A Heterogeneous Network Game Perspective of
Fashion Cycle ∗
Zhi-Gang Cao
†
Cheng-Zhong Qin‡
Xiao-Guang Yang§
Bo-Yu Zhang¶
May 3, 2013
Abstract
Fashion, as a “second nature” of human being, has great economic impacts.
In this paper, we apply a heterogenous network game to analyze the existence of
a fashion cycle. There are two types of agents in the network game, conformists
and rebels. Conformists prefer to match the action taken by the majority of her
neighbors while rebels like to mismatch. Our results imply that social interaction
structures play a crucial role in the evolution of fashion, especially the emergence
of fashion cycles. For example, with a social network composed of a line or a ring,
fashion will almost always evolve into a steady state where no flux is possible
and thus can be deemed as disappeared. When the network is a star, however,
for approximately half of all the configurations of agents’ types, a fashion cycle
of length 4 will always emerge, regardless of the initial action profiles.
∗
This is a sister working paper with:
Bo-yu Zhang, Zhi-gang Cao, Cheng-zhong
Qin, Xiao-guang Yang, Fashion and Homophily (April 14, 2013).
Available at SSRN:
http://ssrn.com/abstract=2250898 or http://dx.doi.org/10.2139/ssrn.2250898. The research of the
first and third authors are supported by the 973 Program (2010CB731405) and National Natural
Science Foundation of China (71101140). We gratefully acknowledge helpful discussions with Xujin
Chen, Zhiwei Cui, Xiaodong Hu, Weidong Ma, Changjun Wang, Lin Zhao, and Wei Zhu.
†
Key Laboratory of Management, Decision & Information Systems, Academy of Mathematics
and Systems Science, Chinese Academy of Sciences, Beijing, 100190, China. Email: zhigangcao@amss.ac.cn.
‡
Department of Economics, University of California, Santa Barbara, CA 93106, USA. Email:
qin@econ.ucsb.edu.
§
Key Laboratory of Management, Decision & Information Systems, Academy of Mathematics and
Systems Science, Chinese Academy of Sciences, Beijing, 100190, China. Email: xgyang@iss.ac.cn.
¶
Department of Mathematics, Beijing Normal University, Beijing, 100875, China. Email: zhangboyu5507@gmail.com.
1
KEYWORDS: fashion cycle; network games; coordination; anti-coordination;
matching pennies.
JEL Classification: A14; C62; C72; D72; D83; D85; Z13
1
Introduction
There are two dictionary definitions of fashion that are contradictory to each other.
One states that fashion is a prevailing custom, usage, or style, and the other says
that fashion is a distinctive or peculiar manner or way.1 The former definition implies
that to be fashionable is a conforming activity or practice. In contrast, under the
latter definition, to be fashionable is a rebeling activity. This may be an indication
that different people have different and even opposite opinions on what is fashionable.
Indeed, in the pursuit of fashion, many people prefer to go along with the majority,
while others like to stay in the minority. For instance, Louis Vuitton and Zara are
both regarded as fashionable brands, although the former is luxury and few people can
afford, while the latter is not luxury at all but still eagerly purchased by fashionists.2
When people say that “This year’s fashion color is black, my white skirt is not ‘in’ any
more”, they believe that conforming to the majority is fashion. In contrast, when they
exclaim with admiration that “Lady Gaga is the Godness of fashion”, they accept that
distinction is fashion.
In history, and even today, many people disgusted fashion so much that they behaved and claimed themselves as “anti-fashion”. However, they did not realize or
acknowledge that they were still in the influence sphere of fashion. As argued by
Simmel (1904, p. 143),
“The man who consciously pays no heed to fashion accepts its form just as
much as the dude does, only he embodies it in another category, the former
in that of exaggeration, the latter in that of negation.”
In fact, these people play very special roles as fashion leaders. But as time goes by,
more and more of them approve Simmel’s argument and are proud of being fashion
leaders.3 This change is probably not due to the diffusion of the theory of Simmel or
any other scholar, but mainly driven by the increasing influence of fashion itself.
1
See, for example, Merriam-Webster: http://www.merriam-webster.com/dictionary/fashion.
Although Zara is most famously known for its “fast fashion”, “cheap fashion” fits it very well too.
Other examples of cheap fashion include H & M, Uniqlo, GAP, The Limited, and Vancl etc.
3
But even today, the two words “fashionist” and “fashionista” both refer only to fashion trend
followers.
2
2
Fashion is now a daily topic in newspapers, magazines, TVs, the Internet, etc.
Fashion is also doubtlessly a huge industry and its economic impacts are extremely
significant.4 Many people mistakenly assume that fashion is restricted only to costume
and adornment. However, the influence of fashion goes far beyond this.5 Blumer (1969)
listed the following fields where anecdotal evidences can be easily found showing the
operation of fashion: pure and applied arts (painting, sculpture, music, architecture
etc.), entertainment and amusement, medicine, business management, mortuary practice, literature, political doctrine, and even science and academic research.6 As claimed
by Svendsen (2006),
“Fashion has been one of the most influential phenomena in Western civilization since the Renaissance. It has conquered an increasing number of
modern man’s fields of activity and has become almost ‘second nature’ to
us ”.
The “second nature” viewpoint has already been implied, and eloquently proved, in
the essay of Simmel (1904), although he did not use the term explicitly. In particular,
Simmel claimed that people’s conforming and rebeling activities in fashion have a
rather stylized pattern; that is, upper class people are fashion leaders and lower class
people are fashion followers.7 As remarked in Blumer (1969):
“For him, fashion arose as a form of class differentiation in a relatively open
class society. In such a society the elite class seeks to set itself apart by
observable marks or insignia, such as distinctive forms of dress. However,
members of immediately subjacent classes adopt these insignia as a means
of satisfying their striving to identify with a superior status. They, in turn,
4
For instance, the global market size of luxury industry is around $ 200 billion US dollars (Okonkwo,
2007; Euromonitor International, June 2011). Only in US, $ 300 billion expenditure are driven by
fashion rather than by (physical) need (Yoganarasimhan, 2012b).
5
This is one of the important reasons why theoretical studies of fashion are seriously lacking, as
pointed out by Blumer (1969), whose arguments on this point are still largely valid today (see Andreozzi and Bianch, 2007). A second reason, also pointed out by Blumer (1969), is that scholars view
fashion as an “aberrant and irrational social happening”. After Gary Becker’s pioneering, convincing, and influential explorations in the economic studies of addiction, crime, family, and tastes etc.,
economists (and sociologists) today should have no obstruction in taking fashion as a behavior that
is rational at least to some degree. This point will be revisited in Footnote 20.
6
More recent rigorous empirical studies can be found in Lieberson and Bell (1992), Schweitzer and
Mach (2008), Yoganarasimhan (2012b).
7
Generally, there can be more than two classes, but the essence is the same: people try to imitate
the behavior of those belonging to a class that is immediately upper than them, by doing this, they
differentiate themselves from people in the same class and the lower classes.
3
are copied by members of classes beneath them. In this way, the distinguishing insignia of the elite class filter down through the class pyramid. In
this process, however, the elite class loses these marks of separate identity.
It is led, accordingly, to devise new distinguishing insignia which, again,
are copied by the classes below, thus repeating the cycle. This, for Simmel,
was the nature of fashion and the mechanism of its operation.”
In the economic literature, the above logic of Simmel is largely identical to the
“conspicuous consumption” theory of Veblen (1899), which is, as its name indicates,
usually restricted to people’s consumption behavior.8 More precisely, Veblen distinguished between “invidious comparison” and “pecuniary emulation”, which correspond
to the rebeling behavior of the upper class and the conforming behavior of the lower
class in Simmel’s theory, respectively (see also Bagwell and Bernheim, 1996).
Subsequent literature (e.g., Pesendorfer, 1995; Bagwell and Bernheim, 1996) are
mostly based on the above theory of Simmel and Veblen.9 However, it is by no means
unchallenged. The most cogent queries on the theory of Simmel and Veblen come
from many empirical evidences where fashion is not initialized by the upper class.
The most famous (counter-)example is blue jeans, which originated from the humble
cowboys and miners (Fine and Leopold, 1993; Trigg, 2001; Benvenuto, 2000; Yoganarasimhan, 2012b). Other examples include slang (Benvenuto, 2000), and hip-hop
(Yoganarasimhan, 2012b), etc.. Similar to the example of slang, a prominent new example in the era of Internet is that most fashionable expressions in the virtual world
are coined by anonymous “grassroots”. To remedy the theory of conspicuous consumption, Bourdieu (1984) presents a “trickle-round” model.10 Although the usual attack
8
This is also known as the Veblen effect. Strictly speaking, Veblen effect refers to the phenomenon
that as the price of a consumers’ good goes up, the demand increases rather than decreases. A similar
concept is the snob effect, which says that an individual’s demand for a consumers’ good decreases as
more people consume it. The two effects are two sides of the same coin, and their difference is that the
former is a function of price while the latter is a function of other people’s consumption. An opposite
phenomenon of the snob effect is the bandwagon effect, which says that an individual’s demand for
a consumers’ good increases as more people consume it. See the detailed discussions in Leibenstein
(1950).
9
Neither Simmel nor Veblen is the first scholar who considered fashion seriously. As reviewed by
Leibenstein (1950), the first one should probably be attributed to John Rae (before 1834), who gave
an extensive analysis of conspicuous consumption, which is subsequently known as the Veblen effect.
See Leibenstein for more early literature.
10
In literature, the fashion diffusion process of Simmel and Veblen is also summarized as “trickledown”, and the opposite bottom-up process is also called “trickle-up” (see Trigg, 2001). In the
“trickle-round” model of Bourdieu (1984), there are (at least) three classes, namely the upper class,
the middle class and the working class. Besides the “trickle-down” flow of fashion diffusion, fashion
can also spread from the working class to the upper class, making a complete directed cycle.
4
loses its power on this new theory, it is still based on the differentiation of classes.
We argue in the next subsection that a “personality differentiation” theory might
be more appropriate in modern time. The fundamental reason is that the world is
becoming flatter and flatter and hence the upper class people are losing more and
more of their advantages in the game of fashion.
1.1
From class differentiation to personality differentiation
Instead of providing more counter-examples, we need first to understand the logic of
Simmel (1904) why upper class people are fashion leaders and lower class people are
followers. This is necessary in order to see why this logic worked well in the age of
Simmel, why it is failing to work today, and why it might be replaced by a classless
theory in the future.
Simmel (1904) argued that two antagonistic tendencies, imitation and differentiation, union and segregation, etc., are simultaneously deeply rooted in the soul of every
individual. For instance, he observed that children “insist upon the repetition of facts”
and “they will object to the slightest variation in the telling of a story they have heard
twenty times”. He argued that “In this imitation and in exact adaption to the past
the child first rises above its momentary existence [...]” (p. 133).
The other fact about children Simmel did not point out, at least not explicitly,
is that they are also curious about new things and willing to accept challenges, although this curiosity is frequently mingled with anxiety and fear.11 What’s more, even
for grown-ups, this mixture of curiosity and fear gives them a feeling of venture and
exploration and consequently, a special kind of pleasure. From the perspective of the
evolutionary theory, this subtle psychology can be explained easily as a result of natural
selection. Only possessing this state of psychology can we inherit the experience and
knowledge of our ancestors, avoid being too risky, and at the same time properly deal
with uncertainty and the unknown and create new knowledge.12 And as the children
11
Hide-and-seek, a popular game loved by almost all small babies, is a good evidence. Below is
another more interesting evidence. Many new parents have the experience that when babies, around
the age of one year old, learn how to walk, they are fond of practicing it time and time again. During
this process, they enjoy the challenge a lot and usually hate the holding of adults. However, once they
master the skills and walking is not challenging any longer (at around one and a half), they become
“lazy” and tend to “conserve energy”!
12
As is well known, human being, just like any other species, has two basic tendencies, namely seek
pleasure and avoid pain. To make right decisions, we need to have the knowledge of which brings
us pleasure and which brings pain. Thus in a deterministic environment, we can always use pure
strategies to maximize our fitness. In dealing with uncertainty, however, no gene coded pure strategy,
neither “seek” nor “avoid”, is universally wise, because the former may lead to disasters and the latter
5
grow up, they are more and more used to new things, and fear becomes more and more
feeble.13 This logic on every single person can be extended to the human being as a
species, where the process of growing up is compared with the process of civilization
(see Simmel, 1904, pp. 137-138).
Also, lower class people can be compared with children and upper class people
with grow-ups, because upper class people are usually more civilized through better
education. This is the first reason why the second tendency is embodied more strongly
in the upper class than in the lower class. The second reason is that upper class people,
in general, have more access to new things. As quoted from Simmel (1904, p. 138),
“[...] the lower classes possess very few modes and those they have are seldom specific”.
The above two advantages of the upper class are becoming less and less sharp, with
the astonishingly fast development of economy and consequent consumption revolutions. This trend, not surprisingly, has been noticed by many scholars in history. For
instance, Galbraith (1958, pp. 72-73) stated that “lush expenditure could be afforded
by so many that it ceased to be useful as a mark of distinction.” Canterbery (1998, p.
148) observed that, “the middle class could now emulate the rich in dress and even in
automobiles, especially as the rich downsized to Volvos”.14
Although the distinction between upper and lower classes in fashion could be blurred
by the development of the economy, the elementary logic of Simmel (1904) that the
two antagonistic tendencies are coded in every individual, however, is still largely
true and will be valid even in the expected future. Time erodes this really slowly.
Also, it is reasonable to assume that, in general, the two tendencies are not equal
for each individual. After all, diversity is universal and one of the ultimate sources
for the development of the society. For those the first tendency is stronger, we call
them conformists; and for those where the opposite is true, we call them rebels. This
distinction reflects people’s different personalities. As also noticed by Simmel (1904,
p.131), in the class-based model, “[fashion] signalizes the lack of personal freedom”.
We believe that in a world where people’s personalities are more and more freed, the
“personality differentiation” model will be more and more suitable than the “class
may miss opportunities. They are both too risky. Natural selection determines that we must take
some strategy in-between, i.e. a “mixed” strategy, when dealing with uncertainty. Thus the mingled
feeling of both fear and excitement towards uncertainty is evolved. The above viewpoint may serve as
a new defence for the widely used, yet not always very clear and sometimes even debatable, concept
of mixed strategy in game theory.
13
Another observation that supports this argument is that in the minds of grown-ups, “interesting”
implies new and unexpected. When someone says that something is interesting, this thing is almost
always new to him/her. This is especially true in the academical world and when we talk jokes.
14
The two examples are both re-quoted from Trigg (2001, p. 103).
6
differentiation” model for the study of fashion.15
1.2
What causes fashion cycle?
Fashion cycle refers to the phenomenon that the most popular choice in a specific
field periodically changes and reappears. Anecdotal observations on the existence of a
fashion cycle include fashion color and length of girls’ skirts, can be found throughout
history.16 Fashion cycle is regarded as the most prominent and the most mysterious
fact about fashion, and it is one of the key focuses in previous studies (e.g., Pesendorfer,
1995; Young, 2001; Yoganarasimhan, 2012a).
It is safe to argue that fashion cycle is caused ultimately by a combined force of
consumers and producers. However, the concrete process may be quite complicated,
and the task of theoretical as well as empirical studies is to reveal more details of
this process. In the model of Pesendorfer (1995), fashion cycle is determined and
completely controlled by a monopolist producer. The problem Pesendorfer attacked
is not on causes of fashion cycle, but how to control it optimally. Although the main
assumption that everything is decided by a powerful producer is obviously unrealistic,
because very rarely can a producer be so powerful17 , this theory provides us new
understandings of the fashion cycle on the producer side. On the consumer side,
motivationally, fashion cycle may come out of the needs for class signaling (Simmel,
1904), wealth signaling (Veblen, 1899), cultural capital signaling (Bourdieu, 1984),
taste signaling (Yoganarasimhan, 2012b), or personality signaling.
1.3
Approach of this paper
In this paper, we try to analyze the causes of fashion cycle from a structural point of
view. Since fashion is generated through people’s interactions and these interactions
are usually restricted within friends (colleagues, neighbors), it is natural to analyze
fashion through a social network. After all, few people care what a stranger wears
15
It should be emphasized that in the classless model, people have no absolute freedom either. It
seems that fashion will not disappear as long as the society exists, and hence people will always affect
each other and there will be no absolute freedom. However, in a classless model, people’s choices,
w.r.t. chasing fashion, are not affected by their social statuses. In this sense, they are more free than
in the class-based model.
16
However, as pointed out by Yoganarasimhan (2012b), “there exists no systematic empirical investigation of the phenomenon of fashion. Indeed, the even more basic question of how to detect fashion
cycles in data has not been studied.”
17
A well-known example in history is the vain effort in 1922 by not only manufacturers but also the
media to reverse the trend towards shorter skirts. See e.g., Blumer (1969).
7
and what they do in pursuing fashion are mainly for setting a particular image or
personality in the eyes of their friends.
Our model is a fashion game adapted from Young (2001, p. 38) and Jackson (2008,
p. 271). We focus on the evolution of fashion from the perspective of consumers only.
There are two types of agents, conformists and rebels, located on a fixed social network.
Conformists like to match the majority action of her neighbors while rebels prefer to
mismatch. Each agent has two available actions, named 0 and 1, such as white or
black skirt, buying or not buying an iphone, etc. With these simplifying assumptions,
interactions between agents can be conveniently modeled by certain familiar two-player
stage games (see Section 2 for formal description).
To study fashion cycles, there must be an underling dynamic process. In Andreozzi
and Bianchi (2007), the process is determined by the log linear response and in Young
(2001), it is determined by fictitious play. In this paper, we adopt another simple
dynamic process; namely, the best response dynamic (BRD in short).18 Under this
dynamic, agents are naive and myopic, in the sense that they only have one step
memory and do not look forward.19 This dynamic is consistent with the empirical
experience that people’s behaviors in fashion are usually not so complicated.20
We consider three kinds of network structures, namely lines, rings, and stars. Our
results show that the emergence of a fashion cycle relies crucially on the network
structures. Specifically, when the social network is a line or a ring, the BRD will
almost always converge to a Nash equilibrium. The convergence is fast and thus, when
the interaction structure is a line or a ring, the system will evolve to a steady state in a
very short time where no flux or fashion cycle is possible. When the network is a star,
however, for approximately half of all the configurations of agents’ types, a fashion
cycle of length 4 will always emerge, regardless of the initial action profiles.
Our research falls into the booming field of network games (see Jackson and Wolinsky, 1996; Goyal, 2007; Vega-Redondo, 2007; Jackson, 2008; Jackson and Zenou, 2014).
18
Both sequential BRD and simultaneous BRD are considered.
We remind the reader that in fictitious play, each agent should remember all the history that she
is involved. Actually, BRD is exactly the one-step-memory simplification of fictitious play.
20
In fact, irrational and even crazy are often associated with fashion, as pointed out by Blumer
(1969). We believe that there is a mixture of rationality and irrationality in the phenomenon of
fashion. However, we are not able to study irrationality that has no patterns. In this sense, a
bounded rationality model, which assumes neither absolute rationality nor patternless irrationality, is
quite proper. In literature, herding models are also used to study the phenomenon of fashion. The word
“herding”, historically, indicates irrational, although the most famous such models assume rationality
of agents (in fact, they are Bayesian learning models, and hence agents must be extremely smart).
See Banerjee (1992) and Bikchandani et al. (1992). It should be noted that only the conforming side
of fashion can be explained in the above two models.
19
8
It can be taken as a new application of this field.21 In addition, the fashion game also
has its own noval theoretical features; namely, it is a mixed model of a network game
with strategic substitutes and a network game with strategic complements, the two most
extensively studied classes of network game models.22 Indeed, for conformists, it exhibits a simple form of strategic complement and for rebels a simple form of strategic
substitute works. A recent model of Hernández et al. (2013) has also strategic complements and strategic substitutes. However, for a particular set of parameters, only
strategic substitute or complement but not both works. In comparison, except for two
extreme cases, both work simultaneously in our model. For fashion games with conformists only, Goles (1987) shows that simultaneous BRD always converges to a cycle
of length at most 2 (see also Berninghaus and Schwalbe, 1996). This result is recently
extended to more general models with arbitrary thresholds (Adam et al., 2013). Cannings (2009) notices that the result of Goles holds for fashion games with rebels only
too. He also observes that when the network structure is complete and the underlying
dynamic is simultaneous BRD, the only possible limit cycle lengths are 1, 2 and 4.
The computational properties of the fashion game adopted in this paper has been
recently investigated in Cao and Yang (2011). They show that testing whether a
fashion game possesses a pure Nash equilibrium is in general NP-hard23 , meaning that
predicting fashion trends might be impossible.24
The rest of the paper is organized as follows. The next section introduces the model
formally. Section 3 discusses briefly the factor of heterogeneity. Section 4 presents main
results of this paper. Section 5 concludes this paper with several further discussions.
Proofs of the main results are organized in Append A and two technical lemmas used
in the proofs are presented in Appendix B.
21
Other notable applications include employment (Calvó-Armengol and Jackson, 2004, 2007, 2009),
education (Calvó-Armengol et. al., 2009), crime (Calvó-Armengol et. al., 2007), and R & D (Goyal
and Moraga-Gonzales, 2001) etc. See also the review of Jackson and Zenou, 2014.
22
See e.g. Bramoullé (2007), Bramoullé and Kranton (2007), Hernández et al. (2013). See also the
review in Jackson (2008), and Jackson and Zenou (2014).
23
Definition of this concept can be found in any textbook about computational complexity. Roughly
speaking, a problem is NP-hard means that there seems to be no way to compute a solution in polynomial time. Since only polynomial time algorithms can be implemented efficiently by the computer,
an NP-hard problem can be equivalently taken as unsolvable when the problem size is large enough.
Thus the result of Cao and Yang (2011) tells us that when the social network is moderately large, it
is impossible for us to determine whether it possess a PNE or not.
24
This conjecture is supported by industrial practices from fast fashion. In Zara, for example,
“leading” or even “predicting ” the fashion trends has already been given up. Hundreds of its designers
fly around the world, watch the latest hot factors in people’s wearing in Paris, New York etc., and
merge these factors into their designs in extremely short times. This kind of “quick response” is one
of its core competencies.
9
0
1
0
1, 1
−1, −1
1 −1, −1
1, 1
0
1
0 1, −1 −1, 1
1 −1, 1 1, −1
0
1
0 −1, −1
1, 1
1
1, 1
−1, −1
Figure 1: Three base games. (1) (Pure) Coordination Game (left): conformist V.S.
conformist; (2) (Pure) Anti-coordination Game (right): rebel V.S. rebel; (3) Matching
Pennies Game (middle): conformist V.S. rebel.
2
The Fashion Game
Suppose two conformists meet. Since they both prefer to conform, i.e. match the action
of their respective opponents, they naturally play the coordination game. Similarly,
when two rebels meet, they both like to mismatch, and hence they play the anticoordination game.25 The most interesting case is when a conformist meets a rebel.
It is worth emphasizing that this case can be modeled by the matching pennies game.
The Row player prefers to match the Column player’s action and hence he can be
regarded as a conformist. In contrast, the Column player likes to mismatch the Row
player’s action and so he can be treated as a rebel. Throughout the rest of this paper,
we assume that those three games are all symmetric, with payoffs as shown in Figure
1.26
We assume that each agent takes the same action in all the local games she plays.
For instance, if a conformist agent has five neighbors, three of whom are conformists
and the other two are rebels, then she plays three pure coordination games and two
matching pennies games. In all the five games, however, either she always chooses
action 0 or always action 1 (we do not consider mixed strategies in this paper). This
restriction is quite common in the study of network games. Intuitively, it says that
each agent should show the same image in front of all her neighbors. The total payoff
of each agent is simply the sum of the payoffs that she receives in all the games she
plays.
An alternative way to understand our network game is that the payoff that each
25
In evolutionary game theory, it is usually called the hawk-dove game. In the field of social
physics, a closely related and extensively studied model is the “minority game” (Challet and Zhang,
1997; Challet et. al., 2004). This model, also known as the El Farol Bar problem, is proposed
by economist Arthur (1994). Though minority games have been applied in the study of financial
markets early enough, Marsili (2001) noticed that people do not necessarily play any minority game in
stock markets, because except for a few “market fundamentalists”, most people are “trend followers”.
Instead, a minority-majority game is more appropriate.
26
Our main results can be easily extended to more general payoff settings.
10
agent receives is the social value of the corresponding action, where the social value of
a particular action for an agent is defined as the number of neighbors who choose an
action that she likes minus the number of neighbors whose action she dislikes (see the
companion paper for more discussions). Note that for a conformist who chooses action
0, neighbors choosing action 0 are liked by her and those choosing action 1 are disliked
by her. In contrast, for a rebel who chooses action 0, action 1 neighbors are liked and
action 0 ones are disliked.
2.1
Formal description
Throughout this paper, we use {C, R} to denote the type set. For each agent i,
Ti ∈ {C, R} is her type: Ti = C means that she is a conformist and Ti = R a rebel.
A fashion game is represented by a triple G = (N, E, T ), where
• N = {1, 2, · · · , n} is the set of agents;
• E ⊆ N × N is the set of links;
• T = (T1 , T2 , · · · , Tn ) ∈ {C, R}N is the configuration of agents’ types.
The underlying graph g = (N, E) is undirected. That is, ij and ji represent the same
link. We say that g is a complete graph if and only if ij ∈ E for all i, j ∈ N and i 6= j.
We say that g is a line if agents can be indexed such that E = {12, 23, · · · , (n − 1)n},
a ring if E = {12, 23, · · · , (n − 1)n, n1}. g is called a star network if there exists an
agent i such that E = {ij : j ∈ N, j 6= i}, in which case i is called the central agent
and others are called peripheral ones.
Two agents i, j ∈ N are neighbors to each other if and only if ij ∈ E. We also use
Ni to denote the set of neighbors of agent i:
Ni = {j ∈ N : ij ∈ E}.
In simultaneous BRD, time elapses discretely. Players’ types do not change over
time, but their actions may change. We use
x(0) = (x1 (0), x2 (0), xn (0)) ∈ {0, 1}N
11
to denote the initial action profile, and define xi (t) ∈ {0, 1} as the action of agent i at
time t.27 Li (x(t)) ⊆ Ni is the subset of neighbors at time t that are liked by agent i.
Since conformists like those taking the same action, and rebels the opposite, we have
8
<
Li (x(t)) = :
{j ∈ Ni : xj (t) = xi (t)} if Ti = C
.
{j ∈ Ni : xj (t) 6= xi (t)} if Ti = R
Set Di (x(t)) = Ni \ Li (x(t)), which is the subset of neighbors that are disliked by
agent i at time t. According to the payoff settings in Figure 1, agent i gains 1 unit from
interacting with each neighboring agent in Li (x(t)) but loses 1 unit from interacting
with each of those in Di (x(t)). Thus, given action profile x(t) = (x1 (t), · · · , xn (t)) at
time t, the time-t utility of agent i is given by
ui (x(t)) = |Li (x(t))| − |Di (x(t))|.
2.2
Fashion cycle
Under simultaneous BRD, agent i switches her action at time t + 1 if and only if
ui (x(t)) < 0, because her utility will be −ui (x(t)) if she switches, assuming that her
neighbors do not change their actions. That is, simultaneous BRD evolves with time
as follows:
8
< 1 − x (t), if u (x(t)) < 0
i
i
xi (t + 1) = :
.
xi (t),
otherwise
Under simultaneous BRD, we use the limit cycle, a basic concept from discrete
dynamical systems, to study the phenomenon of fashion cycles. Intuitively, a limit
cycle is an ordered set of action profiles such that starting from each of these profiles
the system will evolve into its immediate successor. Since there are a total of finitely
many action profiles in the fashion game and simultaneous BRD is deterministic, after
finitely many steps some profile will reappear. From that point on, everything will be
repeated just as that state is reached for the first time. Therefore, limit cycle always
exists and it is unique for any given fashion game G = (N, E, T ) with a fixed initial
action profile x(0). In the rest of this paper, we also use I = (N, E, T, x(0)) to denote
an initialized fashion game. Below is the mathematical definition of fashion cycle.
27
In this paper, agents’ strategies are simply their pure actions. Neither mixed actions nor other
sophisticated strategies are allowed. And hence we use “action” and “action profile” in replace of
“strategy” and “strategy profile”, respectively.
12
Figure 2: Limit cycle of the matching pennies game. Circle stands for conformist,
triangle for rebel. Black color indicates that the agent chooses 1 and white for 0.
Definition 1 Let I = (N, E, T, x(0)) be an initialized fashion game, and t(I) ≥ 1 be
the first time such that
x(t(I)) ∈ {x(0), x(1), · · · , x(t(I) − 1)},
and
x(t) ∈
/ {x(0), x(1), · · · , x(t − 1)}, ∀1 ≤ t < t(I).
Let r(I) < t(I) be the time such that x(t(I)) = x(r(I)). We call the set of states
{x(r(I)), x(r(I) + 1), · · · , x(t(I) − 1)} the limit cycle of I.
In the rest of this paper, we use l(I) = t(I) − r(I) to denote the length of the limit
cycle of I. When time goes beyond r(I), we say that simultaneous BRD enters the
limit cycle.
For an initialized fashion game I, it is clear that a pure strategy Nash equilibrium
(PNE for short) will be eventually reached via simultaneous BRD if and only if l(I) = 1,
in which case we also say that simultaneous BRD converges. We show in Section 4
that if the underlying network g = (N, E) is a line or a ring, then for almost all
configurations of agents’ types T , simultaneous BRD converges for all initial action
profiles. That is, fashion will eventually stop evolving. We show that l(I) ∈ {1, 2, 4}
if g = (N, E) is a line or a star network.
2.3
Several simple facts
As illustrated in Figure 2, the matching pennies game always has a fashion cycle of
length 4 under simultaneous BRD. Proposition 1(c) in Subsection 4.2 can be taken as
a generalization of the above simple fact.
Before proceeding to the next section, we introduce several more simple facts about
the fashion game. As is well-known, the matching pennies game does not have any
13
PNE, but it has a unique mixed Nash equilibrium, in which each agent plays actions
1 and 0 with equal probability. It turns out that playing actions 1 and 0 with equal
probability is also a mixed strategy equilibrium for the fashion game. Jackson (2008)
points out that the case where each agent has no less conformist neighbors than rebel
ones always has a PNE.28 If we only require that each conformist has at least as many
conformist neighbors as rebel ones, PNE can still be guaranteed (Zhang et. al., 2013).
3
The Role of Heterogeneity
Given a fashion game G = (N, E, T ), if Ti = Tj for all i, j ∈ N , i.e. either all agents are
conformists or all agents are rebels, we say that G is homogeneous. In the conformist
homogeneous case, PNE is obviously guaranteed. All agents choosing 1 and all agents
choosing 0 are the two most natural ones.29 The other homogeneous case, where all
agents are rebels, have PNEs as well.30 It can be shown that the two homogeneous
cases are both exact potential games, with the potential function given by the sum
of agents’ payoffs.31 Hence, according to the results of Monderer and Shapley (1996),
PNE will eventually be reached through a sequential BRD (with arbitrary orders).
This means that under the dynamic of sequential BRD, fashion will eventually come
to an end, unless the two types of agents both exist, implying that the heterogeneity
of consumers is a critical factor to sustain the emergence of fashion cycles.
In fact, the necessity of the heterogeneity of agents has already been noted by
Simmel. To be exact, Simmel (1904, p. 138) wrote32
“Two social tendencies are essential to the establishment of fashion, namely,
28
It can be checked easily that all conformists choosing 1 and all rebels choosing 0 is a PNE.
But it should be noted that there may well be other ones. Actually, it can be proved easily that
the set of PNEs for the homogeneous fashion game with only conformists form a lattice (see Exercise
9.5 of Jackson, 2008). Thus the maximum PNE and the minimum PNE are the two natural focal
equilibria. The other way to refine them in previous related studies in social learning is is to use
the concept of stochastic stability instead of Nash equilibrium (Foster and Young, 1990). Since the
forming of conventions can be taken as a process of coordination, this learning theory is also widely
used to study the evolution of conventions. See e.g. the classical paper of Young (1993).
30
In the rebel homogeneous fashion game, its PNE corresponds to a well studied concept in graph
theory, namely the “unfriendly partition” (see Aharoni et. al., 1990).
31
This can be verified easily. For the general anti-coordination game, which is an extension of the
fashion game with only rebels, it is still the case. This is a well known result (see Bramoullé, 2007).
32
This viewpoint is also often emphasized by subsequent scholars. For instance, Benvenuto (2000)
argued that, “Undoubtedly, some of us tend more towards imitation (and thus to conformism) while
others tend to distinction (and thus to eccentricity and dissidence), but fashion’s flux needs both of
these contradictory tendencies in order to work.”
29
14
the need of union on the one hand and the need of isolation on the other.
Should one of these be absent, fashion will not be formed—-its sway will
abruptly end.”
Should the insightful observation of Simmel be verified rigorously, our understanding of the two forces of conforming and rebeling would be significantly improved: On
the one hand, it is intuitive that conforming, just like positive feedback and contagion,
is likely to lead a system to extreme states. On the other hand, rebeling, just like
negative feedback, tends to lead a system to the middle, or “doctrine of the mean”.33
Each one alone will make a system eventually stable. It is the combination of the
two forces that makes a system keep in flux. In the companion paper (Zhang et. al.,
2013), we further investigate the above logic by showing that for the emergence of
a fashion cycle, these two forces may have to be “fully mixed”, namely the interactions between conformists and rebels (the matching pennies games played) should be
frequent enough.
Complication comes when we consider simultaneous BRD, because neither the rebel
homogeneous case nor the conformist homogeneous case will definitely reach a PNE,
although the limit cycles can be of length at most two (Goles, 1987; Berninghaus and
Schwalbe, 1996; Cannings, 2009). It is not yet clear to us as to how likely a limit cycle of
length two can be reached, which we will carry out in a future project. However, even if
it can be shown that PNE is reached almost always under simultaneous BRD when the
fashion game is homogeneous, there is still great space to defend Simmel’s observation,
because it is very unlikely in reality for all the agents to always move simultaneously,
and our definition of fashion cycle through limit cycle is too restrictive. For instance,
when all but two agents are always satisfied in the limit cycle, it should be taken as
that fashion cycle does not occur. See Subsection 5.1 for more discussions.
4
The Role of Interaction Structures
To ease the presentation of our main results, we need several new terms.
33
See Cao et al. (2011) for the exact result on a continuous learning model. A similar effect is also
found by Galam (2004) in a voting model, where the term “contrarian” is used instead of “rebel”.
15
4.1
Three more terms
Definition 2 Given a fashion game G = (N, E, T ), if Ti 6= Tj for all j ∈ Ni and
i ∈ N , i.e. the neighbors of any conformist are all rebels and the neighbors of any rebel
are all conformists, we say that G is cross-heterogeneous.
Obviously, if G is cross-heterogeneous, then the underlying network g = (N, E)
is a bipartite graph with one side all conformists and the other side all rebels.34 If
g = (N, E) is a line or a ring network, then cross-heterogeneous means that conformists
and rebels are alternate. The concept of cross-heterogeneous, as well as homogeneous,
obviously can also be derived naturally from the homophily index. See the companion
working paper (Zhang et. al., 2013) for extensive discussions.
Definition 3 Let G = (N, E, T ) be a fashion game, and the underlying network g =
(N, E) be a line. A cluster of G is defined as any of the following components:
(i) the left most two agents that are of the same type;
(ii) the right most two agents that are of the same type;
(iii) three adjacent agents that are of the same type.
We say that G is clustered if it has at least one cluster. Otherwise, we call it
un-clustered.
For rings, we can define the concept of cluster and clusteredness similarly. The only
difference is that there are no endpoints in rings and thus the first two kinds of clusters
will never occur.
Suppose G = (N, E, T ) is a fashion game, and the underlying network g = (N, E)
is a line or a ring. Since when G is homogeneous, PNE is always guaranteed, it is
intuitive that clusters may be in favor of the convergence of BRD. This is indeed the
case. Actually, agents in a cluster are able to help each other and hence have no
incentive to deviate from the current state no matter what the other agents choose.
Clusters are like “anchors” in the evolution of fashion and their effects are “contagious”,
so that eventually all agents will stop switching and the system converges (Proposition
1(b)).
Definition 4 Given an initialized fashion game I = (N, E, T, x(0)), we say that I is
badly initialized if ui (x(0)) < 0 holds for all i ∈ N .
34
A graph g = (N, E) is called bipartite, if we can partition the node set N into two parts, N 1 and
N , such that the two nodes of each edge are from different parts. That is, g = (N, E) is bipartite if
and only if E ⊆ N 1 × N 2 . This concept can be found in any textbook about graph theory.
2
16
L1
L2
L3
L4
L5
L6
Figure 3: Illustration of homogeneousness, cross-heterogeneousness, clusteredness, and bad initialization. Circles stand for conformists, and triangles for
rebels. Black color indicates that the agent chooses 1, and white 0. Gray means that initial action is not given. L1: homogeneous and badly initialized; L2: homogeneous but
not badly initialized; L3, L4: clustered but not homogeneous; L5: cross-heterogeneous;
L6: un-clustered but not cross-heterogeneous.
Intuitively, a bad initialization is the most terrible scenario we can imagine. No
agent is satisfied initially, so they switch to the other actions simultaneously, leading to
a new state where no agent is satisfied either. In the next round, all agents switch back
to the initial state. The system keeps oscillating between the two states, all agents are
in eternal flux and no agent is ever satisfied (Proposition 1(a)). Evidently, coordination
failure is caused by simultaneous moves.
Straightforwardly, a line or a ring must be homogeneous to be badly initialized.35
For lines and rings, homogeneousness is a special case of clusteredness, cross-heterogeneousness
implies un-clusteredness, and a badly initialized fashion game must be homogeneous.
See Figure 3 for an illustration of these concepts.
4.2
4.2.1
Simultaneous BRD
Lines
Clearly, limit cycles are always of length two for badly initialized fashion games. For
line structures, the following proposition indicates that this is the only case. That is,
if the underlying network is a line, then the fashion game has a limit cycle of length
35
Note that this cannot be extended to general structures.
17
two if and only if it is badly initialized.36
Proposition 1 Let G = (N, E, T ) be a fashion game with the underlying network
g = (N, E) being a line. x(0) is an initial action profile, and I = (N, E, T, x(0)) the
corresponding initialized fashion game.
(a) If I is badly initialized, then l(I) = 2;
(b) If G is clustered and I not badly initialized, then l(I) = 1;
(c) If G is cross-heterogeneous, then l(I) = 4 for all x(0);
(d) If G is un-clustered, but not cross-heterogeneous, then l(I) ∈ {1, 4} for all x(0).
Proof to the above proposition is presented in Appendix A.1. When the underlying
network g = (N, E) is a line, we know from the above proposition that to have a limit
cycle of length two, it must be that the limit cycle is entered immediately at the very
beginning of the evolution. There is no possibility to evolve for certain periods of time
and then enter a length two limit cycle.37 For general network structures, when the
system evolves into a limit cycle of length two, there is still a possibility that some, and
even many, players are always satisfied and never deviate in the limit cycle. Proposition
1 tells us that this never happens for the line structures.
Below is an immediate corollary of Proposition 1, stating that the lengths of limit
cycles of the fashion game are upper-bounded when the underlying network is a line.
Corollary 1 Let G = (N, E, T ) be a fashion game. If the underlying network g =
(N, E) is a line, then l(I) ∈ {1, 2, 4} for all x(0) ∈ {0, 1}n .
To put the above corollary another way, no matter how many agents there are in
the fashion game, the lengths of limit cycles under simultaneous BRD cannot exceed
4, as long as the interaction structure is a line. It is worth noting that we are not able
to extend this result to general structures.38
As shown in Cao and Yang (2011), when the underlying network g = (N, E) is a
line, PNE exists for G = (N, E, T ) if and only if the line is not cross-heterogeneous. The
structure of cross-heterogeneous lines is a natural extension of the matching pennies
game, a dyad composed of a conformist and a rebel. Under this structure, matching
36
It seems that this may also be true for ring structures. See Proposition 2 and discussions in
Subsection 5.2.
37
For general networks, however, this may well happen.
38
Not even possible for ring structures. See Subsection 5.2 for discussions.
18
pennies is the only base game. Hence Proposition 1(c) can be taken as a generalization
of the simple fact that the matching pennies game always has a limit cycle of 4 under
simultaneous BRD.
Combining Proposition 1 and the result of Cao and Yang (2011) that is introduced
in the previous paragraph, we have an immediate interesting corollary.
Corollary 2 Let G = (N, E, T ) be a fashion game with the underlying network g =
(N, E) being a line. If G does not possess a PNE, then l(I) = 4 for all x(0).
Suppose that each agent is randomly assigned a type (without bias), then it follows
easily that the probability that a line is clustered goes to 1 as n goes to infinity.39
Since the probability that a line is homogeneous is 2/2n , Proposition 1(b) implies that
for almost all configurations of agents’ types on lines, simultaneous BRD converges to
a PNE, regardless of the initial action profiles (notice again that badly initialized is
stronger than homogeneous). We present this result formally as a corollary of Proposition 1. Note that the implicit probability space is determined by unbiased coins
flipping.
Corollary 3 Let G = (N, E, T ) be a fashion game. If the underlying network g =
(N, E) is a line, then
lim Prob. [G : l(I) = 1, ∀x(0) ∈ {0, 1}n ] = 1,
n→∞
where I = (N, E, T, x(0)) is the initialized fashion game with an initial action profile
x(0).
4.2.2
Rings
The following proposition for the ring structure is parallel to part (a) and (b) of Proposition 1.
Proposition 2 Let G = (N, E, T ) be a fashion game with the underlying network
g = (N, E) being a ring. x(0) is an initial action profile, and I = (N, E, T, x(0)) the
corresponding initialized fashion game.
(a) If I is badly initialized, then l(I) = 2;
(b) If G is clustered but I not badly initialized, then l(I) = 1.
39
This is true because the probability that there are three or more adjacent agents that are of the
same type goes to 1, as n tends to infinity.
19
Proof to the above proposition is also presented in Appendix A.1. It is worth noting
that when g = (N, E) is a ring, we cannot derive that l(I) ∈ {1, 2, 4}. See Subsection
5.2 for more discussions. However, the following corollary is still obviously true from
Proposition 2(b).
Corollary 4 Let G = (N, E, T ) be a fashion game. If the underlying network g =
(N, E) is a ring, then
lim Prob. [G : l(I) = 1, ∀x(0) ∈ {0, 1}n ] = 1.
n→∞
4.2.3
Stars
Combining Corollary 3 and Corollary 4, we know that when the underlying network
is a line or a ring, then almost always fashion evolves into a steady state through
simultaneous BRD. When the network is a star, however, the above statement is not
true. First of all, let’s see the following complete description for the simultaneous BRD
on star networks.
Proposition 3 Let G = (N, E, T ) be a fashion game with the underlying network
g = (N, E) being a star. x(0) is an initial action profile, and I = (N, E, T, x(0)) the
corresponding initialized fashion game.
(a) If more than half of the peripheral agents are of the different type as the central
agent, and initially the central agent is unsatisfied, then l(I) = 2;
(b) If more than half of the peripheral agents are of the same type as the central
agent, then l(I) = 4;
(c) In all the other cases, l(I) = 1.
Proof to the above proposition is presented in Subsection A.2.
Parallel to Corollary 1 for the line networks, we have the following corollary for the
star networks, which is immediate from Proposition 3.
Corollary 5 Let G = (N, E, T ) be a fashion game. If the underlying network g =
(N, E) is a star, then l(I) ∈ {1, 2, 4} for all x(0) ∈ {0, 1}n .
As before, suppose each agent is randomly assigned a type, being a conformist and
a rebel equally likely. From the above proposition, it can be calculated easily that the
20
probability that the fashion game with a star network always has a limit cycle of length
4 (case (c) and case (d)) is approximately 1/2, when n is moderately large. Whether
the limit cycle has a length of 1 or 2, however, depends on both the configuration
of types and the initial actions.40 If we assume further that the initial actions are
chosen randomly (without bias) by all agents, then it can be shown easily that with a
probability approximately 1/4 the fashion game with a star network goes into a limit
cycle of length 2, and it converges with a probability approximate to 1/4.
Corollary 6 Let G = (N, E, T ) be a fashion game. If the underlying network g =
(N, E) is a star, then
lim Prob. [G : l(I) = 4, ∀x(0) ∈ {0, 1}n ] = 1/2,
n→∞
lim Prob. [I : l(I) = 2] = 1/4,
n→∞
lim Prob. [I : l(I) = 1] = 1/4.
n→∞
Before presenting the result parallel to Corollary 2, we need a simple lemma.
Lemma 1 Let G = (N, E, T ) be a fashion game. If the underlying network g = (N, E)
is a star, then G possesses a PNE if and only if either (i) the central agent is a
conformist and at least half of the peripheral ones are also conformists, or (ii) the
central agent is a rebel and at least half of the peripheral ones are also rebels.
Proof to the above lemma goes as follows. If condition (i) in the above lemma
holds, we let all the conformists choose 0 and all rebels choose 1. It can be checked
easily that this is a PNE. In fact, this is a special case of that considered by Jackson
(2008). If condition (ii) is valid, we let the central rebel choose 0, the peripheral rebels
choose 1, and all the conformists choose 0, then it can be seen that this is a PNE.
There are two left cases: (iii) the central agent is a conformist and more than half of
the peripheral agents are rebels, and (iv) the central agent is a rebel and more than half
of the peripheral agents are conformists. Case (iii) cannot have any PNE, because we
are not able to make both the central conformist and all the peripheral rebels happy.
Similarly, case (iv) does not admit a PNE either. This completes the proof.
40
Except for the case that there are equal number of peripheral rebels and peripheral conformists,
which always leads to l(I) = 1. See the proof to (c) in Subsection A.2. It is notable that this can
only happen when n is odd. And as n tends to infinity, the probability that this happens is zero.
21
Combining Lemma 1 and Proposition 3, we immediately have the following corollary.
Corollary 7 Let G = (N, E, T ) be a fashion game with the underlying network g =
(N, E) being a star. If G does not possess a PNE, then l(I) = 4 for all x(0).
4.3
Sequential BRD
In this subsection, we discuss properties of the sequential BRD. Let I be a given
initialized fashion game. In each time step, as long as PNE is not reached, we let an
arbitrary unsatisfied agent switch her action. We say that sequential BRD converges if
and only if PNE can be definitely reached for any realized deviation order. That is, we
consider exactly the same property as for potential games, although our proofs do not
rely on explicit constructions of potential functions. We use this convergence concept
to study the fashion game on lines and rings. For the star networks, we use a slightly
weaker concept, which will be introduced soon.
Notice first that the fashion game, in general, is not a potential game for the line
network or the ring network or the star network, because in none of the three cases
can PNE be guaranteed, while a potential game always possesses a PNE. However, for
a very wide class of situations, the fashion game with one of the three kinds of network
structures is indeed a potential game.
4.3.1
Lines and rings
Proposition 4 Let G = (N, E, T ) be a fashion game with the underlying network
g = (N, E) being a line. Then sequential BRD converges for all initial action profiles
if and only if G is not cross-heterogeneous.
Proposition 5 Let G = (N, E, T ) be a fashion game with the underlying network
g = (N, E) being a ring. If G is not cross-heterogeneous, then sequential BRD converges
for all initial action profiles.
Proofs to the above two propositions are presented in Appendix A.3. Note that the
condition that G is not cross-heterogeneous is equivalent to that there are two adjacent
that are of the same type. This kind of component is weaker than a cluster. In the
previous subsection, we have seen that clusters play a critical role in the convergence
of a simultaneous BRD when the underlying network is a line or a ring. From the
22
above two propositions, it is natural to guess that sequential BRD is easier to converge
than simultaneous BRD. It is indeed the case for the cases we consider. See the next
subsection for more discussions.
The following corollary, which is similar to Corollary 3 and Corollary 4, is an
immediate result of Proposition 4 and Proposition 5.
Corollary 8 Let G = (N, E, T ) be a fashion game with the underlying network g =
(N, E) being a line or a ring. Then
lim Prob. [G : sequential BRD converges for all initial action profiles] = 1.
n→∞
We also have the following nice characterization for sequential BRD on lines.
Corollary 9 Let G = (N, E, T ) be a fashion game with the underlying network g =
(N, E) being a line, then the following statements are equivalent.
(a) PNE exists for G;
(b) G is not cross-heterogeneous;
(c) Sequential BRD converges for all initial action profiles;
(d) Sequential BRD converges for some initial action profile.
Proof to the above corollary is easy. In fact, the equivalence of (a) and (b) is the
result of Cao and Yang (2011) that we have used several times. The equivalence of (a)
and (c) is exactly the content of Proposition 4. That (c) implies (d) is straightforward.
When (d) holds, we must have PNE exists for G, i.e. (a) is valid, hence (c) is also true.
This completes the whole proof. When the network is a ring, unfortunately, we do not
have so nice a characterization.
4.3.2
Stars
Let I be a given initialized fashion game. In each time step, as long as PNE is
not reached, we uniformly choose an arbitrary unsatisfied agent, and let her switch
her action. We say that sequential BRD converges almost surely if and only if the
probability that PNE can be eventually reached is one. Obviously, this convergence
concept is weaker than that we used for lines and rings.
Proposition 6 Let G = (N, E, T ) be a fashion game. If the underlying network g =
(N, E) is a star, then sequential BRD converges almost surely for all initial action
profiles if and only if at least half of the peripheral agents are of the same type as the
central agent.
23
Proof to the above proposition, which is a direct application of absorbing Markov
chains, is provided in Appendix A.4. Given a star network g = (N, E), let each agent
uniformly choose a type, then the probability that at least half of the peripheral agents
are of the same type as the central agent tends to 1/2 as n goes to infinity. Combining
this fact with Proposition 6, we have the following corollary.
Corollary 10 Let G = (N, E, T ) be a fashion game with the underlying network g =
(N, E) being a star. Then
lim Prob. [G : sequential BRD converges almost surely for all initial action profiles] = 1/2.
n→∞
Similar to Corollary 9, we have the following nice characterization for sequential
BRD on star networks. The proof is trivial and thus omitted.
Corollary 11 Let G = (N, E, T ) be a fashion game with the underlying network g =
(N, E) being a star, then the following statements are equivalent.
(a) PNE exists for G;
(b) At least half of the peripheral agents are of the same type as the central agent;
(c) Sequential BRD converges almost surely for all initial action profiles;
(d) Sequential BRD converges almost surely for some initial action profile.
4.4
Simultaneous BRD V.S. sequential BRD
Below are the formal statements that convergence of simultaneous BRD is stronger
than that of sequential BRD when the underlying network is a line or a star.
Corollary 12 Let G = (N, E, T ) be a fashion game with the underlying network g =
(N, E) being a line. If simultaneous BRD converges for some initial action profile, then
sequential BRD converges for all initial action profiles.
Corollary 13 Let G = (N, E, T ) be a fashion game with the underlying network g =
(N, E) being a star. If simultaneous BRD converges for some initial action profile,
then sequential BRD converges almost surely for all initial action profiles.
Proofs to the above two corollaries are as follows. According to Corollary 9 and
Corollary 11, convergence of sequential BRD for lines is equivalent to the existence of
PNE, and almost sure convergence of sequential BRD for stars is also equivalent to the
existence of a PNE. However, convergence of simultaneous BRD implies the existence
of a PNE. Hence the two corollaries are valid.
24
5
Discussions
In this paper, we have investigated fashion through a heterogeneous network game
model. It is shown that the emergence of a fashion cycle relies crucially on the network
structures of social interactions.
5.1
Two weaknesses
In order to prove our results, we have defined the fashion cycle as the limit cycle in the
corresponding discrete dynamical system. This definition seems to be too restricted.
Empirically, fashion cycle describes the phenomenon that the same type of item appears as the most fashionable one again and again after regular periods of time. Only
the macro fact that “it becomes hot again” is needed, the micro requirement that
“everything comes back again” is excessive. Thus it might be more appropriate to
only look at the time series of the hottest action and define fashion cycle accordingly.
It should be noted that this new definition does not change our result that fashion
will evolve into a dead state almost surely for the line and ring structures, because
micro equilibrium always implies a macro steady state. Nevertheless, if we use this
new definition of a fashion cycle to study the impact of heterogeneity, as discussed in
Subsection 3, then Simmel’s viewpoint may well be favored.
The other weakness of the current paper is that we did not take into account errors
or mutations. That is, when people respond in the fashion dynamic, we assume that
they never make mistakes. This is obviously unrealistic, and more importantly, makes
us unable to refine the equilibria as done in standard stochastic stability analysis. Since
most of our main results (Proposition 1(b), Proposition 2(b), Proposition 4, Proposition
3 (c)(d)) do not depend on initial action profiles, once an agent makes an error during
the BRD process, at worst case we can take it as a restart of the evolution. When the
erring probability is tiny, the new evolution can reach a new limit cycle, before another
error is made, because it can be seen from our proof that the convergence is really fast.
More luckily, for the line and ring networks, the influence of an error can be blocked
by nearby clusters, and even clusters are all far away from the erring agent, it needs a
process to diffuse the impact. When errors occur after the convergence, obviously their
effect is even smaller and can be restored even faster. Hence mistakes can largely be
taken as local and are likely to have very small effect on our main convergence results.
Further studies are needed as to how errors may affect the selection of PNEs,.
25
5.2
Open problems
An obvious technical problem for further research is to give more concrete characterizations of part (d) of Proposition 1, i.e. to distinguish between the instances with
length 1 fashion cycles and those with length 4 ones when the lines are un-clustered
but not cross-heterogeneous.
The second technical problem is to give more complete description for the simultaneous BRD of the fashion game on rings. The main difficulty for rings is that there
are no ending nodes, and thus we cannot expect to have a result parallel to the critial
Lemma 8 (in Appendix B). A similar “cutting” technique (see Appendix A) can still
be applied, so the only case remains unsolved is the cross-heterogeneous one. This is
also the complicated case for lines (see Appendix A). When the number of agents n
is odd, the ring cannot be cross-heterogeneous. However, when n is even, simulations
show that simultaneous BRD on a cross-heterogeneous ring behaves very differently
with that on a cross-heterogeneous line. To be specific, (1) when n/2 is also even, all
the possible lengths of fashion cycles of simultaneous BRD are 1, 2, 4 and n, (2) when
n/2 is odd, all the possible lengths of fashion cycles of simultaneous BRD are 1, 2,
4 and 2n. Figure 4 illustrates two fashion cycles with lengths n and 2n, respectively.
This shows a great difference between simultaneous BRD on lines and simultaneous
BRD on rings. Although the latter structure is but slightly modified from the former,
their properties in the extreme case may vary a lot. To be exact, lengths of limit cycles
of simultaneous BRD on rings cannot be upper-bounded as on lines. In worst cases,
they can be as long as possible, as the number of agents n goes to infinity. In that case,
it is farfetched to say that fashion cycle still exists, or even that there is any regularity
in the evolution of fashion. This may also add to the mystery of fashion.
The third technical problem is to prove or disprove that when the underlying network is a star such that at least half of the peripheral agents are of the same type as
the central agent (recall from Lemma 1 that this is the sufficient and necessary condition for the existence of a PNE), then sequential BRD converges for all initial action
profiles, or equivalently, the corresponding fashion games are potential games. Note
that in this paper we have used a weaker concept, the almost sure convergence.
5.3
Other perspectives
Social learning. Our research is also closely related with the social learning theory,
where BRD and the ring structure are frequently used. Mathematically, many of the
26
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/
Figure 4: Fashion cycles on cross-heterogeneous rings. Circles stand for conformists, and triangles for rebels. Black color indicates that the agent chooses 1, white
for 0. In the left example, the ring has 8 nodes, and the fashion cycle is of length 8. In
the right example, the ring has 6 nodes, and the fashion cycle is of length 12. Unhappy
faces indicate that the corresponding nearby agents are unsatisfied.
social learning models are similar to the conforming side of the fashion game (or its
natural extensions). Accordingly, our paper contributes to the already well-studied
field.41 In addition, our model can be taken as an extension of the famous threshold
model of Granovetter (1978).42 Under the payoff settings of this paper, a conformist
prefers an action to the other if and only if it has been taken by one half or more of
her neighbors. In contrast, a rebel prefers an action if and only if at most one half of
her neighbors has chosen it. Hence the “thresholds” are 0.5 for all agents. The exact
meaning of threshold for conformists is the same as in Granovetter (1978). For rebels,
however, it is actually an “anti-threshold”.
Cellular automaton. Mathematically, simultaneous BRD of the fashion game is
a nonlinear discrete dynamical system and a boolean network, which in general are
notoriously hard to analyze. For the ring structure, the corresponding (simultaneous)
BRD is equivalent to a one-dimensional heterogeneous cellar automaton. More precisely, the evolution is governed by a combination of Rules 77 and Rule 232.43 Although
it is known to be very difficult in general to establish anything meaningful about the
41
See e.g. Ellison (1993), Kandori et. al. (1993), Anderlini and Ianni (1996), Morris (2000), Young
(2009), Chen et. al. (2012) for references.
42
In the fields of computer science and complex systems, the network extension of the original model
of Granovetter (1978) is known as the threshold automata network, see e.g. Goles (1987), Berninghaus
and Schwalbe (1996).
43
These rules are indexed by Wolfram (2002).
The two specific ones can
also be found at the websites http://www.wolframalpha.com/input/?i=rule+77 and
http://www.wolframalpha.com/input/?i=rule+232.
27
evolution of a cellar automaton (e.g, Ilachinski, 2001, p. 225), we have shown that the
corresponding cellar automatons in this paper behave nicely and the analysis is not
that formidably hard.
Strategic cooperation. It is widely accepted that competition and cooperation
are the two eternal topics in game theory. Zero-sum games are polar examples for competition, whereas common interest games are polar examples for cooperation. Among
the three base games of the fashion game, the pure coordination game and the pure
anti-coordination game are both common interest games, while matching pennies is
a zero-sum game. For general normal form games, competition and cooperation are
both embodied. This is a kind of vertical hybrid and is the key observation of Kalai
and Kalai (2012) in their definition of the coco value, a new and promising solution for
two player strategic games. The fashion game, on the other hand, is a kind of horizontal hybrid of competition and cooperation. Through simulations, Cao et. al. (2013)
find that in general people can reach an extraordinarily high level of cooperation in
small-world networks via the simple BRD. Rigorous analysis is still needed.
Matching pennies. We end this paper by remarking that our work shows a novel
value for the elementary game of matching pennies. Most other elementary two player
games, say prisoners’ dilemma, hawk and dove, and game of pig, etc., have found useful interpretations and far-ranging applications, and attracted numerous attention. In
contrast, much less attention has been paid to the matching pennies game. One of the
most important reasons is that for a long time in history, there existed no convincing and fundamental interpretations for this model.44 The insightful observations of
Young (2001) and Jackson (2008) that it closely relates to the phenomenon of fashion,
hopefully can open a door for reconsidering this model. It is by no means a toy model!
For more recent work on matching pennies from the perspective of behavioral science,
we refer the reader to Palacios-Huerta (2003), and Eliaza and Rubinstein (2011).
44
Another reason might be that matching pennies is not symmetric. Only symmetric two-player
games can be straightforwardly extended to multi-player and network scenarios. In evolutionary game
theory, this corresponds to single population games. To extend a two-player asymmetric game to the
multi-player scenario, there must be two populations. However, the inner-population game is usually
hard to define naturally. With the interpretation of coordination and anti-coordination, the above
difficulty is not a barrier for the matching pennies game.
28
A
Main Proofs
It is obvious that when the underlying network g = (N, E) is a line or a ring, ui (x(t))
can only possibly take five values, 2, 0, −2, 1 and −1. Precisely, when i ∈ N \ {1, n}
is a conformist,
8
>
>
<
ui (x(t)) = >
>
:
2 if xi−1 (t) = xi+1 (t) = xi (t)
,
0 if xi−1 (t) + xi+1 (t) = 1
−2 if xi−1 (t) = xi+1 (t) = 1 − xi (t)
and when i ∈ N \ {1, n} is a rebel,
8
>
>
<
ui (x(t)) = >
>
:
2 if xi−1 (t) = xi+1 (t) = 1 − xi (t)
.
0 if xi−1 (t) + xi+1 (t) = 1
−2 if xi−1 (t) = xi+1 (t) = xi (t)
The utility of agent 1 and agent n on a line can only take a value from {1, −1}.
For agents 1 and n on a ring, their utilities can also be characterized similarly. Only
notice in this case that agent 1 and agent n are neighbors of each other.
A.1
Proofs to Proposition 1 and Proposition 2
The biggest complication in this subsection comes mainly from part (c) and part (d)
of Proposition 1. Part (a) and part (b) of Proposition 1, as well as Proposition 2, are
relatively easy. So let’s prove the easier parts first. Before that, we need a new term
to ease our presentation.
Definition 5 Given an initialized fashion game I = (N, E, T, x(0)), if xi (t) keeps
unchanged when simultaneous BRD enters the fashion cycle, we say that agent i is
immune w.r.t. I.
The lemma below characterizes the convergence case.
Lemma 2 Given an initialized fashion game I = (N, E, T, x(0)), and suppose the underlying structure g = (N, E) is either a line or a ring. The following three statements
are equivalent.
(a) l(I) = 1;
(b) All agents are immune w.r.t. I;
(c) There exists one agent who is immune w.r.t. I.
29
Proof. The equivalence of (a) and (b) is obvious. It is trivial that (b) implies (c), so
we are left to show that (c) implies (b). Suppose that i is immune. We claim that
j ∈ Ni , an arbitrary neighbor of i, is also immune.
Suppose not, then there must exist some time t0 ≥ r(I) such that uj (x(t0 )) < 0.
So the action of i is not liked by agent j at time t0 , and j will switch at step t0 ,
i.e. xj (t0 + 1) = 1 − xj (t0 ). Since i is immune, we know that she never switches, i.e.
xi (t0 + 1) = xi (t0 ). Therefore, the action of i is liked by agent j at time t0 + 1, and
hence uj (x(t0 + 1)) ≥ 0. Furthermore, the utility of j keeps nonnegative after time
t0 + 1, because i is immune, she never switches in the limit cycle, and her action is
always liked by j after t0 . Therefore agent j will never switch any more. Thus the
action xj (t0 ) will never be taken again by j after t0 , contradicting the fact that fashion
cycle is entered at time t0 .
Notice that the above lemma relies crucially on the fact that in lines and rings,
each agent has at most two neighbors. As long as one of her two neighbors is liked by
her, she will be satisfied and will not deviate. This is not true other structures and
thus results in the above lemma cannot be extended. So are other lemmas.
The following lemma characterizes length two limit cycles.
Lemma 3 Given an initialized fashion game I = (N, E, T, x(0)), and suppose the
underlying structure g = (N, E) is either a line or a ring. The following statements
are equivalent.
(a) l(I) = 2;
(b) ∃i ∈ N , ui (x(t)) < 0 holds for all t ≥ r(I);
(c) ∃t ≥ 0, ui (x(t)) < 0 holds for all i ∈ N ;
(d) I is badly initialized.
Proof. (a) implies (b) obviously. Suppose (b) is true, then agent i keeps switching her
action in the fashion cycle. This can only happen when her neighbors keep switching
actions. Using this argument again and again, we know that no agent is satisfied in
any step of the fashion cycle, thus (c) is valid. Suppose now (c) is true, i.e. in some
time t no agent is satisfied. This can happen only when all agents are of the same type,
because for any two adjacent agents of different types, it is impossible for both of them
to be unsatisfied. Also, no agent should be satisfied initially. Suppose not, then there
exist i ∈ N s.t. ui (0) ≥ 0, then she likes at least one of her neighbors’ action initially,
and that neighbor likes her action too, because they are of the same type. Therefore,
30
the two agents are both immune, contradicting (c). (d) indicates (a) obviously, and
hence the lemma.
Notice that the equivalence of (a) and (d) in Lemma 3 has already verified part (a)
of Proposition 1 and part (a) of Proposition 2.
Intuitively, agents in a cluster are very likely to be immune, because they can help
each other. Therefore clustered fashion games are supposed to be able to reach PNEs
through BRD, unless the fashion game is badly initialized. Below is the rigorous proof
to part (b) of Proposition 1.
Proof. The condition that I is clustered says that at least one of the following must
hold, (i) T1 = T2 ; (ii) Tn−1 = Tn ; (iii) there are three adjacent agents that are of the
same type. Due to Lemma 3, it suffices to show that l(I) ∈ {1, 2}.
Suppose that T1 = T2 and they are both conformists. If at some step they take the
same action, then they will both be immune. By Lemma 2, this means that l(I) = 1
and we are done. So we suppose now they always take different actions. Since agent 1
has only one neighbor, this means that she will keep switching actions. By Lemma 3
we know that l(I) = 2. If they are both rebels, it can be shown by a similar argument.
By symmetry, it is also valid for the case that Tn−1 = Tn .
Suppose now there are three agents i, i + 1, i + 2, such that Ti = Ti+1 = Ti+2 = C.
If there exists some time t such that either xi (t) = xi+1 (t) or xi+1 (t) = xi+2 (t), then
there will be immune agents and l(I) = 1. So we assume that xi (t) = xi+2 (t) 6= xi+1 (t)
holds for all time t. Therefore the middle agent i + 1 will switch her action at all steps.
Therefore by Lemma 3 we know that l(I) = 2.
Suppose now there are three agents i, i + 1, i + 2, such that Ti = Ti+1 = Ti+2 = R.
If there exists some time t such that either xi (t) 6= xi+1 (t) or xi+1 (t) 6= xi+2 (t), then
there will be immune agents and l(I) = 1. So we assume that xi (t) = xi+2 (t) = xi+1 (t)
holds for all time t. Therefore the middle agent i + 1 will switch her action at all steps.
Therefore by Lemma 3 we know that l(I) = 2.
It can be observed from the above proof that part (b) of Proposition 2 is also true.
We are left, in this subsection, to prove part (c) and (d) of Proposition 1. We need
first to prove a property of the limit cycles for the case l(I) ≥ 3.
Lemma 4 Given an initialized fashion game I = (N, E, T, x(0)), and suppose the
underlying structure g = (N, E) is either a line or a ring. If l(I) ≥ 3, then no
agent will switch in two consecutive steps, i.e. there is no time t such that ui (x(t)) =
ui (x(t + 1)) < 0.
31
Proof. We consider first the case that g = (N, E) is a line. Suppose on the contrary
that ui (x(t)) = ui (x(t + 1)) < 0. It can be shown easily that in the case i ∈ {1, n}, i
must be of the same type as her neighbor, and in the case that i ∈
/ {1, n}, i and her
two neighbors must have the same type. Hence I is clustered. Due to Proposition
1(a), we know that l(I) ≤ 2. A contradiction with the hypothesis that l(I) ≥ 3. The
case that (N, E) is a ring can be proved in exactly the same approach.
The key method we use to deal with part (c) and (d) of Proposition 1 is the “cutting”
technique explored by us. To ease the presentation, we need a further new term.
Given an initialized fashion game I = (N, E, T, x(0)), and suppose that g = (N, E)
is a line. For any agent i, 2 ≤ i ≤ n−1, let I 0 (i) = (N 0 , E 0 , T 0 , x0 (0)) be a new (smaller)
initialized fashion game constructed by the left i agents of I in the following natural
way:
• N 0 = {1, 2, · · · , i};
• E 0 = {ij : i, j ∈ N 0 , ij ∈ E};
• ∀1 ≤ j ≤ i, Tj0 = Tj ;
• ∀1 ≤ j ≤ i, x0j (0) = xj (0).
Definition 6 We say that I 0 (i) is a complete sub-instance of I iff x0j (t) = xj (t) holds
for all t ≥ 1 and 1 ≤ j ≤ i.
Lemma 5 Given an initialized fashion game I = (N, E, T, x(0)), and suppose the
underlying structure is a cross-heterogeneous line. Suppose also that l(I) ≥ 3 and
agents i and i + 1 are of the same type (2 ≤ i ≤ n − 2). Then I 0 (i) is a complete
sub-instance of I.
Proof. Suppose w.l.o.g. that i and i + 1 are both conformists. In the simultaneous
BRD of I, i and i + 1 will always take the opposite actions, because otherwise they
will both be immune and the hypothesis that l(I) ≥ 3 will be violated. Therefore,
whether i will switch her action or not in step t is completely determined by the action
of agent i − 1. This is also true for the simultaneous BRD of I 0 (i), where i − 1 is the
only neighbor of i. Hence the lemma.
With the assistance of two technical lemmas, Lemma 7 and Lemma 8, that are
presented in Appendix B, we are ready to prove the remaining parts of Proposition 1.
32
Suppose now I = (N, E, T, x(0)) is an initialized fashion game, and the underlying
structure is a cross-heterogeneous line. First recall that PNE does not exist in this
case. Since none of the agents has a utility of zero in the fashion cycle (Lemma 8),
and in any utility sequence 2 will never be adjacent with 2, neither will -2 be adjacent
with -2 (Lemma 7(a)), the whole sequence of utilities will consist of alternate 2 and -2
(except for the ending agents, whose utilities are 1 or -1). At any step of the fashion
cycle, either none conformist is satisfied but all the rebels are, or none rebel is satisfied
but every conformist is. Based on the above discussion, it is easy to check that the
length of any fashion cycle is exactly 4. Hence part (c) of Proposition 1 is valid.
Suppose now I = (N, E, T, x(0)) is an initialized fashion game, and the underlying
structure is an un-clustered line but not cross-heterogeneous. Since g = (N, E) is
un-clustered, by Lemma 3 we know that l(I) 6= 2 (note that un-clustered implies not
homogeneous and further not badly-configured). To prove part (d) of Proposition 1, it
suffices to show that if l(I) ≥ 3, then it is only possible that l(I) = 4.
Suppose now l(I) ≥ 3. We cut the line at an arbitrary conformist-conformist edge
or rebel-rebel edge into two pieces. Notice that un-clustered guarantees that the above
cutting will not produce isolated agents. Due to Lemma 5, they are both complete
sub-instances. If the two pieces are both cross-homogeneous, then each of them has
a fashion cycle of length 4, regardless of initial configuration of actions (part (c) of
Proposition 1). By the definition of complete sub-instance, piecing them together
gives part (d) of Proposition 1 immediately. Otherwise, we can cut them again into
even smaller pieces, and use the same argument. Since the line is finite, this process
will terminate and the proof is done.
A.2
Proof to Proposition 3
Before the main proof, we need a lemma that has a similar essence to Lemma 2 and
Lemma 3 for the line and ring structures.
Lemma 6 Let I = (N, E, T, x(0)) be an initialized fashion game and suppose the
underlying network g = (N, E) is a star. The following statements are true.
(a) At any step of the limit cycle, all the peripheral conformists choose the same
action and so do all the peripheral rebels;
(b) l(I) = 1 if and only if there is some time when all agents are satisfied, if and
only if the central agent is always satisfied in the limit cycle.
33
(c) l(I) = 2 if and only if the central agent is always unsatisfied in the limit cycle,
if and only if there is some time when no agent is satsified.
Proof. (a) We only prove for conformists, because the rebel half is analogous. Suppose
the contrary, i.e. at some step of the limit cycle, part of the peripheral conformists
choose action 1 and the others choose action 0. Since each peripheral agent has the
same single neighbor, namely the central agent, it must be true that either all the
action 1 peripheral conformists are satisfied but the action 0 ones are unsatisfied, or
the opposite. Suppose w.l.o.g. the former case occurs. Then the action 1 peripheral
conformists switch to action 0 while the action 0 ones keep their actions fixed at 0.
Hence at the next step all the peripheral conformists have an action of 0. From that
time on, either they switch simultaneously or keep unchanged simultaneously. So in
any of the following step they always share the same action, and hence the state we
assume that part of the peripheral conformists choose action 1 and the others choose
action 0 will never be reached again, contradicting our assumption that it is in the
limit cycle.
(b) The first part is straightforward, we only show the second part. When l(I) = 1,
every agent is always satisfied in the limit cycle, and thus this is true for the central
agent. On the other hand, when the central agent is always satisfied in the limit cycle,
she never switches, and it must be true that all the peripheral ones never switch either,
because if any of them ever switches, she keeps being satisfied after switching (note
again that she has only one neighbor, the central one, who never switches in the limit
cycle).
(c) When l(I) = 2, any agent, either the central one or a peripheral one, must
be either always satisfied and never switches in the limit cycle or always unsatisfied
and hence switches at any step of the limit cycle. By (b) we know that the central
agent must be always unsatisfied. Suppose now the central agent is always unsatisfied
in the limit cycle, we prove that no agent is ever satisfied in the limit cycle. This
is true because otherwise some peripheral agent will always be satisfied in the limit
cycle, which is obviously impossible since her only neighbor, the central agent, keeps
switching in the limit cycle. Suppose now there is some time such that no agent is
satisfied, then all agents will keep switching from that time on, and thus l(I) = 2.
Piecing the above three implications together, we know that (c) is correct.
Now we are ready to prove Proposition 3.
Proof. (a) Let’s first consider the case that the central agent is a rebel. We claim
that limit cycle is entered at step 2, where all the rebels choose the same action, the
34
Figure 5: Illustrations of length-4 limit cycles for the star network. Circles
stand for conformists, and triangles for rebels. Black color indicates that the agents
chooses 1, white for 0.
other action is taken by the conformists, and hence no agent is satisfied and each one
keeps switching from that time on. Suppose w.l.o.g. that the central rebel chooses
action 1 initially. Since she is initially unsatisfied, at step 2 she will switch to action
0. So do the peripheral rebels who initially take action 1. For the peripheral rebels
who initially take action 0, they are initially satisfied and do not switch at the first
step. Hence at the second step, all the peripheral rebels have an action of 0. For the
peripheral conformists who take action 1 initially, they are satisfied at the first step,
and hence still have an action 1 at the second step. For the peripheral conformists who
take action 0 initially, they are unsatisfied at the first step, and hence switch to action
1 at the second step. This proves our claim.
Proof to the other case that the central agent is a conformist is quite similar. The
only difference is that in the limit cycle, all the peripheral conformists choose the same
action, which is opposite to that of the central conformist, and all the peripheral rebels
choose the same action as the central conformist.
(b) Consider the case that the central agent is a rebel. First of all, it must be
true that l(I) ≥ 2, because it is impossible for the central rebel and all the peripheral
conformists to be simultaneously satisfied. Recall that in the limit cycle, all the conformists choose the same action, and so do all the peripheral rebels (Lemma 6 (a)).
So it is impossible that l(I) = 2, because at any step of the limit cycle, either all the
conformists are all satisfied (when they take the same action as the central agent) or
the central rebel is satisfied (when the peripheral conformists take a different action
with the central rebel). And there must exist some time t in the limit cycle such that
the central rebel is satisfied (say she chooses action 0). Since there are more peripheral conformists than peripheral rebels, we know that all peripheral conformists should
35
choose action 1 at time t (hence they are satisfied), because otherwise the central rebel
would be unsatisfied. We discuss in two cases about the peripheral rebels.
• All the peripheral rebels are unsatisfied at time t.
That is, they all choose action 0 too. Then at time t + 1, all the peripheral agents
switch actions, making themselves happy but the central rebel unhappy. So at
time t + 2, the central rebel switches to action 1, making herself happy and all
the peripheral agents unhappy. At time t + 3, all the conformists hold action 1,
all the peripheral rebels choose action 0, and the central rebel has action 1, thus
the only unsatisfied agent is the central rebel. By letting the central rebel switch
to action 0, we can see that the state at time t + 4 is exactly the same as at time
t. Therefore in this case we have l(I) = 4. This limit cycle is illustrated in the
left part of Figure 5.
• All the peripheral rebels are satisfied at time t.
That is, they all choose action 0. It can be checked that the state at time t + 1 in
this case is exactly that at time t + 1 in the previous case. However, the time t
state in this case is none of the four states in the limit cycle of the previous case,
so it will never be reached again, contradicting our assumption that limit cycle
has been entered at time t. Therefore this case never occurs at all.
Consider the other case that the central agent is a conformist. For the same reason
as in the above proof, there must be some time t, after the entrance of limit cycle, such
that the central conformist is satisfied. Suppose w.l.o.g. that the central conformist
chooses action 0 at time t. First of all, all the peripheral rebels must choose action
0 at time t, and they are unsatisfied. We discuss in two cases about the peripheral
conformists.
• All the peripheral conformists are unsatisfied at time t.
That is, they all choose action 1. Then at time t + 1, all the peripheral agents
switch actions, making themselves happy but the central conformist unhappy. So
at time t + 2, the central conformist switches to action 1, making herself happy
and all the peripheral agents unhappy. At time t + 3, all the conformists hold
action 1, all the peripheral rebels choose action 0, and the central rebel has action
1, thus the only unsatisfied agent is the central conformist. By letting the central
conformist switch to action 0, we can see that the state at time t + 4 is exactly
36
the same as at time t. Therefore in this case we have l(I) = 4. This limit cycle
is illustrated in the right part of Figure 5.
• All the peripheral conformists are satisfied at time t.
That is, they all choose action 0. Using exactly the same logic as in the proof to
(c), we know that this case never occurs.
(c) Naturally, we discuss in four cases.
• The central agent is a rebel, more than a half of the peripheral agents are also
rebels, and initially the central rebel is satisfied.
Since the central rebel is initially satisfied, her action at the second step will be
the same as in the first step. For the peripheral rebels, the ones initially taking
the opposite action stay with this action, and the other peripheral rebels will
switch to this action too at the second step. So at the second step each of the
peripheral rebels takes an opposite action with the central rebel. Since more
than a half of the central rebel’s neighbors are rebels, we know that all the rebels
will be always satisfied from the second step on. So are the conformists. Hence
l(I) = 1.
• The central agent is a conformist, more than a half of the peripheral agents are
also conformists, and initially the central conformist is satisfied.
This case can be verified by using an analogous logic as in the previous case.
• The central agent is a rebel, and half of the peripheral agents are rebels and the
other half are conformists.
Suppose on the contrary that l(I) ≥ 2, then there is some time t in the limit
cycle that the central rebel is unsatisfied (recall Lemma ?? (b)). This must be
the case that all the agents take the same action (say action 1), because when
the peripheral conformists and the peripheral rebels take different actions, the
central rebel is satisfied. So at step t + 1, all the rebels will switch to action 0,
and the conformists still hold action 1. Therefore the central rebel is the only
satisfied agent at step t + 1. At time t + 2, the central rebel still has action 0,
the peripheral rebels have action 1, and the conformists have action 0. It can be
seen that all the agents are satisfied at step t + 2, contradicting the assumption
that l(I) ≥ 2.
37
• The central agent is a conformist, and half of the peripheral agents are rebels
and the other half are conformists.
This case can be verified by using an analogous logic as in the previous case.
This completes the proof to the whole proposition.
A.3
Proofs to Proposition 4 and Proposition 5
Proof. Recall that when the underlying network is a line, then G possesses a PNE
if and only if it is not cross-heterogeneous. So the “only if” part of Proposition 4
is straightforward. Next, we prove Proposition 5 and the “if” part of Proposition 4
simultaneously. Our proof is not based on the construction of a potential function, as
is usually done, but in a more direct way. First of all, non-cross-heterogeneous implies
that there are at least two agents, say i and i + 1, who are adjacent and are of the
same type. Then i and i + 1 can serve a similar “anchor” role as a cluster does in the
analysis of Proposition 1(b) for simultaneous BRD.
We discuss only the case that i and i + 1 are both conformists. The other case that
i and i + 1 are both rebels can be proved using the same logic. We claim that there
must be some time t0 , such that i and i + 1 are both satisfied from that time on, i.e.
ui (x(t)) ≥ 0, ui+1 (x(t)) ≥ 0, ∀t ≥ t0 .
W.l.o.g., we assume that xi (0) 6= xi+1 (0), because otherwise both of them would be
initially satisfied and support each other forever, that is, t0 = 0. Using the same logic,
it can be seen that as long as one of them deviates at some time, then the actions of i
and i + 1 will always be the same from that time on. So the only remaining possibility
is that none of them deviates, which implies that they are always satisfied, verifying
our claim.45
After time t0 , the respective neighbors of i and i + 1, i.e. i − 1 and i + 2, will
deviate for at most one time too. So there exists some time t1 , after which no agent in
45
The rough logic is that according to the sequential moving rule, either deterministic or random,
if one of them is ever unsatisfied, she will have a chance to deviate. If we think really carefully, then
one subtle worry may come out. That is, for certain deterministic moving orders, i may be unsatisfied
at some time, and at that time it is not her turn to move. However, when it is time for her turn, she
becomes satisfied, by the previous deviation of her neighbor i − 1. The worry is that this may cycle
again and again. A second thought tells us that this will never happen, because the hypothesis that
i never deviates implies that i − 1 can only deviate for one time.
38
{i − 1, i, i + 1, i + 2} will deviate. Using this argument again and again, we know that
eventually all agents will be satisfied, and hence the sequential BRD converges.
A.4
Proof to Proposition 6
Proof. Taking all the action profiles as states, and letting the transition probabilities
be determined by the underlying dynamic (let a uniformly chosen unsatisfied agent
switch), we have a well-defined Markov chain. According to Lemma 1, the condition
that at least half of the peripheral agents are of the same type as the central agent
implies that PNE exists. In fact, it can be seen that there are exact two PNEs. What’s
more, the two PNEs correspond to the only absorbing states of the Markov chain.
What’s more, this Markov chain is absorbing: every state can reach one of the two
absorbing states. This can be shown easily from the following observation: if we let
the unsatisfied peripheral agents that are of the same type deviate first (one by one in
an arbitrary order), and the other unsatisfied peripheral agents deviate subsequently
(one by one in an arbitrary order), then we will arrive at one of the two PNEs. Since in
absorbing Markov chains, all non-absorbing states are transient, we know immediately
that PNEs will be reached with probability one, and this completes the proof.
B
Two Technical Lemmas
The two lemmas provided in this appendix are used in the proofs to Proposition 1(c)(d),
the most difficult parts of that proposition. Since their proofs, mainly the latter one,
are rather tedious, we present them in this separate section.
In this appendix, we use ui (t) for short of ui (x(t)). This is reasonable, because
as long as the initial actions are set, utility of each agent is a function of the single
variable of time t.
We call [a1 , a2 , · · · , ak ] ∈ {−2, 2, 0, −1, 1}k a utility sequence, if there exists a time t
and an agent i such that [ui (t), ui+1 (t), · · · , ui+k−1 (t)] = [a1 , a2 , · · · , ak ]. To study the
cross-heterogeneous case, it is more convenient to deal with utility sequences instead
of action sequences.
Lemma 7 Given an initialized fashion game I = (N, E, T, x(0)), and suppose g =
(N, E) is a cross-heterogeneous line.
(a) The following utility sequences will never occur: [2, 2], [−2, −2], [2, 0, 2], [2, 0, · · · , 0, 2],
[2, 0, 1], [2, 0, · · · , 0, 1], [−2, 0, −2], [−2, 0, · · · , 0, −2], [−2, 0, −1], [−2, 0, · · · , 0, −1],
39
[1, 0, · · · , 0, 1], [−1, 0, · · · , 0, −1];
(b) If ui (t) = −2, then ui (t + 1) = 2;
(c) If ui (t) = 0, then ui (t + 1) ∈ {−2, 0}.
Proof. One beautiful property of the case under focus is that if one likes the action
of her neighbor, then her neighbor must hate the action of her, and vice versa. Using
this property, it is easy to check that none of the sequences in (a) will occur.
Since no pair of adjacent agents can be simultaneous unsatisfied, we know that if
agent i switches her action at some step, then none of her neighbor switches action at
the same step. Using this fact, (b) is obvious.
If ui (t) = 0, then i likes the action of exactly one of her neighbors {i − 1, i + 1},
and hates the action of the other, at time t. Suppose w.l.o.g. that i − 1 ∈ Li (x(t)) and
i + 1 ∈ Di (x(t)) (recall the notations defined in Section 2). Accordingly, i ∈ Hi−1 (x(t))
and i ∈ Li+1 (x(t)). Therefore, ui−1 (t) ≤ 0 and ui+1 (t) ≥ 0. Since i + 1 is satisfied at
time t, she will not switch. In the case that ui−1 (t) < 0, it is evident that ui (t+1) = −2.
And in the case that ui−1 (t) = 0, we have ui (t + 1) = 0. Hence the lemma.
Given an initialized fashion game I = (N, E, T, x(0)), let Z(t) be the set of agents
whose utilities are zero at time t, i.e.
Z(t) = {i ∈ N : ui (t) = 0}.
The following lemma is critical.
Lemma 8 Given an initialized fashion game I = (N, E, T, x(0)), suppose the underlying structure g = (N, E) is a cross-heterogeneous line. Then Z(t) = ∅ for all t ≥ r(I).
Proof. Above all, the result of Cao and Yang (2011), as introduced in Subsection 4,
tells us that PNE does not exist for I, therefore L(I) 6= 1. Due to Lemma 3, we have
obviously that L(I) 6= 2. Hence
L(I) ≥ 3,
(1)
and the results in Lemma 4, Lemma 5, Lemma 7, are all valid.
We prove in two steps.
(i) We prove first that the cardinality of Z(t) is non-increasing in simultaneous
BRD, i.e.
|Z(t + 1)| ≤ |Z(t)|.
(2)
40
Let
Z + (t) = Z(t + 1) \ Z(t),
and
Z − (t) = Z(t) \ Z(t + 1),
we only need to show that
|Z + (t)| ≤ |Z − (t)|.
If Z + (t) is empty, we are done. Suppose not, take an arbitrary agent i from Z + (t),
i.e. ui (t + 1) = 0 and ui (t) 6= 0. Then by Lemma 7(b), it can only be that ui (t) = 2.
And it is obvious that i ∈
/ {1, n}. So agent i does not switch her action at step t, and
her utility changes from 2 at step t to 0 at step t + 1. This can only happen when
exactly one neighbors of i is unsatisfied at time t. Due to Lemma 7(a), utility sequence
[2, 2] will never occur, so the neighbor of agent i who is satisfied at time t must have a
utility of 0. Therefore, {ui−1 (t), ui+1 (t)} = {−2, 0} (we assume w.l.o.g. that i − 1 6= 1
and i + 1 6= n).
We consider first the case that [ui−1 (t), ui (t), ui+1 (t)] = [−2, 2, 0]. We claim that
Z − (t) ∩ {i + 1, i + 2, · · · , n} 6= ∅.
(3)
Suppose the opposite. Then i + 1 ∈
/ Z − (t). Since i + 1 ∈ Z(t), this implies that
i + 1 ∈ Z(t + 1), i.e. ui+1 (t + 1) = 0. As agent i + 1 does not switch her action at step
t, her left neighbor, agent i, does not switch either, and the utility of i + 1 also keeps
unchanged, we know that the right neighbor of agent i + 1, i + 2, should not switch
at step t either. Therefore ui+2 (t) ≥ 0. By Lemma 7(a), utility sequence [2, 0, 2] and
[2, 0, 1] will never occur, so it is impossible that ui+2 (t) ∈ {2, 1}. It can only be that
ui+2 (t) = 0.
Also, Z − (t) ∩ {i + 1, i + 2, · · · , n} = ∅ implies that i + 2 ∈
/ Z − (t). ui+2 (t) = 0 tells us
that i + 2 ∈ Z(t). Using exactly the same argument as in the last paragraph, we know
that ui+3 (t) = 0. But this cannot continue for ever, because for any time t, it always
holds that un (t) ∈ {1, −1}. Hence a contradiction. Therefore in this case the claim
(3) is valid. For the case that [ui−1 (t), ui (t), ui+1 (t)] = [0, 2, −2], (3) can be similarly
shown to be true.
Based on the above discussion, the following definition for each i ∈ Z + (t) is valid:
f (i) =
8
¦
©
< min j : j ∈ Z − (t) ∩ {i + 1, i + 2, · · · , n}
¦
©
: max j : j ∈ Z − (t) ∩ {1, 2, · · · , i − 1}
41
if [ui−1 (t), ui (t), ui+1 (t)] = [−2, 2, 0]
if [ui−1 (t), ui (t), ui+1 (t)] = [0, 2, −2]
.
The facts below can also be observed from the above discussion. ∀i ∈ Z − (t),
f (i) > i implies [ui−1 (t), · · · , uf (i)+1 (t)] = [−2, 2, 0, · · · , 0, −2],
(4)
f (i) < i implies [uf (i)−1 (t), · · · , ui+1 (t)] = [−2, 0, · · · , 0, 2, −2],
(5)
where the “· · · ”s on the right hand sides denote both a series of 0s.
If we can show that f (i1 ) 6= f (i2 ) holds for all i1 , i2 ∈ Z + (t), i1 6= i2 , then the
main result we want to prove in this step that |Z + (t)| ≤ |Z − (t)| will be valid. Assume
w.l.o.g. that i1 < i2 . We discuss in four cases.
• f (i1 ) < i1 and f (i2 ) > i2 .
Since f (i2 ) > i2 > i1 > f (i1 ), this case is trivial;
• f (i1 ) > i1 and f (i2 ) > i2 .
It can be seen from (4) that uj (t) = 0, ∀i1 < j < f (i1 ). i2 > i1 and ui2 (t) 6= 0
imply that i2 > f (i1 ). Hence f (i2 ) > i2 > f (i1 );
• f (i1 ) < i1 and f (i2 ) < i2 .
By symmetry with the previous case, we have f (i1 ) < i1 < f (i2 );
• f (i1 ) > i1 and f (i2 ) < i2 .
It can be seen from (4) that uf (i1 )+1 (t) = −2, but from (5) we know that
uf (i2 )+1 (t) ∈ {0, 2}, hence f (i1 ) =
6 f (i2 ).
(ii) By the monotonicity property (2) we know immediately that the cardinality of
Z(t) keeps unchanged in the fashion cycle, i.e.
|Z(t + 1)| = |Z(t)|, ∀t ≥ r(I).
(6)
Consequently, we know that
∀t ≥ r(I), ∀j ∈ Z − (t), there exists exactly one i ∈ Z + (t) s.t. f (i) = j.
(7)
Now, we are ready to prove the final result. We use the minimum counter-example
argument. Suppose the lemma is not valid, and I = (N, E, T, x(0)) is a counterexample with the smallest number of agents. Let j ∗ be the agent with the largest
42
index among all the agents having a utility of 0 for at least one time in the fashion
cycle.
First of all, it will be verified that j ∗ = n − 1, because otherwise we could construct
a smaller counter-example. In fact, the choice of j ∗ tells us that if j ∗ 6= n − 1, then
xj ∗ +1 (t) ∈ {2, −2} holds for all t ≥ r(I). In the fashion cycle, if agent j ∗ + 1 hates
the action of j ∗ , then her utility will be −2, because 0 is never attained, and she
will switch her action. And when agent j ∗ + 1 likes the action of j ∗ , evidently she
will not switch. Therefore, whether j ∗ + 1 switches her action at time t ≥ r(I) is
completely determined by the action of j ∗ . Using the same argument as in the proof
to Lemma 5, we know that I 0 (j ∗ + 1) = (N 0 , E 0 , T 0 , x0 (0)) is also a qualified counterexample (fashion cycle is entered at the first step), where N 0 = {1, 2, · · · , j ∗ + 1}, T 0
the corresponding sub-vector of T , E 0 the corresponding left part of E, and x0 (0) =
(x1 (r(I)), x2 (r(I), · · · , xj ∗ +1 (r(I)))). Since I is the smallest counter-example, this is
impossible, and thus it can only be that j ∗ = n − 1.
Suppose uj ∗ (t0 ) = 0 for some t0 ≥ r(I), then there must exist t∗ ≥ t0 such that
uj ∗ (t∗ ) = 0 and uj ∗ (t∗ + 1) 6= 0. Because otherwise, the utility of j ∗ will keep at 0
from time t0 on, and thus she will be immune. As a result, l(I) = 1 and PNE will be
reached by simultaneous BRD. This is a contradiction with (1).
Therefore, j ∗ ∈ Z − (t∗ ). Since t∗ ≥ t0 ≥ r(I), by (7) we know that there exists
exactly one i∗ ∈ Z + (t∗ ) such that f (i∗ ) = j ∗ . Because j ∗ = n − 1 we know that i∗ < j ∗ .
Therefore, by (5) we know that
[ui∗ −1 (t∗ ), · · · , uj ∗ (t∗ ), un (t∗ )] = [−2, 2, 0, · · · , 0, −1].
By Lemma 7(c), it must hold that uj ∗ (t∗ + 1) = −2. Therefore,
[ui∗ −1 (t∗ + 1), · · · , un (t∗ + 1)] = [2, 0, 0, · · · , −2, 1].
(8)
It can be observed from (8) that j ∗ − 1 ∈ Z − (t∗ + 1). Again, using (7), we know
that there is an agent k ∗ ∈ Z + (t∗ + 1) such that f (k ∗ ) = j ∗ − 1. It must hold that
k ∗ < j ∗ − 1, because (8) tells us that the sequence [0, 2, −2] never occurs on the right
side of j ∗ − 1 and (5) says that if k ∗ > j ∗ − 1 we would have a sequence [0,2,-2] on
the right side of j ∗ − 1 (notice by definition that for any i ∈ Z + (t), we always have
f (i) 6= i). Furthermore, comparing (4) and (8) we know that k ∗ = i∗ − 1, which can
only happen when ui∗ −2 (t∗ + 1) ∈ {−2, −1}.
ui∗ −2 (t + 1) = −1, i.e. i∗ − 2 = 1, is not true. Because if so, we would have
43
[ui∗ −2 (t∗ + 2), · · · , un (t∗ + 2)] = [1, 0, 0, 0, · · · , −2, 2, −1], which has n − 4 0s in total.
However, it can be checked that we would further have [ui∗ −2 (t∗ + 3), · · · , un (t∗ + 3)] =
[1, 0, 0, 0, · · · , −2, 2, −2, 1], which has n − 5 0s in total, a contradiction with (6).
Therefore, we have ui∗ −2 (t + 1) = −2 and
[ui∗ −2 (t∗ + 2), · · · , un (t∗ + 2)] = [2, 0, 0, 0, · · · , −2, 2, −1].
(9)
It can be observed from (9) that j ∗ − 2 ∈ Z − (t∗ + 2). Again, there is no [0, 2, −2]
on the right side of j ∗ − 2, and so we can only have f (j ∗ − 2) = i∗ − 2. This can only
happen when ui∗ −3 (t∗ + 2) ∈ {−2, −1}. Again, we can deduce that ui∗ −3 (t∗ + 2) = −2
and
[ui∗ −3 (t∗ + 3), · · · , un (t∗ + 3)] = [2, 0, 0, 0, · · · , −2, 2, −2, 1].
(10)
So this process will be repeated infinitely. But this is obviously impossible, because
the line is finite. Hence the lemma.
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