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Structural holes in social networks: A remark
Journal of Economic Theory 144 (2009) 422–431 www.elsevier.com/locate/jet
Note
Structural holes in social networks: A remark Nicolas Houy CNRS, Laboratoire d’Econométrie-Ecole Polytechnique, 5 rue Descartes, 75005 Paris, France Received 18 April 2007; final version received 7 May 2008; accepted 27 May 2008 Available online 1 July 2008
Abstract In a recent article, [S. Goyal, F. Vega-Redondo, Structural holes in social networks, J. Econ. Theory 137 (1) (2007) 460–492] the authors (GVR) showed the importance of stars and cycles in a given network formation game. Implicitly, in their article, a network is called an equilibrium if it is generated by an equilibrium strategy. We extend the results of GVR to the case of a stronger requirement: namely, that a network can be called an equilibrium only if all the strategies generating it are equilibria. We also show, in a dynamic framework, that both definitions differ in crucial ways. © 2008 Published by Elsevier Inc. JEL classification: C72; C78; D85 Keywords: Networks; Network formation; Structural holes; Intermediation
1. Introduction In a recent article, [1] Goyal and Vega-Redondo (GVR, henceforth) showed that empty networks, stars and cycles are natural equilibria of a given network formation game. In the game they consider, a player increases his payoff (1) when he is linked to other players (possibly indirectly), and (2) when he is essential (i.e. necessary) for linking pairs of players. Moreover, a player’s payoff decreases (1) when he creates direct costly links, and (2) when he rewards essential players between him and other players. E-mail address: [email protected]. 0022-0531/$ – see front matter © 2008 Published by Elsevier Inc. doi:10.1016/j.jet.2008.05.007
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Implicitly, in GVR, a network is called an equilibrium if there exists a strategy profile generating it and that satisfy some equilibrium conditions. We propose a stronger definition of an equilibrium: a network is an equilibrium only if all the strategy profiles generating it satisfy the equilibrium conditions. While the first definition characterizes the networks that can be observed at equilibrium, the second definition characterizes the networks that, when observed, are necessarily equilibria. We claim that the second definition can lead to different and important results. First, if we assume that networks are observable but strategies are not, as it is often assumed, then, when using GVR’s definition and observing a network satisfying an equilibrium property, we cannot conclude that the society has reached an equilibrium state. On the contrary, with the definition we propose, drawing this conclusion is logically possible. Second, as we will show in the last section, GVR’s definition hardly implies any stability feature in the simplest dynamic setting. The definition we propose does. In the first section, we will show that in the case of the empty network, extending GVR’s results to the stronger definition of a network equilibrium is not possible whatever the values of the parameters if the number of players is large enough. In the case of stars, generalizing GVR’s result is possible if the parameters are modified. In the second section, we will show that in the case of cycles, extending GVR’s result is possible if the definition of an equilibrium is slightly modified. These results make the interpretation of GVR, emphasizing the role of stars and cycles, even more justified. In the last section, we show how crucial the difference between the two definitions of an equilibrium can be in some dynamic framework. 2. Notation A population is composed of a finite set of ex-ante identical agents, N = {1, 2, . . . , n} with n 3. These agents play a network formation game in which every one of them makes a simultaneous announcement of intended links. The strategy profile s = {sij }i,j ∈N is the matrix of all announcements of intended links where sij = 1 if individual i announces a wish to form a link with j and sij = 0 if individual i does not announce a wish to form a link with j . Let us denote by Si the set of all possible strategies for individual i. Let S be the set of all strategy profiles. An undirected link is formed between individuals i and j if and only if sij = sj i = 1. Let us denote by g(s) = {gij (s)}i,j ∈N the network resulting from announcements s, where gij (s) = gj i (s) = 1 if a link is formed between individuals i and j and gij (s) = gj i (s) = 0 if no link is formed between individuals i and j . For a player i, let ηi (g(s)) denote the number of players i has a link with, i.e. ηi (g(s)) = #{j ∈ N \ {i}, gij (s) = 1}. In the network g, there exists a path between i and j (i = j ) if, either gij = 1 or if there is a sequence of players (i1 , . . . , im ) such that gi,i1 = gi1 ,i2 = gi2 ,i3 = · · · = gim ,j = 1. Let us denote by Ci (g) the set of players j such that there is a path between i and j in g. A player i ∈ / {j, k} is essential for j and k in g if i lies on every path between j and k in the network g. Let us denote by E(j, k; g) the set of players who are essential for j and k in g and let e(j, k; g) = #E(j, k; g) be the cardinality of E(j, k; g). Finally, for a strategy profile s, the payoff earned by player i is given by 1 1 Πi (s) = + − ηi g(s) c e(i, j ; g(s)) + 2 e(j, k; g(s)) + 2 j ∈Ci (g)
j,k∈N, i∈E(j,k;g(s))
where c is the cost of forming a link. It should be noted that there is no cost of proposing a link. A cost is paid only when a link is formed.
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3. Results and comments 3.1. Strict bilateral equilibria Let us give the definition of a Strict Bilateral Equilibrium as stated in GVR. Definition 1. A strategy profile s ∗ is a Strict Bilateral Equilibrium (SBE) if the following two conditions hold: ∗ ) = g(s ∗ ), Π (s ∗ ) > Π (s , s ∗ ); (1) for any i ∈ N and every si ∈ Si such that g(si , s−i i i i −i ∗ ) = g(s ∗ ), (2) for any pair of players i, j ∈ N and every strategy pair (si , sj ) with g(si , sj , s−i−j ∗ ∗ Πi (s ∗ ) ⇒ Πj si , sj , s−i−j < Πj (s ∗ ). Πi si , sj , s−i−j
Then, a strategy profile is an SBE if and only if (1) no player has a (weak) incentive to deviate and (2) no pair of players can be (weakly) better off by deviating. Obviously the notion of equilibrium applies to strategies, not to networks. In order to avoid any ambiguity, we will say that a network g is SBE supported if and only if there exists an SBE s such that g = g(s). Stronger, we will say that a network g is fully SBE supported if and only if any s such that g = g(s) is an SBE. Notice that, implicitly, GVR use only the definition of SBE supported networks. The first theorem of GVR is the following. Theorem 1. If c > 1/2, the empty network is SBE supported. Then, the empty network can be observed at equilibrium when c > 1/2. However, it might be possible that not all the strategies generating the empty network be SBEs. Indeed, the next example shows that for 5/6 c > 1/2, there exists a strategy profile s such that g(s) is empty and still s is not an SBE. / Example 1. Let us have n 3, and s such that s2,1 = s2,3 = 1 and sij = 0, ∀(i, j ) ∈ {(2, 1), (2, 3)}. By definition, gij (s) = 0, ∀i, j ∈ N . Then, g(s) is empty and Πi (s) = 0, ∀i ∈ N . Let us show that s is not an SBE if 5/6 c > 1/2. Let us consider the strategy profile s such / {(1, 2), (2, 1), (2, 3), (3, 2)}. By defithat s 2,1 = s 1,2 = s 2,3 = s 3,2 = 1 and s ij = 0, ∀(i, j ) ∈ / {(1, 2), (2, 1), (2, 3), (3, 2)}. Said, differently, s is nition, g12 = g23 = 1 and gij (s) = 0, ∀(i, j ) ∈ obtained from s when players 1 and 3 change their strategies to propose a link to player 2. These links are formed since in s, 2 is unilaterally announcing a wish to form a link with players 1 and 3. Then, Π1 (s ) = Π3 (s ) = 5/6 − c. Then, players 1 and 3 can benefit from a change of their strategies. Hence, s is not an SBE since condition (2) of Definition 1 does not hold. Then, the condition c > 1/2 is not sufficient to state that the empty network is fully SBE supported. The next theorem shows that, for three individuals, the condition c > 5/6 is necessary and sufficient to state that the empty network is necessarily generated by an SBE. Theorem 2. Let n = 3. The empty network is fully SBE supported if and only if c > 5/6. Generalizing Theorem 1 for any number of players is not straightforward. Indeed, for any cost c, there exists a large enough number of players such that the empty network is not fully SBE supported.
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Theorem 3. Let c be given. If n is sufficiently large then, the empty network is not fully SBE supported. Theorem 3 shows that only some of the strategy profiles generating the empty network are SBEs when n grows large. This result minimizes the role played by the empty network in the game considered by GVR. A result similar to Theorem 1 is given by GVR for stars when c > 1/6 and n is large enough. In order to generalize this result to all the strategy profiles generating stars, it is enough to consider a cost greater than 2/3. This result strengthens GVR’s interpretation that stars arise as natural equilibria of the studied network formation game. Theorem 4. Let n be arbitrarily large. Any star is fully SBE supported if and only if c > 2/3. 3.2. Bilateral proof equilibria Another important result of GVR consists in showing that cycles containing all the players can arise as equilibria. In order to obtain this result, one needs to make sure that in the case of a bilateral deviation, the players involved will actually implement the deviation. Such an equilibrium is defined as a Strict Bilateral-Proof Equilibrium (SBPE). Definition 2. A strategy profile s ∗ is a Strict Bilateral-Proof Equilibrium (SBPE) if the following two conditions hold: ∗ ) = g(s ∗ ), Π (s ∗ ) > Π (s , s ∗ ); (1) for any i ∈ N and every si ∈ Si such that g(si , s−i i i i −i ∗ (2) for any pair of players i, j ∈ N and every strategy pair (si , sj ) with g(si , sj , s−i−j ) = g(s ∗ ) ∗ ∗ and Πi (si , sj , s−i−j ) Πi (s ), one of the following holds: ∗ ) < Πj (s ∗ ), (a) Πj (si , sj , s−i−j ∗ ∗ ) > Πk (sk , sl , s−k−l ). (b) ∃k, l ∈ {i, j }, k = l and some s˜k ∈ Sk , such that Πk (˜sk , sl , s−k−l
As above, we will say that a network g is SBPE supported if and only if there exists an SBPE s such that g = g(s). We will say that a network g is fully SBPE supported if and only if any s such that g = g(s) is an SBPE. It is shown in GVR that for any cost c, if there are sufficiently many players, any cycle containing all players is SBPE supported. However, we can that, if n 4, this result does not hold if we consider fully SBPE supported cycles. The following definition permits to generalize GVR’s result. Definition 3. A strategy profile s ∗ is a Bilateral-Proof Equilibrium (BPE) if the following two conditions hold: ∗ ) = g(s ∗ ), Π (s ∗ ) > Π (s , s ∗ ); (1) for any i ∈ N and every si ∈ Si such that g(si , s−i i i i −i ∗ (2) for any pair of players i, j ∈ N and every strategy pair (si , sj ) with g(si , sj , s−i−j ) = g(s ∗ ) ∗ ) > Πi (s ∗ ), one of the following holds: and Πi (si , sj , s−i−j ∗ (a) Πj (si , sj , s−i−j ) < Πj (s ∗ ), ∗ ∗ ) > Πk (sk , sl , s−k−l ). (b) ∃k, l ∈ {i, j }, k = l and some s˜k ∈ Sk , such that Πk (˜sk , sl , s−k−l
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Obviously, a BPE is an SBPE. The only difference arises when the two players of a pair earn an unchanged payoff when modifying their strategies. Now, we could show that any cycle containing all players is fully BPE supported if there are sufficiently many players.1 Theorem 5. For any c > 0, there exists a sufficiently large number of players such that any cycle containing all players is fully BPE supported. 4. A dynamic framework In this section, we show how crucial the difference is between a supported equilibrium network and a fully supported equilibrium network. In a simple dynamic framework, we will show that it is important that all strategies generating a given network be considered in order to assess that this network has a stability feature. For the sake of simplicity, let us consider the network formation game and the notions given in Section 3.1 with only three players. In accordance with the definition of a strict bilateral equilibrium, we will set the following stochastic dynamic process. First, let s, s be two strategy profiles. We say that s is attainable from s if and only if: • #{i ∈ N, si = si } 2 and, • ∀i ∈ N, si = si ⇒ Πi (s ) Πi (s). The corresponding dynamic process can be given as follows. From an initial strategy profile s, we draw by chance a set of individuals with cardinality 1 or 2. New strategies are proposed to these individuals, forming the new strategy profile s . For the sake of simplicity, we will lack rigor in not specifying the probability laws. We only need to assume that there exists a real > 0 such that any set of 1 or 2 individuals be drawn with a probability greater than in the first step and any strategy or pair of strategies be drawn with a probability greater than in the second step. If one of the individuals drawn by chance gets a strictly lower payoff in s , then he refuses the change of state and s is not attainable from s. Otherwise, s is attainable from s. Let us make two remarks about this dynamic process. First, individuals are myopic. They only consider the difference in payoff when evaluating the possibility of a new strategy profile. They do not anticipate any further change that could decrease their payoff. Second, we suppose that a change of strategy profile can occur even if there is no strict increase of the payoffs of the individuals involved. In doing so, we are consistent with Definition 1 of a strict bilateral equilibrium.2 Now we can define the binary relation R ∈ S × S as follows: an ordered pair of strategy profiles (s, s ) is an element of R if and only if s is attainable from s. We will call R T the transitive closure of R. In order to characterize the acceptable solutions of this abstract problem (a set endowed with a binary relation), we use the concept of absorbing sets introduced and studied by [2,3] and [4]. Let V ∈ 2S be a set of strategy profiles, we say that V is absorbing if and only if: (1) ∀s, s ∈ V , (s, s ) ∈ R T , and (2) ∀s ∈ V , ∀s ∈ S \ V , not[(s, s ) ∈ R T ]. We can interpret absorbing sets as follows: once an element of an absorbing set is reached, when repeating the dynamic process, all the elements of the same absorbing set will be reached an 1 The proof is omitted and available on request. 2 This has been noted by GVR, p. 465.
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infinite number of times while all the elements not in the absorbing set will never be reached again. Moreover, the probability of reaching an element of an absorbing goes asymptotically to 1 when the dynamic process is repeated an infinite number of times. It is well known that an absorbing set always exists but that it is not necessarily unique. However, when it is unique, the probability of reaching any element of the absorbing set in the future goes asymptotically to 1 when the dynamic process is repeated an infinite number of times. The following proposition characterizes the absorbing set for any value of the cost c.3 Proposition 1. Let N = 3. For any cost of forming a link c, the absorbing set is unique. Moreover, (1) if 0 < c < 1/6, s is an element of the absorbing set if and only if it is complete4 ; (2) if 1/6 < c < 2/3, s is an element of the absorbing set if and only if the network generated by s shows two links; (3) if 2/3 < c < 5/6, s is an element of the absorbing set if and only if the network generated by s shows two links or is empty; (4) if 5/6 < c, s is an element of the absorbing set if and only if the network generated by s is empty. Obviously, for all strategy profiles s and s such that g(s) and g(s ) are empty, we have (s, s ) ∈ R T . Then, by definition, all the strategy profiles generating the empty network are in an absorbing set if and only if one strategy profile generating the empty network is an element of an absorbing set. Theorem 2 states that for any strategy profile s such that g(s ) is not empty, we do not have (s, s ) ∈ R whenever c > 5/6 and whenever s is any strategy profile such that g(s) is empty. Hence, we can take as granted that, when c > 5/6, the set of strategy profiles generating the empty network is absorbing. However, the same reasoning cannot be done if we consider the GVR’s notion of an equilibrium. Indeed, assume that there exists a strategy profile s such that g(s) is empty and such that for any strategy profile s such that g(s ) is not empty, we do not have (s, s ) ∈ R. This assumption is not enough to claim that for any strategy profile s such that g(s) is empty and for any strategy profile s such that g(s ) is not empty, we do not have (s, s ) ∈ R. Hence, it is not necessarily the case that s is an element of an absorbing set. This is proved by considering Theorem 1 and Proposition 1. For 1/2 < c < 2/3, there exists an SBE s such that g(s) is empty and still, s is not in any absorbing set. This example shows that in the simplest dynamic model, it might be more interesting to consider all the strategies generating a given network in order to assess its equilibrium properties. We do not claim that GVR’s Theorem 1 is of no interest. However, we claim that it relies on a very important implicit assumption, namely that it is enough that one strategy be an equilibrium for the network generated by it to be also called an equilibrium. This last section shows that this definition of an equilibrium might hide an important feature: a network can be SBE supported (as in GVR) and still not be an element of any absorbing set even when the deviations allowed in the dynamic process are those allowed in the definition of an equilibrium. Said differently, a network can be SBE supported and still not be observed as the endpoint of the simplest dynamic model allowing deviations consistent with the definition of an SBE. Considering fully SBE sup3 The proof is simple but tedious. Hence, it is omitted. However, it is available upon request. 4 I.e. ∀i, j ∈ N, s = 1. ij
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ported networks makes this criticism impossible. Indeed, any fully SBE supported network is an acceptable endpoint of the same dynamic model. Appendix A. Proof of Theorem 2 “Only if”: shown by Example 1. “If”: Let c > 5/6. Let us show that for any strategy s such that g(s) is empty, condition (1) of Definition 1 is satisfied. Let us denote by γ i the strategy profile defined by: • γiji = 0, ∀j ∈ N , • γjii = 1, ∀j ∈ N , • γjik = 0, ∀j, k ∈ N \ {i}. By definition, gj k (γ i ) = 0, ∀j, k ∈ N . Obviously, condition (1) of Definition 1 is satisfied for any strategy profile and any player deviating if and only if it is satisfied for γ i and i as the deviating player, since γ i is the strategy profile that offers the most link formation possibilities to i and still generates the empty network. Then, i has the possibility to form a link with one of the other players or both of them. Then, he has a payoff ηi (1/2 − c) if he creates ηi links. This is strictly negative when c > 5/6 > 1/2. Let us show that for any strategy s such that g(s) is empty, condition 2 of Definition 1 is satisfied. Let us have i, j, k ∈ {1, 2, 3} mutually different. Let us denote by Γ ij the strategy profile defined by: ij
ij
ij
ij
ij
ij
• Γij = Γj i = 0, • Γki = Γkj = 1, • Γik = Γj k = 0. By definition, gij (γ ij ) = gik (γ ij ) = gj k (γ ij ) = 0. Obviously, condition (2) of Definition 1 is satisfied for any strategy profile and any pair of players deviating if and only if it is satisfied for Γ ij and i and j as deviating players, since Γ ij is the strategy profile that offers the most link formation possibilities to i and j and still generates the empty network. It is straightforward to check that the most profitable moves for both players i and j coincide in forming a link with k. In this case, their payoffs are 5/6 − c. They are both strictly positive when c > 5/6. 5. Proof of Theorem 3 Let c be given. Let us have n = 2k + 2 for some integer k. Let us have s defined by: • ∀i ∈ {1, . . . , k}, si,k+1 = 1, • ∀i ∈ {k + 3, . . . , 2k + 2}, si,k+2 = 1, • sij = 0 for any other ordered pair of players.
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By definition, g(s) is empty. Then, ∀i ∈ N , Πi (s) = 0. However, let us consider the deviations of players k + 1 and k + 2 consisting in forming s such that: • • • •
∀i ∈ {1, . . . , k}, si,k+1 = 1, sk+1,i = 1, ∀i ∈ {k + 3, . . . , 2k + 2}, si,k+2 = 1, sk+2,i = 1, sk+1,k+2 = 1, sk+2,k+1 = 1, sij = 0 for any other ordered pair of players.
Then, g(s ) is given by: • • • •
∀i ∈ {1, . . . , k}, gi,k+1 (s ) = 1, ∀i ∈ {k + 3, . . . , 2k + 2}, gi,k+2 (s ) = 1, gk+1,k+2 (s ) = 1, gij (s ) = 0 for any other pair of players. 2
k+1 + k4 + 2k Then, k + 1’s payoff is given by Πk+1 (s ) = k(k−1) 3 3 + 2 − (k + 1)c. The first term corresponds to k + 1 being essential for pairs of players in {1, . . . , k}. The second term corresponds to k + 1 being essential for pairs of players in {1, . . . , k} and players in {k + 3, . . . , 2k + 2}. The third term corresponds to paths between k + 1 and players in {k + 3, . . . , 2k + 2} and between k + 2 and players in {1, . . . , k}. The fourth term corresponds to direct links between k + 1 and players in {1, . . . , k} ∪ {k + 2}. The last term corresponds to the cost to pay for having links with players in {1, . . . , k} ∪ {k + 2}. For symmetry reasons, Πk+1 (s ) = Πk+2 (s ). Hence, for any c, it is possible to have k large enough so that Πk+1 (s ) = Πk+2 (s ) > 0.
6. Proof of Theorem 4 “Only if”: Let us have c 2/3. Let us denote by γ the strategy profile defined by: • γ1,j = γj,1 = 1, ∀j ∈ N \ {1}, • γ2,3 = γ4,3 = γ4,5 = γ6,5 = 1, • γj,k = 0 for any other ordered pair of players. Clearly, by definition, g(γ ) is a star with 1 in the center. Hence, Π3 (γ ) = Π5 (γ ) = 1/2 + (n − 2)/3 − c. Now, let us consider the deviation by players 3 and 5, leading to the strategy profile γ defined by: • • • •
= 1, ∀j ∈ N \ {1}, γ1,j = 1, ∀j ∈ N \ {1, 3, 5}, γj,1 = γ = γ = γ = γ = γ = γ = γ = 1, γ2,3 3,2 3,4 4,3 4,5 5,4 6,5 5,6 = 0 for any other ordered pair of players. γj,k
By definition, g(γ ) is such that: • g1,j (γ ) = 1, ∀j ∈ N \ {1, 3, 5}, • g2,3 (γ ) = g3,4 (γ ) = g4,5 (γ ) = g5,6 (γ ) = 1, • gi,j (γ ) = 0 for any other pair of players.
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Then, by definition, Π2 (γ ) = Π5 (γ ) = 5/2 + (n − 6)/3 − 2c. Hence, Π2 (γ ) = Π5 (γ ) Π2 (γ ) = Π5 (γ ) when c 2/3. “If”: Let c > 2/3 and let s be a strategy profile such that g(s) is a star and such that there can be unilateral intended links announcements. Let us study i’s incentive to deviate. It is straightforward to check that if i is the center of the star, then he has no incentive to deviate unilaterally in s if n is large enough (the proof is the same as in GVR). Let us show that if i is a peripheral player, he has no incentive to deviate unilaterally in s. Let us consider that in s, ∀k ∈ N \ {i}, ski = 1. Of course, i has no incentive to deviate in this strategy profile if and only if he has no incentive to deviate in any strategy profile such that the resulting network is a star. In the star, Πi (s) = 1/2 + (n − 2)/3 − c. If i deletes all his links, his payoff is null, hence he has no incentive to do that if n is large enough. If i deviates and creates m 1 links, then, the best payoff he can have is m/2 + (n − m − 1)/3 − mc since the center of the star keeps being essential for i and the players he does not create a link with. Moreover, i cannot be essential for any pair of players. Then, i decreases his payoff by deviating when c > 2/3 > 1/6. Let us show that two players cannot deviate and both (weakly) increase their payoffs. Let us have i and j the two players deviating. I. Assume that i is the center of the star and j is a peripheral player in g(s). Let us consider that in s, ∀k ∈ N \ {j }, skj = 1. Of course, i and j have no incentive to deviate in this strategy profile if and only if they have no incentive to deviate in any strategy profile such that the resulting network is a star. Let s be the strategy profile after deviation by i and j . Let A = {k ∈ N \ {i, j }, gki (s ) = gkj (s ) = 1}, B = {k ∈ N \ {i, j }, gki (s ) = 1, gkj (s ) = 0}, C = {k ∈ N \ {i, j }, gki (s ) = 0, gkj (s ) = 1}, D = {k ∈ N \ {i, j }, gki (s ) = gkj (s ) = 0}. By definition, #A + #B + #C + #D = n − 2. (I.1) Let us first assume that gij (s ) = 1. Then, by definition, Πi (s ) = 1/2 + (#A + #B)/2 + (#B + #C)/3 + #B(#B − 1)/6 + #A#B/3 + #B#C/4 − (#A + #B + 1)c and Πj (s ) = 1/2 + (#A + #C)/2 + (#B + #C)/3 + #C(#C − 1)/6 + #A#C/3 + #B#C/4 − (#A + #C + 1)c. To have Πi (s ) Πi (s) = (n − 1)/2 + (n − 1)(n − 2)/6 − (n − 1)c, we necessarily have #B = O(n). Obviously, when #B = n − 2, #A = #C = #D = 0, Πi (s ) = Πi (s). Let us write, with no loss of generality Πi (s ) = Πi (#A, #B, #C, #D). Let us compute: Πi (#A + 1, #B, #C, #D) − Πi (#A, #B, #C, #D) = 1/2 + #B/3 − c,
(A.1)
Πi (#A, #B + 1, #C, #D) − Πi (#A, #B, #C, #D) = 5/6 + (#B + #A)/3 + #C/4 − c,
(A.2)
Πi (#A, #B, #C + 1, #D) − Πi (#A, #B, #C, #D) = 1/3 + #B/4,
(A.3)
Πi (#A, #B, #C, #D + 1) − Πi (#A, #B, #C, #D) = 0.
(A.4)
Then, it is straightforward to check that a decrease of #B by one unit and an increase of #A, #C or #D by one unit strictly lowers i’s payoff when #B = O(n) and n is large enough. Then, any other combination different from #A = 0, #B = n − 2, #C = 0, #D = 0 would decrease i’s payoff. Hence, any strategy profile different from s would decrease i’s payoff. (I.2) Second, let us assume that gij (s ) = 0. (a) Let us assume that #A = 0. The same proof as the proof of condition 1 applies and i’s payoff necessarily decrease. (b) Let us assume that #A = 1. Then, Πi (s ) = #B/2 + 1/2 + 1/3 + #C/4 + #B/3 + #B/4 + #B#C/5 + #B(#B − 1)/6 − (#B + 1)c. The same reasoning as above would show that #B = O(n) and that the maximum payoff for i would be obtained for #A = 1 (by assumption), #B = n − 3, #C = 0, #D = 0.
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However, with s such that #A = 1, #B = n − 3, #C = 0, #D = 0, it is straightforward to show that Πi (s ) < Πi (s) when n is large enough. c) Let us assume that #A 2. Then, we have Πi (s ) = 1/2 + (#A + #B)/2 + (#B + #C)/3 + #B(#B − 1)/6 + #A#B/3 + #B#C/4 − (#A + #B)c. Πj (s ) = 1/2 + (#A + #C)/2 + (#B + #C)/3 + #C(#C − 1)/6 + #A#C/3 + #B#C/4 − (#A + #C)c. The same reasoning as above would show that #B = O(n). Notice that Eqs. (A.1)– (A.4) still apply. Then, it is straightforward to show that a decrease of one unit of #D and an increase of one unit of #C increases both i and j ’s payoff. Then, if s exists such that Πi (s) Πi (s ) and Πj (s) Πj (s ), then, such s exists with #D = 0. With the same reasoning as above, the maximum payoff for i is obtained when #A = 2 (by assumption), #B = n − 4, #C = 0, #D = 0. This maximum payoff is Πi (2, n − 4, 0, 0) = (n − 1)/2 + (n + 1)(n − 4)/6 − (n − 2)c = Πi (s) + c − 1. Then, we necessarily have #C = 0 when n is large enough, since an increase of one unit of #B and an increase of one unit of #C implies a loss of payoff of the order of #B and then, of n for i. Then, #A + #B = n − 2. Hence, j ’s payoff is given by Πj (#A, n − 2 − #A, 0, 0) = (#A + 1)/2 + (n − 2 − #A)/3 − #Ac. Then, it is straightforward to check that Πj (#A, n − 2 − #A, 0, 0) < Πj (0, n, 0, 0) = 1/2 + (n − 2)/3 − c if #A 2 and c > 2/3. Then, it is not possible to have s with #A 2 such that Πi (s) Πi (s ) and Πj (s) Πj (s ). II. Assume that i and j are peripheral players. Let us consider that in s, ∀k ∈ N \ {i, j }, skj = ski = 1. Of course, i and j have no incentive to deviate in this strategy profile if and only if they have no incentive to deviate in any strategy profile such that the resulting network is a star. Let s be the strategy profile after deviation by i and j . Let l ∈ N \ {i, j } be the center of the star. Let A = {k ∈ N \ {i, j, l}, gki (s ) = gkj (s ) = 1}, B = {k ∈ N \ {i, j, l}, gki (s ) = 1, gkj (s ) = 0}, C = {k ∈ N \ {i, j, l}, gki (s ) = 0, gkj (s ) = 1}, D = {k ∈ N \ {i, j, l}, gki (s ) = gkj (s ) = 0}. By definition, #A + #B + #C + #D = n − 3. (II.1) Let us assume that gil (s ) = gj l (s ) = 1. (a) Let us assume that gij (s ) = 0 and #A 1. By definition, Πi (s ) = (#A + #B + #C + 2)/2 + #D/3 − (#A + #B + 1)c and Πj (s ) = (#A + #B + #C + 2)/2 + #D/3 − (#A + #C + 1)c. Then, it is straightforward to check that Πi (s ) + Πj (s ) is maximum when #A = 1, #B = 0, #C = 0, #D = n − 4. It is straightforward to compute Πi (#A = 1, #B = 0, #C = 0, #D = n − 4) + Πj (#A = 1, #B = 0, #C = 0, #D = n − 4) = 3 + 2(n − 4)/3 − 4c. Then, we have Πi (s ) + Πj (s ) < Πi (s) + Πj (s). Hence, i and j cannot have both a higher payoff in s than in s. (b) The proof is similar if gij (s ) = 1 and whatever the value of #A. (c) Let us assume that gij (s ) = 0 and #A = 0. The proof is identical to the one given above showing that no peripheral player has an incentive to deviate unilaterally. (II.2) The proofs are similar when gil (s ) = gj l (s ) = 0 and all the cases are considered. (II.3) The proofs are similar when gil (s ) = 1 and gj l (s ) = 0 and all the cases are considered. References [1] S. Goyal, F. Vega-Redondo, Structural holes in social networks, J. Econ. Theory 137 (1) (2007) 460–492. [2] E. Kalai, E.A. Pazner, D. Schmeidler, Admissible outcomes of social bargaining processes as collective choice correspondence, Econometrica 63 (1976) 299–325. [3] L. Shenoy, On coalition formation: A game theoretical approach, Int. J. Game Theory 8 (1979) 133–164. [4] T. Schwartz, Notes on the Abstract Theory of Collective Choice, School of Urban and Public Affairs, Carnegie– Mellon University, 1974.
Note
Structural holes in social networks: A remark Nicolas Houy CNRS, Laboratoire d’Econométrie-Ecole Polytechnique, 5 rue Descartes, 75005 Paris, France Received 18 April 2007; final version received 7 May 2008; accepted 27 May 2008 Available online 1 July 2008
Abstract In a recent article, [S. Goyal, F. Vega-Redondo, Structural holes in social networks, J. Econ. Theory 137 (1) (2007) 460–492] the authors (GVR) showed the importance of stars and cycles in a given network formation game. Implicitly, in their article, a network is called an equilibrium if it is generated by an equilibrium strategy. We extend the results of GVR to the case of a stronger requirement: namely, that a network can be called an equilibrium only if all the strategies generating it are equilibria. We also show, in a dynamic framework, that both definitions differ in crucial ways. © 2008 Published by Elsevier Inc. JEL classification: C72; C78; D85 Keywords: Networks; Network formation; Structural holes; Intermediation
1. Introduction In a recent article, [1] Goyal and Vega-Redondo (GVR, henceforth) showed that empty networks, stars and cycles are natural equilibria of a given network formation game. In the game they consider, a player increases his payoff (1) when he is linked to other players (possibly indirectly), and (2) when he is essential (i.e. necessary) for linking pairs of players. Moreover, a player’s payoff decreases (1) when he creates direct costly links, and (2) when he rewards essential players between him and other players. E-mail address: [email protected]. 0022-0531/$ – see front matter © 2008 Published by Elsevier Inc. doi:10.1016/j.jet.2008.05.007
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Implicitly, in GVR, a network is called an equilibrium if there exists a strategy profile generating it and that satisfy some equilibrium conditions. We propose a stronger definition of an equilibrium: a network is an equilibrium only if all the strategy profiles generating it satisfy the equilibrium conditions. While the first definition characterizes the networks that can be observed at equilibrium, the second definition characterizes the networks that, when observed, are necessarily equilibria. We claim that the second definition can lead to different and important results. First, if we assume that networks are observable but strategies are not, as it is often assumed, then, when using GVR’s definition and observing a network satisfying an equilibrium property, we cannot conclude that the society has reached an equilibrium state. On the contrary, with the definition we propose, drawing this conclusion is logically possible. Second, as we will show in the last section, GVR’s definition hardly implies any stability feature in the simplest dynamic setting. The definition we propose does. In the first section, we will show that in the case of the empty network, extending GVR’s results to the stronger definition of a network equilibrium is not possible whatever the values of the parameters if the number of players is large enough. In the case of stars, generalizing GVR’s result is possible if the parameters are modified. In the second section, we will show that in the case of cycles, extending GVR’s result is possible if the definition of an equilibrium is slightly modified. These results make the interpretation of GVR, emphasizing the role of stars and cycles, even more justified. In the last section, we show how crucial the difference between the two definitions of an equilibrium can be in some dynamic framework. 2. Notation A population is composed of a finite set of ex-ante identical agents, N = {1, 2, . . . , n} with n 3. These agents play a network formation game in which every one of them makes a simultaneous announcement of intended links. The strategy profile s = {sij }i,j ∈N is the matrix of all announcements of intended links where sij = 1 if individual i announces a wish to form a link with j and sij = 0 if individual i does not announce a wish to form a link with j . Let us denote by Si the set of all possible strategies for individual i. Let S be the set of all strategy profiles. An undirected link is formed between individuals i and j if and only if sij = sj i = 1. Let us denote by g(s) = {gij (s)}i,j ∈N the network resulting from announcements s, where gij (s) = gj i (s) = 1 if a link is formed between individuals i and j and gij (s) = gj i (s) = 0 if no link is formed between individuals i and j . For a player i, let ηi (g(s)) denote the number of players i has a link with, i.e. ηi (g(s)) = #{j ∈ N \ {i}, gij (s) = 1}. In the network g, there exists a path between i and j (i = j ) if, either gij = 1 or if there is a sequence of players (i1 , . . . , im ) such that gi,i1 = gi1 ,i2 = gi2 ,i3 = · · · = gim ,j = 1. Let us denote by Ci (g) the set of players j such that there is a path between i and j in g. A player i ∈ / {j, k} is essential for j and k in g if i lies on every path between j and k in the network g. Let us denote by E(j, k; g) the set of players who are essential for j and k in g and let e(j, k; g) = #E(j, k; g) be the cardinality of E(j, k; g). Finally, for a strategy profile s, the payoff earned by player i is given by 1 1 Πi (s) = + − ηi g(s) c e(i, j ; g(s)) + 2 e(j, k; g(s)) + 2 j ∈Ci (g)
j,k∈N, i∈E(j,k;g(s))
where c is the cost of forming a link. It should be noted that there is no cost of proposing a link. A cost is paid only when a link is formed.
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3. Results and comments 3.1. Strict bilateral equilibria Let us give the definition of a Strict Bilateral Equilibrium as stated in GVR. Definition 1. A strategy profile s ∗ is a Strict Bilateral Equilibrium (SBE) if the following two conditions hold: ∗ ) = g(s ∗ ), Π (s ∗ ) > Π (s , s ∗ ); (1) for any i ∈ N and every si ∈ Si such that g(si , s−i i i i −i ∗ ) = g(s ∗ ), (2) for any pair of players i, j ∈ N and every strategy pair (si , sj ) with g(si , sj , s−i−j ∗ ∗ Πi (s ∗ ) ⇒ Πj si , sj , s−i−j < Πj (s ∗ ). Πi si , sj , s−i−j
Then, a strategy profile is an SBE if and only if (1) no player has a (weak) incentive to deviate and (2) no pair of players can be (weakly) better off by deviating. Obviously the notion of equilibrium applies to strategies, not to networks. In order to avoid any ambiguity, we will say that a network g is SBE supported if and only if there exists an SBE s such that g = g(s). Stronger, we will say that a network g is fully SBE supported if and only if any s such that g = g(s) is an SBE. Notice that, implicitly, GVR use only the definition of SBE supported networks. The first theorem of GVR is the following. Theorem 1. If c > 1/2, the empty network is SBE supported. Then, the empty network can be observed at equilibrium when c > 1/2. However, it might be possible that not all the strategies generating the empty network be SBEs. Indeed, the next example shows that for 5/6 c > 1/2, there exists a strategy profile s such that g(s) is empty and still s is not an SBE. / Example 1. Let us have n 3, and s such that s2,1 = s2,3 = 1 and sij = 0, ∀(i, j ) ∈ {(2, 1), (2, 3)}. By definition, gij (s) = 0, ∀i, j ∈ N . Then, g(s) is empty and Πi (s) = 0, ∀i ∈ N . Let us show that s is not an SBE if 5/6 c > 1/2. Let us consider the strategy profile s such / {(1, 2), (2, 1), (2, 3), (3, 2)}. By defithat s 2,1 = s 1,2 = s 2,3 = s 3,2 = 1 and s ij = 0, ∀(i, j ) ∈ / {(1, 2), (2, 1), (2, 3), (3, 2)}. Said, differently, s is nition, g12 = g23 = 1 and gij (s) = 0, ∀(i, j ) ∈ obtained from s when players 1 and 3 change their strategies to propose a link to player 2. These links are formed since in s, 2 is unilaterally announcing a wish to form a link with players 1 and 3. Then, Π1 (s ) = Π3 (s ) = 5/6 − c. Then, players 1 and 3 can benefit from a change of their strategies. Hence, s is not an SBE since condition (2) of Definition 1 does not hold. Then, the condition c > 1/2 is not sufficient to state that the empty network is fully SBE supported. The next theorem shows that, for three individuals, the condition c > 5/6 is necessary and sufficient to state that the empty network is necessarily generated by an SBE. Theorem 2. Let n = 3. The empty network is fully SBE supported if and only if c > 5/6. Generalizing Theorem 1 for any number of players is not straightforward. Indeed, for any cost c, there exists a large enough number of players such that the empty network is not fully SBE supported.
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Theorem 3. Let c be given. If n is sufficiently large then, the empty network is not fully SBE supported. Theorem 3 shows that only some of the strategy profiles generating the empty network are SBEs when n grows large. This result minimizes the role played by the empty network in the game considered by GVR. A result similar to Theorem 1 is given by GVR for stars when c > 1/6 and n is large enough. In order to generalize this result to all the strategy profiles generating stars, it is enough to consider a cost greater than 2/3. This result strengthens GVR’s interpretation that stars arise as natural equilibria of the studied network formation game. Theorem 4. Let n be arbitrarily large. Any star is fully SBE supported if and only if c > 2/3. 3.2. Bilateral proof equilibria Another important result of GVR consists in showing that cycles containing all the players can arise as equilibria. In order to obtain this result, one needs to make sure that in the case of a bilateral deviation, the players involved will actually implement the deviation. Such an equilibrium is defined as a Strict Bilateral-Proof Equilibrium (SBPE). Definition 2. A strategy profile s ∗ is a Strict Bilateral-Proof Equilibrium (SBPE) if the following two conditions hold: ∗ ) = g(s ∗ ), Π (s ∗ ) > Π (s , s ∗ ); (1) for any i ∈ N and every si ∈ Si such that g(si , s−i i i i −i ∗ (2) for any pair of players i, j ∈ N and every strategy pair (si , sj ) with g(si , sj , s−i−j ) = g(s ∗ ) ∗ ∗ and Πi (si , sj , s−i−j ) Πi (s ), one of the following holds: ∗ ) < Πj (s ∗ ), (a) Πj (si , sj , s−i−j ∗ ∗ ) > Πk (sk , sl , s−k−l ). (b) ∃k, l ∈ {i, j }, k = l and some s˜k ∈ Sk , such that Πk (˜sk , sl , s−k−l
As above, we will say that a network g is SBPE supported if and only if there exists an SBPE s such that g = g(s). We will say that a network g is fully SBPE supported if and only if any s such that g = g(s) is an SBPE. It is shown in GVR that for any cost c, if there are sufficiently many players, any cycle containing all players is SBPE supported. However, we can that, if n 4, this result does not hold if we consider fully SBPE supported cycles. The following definition permits to generalize GVR’s result. Definition 3. A strategy profile s ∗ is a Bilateral-Proof Equilibrium (BPE) if the following two conditions hold: ∗ ) = g(s ∗ ), Π (s ∗ ) > Π (s , s ∗ ); (1) for any i ∈ N and every si ∈ Si such that g(si , s−i i i i −i ∗ (2) for any pair of players i, j ∈ N and every strategy pair (si , sj ) with g(si , sj , s−i−j ) = g(s ∗ ) ∗ ) > Πi (s ∗ ), one of the following holds: and Πi (si , sj , s−i−j ∗ (a) Πj (si , sj , s−i−j ) < Πj (s ∗ ), ∗ ∗ ) > Πk (sk , sl , s−k−l ). (b) ∃k, l ∈ {i, j }, k = l and some s˜k ∈ Sk , such that Πk (˜sk , sl , s−k−l
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Obviously, a BPE is an SBPE. The only difference arises when the two players of a pair earn an unchanged payoff when modifying their strategies. Now, we could show that any cycle containing all players is fully BPE supported if there are sufficiently many players.1 Theorem 5. For any c > 0, there exists a sufficiently large number of players such that any cycle containing all players is fully BPE supported. 4. A dynamic framework In this section, we show how crucial the difference is between a supported equilibrium network and a fully supported equilibrium network. In a simple dynamic framework, we will show that it is important that all strategies generating a given network be considered in order to assess that this network has a stability feature. For the sake of simplicity, let us consider the network formation game and the notions given in Section 3.1 with only three players. In accordance with the definition of a strict bilateral equilibrium, we will set the following stochastic dynamic process. First, let s, s be two strategy profiles. We say that s is attainable from s if and only if: • #{i ∈ N, si = si } 2 and, • ∀i ∈ N, si = si ⇒ Πi (s ) Πi (s). The corresponding dynamic process can be given as follows. From an initial strategy profile s, we draw by chance a set of individuals with cardinality 1 or 2. New strategies are proposed to these individuals, forming the new strategy profile s . For the sake of simplicity, we will lack rigor in not specifying the probability laws. We only need to assume that there exists a real > 0 such that any set of 1 or 2 individuals be drawn with a probability greater than in the first step and any strategy or pair of strategies be drawn with a probability greater than in the second step. If one of the individuals drawn by chance gets a strictly lower payoff in s , then he refuses the change of state and s is not attainable from s. Otherwise, s is attainable from s. Let us make two remarks about this dynamic process. First, individuals are myopic. They only consider the difference in payoff when evaluating the possibility of a new strategy profile. They do not anticipate any further change that could decrease their payoff. Second, we suppose that a change of strategy profile can occur even if there is no strict increase of the payoffs of the individuals involved. In doing so, we are consistent with Definition 1 of a strict bilateral equilibrium.2 Now we can define the binary relation R ∈ S × S as follows: an ordered pair of strategy profiles (s, s ) is an element of R if and only if s is attainable from s. We will call R T the transitive closure of R. In order to characterize the acceptable solutions of this abstract problem (a set endowed with a binary relation), we use the concept of absorbing sets introduced and studied by [2,3] and [4]. Let V ∈ 2S be a set of strategy profiles, we say that V is absorbing if and only if: (1) ∀s, s ∈ V , (s, s ) ∈ R T , and (2) ∀s ∈ V , ∀s ∈ S \ V , not[(s, s ) ∈ R T ]. We can interpret absorbing sets as follows: once an element of an absorbing set is reached, when repeating the dynamic process, all the elements of the same absorbing set will be reached an 1 The proof is omitted and available on request. 2 This has been noted by GVR, p. 465.
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infinite number of times while all the elements not in the absorbing set will never be reached again. Moreover, the probability of reaching an element of an absorbing goes asymptotically to 1 when the dynamic process is repeated an infinite number of times. It is well known that an absorbing set always exists but that it is not necessarily unique. However, when it is unique, the probability of reaching any element of the absorbing set in the future goes asymptotically to 1 when the dynamic process is repeated an infinite number of times. The following proposition characterizes the absorbing set for any value of the cost c.3 Proposition 1. Let N = 3. For any cost of forming a link c, the absorbing set is unique. Moreover, (1) if 0 < c < 1/6, s is an element of the absorbing set if and only if it is complete4 ; (2) if 1/6 < c < 2/3, s is an element of the absorbing set if and only if the network generated by s shows two links; (3) if 2/3 < c < 5/6, s is an element of the absorbing set if and only if the network generated by s shows two links or is empty; (4) if 5/6 < c, s is an element of the absorbing set if and only if the network generated by s is empty. Obviously, for all strategy profiles s and s such that g(s) and g(s ) are empty, we have (s, s ) ∈ R T . Then, by definition, all the strategy profiles generating the empty network are in an absorbing set if and only if one strategy profile generating the empty network is an element of an absorbing set. Theorem 2 states that for any strategy profile s such that g(s ) is not empty, we do not have (s, s ) ∈ R whenever c > 5/6 and whenever s is any strategy profile such that g(s) is empty. Hence, we can take as granted that, when c > 5/6, the set of strategy profiles generating the empty network is absorbing. However, the same reasoning cannot be done if we consider the GVR’s notion of an equilibrium. Indeed, assume that there exists a strategy profile s such that g(s) is empty and such that for any strategy profile s such that g(s ) is not empty, we do not have (s, s ) ∈ R. This assumption is not enough to claim that for any strategy profile s such that g(s) is empty and for any strategy profile s such that g(s ) is not empty, we do not have (s, s ) ∈ R. Hence, it is not necessarily the case that s is an element of an absorbing set. This is proved by considering Theorem 1 and Proposition 1. For 1/2 < c < 2/3, there exists an SBE s such that g(s) is empty and still, s is not in any absorbing set. This example shows that in the simplest dynamic model, it might be more interesting to consider all the strategies generating a given network in order to assess its equilibrium properties. We do not claim that GVR’s Theorem 1 is of no interest. However, we claim that it relies on a very important implicit assumption, namely that it is enough that one strategy be an equilibrium for the network generated by it to be also called an equilibrium. This last section shows that this definition of an equilibrium might hide an important feature: a network can be SBE supported (as in GVR) and still not be an element of any absorbing set even when the deviations allowed in the dynamic process are those allowed in the definition of an equilibrium. Said differently, a network can be SBE supported and still not be observed as the endpoint of the simplest dynamic model allowing deviations consistent with the definition of an SBE. Considering fully SBE sup3 The proof is simple but tedious. Hence, it is omitted. However, it is available upon request. 4 I.e. ∀i, j ∈ N, s = 1. ij
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ported networks makes this criticism impossible. Indeed, any fully SBE supported network is an acceptable endpoint of the same dynamic model. Appendix A. Proof of Theorem 2 “Only if”: shown by Example 1. “If”: Let c > 5/6. Let us show that for any strategy s such that g(s) is empty, condition (1) of Definition 1 is satisfied. Let us denote by γ i the strategy profile defined by: • γiji = 0, ∀j ∈ N , • γjii = 1, ∀j ∈ N , • γjik = 0, ∀j, k ∈ N \ {i}. By definition, gj k (γ i ) = 0, ∀j, k ∈ N . Obviously, condition (1) of Definition 1 is satisfied for any strategy profile and any player deviating if and only if it is satisfied for γ i and i as the deviating player, since γ i is the strategy profile that offers the most link formation possibilities to i and still generates the empty network. Then, i has the possibility to form a link with one of the other players or both of them. Then, he has a payoff ηi (1/2 − c) if he creates ηi links. This is strictly negative when c > 5/6 > 1/2. Let us show that for any strategy s such that g(s) is empty, condition 2 of Definition 1 is satisfied. Let us have i, j, k ∈ {1, 2, 3} mutually different. Let us denote by Γ ij the strategy profile defined by: ij
ij
ij
ij
ij
ij
• Γij = Γj i = 0, • Γki = Γkj = 1, • Γik = Γj k = 0. By definition, gij (γ ij ) = gik (γ ij ) = gj k (γ ij ) = 0. Obviously, condition (2) of Definition 1 is satisfied for any strategy profile and any pair of players deviating if and only if it is satisfied for Γ ij and i and j as deviating players, since Γ ij is the strategy profile that offers the most link formation possibilities to i and j and still generates the empty network. It is straightforward to check that the most profitable moves for both players i and j coincide in forming a link with k. In this case, their payoffs are 5/6 − c. They are both strictly positive when c > 5/6. 5. Proof of Theorem 3 Let c be given. Let us have n = 2k + 2 for some integer k. Let us have s defined by: • ∀i ∈ {1, . . . , k}, si,k+1 = 1, • ∀i ∈ {k + 3, . . . , 2k + 2}, si,k+2 = 1, • sij = 0 for any other ordered pair of players.
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By definition, g(s) is empty. Then, ∀i ∈ N , Πi (s) = 0. However, let us consider the deviations of players k + 1 and k + 2 consisting in forming s such that: • • • •
∀i ∈ {1, . . . , k}, si,k+1 = 1, sk+1,i = 1, ∀i ∈ {k + 3, . . . , 2k + 2}, si,k+2 = 1, sk+2,i = 1, sk+1,k+2 = 1, sk+2,k+1 = 1, sij = 0 for any other ordered pair of players.
Then, g(s ) is given by: • • • •
∀i ∈ {1, . . . , k}, gi,k+1 (s ) = 1, ∀i ∈ {k + 3, . . . , 2k + 2}, gi,k+2 (s ) = 1, gk+1,k+2 (s ) = 1, gij (s ) = 0 for any other pair of players. 2
k+1 + k4 + 2k Then, k + 1’s payoff is given by Πk+1 (s ) = k(k−1) 3 3 + 2 − (k + 1)c. The first term corresponds to k + 1 being essential for pairs of players in {1, . . . , k}. The second term corresponds to k + 1 being essential for pairs of players in {1, . . . , k} and players in {k + 3, . . . , 2k + 2}. The third term corresponds to paths between k + 1 and players in {k + 3, . . . , 2k + 2} and between k + 2 and players in {1, . . . , k}. The fourth term corresponds to direct links between k + 1 and players in {1, . . . , k} ∪ {k + 2}. The last term corresponds to the cost to pay for having links with players in {1, . . . , k} ∪ {k + 2}. For symmetry reasons, Πk+1 (s ) = Πk+2 (s ). Hence, for any c, it is possible to have k large enough so that Πk+1 (s ) = Πk+2 (s ) > 0.
6. Proof of Theorem 4 “Only if”: Let us have c 2/3. Let us denote by γ the strategy profile defined by: • γ1,j = γj,1 = 1, ∀j ∈ N \ {1}, • γ2,3 = γ4,3 = γ4,5 = γ6,5 = 1, • γj,k = 0 for any other ordered pair of players. Clearly, by definition, g(γ ) is a star with 1 in the center. Hence, Π3 (γ ) = Π5 (γ ) = 1/2 + (n − 2)/3 − c. Now, let us consider the deviation by players 3 and 5, leading to the strategy profile γ defined by: • • • •
= 1, ∀j ∈ N \ {1}, γ1,j = 1, ∀j ∈ N \ {1, 3, 5}, γj,1 = γ = γ = γ = γ = γ = γ = γ = 1, γ2,3 3,2 3,4 4,3 4,5 5,4 6,5 5,6 = 0 for any other ordered pair of players. γj,k
By definition, g(γ ) is such that: • g1,j (γ ) = 1, ∀j ∈ N \ {1, 3, 5}, • g2,3 (γ ) = g3,4 (γ ) = g4,5 (γ ) = g5,6 (γ ) = 1, • gi,j (γ ) = 0 for any other pair of players.
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Then, by definition, Π2 (γ ) = Π5 (γ ) = 5/2 + (n − 6)/3 − 2c. Hence, Π2 (γ ) = Π5 (γ ) Π2 (γ ) = Π5 (γ ) when c 2/3. “If”: Let c > 2/3 and let s be a strategy profile such that g(s) is a star and such that there can be unilateral intended links announcements. Let us study i’s incentive to deviate. It is straightforward to check that if i is the center of the star, then he has no incentive to deviate unilaterally in s if n is large enough (the proof is the same as in GVR). Let us show that if i is a peripheral player, he has no incentive to deviate unilaterally in s. Let us consider that in s, ∀k ∈ N \ {i}, ski = 1. Of course, i has no incentive to deviate in this strategy profile if and only if he has no incentive to deviate in any strategy profile such that the resulting network is a star. In the star, Πi (s) = 1/2 + (n − 2)/3 − c. If i deletes all his links, his payoff is null, hence he has no incentive to do that if n is large enough. If i deviates and creates m 1 links, then, the best payoff he can have is m/2 + (n − m − 1)/3 − mc since the center of the star keeps being essential for i and the players he does not create a link with. Moreover, i cannot be essential for any pair of players. Then, i decreases his payoff by deviating when c > 2/3 > 1/6. Let us show that two players cannot deviate and both (weakly) increase their payoffs. Let us have i and j the two players deviating. I. Assume that i is the center of the star and j is a peripheral player in g(s). Let us consider that in s, ∀k ∈ N \ {j }, skj = 1. Of course, i and j have no incentive to deviate in this strategy profile if and only if they have no incentive to deviate in any strategy profile such that the resulting network is a star. Let s be the strategy profile after deviation by i and j . Let A = {k ∈ N \ {i, j }, gki (s ) = gkj (s ) = 1}, B = {k ∈ N \ {i, j }, gki (s ) = 1, gkj (s ) = 0}, C = {k ∈ N \ {i, j }, gki (s ) = 0, gkj (s ) = 1}, D = {k ∈ N \ {i, j }, gki (s ) = gkj (s ) = 0}. By definition, #A + #B + #C + #D = n − 2. (I.1) Let us first assume that gij (s ) = 1. Then, by definition, Πi (s ) = 1/2 + (#A + #B)/2 + (#B + #C)/3 + #B(#B − 1)/6 + #A#B/3 + #B#C/4 − (#A + #B + 1)c and Πj (s ) = 1/2 + (#A + #C)/2 + (#B + #C)/3 + #C(#C − 1)/6 + #A#C/3 + #B#C/4 − (#A + #C + 1)c. To have Πi (s ) Πi (s) = (n − 1)/2 + (n − 1)(n − 2)/6 − (n − 1)c, we necessarily have #B = O(n). Obviously, when #B = n − 2, #A = #C = #D = 0, Πi (s ) = Πi (s). Let us write, with no loss of generality Πi (s ) = Πi (#A, #B, #C, #D). Let us compute: Πi (#A + 1, #B, #C, #D) − Πi (#A, #B, #C, #D) = 1/2 + #B/3 − c,
(A.1)
Πi (#A, #B + 1, #C, #D) − Πi (#A, #B, #C, #D) = 5/6 + (#B + #A)/3 + #C/4 − c,
(A.2)
Πi (#A, #B, #C + 1, #D) − Πi (#A, #B, #C, #D) = 1/3 + #B/4,
(A.3)
Πi (#A, #B, #C, #D + 1) − Πi (#A, #B, #C, #D) = 0.
(A.4)
Then, it is straightforward to check that a decrease of #B by one unit and an increase of #A, #C or #D by one unit strictly lowers i’s payoff when #B = O(n) and n is large enough. Then, any other combination different from #A = 0, #B = n − 2, #C = 0, #D = 0 would decrease i’s payoff. Hence, any strategy profile different from s would decrease i’s payoff. (I.2) Second, let us assume that gij (s ) = 0. (a) Let us assume that #A = 0. The same proof as the proof of condition 1 applies and i’s payoff necessarily decrease. (b) Let us assume that #A = 1. Then, Πi (s ) = #B/2 + 1/2 + 1/3 + #C/4 + #B/3 + #B/4 + #B#C/5 + #B(#B − 1)/6 − (#B + 1)c. The same reasoning as above would show that #B = O(n) and that the maximum payoff for i would be obtained for #A = 1 (by assumption), #B = n − 3, #C = 0, #D = 0.
Notes / Journal of Economic Theory 144 (2009) 422–431
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However, with s such that #A = 1, #B = n − 3, #C = 0, #D = 0, it is straightforward to show that Πi (s ) < Πi (s) when n is large enough. c) Let us assume that #A 2. Then, we have Πi (s ) = 1/2 + (#A + #B)/2 + (#B + #C)/3 + #B(#B − 1)/6 + #A#B/3 + #B#C/4 − (#A + #B)c. Πj (s ) = 1/2 + (#A + #C)/2 + (#B + #C)/3 + #C(#C − 1)/6 + #A#C/3 + #B#C/4 − (#A + #C)c. The same reasoning as above would show that #B = O(n). Notice that Eqs. (A.1)– (A.4) still apply. Then, it is straightforward to show that a decrease of one unit of #D and an increase of one unit of #C increases both i and j ’s payoff. Then, if s exists such that Πi (s) Πi (s ) and Πj (s) Πj (s ), then, such s exists with #D = 0. With the same reasoning as above, the maximum payoff for i is obtained when #A = 2 (by assumption), #B = n − 4, #C = 0, #D = 0. This maximum payoff is Πi (2, n − 4, 0, 0) = (n − 1)/2 + (n + 1)(n − 4)/6 − (n − 2)c = Πi (s) + c − 1. Then, we necessarily have #C = 0 when n is large enough, since an increase of one unit of #B and an increase of one unit of #C implies a loss of payoff of the order of #B and then, of n for i. Then, #A + #B = n − 2. Hence, j ’s payoff is given by Πj (#A, n − 2 − #A, 0, 0) = (#A + 1)/2 + (n − 2 − #A)/3 − #Ac. Then, it is straightforward to check that Πj (#A, n − 2 − #A, 0, 0) < Πj (0, n, 0, 0) = 1/2 + (n − 2)/3 − c if #A 2 and c > 2/3. Then, it is not possible to have s with #A 2 such that Πi (s) Πi (s ) and Πj (s) Πj (s ). II. Assume that i and j are peripheral players. Let us consider that in s, ∀k ∈ N \ {i, j }, skj = ski = 1. Of course, i and j have no incentive to deviate in this strategy profile if and only if they have no incentive to deviate in any strategy profile such that the resulting network is a star. Let s be the strategy profile after deviation by i and j . Let l ∈ N \ {i, j } be the center of the star. Let A = {k ∈ N \ {i, j, l}, gki (s ) = gkj (s ) = 1}, B = {k ∈ N \ {i, j, l}, gki (s ) = 1, gkj (s ) = 0}, C = {k ∈ N \ {i, j, l}, gki (s ) = 0, gkj (s ) = 1}, D = {k ∈ N \ {i, j, l}, gki (s ) = gkj (s ) = 0}. By definition, #A + #B + #C + #D = n − 3. (II.1) Let us assume that gil (s ) = gj l (s ) = 1. (a) Let us assume that gij (s ) = 0 and #A 1. By definition, Πi (s ) = (#A + #B + #C + 2)/2 + #D/3 − (#A + #B + 1)c and Πj (s ) = (#A + #B + #C + 2)/2 + #D/3 − (#A + #C + 1)c. Then, it is straightforward to check that Πi (s ) + Πj (s ) is maximum when #A = 1, #B = 0, #C = 0, #D = n − 4. It is straightforward to compute Πi (#A = 1, #B = 0, #C = 0, #D = n − 4) + Πj (#A = 1, #B = 0, #C = 0, #D = n − 4) = 3 + 2(n − 4)/3 − 4c. Then, we have Πi (s ) + Πj (s ) < Πi (s) + Πj (s). Hence, i and j cannot have both a higher payoff in s than in s. (b) The proof is similar if gij (s ) = 1 and whatever the value of #A. (c) Let us assume that gij (s ) = 0 and #A = 0. The proof is identical to the one given above showing that no peripheral player has an incentive to deviate unilaterally. (II.2) The proofs are similar when gil (s ) = gj l (s ) = 0 and all the cases are considered. (II.3) The proofs are similar when gil (s ) = 1 and gj l (s ) = 0 and all the cases are considered. References [1] S. Goyal, F. Vega-Redondo, Structural holes in social networks, J. Econ. Theory 137 (1) (2007) 460–492. [2] E. Kalai, E.A. Pazner, D. Schmeidler, Admissible outcomes of social bargaining processes as collective choice correspondence, Econometrica 63 (1976) 299–325. [3] L. Shenoy, On coalition formation: A game theoretical approach, Int. J. Game Theory 8 (1979) 133–164. [4] T. Schwartz, Notes on the Abstract Theory of Collective Choice, School of Urban and Public Affairs, Carnegie– Mellon University, 1974.