Ge Jiang

# Abstract

In this paper, we consider a model of social coordination and network formation, where players of two groups play a 2 × 2 coordination game when they are connected. Players in one group actively decide on whom they play with and on the action in the game, while players in the other group decide on the action in the game only and passively accept all the connections from the active group. The players in the active group can connect to a limited number of opponents in the other group. We find that the selection of long-run outcomes is determined by the population size of each group, not the overall population size of them. If either group’s population size is small in comparison to the number of maximally allowed links, all players will choose the risk-dominant equilibrium, while when both groups are sufficiently large with respect to the number of maximally allowed links, the players of the two groups will coordinate on the payoff-dominant action.

Corresponding author: Ge Jiang, Department of Economics, Business School, Nanjing University, No. 22 Hankou Road, Nanjing, 210093, China, E-mail:

Funding source: National Natural Science Foundation of China

Award Identifier / Grant number: 71671010

## Appendix A Review of Techniques

We have denoted by s a state of the model, which specifies how M-players and F-players are connected and choose their actions in graph g, and we have denoted by S the set of states. An absorbing state is a state in which no alternative state s can be reached from s without mutations.

We will rely on the characterization of the set of LRE developed by Ellison (2000). Given two absorbing set X and Y, denote c(X,Y) as the minimal number of mutations necessary for a direct transition from X to Y, i.e., the transition cost from X to Y. This direct transition does not go through any other absorbing set, and c(X,Y)>0. Define a path from X to Y as a finite sequence of absorbing sets P={X=S0,S1,SL(P)=Y}, where L(P) is the length from X to Y, i.e., the number of elements of the sequence minus 1. Let W(X,Y) be the set of all paths from X to Y. We extend the cost function to paths by c(P)=k=1L(p)c(Sk1,Sk), then the minimal number of mistakes require for a transition, direct or indirect, from X to Y is given by:

C(X,Y)=minPW(X,Y)c(P).

The result of Ellison (2000) can be summarized as follow: The radius of an absorbing set X is defined as R(X)=min{C(X,Y)|Y is an absorbing set, YX}, i.e., the minimal number of mistakes necessary for leaving X.

We define the coradius of X as the maximal number of mistakes necessary for every other absorbing set to enter the basin of attraction of X, formally:CR(X)=max{C(Y,X)|Y is an absorbing set, YX}.

Ellison (2000) provides a powerful result that if R(X)>CR(X) for a given absorbing set X, then X is the unique stochastically stable set.

Lemma A1: (Ellison, 2000) Let X be an absorbing set, if R(X)>CR(X), then the LRE are the states in X.

This is the Radius-Coradius Theorem of Ellison (2000). Note that if there are only two absorbing set, we have C(X,Y)=R(X)=CR(Y) and C(Y,X)=R(Y)=CR(X); when C(X,Y)=C(Y,X), both absorbing sets are LRE.

Appendix B

The switching thresholds in Table 1: Depending on the relationship between , f and fA, we have four cases:

1. (i)

fA and ffA

In this case, both A-players and B-players may fill up all their links with F-players of their own kind. Since (B,B) is the payoff dominant equilibrium, M-players will earn a higher payoff by adopting B, so B is the optimal action in this case.

1. (ii)

fA< and ffA

In this case, B-players can fill up all their links to F-players of their own kind, but A-players do not find sufficiently many F-players of their own kind to fill up all the slots. Again B is the optimal action in this case.

1. (iii)

fA and ffA<

In this case, A-players can fill up all their links to F-players of their own kind, but B-players cannot. An M-player will choose A if V(A,fA)>V(B,fA), which turns to

a>d(f+fA)+b(ffA),

rearranging terms yields

Conversely, note that if when fA<f(ad)bd, V(B,fA)>maxV(A,fA)=a, i.e., if the number of B-players in F-group is larger than (ad)bd, it is always optimal for M-players to play B. We denote by ψ1=(ad)bd.

1. (iv)

fA< and ffA<

In this case, neither A- nor B-players in the M-group will fill up all their links with F-players of their own kind. The M-players will choose A with probability one if

fA>(bd)f(cd)a+bcd.

We denote by ψ2=f(bd)f(cd)a+bcd.

## Appendix C Proofs

Proof of Lemma 2: First, we consider the transition from Ag to Bg via the states where aM=B.

We start with the case where the mutations only occur among the F-players. Let mAB be the minimal number of F-players switching from A to B, such that every M-player will choose B when given revision opportunity. According to Table 1, the M-player will choose B with probability 1 in the following three cases:

1. ffA. When the number of B-players in F-group is greater than or equal to , then M-player will choose B over A. To reach a state in this case, we need at least F-players switch from A to B. So, we have mAB=;

2. fA and ψ1<ffA<. We have at least A-players in F-group, an M-player will switch to B if at least ψ1F-players are choosing B. In this case, we need at least ψ1F-players switch from A to B. That is mAB=ψ1. Recalling that we need fA and mAB< in this case, which means that in the F-group the number of A-players is great than or equal to , and the number of B-player is less than . that is fψ1 and ψ1<. The first condition yields bdb+a2df, and ψ1< is always true under this condition.

3. fA< and ψ2<ffA<. When both A-players and B-players are less than in F-group, an M-player will switch to B if at least ψ2F-players are choosing B. So we have mAB=ψ2 in this case. Note that we need fA and mAB< in this case. The first condition yields bdb+a2df<<f, and ψ2< is always true under this condition.

To sum up, we have:

mAB={ψ1ifbdb+a2dfψ2ifbdb+a2df<<f.

Now we know how many F-players switch to B will lead all the M-players choose B as best response. Then, we wonder if there exist any paths of process with fewer mutations. Recall that F-players with incoming links choose their action based on the action distribution in their neighborhood. Think of an F-player has only one incoming link, her action choice only depends on the action of her sole opponent. Then, in particular, consider a set of F-players, such that each of these F-players has only one incoming link from the same M-player. If this M-player switches, all F-players link to him switch to the same action as he chooses with positive probability. This observation will accelerate the transition by changing F-players’ action with only one mutation.

In this spirit, we consider the following interaction structure, under which one M-player can influence most possible F-players. Let (m1)M-players have all their links to a subset of F-players. It will leave (f)F-players for the remaining M-player to connect to. So, the number of F-players that only connect to this M-player is given by min{,f}. We refer to such an interaction structure as the M-player influence structure. In Figure 4, we provide an example of how the transition under the M-player influence structure takes place, and why it needs fewer mutations.

Then, we calculate the transition cost from Ag to Bg via the states where aM=B under an M-player influence structure. We already know that under an M-player influence structure, the number of F-players with one incoming link is min{,f}. Thus, we consider two cases when f2 and when 2>f>.

First, if f2, under an M-player influence structure, one M-player can support all his links to the F-players, each of whom has the only one incoming link from the M-player. Thus, if the M-player switches to B, there will be F-players choosing B with positive probability. We know that when f+(ad)bd, we need mAB=ψ1F-players to choose B so that B is the best reply for every M-player. Since it is always true that >ψ1, when f2, we are able to reach a state in which aM=B with just one mutation, that is PM(A,B)=1.

Second, if 2>f>, under an M-player influence structure, one M-player can link to (f)F-players, each of whom has the only incoming link from the M-player. We know that the transition from Ag to a state where aM=B requires at least mABF-players to switch from A to B, so we distinguish two subcases: i) 2>f+(ad)bd and ii) +(ad)bd>f>.

1. When 2>f+(ad)bd, we need ψ1F-players to choose B so that every M-player will prefer B over A. Consider an M-player influence structure (as Figure 3(a)), under which one M-player connects to the highest number of F-players, each of whom has the only incoming link from this M-player. If this M-player switches to B, (f)F-players will switch to B with positive probability. Since it is always true that fψ1 in this case, we are able to reach a state in which aM=B with just one mutation, that is PM(A,B)=1.

2. When +(ad)bd>f>, we need ψ2F-players to choose B so that every M-player will prefer B over A. Given an M-player influence structure, if one M-player switches to B, (f)F-players will switch to B with positive probability. Since f<ψ2 in this case, in addition to the M-player, we still need that [ψ2(f)]F-players who are not only connected to the M-player to also switch to B by mistakes. Thus, when +(ad)bd>f>, we find that PM(A,B)=ψ2(f)+1.

To sum up, the transition cost from Ag to Bg via the states where aM=B is given by:

Next, we consider the transition from Bg to Ag via the states where aM=A. We denote by mBA the minimal number of F-players switching from B to A, such that every M-player will choose A when given revision opportunity. According to Table 1, the M-player will choose A when the number of B-players in the F-group is greater than or equal to . So, in the path of transition to Ag, we need ffA< to be true. Furthermore, M-player will choose A with probability 1 in the following two cases:

1. fA and ffA<ψ1 When the M-players choosing A can fill up all of their links to the A-players in F-group (fA), we can infer from Table 1 that mBA=fψ1 in this case. This case happens when the number of F-players choosing A is greater than or equal to , or fψ1, which holds if fψ1, hence we have f+(ad)bd.

2. fA< and ffA<ψ2 When the M-players choosing A cannot fill up all of their links to the A-players in F-group (fA<), according to Table 1, we have mBA=fψ2 in this case. This case happens if the number of F-players choosing A is less than , that is fψ2<, which holds if fψ2<. We can translate this into f<+(ad)bd.

Recall that in both cases, the remaining number of F-players choosing B should be less than , that is fmBA<. When f+(ad)bd, it must be the case that fmBA<. When f<+(ad)bd, we need that ffψ2<. From the analysis above, we know that ψ2< must be true, thus, we always have fmBA< when f<+(ad)bd. Thus, we have:

To complete the transition from Bg to Ag via the states where aM=A with the minimal number of mutations, again we consider the M-player influence structures. Under such an interaction structure, if one M-player switches to A, min{,f}F-players will choose A with positive probability. We distinguish two cases when 12f and when 12f<<f.

First, if 12f, under an M-player influence structure, one M-player can support all his links to the F-players, each of whom has the only incoming link from the M-player. We know that when bdb+a2df, we need mAB=fψ1F-players to choose A so that A is the best reply for every M-player. Once the M-player switches to A by mistakes, there will be F-players choosing A with positive probability. Since it must be the case that fψ1> when 12f, to make A the best reply for every M-player, in addition to the M-player, we need (fψ1)F-players who are not connected to the M-player to switch to A. Thus, we find that PM(B,A)=fψ1+1 in this case.

Second, if 12f<<f, under an M-player influence structure, one M-player can link to (f)F-players, each of whom has only one incoming link from the M-player. We know that the transition from Bg to a state where aM=A requires at least mBAF-players to choose A, so we distinguish two subcases i) 12f<bdb+a2df and ii) bdb+a2df<<f.

1. When 12f<bdb+a2df , every M-player will prefer A over B if there are fψ1F-players choosing A. Once the M-player switches to A by mistakes, there will be (f)F-players choosing A with positive probability. Since it must be the case that fψ1>f when 12f<bdb+a2df , to make A the best reply for every M-player, in addition to the M-player, we need [fψ1(f)]F-players who are not only connected to the M-player to switch to A. So we have PM(B,A)=fψ1(f)+1 in this case.

2. When bdb+a2df<<f, every M-player will prefer A over B if there are fψ2F-players choosing A. Once the M-player switches to A by mistakes, there will be (f)F-players choosing A with positive probability. Since it must be the case that fψ2>f when bdb+a2df<<f, in addition to the M-player, we need [fψ2(f)]F-players who are not only connected to the M-player to switch to A, so that every M-player prefer A over B. Thus, we have PM(B,A)=fψ2(f)+1 in this case.

To sum up, the transition cost from Bg to Ag via the states where aM=B is given by:

PM(B,A)={fψ1+1if12ffψ1(f)+1if12f<bdb+a2dffψ2(f)+1ifbdb+a2df<<f.
Proof of Lemma 3: First, we consider the transition from Ag to Bg via the states where aF=B. That is the case where the mutations occur among the M-players. We denote by fAB the minimal number of M-players switching from A to B, such that every F-player will choose B when given revision opportunity. We claim that in this case, fABm(1q), and prove it via contradiction.

Assume that fAB=γ and γ<m(1q), that is when γ M-players switch to B by mistakes, every F-player will choose B with positive probability. Recall that if an F-player i has at least one incoming link (jMgji>0), she will switch to B when at least (1q) of her neighbors choose B; if she has no incoming link (jMgji=0), she will switch to B when at least (1q) of M-players choose B. Since γ<m(1q), an F-player with no incoming link will not switch to B when fAB=γ. Thus, we consider an interaction structure where every F-player has at least one incoming link. Under this interaction structure, the players from M-group support m links in all, and γ of these links are supported by B-players. To make sure that every F-player prefers B over A, we need that for every F-player, at least (1q) of her neighbors choose B, that is we need niBni(1q) for iF. Summing up all the F-players, we need

iFniBiFni(1q)

For the right hand side of the inequality, according to the property of ceiling function, x+yx+y, we have iFni(1q)(1q)iFni=(1q)m. We also know that the left hand side of the inequality represents the total number of links supported by B-players, so we have iFniB=γ. Thus, the inequality implies that γm(1q), which contradicts γ<m(1q). So we prove the claim that fABm(1q).

As in Lemma 2, we need to find an interaction structure, under which we can reach a state where aF=B with the minimal number of mutations. We consider the following interaction structure: All the M-players support their links to the same set of F-players, and the rest of the F-players are left unlinked (See Figure 3(b) for instance). We refer to such an interaction structure as the F-player influence structure. Given this interaction structure, every F-player with incoming links will choose B with positive probability when at least m(1q)M-players switch to B by mistakes. And every F-players with no incoming links will also choose B as the best reply to the action distribution in M-group. Thus, the minimal transition cost from Ag to Bg via the states in which aF=B is m(1q), establishing that PF(A,B)=m(1q).

Next, we consider the transition from Bg to Ag via the states where aF=A. To reach a state where aF=A with the minimal number of mutations, we still consider the F-player influence structure as Figure 2(b). Similar to the previous analysis, every F-player will choose A if at least mqM-players switch to A. Thus, we can obtain that PF(B,A)=mq. □

Proof of Proposition 1: We already know that CR(Bg)=R(Ag)=min{m,f}(1q) and CR(Ag)=R(Bg)=min{m,f}q when f=. By risk-dominance, we know that q<12, so we have min{m,f}q<min{m,f}(1q), which implies that min{m,f}qmin{m,f}(1q). Thus, according to Lemma A1, when f=, the sets of LRE of the model are given by S=Ag if min{m,f}q<min{m,f}(1q); S=AgBg if min{m,f}q=min{m,f}(1q).□

Proof of Proposition 2: Part (1): In this part, we will identify the set of LRE when f>>bdb+a2df. By Lemma 2 and Lemma 3, we have PF(A,B)=m(1q) , PM(A,B)=ψ2(f)+1, PF(B,A)=mq, and PM(B,A)=fψ2(f)+1. According to Lemma A1, C(A,B)=min{PM(A,B),PF(A,B)} and C(B,A)=min{PM(B,A),PF(B,A)}, the set of LRE is the set of states with least transition cost to leave. Therefore, we need to compare the transition costs of four different paths. Note that, by risk dominance, we always have PF(A,B)PF(B,A). By Lemma 3, we know that if the path in which the F-group chooses the same action first needs fewer mutations, the transition cost is only determined by m. Then, we distinguish the following two cases a) when m<m̲, or m is sufficiently small, we compare the transition cost of PF(A,B) and PF(B,A), and b) when m>m̲, or m is sufficiently large, we compare the transition cost of PM(A,B) and PM(B,A).

1. Let m̲=min{ψ2f++11q,ψ2+1q}. When m<m̲, we always have PF(A,B)<PM(A,B) and PF(B,A)<PM(B,A), which implies that the path via the states where every F-player chooses the same action requires fewer mistakes to complete the transition. Thus, the radius and coradius of the absorbing sets are given by: CR(Bg)=R(Ag)=PF(A,B)=m(1q) and CR(Ag)=R(Bg)=PF(B,A)=mq. By risk-dominance, we always have mqm(1q). Thus, the set of LRE are S=Ag if mq<m(1q); S=AgBg if mq=m(1q).

2. Let m¯=max{ψ2f++11q,ψ2+1q}. When m>m¯, we always have PF(A,B)>PM(A,B) and PF(B,A)>PM(B,A), which implies that the path via the states where every M-player chooses the same action requires fewer mistakes to complete the transition. Thus, the radius and coradius of the absorbing sets are given by: CR(Bg)=R(Ag)=ψ2(f)+1 and R(Bg)=CR(Ag)=fψ2(f)+1. To determine the set of LRE, we compare CR(Bg) (or R(Ag)) and CR(Ag) (or R(Bg)), the comparison boils down to confirm the relation between ψ2 and fψ2.

First, when ψ2>fψ2, we have R(Ag)>CR(Ag), establishing that S=Ag. If ψ2, the condition holds if ψ2>fψ2, which translates into >(ba+cd)f2(cd). If ψ2, the condition holds if ψ2>12(f1), which yields >(ba+cd)f(a+bcd)2(cd). Recall that we need f>>bdb+a2df in this case. It turns out that if ψ2, and f>>(ba+cd)f2(cd), or ψ2, and f>>(ba+cd)f(a+bcd)2(cd), the set of LRE is S=Ag.

Next, when ψ2<fψ2, we have R(Bg)>CR(Bg), establishing that S=Bg in this case. If ψ2, the condition holds if ψ2<fψ2, which translates into <(ba+cd)f2(cd). If ψ2, the condition holds if ψ2<12(f+1), which yields <(ba+cd)f+(a+bcd)2(cd). Recall that we need f>>bdb+a2df in this case. It turns out that if ψ2, and (ba+cd)f2(cd)>>bdb+a2df, or ψ2, and (ba+cd)f+(a+bcd)2(cd)>>bdb+a2df, the set of LRE is S=Bg.

Last, when ψ2=fψ2, we have S=AgBg. If ψ2, the condition holds when ψ2=12f, which yields f=2(cd)b+cad. If ψ2, the condition holds only if f is odd and ψ2=12(f+1), we can translate the equation into (ba+cd)f+(a+bcd)2(cd)>(ba+cd)f(a+bcd)2(cd). Thus, the set of LRE is S=AgBg when ψ2 and l=(ba+cd)f2(cd), or ψ2 and (ba+cd)f+(a+bcd)2(cd)>(ba+cd)f(a+bcd)2(cd).

Part (2): In this part, we will identify the set of LRE when bdb+a2df. When bdb+a2df, we have PF(A,B)=m(1q) , PM(A,B)=1, PF(B,A)=mq, and PM(B,A)=fψ1min{f,}. Since the transition cost between absorbing sets is a natural number and should be no less than one, it must be the case that min{PM(B,A),PF(B,A)}1 and m(1q)1. So, we have

CR(Ag)=R(Bg)=min{PM(B,A),PF(B,A)}1

and

CR(Bg)=R(Ag)=min{PM(A,B),PF(A,B)}=1.

Since PM(B,A)=fψ1min{f,}+1>1 must be true when bdb+a2df, both absorbing sets are LRE only if mq=1, which implies that 1m1q. Bg is the unique LRE if mq>1, which implies that m>1q. Thus, we have S=Bg if mq>1, and S=AgBg if m1q. □

# Acknowledgments

I thank Simon Weidenholzer and Friederike Mengel for helpful comments and suggestions. I am also grateful to the editor and two anonymous referees for suggesting ways to improve both the content and exposition of this paper. This work was financially supported by the National Science Foundation of China (No. 71671010).

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