Coalition Formation in Nondemocracies?
New Economic School
We study the formation of a ruling coalition in nondemocratic societies where institutions do not
enable political commitments. Each individual is endowed with a level of political power. The ruling
coalition consists of a subset of the individuals in the society and decides the distribution of resources. A
ruling coalition needs to contain enough powerful members to win against any alternative coalition that
may challenge it and it needs to be self-enforcing, in the sense that none of its subcoalitions should be able
to secede and become the new ruling coalition. We present both an axiomatic approach that captures
these notions and determines a (generically) unique ruling coalition and the analysis of a dynamic game
of coalition formation that encompasses these ideas. We establish that the subgame perfect equilibria
of the coalition formation game coincide with the set of ruling coalitions resulting from the axiomatic
approach. A key insight of our analysis is that a coalition is made self-enforcing by the failure of its
winning subcoalitions to be self-enforcing. This is most simply illustrated by the following example: with
“majority rule,” two-person coalitions are generically not self-enforcing and consequently, three-person
coalitions are self-enforcing (unless one player is disproportionately powerful). We also characterize the
structure of ruling coalitions. For example, we determine the conditions under which ruling coalitions
are robust to small changes in the distribution of power and when they are fragile. We also show that
when the distribution of power across individuals is relatively equal and there is majoritarian voting, only
certain sizes of coalitions (e.g., with “majority rule,” coalitions of size 3, 7, 15, 31, etc.) can be the ruling
Keywords: coalition formation, political economy, self-enforcing coalitions, stability.
JEL Classi…cation: D71, D74, C71.
?We thank Attila Ambrus, Salvador Barbera, Jon Eguia, Irina Khovanskaya, Eric Maskin, Benny Moldovanu,
Victor Polterovich, Andrea Prat, Debraj Ray, Muhamet Yildiz, three anonymous referees, and seminar par-
ticipants at the Canadian Institute of Advanced Research, MIT, the New Economic School, the Institute for
Advanced Studies, and University of Pennsylvania PIER, NASM 2007, and EEA-ESEM 2007 conferences for
useful comments. Acemoglu gratefully acknowledges …nancial support from the National Science Foundation.
We study the formation of a ruling coalition in a nondemocratic (“weakly institutionalized”)
environment. A ruling coalition must be powerful enough to impose its wishes on the rest of the
society. A key ingredient of our analysis is that because of the absence of strong, well-functioning
institutions, binding agreements are not possible.1This has two important implications: …rst,
members of the ruling coalition cannot make binding o¤ers on how resources will be distributed;
second, and more importantly, members of a candidate ruling coalition cannot commit to not
eliminating (sidelining) fellow members in the future. Consequently, there is always the danger
that, once a particular coalition has formed and has centralized power in its hands, a subcoalition
will try to remove some of the original members of the coalition in order to increase the share
of resources allocated to itself. Ruling coalitions must therefore not only be powerful enough to
be able to impose their wishes on the rest of the society, but also self-enforcing so that none
of their subcoalitions are powerful enough and wish to split from or eliminate the rest of this
coalition. These considerations imply that the nature of ruling coalitions is determined by a
tradeo¤ between “power” and “self-enforcement”.
More formally, we consider a society consisting of an arbitrary number of individuals with
di¤erent amount of political or military powers (“guns”). Any subset of these individuals can
form a coalition and the power of the coalition is equal to the sum of the powers of its members.
We formalize the interplay between power and self-enforcement as follows: a coalition with
su¢cient power is winning against the rest of the society and can centralize decision-making
powers in its own hands (for example, eliminating the rest of the society from the decision-
making process). How powerful a coalition needs to be in order to be winning is determined
by a parameter ?. When ? = 1=2, this coalition simply needs to be more powerful than the
rest of the society, so this case can be thought of as “majority rule.” When ? > 1=2, the
coalition needs “supermajority” or more than a certain multiple of the power of the remainder
of the society. Once this …rst stage is completed, a subgroup can secede from or sideline the
rest of the initial winning coalition if it has enough power and wishes to do so. This process
continues until a self-enforcing coalition, which does not contain any subcoalitions that wish to
engage in further rounds of eliminations, emerges. Once this coalition, which we refer to as the
ultimate ruling coalition (URC), is formed, the society’s resources are distributed according to
some pre-determined rule (for example, resources may be distributed among the members of this
1Acemoglu and Robinson (2006) provide a more detailed discussion and various examples of commitment
problems in political-decision making. The term weakly-institutionalized polities is introduced in Acemoglu,
Robinson, and Verdier (2004) to describe societies in which institutional rules do not constrain political interactions
among various social groups or factions.
coalition according to their powers). This simple game formalizes the two key consequences of
weak institutions mentioned above: (1) binding agreements on how resources will be distributed
are not possible; (2) subcoalitions cannot commit to not sidelining their fellow members in a
Our main results are as follows. First, we characterize the equilibria of this class of games
under general conditions. We show that a ruling coalition always exists and is “generically”
unique. Moreover, the equilibrium always satis…es some natural axioms that are motivated
by the power and self-enforcement considerations mentioned above. Therefore, our analysis
establishes the equivalence between an axiomatic approach to the formation of ruling coalitions
(which involves the characterization of a mapping that determines the ruling coalition for any
society and satis…es a number of natural axioms) and a noncooperative approach (which involves
characterizing the subgame perfect equilibria of a game of coalition formation). We also show
that the URC can be characterized recursively. Using this characterization, we establish the
following results on the structure of URCs.
1. Despite the simplicity of the environment, the URC can consist of any number of players,
and may include or exclude the most powerful individuals in the society. Consequently, the
equilibrium payo¤ of an individual is not monotonic in his power. The most powerful will
belong to the ruling coalition only if he is powerful enough to win by himself or weak enough to
be a part of a smaller self-enforcing coalition.
2. An increase in ?, that is, an increase in the degree of supermajority needed to eliminate
opponents, does not necessarily lead to larger URCs, because it stabilizes otherwise non-self-
enforcing subcoalitions, and as a result, destroys larger coalitions that would have been self-
enforcing for lower values of ?.
3. Self-enforcing coalitions are generally “fragile.” For example, under majority rule (i.e.,
? = 1=2), adding or subtracting one player from a self-enforcing coalition necessarily makes it
4. Nevertheless, URCs are (generically) continuous in the distribution of power across indi-
viduals in the sense that a URC remains so when the powers of the players are perturbed.
5. Coalitions of certain sizes are more likely to emerge as the URC. For example, with
majority rule (? = 1=2) and a su¢ciently equal distribution of powers among individuals, the
URC must have size 2k? 1 where k is an integer (i.e., 1, 3, 7, 15,...). A similar formula for the
size of the ruling coalition applies when ? > 1=2.
2The game also introduces the feature that once a particular group of individuals has been sidelined, they
cannot be brought back into the ruling coalition. This feature is adopted for tractability.
We next illustrate some of the main interactions using a simple example.
Example 1 Consider two agents A and B. Denote their powers ?A> 0 and ?B> 0 and assume
that the decision-making rule requires power-weighted majority, that is, ? = 1=2. This implies
that if ?A> ?B, then starting with the coalition fA;Bg, the agent A will form a majority by
himself. Conversely, if ?A< ?B, then agent B will form a majority. Thus, “generically” (i.e.,
as long as ?A6= ?B), one of the members of the two-person coalition can secede and form a
subcoalition that is powerful enough within the original coalition. Since each agent will receive
a higher share of the scarce resources in a coalition that consists of only himself than in a
two-person coalition, two-person coalitions are generically not self-enforcing.
Now, consider a coalition consisting of three agents, A, B and C with powers ?A, ?Band
?C, and suppose that ?A< ?B< ?C< ?A+ ?B. Clearly, no two-person coalition is self-
enforcing. The lack of self-enforcing subcoalitions of fA;B;Cg implies that fA;B;Cg is itself
self-enforcing. To see this, suppose, for example, that fA;Bg considers seceding from fA;B;Cg.
They can do so since ?A+ ?B> ?C. However, we know from the previous paragraph that the
subcoalition fA;Bg is itself not self-enforcing, since after this coalition is established, agent B
would secede or eliminate A. Anticipating this, agent A would not support the subcoalition
fA;Bg. A similar argument applies for all other subcoalitions. Moreover, since agent C is
not powerful enough to secede from the original coalition by himself, the three-person coalition
fA;B;Cg is self-enforcing and will be the ruling coalition.
Next, consider a society consisting of four individuals, A;B;C and D. Suppose that we
have ?A= 3;?B= 4;?C= 5; and ?D= 10. D’s power is insu¢cient to eliminate the coalition
fA;B;Cg starting from the initial coalition fA;B;Cg. Nevertheless, D is stronger than any
two of A;B;C. This implies that any three-person coalition that includes D would not be self-
enforcing. Anticipating this, any two of fA;B;Cg would decline D’s o¤er to secede. However,
fA;B;Cg is self-enforcing, thus the three agents would be happy to eliminate D. Therefore, in
this example, the ruling coalition again consists of three individuals, but interestingly excludes
the most powerful individual D.
The most powerful individual is not always eliminated. Consider the society with ?A=
2;?B= 4;?C= 7 and ?D= 10. In this case, among the three-person coalitions only fB;C;Dg
is self-enforcing, and it will eliminate the weakest individual, A, and become the ruling coalition.
This example also illustrates why three-person coalitions (22? 1 = 3) may be more likely than
two-person (and also four-person) coalitions.3
3It also shows that in contrast to approaches with unrestricted side-payments (e.g., Riker, 1962), the ruling
coalition will not generally be a minimal winning coalition (the unique minimum winning coalition is fA;Dg,
Although our model is abstract, it captures a range of economic forces that appear salient
in nondemocratic, weakly-institutionalized polities. The historical example of Stalin’s Soviet
Russia illustrates this in a particularly clear manner. The Communist Party Politburo was the
highest ruling body of the Soviet Union. All top government positions were held by its members.
Though formally its members were elected at Party meetings, for all practical purposes the
Politburo determined the fates of its members, as well as those of ordinary citizens. Soviet
archives contain execution lists signed by Politburo members; sometimes a list would contain
one name, but some lists from the period of 1937-39 contained hundreds or even thousands
names (Conquest, 1968).
Of 40 Politburo members (28 full, 12 non-voting) appointed between 1919 and 1952, only 12
survived through 1952. Of these 12, 11 continued to hold top positions after Stalin’s death in
March 1953. There was a single Politburo member (Petrovsky) in 33 years who left the body
and survived. Of the 28 deaths, there were 17 executions decided by the Politburo, 2 suicides,
1 death in prison immediately after arrest, and 1 assassination.
To interpret the interactions among Politburo members through the lenses of our model,
imagine that the Politburo consists of …ve members, and to illustrate our main points, suppose
that their powers are given by f3;4;5;10;20g. It can be veri…ed that with ? = 1=2, this …ve-
member coalition is self-enforcing. However, if either of the lower power individuals, 3;4;5; or
10; dies or is eliminated, then the ruling coalition consists of the singleton, 20. If, instead, 20
dies, the ultimate ruling coalition becomes f3;4;5g and eliminates the remaining most powerful
individual 10. This is because 10 is unable to form an alliance with less powerful players. While
the reality of Soviet politics in the …rst half of the century is naturally much more complicated,
this simple example sheds light on three critical episodes.
The …rst episode is the suicides of two members of the Politburo, Tomsky and Ordzhonikidze,
during 1937-38. An immediate implication of these suicides was a change in the balance of power,
something akin to the elimination of 5 in the f3;4;5;10;20g example above. In less than a year,
11 current or former members of Politburo were executed. Consistent with the ideas emphasized
in our model, some of those executed in 1939 (e.g., Chubar, Kosior, Postyshev, and Ezhov) had
earlier voted for the execution of Bukharin and Rykov in 1937. The second episode followed
the death of Alexei Zhdanov in 1948 from a heart attack. Until Zhdanov’s death, there was a
period of relative “peace”: no member of this body had been executed in nine years. Monte…ore
(2003) describes how the Zhdanov’s death immediately changed the balance in the Politburo.
The death gave Beria and Malenkov the possibility to have Zhdanov’s supporters and associates
which has the minimum power among all winning coalitions).
in the government executed.4The third episode followed the death of Stalin himself in March
1953. Since the bloody purge of 1948, powerful Politburo members conspired in resisting any
attempts by Stalin to have any of them condemned and executed. When in the Fall of 1952,
Stalin charged two old Politburo members, Molotov and Mikoian, with being the “enemies of
the people,” the other members stood …rm and blocked a possible trial (see Monte…ore, 2003, or
Gorlizki and Khlevniuk, 2004). After Stalin’s death, Beria became the most powerful politician
in Russia. He was immediately appointed the …rst deputy prime-minister as well as the head
of the ministry of internal a¤airs and of the ministry of state security, the two most powerful
ministries in the USSR. His ally Malenkov was appointed prime-minister, and no one succeeded
Stalin as the Secretary General of the Communist Party. Yet in only 4 months, the all-powerful
Beria fell victim of a military coup by his fellow Politburo members, was tried and executed. In
terms of our simple example with powers f3;4;5;10;20g, Beria would correspond to 10. After
20 (Stalin) is out of the picture, f3;4;5g becomes the ultimate ruling coalition, so 10 must be
Similar issues arise in other dictatorships when top …gures were concerned with others becom-
ing too powerful. These considerations also appear to be particularly important in international
relations, especially when agreements have to be reached under the shadow of the threat of
war (e.g., Powell, 1999). For example, following both World Wars, many important features of
the peace agreements were in‡uenced by the desire that the emerging balance of power among
states should be self-enforcing. In this context, small states were viewed as attractive because
they could combine to contain threats from larger states but they would be unable to become
dominant players. Similar considerations were paramount after Napoleon’s ultimate defeat in
1815. In this case, the victorious nations designed the new political map of Europe at the
Vienna Congress, and special attention was paid to balancing the powers of Britain, Germany
and Russia, to ensure that “... their equilibrium behaviour... maintain the Vienna settlement”
Our paper is related to models of bargaining over resources, particularly in the context of
political decision-making (e.g., models of legislative bargaining such as Baron and Ferejohn,
1989, Calvert and Dietz, 1996, Jackson and Moselle, 2002). Our approach di¤ers from these
papers, since we do not impose any speci…c bargaining structure and focus on self-enforcing
4In contrast to the two other episodes from the Soviet Politburo we discuss here, the elimination of the
associates of Zhadanov could also be explained by competition between two groups within the Politburo rather
than by competition among all members and lack of commitment, which are the ideas emphasized by our model.
5Other examples of potential applications of our model in political games are provided in Pepinsky (2007),
who uses our model to discuss issues of coalition formation in nondemocratic societies.
More closely related to our work are the models of on equilibrium coalition formation, which
combine elements from both cooperative and noncooperative game theory (e.g., Peleg, 1980,
Hart and Kurz, 1983, Greenberg and Weber, 1993, Chwe, 1994, Bloch, 1996, Mariotti, 1997,
Ray, 2007, Ray and Vohra, 1997, 1999, 2001, Seidmann and Winter, 1998, Konishi and Ray, 2001,
Maskin, 2003, Eguia, 2006, Pycia, 2006). The most important di¤erence between our approach
and the previous literature on coalition formation is that, motivated by political settings, we
assume that the majority (or supermajority) of the members of the society can impose their
will on those players who are not a part of the majority. This feature both changes the nature
of the game and also introduces “negative externalities” as opposed to the positive externalities
and free-rider problems upon which the previous literature focuses (Ray and Vohra, 1999, and
Maskin, 2003). A second important di¤erence is that most of these works assume the possibility
of binding commitments (Ray and Vohra, 1997, 1999), while we suppose that players have
no commitment power. Despite these di¤erences, there are important parallels between our
results and the insights of this literature. For example, Ray (1979) and Ray and Vohra (1997,
1999) emphasize that the internal stability of a coalition in‡uences whether it can block the
formation of other coalitions, including the grand coalition. In the related context of risk-sharing
arrangements, Bloch, Genicot, and Ray (2006) show that stability of subgroups threatens the
stability of a larger group.7Another related approach to coalition formation is developed by
Moldovanu and Winter (1995), who study a game in which decisions require appoval by all
members of a coalition and show the relationship of the resulting allocations to the core of a
related cooperative game.8
Finally, Skaperdas (1998) and Tan and Wang (1999) investigate
coalition formation in dynamic contests. Nevertheless, none of these papers study self-enforcing
coalitions in political games without commitment, or derive existence, generic uniqueness and
characterization results similar to those in our paper.
The rest of the paper is organized as follows. Section 2 introduces the formal setup. Section
3 provides our axiomatic treatment. Section 4 characterizes subgame perfect equilibria of the
6See also Perry and Reny (1994), Moldovanu and Jehiel (1999), and Gomes and Jehiel (2005) for models of
bargaining with a coalition structure.
7In this respect, our paper is also related to work on “coalition-proof” Nash equilibrium or rationalizability,
e.g., Bernheim, Peleg, and Whinston (1987), Moldovanu (1992), Ambrus (2006). These papers allow deviations
by coalitions in noncooperative games, but impose that only stable coalitions can form.
considerations are captured in our model by the game of coalition formation and by the axiomatic analysis.
8Our game can also be viewed as a “hedonic game” since the utility of each player is determined by the
composition of the ultimate coalition he belongs to. However, it is not a special case of hedonic games de…ned
and studied in Bogomolnaia and Jackson (2002), Banerjee, Konishi, and Sonmez (2001), and Barbera and Gerber
(2007), because of the dynamic interactions introduced by the self-enforcement considerations. See Le Breton,
Ortuno-Ortin, and Weber (2008) for an application of hedonic games to coalition formation.
In contrast, these
extensive-form game of coalition formation. It then establishes the equivalence between the
ruling coalition of Section 3 and the equilibria of this extensive-form game. Section 5 contains
our main results on the nature and structure of ruling coalitions in political games. Section 6
concludes. The Appendix contains the proofs of all the results presented in the text.
2The Political Game
Let I denote the collection of all individuals, which is assumed to be …nite. The non-empty
subsets of I are coalitions and the set of coalitions is denoted by C. In addition, for any X ? I,
CXdenotes the set of coalitions that are subsets of X and jXj is the number of members in X.
In each period there is a designated ruling coalition, which can change over time. The game
starts with ruling coalition N, and eventually the ultimate ruling coalition (URC) forms. We
assume that if the URC is X, then player i obtains baseline utility wi(X) 2 R. We denote
w(?) ? fwi(?)gi2I.
Our focus is on how di¤erences in the powers of individuals map into political decisions. We
de…ne a power mapping to summarize the powers of di¤erent individuals in I:
? : I ! R++;
where R++ = R+n f0g. We refer to ?i? ? (i) as the political power of individual i 2 I.
In addition, we denote the set of all possible power mappings by R and a power mapping ?
restricted to some coalition N ? I by ?jN(or by ? when the reference to N is clear). The power
of a coalition X is ?X?P
Coalition Y ? X is winning within coalition X if and only if ?Y> ??X, where ? 2 [1=2;1)
is a …xed parameter referring to the degree of (weighted) supermajority. Naturally, ? = 1=2
corresponds to majority rule. Moreover, since I is …nite, there exists a large enough ? (still less
than 1) that corresponds to unanimity rule. We denote the set of coalitions that are winning
within X by WX. Since ? ? 1=2, if Y;Z 2 WX, then Y \ Z 6= ?.
The assumption that payo¤s are given by the mapping w(?) implies that a coalition cannot
commit to a redistribution of resources or payo¤s among its members (for example, a coalition
consisting of two individuals with powers 1 and 10 cannot commit to share the resource equally
if it becomes the URC). We assume that the baseline payo¤ functions, wi(X) : I ? C ! R for
any i 2 N, satisfy the following properties.
Assumption 1 Let i 2 I and X;Y 2 C. Then:
(1) If i 2 X and i = 2 Y , then wi(X) > wi(Y ) [i.e., each player prefers to be part of the
(2) For i 2 X and i 2 Y , wi(X) > wi(Y ) () ?i=?X> ?i=?Y( () ?X< ?Y) [i.e., for
any two URCs that he is part of, each player prefers the one where his relative power is greater].
(3) If i = 2 X and i = 2 Y , then wi(X) = wi(Y ) ? w?
URCs he is not part of].
i[i.e., a player is indi¤erent between
This assumption is natural and captures the idea that each player’s payo¤ depends positively
on his relative strength in the URC. A speci…c example of function w(?) that satis…es these
requirements is sharing of a pie between members of the ultimate ruling coalition proportional
to their power:
if i 2 X
if i = 2 X
The reader may want to assume (1) throughout the text for interpretation purposes, though this
speci…c functional form is not used in any of our results or proofs.
We next de…ne the extensive-form complete information game ? = (N;?jN;w(?);?), where
N 2 C is the initial coalition, ? is the power mapping, w(?) is a payo¤ mapping that satis…es
Assumption 1, and ? 2 [1=2;1) is the degree of supermajority; denote the collection of such
games by G. Also, let " > 0 be su¢ciently small such that for any i 2 N and any X;Y 2 C, we
wi(X) > wi(Y ) =) wi(X) > wi(Y ) + 2"(2)
(this holds for su¢ciently small " > 0 since I is a …nite set). This immediately implies that for
any X 2 C with i 2 X, we have
wi(X) ? w?
The extensive form of the game ? = (N;?jN;w(?);?) is as follows. Each stage j of the game
starts with some ruling coalition Nj (at the beginning of the game N0= N). Then the stage
game proceeds with the following steps:
1. Nature randomly picks agenda setter aj;q2 Njfor q = 1.
2. [Agenda-setting step] Agenda setter aj;qmakes proposal Pj;q2 CNj, which is a subcoalition
of Njsuch that aj;q2 Pj;q(for simplicity, we assume that a player cannot propose to eliminate
3. [Voting step] Players in Pj;qvote sequentially over the proposal (we assume that players in
Njn Pj;qautomatically vote against this proposal). More speci…cally, Nature randomly chooses
the …rst voter, vj;q;1, who then casts his vote vote ~ v (vj;q;1) 2 f~ y; ~ ng (Yes or No), then Nature
chooses the second voter vj;q;26= vj;q;1, etc. After all jPj;qj players have voted, the game proceeds
to step 4 if players who supported the proposal form a winning coalition within Nj (i.e., if
fi 2 Pj;q: ~ v (i) = ~ yg 2 WNj), and otherwise it proceeds to step 5.
4. If Pj;q = Nj, then the game proceeds to step 6. Otherwise, players from Njn Pj;q are
eliminated and the game proceeds to step 1 with Nj+1= Pj;q (and j increases by 1 as a new
transition has taken place).
5. If q < jNjj, then next agenda setter aj;q+12 Nj is randomly picked by Nature among
members of Njwho have not yet proposed at this stage (so aj;q+16= aj;rfor 1 ? r ? q), and the
game proceeds to step 2 (with q increased by 1). If q = jNjj, the game proceeds to step 6.
6. Njbecomes the ultimate ruling coalition. Each player i 2 N receives total payo¤
where If?gis the indicator function taking the value of 0 or 1.
The payo¤ function (4) captures the idea that an individual’s overall utility is the di¤erence
Ui= wi(Nj) ? "
between the baseline wi(?) and disutility from the number of transitions (rounds of elimination)
this individual is involved in.The arbitrarily small cost " can be interpreted as a cost of
eliminating some of the players from the coalition or as an organizational cost that individuals
have to pay each time a new coalition is formed. Alternatively, " may be viewed as a means
to re…ne out equilibria where order of moves matters for the outcome. Note that ? is a …nite
game: the total number of moves, including those of Nature, does not exceed 4jNj3. Notice
also that this game form introduces sequential voting in order to avoid issues of individuals
playing weakly-dominated strategies. Our analysis below will establish that the main results
hold regardless of the speci…c order of votes chosen by Nature.9
Before characterizing the equilibria of the dynamic game ?, we take a brief detour and introduce
four axioms motivated by the structure of the game ?. Although these axioms are motivated
by game ?, they can also be viewed as natural axioms to capture the salient economic forces
discussed in the introduction. The analysis in this section identi…es an outcome mapping ? : G ?
C that satis…es these axioms and determines the set of (admissible) URCs corresponding to each
game ?. This analysis will be useful for two reasons. First, it will reveal certain attractive
features of the game presented in the previous section. Second, we will show in the next section
that equilibrium URCs of this game coincide with the outcomes picked by the mapping ?.
More formally, consider the set of games ? = (N;?jN;w(?);?) 2 G. Holding ?;w and ?
…xed, consider the correspondence ? : C ? C de…ned by ?(N) = ?(N;?jN;w;?) for any N 2 C.
9See Acemoglu, Egorov, and Sonin (2006) both for the analysis of a game with simultaneous voting and a
stronger equilibrium notion, and for an example showing how, in the absence of the cost " > 0, the order of moves
We adopt the following axioms on ? (or alternatively on ?).
Axiom 1 (Inclusion) For any X 2 C, ?(X) 6= ? and if Y 2 ?(X), then Y ? X.
Axiom 2 (Power) For any X 2 C, Y 2 ?(X) only if Y 2 WX.
Axiom 3 (Self-Enforcement) For any X 2 C, Y 2 ?(X) only if Y 2 ?(Y ).
Axiom 4 (Rationality) For any X 2 C, for any Y 2 ?(X) and for any Z ? X such that
Z 2 WXand Z 2 ?(Z), we have that Z = 2 ?(X) () ?Y< ?Z.
Motivated by Axiom 3, we de…ne the notion of a self-enforcing coalition as a coalition that
“selects itself”. This notion will be used repeatedly in the rest of the paper.
De…nition 1 Coalition X 2 P (I) is self-enforcing if X 2 ?(X).
Axiom 1, inclusion, implies that ? maps into subcoalitions of the coalition in question (and
that it is de…ned, i.e., ?(X) 6= ?). It therefore captures the feature introduced in ? that players
that have been eliminated (sidelined) cannot rejoin the ruling coalition. Axiom 2, the power
axiom, requires a ruling coalition be a winning coalition. Axiom 3, the self-enforcement axiom,
captures the key interactions in our model. It requires that any coalition Y 2 ?(X) should be
self-enforcing according to De…nition 1. This property corresponds to the notion that in terms
of game ?, if coalition Y is reached along the equilibrium path, then there should not be any
deviations from it. Finally, Axiom 4 requires that if two coalitions Y;Z ? X are both winning
and self-enforcing and all players in Y \Z strictly prefer Y to Z, then Z = 2 ?(X) (i.e., Z cannot
be the selected coalition). Intuitively, all members of winning coalition Y (both those in Y \ Z
by assumption and those in Y nZ because they prefer to be in the URC) strictly prefer Y to Z;
hence, Z should not be chosen in favor of Y . This interpretation allows us to call Axiom 4 the
Rationality Axiom. In terms of game ?, this axiom captures the notion that, when he has the
choice, a player will propose a coalition in which his payo¤ is greater.
At the …rst glance, Axioms 1–4 may appear relatively mild. Nevertheless, they are strong
enough to pin down a unique mapping ?. Moreover, under the following assumption, these
axioms also imply that this unique mapping ? is single valued.
Assumption 2 The power mapping ? is generic in the sense that if for any X;Y 2 C, ?X= ?Y
implies X = Y . We also say that coalition N is generic or that numbers f?igi2Nare generic if
mapping ?jN is generic.
Intuitively, this assumption rules out distributions of powers among individuals such that two
di¤erent coalitions have exactly the same total power. Notice that mathematically, genericity
assumption is without much loss of generality since the set of vectors f?igi2I2 RjIj
not generic has Lebesgue measure 0 (in fact, it is a union of a …nite number of hyperplanes in
Theorem 1 Fix a collection of players I, a power mapping ?, a payo¤ function w(?) such that
Assumption 1 holds, and ? 2 [1=2;1). Then:
1. There exists a unique mapping ? that satis…es Axioms 1–4. Moreover, when ? is generic
(i.e. under Assumption 2), ? is single-valued.
2. This mapping ? may be obtained by the following inductive procedure. For any k 2 N, let
Ck= fX 2 C : jXj = kg. Clearly, C = [k2NCk. If X 2 C1, then let ?(X) = fXg. If ?(Z) has
been de…ned for all Z 2 Cnfor all n < k, then de…ne ?(X) for X 2 Ckas
M(X) = fZ 2 CXn fXg : Z 2 WXand Z 2 ?(Z)g.(6)
Proceeding inductively ?(X) is de…ned for all X 2 C.
The intuition for the inductive procedure is as follows. For each X, (6) de…nes M(X) as
the set of proper subcoalitions which are both winning and self-enforcing. Equation (5) then
picks the coalitions in M(X) that have the least power. When there are no proper winning and
self-enforcing subcoalitions, M(X) is empty and X becomes the URC), which is captured by
(5). The proof of this theorem, like all other proofs, is in the Appendix.
Theorem 1 establishes not only that ? is uniquely de…ned, but also that when Assumption
2 holds, it is single-valued. In this case, with a slight abuse of notation, we write ?(X) = Y
instead of ?(X) = fY g.
Corollary 1 Take any collection of players I, power mapping ?, payo¤ function w(?), and
? 2 [1=2;1). Let ? be the unique mapping satisfying Axioms 1–4. Then for any X;Y;Z 2 C,
Y;Z 2 ?(X) implies ?Y= ?Z. Coalition N is self-enforcing, that is, N 2 ?(N), if and only if
there exists no coalition X ? N, X 6= N, that is winning within N and self-enforcing. Moreover,
if N is self-enforcing, then ?(N) = fNg.
Corollary 1, which immediately follows from (5) and (6), summarizes the basic results on self-
enforcing coalitions. In particular, Corollary 1 says that a coalition that includes a winning and
self-enforcing subcoalition cannot be self-enforcing. This captures the notion that the stability
of smaller coalitions undermines the stability of larger ones.
As an illustration to Theorem 1, consider again three players A, B and C and suppose that
? = 1=2. For any ?A< ?B< ?C< ?A+ ?B, Assumption 2 is satis…ed and it is easy to see
that fAg, fBg, fCg, and fA;B;Cg are self-enforcing coalitions, whereas ?(fA;Bg) = fBg,
?(fA;Cg) = ?(fB;Cg) = fCg. In this case, ?(X) is a singleton for any X. On the other hand,
if ?A= ?B= ?C, all coalitions except fA;B;Cg would be self-enforcing, while ?(fA;B;Cg) =
ffA;Bg;fB;Cg;fA;Cgg in this case.
4 Equilibrium Characterization
We now characterize the Subgame Perfect Equilibria (SPE) of game ? de…ned in Section 2 and
show that they correspond to the ruling coalitions identi…ed by the axiomatic analysis in the
previous section. The next subsection provides the main results. We then provide a sketch of
the proofs. The formal proofs are contained in the Appendix.
The following two theorems characterize the Subgame Perfect Equilibrium (SPE) of game ? =
(N;?jN;w;?) with initial coalition N. As usual, a strategy pro…le ? in ? is a SPE if ? induces
continuation strategies that are best responses to each other starting in any subgame of ?,
denoted ?h, where h denotes the history of the game, consisting of actions in past periods
(stages and steps).
Theorem 2 Suppose that ?(N) satis…es Axioms 1-4 (cfr. (5) in Theorem 1). Then, for any
K 2 ?(N), there exists a pure strategy pro…le ?K that is an SPE and leads to URC K in at
most one transition. In this equilibrium player i 2 N receives payo¤
Ui= wi(K) ? "Ifi2KgIfN6=Kg: (7)
This equilibrium payo¤ does not depend on the random moves by Nature.
Theorem 2 establishes that there exists a pure strategy equilibrium leading to any coalition
that is in the set ?(N) de…ned in the axiomatic analysis of Theorem 1.10This is intuitive in view
of the analysis in the previous section: when each player anticipates members of a self-enforcing
ruling coalition to play a strategy pro…le such that they will turn down any o¤ers other than K
10It can also be veri…ed that Theorem 2 holds even when " = 0. The assumption that " > 0 is used in Theorem
and they will accept K, it is in the interest of all the players in K to play such a strategy for any
history. This follows because the de…nition of the set ?(N) implies that only deviations that
lead to ruling coalitions that are not self-enforcing or not winning could be pro…table. But the
…rst option is ruled out by induction while a deviation to a non-winning URC will be blocked
by su¢ciently many players. The payo¤ in (7) is also intuitive. Each player receives his baseline
payo¤ wi(K) resulting from URC K and then incurs the cost " if he is part of K and if the
initial coalition N is not equal to K (because in this latter case, there will be one transition).
Notice that Theorem 2 is stated without Assumption 2 and does not establish uniqueness. The
next theorem strengthens these results under Assumption 2.
Theorem 3 Suppose Assumption 2 holds and suppose ?(N) = K. Then any (pure or mixed
strategy) SPE results in K as the URC. The payo¤ player i 2 N receives in this equilibrium is
given by (7).
Since Assumption 2 holds, the mapping ? is single-valued (with ?(N) = K). Theorem 3
then shows that even though the SPE may not be unique in this case, any SPE will lead to
K as the URC. This is intuitive in view of our discussion above. Because any SPE is obtained
by backward induction, multiplicity of equilibria results only when some player is indi¤erent
between multiple actions at a certain nod. However, as we show, this may only happen when a
player has no e¤ect on equilibrium play and his choice between di¤erent actions has no e¤ect on
URC (in particular, since ? is single-valued in this case, a player cannot be indi¤erent between
actions that will lead to di¤erent URCs).
It is also worth noting that the SPE in Theorems 2 and 3 is “coalition-proof”. Since the
game ? incorporates both dynamic and coalitional e¤ects and is …nite, the relevant concept of
coalition-proofness is Bernheim, Peleg, and Whinston’s (1987) Perfectly Coalition-Proof Nash
Equilibrium (PCPNE). This equilibrium re…nement requires that the candidate equilibrium
should be robust to deviations by coalitions in all subgames when the players take into account
the possibility of further deviations. Since ? introduces more general coalitional deviations
explicitly, it is natural to expect the SPE in ? to be PCPNE. Indeed, if Assumption 2 holds, it
is straightforward to prove that the set of PCPNE coincides with the set of SPE.11
4.2Sketch of the Proofs
We now provide an outline of the argument leading to the proofs of the main results presented
in the previous subsection and we present two key lemmas that are central for these theorems.
11A formal proof of this result follows from Lemma 2 below and is available from the authors upon request.
Consider the game ? and let ? be as de…ned in (5). Take any coalition K 2 ?(N). We will
outline the construction of the pure strategy pro…le ?Kwhich will be a SPE and lead to K as
Let us …rst rank all coalitions so as to “break ties” (which are possible, since we have not yet
imposed Assumption 2). In particular, n : C !?1;:::;2jIj? 1?be a one-to-one mapping such
that for any X;Y 2 C, ?X> ?Y) n(X) > n(Y ), and if for some X 6= K we have ?X= ?K,
then n(X) > n(K) (how the ties among other coalitions are broken is not important). With
this mapping, we have thus ranked (enumerated) all coalitions such that stronger coalitions are
given higher numbers, and coalition K receives the smallest number among all coalitions with
the same power. Now de…ne the mapping ? : C ! C as
?(X) = argmin
n(Y ). (8)
Intuitively, this mapping picks an element of ?(X) for any X and satis…es ?(N) = K. Also,
note that ? is a projection in the sense that ?(?(X)) = ?(X). This follows immediately since
Axiom 3 implies ?(X) 2 ?(?(X)) and Corollary 1 implies that ?(?(X)) is a singleton.
The key to constructing a SPE is to consider o¤-equilibrium path behavior. To do this,
consider a subgame in which we have reached a coalition X (i.e., j transitions have occurred
and Nj = X) and let us try to determine what the URC would be if proposal Y is accepted
starting in this subgame. If Y = X, then the game will end, and thus X will be the URC. If,
on the other hand, Y 6= X, then the URC must be some subset of Y . Let us de…ne the strategy
pro…le ?Ksuch that the URC will be ?(Y ). We denote this (potentially o¤-equilibrium path)
URC following the acceptance of proposal Y by X(Y ), so that
X(Y ) =
if Y 6= X;
By Axiom 1 and equations (8) and (9), we have that
X = Y () X(Y ) = X: (10)
We will introduce one …nal concept before de…ning pro…le ?K. Let FX(i) denote the “fa-
vorite” coalition of player i if the current ruling coalition is X. Naturally, this will be the weakest
coalition among coalitions that are winning within X, that are self-enforcing and that include
player i. If there are several such coalitions, the de…nition of FX(i) picks the one with the
smallest n, and if there are none, it picks X itself. Therefore,
Similarly, we de…ne the “favorite” coalition of players Y ? X starting with X at the current
stage. This is again the weakest coalition among those favored by members of Y , thus
FX(Y ) =
n(Z) if Y 6= ?;
Equation (12) immediately implies that
For all X 2 C : FX(?) = X and FX(X) = ?(X). (13)
The …rst part is true by de…nition. The second part follows, since for all i 2 ?(X), ?(X)
is feasible in the minimization (11), and it has the lowest number n among all winning self-
enforcing coalitions by (8) and (5) (otherwise there would exist a self-enforcing coalition Z that
is winning within X and satis…es ?Z< ??(X), which would imply that ? violates Axiom 4).
Therefore, it is the favorite coalition of all i 2 ?(X) and thus FX(X) = ?(X).
Now we are ready to de…ne pro…le ?K. Take any history h and denote the player who is
supposed to move after this history a = a(h) if after h, we are at an agenda-setting step, and
v = v (h) if we are at a voting step deciding on some proposal P (and in this case, let a be the
agenda-setter who made proposal P). Also denote the set of potential agenda setters at this
stage of the game by A. Finally, recall that ~ n denotes a vote of “No” and ~ y is a vote of “Yes”.
Then ?Kis the following simple strategy pro…le where each agenda setter proposes his favorite
coalition in the continuation game (given current coalition X) and each voter votes “No” against
proposal P if the URC following P excludes him or he expects another proposal that he will
prefer to come shortly.
In particular, notice that v 2 FX(A) and P 6= FX(A [ fag) imply that voter v is part of a
di¤erent coalition proposal that will be made by some future agenda setter at this stage of
agenda-setter a proposes P = FX(a);
voter v votes
if either v = 2 X(P) or
v 2 FX(A), P 6= FX(A [ fag), and ?FX(A)? ? X(P);
the game if the current voting fails, and ?FX(A)? ? X(P)implies that this voter will receive
weakly higher payo¤ under this alternative proposal. This expression makes it clear that ?Kis
similar to a “truth-telling” strategy; each individual makes proposals according to his preferences
(constrained by what is feasible) and votes truthfully.
With the strategy pro…le ?K de…ned, we can state the main lemma, which will essentially
constitute the bulk of the proof of Theorem 2. For this lemma, also denote the set of voters
who already voted “Yes” at history h by V+, the set of voters who already voted “No” by V?.
Then, V = P n (V+[ V?[ fvg) denotes the set of voters who will vote after player v.
Lemma 1 Consider the subgame ?hof game ? after history h in which there were exactly jh
transitions and let the current coalition be X. Suppose that strategy pro…le ?K de…ned in (14)
is played in ?h. Then:
(a) If h is at agenda-setting step, the URC is R = FX(A [ fag); if h is at voting step
and V+[ fi 2 fvg [ V : i votes ~ y in ?Kg 2 WX, then the URC is R = X(P); and otherwise
R = FX(A).
(b) If h is at the voting step and proposal P will be accepted, player i 2 X receives payo¤
Ui= wi(R) ? "?jh+ IfP6=Xg
Otherwise (if proposal P will be rejected or if h is at agenda-setting step), then player i 2 X
Ui= wi(R) ? "?jh+ IfR6=XgIfi2Rg
The intuition for the results in this lemma is straightforward in view of the construction
of the strategy pro…le ?K. In particular, part (a) de…nes what the URC will be. This follows
immediately from ?K. For example, if we are at an agenda-setting step, then the URC will be the
favorite coalition of the set of remaining agenda setters, given by A[fag. This is an immediate
implication of the fact that according to the strategy pro…le ?K, each player will propose his
favorite coalition and voters will vote ~ n (“No”) against current proposals if the strategy pro…le
?K will induce a more preferred outcome for them in the remainder of this stage. Part (b)
simply de…nes the payo¤ to each player as the di¤erence between the baseline payo¤, wi(R), as
a function of the URC R de…ned in part (a), and the costs associated with transitions.
Given Lemma 1, Theorem 2 then follows if strategy pro…le ?Kis a SPE (because in this case
URC will be K and it will be reached with at most one transition). With ?Kde…ned in (14), it
is clear that no player can pro…tably deviate in any subgame.
The next lemma strengthens Lemma 1 for the case in which Assumption 2 holds by estab-
lishing that any SPE will lead to the same URC and payo¤s as those in Lemma 1.
Lemma 2 Suppose Assumption 2 holds and ?(N) = fKg. Let ?K be de…ned in (14). Then
for any SPE ? (in pure or mixed strategies) and for any history h, the equilibrium plays induced
by ? and by ?Kin the subgame ?hwill lead to the same URC and to identical payo¤s for each
Since ?(N) = fKg, Theorem 3 follows as an immediate corollary of this lemma (with h = ?).
5 The Structure of Ruling Coalitions
In this section, we present several results on the structure of URCs. Given the equivalence
result (Theorems 2 and 3), we will make use of the axiomatic characterization in Theorem
1. Throughout, unless stated otherwise, we …x a game ? = (N;?;w(?);?) with w satisfying
Assumption 1 and ? 2 [1=2;1). In addition, to simplify the analysis in this section, we assume
throughout that Assumption 2 holds and we also impose:
Assumption 3 For no X;Y 2 C such that X ? Y the equality ?Y= ??Xis satis…ed.
Assumption 3 guarantees that a small perturbation of a non-winning coalition Y does not
make it winning. Similar to Assumption 2, this assumption fails only in a set of Lebesgue
measure 0 (in fact, it coincides with Assumption 2 when ? = 1=2).All proofs are again
provided in the Appendix.
We start with the result that the set of self-enforcing coalitions is open (in the standard topol-
ogy); this is not only interesting per se but facilitates further proofs.Note that for game
? = (N;?;w(?);?), a power mapping ? (or more explicitly ?jN) is given by a jNj-dimensional
vector f?igi2N? RjNj
by A(N), and the subset of A(N) for which ??N;f?igi2N;w;??= N (i.e., the subset of power
distributions for which coalition N is stable) by S (N) and let N (N) = A(N) n S (N).
++. Denote the subset of vectors f?igi2Nthat satisfy Assumptions 2 and 3
Lemma 3 1. The set of power allocations that satisfy Assumptions 2 and 3, A(N), its subsets
for which coalition N is self-enforcing, S (N), and its subsets for which coalition N is not self-
enforcing, N (N), are open sets in RjNj
2. Each connected component of A(N) lies entirely within either S (N) or N (N).
++. The set A(N) is also dense in RjNj
An immediate corollary of Lemma 3 is that if the distributions of powers in two di¤erent
games are “close,” then these two games will have the same URC and also that the inclusion of
su¢ciently weak players will not change the URC. To state and prove this proposition, endow
the set of mappings ?, R, with the sup-metric, so that (R;?) is a metric space with ?(?;?0) =
ij. A ?-neighborhood of ? is f?02 R : ?(?;?0) < ?g.
Proposition 1 Consider ? = (N;?;w(?);?) with ? 2 [1=2;1). Then:
1. There exists ? > 0 such that if ?0: N ! R++ lies within ?-neighborhood of ?, then
?(N;?;w;?) = ?(N;?0;w;?).
2. There exists ?0> 0 such that if ?02 [1=2;1) satis…es j?0? ?j < ?0, then ?(N;?;w;?) =
3. Let N = N1[ N2with N1and N2disjoint. Then, there exists ? > 0 such that for all N2
with ?N2< ?, ?(N1) = ?(N1[ N2).
This proposition is intuitive in view of the results in Lemma 3. It implies that URCs have
some degree of continuity and will not change as a result of small changes in power or in the
rules of the game.
5.2Fragility of Self-Enforcing Coalitions
Although the structure of ruling coalitions is robust to small changes in the distribution of power
within the society, it may be fragile to more sizeable shocks. The next proposition shows that
in fact the addition or the elimination of a single member of the self-enforcing coalition turns
out to be such a sizable shock when ? = 1=2.
Proposition 2 Suppose ? = 1=2 and …x a power mapping ? : I ! R++. Then:
1. If coalitions X and Y such that X \ Y = ? are both self-enforcing, then coalition X [ Y
is not self-enforcing.
2. If X is a self-enforcing coalition, then X [ fig for i = 2 X and X n fig for i 2 X are not
The most important implication is that, under majority rule ? = 1=2, the addition or the
elimination of a single agent from a self-enforcing coalition makes this coalition no longer self-
enforcing. This result motivates our interpretation in the Introduction of the power struggles in
Soviet Russia following random deaths of Politburo members.
5.3Size of Ruling Coalitions
Proposition 3 Consider ? = (N;?;w(?);?).
1. Suppose ? = 1=2, then for any n and m such that 1 ? m ? n, m 6= 2, there exists a set of
players N, jNj = n, and a generic mapping of powers ? such that j?(N)j = m. In particular,
for any m 6= 2 there exists a self-enforcing ruling coalition of size m. However, there is no
self-enforcing coalition of size 2.
2. Suppose that ? > 1=2, then for any n and m such that 1 ? m ? n,there exists a set of
players N, jNj = n, and a generic mapping of powers ? such that j?(N)j = m.
These results show that one can say relatively little about the size and composition of URCs
without specifying the power distribution within the society further (except that when ? = 1=2,
coalitions of size 2 are not self-enforcing). However, this is largely due to the fact that there can
be very unequal distributions of power. For the potentially more interesting case in which the
distribution of power within the society is relatively equal, much more can be said about the
size of ruling coalitions. In particular, the following proposition shows that, as long as larger
coalitions have more power and there is majority rule (? = 1=2), only coalitions of size 2k? 1
for some integer k (i.e., coalitions of size 3, 7, 15, etc.) can be the URC (Part 1). Part 2 of the
proposition provides a su¢cient condition for this premise (larger coalitions are more powerful)
to hold. The rest of the proposition generalizes these results to societies with values of ? > 1=2.
Proposition 4 Consider ? = (N;?;w(?);?) with ? 2 [1=2;1).
1. Let ? = 1=2 and suppose that for any two coalitions X;Y 2 C such that jXj > jY j we have
?X> ?Y(i.e., larger coalitions have greater power). Then ?(N) = N if and only if jNj = km
where km= 2m? 1, m 2 Z. Moreover, under these conditions, any ruling coalition must have
size km= 2m? 1 for some m 2 Z.
2. For the condition 8X;Y 2 C : jXj > jY j ) ?X> ?Yto hold, it is su¢cient that there
exists some ? > 0 such that
3. Suppose ? 2 [1=2;1) and suppose that ? is such that for any two coalitions X ? Y ? N
such that jXj > ?jY j (jXj < ?jY j; resp.) we have ?X> ??Y(?X< ??Y; resp.). Then
?(N) = N if and only if jNj = km;? where k1;?= 1 and km;?= bkm?1;?=?c + 1 for m > 1,
where bzc denotes the integer part of z.
4. There exists ? > 0 such that maxi;j2N
?jY j; resp.) whenever ?X> ??Y(?X< ??Y; resp.). In particular, coalition X 2 C is self-
enforcing if and only if jXj = km;?for some m (where km;?is de…ned in Part 3).
??? < 1: (17)
?< 1 + ? implies that jXj > ?jY j (jXj <
This proposition shows that although it is impossible to make any general claims about
the size of coalitions without restricting the distribution of power within the society, a tight
characterization of the structure of the URC is possible when individuals are relatively similar
in terms of their power.
5.4Power and the Structure of Ruling Coalitions
One might expect that an increase in ?—the supermajority requirement—cannot decrease the
size of the URC. One might also expect that if an individual increases his power (either exoge-
nously or endogenously), this should also increase his payo¤. However, both of these are generally
not true. Consider the following simple example: let w(?) be given by (1). Then coalition (3;4;5)
is self-enforcing when ? = 1=2; but is not self-enforcing when 4=7 < ? < 7=12, because (3;4)
is now a self-enforcing and winning subcoalition. Next, consider game ? with ? = 1=2 and …ve
players A;B;C;D;E with powers ?A= ?0
?E= 21, and ?0
A= 2, ?B= ?0
B= 10, ?C= ?0
C= 15, ?D= ?0
E= 40. Then ?(N;?;w;?) = fA;D;Eg, while ?(N;?0;w;?) = fB;C;Dg, so
player E, who is the most powerful player in both cases, belongs to ?(N;?;w;?) but not to
We summarize these results in the following proposition (proof omitted).
Proposition 5 1. An increase in ? may reduce the size of the ruling coalition. That is, there
exists a society N, a power mapping ? and ?;?02 [1=2;1), such that ?0> ? but for all X 2
?(N;?;w;?) and X02 ?(N;?;w;?0), jXj > jX0j and ?X> ?X0.
2. There exists a society N, ? 2 [1=2;1), two mappings ?;?0: N ! R++satisfying ?i= ?0
for all i 6= j, ?j< ?0
applies even when j is the most powerful player in both cases, i.e. ?0
jsuch that j 2 ?(N;?;w;?), but j = 2 ?(N;?0;w;?). Moreover, this result
i= ?i< ?j< ?0
i 6= j.
Intuitively, higher ? turns certain coalitions that were otherwise non-self-enforcing into self-
enforcing coalitions. But this implies that larger coalitions are now less likely to be self-enforcing
and less likely to emerge as the ruling coalition. This, in turn, makes larger coalitions more
stable. The …rst part of the proposition therefore establishes that greater power or “agreement”
requirements in the form of supermajority rules do not necessarily lead to larger ruling coalitions.
The second part implies that being more powerful may be a disadvantage, even for the most
powerful player. This is for the intuitive reason that other players may prefer to be in a coalition
with less powerful players so as to receive higher payo¤s.
This latter result raises the question of when the most powerful player will be part of the
ruling coalition. This question is addressed in the next proposition.
Proposition 6 Consider the game ?(N;?;w(?);?) with ? 2 [1=2;1), and suppose that
?1;:::;?jNjis an increasing sequence. If ?jNj2
then either coalition N is self-enforcing or the most powerful individual, jNj, is not a part of the
?j=(1 ? ?);?PjNj?1
?j=(1 ? ?)
6 Concluding Remarks
We presented an analysis of political coalition formation in nondemocratic societies. The ab-
sence of strong institutions regulating political decision-making in such societies implies that
individuals competing for power cannot make binding promises (for example, they will be un-
able to commit to a certain distribution of resources in the future) and they will also be unable
to commit to abide by the coalitions they have formed. This latter feature implies that once
a particular ruling coalition has formed, a subcoalition can try to sideline some of the original
members. These considerations imply that ruling coalitions in nondemocratic societies should
be self-enforcing, in the sense that there should not exist a self-enforcing subcoalition that can
sideline some of the members of this ruling coalition. This implies that coalition formation in
such political games must be forward-looking; at each point, individuals have to anticipate how
future coalitions will behave. Despite this forward-looking element, we showed that self-enforcing
ruling coalitions can be determined in a relatively straightforward manner. In particular, we
presented both an axiomatic analysis and a noncooperative game of coalition formation, and
established that both approaches lead to the same set of self-enforcing ruling coalitions. More-
over, because such coalitions can be characterized recursively (by induction), it is possible to
characterize the key properties of self-enforcing ruling coalitions in general societies.
Our main results show that such ruling coalitions always exist and that they are generically
unique. Moreover, a coalition will be a self-enforcing ruling coalition if and only if it does not
possess any subcoalition that is su¢ciently powerful and self-enforcing. We also demonstrated
that, although equilibrium ruling coalitions are robust to small perturbations, the elimination
of a member of a self-enforcing coalition corresponds to a “large” shock and may change the
nature of the ruling coalition dramatically. This result provides us with a possible interpretation
for the large purges that took place in Stalin’s Politburo following deaths of signi…cant …gures.
Finally, we showed that although ruling coalitions can, in general, be of any size, once we
restrict attention to societies where power is relatively equally distributed, we can make strong
predictions on the size of ruling coalitions (for example, with majority rule, ? = 1=2, the ruling
coalition must be of the size 2k? 1, where k is an integer).
Naturally, the result that the ultimate ruling coalition always exists and is genetically unique
depends on some of our assumptions. In particular, the assumption that there is no commitment
to future divisions of resources is crucial both for the uniqueness and the characterization results.
Other assumptions can be generalized, however, without changing the major results in the paper.
For instance, the payo¤ functions can be generalized so that individuals may sometimes wish to
be part of larger coalitions, without a¤ecting our main results.
Other interesting areas of study in this context relate to some of the results presented in
Section 5. For example, we saw that individuals with greater power may end up worse o¤. This
suggests that individuals may voluntarily want to relinquish their power (for example, their
guns) or they may wish to engage in some type of power exchange before the game is played.
Some of these issues were discussed in the working paper version of our paper, Acemoglu, Egorov
and Sonin (2006), and developing these themes in the context of more concrete problems appears
to be a fruitful area for future research. The two most important challenges in future research
are to extend these ideas to games that are played repeatedly and are subject to shocks, and to
relax the assumption that individuals that are sidelined have no say in future decision-making.
Relaxing the latter assumption is particularly important to be able to apply similar ideas to
political decision-making in democratic societies.
Proof of Theorem 1: Consider …rst the properties of the set M(X) in (6) and the mapping
?(X) in (5) (Step 1). We then prove that ?(X) satis…es Axioms 1–4 (Step 2) and that this is
the unique mapping satisfying Axioms 1–4 (Step 3). Finally, we establish that when Assumption
2 holds, ? is single valued (Step 4). These four steps together prove both parts of the theorem.
Step 1: Note that at each step of the induction procedure, M(X) is well-de…ned because
Z in (6) satis…es jZj < jXj and thus ? has already been de…ned for Z. The argmin set in (5) is
also well de…ned, because it selects the minimum of a …nite number of elements (this number is
smaller than 2jXj; X is a subset of I, which is …nite). Non-emptiness follows, since the choice
set includes X. This implies that this procedure uniquely de…nes some mapping ? (which is
uniquely de…ned, but not necessarily single-valued).
Step 2: Take any X 2 C. Axiom 1 is satis…ed, because either ?(X) = X (if jXj = 1) or is
given by (5), so ?(X) contains only subsets of X such that ?(X) 6= ?. Furthermore, in both
cases ?(X) contains only winning (within X) coalitions, and thus Axiom 2 is satis…ed.
To verify that Axiom 3 is satis…ed, take any Y 2 ?(X). Either Y = X or Y 2 M(X). In
the …rst case, Y 2 ?(X) = ?(Y ), while in the latter, Y 2 ?(Y ) by (6).
Finally, Axiom 4 holds trivially when jXj = 1, since there is only one winning coalition. If
jXj > 1, take Y 2 ?(X) and Z ? X, such that Z 2 WX and Z 2 ?(Z). By construction of
?(X), we have that
Note also that Z 2 M(X) [ fXg from (6). Then, if
Z = 2argmin
we must have ?Z> ?Y, and vice versa, which completes the proof that Axiom 4 holds.
Step 3: We next prove that Axioms 1–4 de…ne a unique mapping ?. Suppose that there
exist ? and ?06= ? that satisfy these axioms. Then, Axioms 1 and 2 imply that if jXj = 1, then
?(X) = ?0(X) = X; this is because ?(X) 6= ? and ?(X) ? CXand the same applies to ?0(X).
Therefore, there must exist k > 1 such that for any A with jAj < k, we have ?(A) = ?0(A),
and there exists X 2 C, jXj = k, such that ?(X) 6= ?0(X). Without loss of generality, suppose
Y 2 ?(X) and Y = 2 ?0(X). Take any Z 2 ?0(X) (such Z exists by Axiom 1 and Z 6= Y by
hypothesis). We will now derive a contradiction by showing that Y = 2 ?(X).
We …rst prove that ?Z< ?Y. If Y = X, then ?Z< ?Yfollows immediately from the
fact that Z 6= Y and Z ? X (by Axiom 1). Now, consider the case Y 6= X, which implies
jY j < k (since Y ? X). By Axioms 2 and 3, Y 2 ?(X) implies that Y 2 WXand Y 2 ?(Y );
however, since jY j < k, we have ?(Y ) = ?0(Y ) (by the hypothesis that for any A with jAj < k,
?(A) = ?0(A)) and thus Y 2 ?0(Y ). Next, since Z 2 ?0(X), Y 2 WX, Y 2 ?0(Y ) and
Y = 2 ?0(X), Axiom 4 implies that ?Z< ?Y.
Note also that Z 2 ?0(X) implies (from Axioms 2 and 3) that Z 2 WX and Z 2 ?0(Z).
Moreover, since ?Z< ?Y, Z 6= X and therefore jZj < k (since Z ? X). This again yields
Z 2 ?(Z) by hypothesis. Since Y 2 ?(X); Z 2 ?(Z); Z 2 WX; ?Z< ?Y, Axiom 4 implies
that Z 2 ?(X). Since Z 2 ?(X); Y 2 ?(Y ); Y 2 WX; ?Z< ?Y, Axiom 4 implies that
Y = 2 ?(X), yielding a contradiction. This completes the proof that Axioms 1–4 de…ne at most
Step 4: Suppose Assumption 2 holds. If jXj = 1, then ?(X) = fXg and the conclusion
follows. If jXj > 1, then ?(X) is given by (5); since under Assumption 2 there does not exist
Y;Z 2 C such that ?Y= ?Z,
must be a singleton. Consequently, for any jXj, ?(X) is a singleton and ? is single-valued. This
completes the proof of Step 4. ?
Proof of Lemma 1: This lemma is proved by induction on the maximum length of histories
of ? (the number of steps in subgame ?h).
Base. If ?hincludes one last step only, this means that the current coalition is some X and
the current step is voting by the last voter v is voting over the last agenda setter’s proposal
P = X. In this case, ? implies that the URC must be R = X = X(X) = FX(?) and it does
not depend on v’s vote. Moreover, there are no more eliminations, hence each player i who was
not eliminated before receives wi(R) ? "jh, which coincides with both (15) and (16).
Step. Suppose that the result is proven for all proper subgames of ?h. Consider two cases.
Case 1: The current step is voting. Then, proposal P will be accepted if and only if
V+[fi 2 fvg [ V : i votes ~ y in ?Kg 2 WX(recall the de…nitions of V , V?and V+from the text
as the set of future voters, the set of those that have voted ~ n and the set of those that have voted
~ y respectively). If P is accepted, the URC will be X if P = X, while if P 6= X, the game will have
a transition to P, after which the URC will be FP(P) = ?(P) by induction (recall that after
transition the game proceeds to an agenda-setting step). In both cases, the URC R = X(P).
If P = X, player i 2 X gets wi(X) ? "jh, which equals (15). If P 6= X, player i 2 P receives
wi(R)?"?jh+ 1 + IfR6=PgIfi2Rg
?, which in this case equals (15), while player i 2 X nP obtains
i? "jh, which again equals (15) because i = 2 ?(P) ? P. On the other hand, if proposal P is
rejected, then the game ends when the voting ends if A = ? (then R = X = FX(?) = FX(A)
and each player i 2 X gets wi(R) ? "jh, which equals (16)) or, if A 6= ?, the game continues
with some b 2 X as agenda-setter and the remaining set of agenda-setters being B = A n fbg.
In the latter case, we know by induction that R = FX(B [ fbg) = FX(A) will be the URC, and
the payo¤ player i 2 X receives is given by (16).
Case 2: The current step is agenda-setting; suppose player a is to propose P = FX(a).
Note …rst that such P satis…es X(P) = P. Indeed, this automatically holds if P = X, while
if P 6= X, then X(P) = ?(P), but for P = FX(a) 6= X we must have ?(P) = P by (11), so
X(P) = P. Consider two subpossibilities. First, suppose P = FX(A [ fag). Then, as (14)
suggests, each player i 2 X(P) will vote ~ y. Note that X(P) is necessarily winning within
X: if P = X it follows from X = X(P) 2 WX, while if P 6= X, then, as we just showed,
X(P) = P = FX(a) which is winning by (11). This means that if proposal P = FX(A [ fag),
it is accepted, and R = P = FX(A [ fag) both in the case P = X and P 6= X (in the latter case,
R = X(P) by induction, and X(P) = P). Payo¤s in this case are given by (16) because there
are no more transitions if P = X and exactly one more transition if P 6= X, and only players in
P get the additional ?". Second, consider the case P 6= FX(A [ fag). Then ?FX(A)? ? X(P),
for ?FX(A)> ? X(P)would imply that minimum in (12) for Y = A [ fag 6= ? is reached at
FX(a) = X(P) = P and thus P = FX(A [ fag) which leads to a contradiction. But then,
as (14) suggests, all players in FX(A) 2 WXwill vote against proposal P, and thus P will be
rejected. We know by induction that then R = FX(A) and the payo¤s are given by (16). This
completes the proof of Lemma 1. ?
Proof of Theorem 2: Pro…le ?Kinvolves only pure strategies. Applying Lemma 1 to the
…rst stage where h = ?, we deduce that the URC under ?K is FN(N) = ?(N) = K, and
payo¤s are given by (16) which equals (7) because jh= 0 (there were no eliminations before
and N = X, K = R). The theorem is therefore proved if we establish that pro…le ?Kis a SPE.
To do this we show that there is no pro…table one-shot deviation, which is su¢cient since ? is
Suppose, to obtain a contradiction, that there is a one-shot pro…table deviation after history
h; since only one player moves at each history, this is either voter v or agenda-setter a. Let us
start with the former case, which is then subdivided into two subcases.
Subcase 1: suppose voter v votes ~ y in ?K (this means v 2 X(P) ? P), but would be
better o¤ if he voted ~ n. In pro…le ?K, the votes of players who vote after voter v (those in
V ) do not depend on the vote of player v. Hence, if proposal P is accepted in equilibrium,
deviating to ~ n can result in rejection, but not vice versa. This deviation may only be pro…table
if voter v is pivotal, so we restrict attention to this case. From Lemma 1, the URC will be
X(P) if proposal P is accepted and FX(A) if P is rejected; the number of transitions will be
between 0 and 2 (including transition from X to P if P 6= X) in the …rst case and either 0 or
1 in the second case, so the number of transitions matters only if wv( X(P)) = wv(FX(A))
(see (2)). Let us prove that v 2 FX(A), ?FX(A)? ? X(P), and P 6= FX(A [ fag). First,
since deviation is pro…table, v 2 FX(A) (recall that v 2 X(P) simply because v votes ~ y in
?K). Second, if instead ?FX(A)> ? X(P), this would imply wv(FX(A)) < wv( X(P)) due to
Assumption 1. Third, if instead P = FX(A [ fag), then X(P) = P (for in this case either
P = X or P = FX(i) for some i 2 X; the …rst case is trivial while the second is considered in
the proof of Lemma 1). But we just showed that either ?FX(A)< ? X(P)or ?FX(A)= ? X(P).
In the …rst case, the minimum in (12) for Y = A [ fag cannot be achieved at X(P) =
P because n(FX(A)) < n( X(P)) and FX(A) is feasible in this minimization problem, so
P 6= FX(A [ fag) which is a contradiction. In the second case, wv(FX(A)) = wv( X(P)),
and since FX(A) 6= X if and only if X(P) 6= X (FX(A) = X 6= X(P) would contradict
?FX(A)= ? X(P), and so would FX(A) 6= X = X(P)), the number of additional transitions is
the same. Hence, deviation to ~ n is not pro…table because with or without this deviation player v
would get wv(FX(A))?"?jh+ IfFX(A)6=Xg
This, together with v 2 FX(A) and ?FX(A)? ? X(P)implies that voter v must vote ~ n in pro…le
?K, which contradicts the assumption that he votes ~ y.
?. This contradiction proves that P 6= FX(A [ fag).
Subcase 2: suppose that voter v votes ~ n in ?K, but would be better o¤ voting ~ y. Again,
this deviation does not change other voters’ votes, it can only change the URC from FX(A)
to X(P) and is only pro…table if v is pivotal. Consider two possible cases. If v = 2 X(P),
voting ~ y gives v exactly w?
v? "(jh+ 1) (v 2 P, so v is part of transition from X to P, and
Lemma 1 implies that transition from P to X(P) 6= P will proceed in one step, so v will
participate in exactly one more transition). Voting ~ n will then result in at most one additional
transition, so v obtains a payo¤ no less than wv(FX(A)) ? "(jh+ 1). This implies that the
payo¤ of player v from voting ~ n is at least as large as his payo¤ from deviating to ~ y, thus
deviation is not pro…table. On the other hand, if v 2 X(P), then, as implied by equation
(14), v 2 FX(A), ?FX(A)? ? X(P), and P 6= FX(A [ fag). By Lemma 1, if v votes ~ n, the
URC is FX(A) and he receives payo¤ wv(FX(A)) ? "?jh+ IfFX(A)6=Xg
?; if he votes ~ y, the
URC is X(P) and he receives wv( X(P)) ? "?jh+ IfP6=Xg
? X(P)implies wv(FX(A)) ? wv( X(P)), so the deviation could only be pro…table for v if
X. In this case, however, strict inequality ?FX(A)< ? X(P)holds, and therefore wv(FX(A)) >
wv( X(P)). Then (2) implies that deviation for v is not pro…table. This completes the proof
?1 + If X(P)6=Pg
??. But ?FX(A)?
?1 + If X(P)6=Pg
?. This can only be true if FX(A) 6= X and X(P) = P =
that no one-shot deviation by a voter may be pro…table.
The remaining case is where agenda-setter a has a best response Q 6= P, and P = FX(a)
does not belong to the best response set. Consider two subcases. Subcase 1: suppose coalition
P is accepted if proposed; in that case, Q cannot be rejected under pro…le ?K. The reason is
as follows: if Q were rejected, then the URC would be FX(A). If i 2 FX(A), then coalition
FX(A) is feasible in minimization problem (11), which means that wa(FX(A)) ? wa(P). If
this inequality is strict, so wa(FX(A)) < wa(P), then deviation to Q is not pro…table; if
wa(FX(A)) = wa(P), then the either FX(A) = P = X or neither FX(A) nor P equals X, but
in both cases a participates in the same number (0 or 1, respectively) of extra transitions, so
utility from proposing P and Q is the same and the deviation is not pro…table either. If, however,
i = 2 FX(A), then proposing Q will give a payo¤ w?
wa(P)?"(jh+ 1) (again, X(P) = P for P = FX(a)), so deviation is again not pro…table. This
proves that Q must be accepted, which immediately implies that X(Q) 2 WX(only members of
X(Q) vote for Q in ?K, see (14)), and then Q 2 WXbecause X(Q) ? Q. We next prove that
X(Q) = Q. Suppose, to obtain a contradiction, that X(Q) 6= Q; this immediately implies
Q 6= X and thus X(Q) = ?(Q). If a proposed X(Q) instead of Q, it would be accepted, too.
Moreover, the fact that ? is a projection implies that X( X(Q)) = X(?(Q)) = ?(?(Q)) =
a? "jhwhile proposing P will give at least
?(Q) = X(Q). In addition, any player i who votes ~ y if Q is proposed is part of X(Q), and
therefore would participate in voting for X(Q); moreover, he would vote ~ y under ?Kin that
case, too, because X(Q) 6= FX(A [ fag) implies Q 6= FX(A [ fag) (otherwise Q would satisfy
X(Q) = Q), and from (14) one can see that anyone who votes ~ n if X(Q) is proposed would
also vote ~ n if Q were proposed. Therefore, X(Q) would be accepted if proposed, but proposing
X(Q) would result in only one transition while proposing Q would result in two. Agenda-setter
a must be in X(Q), so proposing X(Q) is better than proposing Q, which contradicts the
assumption that Q is a best response for a and establishes that X(Q) = Q. Finally, for a to
propose Q, a 2 Q must hold. We have proved that coalition Q is feasible in minimization (11)
for i = a, and Q 6= P implies n(P) < n(Q). But in that case either ?P< ?Q(then a prefers
having P instead of Q as the URC, even if it means an extra transition) or ?P= ?Q(then
a is indi¤erent, because the number of transitions is the same because both P = X(P) and
Q = X(Q)). These arguments together imply that deviation to Q is not pro…table for a when
P will be accepted under ?K.
Subcase 2: suppose coalition P is rejected if proposed. Clearly, Q must be accepted, for
otherwise the payo¤s under the two proposals are identical and Q is not a pro…table deviation.
Since P = X(P) is winning within X, but is not accepted, then, from (14), ?FX(A)? ? X(P)
and P 6= FX(A [ fag). As in the previous case we can show that Q 2 WX, X(Q) = Q, and
a 2 Q. Since Q is accepted, (14) implies that either ?FX(A)> ? X(Q)or Q = FX(A [ fag),
for otherwise members of winning coalition FX(A) would vote against Q in ?K and Q would
be rejected. In both cases, n(Q) < n(P) (in the …rst case because ?Q= ? X(Q)< ?FX(A)?
? X(P)= ?P, and in the second case because P = FX(a) is feasible in minimization (12) for
Y = A[fag, and P 6= Q). This means, however, that P cannot be the outcome in minimization
(11) for i = a because Q is also feasible (Q 2 WX, a 2 Q, and ?(Q) = Q because X(Q) = Q
and Q 6= X where the latter follows from n(Q) < n(P)). This, however, contradicts that
P = FX(a) by construction of pro…le ?Kin (14). Therefore, there is no pro…table deviation at
the agenda setting step either. This completes the proof of Theorem 2. ?
Proof of Lemma 2: This proof also uses induction on the number of steps in ?h.
Base: If only one step remains, then the current ruling coalition is some X, and this step
must be voting by the last voter v over proposal P = X made by the last agenda-setter.
Regardless of the vote (and therefore in either pro…le), coalition X will be the URC, and each
player i 2 X will receive payo¤ wi(X)?"jh; each players in N nX will receive the same payo¤
under both pro…les, because the intermediate coalitions and the number of transitions each
player faced is the same because histories up to the last step are identical.
Step. Take any history h and denote the …rst player to act in subgame ?hby b and the
payo¤ to player i when b plays action ? by Ui(?). By induction this value is the same both if
pro…le ? and ?K is played thereafter. Consequently, if some action is optimal for player b if
pro…le ? is played in subgames of ?h, the same is true if pro…le ?K is played, and vice versa.
Let ?Kbe the action played by b in pro…le ?K and ?0be an action played in pro…le ? with a
non-zero probability. Then both ?Kand ?0must yield the same payo¤ for b because both are
optimal when ?Kis played thereafter. Thus Ub(?K) = Ub(?0).
It therefore su¢ces to show that both action ?0followed by equilibrium play of pro…le ?K
and action ?Kfollowed by equilibrium play of the same pro…le ?Kresult in the same URC and
the same payo¤ for all players i 2 N (then by induction, action ?0followed by equilibrium play
of pro…le ? will result in the same URC and payo¤s). This is clearly true when ?0= ?K. Now
consider the case where ?06= ?K. Both action ?Kand action ?0, accompanied by equilibrium
play of pro…le ?K, will result in 0, 1, or 2 additional eliminations, as follows from Lemma 1:
after the …rst elimination, if any, equilibrium play will have at most one more elimination. This,
together with (2), implies that jwb(R0) ? wb(RK)j ? 2" and wb(R0) = wb(RK), where R0and
RKare URCs if ?0and ?Kare played by b, respectively. There are two possibilities.
First, consider the case wb(R0) = wb(RK) 6= w?
b, then b 2 R0, b 2 RK, hence ?R0= ?RK,
and by Assumption 2, R0= RK, so that the URC is the same in both cases. The number of
transitions is also the same (because b participates in all transitions and is indi¤erent between
?0and ?K). If there are no more transitions, then each player i 2 X obtains utility wi(R0)?"jh
for both actions. If there is exactly one transition from X to R0, then player i 2 R0 gets
wi(R0) ? "(jh+ 1) and player i 2 X n R0gets w?
exactly two transitions both after ?0and ?K. This cannot be the case if the …rst step of ?his
i? "jh. Consider the case where there are
agenda-setting, for in that case Lemma 1 implies that if action ?Kplayed under pro…le ?K, the
equilibrium play involves only one more transition. Then the …rst step of ?his voting over some
proposal P; moreover, both action ?0and ?Kwill result in acceptance of this proposal, for a
rejection, again by Lemma 1, would lead to only one extra transition. But in that case the two
additional transitions are from X to P and from P to X(P) 6= P. This establishes that each
player i 2 X receives the same payo¤ regardless of whether b plays action ?0or ?K.
Second, consider the case wb(R0) = wb(RK) = w?
b. Suppose …rst that b is agenda-setter;
then action ?Kcorresponds to making proposal P = FX(b). Then, as implied by Lemma 1, the
URC is FX(b) that b is part of (this happens if P is accepted) or FX(A) which b may or may
not be part of (this happens if P is rejected); here A is the set of would-be agenda-setters. In
the case under consideration b = 2 RK, hence RK= FX(A) and Ub(?K) = w?
the proposal of some coalition Q 6= P such that b 2 Q. If Q is accepted, but b = 2 R0, then b has
an extra transition to Q but is eventually eliminated, so he receives Ub(?0) = w?
b?"jh. Action ?0is
b? "(jh+ 1)
and this contradicts Ub(?K) = Ub(?0). Therefore, Q must be rejected, the URC must be
R0= FX(A) = RK, and each player i 2 R0will receive wi(R0) ? "(jh+ 1) while i 2 X n R0
gets wi(R0)?"jhin the case of either action. Now suppose that b is voting over some proposal
P, then b 2 P. Then one of actions ?0and ?Kis ~ y and the other ~ n. For the action to matter,
proposal must be accepted if ~ y is played and rejected if ~ n is played (or vice versa, but this is
impossible under ?K). But recall that voter b is not a member of R0 and RK. Therefore, b
votes ~ y, he receives w?
i?"(jh+ 1) (he does not participate in the second transition, which will
happen under ?Kbecause b 2 P and b = 2 R0, b = 2 Rk). On the other hand, if b votes ~ n, then by
Lemma 1 he receives w?
i?"jh, because there is only one transition in which b is eliminated. But
this means that Ub(~ y) 6= Ub(~ n), so Ub(?0) 6= Ub(?K), which implies that b cannot be indi¤erent
between the two actions ?0and ?K, thus yielding a contradiction.
We have therefore proved that after either of the two actions ?0and ?Kis played, the URC
is the same and each player i 2 X is indi¤erent. But any player i 2 N n X is indi¤erent, too,
because in this case the payo¤s are entirely determined by history h. This completes the proof
of the step of induction, and therefore of Lemma 2. ?
Proof of Theorem 3: The proof follows immediately from the application of Lemma 2 to
the entire game ?, which is starting with history h = ?. The lemma then implies that the URC
in any SPE must coincide with that under the strategy pro…le ?K, i.e. K, and payo¤s must be
given by (7), as implied by Lemma 2. ?
Proof of Lemma 3: (Part 1) The set A(N) may be obtained from RjNj
a …nite number of hyperplanes given by equations ?X= ?Yfor all X;Y 2 P (N) such that
X 6= Y and by equations ?Y= ??Xfor all X;Y 2 P (N) such that X ? Y . These hyperplanes
are closed sets (in the standard topology of RjNj
++), hence, a small perturbation of powers of a
generic point preserves this property (genericity). This ensures that A(N) is an open set; it is
dense because hyperplanes have dimension lower than jNj. The proofs for S (N) and N (N) are
by induction. The base follows immediately since S (N) = R++and N (N) = ? are open sets.
Now suppose that we have proved this result for all k < jNj. For any distribution of powers
f?igi2N, N is self-enforcing if and only if there are no proper winning self-enforcing coalitions
within N. Now take some small (in the sup-metric) perturbation of powers f?0
perturbation is small, then the set of winning coalitions is the same, and, by induction, the
igi2N. If this
set of proper self-enforcing coalitions is the same as well. Therefore, the perturbed coalition
completes the induction step.
ig is self-enforcing if and only if the initial coalition with powers f?ig is self-enforcing; which
(Part 2) Take any connected component A ? A(N). Both S (N) \ A and N (N) \ A are
open in A in the topology induced by A(N) (and, in turn, by RjNj
topology. Also, (S (N) \ A) \ (N (N) \ A) = ? and (S (N) \ A) [ (N (N) \ A) = A, which,
given that A is connected, implies that either S (N) \ A or N (N) \ A is empty. Hence, A lies
either entirely within S (N) or N (N). This completes the proof. ?
Proof of Proposition 1: The …rst two parts follow by induction. If N = 1, for any ? and
++) by de…nition of induced
?, ?(N;?;w;?) = fNg. Now suppose that this is true for all N with jNj < n; take any society
N with jNj = n. We then use the inductive procedure for determining ?(N;?;w;?), which is
described in Theorem 1. In particular, Assumptions 2 and 3 imply that the set M(N) in (6)
is identical for ?(N;?;w;?), ?(N;?0;w;?), and ?(N;?;w;?0), provided that ? is su¢ciently
small (the result self-enforcing coalitions remain self-enforcing after perturbation follows from
Lemma 3). Moreover, if ? is small, then ?X> ?Yis equivalent to ?0
implies that ?(N;?;w;?) = ?(N;?0;w;?) = ?(N;?;w;?0). This completes the proof of parts
Y. Therefore, (5)
1 and 2.
The proof of part 3 is also by induction. Let jN1j = n. For n = 1 the result follows
straightforwardly. Suppose next that the result is true for n. If ? is small enough, then ?(N1) is
winning within N = N1[N2; we also know that it is self-enforcing. Thus we only need to verify
that there exists no X ? N1[ N2such that ?(X) = X (i.e., X that is self-enforcing, winning
in N1[ N2and has ?X< ??(N1)). Suppose, to obtain a contradiction, that this is not the case
(i.e., that the minimal winning self-enforcing coalition X 2 P (N1[ N2) does not coincide with
?(N1)). Consider its part that lies within N1, X \ N1. By de…nition, ?N1? ??(N1)> ?X?
?X\N1, where the strict inequality follows by hypothesis. This string of inequalities implies that
X \N1is a proper subset of N1, thus must have fewer elements than n. Then, by induction, for
small enough ?, ?(X \ N1) = ?(X) = X (since X is self-enforcing). However, ?(X \ N1) ? N1,
and thus X ? N1. Therefore, X is self-enforcing and winning within N1(since it is winning
within N1[ N2). This implies that ??(N1)? ?X(since ?(N1) is the minimal self-enforcing
coalition that is winning within N1). But this contradicts the inequality ??(N1)> ?Xand
implies that the hypothesis is true for n + 1. This completes the proof of part 3. ?
Proof of Proposition 2: (Part 1) Either X is stronger than Y or vice versa. The stronger
of the two is a winning self-enforcing coalition that is not equal to X [ Y . Therefore, X [ Y is
not the minimal winning self-enforcing coalition, and so it is not the URC in X [ Y .
(Part 2) For the case of adding, it follows directly from Part 1, since coalition of one player
is always self-enforcing. For the case of elimination: suppose that it is wrong, and the coalition
is self-enforcing. Then, by Part 1, adding this person back will result in a non-self-enforcing
coalition. This is a contradiction which completes the proof of Part 2. ?
Proof of Proposition 3: (Part 1) Given Part 3 of Proposition 1, it is su¢cient to show
that there is a self-enforcing coalition M of size m (then adding n ? m players with negligible
powers to form coalition N would yield ?(N) = ?(M) = M). Let i 2 M = f1;:::;mg be
the set of players. If m = 1, the statement is trivial. Fix m > 2 and construct the following
sequence recursively: ?1= 2, ?k>Pk?1
straightforward to check that numbers f?igi2Mare generic. Let us check that no proper winning
coalition within M is self-enforcing. Take any proper winning coalition X; it is straightforward
j=1?jfor all k = 2;3;:::;m?1, ?m=Pm?1
j=1?j?1. It is
to check that jXj ? 2, for no single player forms a winning coalition. If coalition X includes
player m (with power ?m), then it excludes some player k with k < m; his power ?k? 2 by
which means that player m is stronger than the rest, and thus coalition M is non-self-enforcing.
?j? 1 >
?j? ?k? ?Xnfmg;
If X does not include ?m, then take the strongest player in X; suppose it is k, k ? m ? 1.
However, by construction he is stronger than all other players in X, and thus X is not self-
enforcing. This proves that M is self-enforcing. However, if jXj = 2 and Assumption 2 holds,
then one of the players, say player i, is stronger than the other one, and thus fig is a winning
self-enforcing coalition. But then, by Corollary 1, X cannot be self-enforcing.
(Part 2) The proof is identical to Part 1. The recursive sequence should be constructed as
follows: ?1= 2, ?k> ?Pk?1
Proof of Proposition 4: (Part 1) This part follows as a special case of Part 3. To see
this, note that the condition in Part 4 is satis…ed , since for any X ? Y ? N, jXj ? ?jY j ()
jXj ? jY n Xj =) ?X? ?Y nX
km’s in Part 1 and in Part 3 are equal since k1= 21? 1 = 1, and if km?1= 2m?1? 1 then
km= 2m? 1 = b2km?1c + 1 and thus the desired result follows by induction.
(Part 2) Suppose, to obtain a contradiction, that the claim is false, i.e., that for some
j=1?jfor all k = 2;3;:::;m ? 1, ?m= ?Pm?1
j=1?j? 1. ?
() ?X? ??Yfor ? = 1=2. Moreover, the sequences of
X;Y ? N such that jXj > jY j we have ?X? ?Y. Then the same inequalities hold for X0=
X n(X \ Y ) and Y0= Y n(X \ Y ), which do not intersect, so thatP
Rearranging, we have
??j=? ? 1?+jY0j.
j2Y0?j=?, and thusP
??j=? ? 1?+jX0j ?P
However, X0and Y0do not intersect, and therefore this violates (17). This contradiction com-
pletes the proof of Part 2.
(Part 3) The proof is by induction. The base is trivial: a one-player coalition is self-
enforcing, and jNj = k1= 1. Now assume the claim has been proved for all q < jNj, let us prove
it for q = jNj. If jNj = kmfor some m, then any winning (within N) coalition X must have
size at least ?(bkm?1=?c + 1) > km?1(if it has smaller size then ?X< ??N). By induction, all
such coalitions are not self-enforcing, and this means that the grand coalition is self-enforcing.
If jNj 6= kmfor any m, then take m such that km?1< jNj < km. Now take the coalition of
the strongest km?1individuals. This coalition is self-enforcing by induction. It is also winning
(this follows since km?1? ?bkm?1=?c = ?(km? 1) ? ?jNj, which means that this coalition
would have at least ? share of power if all individuals had equal power, but since this is the
strongest km?1individuals, the inequality will be strict). Therefore, there exists a self-enforcing
winning coalition, di¤erent from the grand coalition. This implies that the grand coalition is
non-self-enforcing, completing the proof.
(Part 4) This follows from Part 3 and Proposition 3. ?
Proof of Proposition 6: Inequality ?jNj> ?Pn?1
that includes jNj, but excludes even the weakest player will not be self-enforcing. The inequality
j=2?j=(1 ? ?) implies that any coalition
j=2?j=(1 ? ?) implies that player jNj does not form a winning coalition by himself.
Therefore, either N is self-enforcing or ?(N) does not include the strongest player. ?
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