \ ht; pi: Then let w0 and w00 be the unique gambles for which fw0 g ¼ hq; t0 i \ hqgs; pdri and fw00 g ¼ hq; t00 i \ hqgs; pri: Since qgs; w0 ; w00 2 P0 and pdr; pr 2 Pþ ;A4 gives hw0 ; w00 i P0 : Since hp; q; ri [ hp; r; ti Pþ ; we obtain that hq; t0 ; t00 i Pþ : By construction, hw0 ; w00 i hq; t0 ; t00 i; a contradiction. Hence q must be threshold-free. & Step 2: Recall that T þ \ T is the set of all thresholdfree gambles. Let I ¼ P0 nðT þ \ T Þ: We define a binary relation on I as follows: for all p; q 2 GX ; p q 3 ½p; q I: The aim of the step is to prove the following claim. Claim A.2. on I is an equivalence relation. To prove the claim, we need the following three lemmas. Lemma A.2.1. Let 2 fþ; g: If hp; qi P and ½p; r [ ½r; q P0 ; then there is an 0pao1 such that either 0
and
hp; q; qari P ;
0
and
hp; q; pari P :
½p; r; qar P or
½q; r; par P
Proof (See Fig. 5 for illustration). Suppose that hp; qi Pþ and ½p; r [ ½r; q P0 : When hp; qi P ; the proof is similar. By Lemma 3.2(1), there is a real-valued function f on ð0; 1Þ such that, for all 0ogo1; 0pfðgÞo1; ½r; ðpgqÞfðgÞr P0 ; and hðpgqÞfðgÞr; pgqi Pþ : By A2, either fðgÞ ¼ 0 for all 0ogo1 or fðgÞ40 for all
Fig. 5. An illustration of the proof of Lemma A.2.1.
393
0ogo1: The former case gives the desired result. Thus, we assume that fðgÞ40 for all 0ogo1: It follows from A4 that, for all g0 ; g00 2 ð0; 1Þ; ½r; ðpg0 qÞfðg0 Þr; ðpg00 qÞfðg00 Þr P0 ; hðpg0 qÞfðg0 Þr; ðpg00 qÞfðg00 Þr; pg0 q; pg00 qi Pþ : This implies that there are 0oap1 and 0obp1 such that abo1; ½r; par; qbr P0 ; and hp; q; par; qbri Pþ : It suffices to show that either a ¼ 1 or b ¼ 1: Suppose that ao1 and bo1: Then by Claim A.1, par and qbr are threshold-free. Take any s 2 P : Then hs; par; qbri P : Since pgq 2 Pþ for 0ogo1; A1 implies that t ¼ ðpgqÞds 2 Pþ for some 0odo1: As illustrated in Fig. 5, hr; ti \ hs; par; qbria;: This contradicts A3. & Lemma A.2.2. If p; q; r 2 P0 and hp; qi [ hq; ri P ; then hp; ri P0 : Proof (See Fig. 6 for illustration). Suppose that p; q; r 2 P0 and hp; qi [ hq; ri Pþ : When hp; qi [ hq; ri P the proof is similar. Take any s 2 P : By A1 and A2, ½p; par [ ½pbr; r P0 and hpar; pbri Pþ for some 0pboap1: Assume that a ¼ 1 and b ¼ 0; so that hp; ri Pþ : By Lemma A.1.1 and Claim A.1, there are threshold-free gambles t 2 hq; r; si and w 2 hp; r; si; where, for some a; b; g 2 ð0; 1Þ; gð1 bÞ ðqbsÞ; 1 bg gð1 aÞ ðpasÞ: w¼r 1 ag t¼r
Since any gamble in ½t; w is threshold-free, we take a w0 2 ht; wi: Then, as illustrated in Fig. 6, let a gamble t0 2 hqbs; pasi be such that w0 2 ht0 ; ri: Note that t0 2 P ; since, by Lemma A.1.1, hqbs; pasi P : Let 0odo1 satisfy t0 2 hpdq; si: Such a d also satisfies that w0 2 hpdq; rgsi: By A1, there is a gamble t0 ðpdqÞ; denoted t00 ;
Fig. 6. An illustration of the proof of Lemma A.2.2.
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for some 0oo1 that is contained in P : Let fw00 g ¼ ht00 ; ri \ hpdq; rgsi: Since pdq 2 Pþ and w0 is thresholdfree, we must have w00 2 Pþ :This contradicts A3. When 0ob and ao1; similar contradictions obtain. Hence hp; ri P0 : & Lemma A.2.3. If p is not threshold-free and ½p; r P0 ; then there exists at most one threshold-free gamble in ½p; r: Moreover, letting q 2 P and s 2 Pþ ; we have (1) if ½p; r contains no threshold-free gamble, then there are a; a0 ; b; b0 2 ½0; 1 such that a þ a0 o2; b þ b0 o2; ½p; paq; rbq; r [ ½p; pa0 s; rb0 s; r P0 ; hpaq; rbq; qi P
and
hpa0 s; rb0 s; ri Pþ :
(2) if ½p; r contains a threshold-free gamble pgr for some 0pgo1; then there are a; a0 ; b; b0 2 ½0; 1 such that a þ a0 o2; b þ b0 ¼ 2 if g ¼ 0; b þ b0 o2 if 0og; ½p; paq; pgr [ ½pgr; rbq; r [ ½p; pa0 s; pgr [ ½pgr; rb0 q; s P0 ; hq; paq; pgr; rbqi [ hq; paqi [ hq; rbqi P ; hs; pa0 s; pgr; rb0 si [ hs; pa0 si [ hs; rb0 si Pþ :
Proof (See Fig. 7 for illustration). Suppose that p is not threshold-free, ½p; r P0 ; q 2 P ; and s 2 Pþ : Then ½p; paq [ ½p; pa0 s P0 ; hpaq; qi P ; and hpa0 s; si Pþ for some 0oap1 and 0oa0 p1 with a þ a0 o2: Assume that ao1: When a0 o1; the proof is similar. Suppose that there are at least two threshold-free gambles in ½p; r: With no loss of generality, let r and pgr for some 0ogo1 be threshold-free gambles, so that any gamble in hr; pgri are threshold-free. It follows from A2 and Lemma 3.2(2) that hq; r; pgr; paq; sqi P and ½p; paq; pgr; sq [ ½r; pgr; paq P0 for some 0oo1: We have ðpgrÞðaKÞðsqÞ ¼ ðpaqÞðgKÞðsLrÞ;
Fig. 7. An illustration of the proof of Lemma A.2.3.
where K ¼ ð1 Þ=ðað1 Þ þ gð1 aÞÞ and L¼
gð1 aÞ : ð1 Þða gÞ þ gð1 aÞ
Let t ¼ ðpgrÞðaKÞðsqÞ: For 0odoL; let ft0 g ¼ hsdr; paqi \ ht; ri as illustrated in Fig. 7. Since ht0 ; paqi P ; paq 2 P0 ; and sdr 2 Pþ ; A3 gives a contradiction. Thus r cannot be threshold-free. Hence hp; ri contains at most one threshold-free gamble. Suppose that ½p; r contains at most one threshold-free gamble. There is a 0obp1 such that hq; rbqi P and ½rbq; r P0 : Assume first that ½p; r contains no threshold-free gamble. Then, by Claim A.1, ½paq; rbq P0 : Thus (1) obtains. Assume next that ½p; r contains one threshold-free gamble pgr for some 0pgo1: Then, by Claim A.1, ½paq; pgr [ ½pgr; rbq P0 : Hence (2) obtains. & Proof of Claim A.2. (See Fig. 8 for illustration). Since ½p; q I iff ½q; p I; is reflexive and symmetric. To show that is transitive, assume that p q and q r: We have the following two cases to examine, Case 1: either hp; ri \ Pþ a; or hp; ri \ P a;; Case 2: ½p; r P0 and ½p; rnIa;: In each case, we obtain a contradiction, so that ½p; r I; i.e., p r: Case 1: Suppose that hp; ri \ Pþ a;: When hp; ri \ P a;; the proof is similar. By Lemma 3.2(1), there are 0paobp1 such that ½r; par [ ½p; pbr P0 and hpar; pbri Pþ : First we assume that a ¼ 0 and b ¼ 1: Then by Lemma A.2.1, there is an 0ogp1 such that either (a) or (b) holds: (a) ½p; q; qgr P0 and hp; r; qgri Pþ ; (b) ½q; r; qgp P0 and hp; r; qgpi Pþ : By Claim A.1, qgr is threshold-free for (a) and qgp is threshold-free for (b). They are contradictions. Next we assume that a ¼ 0 and ba1: When If aa0 and b ¼ 1; the proof is similar. If ½q; pbr P0 ; then it follows from Lemma A.2.1 and A4 that ½q; pbr; qgr P0 and hpbr; r; qgri Pþ for some 0ogp1: Thus by Claim A.1, qgr is threshold-free, a contradiction.
Fig. 8. An illustration of the proof of Claim A.2..
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If hq; pbri \ Pþ a;; then ½q; qgðpbrÞ P0 and hqgðpbrÞ; pbri Pþ for some 0ogo1 as illustrated in Fig. 8(a), where the shaded areas are shown to be contained in P0 : Then by the similar argument of the preceding paragraph that ½qgðpbrÞ; r P0 : We note qgðpbrÞ ¼ ðqKpÞ
g r; K
where K ¼ g=ðb þ g bgÞ: By A1 and A2, ½qKp; qgðpbrÞ P0 : It follows from the similar argument of the preceding paragraph and A2 that ½qKp; pbr P0 : Thus, by Claim A.1, qKp is thresholdfree, a contradiction. Last we assume that aa0 and ba1: If ½q; pbr [ ½q; par P0 ; then by Lemma A.2.1 and A4, hq; par; pbri Pþ : Thus by Claim A.1, q is thresholdfree, a contradiction. Therefore, we assume that hq; pari \ Pþ a; or hq; pbri \ Pþ a;: Then a similar argument of the preceding paragraph gives that hr; qi (respectively, hp; qi) contains a threshold-free gamble if hq; pari \ Pþ a; (respectively, hq; pbri \ Pþ a;), a contradiction. Case 2: Suppose that ½p; r P0 and hp; ri includes a threshold-free gamble t. Let s 2 Pþ : As illustrated in Fig. 8(b), it follows from Lemma A.2.3 that there are a; b; g 2 ð0; 1Þ such that ½p; t; pgs [ ½r; t; rbs P0 ; ½p; q; qas; pgs [ ½q; r; qas; rbs P0 ; þ
hs; pgs; rbs; ti P ; hs; qas; pgsi [ hs; qas; rbsi Pþ : Let 0odog and 0oob: Then pds; rs 2 Pþ : Let ft0 g ¼ ½pgs; qas \ hq; pdsi and ft00 g ¼ ½rbs; qas \ hq; rsi: Then t0 ; t00 2 P0 : By A4, ht0 ; t00 i P0 : Since d and is arbitrary in open intervals ð0; gÞ and ð0; bÞ; respectively, we obtain that hpgs; rbs; qasi P0 : However, hpgs; rbsi Pþ which contradicts A1. & Step 3: We say that the empty set is convex. This step proves the following claim. Claim A.3. There are two disjoint convex sets, I 1 and I 2 ; such that I ¼ I 1 [ I 2 :
where gð1 b dÞ ðqdrÞ gð1 a dÞ þ að1 bÞ að1 b dÞ ¼ ðpgrÞ ðrbqÞ: gð1 a dÞ þ að1 bÞ
s ¼ ðpaqÞ
Proof (See Fig. 9 for illustration). Suppose that p 2 P0 ; q 2 P ; and r 2 Pþ : Then by Lemma 3.2, there are 0oap1; 0obo1; 0ogp1; and 0odo1 such that b þ dp1; hq; paqi [ hq; rbqi P ; hr; pgri [ hr; qdri Pþ ; and hpaq; pi [ hp; pgri [ hrbq; qdri P0 : We show that hpaq; qdri [ hpgr; rbqi P0 : To show that hpaq; qdri P0 : Assume that there is a gamble t 2 hpaq; qdri such that t 2 Pþ : A similar proof applies to obtain that hpgr; rbqi P0 : By A1, there is a gamble w 2 hq; ti such that w 2 Pþ : As illustrated in Fig. 9, w 2 hq; pri for some 0ooa: Since pr 2 P ; Lemma 3.2(1) gives wePþ ; a contradiction. Therefore, tePþ : Similarly, teP : Hence hpaq; qdri P0 : Let s be a gamble defined in the lemma. Then s 2 ½p; qKr; where K¼
ð1 bÞðgð1 aÞð1 dÞ þ ð1 gÞadÞ : að1 bÞð1 gÞ þ gð1 aÞð1 dÞ
Since ½p; qKr P0 ; it follows from Lemma A.2.3 and A3 that either (1) or (2) holds. & Lemma A.3.2. If p 2 P ; q 2 P0 ; and paq 2 P0 for 0oao1; then ½q; paq I and hq; paqi InðT [ T þ Þ: Proof. Suppose that p 2 Pþ ; q 2 P0 ; and paq 2 P0 for 0oao1: When p 2 P ; the proof is similar. Then by Lemma 3.2, ½q; pbp 2 P0 and hpbq; pi Pþ for some 0obo1: Thus apb and q is not threshold-free. Suppose on the contrary that notðq paqÞ: Then paq must be threshold-free. Take any s 2 P ; so that hpaq; si P : By A1, pgs 2 Pþ for some 0ogo1: We note that a g q ¼ ðpaqÞ s: ðpgsÞ að1 gÞ þ g bð1 gÞ þ g By Lemma 3.2(1), this measure does not belong to P ; a contradiction. Therefore, paq is not lower threshold-free.
We need the following two lemmas to prove the claim. Lemma A.3.1. If p 2 P0 ; q 2 P ; and r 2 Pþ ; then there are 0oap1; 0obo1; 0ogp1; and 0odo1 such that b þ dp1; hq; paqi [ hq; rbqi P ; hr; pgri [ hr; qdri Pþ ; and either one of (1) and (2) holds. (1) ½p; paq; rbq; qdr; pgr P0 ; hq; paq; rbqi P ; and hr; pgr; qdri Pþ ; (2) ½p; paq; s; pgr [ ½s; rbq; qdr P0 ; hq; paq; s; rbqi P ; and hr; pgr; s; qdri Pþ ;
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Fig. 9. An illustration of the proof of Lemma A.3.1.
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Fig. 10. An illustration of the proof of Claim A.3.
Let t 2 Pþ : Then, by Lemma 3.2(1), ½q; pgt P0 and hpgt; ti Pþ for some 0pgo1: By A2 and A4, ½q; pgt; pbq P0 : Since b40; A2 implies that g40: Thus paq is not upper threshold-free. Hence ½q; paq I and hq; paqi InðT [ T þ Þ: &
Proof of Claim A.3. (See Fig. 10 for illustration). We are to show that if not ðp qÞ and not ðq rÞ for p; q; r 2 I; then p r: We have the following four cases to examine: Case 1: ½p; q \ P a; and ½q; r \ P a; for 2 fþ; g; Case 2: ½p; q \ Pþ a; and ½q; r \ P a;; Case 3: ½p; q \ P a; for 2 fþ; g and ½q; r P0 ; ½q; rnIa;; Case 4: ½p; q [ ½q; r P0 ; ½p; qnIa;; and ½q; rnIa;: Case 1: Suppose that ½p; q \ P a; and ½q; r \ P a; for 2 fþ; g: Assume the case of ¼ þ: A proof is similar for ¼ : By Lemma 3.2(1), there are 0paogp1 and 0pbodp1 such that ½q; paq [ ½p; pgq [ ½q; rbq [ ½r; rdq P0 and hpaq; pgqi [ hrbq; rdqi Pþ : By A1 and A2, hpaq; rbqi P0 [ Pþ : If hpaq; rbqi \ þ P a;; then by A1 and A2, hpaq; rbqi Pþ : Thus by Claim A.1, q is threshold-free, a contradiction. Hence, hpaq; rbqi P0 :
By A2 and A3, hp; ri P0 [ Pþ : If hp; ri \ Pþ a;; then a similar argument of the preceding paragraph gives that p and r are threshold-free, a contradiction. Hence hp; ri P0 : By A1 and A2, hpgq; rdqi P0 [ Pþ : If hpgq; rdqi \ þ P a;; then a similar argument of Lemma A.2.3 gives that ½p; pgq; pr [ ½r; rdq; pr P0 and hpgq; rdq; pri Pþ for some 0oo1 as illustrated in Fig. 10(a). Take any s 2 hpaq; rbqi; t 2 hpgq; pri; and w 2 hrdq; pri; so that s; t; w 2 P0 and hs; ti [ ht; wi [ hw; si Pþ : This is prohibited by Lemma A.2.2. Hence ½pgq; rdq P0 : If g þ da2; then any gamble in hp; ri is not thresholdfree. Therefore, p r: Assume that g ¼ d ¼ 1; and pr is threshold-free for some 0oo1: Take any gamble t 2 hq; pri for which t 2 Pþ : Let s 2 P : Then by A1, there is a gamble t0 2 ht; si for which t0 2 Pþ : Let w be a gamble such that hw; qi contains t0 as illustrated in Fig. 10(b). By A3, w 2 Pþ [ P0 : Since pr is threshold-free, we must have w 2 P : This is a contradiction. Hence hp; ri contains no threshold-free gamble, so that p r: Case 2: Suppose that ½p; q \ Pþ a; and ½r; q \ P a;: By Lemma 3.2(1), there are 0paogp1 and 0pbodp1 such that ½q; paq [ ½p; pgq [ ½q; rbq [ ½r; rdq P0 ; hpaq; pgqi Pþ ; and hrbq; rdqi P : By Case 1, it is prohibited that hp; ri \ P a; for 2 fþ; g: Thus hp; ri P0 : Suppose that pr is thresholdfree for some 0oo1: Then hpr; paq; pgqi Pþ and
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hpr; rbq; rdqi P : Take any s 2 hrbq; rdqi: By A3, hp; si P0 [ P : However, as illustrated in Fig. 10(c), there is a gamble t 2 hp; si for which t 2 hpr; paq; pgqi; a contradiction. Hence hp; ri contains no threshold-free gamble, so p r: Case 3: Suppose that ½q; r P0 ; ½q; rnIa;; and ½p; q \ P a; for 2 fþ; g: Assume the case of ¼ þ: A proof is similar for ¼ : By Lemma 3.2(1), there are 0paogp1 such that ½q; paq [ ½p; pgq P0 and hpaq; pgqi Pþ : Let rbq be a threshold-free gamble in hq; ri: By Cases 1 and 2, hp; ri P0 : Suppose that hp; ri contains a threshold-free gamble, say pr for some 0oo1: Then ½p; pgq; pr [ ½q; paq; rbq [ ½r; rbq; pr P0 and hpaq; rbq; pgq; pri Pþ : As illustrated in Fig. 10(d), take any gambles s 2 hpaq; rbqi; t 2 hpgq; pri; and w 2 hrbq; pri; so that hs; ti [ ht; wi [ hw; si Pþ : This is prohibited by Lemma A.2.2. Hence hp; ri contains no threshold-free gamble, so p r: Case 4: Suppose that ½p; q [ ½q; r P0 ; ½p; qnIa;; and ½q; rnIa;: By Case 3, hp; ri \ P ¼ ; for 2 fþ; g: Suppose that ½p; rnIa;: Let paq; qbr; and pgr be, respectively, threshold-free gambles in hp; qi; hq; ri; and hp; ri: Then every gamble in hpaq; qbri is threshold-free. Since hpaq; qbri \ hq; pgria;; hq; pgri contains more than two threshold-free gambles. This contradicts Lemma A.2.3 which says that hq; pgri contains at most one threshold-free gamble, since q is not threshold-free. Hence hp; rinI ¼ ;; so p r: & Step 4. This step completes the sufficiency proof of Theorem 3.1. For fi; jg ¼ f1; 2g we define þ Qþ i ¼ P [ ðI j nT Þ; þ Q i ¼ P [ ðI i nT Þ;
Q0i ¼ ðT þ \ I i Þ [ ðT \ T þ Þ [ ðT \ I j Þ: We need the following lemma Lemma A.4.1. For i ¼ 1; 2 and 2 f; þg; we have (1) Qþ i and Qi are not empty. þ 0 (2) if p 2 Qi and q 2 Q i ; then paq 2 Qi for some 0oao1: (3) if p 2 Qi and q 2 Qi [ Q0i ; then hp; qi Qi : (4) if p; q 2 Q0i ; then hp; qi Q0i :
Proof. (1) This follows from P a; and Pþ a;: 0 (2) We show that if p 2 Qþ 1 and q 2 Q1 ; then paq 2 Q1 for some 0oao1: When i ¼ 2; the proof is similar. We have four cases to examine Case 1: p 2 Pþ ; q 2 P ; Case 2: p 2 Pþ ; q 2 I 1 nT þ ; Case 3: p 2 I 2 nT ; q 2 P ; Case 4: p 2 I 2 nT ; q 2 I 1 nT þ : Case 1: By Lemma 3.2(2), there are 0oapbo1 such that hq; paqi P ; ½paq; pbq P0 ; and hpbq; qi Pþ : We note that paq 2 T ; pbq 2 T þ ; and either
397
½paq; pbq I 1 or ½paq; pbq I 2 or ½paq; pbq ðT \ T þ Þ: Thus we have paq 2 ðT \ I 2 Þ Q01 when ½paq; pbq I 2 ; pbq 2 ðT þ \ I 1 Þ Q01 when ½paq; pbq I 1 ; paq 2 ðT þ \ T þ Þ Q01
when ½paq; pbq ðT \ T þ Þ:
Case 2: By Lemma 3.2(1), there is an 0oao1 such that hp; paqi Pþ and ½q; paq I 1 : Then we have paq 2 T þ \ I 1 Q01 : Case 3: Similar to Case 2. Case 4: Suppose that p 2 I 2 nT and q 2 I 1 nT þ : Assume first that ½p; q \ Pþ a;: When ½p; q \ P a;; the proof is similar. Then, by Lemma 3.2(1), there are 0oaobp1 such that ½q; paq I 1 ; hpaq; pbqi Pþ ; and ½pbq; p I 2 : Note that paq 2 I 1 \ T þ and pbq 2 I 2 \ T þ : Thus paq 2 Q01 : Assume next that ½p; qnIa;; paq 2 ½p; qnI for some 0oao1: Thus paq 2 T þ \ T Q01 : (3) We show that if p 2 Q1 and q 2 Q1 [ Q01 ; then hp; qi Q1 : When i ¼ 2; the proof is similar. We prove the case of ¼ þ: The proof for ¼ is similar. We note that þ p 2 Qþ 1 3 p 2 P [ ðI 2 nT Þ; 0 þ þ q 2 Qþ 1 [ Q1 3 q 2 P [ I 2 [ ðT nI 2 Þ:
We are to show that hp; qi Pþ [ ðI 2 nT Þ: We have two cases to examine: p 2 Pþ ; p 2 I 2 nT : Case 1: ðp 2 Pþ Þ If q 2 Pþ ; then by A2, hp; qi Pþ þ Q1 : If q 2 I 2 ; then by Lemma 3.2(1), ½q; paq P0 and hpaq; qi Pþ for some 0pao1: By Lemma A.3.2, hq; paqi I 2 nðT [ T þ Þ: Since paq 2 T þ ; we obtain that hp; qi Pþ [ ðI 2 nT Þ: If q 2 T þ nI 2 ; then it follows from the definition of T þ that hp; qi Pþ : Case 2: ðp 2 I 2 nT Þ Take any s 2 P ; so that hs; pasi P and ½pas; p P0 for some 0oao1: If q 2 Pþ ; then the desired result follows from Case 1. If q 2 I 2 ; then hp; qi I 2 by the convexity of I 2 : It follows from Lemma A.2.3 that plq is threshold-free when plq 2 T : Since there is no threshold-free measure in hp; qi; we have hp; qi I 2 nT : Suppose that q 2 T þ nI 2 : Let t 2 Pþ : By definition of þ T ; hq; ti Pþ ; since q 2 T þ : It follows from A1 and A3 that hp; qi \ P ¼ ;: Assume first that hp; qi \ Pþ a;: Then hpaq; qi Pþ and ½paq; p P0 for some 0oap1: By Lemma A.3.2, ½paq; p I 2 nT : Hence, hp; qi Pþ [ ðI 2 nT Þ: Assume next that hp; qi P0 : Thus, by Lemma A.2.3, q 2 T þ \ T : Therefore, Lemma A.2.3 implies that any measure in hp; qi is not lower thresholdfree. Hence hp; qi I 2 nT : (4) We show that if p; q 2 Q01 ; then hp; qi Q01 : When i ¼ 2; the proof is similar. Thus, we assume that p; q 2 ðT þ \ I 1 Þ [ ðT \ T þ Þ [ ðT \ I 2 Þ: Take any s 2 P and any t 2 Pþ :
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Suppose that p 2 T þ \ I 1 : When p 2 T \ I 2 ; the proof is similar. If q 2 T þ \ I 1 ; then hp; qi I 1 ; since I 1 is convex. By A2, hp; qi T þ : Thus hp; qi Q01 : Assume next that q 2 T \ I 2 : By A1 and A3, hp; qi 0 P : Since p 2 I 1 and q 2 I 2 ; there must be a thresholdfree measure in hp; qi: It follows from Lemma A.2.3 that, for some a; b; l 2 ð0; 1Þ; ½p; plq; pas [ ½q; plq; qbt P0 ; hs; pas; plq; qi P ; and ht; p; plq; qbti Pþ : Thus hp; plqi T \ I 1 and hq; plqi T \ I 2 : Hence hp; qi Q01 : Assume last that q 2 T \ T þ : By A2, hp; q; ti þ T \ I 1 ; so that hp; qi Q01 : Suppose that p; q 2 T \ T þ : Then hp:qi T \ T þ : Hence hp; qi Q01 : & Proof of Theorem 3.1. It follows from Lemma A.4.1 and Theorem 3.2 that there are linear functionals, u1 and u2 ; on GX such that, for all p 2 GX and i ¼ 1; 2; p 2 Qþ i 3 ui ðpÞ40; p 2 Q0i 3 ui ðpÞ ¼ 0; p 2 Q i 3 ui ðpÞo0: Furthermore, u1 and u2 are unique up to multiplicative transformations by positive constants. Since I 1 and I 2 are disjoint, it follows from the definitions that P ¼
Q1 \ Q2 for 2 fþ; g: Thus þ p 2 Pþ 3p 2 Qþ 1 \ Q2 3u1 ðpÞ40 and u2 ðpÞ40; p 2 P 3p 2 Q 1 \ Q2 3u1 ðpÞo0 and u2 ðpÞo0:
This completes the proof.
&
References Fishburn, P. C. (1982). Nontransitive measurable utility. Journal of Mathematical Psychology, 26, 31–67. Fishburn, P. C. (1988). Nonlinear preference and utility theory. Baltimore: Johns Hopkins University Press. Goldstein, W. M. (1991). Decomposable threshold models. Journal of Mathematical Psychology, 35, 64–79. Karni, E., & Schmeidler, D. (1991). Utility theory with uncertainty. In Hildenbrand, W., & Sonnenschein, H. (Eds.), Handbook of mathematical economics, Vol. 4. Amsterdam: North-Holland. Quiggin, J. (1993). Generalized expected utility theory: the rank dependent model. Dordrecht: Kluwer Academic. Schmidt, U., 1998. Axiomatic utility theory under risk. Lecture notes in economics and mathematical systems (Vol. 461). Berlin: Springer.