Stable sets in three agent pillage games∗

Manfred Kerber † Colin Rowat‡

June 30, 2009

Abstract

Jordan [2006, “Pillage and property”, JET] characterises stable sets for threespecial cases of ‘pillage games’. For anonymous, three agent pillage games weshow that: when the core is non-empty, it must take one of five forms; all suchpillage games with an empty core represent the same dominance relation; whena stable set exists, and the game also satisfies a continuity and a responsivenessassumption, it is unique and contains no more than 15 elements. This result uses athree step procedure: first, if a single agent can defend all of the resources againstthe other two, these allocations belong to the stable set; dominance is then transi-tive on the loci of allocations on which the most powerful agent can, with any ally,dominate the third, adding the maximal elements of this set to the stable set; finally,if any allocations remain undominated or not included, the game over the remain-ing allocations is equivalent to the ‘majority pillage game’, which has a unique sta-ble set [Jordan and Obadia, 2004, “Stable sets in majority pillage games”, mimeo].Non-existence always reflects conditions on the loci of allocations along which themost powerful agent needs an ally. The analysis unifies the results in Jordan [2006]when n = 3.

Key words: pillage, cooperative game theory, core, stable sets, algorithm

JEL classification numbers: C63; C71; P14

This file: 090506comp-pill.tex

1 IntroductionJordan [2006a] introduced ‘pillage games’, a subset of abstract games in which coali-tions’ power increases monotonically in both coalitional membership, and members’resources. The outcome of the contest of power between opposing coalitions representsthe dominance relation usually taken as primitive in an abstract cooperative game.

As this contest of power generates externalities, pillage games are distinct fromgames in characteristic function form, the best known class of cooperative games. Ascoalitions’ power depends not just on their membership, but also on the resources that∗We are grateful to Jim Jordan and Herakles Polemarchakis for useful conversations, to seminar audiences

at ESEM 2008, FEMES 2008, LAMES 2008, and to the ESRC for funding under its World Economy andFinance programme (RES-156-25-0022). Rowat thanks Birkbeck for its hospitality.†School of Computer Science, University of Birmingham; m.kerber@cs.bham.ac.uk‡Department of Economics, University of Birmingham; c.rowat@bham.ac.uk

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they hold, pillage games are also distinct from games in partition function form, thesecond best known class.

In studying contests of power, pillage games may be compared to games in whichcontest success functions convert investments in rent-seeking into a share of the spoilsq.v. [Skaperdas, 1992]. Both approaches are required to make assumptions about howsuch contests occur: in the former case, a power function is axiomatically derived; inthe latter, a contest success function is [Skaperdas, 1996], and a game form specified.Thus, while both make assumptions about the technology of power that largely remainuntested, the latter also makes assumptions about game forms; insofar as real conflictis often characterised by not obeying any pre-specified rules, this may be a source ofnon-robustness.

The perhaps larger difference between the two approaches can be seen in their treat-ment of the costs of contests of power. While the non-cooperative approach assumesthat rent-seeking is costly, the pillage literature (and variants such as Piccione and Ru-binstein [2007]) assume otherwise. In the conflict modeled by pillage games, the morepowerful party is recognised as such, and not resisted by the less powerful. This, there-fore, seems consistent with environments in which there is a high degree of commonknowledge about bargaining strengths; such environments may therefore include thecontests of power undertaken by political factions in advanced democracies.

Jordan [2006a] presented results for both the core (the set of undominated allo-cations) and stable sets (no allocation in the set dominates any other; any allocationoutside the set is dominated by one in it).1 For the former, he derived a general con-dition satisfied by any core allocation in any pillage games. For the latter, he providedresults in a number of special cases: in the first, a unique stable set was characterised;in the second, partial characterisation was possible; in the third, a non-existence resultwas presented.

A number of natural questions were, however, left open. Is an explicit character-isation of the core possible? How are the apparently dissimilar results for stable setsrelated? How many other classes of stable sets are possible? What determines unique-ness or existence? Can any insight be gained into which classes of power functionsrepresent the same dominance relations? Do any of the insights gained here aid analy-sis of games with multiple goods, in which both trade and pillage are possible [Rowat,2009]?

These questions may be difficult to answer given the intractability of stable sets.Shapley [1959] showed that stable sets could be constructed with arbitrary closed com-ponents, by appropriate choice of the abstract game. While von Neumann and Mor-genstern [1944] assumed that stable sets existed for games in characteristic functionform - the focus of their analysis - it took a quarter century for a counterexample to befound [Lucas, 1968].2 However, Jordan [2006a] proved that a set is stable if there isa consistent expectation for which it is the (farsighted) core in expectation.3 Thus, tothe extent that the core and forward looking agents are compelling, the stable sets bearstudy.

In the context of pillage games, there has been cause for optimism. Theoretically,the monotonicity axioms that underpin dominance impose considerable structure on

1Richardson [1953] noted that stable sets appear as a Punktbasis zweiter Art – a point basis of the secondtype – in Konig [1936]. In von Neumann and Morgenstern [1944] they were simply called ‘solutions’.

2For abstract cooperative games, we are grateful to Oleg Itskhoki for noting that Condorcet cycles pro-duce counterexamples: let x, y and z be all possible imputations and x K y K z K x the only dominancerelations on them; then no stable set exists.

3See Anesi [2006] and the references therein for more motivations of stable sets based on farsightedness.

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pillage games relative to general abstract games. Empirically, no examples of multiplestable sets have been found for pillage games. Theoretically, the cardinality of stablesets in pillage games is not only finite [Jordan, 2006a], but bounded by a Ramseynumber [Kerber and Rowat, 2009].

This paper addresses the above questions in the case of anonymous, three agent,one-good pillage games. When the core is non-empty, it must take one of five forms;these include configurations not seen in Jordan’s examples. Turning to stable sets,all power functions yielding empty cores represent the same dominance relation, andtherefore yield the same stable set.

When the core is not empty, it is known from Jordan [2006a] that it must containthe ‘tyrannical’ allocations – those giving all of society’s resources to a single agent;further, those agents are able to defend their allocations, alone, against all other agents.As core allocations necessarily belong to any stable set, these three allocations there-fore seed our candidate stable sets.

As a coalition’s power is monotonic in its resource holdings, decreasing the tyrant’sholdings may eventually yield a balance of power with the other two agents. Whenpower is continuous in resources, allocations along this locus – which may not exist –are only dominated by allocations in turn dominated by tyrannical allocations. Thus,ensuring external stability on this locus requires inclusion of allocations from the locusitself. Further, as dominance is transitive along this locus, those allocations allowingexternal stability are unique. For each agent i, there are up to three of these.

Further decreasing the tyrant’s holdings may then yield allocations for which anytwo agents can defeat a single agent. Over such allocations, dominance is shown to beequivalent to that in the ‘majority pillage game’, in which any two agents can alwaysdefeat a third; when singleton coalitions oppose each other, the wealthier agent wins.As Jordan and Obadia [2004] prove the majority pillage game has a unique stable setof three allocations, there is a unique stable set on this domain.

When power satisfies a responsiveness axiom, in addition to anonymity and con-tinuity, this three step procedure both forces a unique stable set, when one exists, andsets an upper bound of fifteen allocations on it. It also allows us to locate the sourceof non-existence results along the locus of allocations along which the most powerfulagent needs an ally: non-existence either reflects the absence of a maximal elementalong this locus, a new application of an old result [von Neumann and Morgenstern,1944, §65.4.2], or the dominance of maximal elements by a tyrannical allocation. Thenon-existence result in Jordan [2006a] arose from the latter mechanism; the former isonly possible when the power function is discontinuous in resource holdings, a featurenot present in any of Jordan’s examples.

The procedure allows us to present an alternative algorithm to that in Roth [1976],applied to pillage games by Jordan and Obadia [2004]. That was incomplete in anumber of respects: it provided limited guidance on initial conditions; it also did notguarantee an answer to either the question of whether a stable set exists and, if so,what it looks like. By contrast, our algorithm answers both of these questions for anyanonymous, continuous, responsive n = 3 power function. Results for n > 3 willdepend on results for the majority pillage game when n > 3; at present, these arelimited. Developing these further, then, is an avenue for future research.

Finally, the procedure unifies the results in Jordan [2006a]. In doing so, it intro-duces new examples of power functions, revealing new configurations of the core andstable sets, including allocations that lie off of the symmetry axes, something not yet

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seen in the literature.4

Section 2 of this paper introduces Jordan [2006a] pillage games. Section 3 presentsa result characterising the core. Section 4 derives the three step procedure for stablesets. Section 5 first applies the preceding theory to examples satisfying the continu-ity and responsiveness axioms; it then presents results for stable sets violating thoseaxioms. In doing so, it shows that the stable set of the ‘wealth is power’ function in-troduced by Jordan [2006a] generalises to a class of power functions that weaken itsconservatism axiom to a point symmetry axiom; all such power functions represent thesame dominance ordering.

Section 6 concludes, discussing the anonymity, continuity and responsiveness ax-ioms, and compares pillage games to those in characteristic function form. An ap-pendix collects some results of interest that are not used in the main arguments.

2 Pillage gamesLet I = {1, . . . , n} be a finite set of agents. An allocation divides a unit resource amongthem, so that the feasible set is a compact, continuous n − 1 dimensional simplex:

X ≡

{xi}i∈I

∣∣∣∣∣∣∣xi ≥ 0,∑i∈I

xi = 1

.2.1 Power functionsLet ⊂ denote a proper subset, and use ⊆ to allow the possibility of equality. Jordan[2006a] defined a power function over subsets of agents and allocations, so that π :2I × X → R satisfies:

(WC) if C ⊂ C′ ⊆ I then π (C′,x) ≥ π (C,x)∀x ∈ X;

(WR) if yi ≥ xi∀i ∈ C ⊆ I then π (C,y) ≥ π (C,x); and

(SR) if ∅ , C ⊆ I and yi > xi∀i ∈ C then π (C,y) > π (C,x).

Axiom WC requires weak monotonicity in coalitional inclusion; WR requires weakmonotonicity in resources; SR requires strong monotonicity in resources. These ax-ioms imply the following representation:

Lemma 1. Any power function, π (C,x), can be represented by another, π′(C, {xi}i∈C

),

which depends only on the resource holdings of its coalition members.

Proof. Consider arbitrary x,y such that xi = yi∀i ∈ C ⊆ I. Then yi ≥ xi and xi ≥ yi sothat axiom WR requires π (C,y) ≥ π (C,x) ≥ π (C,y). For this to hold, π (C,x) cannotdepend on x j for any j < C. �

The axioms also imply that π (∅,x) is the smallest value that π can take, and isindependent of x.5 Without loss of generality, we normalise π (∅,x) = 0.

The following axioms are not necessary for a function to be a power function, butwill be analytically useful:

4This therefore parallels von Neumann and Morgenstern [1944, §60, 62, 63], which exhaustively charac-terised stable sets for all three agent games in characteristic function form.

5See Beardon and Rowat [2009] for a slightly longer discussion.

4

(AN) if σ : I → I is a 1:1 onto function permuting the agent set, i ∈ C ⇔σ (i) ∈ C′, and xi = x′σ(i) then π (C,x) = π (C′,x′).

(CX) π (C,x) is continuous in x.

(RE) if i < C and π ({i} ,x) > 0 then π (C ∪ {i} ,x) > π (C,x).

Axiom AN is an anonymity axiom which means that power does not depend on theidentity of agents, merely their cardinality and resources.6 Axiom RE is the respon-siveness axiom of Jordan [2007]: the addition of an agent which has power even as asingleton strictly increases the power of its new coalition.

A pillage game is then a triple, (n,X, π).The three power functions defined in Jordan [2006a] are ‘wealth is power’ (WIP),

‘strength in numbers’ (SIN) and ‘Cobb-Douglas’ (CD):

πw (C,x) =∑i∈C

xi; (1)

πv (C,x) = πw (C,x) + v‖C‖ =∑i∈C

(xi + v) ; (2)

πc (C,x) = ‖C‖α · πw (C,x)1−α ; (3)

where v ≥ 0, ‖C‖ denotes the cardinality of C and α ∈ (0, 1). When v > 1, SIN is themajority pillage game of Jordan and Obadia [2004]. These are clearly related:

limv→0

πv (C,x) = πw (C,x) ;

limα→0

πc (C,x) = πw (C,x) ; and

limα→1

πc (C,x) = limv→∞

1vπv (C,x) = ‖C‖.

All three power functions satisfy additional axioms AN, CX and RE.

2.2 DominanceAn allocation y dominates an allocation x, written y K x iff

π (W,x) > π (L,x) ;

where W ≡ {i |yi > xi } and L ≡ {i |xi > yi }. By the strict inequality, domination isirreflexive; by axiom SR, it is asymmetric.

We shall use the following immediate result later:

Lemma 2. Let x,y ∈ X such that W = {i |yi > xi } = {1} and L = {i |yi < xi } = {2}.Then, for any power function satisfying axiom AN, y K x⇔ x1 > x2.

Multiple power functions may therefore represent the same dominance relation, inthe same way that multiple utility functions may represent the same preference relation.Monotonic transformations of power functions clearly represent the same dominancerelations [Jordan and Obadia, 2004];7 we would like to know how much broader theseequivalence classes are. Theorem 11, below, proves that not all pillage games can be

6Jordan and Obadia [2004] called this axiom ‘symmetry’.7Thus, without loss of generality, we may restrict π (C,x) ≥ 0.

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represented by power functions satisfying a continuity axiom. As dominance, a binaryoperator, is necessarily discontinuous in both the relative power and composition of Wand L, this may be surprising.8

For Y ⊂ X, letD (Y) ≡ {x ∈ X |∃y ∈ Y s.t. y K x } ;

be the dominion ofY, the set of allocations dominated by an allocation inY. Similarly,U (Y) = X\D (Y), the set of allocations undominated by any allocation in Y.

3 The coreThe core, K , is the set of undominated allocations, U (X) = X\D (X). FollowingJordan [2006a], let ti ∈ X be a tyrannical allocation such that ti

i = 1 and tij = 0∀ j , i ∈

I. Then:

Theorem 1 (Jordan [2006a], 2.6). The core is the set

K = {x ∈ X |xi > 0⇔ π ({i} ,x) ≥ π (I\ {i} ,x) } .

In particular,

(a) for each i ∈ I, ti ∈ K iff π({i} , ti

)≥ π

(I\ {i} , ti

); and

(b) if π({i} , ti

)< π

(I\ {i} , ti

)∀i ∈ I then K = ∅.

The inequality in item (b), above, was called the no-tyranny condition in Jordan[2006b]. The main result presented in this subsection, Theorem 2, will be an explicitenumeration of all possible cores. To present this, we first proceed with some defini-tions.

Let s jk ∈ X be a split allocation with s jki = 0 and s jk

j = s jkk = 1

2 for all distinct i, jand k. Some basic implications follow immediately:

Lemma 3. When n = 3:

1. K = ∅ implies ti ∈ D(s jk

); and

2. axiom AN implies that s jk < D(ti);

for distinct i, j, k ∈ I.

Proof. 1. As K = ∅, no agent can defend its holdings against both others, so thatπ({i} , ti

)< π

({ j, k} , ti

)for distinct i, j and k. As { j, k} prefers s jk to ti, this

ensures that s jk K ti.

2. Suppose the contrary. Then, for disjoint i, j and k,

π({ j, k} , s jk

)≥ π

({ j} , s jk

)> π

({ j} , sik

)= π

({i} , s jk

)> π

({ j, k} , s jk

);

by axioms WC, SR and AN, and assumption, respectively. This is a contradic-tion.

�

8See Bridges and Mehta [1995, Chapter 8] for a treatment of continuous utility function representationsof preference relations.

6

Following Jordan [2006a], define:

Definition 1. A scalar xi ∈ [0, 1] is dyadic if xi = 0 or x = 2−k for some nonnegativeinteger k. An allocation x = (x1, . . . , xn) is dyadic if each xi component is dyadic. LetD denote the set of dyadic allocations. For each nonnegative integer k, let

Dk ≡{x ∈ X

∣∣∣x ∈ D, for each i, xi > 0⇒ xi ≥ 2−k}.

ThenDk ⊂ Dk+1 for each k, andD ≡ ∪kDk.

Thus,D0 ={t1, t2, t3

},D1 =

{t1, t2, t3, s12, s13, s23

}andD1\D0 =

{s12, s13, s23

}.

Finally, define the ‘champion’ power function as

π (C,x) ≡ maxi∈C

xi; (4)

and let the ‘leader’ power function be

π (C,x) ≡{

maxi∈C xi if maxi∈C xi ≤13∑

i∈C [(1 − v) xi + v] otherwise

}; (5)

where v ∈ (0, 1). The leader power function therefore functions as the champion whenno agent holds more than a third of the resources; once an agent’s holdings increasebeyond this level, the leader power function behaves according to strength in numbers.Satisfying axiom SR requires v < 1; were v = 0, the second expression would reduceto wealth is power. Both π and π satisfy the anonymity axiom AN, neither satisfy theresponsiveness axiom RE; while π satisfies the continuity axiom, CX, π generally doesnot.

To derive the core for the champion power function, we denote by p =(

13 ,

13 ,

13

)the

simplex’ centroid:

Lemma 4. Under π (C,x) ,K = D1 ∪ {p}.

Proof. First use Theorem 1 to establish that D1 ∪ {p} ⊆ K : π({1} , t1

)= 1 > 0 =

π({2, 3} , t1

); π

({1} , s12

)= 1

2 = π({2, 3} , s12

); π ({1} , p) = 1

3 = π ({2, 3} , p).Now establish that no other allocation is undominated. Any allocation (x1, x2, x3)

whose maximal element, x1 without loss of generality, is greater than its next largestlies in D

(t1

). This leaves only the following allocations, x = (x, x, 1 − 2x) where

x ∈(

13 ,

12

); in these cases, x > 1 − 2x so that x ∈ D

(s12

). �

Lemma 5. Under π (C,x) ,K = D0 ∪ {p}.

Proof. First add allocations to the core: π({1} , t1

)= 1 > 0 = π

({2, 3} , t1

); π ({1} , p) =

13 = π ({2, 3} , p).

Now show that no other allocations may belong to the core: let x = (x1, x2, x3) suchthat, without loss of generality, x1 ≥ x2 ≥ x3. Then, as we need not consider p again,x1 >

13 ; similarly, as t1 has already been considered, x1 < 1, so that x2 > 0. There are

then one of three possibilities:

1. π ({1} ,x) > π ({2, 3} ,x), implying that x is dominated by t1, excluding it fromthe core.

2. π ({1} ,x) = π ({2, 3} ,x) which, as π ({1, 3} ,x) > π ({1} ,x) when x1 >13 , and

π ({2} ,x) ≤ π ({2, 3} ,x) by axiom WC, implies that π ({1, 3} ,x) > π ({2} ,x), sothat agent 2 cannot defend its non-zero holdings, violating the core condition.

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3. π ({1} ,x) < π ({2, 3} ,x), so that it is now agent 1 that cannot defend its positiveholdings.

�

We may now extend Theorem 1 to fully characterise the core for all anonymouspower functions when n = 3. The examples of power functions considered in Jordan[2006a] favour concentrated core allocations, dividing resources between one or twoagents. The next result shows that, when n = 3 and the anonymity axiom holds, thereis one other possible allocation, and two other possible cores.

Theorem 2. When n = 3 and axiom AN holds, the only possibilities for K are: K =

∅,K = D0,K = D0 ∪ {p} ,K = D1 and K = D1 ∪ {p}.

Thus, all allocations in the core, x ∈ K , lie on symmetry axes, so that x j = xk forj , k. We shall later see that this need not be true of allocations in stable sets.

Proof. First show that the cases listed are, in fact, feasible:

1. the ‘strength in numbers’ power function, defined in equation 2, yields an emptycore [Jordan, 2006a, Proposition 4.2] when v > 1.

2. for the same power function, v ∈ (0, 1) yields K = D0 [Jordan, 2006a, Proposi-tion 4.2].

3. the ‘leader’ power function, defined in equation 5, yields K = D0 ∪ {p} byLemma 5.

4. the ‘wealth is power’ power function, defined in equation 1, yields K = D1[Jordan, 2006a, Proposition 3.2].

5. the ‘champion’ power function, defined in equation 4, yields K = D1 ∪ {p} byLemma 4.

Now show that no other allocations than the ti, the s jk and p may be in the core.Trivially, the only x ∈ X such that x j = xk = 0 are the ti, which have already beenconsidered. Of those x ∈ X such that xi, x j > 0, xk = 0, only those setting π ({i} ,x) =

π ({ j} ,x) may be in K ; by axiom AN, these are the s jk. Finally, consider the strictlypositive x ∈ X such that x > 0. In order to guard against the richer dominationpossibilities, these must satisfy π ({i} ,x) = π ({ j} ,x) = π ({k} ,x) = π ({ j, k} ,x); byaxiom AN, this uniquely implies p.

Finally, show no other combination of these allocations is permissible. By the firstclause in Theorem 1, a non-empty core must contain the ti. This ends the proof. �

When n > 3, the ‘champion’ power function’s core contains the generalisationof p, namely,

(1n , . . . ,

1n

). Usually, core allocations do not grant all agents positive

resources as this requires that the allocation must be undominated by perturbations inany direction; under the champion power function, agents with positive resources neednot contribute to their coalition’s power.

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4 Stable setsA set of allocations, S ⊆ X, is a stable set iff it satisfies internal stability,

S ∩ D (S) = ∅; (IS)

and external stability,S ∪ D (S) = X. (ES)

The conditions combine to yield S = X\D (S). While stable sets may not exist, or maybe non-unique in general cooperative games, the core necessarily belongs to any stableset; when the core also satisfies external stability, it is the unique stable set [Jordan,2006a, Proposition 2.4].9

As stable sets are more complex objects than the core, Jordan [2006a] derived re-sults only for the WIP, SIN and CD power functions defined in equations 1 to 3, above.WIP has a unique stable set; a stable set for some instances of SIN is derived; no stableset exists for Cobb-Douglas. All three power functions satisfy anonymity axiom AN.

Deriving more general results may benefit from an explicitly algorithmic approach.However, the complexity of stable sets is such that results only exist for special cases.Brandt et al. [2007] considered irreflexive, asymmetric dominance relations on abstractgames containing finite sets of allocations. By reducing the problem of whether an al-location belongs to a stable set (assumed to exist) to the Boolean satisfiability problem,which is known to be hard (technically, NP-complete), they proved that the stable setproblem is hard in the same way, requiring exponentially increasing resources (in nand ‖X‖) to solve. Thus, when the cardinality of X is infinite, as here, the approach ofBrandt et al. [2007] does not provide a general algorithm. On the question of whethera stable set even exists, Deng and Papadimitriou [1994] warned that the question “isnot even known to be decidable” as it “is not obvious at all that there is any algorithm(however slow) for deciding this!”

The existing algorithm for constructing stable sets in pillage games owes to Roth[1976] via Jordan and Obadia [2004].10 Before introducing it, define a weaker versionof condition ES, namely self-protection:

S ⊆ U2 (S) ; (SP)

where U2 (S) = U (U (S)). Whereas external stability requires that all allocationsoutside the set in question be dominated by one inside it, self-protection only requiresthis of those external allocations that dominate allocations within the set.

Algorithm 1 takes as input a non-empty set satisfying conditions IS and SP; if thecore is non-empty, this is the natural candidate to use. Each iteration of the algorithmgenerates weakly larger sets S0 ⊆ S1 ⊆ . . . satisfying both IS and SP. As, by Jordan[2006a], stable sets are finite, the algorithm will terminate after finitely many itera-tions. If its final iterate also satisfies condition ES, then it is the unique stable set. Thealgorithm may, however, terminate before this point without finding a stable set.11

The algorithm is therefore incomplete in four respects. First, it provides no meansfor calculating the core. Second, when the core is empty, it provides no clue for findingan internally stable and self-protecting initial iterate, S0. Indeed, for some n > 2pillage games the only such sets are themselves stable [Jordan and Obadia, 2004], in

9Asilis and Kahn [1992] expressed this result in terms of transparency.10See also Asilis and Kahn [1992] and Jordan [2006b, Proposition 3.9].11If S0 = ∅, which trivially satisfies IS and SP, then the algorithm will terminate immediately.

9

Algorithm 1 The Roth-Jordan algorithm1: S0 ≡ K

2: i = 13: repeat4: Si ≡ U2 (Si−1)5: i = i + 16: until Si = Si−17: if Si satisfies ES then8: S = Si

9: end if

which case the algorithm cannot help find them. Third, it does not specify an efficientway to compute U2 (Si−1). Finally, it is not clear whether terminating at an Si which isnot externally stable means that no stable set exists, or merely that further steps mustbe taken independently of the algorithm.

Thus, the Roth-Jordan algorithm leaves open the Deng and Papadimitriou [1994]question about the decidability of stable set problems, even in the context of pillagegames.12 In Examples 2 and 9, below, the algorithm makes no progress beyond S0;in the first case, no stable set exists; in the second case, one does. In Example 3, itterminates at S2, without finding the unique stable set.

We ultimately seek to identify all possible stable sets for n = 3 pillage games. Togain intuition into this general problem, we first consider those whose power functionssatisfy the anonymity axiom, AN. This allows us to return to the algorithmic ques-tion, proposing an alternative to the above that both decides the stable set question foranonymous n = 3 pillage games, and allows computation of stable sets.

The following two subsections address two complementary situations. First, anony-mous, n = 3 pillage games with an empty core; this subsection shows all such games torepresent the same dominance relation. Second, those games with a non-empty core.While these must form part of any stable set, we differ from Roth-Jordan by distin-guishing between different types of core allocations.

4.1 Empty coreWhen the core is empty, the Roth-Jordan algorithm can obviously not use the core asits first iterate. This subsection demonstrates that these cases do not actually pose aproblem: any anonymous three agent pillage game with an empty core yields a uniquestable set, S = D1\D0. The argument proceeds in two steps: an example of the preced-ing is found; Theorem 3 then shows that all other anonymous three agent pillage gameswith an empty core represent the same dominance ordering as that in the example. In-tuitively, the empty core leaves D

(s jk

)defined only by the coalitional membership

inequalities x j, xk <12 , which does not distinguish between power functions.

Our example shall be the SIN power function defined in equation 2. It is easy toderive that

D(si j

)=

{x ∈ X

∣∣∣∣∣∣(

12> xi, x j; xi + x j >

1 − v2

)∪

(12

= xi > x j > xk

)∪

(12

= x j > xi > xk

)}.

12A decision problem may be answered by a ‘yes’ or a ‘no’; an algorithm decides a decision problem if,for every instance of the problem, it returns ‘yes’ if the input represents a ‘yes’-instance, and a ‘no’ otherwise[Kellerer et al., 2004].

10

Jordan [2006a, Proposition 4.2] proved a general result which, in the n = 3 case,implies that: when v > 1, so that the core is empty,D1\D0 is stable. This may be seenin Figure 1: for n = 3,D (D1\D0) = X\ (D1\D0).

t1 t2

t3

s23

D(s23

)

Figure 1: SIN when v > 1: S = D1\D0

Theorem 3 (SIN equivalence). When n = 3, all power functions satisfying axiom ANand yielding an empty core represent the same dominance ordering.

Proof. Fix an x and a y, in turn fixing W and L. By axiom AN and Lemma 2, thesepower functions do not discriminate between dominance orderings when the inducedW and L are singletons.

For the non-singleton W and L, make use of K = ∅. First consider W = {1} andL = {2, 3}. There are no y and x for which y K x as

π ({1} ,x) ≤ π({1} , t1

)< π

({2, 3} , t1

)≤ π ({2, 3} ,x)

for all x ∈ X, with the strict inequality following fromK = ∅. Reading the inequalitiesabove from right to left yields the case of W = {2, 3} and L = {1}. This always setsy K x without further conditions, just as for SIN. �

Corollary 1. When n = 3, all power functions satisfying axiom AN and yielding anempty core haveD1\D0 as their unique stable set.

Proof. Theorem 3.4 of Jordan and Obadia [2004] proved that D1\D0 is the uniquestable set of SIN when v > 1 and n = 3. As this pillage game has an empty core, theresult follows from the theorem. �

4.2 Non-empty core

When the core is non-empty, D(s jk

)will be defined not just by the coalitional mem-

bership inequalities, but by particular features of the power function. Thus, unlikethe previous case, power functions admitting non-empty cores cannot generally be ex-pected to represent the same dominance orderings, or even to yield the same stable sets.In this subsection, we show that all anonymous n = 3 pillage games with non-emptycores whose power functions also satisfy a continuity and a responsiveness axiom ei-ther have no stable set, or have a unique stable set. In the former case, we isolate thesource of non-existence; analysis of both cases yields an algorithm for computing aswell as deciding the stable set question in these games.

11

First define the locus of allocations lying between D(ti)

and D(s jk

):

Bi ≡ {x ∈ X |π ({i} ,x) = π ({ j, k} ,x) } .

Note that Bi may be empty. When it is non-empty, a number of objects are often ofinterest. The first are the qi ∈ Bi, the midpoint of Bi, so that qi

j = qik. The second is the

set of maximal elements from agent i’s point of view:

Mi ≡

{x ∈ Bi

∣∣∣∣∣∣xi = maxy∈Bi

yi

}with elements mi ∈ Mi. Finally, there are the elements of the maximal set mostfavourable to each of the other two agents,

Ri ≡{ri

j, rik ∈ Mi

∣∣∣rij j ≥ x j, ri

kk ≥ yk∀x,y ∈ Mi}

;

where rij j is the jth coordinate of ri

j. Let B ≡{Bi

}3

i=1,M ≡

{Mi

}3

i=1,R ≡

{Ri

}3

i=1and

Q ={qi

}3

i=1.

Clearly, when Bi = ∅,Mi and Ri are also empty, and qi does not exist. These objectsmay also be empty or fail to exist if Bi is open in the right way, as will be demonstratedin Examples 8 and 9. Finally, it may also be the case that qi = ri

j = rik, or that ri

j = si j.Before proceeding, we establish a number of conditions satisfied along the Bi, when

they exist.

Lemma 6. When n = 3, all power functions satisfying axiom AN yield

Bi ⊂

{x ∈ X

∣∣∣∣xi ≥ max{x j, xk

} }.

Proof. Suppose otherwise, so that xi < x j for some x ∈ Bi. Then, by axioms SR andAN, π ({i} ,x) < π ({ j} ,x) which, by axiom WC, is weakly less than π ({ j, k} ,x). Thus,x < Bi. �

The champion power function of equation 4, illustrated in Figure 9, therefore ex-emplifies the lowest possible bound of Bi.

Lemma 7. ∀x,y ∈ Bi, if yi > xi then either y j > x j or yk > xk for distinct i, j and k.

Proof. Suppose otherwise, so that y j ≤ x j and yk ≤ xk. Then, by definition of Bi andaxioms WR and SR:

π ({i} ,y) > π ({i} ,x) = π ({ j, k} ,x) ≥ π ({ j, k} ,y) ;

contradicting y ∈ Bi. �

Corollary 2. When n = 3, a ray from ti through the simplex cuts Bi at most once.

Proof. If x and y lie on such a ray then yi > xi implies y j < x j and yk < xk, contradict-ing Lemma 7. �

For the next lemma, define Bi+ ≡

{x ∈ Bi

∣∣∣∣xi > max{x j, xk

} }, the subset of Bi such

that agent i has strictly more power than either of j or k, alone. (In the champion powerfunction, Bi

+ = ∅; in SIN with v ∈ (0, 1) , Bi+ = Bi.) We may now state conditions under

which dominance follows from inequality:

12

Lemma 8. When n = 3 and π satisfies axiom AN, if either

1. x,y ∈ Bi are such that yi > xi ≥ x j > xk and y j > x j; or

2. x ∈ Bi+ and y ∈ Bi are such that yi > xi;

then y K x.

Proof. In the first case, W ≡ {l |yl > xl } = {i, j} and L ≡ {l |xl > yl } = {k} by assump-tion. In the second case, the same W and L can be assumed without loss of generalityby Lemma 7.

In the first case, xi > xk is assumed; in the second, it follows from x ∈ Bi+. In

both cases, axioms AN and SR then ensure that π ({i} ,x) > π ({k} ,x); by axiom WC,it follows that π (W,x) > π (L,x), hence y K x. �

When axiom RE holds, Bi\Bi+ may disappear:

Lemma 9. When n = 3 and π satisfies axioms AN and RE, Bi\Bi+ ⊆ D1\D0.

Proof. First consider the x > 0 in Bi. By axiom SR, π ({i} ,x) , π ({ j} ,x) , π ({k} ,x) >0 so that, by axiom RE, π ({ j, k} ,x) > max {π ({ j} ,x) , π ({k} ,x)}. If x ∈ Bi\Bi

+ thenxi = max

{x j, xk

}so that, by axiom AN, π ({i} ,x) = max {π ({ j} ,x) , π ({k} ,x)} <

π ({ j, k} ,x), so that x < Bi.Now consider the x ∈ Bi that set at least one component to 0. This can’t be xi as

axioms AN and WC would imply π ({i} ,x) < π ({ j, k} ,x). Without loss of generality,then, let xk = 0 so that π ({i} ,x) = π ({ j, k} ,x). Axiom AN then requires that xi = x j =12 . �

Lemma 10. When n = 3 and π satisfies axioms AN and RE, if Ri , ∅ and x > 0 lies inBi\

(Ri ∪

{qi

}), then x ∈ D

(Ri

).

Proof. First consider the case in which x lies in Mi\(Ri ∪

{qi

}), so that agent i is indif-

ferent between allocation x and those in Ri. As x , qi let, without loss of generality,x j > xk, so that, by axiom AN, π ({ j} ,x) > π ({k} ,x) so that ri

j K x, establishing theresult.

Now let x lie in Bi\(Mi ∪

{qi

}). First consider the subcase in which x ∈ Bi

+.The set of agents preferring ri

j to x therefore includes i (by definition of Mi) and at

least one other (by Lemma 7). As x ∈ Bi+, xi > max

{x j, xk

}so that π ({i} ,x) >

π({

arg max{x j, xk

}},x

). By axiom WC, ri

j K x.

Finally, let x ∈ Bi\(Bi

+ ∪ Mi ∪{qi

})so that, without loss of generality, ri

ji > xi =

x j > xk > 0. Then the set of agents preferring rij to x includes i and at least one other;

in the worst case, this is k. As xk > 0 and axiom RE holds, π ({i, k} ,x) > π ({i} ,x) =

π ({ j} ,x) so that rij K x. �

The assumption that Ri , ∅ is made in addition to the axiomatic assumptions as noconditions have been presented as yet for ruling out Bi resembling the topologist’s sinecurve, y = sin 1

x .A set of allocations K i ⊆ Y ⊆ X that satisfies K i = Y\D (Y) is the core on Y.

Similarly, a set of allocations Si ⊆ Y ⊆ X satisfying Si ∩D(Si

)= ∅ and Si ∪D

(Si

)=

Y, is stable on Y.

13

Theorem 4. When n = 3, π satisfies axioms AN and RE, and Ri , ∅, the unique stableset to exist on

(Bi

+ ∪ Ri)

is Si ≡ Ri ∪({

qi}∩ Mi

). When Mi = ∅, there is no stable set

on Bi+.

Proof. Lemma 8 establishes that, on Bi+, dominance is transitive as it inherits the strict

partial order on xi.13

Now consider the restriction of(Bi

+ ∪ Ri)

to, without loss of generality, those allo-cations, x, for which x j > xk. Lemmas 8 and 10 together establish that dominance isa complete ordering on this restricted domain. Thus, if ri

j exists, it is the unique stableset on this restricted domain; otherwise, there is no stable set on it [von Neumann andMorgenstern, 1944, §65.4]. (When Ri =

{qi

}the next paragraph may be omitted.)

As the restriction of(Bi

+ ∪ Ri)

to x j = xk is at most a singleton, qi is trivially theunique stable set on it, when it exists.

Finally, as neither rij K ri

k nor the reverse, both must belong to any stable set on

Bi+; qi will also belong to any stable set on

(Bi

+ ∪ Ri)

iff qi ∈ Mi as it will, otherwise,

belong to D(Ri

). �

The proof may be compared to that in Jordan [2006a][Proposition 4.6], which es-tablished non-existence of a stable set for the Cobb-Douglas power function, πc, andthe SIN power function when v ∈ (0, 1).14 In that case, the proof demonstrated thatany stable set that might exist was necessarily infinite, a contradiction; the proof hereestablishes a complete ordering over Bi

+ ∪ Ri.

Lemma 11. Let Si be the unique stable set on(Bi

+ ∪ Ri). If K , ∅,x ∈ Si, π satisfies

axiom CX and y K x then y ∈ D(ti).

Proof. As x ∈ Si it is also in Bi. If y K x then one of the following three disjointcases applies:

1. y sets yi > xi so that, by axiom SR, π ({i} ,y) > π ({i} ,x). As x ∈ Si it isalso in Mi. Therefore, it cannot be that π ({i} ,y) = π ({ j, k} ,y) as then y ∈ Bi

with yi > xi, so that x < Mi, a contradiction. If π ({i} ,y) > π ({ j, k} ,y) theny ∈ D

(ti), the desired result. Finally, suppose that π ({i} ,y) < π ({ j, k} ,y); then,

asK , ∅, π({i} , ti

)≥ π

({ j, k} , ti

)so that axioms SR and CX guarantee that there

exists a z such that 1 ≤ zi > yi and π ({i} , z) = π ({ j, k} , z), again contradictingx ∈ Mi.

2. y sets yi = xi and, permuting indices j and k if necessary, y j > x j > xk > yk.(Without x j > xk, y would not dominate x under axiom AN.) As this pre-cludes x = qi, it follows from Theorem 4 that x = ri

j. It cannot be thatπ ({i} ,y) = π ({ j, k} ,y) as then y is an element of Mi with y j > x j, contra-dicting the assumption that x = ri

j. Finally, π ({i} ,y) < π ({ j, k} ,y) is ruledout by implying the existence of the same z that yielded the contradiction in theprevious case.

3. y sets yi < xi. As x ∈ Bi, π ({i} ,x) = π ({ j, k} ,x), so that y 6K x.

13von Neumann and Morgenstern [1944, §65.3] omit the adjective ‘strict’.14There, the allocations indexed by λ define a generalised version of Bi. The present lemma is obviously

more restrictive insofar as it only applies to the n = 3 case, but more general in applying to a broader classof pillage games.

14

�

The following example shows that, when axiom CX is violated, a y < D(ti)

maybe found to dominate x:

Example 1. Consider a discontinuous version of SIN, so that

π (C, x) ≡∑i∈C

{ 12 xi + v if xi ≤

1−v2

xi + v otherwise

}; (6)

where v ∈(

17 , 1

). Feasibility of x requires that xi >

1−v2 ≥ x j, xk:

Bi =

{x ∈ X

∣∣∣∣∣xi =1 + 2v

3,max

{x j, xk

}≤

1 − v2

}.

Thus, depicted in the simplex, Bi is a line that stops short of the edges. As xi remainsconstant over Bi, Mi = Bi; qi sets qi

i = 1+2v3 and qi

j = qik = 1−v

3 ; finally, Ri ={ri

j, rik

},

where rij sets ri

ji = 1+2v3 , ri

j j = 1−v2 and ri

jk = 1−v6 . There is therefore a y ∈ D

(s jk

)with

yi = riji and y j > ri

j j such that y K x for any x ∈ Bi with xi = riji and x j > xk.

While the lemma therefore does not apply to this example,D1 ∪R ∪Q is, nonethe-less, stable.

Finally, then:

Theorem 5. If K , ∅, π satisfies axioms AN, CX and RE, and Ri , ∅, then any stableset, S, must contain Si, as defined in Theorem 4.

Proof. As Ri ⊆ Bi, the assumed conditions ensure that Bi exists. By Theorem 4, Si isthe unique stable set on it. Finally, as ti must belong to S (as D0 ⊆ K ⊆ S) so thatno allocation in D

(ti)

can belong to S. Lemma 11 then ensures that including Si isnecessary for external stability. This establishes the result. �

The preceding theorem and the existing result that core allocations must belong to astable set now force two types of allocations into any stable set that exists. With the nextresults, a third and final type is forced in. This shall form the basis for our uniquenessresult. We begin with a special case that introduces the required equivalence.

Theorem 6. For the anonymous, n = 3 SIN pillage game with v = 1, the unique stableset is S = D1.

Proof. It is immediate that K = D0 so that U (K) = {x ∈ X |x > 0 ∪ x ∈ D1\D0 }.Adding theD1\D0 to the core therefore produces a stable set, S = D1.

To establish uniqueness, recall that, for v > 1, Jordan and Obadia [2004, Theorem3.4] prove thatD1\D0 is the unique stable set on X. Two steps allow direct applicationof this result. First, when v > 1, dominance on X is equivalent to dominance onU (D0) when v = 1: for any y,x,W ≡ {i |yi > xi } and L {i |yi < xi } ,y K x iff either(‖W‖ = 2 > ‖L‖ = 1) or, if ‖W‖ = ‖L‖ = 1 such that the W member is granted a largerendowment in x than is the L member.

Second, it must be ensured that the geometry of U (D0) allows the argumentsused by Jordan and Obadia [2004] on X. This is ensured by replacing

(34 , 0,

14

)with(

34 , ε,

14 − ε

)for small ε > 0. Thus, D1\D0 is the unique candidate for addition to

K = D0 when v = 1; asD1 is stable, the result follows. �

15

In this degenerate case,{qi

}= Ri = Mi = Bi, the singleton

{ti}. The Roth-Jordan

algorithm fails to make progress beyond S0 = D0, setting U2 (S0) = S0.To establish our final result, define

X− ≡{x ∈ X

∣∣∣x1 > r211, x2 > r3

22, x3 > r133

};

and p jk ≡ 12

(r j

i + rki

), so that P ≡

{p12,p13,p23

}. Allocations in the restricted domain,

X−, are undominated by those in Ri; they form a contraction of the original domain,X. The p jk are the midpoints of the edges of the contracted simplices; they are alsoundominated by allocations in Ri. When the p jk = p, the centroid, P is a degeneratesingleton. As WIP power function shows, Pmay be empty (q.v. the left panel of Figure4).

Theorem 7. When n = 3, π satisfies axioms AN, CX and RE, K , ∅ and Ri , ∅, thenthere exists a unique stable set, S = K ∪

{Si

}3

i=1∪P, where Si is defined in Theorem 4.

Proof. The set of allocations undominated byK∪{Si

}3

i=1is X−∪P. By the same argu-

ment as in the proof of Theorem 6, Jordan and Obadia [2004, Theorem 3.4] establishesthat P is the unique stable set on X− ∪ P. The result then follows from now familiararguments: any stable set must contain D0 ⊆ K , which excludes D

(ti), forcing, in

turn, the inclusion of the Si (by Theorem 5); finally, this leaves X− ∪P, which has justbeen addressed. �

WhenP is empty, the unique Smay therefore be computed in two, rather than threesteps.

Corollary 3. When n = 3, and π satisfies axiom AN, CX and RE, if a stable set, S,exists, then ‖S‖ ≤ 15.

Proof. When the core is empty, ‖S‖ = 3. Otherwise, the three tyrannical allocationsbelong to the core, and therefore to S. Each Si contributes up to three more. Finally,when P is non-empty, it adds either one (if degenerate) or three (otherwise) more allo-cations. �

This bound improves significantly on both the finite bound in Jordan [2006a], andthe Ramsey number bound in Kerber and Rowat [2009]; both of these derived implica-tions of axiom SR alone. Further, the bound is tight, as demonstrated by the followingpower function:

πk (C,x) ≡∑i∈C

[xi +

sin (kπxi)kπ

]; (7)

for k ∈ Z. When k = 9 all allocations enumerated by the Corollary are present.15

To conclude this section, Algorithm 2 presents an alternative approach to generat-ing stable sets in n = 3 pillage games.16 Like Roth-Jordan, the algorithm adds elementsto a small, undominated set in a number of steps. Further, all sets generated are self-protected, and generated by the U2 (·) operation: if S exists, then S1 = U2 (S0) and S,the final iterate, is equal to U2 (S1). Our algorithm differs from Roth-Jordan in beingable to add core elements at all three steps by distinguishing between core allocations

15This class of power functions shall be used in Section 5, below.16The pseudo-code is presented for clarity rather than coding efficiency.

16

at which one agent is strictly more powerful than the other two, and those at which thefirst agent is just as powerful as the others.17 While their S0 = K , ours isD0 ⊆ K .18

The algorithm therefore both decides the stable set question for an n = 3 pillagegame satisfying additional axioms AN, CX and RE, and computes its unique stableset. As such, it overcomes three of the four ways in which the Roth-Jordan algorithmis incomplete, albeit over a narrow class of pillage games: the first two weaknessesare addressed as the new algorithm starts with D0, not the core and, thus, never needsto compute the core; and the last shortcoming is dealt with since the algorithm alwayseither return that no stable set exists or returns the unique stable set. The third problem,that of efficiently computing the interim sets, remains.

Algorithm 2 The stable set in n = 3 pillage games satisfying axioms AN, CX and RE

1: if π({i} , ti

)≥ π

({ j, k} , ti

)then

2: S0 = D03: if Mi = ∅ then4: return “no stable set exists”5: else6: S1 = S0 ∪

{Si

}3

i=17: if S1 ∪ D (S1) , X then8: S2 = S1 ∪ P

9: else10: S = S111: end if12: end if13: else14: S = D115: end if16: return S

The above procedure provides guidance on handling the n > 3 case as well: theremay be a D

(ti)

region around each ti, separated from the rest of X by a Bi. Analysis of

D(ti)

and Bi likely proceeds as in the n = 3 case. On the other side of the Bi, the gamewill again be equivalent to the SIN game with v > 1, for which few results are known[Jordan and Obadia, 2004]. Thus, progress in analysing n > 3 pillage games depends,in general, on progress in analysing these SIN games.

5 ExamplesIn this section we first apply the theory presented above to examples satisfying axiomsAN, CX and RE. We then derive results for cases violating axioms CX and RE.

We first present some more properties that may or may not be satisfied by particularpower functions:

(C0) if xi = 0∀i ∈ C then π (C,x) = 0.

(CN) for all partitions on I,{C1, . . . ,Cm

},∑m

i=1 π(Ci,x

)is constant.

17q.v. the champion power function, displayed in Figure 9.18Thus, Roth-Jordan appears “greedy”, adding more elements in its first iteration; this is irrelevant, though,

as that presented here eventually adds all core allocations.

17

(AD) π (C,x) =∑

i∈C π ({i} ,x).

Axiom CN a conservatism axiom; the additivity axiom, AD, is stronger than axiomRE.

5.1 Jordan examples and variants

t1 t2

q3 = r

t3

s23

D(t1

) D(s23

)

Figure 2: Strictly concave power function, π (C,x) =∑

i∈C√

xi

When π satisfies axioms AN, CX and RE and{qi

}= Ri = Mi ⊂ Bi, a non-

degenerate locus, then, by Theorem 7, S ≡ D1 ∪ Q. The next theorem shows thatany strictly concave π satisfying all additional axioms except CN has this as its uniquestable set:

Theorem 8. When n = 3, π satisfies axioms AN, CX, AD and C0, and is strictlyconcave in xi∀i ∈ C, S ≡ D1 ∪ Q is the unique stable set.

Proof. By Jordan [2006a, 2.6], the core is the set D1, which must therefore belong toany stable set that exists. Furthermore, Bi is non-empty. Thus, the result follows if wecan prove that Mi =

{qi

}.

By the additional axioms and the notational abuse allowed by Lemma 1, allocationsalong Bi may be represented by

π (xi) = π(x j

)+ π

(1 − xi − x j

);

for distinct i, j and k. As π is strictly increasing and concave, it has an inverse, sayρ, that is also strictly increasing. Thus, the allocations in Bi may be representedby xi = ρ

[π(x j

)+ π

(1 − xi − x j

)]so that allocations in Mi solve maxx∈X π

(x j

)+

π(1 − xi − x j

). As the first order conditions produce x j = 1−xi

2 and feasibility requiresxk = 1−xi

2 , the proof is complete. �

Figure 2 illustrates an example of this.19

Jordan [2006a, Proposition 4.6] proves that no stable set exists for SIN with v ∈(0, 1). The next example interprets that result in terms of the present theory:

19In Jordan [2007], power depends on production, a concave function of wealth.

18

t1 t2

t3

s23

D(t1

) D(s23

)

Figure 3: SIN when v ∈ (0, 1)

Example 2. Under the SIN power function with v ∈ (0, 1), illustrated in Figure 3,K = D0. Further, Mi = Bi =

{x ∈ X

∣∣∣x > 0, xi = 1+v2

}. Thus, Bi is an open set as(

1+v2 , 1−v

2 , 0)∈ D

(t1

). Consequently, by Theorem 5, no stable set can exist.

Applying the Roth-Jordan algorithm, S0 = K = D0,U (S0) = D1 ∪X\D (D0) andU2 (S0) = S0. Thus, the algorithm fails to make progress, indicating – in this case –non-existence of S.

In the next example, the unique stable set is not found by the Roth-Jordan algo-rithm; furthermore, and unusually, it does not contain the split allocations, D1\D0 ={s12, s13, s23

}.

Example 3. Letπ (C,x) =

∑i∈C

[√xi + v

]where v ∈ (0, 1). It is easily confirmed that K = D0. By Theorem 4, Q =M = R must

also be elements of any stable set, where qi implicitly solves√

qii =

√2(1 − qi

i

)+ v

and qij = qi

k.Now search for P ⊆ U (D0 ∪ Q) such that P is stable on (D0 ∪ Q) ∪ U (D0 ∪ Q)

and that S ≡ D0∪Q∪P is internally stable. The problem is identical to that addressed

by Theorem 6 when U (D0) is scaled down to U (D0 ∪ Q). LettingP be(

1+qii

4 ,1+qi

i4 ,

1−qii

2

)and its permutations, it follows that the unique stable set is S = D0∪Q∪P. Unusually,the split allocations,D1\D0 do not belong to S as they lie within D (P).

Interpreting the above in light of Roth-Jordan, S0 = D0, so that

U (S0) = D0 ∪ {x ∈ X |π ({i} ,x) < π ({ j, k} ,x) } ;

and S1 = D0 ∪ Q. But U (S1) = U (D0 ∪ Q) and S2 = S1, so that the algorithmterminates before finding S. By contrast to Example 2, however, a unique stable setexists in this case.

We now turn to cases in which power functions are convex in the xi. This allowsa coalition to increase its power by transferring resources from the poorer to the richermember, even if its aggregate resources are diminished:

19

Example 4. Let π (C,x) =∑

i∈C x2i . Then, for all ε ∈

(0, 1

4

),

π

({2, 3} ,

(12,

14,

14

))=

18< π

({2, 3} ,

(12,

14

+ ε,14− ε

))=

18

+ 2ε2.

We may also state a parallel result to Theorem 8 for strictly convex functions:

Theorem 9. When n = 3, π satisfies axioms AN, CX, AD and C0, and is strictly convexin xi for all i in C,D1, the core, is the unique stable set.

Proof. The core isD1. By Jordan [2006a, Proposition 2.4], any stable set contains K ,the core. As

Bi ⊆ D(si j

)∪ D

(sik

)∀i ∈ I;

there exists no x ∈ X\K such that x < D (K). �

Thus, as π (C,x) varies from strictly concave to strictly convex, the interior (non-core) allocations in the unique stable set distort towards the WIP stable set, and thendisappear, leaving just the core allocations.

The right panel of Figure 4 depicts the case when π (C,x) is strictly convex in thexi; for sake of reference, the WIP case is depicted in its left panel.

t1 t2

q3 = r

t3

s23

D(t1

) D(s23

)

t1 t2

t3

s23 = r

D(t1

) D(s23

)

Figure 4: WIP, and strictly convex π (C,x)

Power functions need not to convex or concave over their whole domain to produceidentical stable sets to those that are. The next two examples illustrate this. Both usethe following lemma:

Lemma 12. When n = 3 and π satisfies axioms AN, C0 and AD then, for distinct i, jand k, π

({i} , si j

)= π

({ j, k} , si j

).

Proof. By axioms AD and C0 respectively:

π({ j, k} , si j

)= π

({ j} , si j

)+ π

({k} , si j

)= π

({ j} , si j

).

Axiom AN ensures the result. �

Example 5. Consider

πh (C,x) ≡∑i∈C

haversin ($xi) ≡∑i∈C

1 − cos ($xi)2

;

20

where $ is the constant pi.20 This satisfies all of the additional axioms except CN,conservatism. It is locally convex for all xi ∈

(0, 1

2

)and locally concave for all xi ∈(

12 , 1

). By Lemma 12 and axiom SR, x1 = 1

2 is the maximal allocation for agent 1

satisfying π ({1} ,x) = π ({2, 3} ,x). Thus, π is convex for all x ∈ D(ti), which is then

– as in the right panel of Figure 4 – a convex set. Therefore, asD1 is the unique stableset in that case, it is in this case as well.

Example 6. Parametrically define the power function πp (C,x) ≡∑

i∈C πp ({i} ,x) asfollows:

xi =2ti − sin ti

8$;

πp ({i} ,x (t)) =2ti + sin ti

8$; (8)

for ti ∈ [0, 4$]∀i ∈ I. Thus, πp ({i} ,x) is the sine curve, attenuated and then rotatedcounterclockwise $

4 radians. It satisfies all the additional axioms except conservatism,CN. Again, Lemma 12 holds, but now the dominance regions are as in the left panel ofFigure 4. Thus, this power function belongs to an equivalence class with WIP.

The key feature in both of the above examples is a symmetry condition. This maybe generally stated:

Theorem 10. When n = 3, and π satisfies axioms AN, C0, AD as well as

(QS) π ({ j} ,x) |x j=14−ε

+ π ({k} ,x) |xk= 14 +ε = 1

2 ;

for all ε ∈[0, 1

4

], thenD2 is the unique stable set.

Proof. Property QS and axioms AD and C0 ensure that π ({1} ,x) |x1= 12

= 12 . Axioms

C0 and AD then ensure that π ({2, 3} ,x) |x1= 12

= 12 . Therefore

(12 ,

14 + ε, 1

4 − ε), for

ε ∈[0, 1

4

], defines B1. For all ε , 0 these allocations are dominated by

(12 ,

12 , 0

). As

axiom AN implies that the problem is symmetric in the agents, the result follows. �

Corollary 4 (WIP equivalence). When n = 3, all power functions satisfying axiomsAN, C0, AD and QS represent the same dominance ordering.

Proof. As in Theorem 3, fix an x and a y. As before, when these sets are both single-tons, Lemma 2 notes that all power functions satisfying axiom AN represent the samedominance orderings.

There are two non-singleton W and L. First consider W = {1} and L = {2, 3}. Thusy K x is equivalent to π ({1} ,x) > π ({2, 3} ,x). As, under the assumed axioms, theinequality holds with equality when x1 = 1

2 , dominance is equivalent to x1 >12 for any

power function satisfying these axioms.The case when W = {2, 3} and L = {1} reverses the inequality above. �

As sin x is point symmetric about x = π, the power function defined in Example6 is point symmetric about

(14 ,

14

). Thus, Theorem 10 applies. It also applies to other

transformations of sin x, including the family defined by equation 7 when 14 k ∈ Z.

Modifications that destroy the symmetry about xi = 14 , such as that displayed in Figure

5, destroy the result.20We use $ to denote the constant pi ≈ 3.14 . . . to avoid confusion with the power function.

21

t1 t2

t3

s23

D(t1

) D(s23

)

Figure 5: π3 (C,x) : S = D1 ∪ Q = D1 ∪ R

For completeness, we present an example in which Si ⊂ Bi lacks the central qi

allocations. The asymmetric allocations in R were present when the πk class of powerfunctions were introduced in equation 7, but have not yet been explicitly illustrated.Relative to the cases studied in Jordan [2006a], they are new.

Example 7. For n = 3 and π7 (C,x), the unique stable set is D1 ∪ R, as illustrated inFigure 6.

t1 t2

t3

s23r3

2

D(t1

)

Figure 6: π7 (C,x) : S = D1 ∪ R

These sinusoidal power functions have ‘wealth is power’ as a limit case:

limk→∞

πk (C,x) =∑i∈C

xi.

As k → ∞, the range of the sinusoidal Bi thus collapses.

5.2 Discontinuous examplesWhen axiom CX is violated, the results presented above do not apply, as demonstratedby Example 1. In this subsection, we therefore present two different examples of dis-continuities; in the first case, no stable set exists; in the second, a unique stable set iscomputed.

22

The first example introduces a discontinuity to the power functions considered inTheorem 8 so that the most powerful agent just reaches it at the apex of Bi, droppingout the qi. Reducing the jump point increases the segment missing from Bi.

t1 t2

t3

s23

D(t1

) D(s23

)

Figure 7: Example 8: discontinuity prevents existence of S

Example 8. Consider

π (C, x) ≡∑i∈C

{ 12√

xi if xi <23√

xi otherwise

}. (9)

This satisfies all of the additional axioms except CX, continuity, C0, continuity at zero,and CN, conservatism. This example, for which K = D1, is depicted in Figure 7. Inthis case, Bi is identical to its continuous analog (q.v. the bottom panel of Figure 4)other than excluding the point xi = 2

3 , x j = xk = 16 on the symmetry axis. Consequently,

Mi = ∅ so that the rij do not exist. The qi also do not exist. As the allocation removed

from each Bi now belongs to D(ti), it can no longer be used to dominate allocations in

Bi. Thus, as the core contains both the ti and the s jk, no stable set exists.

The next example introduces a discontinuity so that the most powerful agent alignedagainst i reaches it just when its partner runs out of resources; in turn, this causes Bi

not to exist. By reducing the jump point below 1−v2 , Bi can be extended to the whole

interval; as v→ 0, the example reduces to the continuous Example 2.

Example 9. Add a discontinuity to the power function in Example 2, so that

π (C, x) ≡∑i∈C

{ 12 xi + v if xi <

1−v2

xi + v otherwise

}; (10)

for v ∈ (0, 1). This fails to satisfy the same additional axioms that Example 8 does. Asabove, K = D0. The dominance regions resemble those depicted in Figure 3, with animportant difference: when x > 0 such that xi = 1+v

2 ,

π ({i} ,x) =12

+32

v >14

+74

v = π ({ j, k} ,x) .

When either x j or xk is zero, however, equality is restored. As D(ti)

contains xi = 1+v2

when x > 0, Bi does not exist, so that Mi and Ri do not either. Our algorithm cannot

23

proceed in this case: as Ri does not exist, P does not either. However, the Jordan andObadia [2004] arguments for SIN with v > 1 can be applied on X\D (D0), so that theunique stable set is S = D1.

In terms of Roth-Jordan, S0 = K = D0 and

U (S0) = D0 ∪

{x ∈ X

∣∣∣∣∣∣{

xi <1 + v

2∀i

}∪

{xk = 0⇒ xi = x j for distinct i, j, k

}}.

The non-core allocations in U (S0) all have the property that a two-member W candominate a one-member L; as, for each of these allocations, it is possible to find an-other that dominates it. Thus, S1 = D0 = S0 and the algorithm terminates withoutfinding the stable set.

The above examples provide the intuition for this subsection’s conclusion, namelythat the continuity axiom is restrictive. We state one further lemma before proving this:

Lemma 13. When n = 3, π satisfies axiom CX and Bi , ∅, then Bi is a connected setcontaining sequences

{xh

}and {ym} such that limh→∞ xh

j = 0 and limm→∞ ymk = 0 for

distinct i, j and k.

Proof. Proceed by contradiction, assuming an x > 0 in Bi such that there is no y ∈ Bi

with yk = xk − ε for ε > 0. By axiom WR, we may use Lemma 1 to abuse notation towrite

π (xi) = π(x j, xk

);

π (xi) ≥ π(x j, xk − ε

). (11)

When inequality 11 holds with strict inequality, it generates three subcases:

1. π (xi) < π(x j + ε, xk − ε

)so that, by the intermediate value theorem, there is a δ ∈

(0, ε) such that π (xi + δ) = π(x j + ε − δ, xk − ε

), contradicting the assumption

that xk > 0 is the least value of the kth component.

2. π (xi) = π(x j + ε, xk − ε

), a contradiction immediately.

3. π (xi) > π(x j + ε, xk − ε

). As x ∈ Bi it follows from axiom SR that π (xi − ε) <

π(x j + ε, xk

)so that the IVT ensures the existence of a δ ∈

(0, 1

2ε)

for which

π (xi − ε + 2δ) = π(x j + ε − δ, xk − δ

), again yielding a contradiction.

When inequality 11 holds with equality axiom SR implies π (xi + ε) > π(x j, xk − ε

).

As π (xi) < π(x j + 1

2ε, xk −12ε

), the IVT again ensures the contradiction.

The above argument establishes that no component of Bi is the union of two disjointnon-empty closed sets; thus, Bi is locally connected. By Corollary 2, there can only bea single component, so that Bi is also connected. �

Theorem 11. Not all pillage games can be represented by a power function satisfyingcontinuity axiom CX.

Proof. Example 8 provided an example of a pillage game for which Bi , ∅, but whichwas not a connected set: there was no x ∈ Bi such that x2 = x3. By Lemma 13, thiscannot be the case for a pillage game satisfying axiom CX. �

Even though a continuous power function may not be able to represent the samedominance ordering as a discontinuous one, there is no evidence as yet that discontin-uous power functions can generate stable sets that continuous power functions cannot.

24

5.3 Irresponsive examplesWe now consider examples violating axiom RE. The subsection’s main result is thatany anonymous power function with the same core as the champion power function,has the same unique stable set as it.

D(s23

)D

(t1

)t1 t2

t3

s23 = r

p

D(s23

)D

(t1

)t1 t2

t3

s23 = r

p

Figure 8: π (C,x) when v ∈(0, 1

4

), and when v ∈

[14 , 1

)Theorem 12. When n = 3 and power is defined by the leader power function, π (C,x):

1. for v ∈(0, 1

4

), no stable set exists.

2. for v ∈[

14 , 1

),S = D1 ∪ {p} is the unique stable set.

Proof. From Lemma 5, K = D0 ∪ {p} for the leader power function. It is immediatethat D (p) = ∅.

For any value of v ∈ (0, 1),{x ∈ X

∣∣∣∣∣∣{x1 > x2; x3 = 0} ∪ {x1 > x2; x3 = 0} ∪{

x1 >13≥ x2, x3

}}⊆ D

(t1

):

for the first two sets, t1 K x by the anonymity axiom; the third restricts π ({2, 3} ,x) ≤13 < x1 + (1 − v) x1 = π ({1} ,x). When v ∈

(0, 1

4

), there is an additional compo-

nent to D(t1

)as the SIN effect is no longer masked by the champion effect, namely{

x ∈ X∣∣∣∣x1 >

12(1−v)

}. Thus

1. for v ∈(0, 1

4

), consider

{x ∈ Bi

∣∣∣x j > xk

}. Within this set, π satisfies axioms CX

and RE. The arguments in Lemmas 10 and 11 and Theorem 4 may therefore beeasily modified to prove that, as Bi lacks a maximal element, no stable set exists.

2. for v ∈[

14 , 1

),D

(s23

)is{

x ∈ X

∣∣∣∣∣∣{

x3 =12, x2 > x1

}∪

{x2 =

12, x3 > x1

}∪

{x2 ∈

(13,

12

), x3 <

12

}∪

{x3 ∈

(13,

12

), x2 <

12

}}.

It is then straightforward to show that K ∪ D (K) = D1 ∪ {p} ∪ D (D1) = X.Thus, the core is externally stable, making it the unique stable set [Jordan, 2006a,Proposition 2.4].

25

�

Finally, we demonstrate that any anonymous three agent pillage game whose coreisD1 ∪ {p} has the core as its unique stable set. An example is first presented; it is thenshown that, for any such power function, the core is externally stable, and therefore theunique stable set.

D(t1

)D

(s23

)

t1 t2

t3

s23 = r

p

Figure 9: π (C,x) : S = D1 ∪ {p}

The simplex corresponding to the ‘champion’ power function is depicted in Figure9, with q = p as it lies on the boundary between D

(ti)

and D(s jk

).

Then:

Lemma 14. When n = 3, the unique stable set of the champion power function is

S = K = D1 ∪ {p} .

The lemma might be more properly stated as a corollary to Lemma 4, which estab-lished both thatK = D1∪{p}, and thatK is externally stable; thus, the core is uniquelystable.

Theorem 13 (champion equivalence). When n = 3 all anonymous power functionssuch that K = D1 ∪ {p} yield S = K as their unique stable set.

This is a weaker result than Theorem 3 as it implies nothing about whether x K yfor allocations x,y < S.

Proof. Consider, without loss of generality, the three classes of allocations, x ∈ X:

1. x = (x, 1 − x, 0) with x ∈(

12 , 1

). Then, by axiom AN, x ∈ D

(t1

).

2. x > 0 with x1 > max {x2, x3} so that x1 >13 . Assume that x < D

(t1

). Then

π ({1} , p) < π ({1} ,x) ≤ π ({2, 3} ,x) ≤ π ({2, 3} , p) ;

where the inequalities follow from axiom SR, the assumption, and axiom WR,respectively. Thus, p < K , a contradiction which, in turn, implies that x ∈

D(t1

).

26

3. x = (x, x, 1 − 2x) x ∈(

13 ,

12

). Then

π ({1, 2} ,x) ≥ π ({1} ,x) > π ({3} ,x) ;

by axioms WC, SR and AN, respectively. Thus, x ∈ D(s12

).

As the three classes of allocations above cover all X\K , the core is externally stable,and therefore the unique stable set. �

6 DiscussionThe paper’s main result is that, when a stable set exists in an n = 3 pillage game satis-fying axioms AN, CX and RE, it is unique. None of the examples violating axioms CXand RE subsequently explored violate this conclusion. When anonymity does not hold,we expect that we shall still be interested in the Bi, and that (asymmetric) analogues ofq and r shall play the same roles in the Bi that they do under anonymity, albeit overdistorted domains relative to the anonymous case. However, relaxing anonymity in-troduces dominance regions that do not exist in the anonymous cases considered here;in those, either x ∈ D

(ti)

(so that the strongest agent is stronger than the other two),x ∈ Bi (so that the strongest agent is as strong as the other two), or no agent is asstrong as the other two. Consider now a non-anonymous majority pillage game satis-fying π ({2, 3} ,x) > π ({1} ,x) > π ({2} ,x) π ({3} ,x) for all x ∈ X.21 The SIN powerfunction can generate this with v1 = 10, v2 = 8 and v3 = 6. While all the Bi are emptyon X, the game has no stable set: while, in the anonymous case, the empty Bi implies amajority pillage game on all of X, that is not so here.

We expect that stable sets in anonymous pillage games that violate axiom CX willnot be larger than those in their continuous counterparts: discontinuities may removemaximal elements from Bi but do not otherwise undermine domination’s inheritance ofa complete order. Finally, while the examples violating axiom RE may seem extremelyrestrictive, they are responsible for two of the five types of core in Theorem 2; thus,their ability to generate new phenomena is clear.

For n = 3 games in characteristic function form, “stable sets are typically notunique”, but existence is guaranteed [Lucas, 1992, pp.562-3]. In contrast, n = 3 pillagegames satisfying axioms AN, CX and RE have unique stable sets, but only when theyexist. Given some x in a game in characteristic function form, an infinite subset of theallocations near it may be incomparable, in the sense that neither x K y nor y K x[q.v. Lucas, 1992, Examples 3, 4]. This allows construction of infinite stable sets fromsuch incomparable allocations.

As, in pillage games, power depends monotonically on resource holdings, the do-main of this incomparability is reduced.22 This reduced domain also seems responsiblefor the finitude of stable sets in pillage games. Indeed, as Lucas [1992] notes, only onen = 3 game in characteristic function form has a finite stable set, namely the majoritygame in which y K x iff two agents prefer y to x. In this case, the regions of incompa-rability are 1-dimensional curves in the simplex, which still admit a family of infinitestable sets.23

21We are grateful to Jim Jordan for this suggestion.22Jordan [2006a] proved that stable sets in pillage games are finite by showing the impossibility of a

sequence of four allocations over which W and L remain constant.23See Jordan and Obadia [2004] for a discussion of the majority pillage game’s use of resource holdings

to break ties in the majority game.

27

A Other resultsLemma 15 (convex transitivity). If x K y then x K αx + (1 − α) y K y∀α ∈ (0, 1).

Proof. x K y ⇔ π (W1,y) > π (L1,y), where W1 ≡ {i |xi > yi } and L1 ≡ {i |yi > xi }.Now define

Wα ≡ {i |αxi + (1 − α) yi > yi } ;

and Lα similarly for α ∈ (0, 1). It is then immediate that Wα = W1 (resp. Lα = L1) sothat

αx + (1 − α) y K y.

The second dominance operation follows in the same way. �

Lemma 16. When n = 3, π satisfies axioms AN and CX, and rij ∈ Bi\Bi

+, then rij = si j.

Proof. Suppose otherwise, so that, setting (i, j) = (1, 2) without loss of generality,ri

j = (r, r, 1 − 2r), with r ∈[

13 ,

12

). By Lemma 13, there is a y ∈ Bi with y3 < 1 − 2r;

by Lemma 1, this may be written as π (r + δ) = π (r + ε − δ, 1 − 2r − ε) for some δ ∈[0, ε]. Axiom AN implies, by Lemma 6, that δ > ε − δ so that δ > 1

2ε > 0, so thatπ (r) < π (r + δ), contradicting the requirement that r−j ∈ Mi. �

Lemma 17. When n = 3, and π satisfies axioms AN and CX, Mi is non-empty and qi

exists.

Proof. By Corollary 2, Bi may be represented by a curve indexed by y j

yk, which defines

the Corollary’s rays; the Lemma establishes that y j

yk∈ [0,∞). As Bi is connected, the

curve exists at y j

yk= 1, so qi does as well. As Bi is closed and bounded, Mi is non-

empty. �

Lemma 18. WhenD1 ⊆ K , all x ∈ Bi\Bi+ either belong to the core, or are dominated

by a core allocation.

Proof. Any x ∈ Bi\Bi+ may be written as a permutation of (x, x, 1 − 2x), where x ∈[

13 ,

12

]. As D1 ⊆ K the core allocation s12, which corresponds to x = 1

2 , dominates x

for all x ∈(

13 ,

12

); when s12 does not also dominate p, which corresponds to x = 1

3 , thenp is undominated, and so belongs to the core. �

When axiom RE holds, the lemma is less important, as the s jk are the only possibleallocations in Bi\Bi

+.

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