Social coordination with locally observable types

Abstract

In this paper we study the typical dilemma of social coordination between a risk-dominant convention and a payoff-dominant convention. In particular, we consider a model where a population of agents play a coordination game over time, choosing both the action and the network of agents with whom to interact. The main modeling novelties with respect to the existing literature are: (1) Agents come in two distinct types, (2) the interaction with a different type is costly, and (3) an agent’s type is unobservable prior to interaction. We show that when the cost of interacting with a different type is small with respect to the payoff of coordination, the payoff-dominant convention is the only stochastically stable convention; instead, when the cost of interacting with a different type is large, the only stochastically stable conventions are those where all agents of one type play the payoff-dominant action and all agents of the other type play the risk-dominant action.

Appendix: Proofs of Propositions and Lemmas

1.1 Proof of Proposition 1

Proof

Suppose that agent i is given a revision opportunity at time $t+1$, and suppose ad absurdum that there exists a strategy $s_i = (a_i, g_i)$ that maximizes the interim utility of i and tells him to maintain the link with agent j, i.e., $g_{ij}=1$. We construct another strategy $s_i' = (a_i', g_i')$ such that $a_i' = a_i$ and $g_i'$ is equal to $g_i$ with the only difference that in $g_i'$ the link with agent j is removed and a new link is formed with an agent $\ell$ such that $a_{\ell }^t = a_j^t$. We observe that the assumption that $n(w_i, a_j^t | s^t) \ge k+1$ implies that there exists at least an agent having the same type as i and choosing the same action as j. This in turn implies (1) that a new link can actually be formed with an agent $\ell$ choosing $a_{\ell }^t = a_j^t$, and (2) that the overall change in utility for i by playing $s_i'$ instead of $s_i$ is strictly positive: This is so because the change of utility due to the play of the social game with all neighbors is trivially equal to zero (since $a_{\ell }^t = a_j^t$ and all other neighbors have remained the same, while the change in the expected number of different types is negative (since there is a positive probability that $\ell$ has the same type as i, and agent j is known for sure to be of different type).

We have proven that $u_i(s_i', s^t) > u_i(s_i, s^t)$, and this suffices to show that agent i cannot choose a strategy such that $g^{t+1}_{ij}=1$. $\square$

1.2 Proof of Proposition 2

Proof

Let us suppose, ad absurdum, that strategy $s_i = (a_i, g_i)$ such that $g_{ij}=0$ maximizes i’s interim utility. We first observe that $g_i$ cannot tell i to have less than k links, because otherwise i might increase his utility by simply adding the link with j, obtaining an additional utility of $\pi (a_i, a_j^t)$, which is surely positive because of the assumption that $c < \pi (B,A)$, which implies that every payoff in the social game is positive even after subtracting the maintenance cost. Therefore, $g_i$ tells i to have k links, which in turn implies that a new link with some agent $\ell$ has been formed, since the link with j has been removed. We construct another strategy $s_i' = (a_i', g_i')$ such that $a_i' = a_i$ and $g_i'$ is equal to $g_i$ with the only difference that in $g_i'$ the link with agent j is maintained and the link with agent $\ell$ is not formed. We now argue that $u_i(s_i', s^t) > u_i(s_i, s^t)$. We first note that the expected number of mismatches in types is lower with $s_i'$ than with $s_i$, because $w(i) = w(j)$ for sure while $\ell$’s type is different from i’s type with positive probability: Indeed, if $a_{\ell }^t = a^t_j$, then we have the assumption that $n_{i0}(\lnot w(i), a_j^t | s^t) \ge 1$, while if $a_{\ell }^t \ne a_j^t$ (and hence $a_{\ell }^t = a_i^t$) then we have that $n_{i0}(a_i^t | s^t) \ge 1$ and hence the assumption that $d {\frac{n_{i0}(\lnot w(i), a_i^t | s^t)}{n_{i0}(a_i^t | s^t)}} > \pi (a_i^t, a_i^t) - \pi (a_i^t, a_j^t)$ implies $n_{i0}(\lnot w(i), a_i^t | s^t) \ge 1$.

Therefore, if i obtains the same or a larger utility in the social game by interacting with j than with $\ell$, then we have obtained that $u_i(s_i', s^t) > u_i(s_i, s^t)$. The only case where i obtains a larger utility interacting with $\ell$ than with j is if $a_j^t \ne a_i$ and $a_{\ell }^t = a_i$. Even in such a case, the assumption that $d {\frac{n_{i0}(\lnot w(i), a_i^t | s^t)}{n_{i0}(a_i^t | s^t)}} > \pi (a_i^t, a_i^t) - \pi (a_i^t, a_j^t)$ ensures us $u_i(s_i', s^t) > u_i(s_i, s^t)$.

We can conclude that no strategy $s_i$ such that $g_{ij}=0$ can maximize i’s interim utility, and this means that i will surely maintain the link with j if he has a revision opportunity at time $t+1$ (and no change of course happens if i is not given a revision opportunity). $\square$

1.3 Proof of Proposition 3

Proof

We first prove point (a). Consider a state s that is A-monomorphic, k-regular, and type-segregated. We check that an agent who receives a revision opportunity at s would see his utility reduced if he changes strategy. Indeed, having less than k links is suboptimal, since $\pi (A,A) > c$. Moreover, removing a link and casting a new one brings a neighbor who still plays A (since the state is A-monomorphic) but possibly is of a different type, hence generating an expected loss. Finally, switching from A to B is clearly detrimental, due to $\pi (A,A) > \pi (B,A)$. Therefore we can conclude that no agent will ever change strategy, and hence state s belongs to a singleton absorbing state. An analogous reasoning can be made for a state that is B-monomorphic, k-regular, and type-segregated, where $\pi (B,B) > c$ holds because $\pi (B,B) > \pi (A,A)$.

We now prove point (b). Consider a state s that is type-monomorphic with x on A and y on $B,\,k$-regular and type-segregated. Take an agent of type x who receives a revision opportunity. Maintaining less than k links is suboptimal for him, since $\pi (A,A) > c$ and, in addition, all agents playing A are of type x so that there is no risk of a type mismatch. Replacing an existing link with a new one has no effect on utility if the new link is cast toward an agent who currently plays A, since all agents playing A are of type x and hence there is no risk of a type mismatch. Replacing an existing link with a new one has a negative effect on expected utility if the new link is cast toward an agent playing B, since all agents currently playing B are of type y and $d > \pi (B,B) - \pi (A,A)$, which means that the penalty for the type mismatch is larger than the maximum attainable gain. Finally, switching from A to B without changing neighbors is detrimental because of $\pi (A,A) > \pi (B,A)$. A similar argument holds a fortiori if we consider an agent y who receives a revision opportunity. We can conclude that any revising agent will at most reshuffle his links among agents playing his same action, who are surely of his same type (due to the perfect correlation between actions and types). Therefore, starting from s we can only reach other states that are type-monomorphic with x on A and y on $B,\,k$-regular and type-segregated. This shows that an absorbing set exists, containing type-monomorphic states with x on A and y on B. Clearly, if we invert x with y, we obtain that the same reasoning applies to type-monomorphic states with x on B and y on A.

Then, we prove point (c). Consider a state s where $k+1$ agents of type x play $B,\,k+1$ agents of type y play B, and all other agents play A; also, s is k-regular and type-segregated (which is possible, since $n_x \ge n_y \ge 2k+2$). Any agent playing B will never change strategy, because he is attaining the maximum possible payoff [i.e., $k(\pi (B,B) -c)$], which is not reachable if he switches to A, and changing neighbors comes with the risk of a type mismatch—since some agents currently playing B are of type x and some are of type y. Any agent playing A will never change his strategy as well. Indeed, the maximum gain which can be obtained by removing an existing link and connecting to someone playing B is $\pi (B,B) - \pi (A,A)$, which is lower than the expected cost of a type mismatch (which is equal to d / 2 since $n(x,B|s) = k+1 = n(y,B|s)$) because of the assumption that $d > 2 (\pi (B,B) - \pi (A,A))$. Moreover, removing an existing link and connecting to someone playing A brings no benefit and an expected cost due to type mismatch; deleting any of the k links is suboptimal, since $\pi (A,A) > c$; and switching from A to B without changing neighbors is detrimental because of $\pi (A,A) > \pi (B,A)$. Therefore, state s belongs to a singleton absorbing state.

Finally, we prove point (d). Consider a state s where agent i of type y plays A and maintains no link, while all other agents play B and maintain k links toward agents of the same type different form i. Any agent other than i will never change strategy, because he is attaining the maximum possible payoff (i.e., $k(\pi (B,B) -c)$), which is not reachable if he switches to A, and changing neighbors come with the risk of a type mismatch because some agents currently playing B are of type y. If agent i chooses to connect toward an agent playing B, he will earn at most $\pi (B,B)$, but has to suffer an expected cost of type mismatch equal to $d n_x/(n-1)$, which is larger than $\pi (B,B)$ due to the assumption that $d > \pi (B,B)(n-1)/n_x$. If agent i is isolated, then he is indifferent between playing A and B. Therefore, we have found an absorbing set that is made of states s and $s'$, where $s'$ is identical to s with the only difference that agent i plays B instead of A. $\square$

1.4 Proof of Lemma 1

Proof

The requirement that s is B-monomorphic is trivially a necessary condition for $s \in S^B_B$. We first show that, given (1), if we are not in a state such that also (2) and (3) hold, then it must be the case that with positive probability we reach a state where (1), (2) and (3) hold. Suppose that $a_i = B$ for all $i \in N$, but s is not k-regular and/or not type-segregated. We observe that, for a revising agent, choosing action A would clearly be suboptimal. Moreover, the expected payoff of forming a link with a new neighbor who plays B is both higher than not forming that link at all (because $\pi (B,B) - \pi (A,A) > d$ and $\pi (A,A) > c$ imply $\pi (B,B) - c> d$) and higher than maintaining an existing connection with a type different from one’s own (because the resulting match cannot be worse, and possibly better). Therefore, with positive probability any agent who has less than k links and/or links with agents of a type different from his own will form new links with agents who play B, and with positive probability these new agents are of his own type.

We now show that a state satisfying (1), (2) and (3) forms a singleton absorbing set. To do so, it is enough to observe that any agent who receives a revision opportunity would see his payoff decreased, in expectation, by changing strategy. Indeed, by choosing action A, the agent would obtain a utility that is surely lower than his current utility $k (\pi (B,B) - c)$, and the same is true if he chooses to maintain less than k links; also, substituting an existing link with a new one comes with the risk of linking to an agent of a type different from one’s own, which leads to a lower expected payoff. $\square$

1.5 Proof of Lemma 2

Proof

We provide the proof for point (a) only, as the proof for point (b) is almost identical to the proof of point (a).

By definition, if $s \in S^A_B$, then s is type-monomorphic with x on A and y on B, so (1) is trivially necessary. We first show that, given (1), if we are not in a state such that also (2) and (3) hold, then it must be the case that with positive probability we reach a state where (1), (2) and (3) hold. Suppose that s is type-monomorphic with x on A and y on B, but not k-regular and/or not type-segregated. Consider a revising agent currently playing B. We observe that choosing A is clearly suboptimal since, at most, he can obtain $k(\pi (A,A) - c - d) < 0$ because, by (1), all agents playing A are of a type different from his own and, by assumption, $d> \pi (B,B) (n-1)/n_y > \pi (A,A)$. Moreover, the expected payoff of forming a link with a new neighbor who plays B is the same of keeping an existing link with a neighbor who also plays B [because, again by (1), B is played only by agents of similar type], it is strictly greater than the expected payoff of forming a link with an agent who plays A [because, by (2), A is played only by agents of a different type], and it is strictly greater that not forming that link at all [because $\pi (B,B) > \pi (A,A)$ and $\pi (A,A) > c$ imply $\pi (B,B) - c> 0$]. Consider a revising agent currently playing A. We observe that choosing B is suboptimal since, at most, he can obtain $k(\pi (B,B) - c - d) < 0$ because, by (1), all agents playing B are of a type different from his own and, by assumption, $d> \pi (B,B) (n-1)/n_y > \pi (B,B)$. Moreover, the expected payoff of forming a link with a new neighbor who plays A is the same of keeping an existing link with a neighbor who also plays A [because, by (1), A is played only by agents of similar type], it is strictly greater than the expected payoff of forming a link with an agent who plays B [because, by (1), A is played only by agents of a different type and, by assumption, $d> \pi (B,B) (n-1)/n_y > \pi (A,B)$, so that $\pi (A,B ) - d< 0 < \pi (A,A)$], and it is strictly greater that not forming that link at all (because $\pi (A,A) - c > 0$). Therefore, with positive probability any agent who plays B (respectively, A) and that has less than k links and/or links with agents of a type different from his own will form new links with agents who play B (respectively, A) up to k connections in total, and with certainty these new agents are of his own type.

We now show that the set of states satisfying (1), (2) and (3) forms an absorbing set. To do so, we first observe that, for the same arguments described above, any agent who receives a revision opportunity would see his payoff certainly decreased by changing action, and/or by choosing to maintain less than k links, and/or by linking to new agents who play a different action [since, by (1), they must be of a different type]. Therefore, if we start from a state where conditions (1), (2) and (3) are satisfied, we will always remain in states where those conditions are satisfied. We finally show that, taken any two distinct states s and $s'$ satisfying (1), (2) and (3), we can move from one to the other with positive probability. Indeed, $s = (a,g)$ and $s'=(a',g')$ can only differ because $g_i \ne g_i'$ for some agent i; every such agent can receive with positive probability a revision opportunity, and he can choose with positive probability to reshuffle all his links as long as links are cast toward agents choosing his own action, since by (1) there is no risk of forming a link with an agent of a type different from one’s own. $\square$

1.6 Proof of Lemma 3

Proof

We first show that 1 mutation is not sufficient to move from $S^A_B$ to another absorbing set. Consider a state $s \in S^A_B$, and suppose that a single mutation hits an agent possibly changing both his action and his network of interactions. Suppose that an agent different from the mutant is given a revision opportunity. We claim that such an agent will not change action and will not form new links with agents choosing an action different from his own. To see why this is so, we observe five facts. First, forming new links with an agent who is currently playing a different action is suboptimal, as the expected payoff is negative due to the high penalty for a mismatch in type [because the expected payoff from such a link is at most $\pi (B,B ) - \frac{n_y}{n_y - 1}d$, which is negative due to the assumption that $d > \pi (B,B ) \frac{n-1}{n_y}$]. Second, if the mutant switched from A to B, then any revising agent who is maintaining a connection with the mutant will not switch to B since he has $k-1$ other neighbors playing A and so by switching he would get $(k-1)\pi (A,A)+\pi (A,B) > \pi (B,B) + (k-1)\pi (B,A)$ (the inequality holding because $k \ge 2$ and A is the risk-dominant action). Third, if instead the mutant switched from B to A, then any neighboring revising agent will not switch to A since he can keep playing B, remove the link with the mutant and form a new link with another agent playing B (who exists because $n_y \ge 2k+1$) which gives him $k\pi (B,B)-kc > \pi (A,A) + (k-1)\pi (A,B)-kc$. Fourth, changing action is clearly suboptimal for an agent who is not maintaining a connection with the mutant. Finally, we observe that when the mutant is given a revising opportunity, he will certainly choose to have k links toward agents playing the same action he was playing before the mutation, because only doing so he can avoid to pay the cost $d > \pi (B,B)$ since all other agents’ action is perfectly correlated with their type; given this, it follows that the mutant will also choose to play the action he was playing before the mutation, since this allows him to coordinate. From these five observations it follows that, after that a single mutation has occurred, the system will surely go back to a state where conditions (1), (2) and (3) in point (a) of Lemma 2 are satisfied, hence belonging to $S^A_B$.

A similar reasoning can be applied considering $S^B_A$ in the place of $S^A_B$, thus obtaining that $R(S^A_B) \ge 2$ and $R(S^B_A) \ge 2$. In what follows we show that if in addition we have that:

$$\begin{aligned} {n_y > 2k + \frac{kd}{\pi (B,B) - \max \left\{ \pi (A,A), \pi (A,B)\right\} }}, \end{aligned}$$

(1)

then $R(S^A_B)$ and $R(S^B_A)$ can be computed considering only mutations that hit agents of a single type—either x or y.

We start by providing a sufficient condition to have that m mutations hitting agents of type x are enough for the system to leave $S^A_B$ with positive probability. By Lemma 2, we know that $S^A_B$ is a single absorbing set. So we can choose a specific state in $S^A_B$ to start from, and in particular we can choose a state where there exists a cluster made of $k+1$ agents of type x (i.e., $g_{ij} = 1$ for any i and j in the cluster). If m mutations hit m distinct agents in the cluster inducing them to switch action from A to B (and with no change to their interaction networks), and if the following inequality is satisfied:

$$\begin{aligned} m (\pi (B,B ) - c) + (k - m) \max \left\{ 0, \pi (B,A) - c \right\} > k (\pi (A,A) - c), \end{aligned}$$

(2)

then all non-mutants in the cluster who receive a revision opportunity will find it optimal to choose to perform action B while not changing their interaction networks. So, with positive probability the system reaches a new state that belongs to an absorbing set different from $S^A_B$ (indeed, at least the $k+1$ agents of type x in the cluster will never go back to action A). We also observe that other agents of type x might then find it profitable to switch to action B and that all agents of type y will keep playing action B. Finally, we note that (2) is surely satisfied when $m = k$.

Suppose now that, starting from a state $s \in S^A_B,\,m_x$ mutations hit agents of type $x,\,m_y$ mutations hit agents of type y, so that state $s'$ is reached from which another absorbing set can be reached with positive probability. Since we know that (2) is satisfied when $m = k$, then in the following we focus on the case where $m_x + m_y < k$.

We first observe that at least one of the following two inequalities must be satisfied in $s'$:

$$\begin{aligned}&m_x (\pi (B,B) - c) + (k - m_x) \max \{0, \pi (B,A) - c\} \nonumber \\&\quad \ge k \left[ \pi (A,A) - c - {\frac{d m_y}{n_x - 2k + m_y}}\right] ; \end{aligned}$$

(3)

$$\begin{aligned}&m_y (\pi (A,A) - c) + (k - m_y) \max \{0, \pi (A,B) - c\}\nonumber \\&\quad \ge k \left[ \pi (B,B) - c - {\frac{d m_x}{n_y - 2k + m_x}}\right] . \end{aligned}$$

(4)

To see why, suppose that both (3) and (4) are not satisfied. Then, no agent of type x who is given a revision opportunity finds it profitable to play action B [due to the failure of (3)], and no agent of type y who is given a revision opportunity finds it profitable to play action A [due to the failure of (4)]. For the mutant this is true a fortiori, since he can interact with smaller number of mutants—being himself one of the mutants. Hence, sooner or later all agents of type x will go back to play A and all agents of type y will go back play B. When we have perfect correlation between types and actions, it is obvious [because of the assumption that $d > \pi (B,B) (n-1)/(n_y)$] that revising agents will choose to maintain k links with agents choosing their same action (and hence having their same type). Therefore, if both (3) and (4) are not satisfied, then from $s'$ no other absorbing set can be reached.

We now show that, under the assumption that ${n_y > 2k + \frac{kd}{\pi (B,B) - \max \{\pi (A,A), \pi (A,B)\}}}$, inequality (4) is false. To understand this, it is enough to rewrite (4) with < instead of $\ge$ and to obtain an explicit bound on $n_y$ (getting rid of $m_x$ by making the new inequality harder to be satisfied). Therefore, inequality (3) must hold.

We then observe that, if inequality (3) holds, then inequality (2) is implied by the assumption that ${n_y > 2k + \frac{kd}{\pi (B,B) - \max \{\pi (A,A), \pi (A,B)\}}}$. To understand this, we fix $m = m_x + m_y$, we take the difference between the left-hand side of (2) and the left-hand side of (3), and we set it larger than the difference between the right-hand side of (2) and the right-hand side of (3). Working out such inequality we obtain a bound on $n_x$ that is implied by ${n_y > 2k + \frac{kd}{\pi (B,B) - \max \{\pi (A,A), \pi (A,B)\}}}$ (once it is noted that $n_y < n_x$). This means that, if $m_x$ mutations hitting agents of type x and $m_y$ mutations hitting agents of type y allow the system to leave $S^A_B$ with positive probability, then $m_x + m_y$ mutations hitting agents of type x only are also sufficient for the system to leave $S^A_B$ with positive probability.

We can repeat all the previous arguments with $S^B_A$ in the place of $S^A_B$. The only difference is that $m_x$ and $m_y$ have inverted roles, and the same occurs for $n_x$ and $m_y$. Summing up, it is true that, if ${n_y > 2k + \frac{kd}{\pi (B,B) - \max \{\pi (A,A), \pi (A,B)\}}}$, then we are allowed to focus on mutations hitting only agents of one type, in the attempt to determine $R(S^A_B)$ and $R(S^B_A)$.

Finally, we focus on mutations hitting only one type of agents, and we determine $R(S^A_B)$ and $R(S^B_A)$. We consider a state $s \in S^A_B$, and we observe that we already know that m mutations that hit agents of type x inducing them to switch from A to B are enough to leave $S^A_B$ with positive probability, if inequality (2) is satisfied. We remark that such inequality is also necessary for such an exit from $S^A_B$. Indeed, it is immediate to observe that, if (2) is not satisfied, then no agent of type x who is given a revision opportunity will find it profitable to play action B, and agents of type y will clearly keep playing action A. Furthermore, once a mutant goes back to action A, the gain (potentially negative) of choosing B over A for agents of type x is further reduced, while agents of type y never find it profitable to play A over B. Sooner or later, perfect correlation between types and actions will be restored, and a k-regular and type-segregated state will be reached, belonging to $S^A_B$. We denote with $\overline{m}$ the minimum m such that inequality (2) is satisfied. We note that $\overline{m} \ge k$.

We now consider mutations hitting agents of type y who switch from B to A. As long as at least $k+1$ agents of type B keep choosing B, then mutants will sooner of later go back to B (due to the perfect correlation between B and y, and the fact that B is the payoff-dominant action). Since $n_y \ge 2k+1$, this means that at least $k+1$ mutations hitting agents of type y are required to leave $S^A_B$. Since $\overline{m} \ge k$, we can conclude that $R(S^A_B) = \overline{m}$. We can repeat exactly the same arguments with $S^B_A$ in the place of $S^A_B$, thus obtaining that $R(S^B_A) = \overline{m}$. Therefore, $R(S^A_B) = R(S^B_A)$. $\square$

1.7 Proof of Lemma 4

Proof

The proof begins by showing that, starting from a generic state $s \in Q$, another state $\hat{s}$ can be reached with positive probability such that it satisfies properties that help in the following construction of a path from $\hat{s}$ to an absorbing set $Q_{\ell } \subseteq S^A_A \cup S^B_B \cup S^A_B \cup S^B_A$.

Preliminarily, we define $\beta _x(s) \subseteq N_x$ as the set of agents of type x at state s who are playing A and have at least one best reply strategy where either action B is played, or a new link toward an agent currently playing B is cast, or both. Similarly, define $\alpha _y(s)$ as the set of agents of type y such that, at state s, they are playing B and have at least one best reply strategy where either action A is played, or a new link toward an agent currently playing A is cast, or both. We now show how the system can move with positive probability from state s to a state $\hat{s}$ where $\beta _x(\hat{s}) = \emptyset$ and $\alpha _y(\hat{s}) = \emptyset$.

Starting from state s, with positive probability all and only the agents in $\beta _x(s) \cup \alpha _y(s)$ will receive a revision opportunity and will choose a best reply strategy where either action B (respectively, action A) is played, or a new link toward an agent currently playing B (respectively, A) is cast, or both. Call $s'$ the state reached after these updates. If $\beta _x(s') = \emptyset$ and $\alpha _y(s') = \emptyset$, then we are done; otherwise we iterate the updating process, giving revision opportunities to all and only the agents in $\beta _x(s') \cup \alpha _y(s')$. We observe that this iteration will yield a state $\hat{s}$ with $\beta _x(\hat{s}) = \emptyset$ and $\alpha _y(\hat{s}) = \emptyset$ in a finite number of repetitions. This is so because agents who switch from A to B (respectively, from B to A) exit definitely set $\beta _x(s)$ [respectively, set $\alpha _y(s)$], and agents who cast a new link toward an agent currently playing B (respectively, A) can be part of set $\beta _x(s)$ [respectively, set $\alpha _y(s)$] for at most k times (since k is the maximum number of links that each agent can maintain).

Let us denote with $N_{xA}(\hat{s})$ the set of agents of type x who are playing action A at state $\hat{s}$, and with $N_{yB}(\hat{s})$ the set of agents of type y who are playing action B at state $\hat{s}$. The proof now proceeds by considering 4 possible cases concerning the emptiness/non-emptiness of the sets $N_{xA}(\hat{s})$ and $N_{yB}(\hat{s})$. For each case, we construct the needed sequence of states from $\hat{s} \in Q$ to a state in $Q_{\ell }$ that is either monomorphic or type-monomorphic.

Case 1 Suppose first that $N_{xA}(\hat{s}) \ne \emptyset$ and $N_{yB}(\hat{s}) \ne \emptyset$. We apply the following path-building procedure.

Consider a single mutation that hits an agent j of type x who is playing action B at $\hat{s}$. If no such agent exists, we are done. Otherwise, suppose that after the mutation agent j copies the strategy $\hat{s}_i = (\hat{a}_i, \hat{g}_i)$ of an agent $i \in N_{xA}(\hat{s})$; in particular, j will adopt strategy $s'_j = (a'_j, g'_j)$ such that $a'_j = \hat{a}_i,\,g'_{jh} = \hat{g}_{ih}$ for every $h \ne i, j$, and $g'_{ji} = \hat{g}_{ij}$. We now observe that 3 properties hold for every agent $h \in N_{xA}(s')$: Agent h has no best reply where (1) action B is chosen, or (2) a new link toward an agent playing B is cast, or (3) an existing link toward an agent of type x playing A is removed unless he is certain to find an agent of type x when casting a new link toward an agent playing A. (1) and (2) come from the fact that agents belonging to $N_{xA}(\hat{s})$ have, by construction, no best reply where action B is chosen or a new link toward an agent choosing B is cast; the same holds for the same agents a fortiori at state $s'$ (where agent j has switched from B to A), and it also holds for agent j, who is copying agent i after mutation. (3) comes from the simple observation that, given the optimality of choosing A, it cannot be optimal to remove a link from an agent of type x playing A unless he is certain to find an agent of type x when casting a new link toward an agent playing A.

For similar reasons, analogous properties hold for the agents belonging to $N_{yB}(s')$ (${=}N_{yB}(\hat{s})$): No agent of type y who is playing B has a best reply where action A is chosen, or a new link toward an agent playing A is cast, or an existing link toward an agent of type y playing B is removed unless he is certain to find an agent of type y when casting a new link toward an agent playing A.

The above three properties imply that any state $s''$ that is reachable with positive probability in the next period of the unperturbed dynamic is such that, for every agent $h \in N_{xA}(s')$, (1.) the number of h’s neighbors of type x choosing action B has not increased, i.e., $n_{hxB}(s'') \le n_{hxB}(\hat{s})$, (2.) the number of neighbors of type x choosing action A has not decreased, i.e., $n_{hxA}(s'') \ge n_{hxA}(\hat{s})$, (3.) the probability of mismatch for a new link toward an agent choosing action B has not decreased, i.e., ${\frac{n_{yB}(s'')-n_{hyB}(s'')}{n_{B}(s'')-n_{hB}(s'')} \ge \frac{n_{yB}(\hat{s})-n_{hyB}(\hat{s})}{n_{B}(\hat{s})-n_{hB}(\hat{s})}}$, and (4.) the probability of type mismatch for a new link toward an agent choosing action A has not increased, i.e., ${\frac{n_{yA}(s'')-n_{hyA}(s'')}{n_{A}(s'')-n_{hA}(s'')} \le \frac{n_{yA}(\hat{s})-n_{hyA}(\hat{s})}{n_{A}(\hat{s})-n_{hA}(\hat{s})}}$.

Analogous inequalities, of course appropriately adjusted, hold for agents belonging to $N_{yB}(s')$. Altogether these inequalities imply that, for the agents in $N_{xA}(s') \cup N_{yB}(s')$, the 3 properties holding at state $s'$ also hold at state $s''$. By induction, we can conclude that the same properties will hold forever, and hence an absorbing set must be reached where the number of agents of type x playing A never falls below $n_{xA}(s') = n_{xA}(\hat{s}) + 1$, and the number of agents of type y playing B never falls below $n_{yB}(s)' = n_{yB}(\hat{s})$.

Starting from any state $s'$ in this absorbing set, and following the reasoning done at the beginning of the proof for state s, a state $\hat{s}'$ where $\beta _x(\hat{s}') = \emptyset$ and $\alpha _y(\hat{s}') = \emptyset$ can be reached with positive probability. At state $\hat{s'}$, there exist at least $n_{xA}(\hat{s}) + 1$ agents of type x playing A, and $n_{yB}(\hat{s})$ agents of type y playing B. Then, following the above argument, a single mutation allows to reach another absorbing set where the number of agents of type x playing A is at least $n_{xA}(\hat{s}) + 2$, and the number of agents of type y playing B is at least $n_{yB}(\hat{s})$. Given the finiteness of the set $N_x$, an absorbing set where all agents of type x play A must eventually be reached and the number of agents of type y playing B is at least $n_{yB}(\hat{s})$. This completes the procedure.

The same path-building procedure can now be repeated, constructing a sequence of absorbing sets, with each step requiring a single mutation, and where the minimum number of agents of type y playing B increases by at least 1 at each step, while the minimum number of agents of type x playing A always remains $n_x$. Given the finiteness of the set $N_y$, at the end of this procedure a state is reached where all agents of type x play A, and all agents of type y play B; at such a state, all agents find it optimal to have exactly k links with agents choosing the same action and hence will end up having k connections with agents of the same type; by Lemma 2, we know that the absorbing set $S^A_B$ has been reached.

Case 2 We now suppose that $N_{xA}(\hat{s}) = \emptyset$ and $N_{yB}(\hat{s}) \ne \emptyset$. We apply the procedure described above to agents in $N_{yB}(\hat{s})$, thus obtaining that a single mutation per step is sufficient to move along a sequence of absorbing sets, where the number of agents of type y playing B increases by at least 1 at each step, until all agents of type y play B. Starting from any state $s''$ in the absorbing set that has been reached, and following the reasoning done at the beginning of the proof for state s, a state $\hat{s}''$ where $\beta _x(\hat{s}'') = \emptyset$ and $\alpha _y(\hat{s}'') = \emptyset$ can be reached with positive probability. We know for sure that $N_{yB}(\hat{s}'') = N_y$. If $N_{xA}(\hat{s}'') \ne \emptyset$, then we can apply the path-building procedure to agents in $N_{xA}(\hat{s})'$ and reason analogously to what done for case 1, so reaching the absorbing set $S^A_B$.

If instead $N_{xA}(\hat{s}'') = \emptyset$, then all agents of type x are playing B at state $\hat{s''}$. The only possibility that some agents of type x are indifferent between playing A and playing B is that they are isolated (an agent i is isolated if $g_{ij}=0$ for all $j \in N$). If they want to cast new links with agents choosing B, with positive probability they will do so and will be lucky enough to link to agents of type x. These agents now strictly prefer B over A. If more than one agent of type x remains isolated, then all such agents can jointly switch from B to A with positive probability; in the subsequent period, these agents will find it optimal to connect among themselves as playing A now implies to be an x and there is no risk of type mismatch; this leads with positive probability to a state $\hat{s}'''$ where $\beta _x(\hat{s}''') = \emptyset ,\,\alpha _y(\hat{s}''') = \emptyset$, and $N_{xA} \ne \emptyset$. Then, the path-building procedure described in case 1 can be applied starting from $\hat{s}'''$, and the absorbing set $S^A_B$ is eventually reached. Finally, if at most one agent of type x is isolated and indifferent between A and B, then a single mutation can hit such an agent and let him connect with other agents of type x playing B, so that an absorbing set belonging to $S^B_B$ is reached.

Case 3 Suppose that $N_{xA}(\hat{s}) \ne \emptyset$ and $N_{yB}(\hat{s}) = \emptyset$. This case runs as in case 2, with reversed roles between x and y and, when only one agent of type y is isolated, leading to an absorbing set belonging to $S^A_A$.

Case 4 Finally, we consider the case in which $N_{xA}(\hat{s}) = \emptyset$ and $N_{yB}(\hat{s}) = \emptyset$. All agents of type x find it optimal to choose B and to have k links toward agents playing B (who are surely of type x), while all agents y find it optimal to choose A and to have k links toward agents playing A (who are surely of type y). The absorbing set $S^B_A$ is so necessarily reached. $\square$

1.8 Proof of Lemma 5

Proof

Suppose to start from $S^A_B$. As shown in the proof of Lemma 3, after the formation of a cluster of $k+1$ agents of type x (which happens with positive probability starting from any $s \in S^A_B$) it is enough to have $R(S^A_B)$ mutations hitting the agents in such a cluster (in particular, making them switch from action A to action B while keeping their interaction network fixed) to move the system to a state from which, with positive probability, a new absorbing set $\widetilde{Q}$ is reached where at least those $k+1$ agents of type x choose B, and all agents of type y keep choosing B.

If $\widetilde{Q} \subseteq S^B_B$, we are done. Otherwise, consider a single mutation hitting an agent of type x who currently plays A, making him choose action B and cast all his connections toward agents choosing action B. With positive probability, the mutant casts all his links toward agents of type x (which, by construction, are at least $k+1$). This leads the system to either $S^B_B$ or to another absorbing set where the number of agents of type x playing B has increased by at least 1, while all agents of type y keep playing B. By repeating this argument, $S^B_B$ is surely reached within a finite number of steps each of which requires 1 mutation only.

The same reasoning can be applied to $S^B_A$ in the place of $S^A_B$, completing the proof. $\square$

1.9 Proof of Lemma 6

Proof

We show in the following that, starting from state $s \in S^A_A \cup S^B_B$, we can reach $S^A_B$ following a path of absorbing sets such that a single mutation is sufficient to move from one absorbing set to its successor in the path. The same arguments can be repeated for $S^B_A$ instead of $S^A_B$, completing the proof.

Suppose that to be in state $s \in S^B_B$. Suppose also that a single mutation hits an agent, say i, of type x making him switch to action A and no links. Call this new state $s'$. Since $d > \pi (B,B) \frac{n-1}{n_y}$, at $s'$ agent i does not want to cast new links toward agents playing B, so all states which are reachable with positive probability from $s'$ with one round of revision opportunities are such that i maintains no links. Moreover, if i is the only isolated agent of type x and no other agent wants to switch to action A, then $s'$ must belong to an absorbing set; otherwise, we reach a new state $s''$ which belongs to a new absorbing set where either i forms links with other isolated agents of type x who (with positive probability) switch to play A or some agents currently maintaining a link toward i switch to action A. With a further single mutation, another agent of type x who is currently playing B can be made switch to A, sever all his current links, and connect to and only to agents of type x who are playing A; this leads to a new state $s''$ belonging to a new absorbing set where the number of agents of type x playing A has increased. We can iterate the last passage until we get to some state in $S^A_B$.

Suppose now that $s' \in S^A_A$. We can apply an argument similar to the one just described (with the only difference that mutations affect agents of type y) and draw an analogous conclusion. $\square$