Social Coordination and Network Formation in Bipartite Networks

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 = S 0 ,   S 1 ,   … S L ( 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 = 1 L ( p ) c ( S k − 1 ,   S k ) , then the minimal number of mistakes require for a transition, direct or indirect, from X to Y is given by:

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, Y ≠ X } , 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: C R ( X ) = max { C ( Y ,   X ) | Y is an absorbing set, Y ≠ X } .

Ellison (2000) provides a powerful result that if R ( X ) > C R ( 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 ) > C R ( 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 ) = C R ( Y ) and C ( Y ,   X ) = R ( Y ) = C R ( X ) ; when C ( X ,   Y ) = C ( Y ,   X ) , both absorbing sets are LRE.

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

1. f A ≥ ℓ and f − f A ≥ ℓ

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. f A < ℓ and f − f A ≥ ℓ

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. f A ≥ ℓ and f − f A < ℓ

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 ,   f A ) > V ( B ,   f A ) , which turns to

rearranging terms yields

Conversely, note that if when f A < f − ( a − d ) ℓ b − d , V ( B ,   f A ) > max V ( A ,   f A ) = a ℓ , i.e., if the number of B-players in F-group is larger than ( a − d ) ℓ b − d , it is always optimal for M-players to play B. We denote by ψ 1 = ( a − d ) ℓ b − d .

1. f A < ℓ and f − f A < ℓ

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

We denote by ψ 2 = f − ( b − d ) f − ( c − d ) ℓ a + b − c − d .

Appendix C Proofs

Proof of Lemma 2: First, we consider the transition from A → g to B → g via the states where a M = B → .

We start with the case where the mutations only occur among the F-players. Let m A B 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. f − f A ≥ ℓ . When the number of B-players in F-group is greater than or equal to ℓ , then M-player will choose B over A. To reach a state in this case, we need at least ℓ F-players switch from A to B. So, we have m A B = ℓ ;

2. f A ≥ ℓ and ψ 1 < f − f A < ℓ . We have at least ℓ A-players in F-group, an M-player will switch to B if at least ψ 1 F-players are choosing B. In this case, we need at least ψ 1 F-players switch from A to B. That is m A B = ⌈ ψ 1 ⌉ . Recalling that we need f A ≥ ℓ and m A B < ℓ in this case, which means that in the F-group the number of A-players is great than or equal to ℓ , and the number of B-player is less than ℓ . that is f − ⌈ ψ 1 ⌉ ≥ ℓ and ⌈ ψ 1 ⌉ < ℓ . The first condition yields ℓ ≤ b − d b + a − 2 d f , and ⌈ ψ 1 ⌉ < ℓ is always true under this condition.

3. f A < ℓ and ⌈ ψ 2 ⌉ < f − f A < ℓ . When both A-players and B-players are less than ℓ in F-group, an M-player will switch to B if at least ⌈ ψ 2 ⌉ F-players are choosing B. So we have m A B = ⌈ ψ 2 ⌉ in this case. Note that we need f A ≥ ℓ and m A B < ℓ in this case. The first condition yields b − d b + a − 2 d f < ℓ < f , and ⌈ ψ 2 ⌉ < ℓ is always true under this condition.

To sum up, we have:

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 ℓ 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 ( m − 1 ) M-players have all their links to a subset of ℓ F-players. It will leave ( f − ℓ ) 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 { ℓ ,   f − ℓ } . 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 A → g to B → g via the states where a M = 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 { ℓ ,   f − ℓ } . Thus, we consider two cases when f ≥ 2 ℓ and when 2 ℓ > f > ℓ .

First, if f ≥ 2 ℓ , under an M-player influence structure, one M-player can support all his ℓ 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 ℓ F-players choosing B with positive probability. We know that when f ≥ ℓ + ( a − d ) ℓ b − d , we need m A B = ψ 1 F-players to choose B so that B is the best reply for every M-player. Since it is always true that ℓ > ψ 1 , when f ≥ 2 ℓ , we are able to reach a state in which a M = B → with just one mutation, that is P M ( A , B ) = 1 .

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

1. When 2 ℓ > f ≥ ℓ + ( a − d ) ℓ b − d , we need ⌈ ψ 1 ⌉ F-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 − ℓ ) F-players will switch to B with positive probability. Since it is always true that f − ℓ ≥ ⌈ ψ 1 ⌉ in this case, we are able to reach a state in which a M = B → with just one mutation, that is P M ( A , B ) = 1 .

2. When ℓ + ( a − d ) ℓ b − d > f > ℓ , we need ⌈ ψ 2 ⌉ F-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 − ℓ ) F-players will switch to B with positive probability. Since f − ℓ < ⌈ ψ 2 ⌉ in this case, in addition to the M-player, we still need that [ ψ 2 ⌉ − ( f − ℓ ) ] F-players who are not only connected to the M-player to also switch to B by mistakes. Thus, when ℓ + ( a − d ) ℓ b − d > f > ℓ , we find that P M ( A , B ) = ψ 2 − ( f − ℓ ) + 1 .

To sum up, the transition cost from A → g to B → g via the states where a M = B → is given by:

Next, we consider the transition from B → g to A → g via the states where a M = A → . We denote by m B A 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 ℓ . So, in the path of transition to A → g , we need f − f A < ℓ to be true. Furthermore, M-player will choose A with probability 1 in the following two cases:

1. f A ≥ ℓ and f − f A < ψ 1 When the M-players choosing A can fill up all of their links to the A-players in F-group ( f A ≥ ℓ ), we can infer from Table 1 that m B A = f − ψ 1 in this case. This case happens when the number of F-players choosing A is greater than or equal to ℓ , or f − ψ 1 ≥ ℓ , which holds if f − ψ 1 ≥ ℓ , hence we have f ≥ ℓ + ( a − d ) ℓ b − d .

2. f A < ℓ and f − f A < ψ 2 When the M-players choosing A cannot fill up all of their links to the A-players in F-group ( f A < ℓ ), according to Table 1, we have m B A = f − ψ 2 in this case. This case happens if the number of F-players choosing A is less than ℓ , that is f − ψ 2 < ℓ , which holds if f − ψ 2 < ℓ . We can translate this into f < ℓ + ( a − d ) ℓ b − d .

Recall that in both cases, the remaining number of F-players choosing B should be less than ℓ , that is f − m B A < ℓ . When f ≥ ℓ + ( a − d ) ℓ b − d , it must be the case that f − m B A < ℓ . When f < ℓ + ( a − d ) ℓ b − d , we need that f − f − ψ 2 < ℓ . From the analysis above, we know that ψ 2 < ℓ must be true, thus, we always have f − m B A < ℓ when f < ℓ + ( a − d ) ℓ b − d . Thus, we have:

To complete the transition from B → g to A → g via the states where a M = 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 { ℓ , f − ℓ } F-players will choose A with positive probability. We distinguish two cases when ℓ ≤ 1 2 f and when 1 2 f < ℓ < f .

First, if ℓ ≤ 1 2 f , under an M-player influence structure, one M-player can support all his ℓ links to the F-players, each of whom has the only incoming link from the M-player. We know that when ℓ ≤ b − d b + a − 2 d f , we need m A B = f − ψ 1 F-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 ℓ F-players choosing A with positive probability. Since it must be the case that f − ψ 1 > ℓ when ℓ ≤ 1 2 f , to make A the best reply for every M-player, in addition to the M-player, we need ( f − ψ 1 − ℓ ) F-players who are not connected to the M-player to switch to A. Thus, we find that P M ( B , A ) = f − ψ 1 − ℓ + 1 in this case.

Second, if 1 2 f < ℓ < f , under an M-player influence structure, one M-player can link to ( f − ℓ ) F-players, each of whom has only one incoming link from the M-player. We know that the transition from B → g to a state where a M = A → requires at least m B A F-players to choose A, so we distinguish two subcases i) 1 2 f < ℓ ≤ b − d b + a − 2 d f and ii) b − d b + a − 2 d f < ℓ < f .

1. When 1 2 f < ℓ ≤ b − d b + a − 2 d f , every M-player will prefer A over B if there are f − ψ 1 F-players choosing A. Once the M-player switches to A by mistakes, there will be ( f − ℓ ) F-players choosing A with positive probability. Since it must be the case that f − ψ 1 > f − ℓ when 1 2 f < ℓ ≤ b − d b + a − 2 d f , to make A the best reply for every M-player, in addition to the M-player, we need [ f − ψ 1 − ( f − ℓ ) ] F-players who are not only connected to the M-player to switch to A. So we have P M ( B ,   A ) = f − ψ 1 − ( f − ℓ ) + 1 in this case.

2. When b − d b + a − 2 d f < ℓ < f , every M-player will prefer A over B if there are f − ψ 2 F-players choosing A. Once the M-player switches to A by mistakes, there will be ( f − ℓ ) F-players choosing A with positive probability. Since it must be the case that f − ψ 2 > f − ℓ when b − d b + a − 2 d f < ℓ < f , in addition to the M-player, we need [ f − ψ 2 − ( f − ℓ ) ] 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 P M ( B ,   A ) = f − ψ 2 − ( f − ℓ ) + 1 in this case.

To sum up, the transition cost from B → g to A → g via the states where a M = B → is given by:

Proof of Lemma 3: First, we consider the transition from A → g to B → g via the states where a F = B → . That is the case where the mutations occur among the M-players. We denote by f A B 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, f A B ≥ m ( 1 − q ∗ ) , and prove it via contradiction.

Assume that f A B = γ and γ < m ( 1 − q ∗ ) , 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 ( ∑ j ∈ M g j i > 0 ) , she will switch to B when at least ( 1 − q ∗ ) of her neighbors choose B; if she has no incoming link ( ∑ j ∈ M g j i = 0 ) , she will switch to B when at least ( 1 − q ∗ ) of M-players choose B. Since γ < m ( 1 − q ∗ ) , an F-player with no incoming link will not switch to B when f A B = 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 ℓ links in all, and γ ℓ 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 ( 1 − q ∗ ) of her neighbors choose B, that is we need n i B ≥ n i ( 1 − q ∗ ) for ∀ i ∈ F . Summing up all the F-players, we need

For the right hand side of the inequality, according to the property of ceiling function, x + y ≥ x + y , we have ∑ i ∈ F n i ( 1 − q ∗ ) ≥ ( 1 − q ∗ ) ∑ i ∈ F n i = ( 1 − q ∗ ) m ℓ . We also know that the left hand side of the inequality represents the total number of links supported by B-players, so we have ∑ i ∈ F n i B = γ ℓ . Thus, the inequality implies that γ ℓ ≥ m ℓ ( 1 − q ∗ ) , which contradicts γ < m ( 1 − q ∗ ) . So we prove the claim that f A B ≥ m ( 1 − q ∗ ) .

As in Lemma 2, we need to find an interaction structure, under which we can reach a state where a F = B → with the minimal number of mutations. We consider the following interaction structure: All the M-players support their ℓ 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 ℓ incoming links will choose B with positive probability when at least m ( 1 − q ∗ ) 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 A → g to B → g via the states in which a F = B → is m ( 1 − q ∗ ) , establishing that P F ( A ,   B ) = m ( 1 − q ∗ ) .

Next, we consider the transition from B → g to A → g via the states where a F = A → . To reach a state where a F = 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 m q ∗ M-players switch to A. Thus, we can obtain that P F ( B ,   A ) = m q ∗ . □

Proof of Proposition 1: We already know that C R ( B → g ) = R ( A → g ) = min { m ,   f } ( 1 − q ∗ ) and C R ( A → g ) = R ( B → g ) = min { m ,   f } q ∗ when f = ℓ . By risk-dominance, we know that q ∗ < 1 2 , so we have min { m ,   f } q ∗ < min { m ,   f } ( 1 − q ∗ ) , which implies that min { m ,   f } q ∗ ≤ min { m ,   f } ( 1 − q ∗ ) . Thus, according to Lemma A1, when f = ℓ , the sets of LRE of the model are given by S = A → g if min { m ,   f } q ∗ < min { m ,   f } ( 1 − q ∗ ) ; S = A → g ∪ B → g if min { m ,   f } q ∗ = min { m ,   f } ( 1 − q ∗ ) .□

Proof of Proposition 2: Part (1): In this part, we will identify the set of LRE when f > ℓ > b − d b + a − 2 d f . By Lemma 2 and Lemma 3, we have P F ( A ,   B ) = m ( 1 − q ∗ ) , P M ( A ,   B ) = ψ 2 − ( f − ℓ ) + 1 , P F ( B ,   A ) = m q ∗ , and P M ( B ,   A ) = f − ψ 2 − ( f − ℓ ) + 1 . According to Lemma A1, C ( A ,   B ) = min { P M ( A ,   B ) ,   P F ( A ,   B ) } and C ( B ,   A ) = min { P M ( B ,   A ) ,   P F ( 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 P F ( A ,   B ) ≥ P F ( 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 < m ̲ , or m is sufficiently small, we compare the transition cost of P F ( A ,   B ) and P F ( B ,   A ) , and b) when m > m ̲ , or m is sufficiently large, we compare the transition cost of P M ( A ,   B ) and P M ( B ,   A ) .

1. Let m ̲ = min { ψ 2 − f + ℓ + 1 1 − q ∗ ,   ℓ − ψ 2 + 1 q ∗ } . When m < m ̲ , we always have P F ( A ,   B ) < P M ( A ,   B ) and P F ( B ,   A ) < P M ( 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: C R ( B → g ) = R ( A → g ) = P F ( A ,   B ) = m ( 1 − q ∗ ) and C R ( A → g ) = R ( B → g ) = P F ( B ,   A ) = m q ∗ . By risk-dominance, we always have m q ∗ ≤ m ( 1 − q ∗ ) . Thus, the set of LRE are S = A → g if m q ∗ < m ( 1 − q ∗ ) ; S = A → g ∪ B → g if m q ∗ = m ( 1 − q ∗ ) .

2. Let m ¯ = max { ψ 2 − f + ℓ + 1 1 − q ∗ ,   ℓ − ψ 2 + 1 q ∗ } . When m > m ¯ , we always have P F ( A ,   B ) > P M ( A ,   B ) and P F ( B ,   A ) > P M ( 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: C R ( B → g ) = R ( A → g ) = ψ 2 − ( f − ℓ ) + 1 and R ( B → g ) = C R ( A → g ) = f − ψ 2 − ( f − ℓ ) + 1 . To determine the set of LRE, we compare C R ( B → g ) (or R ( A → g ) ) and C R ( A → g ) (or R ( B → g ) ), the comparison boils down to confirm the relation between ψ 2 and f − ψ 2 .

   First, when ψ 2 > f − ψ 2 , we have R ( A → g ) > C R ( A → g ) , establishing that S = A → g . If ψ 2 ∈ ℤ , the condition holds if ψ 2 > f − ψ 2 , which translates into ℓ > ( b − a + c − d ) f 2 ( c − d ) . If ψ 2 ∉ ℤ , the condition holds if ψ 2 > 1 2 ( f − 1 ) , which yields ℓ > ( b − a + c − d ) f − ( a + b − c − d ) 2 ( c − d ) . Recall that we need f > ℓ > b − d b + a − 2 d f in this case. It turns out that if ψ 2 ∈ ℤ , and f > ℓ > ( b − a + c − d ) f 2 ( c − d ) , or ψ 2 ∉ ℤ , and f > ℓ > ( b − a + c − d ) f − ( a + b − c − d ) 2 ( c − d ) , the set of LRE is S = A → g .

   Next, when ψ 2 < f − ψ 2 , we have R ( B → g ) > C R ( B → g ) , establishing that S = B → g in this case. If ψ 2 ∈ ℤ , the condition holds if ψ 2 < f − ψ 2 , which translates into ℓ < ( b − a + c − d ) f 2 ( c − d ) . If ψ 2 ∉ ℤ , the condition holds if ψ 2 < 1 2 ( f + 1 ) , which yields ℓ < ( b − a + c − d ) f + ( a + b − c − d ) 2 ( c − d ) . Recall that we need f > ℓ > b − d b + a − 2 d f in this case. It turns out that if ψ 2 ∈ ℤ , and ( b − a + c − d ) f 2 ( c − d ) > ℓ > b − d b + a − 2 d f , or ψ 2 ∉ ℤ , and ( b − a + c − d ) f + ( a + b − c − d ) 2 ( c − d ) > ℓ > b − d b + a − 2 d f , the set of LRE is S = B → g .

   Last, when ψ 2 = f − ψ 2 , we have S = A → g ∪ B → g . If ψ 2 ∈ ℤ , the condition holds when ψ 2 = 1 2 f , which yields f = 2 ( c − d ) ℓ b + c − a − d . If ψ 2 ∉ ℤ , the condition holds only if f is odd and ψ 2 = 1 2 ( f + 1 ) , we can translate the equation into ( b − a + c − d ) f + ( a + b − c − d ) 2 ( c − d ) > ℓ ≥ ( b − a + c − d ) f − ( a + b − c − d ) 2 ( c − d ) . Thus, the set of LRE is S = A → g ∪ B → g when ψ 2 ∈ ℤ and l = ( b − a + c − d ) f 2 ( c − d ) ∈ ℤ , or ψ 2 ∉ ℤ and ( b − a + c − d ) f + ( a + b − c − d ) 2 ( c − d ) > ℓ ≥ ( b − a + c − d ) f − ( a + b − c − d ) 2 ( c − d ) .

Part (2): In this part, we will identify the set of LRE when ℓ ≤ b − d b + a − 2 d f . When ℓ ≤ b − d b + a − 2 d f , we have P F ( A ,   B ) = m ( 1 − q ∗ ) , P M ( A ,   B ) = 1 , P F ( B ,   A ) = m q ∗ , and P M ( B ,   A ) = f − ψ 1 − min { f − ℓ ,   ℓ } . 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 { P M ( B ,   A ) ,   P F ( B ,   A ) } ≥ 1 and m ( 1 − q ∗ ) ≥ 1 . So, we have

and

Since P M ( B ,   A ) = f − ψ 1 − min { f − ℓ ,   ℓ } + 1 > 1 must be true when ℓ ≤ b − d b + a − 2 d f , both absorbing sets are LRE only if m q ∗ = 1 , which implies that 1 ≤ m ≤ 1 q ∗ . B → g is the unique LRE if m q ∗ > 1 , which implies that m > 1 q ∗ . Thus, we have S = B → g if m q ∗ > 1 , and S = A → g ∪ B → g if m ≤ 1 q ∗ . □