One-Way Flow Networks with Decreasing Returns to Linking

Abstract

This paper presents a model of noncooperative network formation in which the benefit of new links eventually decreases. Agents link with each other to gain information and update their links according to better-reply dynamics. We show that all equilibrium networks display a center-periphery structure and may be disconnected. The only equilibrium networks that satisfy some strictness conditions are constellations of starred wheels, where central agents form possibly several optimally sized disjoint wheels and peripheral agents are linked to one of the wheels from outside. In the long run, the system settles to such a network architecture. The main features of a constellation of starred wheels are reminiscent of some well-known real-world facts. Collections of smaller disjoint networks connecting only a few agents are more common than global networks connecting all the agents in a community. Also differences within a connected component such as the center and the periphery are often found.

Appendix: Proofs

Proof of Proposition 1

Let $P\le n^{*}$ then it is easy to show that we are in the assumptions of BG, so their proposition 3.1 applies and Nash networks are minimally connected and every player observes P agents. Now let $P>n^{*}$. The rest of the proof contains the following steps.

• 1. In a Nash network, there exists at least one player observing $n^{*}$ agents.

Proof

By contradiction. We show that in every network in which there are no agents observing $n^{*}$ there exists an agent who has a profitable deviation. Suppose g is a Nash network. By the strong incentives to linking assumption we know g is not empty. Additionally assume that in g all agents observe less than $n^{*}$. Take the player who observes most agents, $j_{1}=\arg \max _{j\in P}\mu _{j}$ and take the player who observes fewest agents among those not observed by $j_{1}$, $j_{2}=\arg \min _{j\in P\backslash N\left( j_{1},g\right) }\mu _{j}$. Note that $j_{2}$ exists because $\mu _{j_{1}}<n^{*}<P$ and that $\mu _{j_{2}}\le \mu _{j_{1}}$. Assume that $\mu _{j_{2}}^{d}\ge 1$. Let $j_{2}$ cut his link(s) and link to $j_{1}$: this is a profitable deviation because $u(\mu _{j_{1}}+1,1)>u(\mu _{j_{1}},1)\ge u(\mu _{j_{2}},1)\ge u(\mu _{j_{2}},\mu _{j_{2}}^{d})$. Assume now that $\mu _{j_{2}}^{d}=0$. Then by the strong incentives to linking assumption, linking to $j_{1}$ is a profitable deviation because $u(\mu _{j_{1}}+1,1)\ge u(2,1)>u(1,0)$ since $\mu _{j_{1}}\ge 1$. Assume now that in g all agents observe more than $n^{*}$. Let all agents drop redundant links. Suppose first that there exists an agent, $j_{1}$, who sponsors only one link, i.e., $\mu _{j_{1}}^{d}=1$. Consider the agent, say $j_{2}$, who has a link to $j_{1}$. Since there are no redundant links, $j_{2}$ observes $j_{1}$ only through the link he sponsors, $g_{j_{2}j_{1}}=1$. Then $j_{2}$ may cut his link to $j_{1}$ and link directly to the agent who $j_{1}$ has his only link to: $j_{2}$ would then increase his payoff since he would reduce the number of agents he observes by 1: $u(\mu _{j_{2}}-1,\mu _{j_{2}}^{d})>u(\mu _{j_{2}},\mu _{j_{2}}^{d})$. Consider now the opposite case in which all agents sponsor at least two links. This is impossible since P is finite, every agent observes more than $n^{*}$ and there are no redundant links. Indeed, take any such network and pick an agent, say $j_{1}$. By assumption $j_{1}$ sponsors direct links to (at least) two agents. Consider one of these, call him $j_{2}$: by assumption $\mu _{j_{2}}^{d}>1$. If $j_{2}$ sponsored direct links to any agents also receiving a direct link from $j_{1}$ this would violate non-redundancy of $j_{1}$’s links. So $j_{2}$ can only sponsor links to any agent not receiving links from $j_{1}$ and, in case, to $j_{1}$ as well. Consider an agent, $j_{3}\ne j_{1}$ receiving a direct link from $j_{2}$. By assumption $\mu _{j_{3}}^{d}>1$. If $j_{3}$ sponsored direct links to any agents receiving a direct link from $j_{2}$ or $j_{1}$ this would violate non-redundancy of $j_{2}$’s or $j_{1}$’s links. So $j_{3}$ can only sponsor direct links to any agent not receiving a direct link from $j_{2}$ or $j_{1}$ and, in case, to $j_{2}$ or $j_{1}$ only if $j_{1}\notin N\left( j_{2},g\right)$. Repeating this argument enough times shows that eventually there are no agents to link to, who are not already receiving a direct link from others in the network. This means that there is one last agent, call him $j_{M}$, who can only link to agents observed by $j_{1}$. This violates non-redundancy. So no network may exist in which all agents observe more than $n^{*}$ and all sponsor more than one non-redundant link. Finally, assume that in g some agents observe more than $n^{*}$ and some less than $n^{*}$. Take $j_{1}=\arg \max _{j\in \left\{ k\in P:\mu _{k}>n^{*}\right\} }\mu _{j}$ and $j_{2}=\arg \max _{j\in \left\{ k\in P:\mu _{k}<n^{*}\right\} }\mu _{j}$, so $j_{1}$ is the agent observing the most among those observing more than $n^{*}$ and $j_{2}$ is the agent observing the most among those observing less than $n^{*}$. Obviously $\mu _{j_{1}}\ge \mu _{j_{2}}+2$, since $\mu _{j_{2}}\le n^{*}-1$ and $\mu _{j_{1}}\ge n^{*}+1$. Now two possibilities occur: $u(\mu _{j_{2}},1)\ge u(\mu _{j_{1}},1)$ or $u(\mu _{j_{2}},1)<u(\mu _{j_{1}},1)$. In the first case $j_{1}$ has the profitable deviation of cutting his link(s) and linking to $j_{2}$ since $u(\mu _{j_{2}}+1,1)>u(\mu _{j_{1}},1)\ge u(\mu _{j_{1}},\mu _{j_{1}}^{d})$. In the second case, if there exists $j_{3}$ such that $n^{*}<\mu _{j_{3}}<\mu _{j_{1}}-1$ then $j_{2}$ has the profitable deviation of cutting his link(s) and linking to $j_{3}$ because $u(\mu _{j_{3}}+1,1)>u(\mu _{j_{1}},1)>u(\mu _{j_{2}},1)\ge u(\mu _{j_{2}},\mu _{j_{2}}^{d})$. Otherwise, all agents observing more than $n^{*}$ observe $\mu _{j_{1}}$. There are now two possibilities. Suppose first that there is a subset of $\mu _{j_{1}}>n^{*}$ agents observing all the agents in the subset. Then, applying the argument used above for the network in which all the agents observe more than $n^{*}$, we know that there is an agent who has a profitable deviation. Suppose now that all agents, observed by each agent observing $\mu _{j_{1}}$, observe less than $n^{*}$. Then consider an agent, call him $k_{0}$, observing $\mu _{k_{0}}=\mu _{j_{1}}>n^{*}$ and two agents observed by him who observe less than $n^{*}$ and do not observe each other, i.e., $k_{1},k_{2}\in N\left( k_{0},g\right)$ such that $\mu _{k_{1}}<n^{*}$, $\mu _{k_{2}}<n^{*}$ and $k_{i}\notin N\left( k_{j},g\right)$ for $i=1,2$ and $j\ne i$. Note that $k_{1}$ and $k_{2}$ always exist because $\mu _{k_{0}}>n^{*}$ and $\mu _{\ell }<n^{*}$ for all $\ell \in N\left( k_{0},g\right)$. Without loss of generality, let us posit $\mu _{k_{1}}\ge \mu _{k_{2}}$. Then, $k_{2}$ has the profitable deviation of cutting his link(s) and linking to $k_{1}$ since $u(\mu _{k_{1}}+1,1)>u(\mu _{k_{2}},1)\ge u(\mu _{k_{2}},\mu _{k_{2}}^{d})$. Hence, in any Nash network there exists an agent $\hat{\imath }$ such that $\mu _{\hat{\imath }}=n^{*}$.

• 2. All the agents, observed by a player $\hat{\imath }$ who observes $n^{*}$, observe $n^{*}$ too.

Proof

By contradiction. We show that if there exist an agent $j\in N\left( \hat{\imath },g\right)$ such that $\mu _{j}\ne n^{*}$ then there exists a profitable deviation. Note that $\mu _{j}<n^{*}$ otherwise $\mu _{\hat{\imath }}>n^{*}$. In particular take the agent farthest away from $\hat{\imath }$ among those who are observed by him and observe fewer than $n^{*}$ agents, i.e., $j_{1}=\arg \max _{j\in \left\{ k\in N(\hat{\imath },g):\mu _{k}<n^{*}\right\} }d\left( \hat{\imath },j;g\right)$. If such an agent does not exist the proof is complete. Otherwise, $\hat{\imath }$ does not use any of $j_{1}$’s link to observe anyone else. Then $j_{1}$ can cut all his links (if any) and link to $\hat{\imath }$. This is a profitable deviation because he would then observe $n^{*}$ agents with one link: $u(n^{*},1)>u(\mu _{j_{1}},1)\ge u(\mu _{j_{1}},\mu _{j_{1}}^{d})$. This argument implies that a Nash network may be composed of components “built around” disjoint subsets of $n^{*}$ agents each of whom observes all the $n^{*}$ players in the same subset.

• 3. In a Nash network, there are only agents observing $n^{*}$ or $n^{*}+1$ agents.

Proof

In the previous step, we established that in the network there is at least one subset of $n^{*}$ agents each observing all the $n^{*}$ agents in the same subset. Moreover, by the same argument we can conclude that there are no agents observed by someone observing $n^{*}$ who observe less than $n^{*}$. Since we assumed that $N>n^{*}$ we have to characterize the links of those not observed by anyone observing $n^{*}$. Let M be the set of such agents, i.e., $M=\left\{ k\in P:k\notin N(\hat{\imath },g),\text {for some }\hat{\imath }\text { with }\mu _{\hat{\imath }}=n^{*}\right\}$. Let us proceed by contradiction. Take any agent $j_{1}\in M$, he observes $\mu _{j_{1}}\ne n^{*}$ and sponsors $\mu _{j_{2}}^{d}\ge 1$ links. If no such agent exists then the proof is completed. There are now two possibilities. First assume that there exists another player $j^{\prime }\in M\backslash N\left( j_{1},g\right)$ such that $j_{1}\notin N\left( j^{\prime },g\right)$ and $\mu _{j^{\prime }}=n^{*}-1$. Then $j_{1}$ cuts all his links and links to $j_{2}$: by doing so $j_{1}$ gets to observe $n^{*}$ agents with one link, which is a profitable deviation as $u(n^{*},1)>u(\mu _{j_{1}},\mu _{j_{1}}^{d})$. Now assume that no such agent $j^{\prime }$ exists and consider the case in which $\mu _{j_{1}}>n^{*}+1$. Then $j_{1}$ has the profitable deviation of cutting his link(s) and linking to $\hat{\imath }$. By so doing he earns $u(n^{*}+1,1)>u(n^{*}+1,\mu _{j_{1}}^{d})>u(\mu _{j_{1}},\mu _{j_{1}}^{d})$. Alternatively let $\mu _{j_{1}}<n^{*}$. Since $u(n^{*}+1,1)>u(m,1)$ for $m<n^{*}$, $j_{1}$ has the profitable deviation of cutting his link(s) and linking to $\hat{\imath }$. By so doing he earns $u(n^{*}+1,1)>u(\mu _{j_{1}},1)\ge u(\mu _{j_{1}},\mu _{j_{1}}^{d})$. Finally if in g there are redundant links, then some agent could delete a link and still observe same number of agents increasing his payoff.

This concludes the proof of Proposition 1.

Proof of Proposition 2

We first show that a constellation of starred wheels of dimension $n^{*}$ belongs to the PSSE set of the game. Given the $n^{*}+m$ agents who form one $SW\left( n^{*},m\right)$, none of them wants to individually deviate in a way that alters the starred wheel architecture. Consider first the $n^{*}$ agents who form the $W\left( n^{*}\right)$. They obtain the maximum payoff since they observe $n^{*}$ agents and only pay for one link. So they have no incentive to deviate. Let us now consider the m peripheral agents who are linked to the wheel. None of them can improve his payoff: if one of them cuts his link and links somewhere else to the wheel neither his payoff nor the architecture change. If he links to someone else who is directly linked to the $W\left( n^{*}\right)$ his payoff is reduced since now this agent observes $n^{*}+2$. If a peripheral agent links somewhere outside the starred wheel then the starred wheel still exists. He can only increase his payoff by linking (with only one link) to someone who observes $n^{*}-1.$ This is impossible since the original network was a constellation of starred wheels of dimension $n^{*}$. So a constellation of starred wheels of dimension $n^{*}$ belongs to the PSSE set because every time agents deviate in such a way that the resulting architecture is not a constellation of $SW\left( n^{*},m\right)$, these agents are strictly worse off and agents are indifferent only between strategies that do not alter the architecture. Let us now show the other direction. From Proposition 1 we know that in any Nash network some agents, the central agents, belong to subsets of $n^{*}$ agents observing each other and other agents, the peripheral agents, observe them without being observed by anyone else. Then, to prove that any network in the PSSE set must be a constellation of $SW\left( n^{*}\right)$, we only need to show that the central agents must form a $W\left( n^{*}\right)$. This is very easy as we can directly apply the argument in the proof Proposition 3.2 in BG, to show that each central agent sponsors one link only and receives one link only from another central agent (while he may receive a link from a peripheral one).

Proof of Lemma 2

Let $M=N\left( i_{1},g\right)$. Consider $i_{2}\in \arg \max _{\ell \in M}d\left( i_{1},\ell ;g\right)$ so $i_{2}$ is farthest away from $i_{1}$ among those who are observed by $i_{1}$. Note that $i_{1}$ does not use any of $i_{2}$’s links to observe anyone else. Agent $i_{2}$ can improve his payoff by cutting all his links and linking to $i_{1}$ directly since $\pi _{i_{2}}(g)\le \pi _{i_{1}}(g)\le u(m,1)$ as $i_{2}\in N\left( i_{1},g\right)$. So he does this. We will call this new network $g^{(2)}$. Now $i_{2}$ observes all the agents in M (so m) with one single link and his payoff is exactly $u\left( m,1\right)$ which is the new maximum payoff in M. Moreover, all the agents who observe $i_{2}$ now observe all the agents in M. We proceed by induction. Assume we have done the first $\ell -1$ iterations. Call agent $i_{\ell }$ the player who moved in iteration $\ell -1$ for $\ell >1$. Let $g^{\left( \ell \right) }$ be the network formed at the end of iteration $\ell$, i.e., after $i_{\ell }$ moved. All the agents who moved in the previous round are consecutively linked to each other, i.e., $i_{s}$ has one single link to agent $i_{s-1}$ for $s=2,\ldots ,\ell$. So the distance between $i_{\ell }$ and $i_{1}$ in the network $g^{\left( \ell \right) }$ is $\ell -1$ for $\ell >1$. Moreover $\left\| N\left( i_{s},g^{\left( \ell \right) }\right) \right\| =m$ for $s=1,\ldots ,\ell$ and $\ell >1$. Consider now $i_{\ell +1}\in \arg \max _{j\in M}d\left( i_{\ell },j;g^{\left( \ell \right) }\right)$ so $i_{\ell +1}$ is farthest away from $i_{\ell }$. By the reasoning applied to $i_{2}$ agent $i_{\ell +1}$ has a better reply of cutting all his links and linking to $i_{\ell }$ directly. We now show that eventually $i_{1}$ is selected, so that a wheel $W\left( m\right)$ forms. Consider the iterative procedure above. Since $i_{h}\in M$ and $\left\| N\left( i_{h},g^{\left( h\right) }\right) \right\| =m$ for all $h>2$ eventually $i_{1}$ is selected as M is finite and the distance between $i_{h}$ and $i_{1}$ increases by 1 in each step. Let $i_{k}$ be the last agent who moves before $i_{1}$ is selected. Notice that $i_{k}$ observes $i_{1}$ through a path consisting of all the agents who moved before him. As $i_{k}$ observes m agents and $i_{1}$ is the farthest away from $i_{k}$ it follows that $k=m$. Now $i_{1}$ is the only agent in M who can possibly have more than one link. If so then let $i_{1}$ play. By the same reasoning as $i_{2}$, agent $i_{1}$’s better response is to cut all his links and link to $i_{m}$. Now $i_{1}$ observes m agents with one single link and closes a wheel of dimension m among the agents $i_{1},\ldots ,i_{m}$ with possibly other players observing them.

To prove Proposition 3 we need to establish some preliminary results. The following Lemmas present some relevant cases in which the better-response dynamics (3) leads to the emergence of an agent who observes $n^{*}$ players. Lemma 3 considers the case where more than $n^{*}$ agents observe fewer than $n^{*}$.

Lemma 3

If there exist more than $n^{*}$ agents each of whom observes between $\underline{n}$ and $n^{*}-1$ agents then in finite time there will be at least one agent who observes exactly $n^{*}\$agents, i.e., the better-response dynamics $g^{t}$ is such that for some finite $t^{\prime }$ there exists an agent j for whom $\left\| N\left( j,g^{t^{\prime }}\right) \right\| =n^{*}$.

Proof

Call $\hat{P}$ the set of agents each of whom observes between $\underline{n}$ and $n^{*}-1$ agents in the network g, formally $\hat{P}=\left\{ h\in P:\underline{n}\le \left\| N\left( h,g\right) \right\| \le n^{*}-1\right\}$. Notice that $||\hat{P}||\ge n^{*}+1$. Among the agents belonging to $\hat{P}$ let $i_{1}$ be the agent who observes the most agents, i.e., $i_{1}\in \arg \max _{j\in \hat{P}}\left\| N\left( j,g\right) \right\|$. Let $m_{1}=\left\| N\left( i_{1},g\right) \right\|$. Note that $m_{1}\le n^{*}-1$. Consider $i_{2}\in \arg \max _{\ell \in N\left( i_{1},g\right) }d\left( i_{1},\ell ;g\right)$ so $i_{2}$ is farthest away from $i_{1}$ among those who are observed by $i_{1}$. Recall that $i_{1}$ does not use any of $i_{2}$’s links to observe anyone else. Agent $i_{2}$ can improve his payoff by cutting all his links and linking to $i_{1}$ directly since $\pi _{i_{2}}\left( g\right) \le u\left( m_{1},1\right)$ as $i_{2}\in N\left( i_{1},g\right)$. So he does this and receives a payoff exactly equal to $u\left( m_{1},1\right)$. Call this new network $g^{0}$. Now $N\left( i_{2},g^{0}\right) =N\left( i_{1},g\right)$ and $i_{2}$ is the best-off among the agents he observes. So by Lemma 2 the agents observed by $i_{2}$ form a $W\left( m_{1}\right)$. Call this new network $g^{\prime }$. Assume there exist two agents $k\in \hat{P}\backslash N\left( i_{j},g^{\prime }\right)$ and $i_{j}$ in the $W(m_{1})$, such that $g^{\prime }(k,i_{j})=1$, i.e., agent k is directly linked to agent $i_{j}$ in the wheel without belonging to the wheel. Assume that agent k sponsors one link only, i.e., $\mu _{k}^{d}\left( g^{\prime }\right) =1$. Then as $m_{1}\le n^{*}-1$, agent $i_{j+1}$ in the $W\left( m_{1}\right)$ whose only link is to $i_{j}$ has a better response of cutting his link and linking to k. This makes a wheel of dimension $m_{1}+1$. Assume now that agent k sponsors more than one link, i.e., $\mu _{k}^{d}\left( g^{\prime }\right) >1$. Let $m_{k}=\left\| N\left( k,g^{\prime }\right) \right\|$, note that $m_{k}\ge m_{1}+1$. First, if $\left\| N\left( k,g^{\prime }\right) \right\| =n^{*}$ the proof is complete. Second, if $\left\| N\left( k,g^{\prime }\right) \right\| >n^{*}$ then we need to consider two cases. If $u\left( m_{1},1\right) \ge u\left( m_{k},1\right)$ then k cuts all his other links and remains only with the link to $i_{j}$: we are now back to the case in which agent k has only one link to the wheel. If $u\left( m_{1},1\right) <u\left( m_{k},1\right)$ then agent $i_{j+1}$ cuts his link to $i_{j}$ and links to k and then he and all the other agents in the wheel get to observe $m_{k}>n^{*}$: by Lemma 4 below an agent observing $n^{*}$ agents emerges in finite time. Finally, if $\left\| N\left( k,g^{\prime }\right) \right\| <n^{*}$ then agent $i_{j+1}$ has a better response to link to k instead of $i_{j}$: by doing so he observes $m_{k}$ agents with one link only and hence is the best-off among those he observes and by Lemma 2 a wheel of dimension $m_{k}$ forms around him. Repeat the argument above until an agent gets to observe $n^{*}$ (which completes the proof) or no such k outside the wheel exists anymore. In the latter case call the new network $\tilde{g}$ and $m_{2}\$the dimension of the (enlarged) wheel. So $m_{2}\ge m_{1}$. All the agents in $\hat{P}$ either belong to the wheel $W\left( m_{2}\right)$ or are not in the wheel and hence observe fewer than or equal to $m_{1}$ agents. Call L the subset of $\hat{P}$ of agents who do not belong to the wheel. So $\hat{P}$ is partitioned into the agents belonging to the wheel $W\left( m_{2}\right)$ and those in L. Each agent $\ell \in L$ has a better response of cutting all his links and linking directly to the wheel as $\left\| N\left( \ell ,\tilde{g}\right) \right\| \le m_{2}\le n^{*}-1$. By so doing all the agents in L directly link to the wheel without belonging to it. So we re-apply the argument developed above for an agent k who is linked to the wheel directly without belonging to it. At some point an agent observes $n^{*}$ players as $||\hat{P}||>n^{*}$ by assumption.

In the next lemma, we consider the case of a network which is so highly connected that there is a subset of the population such that all the agents in this subset observe more than $n^{*}$ agents and all the agents observed by them also observe more than $n^{*}$.

Lemma 4

Assume there exists an agent k such that $n^{*}+1\le \left\| N\left( \ell ,g\right) \right\| \le \overline{n}$ for all $\ell \in N\left( k,g\right)$, i.e., such that all the agents observed by him observe between $n^{*}+1$ and $\overline{n}$ agents. Then in finite time there will be at least one agent who observes exactly $n^{*}\$agents.

Proof

Let $\hat{P}=\mathop {\textstyle \bigcup }_{\ell \in N\left( k,g\right) }N\left( \ell ,g\right)$. Note that $||\hat{P}||\ge n^{*}+1$ since $\left\| N\left( i,g\right) \right\| \ge n^{*}+1$ for all $i\in \hat{P}$. Consider now the agent who is best-off in $\hat{P}$ and call him $i_{1}$, so $\pi _{i^{\prime }}\left( g\right) \le \pi _{i_{1}}\left( g\right)$ for all $i^{\prime }\in N\left( i_{1},g\right)$. Let $m_{1}=\left\| N\left( i_{1},g\right) \right\|$ so $m_{1}\ge n^{*}+1$. By Lemma 2 the agents in $N\left( i,g\right)$ form a wheel $W\left( m_{1}\right)$. Call this new network $g^{\prime }$. Now take an agent j belonging to this wheel $W\left( m_{1}\right)$. Agent j has a better response (in fact it is his best response) of cutting his link and linking to agent r in the wheel such that $d\left( r,j;g^{\prime }\right) =n^{*}$. So he does so and observes exactly $n^{*}$ agents improving his payoff. This completes the proof.

The following lemma considers another highly connected network in which an agent $j_{1}$ observes more than $n^{*}$ agents. Moreover, among the agents he observes all those who observe fewer than $n^{*}$ players have a better response to cut all links and link to $j_{1}$ directly.

Lemma 5

Assume there exists an agent $j_{1}$ such that $n^{*}+1\le \left\| N\left( j_{1},g\right) \right\| \le \overline{n}$ and for each agent $\ell \in N\left( j_{1},g\right)$ such that $\left\| N\left( \ell ,g\right) \right\| \le n^{*}-1$ we have $u(\mu _{j_{1}}(g),1)\ge u(\mu _{j_{\ell }}(g),1)$. Then in finite time there will be at least one agent who observes exactly $n^{*}\$agents.

Proof

Let $m_{1}=\left\| N\left( j_{1},g\right) \right\|$, so $m_{1}$ is the number of agents observed by $j_{1}$. Consider $j_{2}\in \arg \max _{\ell \in \left\{ \ell \in N\left( j_{1},g\right) :\left\| N\left( \ell ,g\right) \right\| <n^{*}\right\} }\left\| N\left( \ell ,g\right) \right\|$ so $j_{2}$ is the agent who observes the maximum number of agents among those observed by $j_{1}$ who observe fewer than $n^{*}$. Agent $j_{2}$ exists by assumption. Consider an agent $\hat{\jmath }\in \arg \max _{\ell \in N\left( j_{2},g\right) }d\left( j_{2},\ell ;g\right)$ so $\hat{\jmath }$ is farthest away from $j_{2}$ among those who are observed by $j_{2}$, so $\left\| N\left( \hat{\jmath },g\right) \right\| \le \left\| N\left( j_{2},g\right) \right\| \le n^{*}-1$. Note that $\pi _{\hat{\jmath }}\left( g\right) =u(\mu _{\hat{\jmath }}(g),\mu _{\hat{\jmath }}^{d}(g))\le u(\mu _{\hat{\jmath }}(g),1)\le u(m_{1},1)$. Hence, it is a better response for $\hat{\jmath }$ to cut all his links and link to $j_{1}$ getting to observe $m_{1}$ agents. So he does this and receives a payoff equal to $u(m_{1},1)$. Call this new network $g^{\prime }$. Notice that $\hat{\jmath }$ and all the agents in $N\left( j_{1},g\right)$ who observed $\hat{\jmath }$ in g now observe exactly $m_{1}\ge n^{*}+1$ agents in $g^{\prime }$ as they were all observed by $j_{1}$ in g. However, nothing has changed for the other agents observed by $j_{1}$. So $g^{\prime }$ is identical to g with the exception of the move made by $\hat{\jmath }.$ In particular for each agent $\ell \in N\left( j_{1},g^{\prime }\right)$ such that $\left\| N\left( \ell ,g^{\prime }\right) \right\| \le n^{*}-1$ it is still true that $u(\mu _{j_{1}}(g^{\prime }),1)\ge u(\mu _{j_{\ell }}(g^{\prime }),1)$. We can then replicate this argument until no agent observes fewer than $n^{*}$. Now by Lemma 4 at least one player who observes exactly $n^{*}$ agents emerges.

Proof of Proposition 3

We find a path of the better-reply dynamics (3) such that starting from an arbitrary network it leads in finite time to a constellation of starred wheels of dimension $n^{*}$ through the following steps:

• 1. From an arbitrary network in finite time, we eliminate all agents whose payoff is less that $u\left( 1,0\right)$. In the resulting network, there are stand-alones or terminals or agents who do not observe fewer than $\underline{n}$ or more than $\overline{n}$ agents.

Proof

Take an arbitrary network g. Consider the subset of agents $Q\left( g\right) =\{ i\in P\vert \pi _{i}\left( g\right) <u\left( 1,0\right) \}$. Take the agent with the highest index in $Q\left( g\right)$ and call this agent j. So j’s better response is to cut all his links. Thus j becomes a stand-alone (or a terminal in the case he was observed by someone else) and receives $u\left( 1,0\right)$. Call this new network $g^{\prime }$. For every agent $i\notin Q\left( g\right)$ such that $j\in N\left( i,g\right)$ if j cuts all his links then $\pi _{i}\left( g^{\prime }\right) \ge u\left( 1,0\right)$ since $\underline{n}=2$. So $||Q\left( g^{\prime }\right) ||\le ||Q\left( g\right) ||-1$. Replicate this argument until all the agents receive a payoff greater or equal to $u\left( 1,0\right)$. Then all the agents who are neither stand-alones nor terminals observe between $\underline{n}$ and $\overline{n}$.

• 2. Let us eliminate the terminals from the network in finite time.

Proof

Consider a terminal agent i. By definition there exists an agent $j\in P\backslash \left\{ i\right\}$ who observes i, i.e., $i\in N\left( j,g\right)$. We show that if i links to j then i is better off. Call the new network $g^{\prime }$. When i links to j then $N\left( i,g^{\prime }\right) =N\left( j,g\right)$ as i was a terminal in g. Since i has only one link in $g^{\prime }$, $\left\| N\left( i,g^{\prime }\right) \right\| =\left\| N\left( j,g\right) \right\| \le \overline{n}$ and $\pi _{j}\left( g\right) =u((\mu _{j}(g),\mu _{j}^{d}(g))\ge u(1,0)$ we find $\pi _{i}(g^{\prime })=u(\mu _{i}(g^{\prime }),1))=u(\mu _{j}(g),1))\ge u(\mu _{j}(g),\mu _{j}^{d}(g))\ge u(1,0)$. Replicate this argument until all terminals connect.

• 3. We now show that in finite time stand-alones connect.

  • (a) If there exists an agent $i\in g$ such that $\underline{n}\le \left\| N\left( i,g\right) \right\| \le \overline{n}-1$ , i.e., if there exists an agent i who observes no more than $\overline{n}-1$ and no fewer than $\underline{n}$ then we eliminate all the stand-alones in finite time.

Proof

Any stand-alone has a better response of linking to i. By so doing he observes $\left\| N\left( i,g\right) \right\| +1\le \overline{n}$ and receives a payoff greater than or equal to $u\left( 1,0\right)$. Agent i still observes the same agents as in g.

• (b) If $\left\| N\left( j,g\right) \right\| =\overline{n}$ holds for all $j\in g$ such that $\left\| N\left( j,g\right) \right\| >1$ we eliminate all the stand-alones in finite time.

Proof

Recall that $n^{*}\le \overline{n}-1$ and $\underline{n}=2$. Let us divide the proof into two cases. First, let us assume that there exist an agent i such that $\mu _{i}^{d}\left( g\right) =1$, i.e., i has only one link. Therefore i is the best-off among all the agents he observes. So by Lemma 2 a wheel $W\left( \overline{n}\right)$ forms among all the agents in $N\left( i,g\right)$. Call this new network $g^{\prime }$. Take now an agent j belonging to this wheel $W\left( \overline{n}\right)$. As $n^{*}\le \overline{n}-1$ agent j in the wheel has a better response of cutting his link and linking to agent r in the wheel such that $d\left( r,j;g^{\prime }\right) =\overline{n}-1$. So he does this and observes exactly $\overline{n}-1$ agents improving his payoff. Now we are back to the case considered in part a) of this Step and the proof is complete. Now assume that $\mu _{i}^{d}\left( g\right) \ne 1$ for all i. Fix one agent and call him $i_{1}$. Consider an agent $i_{2}\in \arg \max _{\ell \in N\left( i_{1},g\right) }d\left( i_{1},\ell ;g\right)$ so $i_{2}$ is farthest away from $i_{1}$ among those who are observed by $i_{1}$. Note that $i_{1}$ does not use any of $i_{2}$’s links to observe anyone else. Agent $i_{2}$ improves his payoff by cutting all his links (which are more than one) and linking to $i_{1}$ directly. Now agent $i_{2}$ has one link only and we are back in the previous case of part b).

• 4. Now all the players in the network only observe a number of agents between $\underline{n}$ and $\overline{n}$. From any such network in finite time a player that observes $n^{*}$ agents emerges.

Proof

Take the agent who observes the maximum number of agents $i_{1}\in \arg \max _{j\in P}\left\| N\left( j,g\right) \right\|$. If $\left\| N\left( i_{1},g\right) \right\| =n^{*}$ the proof is complete. Take $i_{2}\in$ $\arg \max _{j\in \left\{ \ell \in P:\left\| N\left( \ell ,g\right) \right\| <n^{*}\right\} }\left\| N\left( j,g\right) \right\|$, so $i_{2}$ is the agent who observes the maximum number of agents among those who observe fewer than $n^{*}$. If $i_{2}$ does not exist then by Lemma 4 at least one player who observes exactly $n^{*}$ agents will emerge. Let $m_{1}=\left\| N\left( i_{1},g\right) \right\|$ and $m_{2}=\left\| N\left( i_{2},g\right) \right\|$ whenever it exists.

Let us consider the following 3 cases.

• (a) Assume that $m_{1}=m_{2}$ . In finite time one player observing $n^{*}$ emerges.

  If $m_{1}=m_{2}$ then all the players observe fewer than $n^{*}$ agents. Hence by Lemma 3 at least one player observing exactly $n^{*}$ agents emerges.

• (b) Assume that $u(m_{1},1)>u(m_{2},1)$ and $m_{1}\ge n^{*}+1$ . Then in finite time one agent observes exactly $n^{*}\$ agents.

  Assume first that some agents who observe fewer than $n^{*}$ are observed by $i_{1}$. For all agents j with $\left\| N\left( j,g\right) \right\| \le n^{*}-1$ we have $\left\| N\left( j,g\right) \right\| \le m_{2}$. In particular for all $\hat{\jmath }\in N\left( i_{1},g\right)$ such that $\hat{m}\le n^{*}-1$ where $\hat{m}=\left\| N\left( \hat{\jmath },g\right) \right\| ,$ we have $u(m_{1},1)>u(\hat{m},1)$. So by Lemma 5 in finite time at least one player observing exactly $n^{*}$ agents emerges. If instead for all j such that $\left\| N\left( j,g\right) \right\| \le n^{*}-1$ we have $j\notin N\left( i_{1},g\right)$ then $\left\| N\left( k,g\right) \right\| \ge n^{*}+1$ for all agents $k\in N\left( i_{1},g\right)$ and hence by Lemma 4 at least one player who observes exactly $n^{*}$ agents will emerge.

• (c) Assume $u(m_{1},1)\le u(m_{2},1)$ and $m_{1}\ge n^{*}+1$ . To show: one agent that observes $n^{*}$ players will emerge in finite time.

  Let $\hat{P}=N\left( i_{1},g\right) \cup N\left( i_{2},g\right)$. Note that $||\hat{P}||\ge m_{1}>n^{*}$. Also $u\left( m_{1},1\right) \le u\left( m_{2},1\right)$ implies $\pi _{i_{1}}\left( g\right) =u(m_{1},\mu _{j_{1}}^{d})\le u(m_{1},1)\le u(m_{2},1)$. So agent $i_{1}$’s better response is to link to $i_{2}$ cutting all his original links. Let him do it and call the new network $g^{\prime }$. Now $N\left( i_{1},g^{\prime }\right) =\left\{ i_{1}\right\} \cup N\left( i_{2},g\right)$ and $\left\| N\left( i_{1},g^{\prime }\right) \right\| =m_{2}+1\le n^{*}$. If $\left\| N\left( i_{1},g^{\prime }\right) \right\| =n^{*}$ then the proof is complete. If $\left\| N\left( i_{1},g^{\prime }\right) \right\| \le n^{*}-1$ then $i_{1}$ is best-off among the agents in $N\left( i_{1},g^{\prime }\right)$. So by Lemma 2 a wheel $W\left( m_{2}+1\right)$ forms possibly with other agents observing it. Call this new network $g^{\prime \prime }$. We now show that the wheel expands so that no agent in $\hat{P}$ can be directly linked to the (enlarged) wheel and observe fewer than $n^{*}$ overall. Take any such agent and call him player i. By definition of i there exists an agent $i^{\prime }$ in the wheel such that $g(i,i^{\prime })=1$. As $m_{2}+1\le n^{*}-1$ agent $i^{\prime \prime }$ in the wheel whose only link is to $i^{\prime }$ has a better response of cutting his link and linking to i. So the wheel expands by 1. Call this new network $g^{\prime \prime \prime }$. Assume further that i observes some agents who do not belong to the $W\left( m_{2}+1\right)$. As $\left\| N\left( i,g^{\prime \prime }\right) \right\| \le n^{*}-1$ also $\left\| N\left( i,g^{\prime \prime \prime }\right) \right\| \le n^{*}-1$. For all j in the wheel $N\left( j,g^{\prime \prime \prime }\right) =N\left( i,g^{\prime \prime \prime }\right)$. In particular agent $i^{\prime }$ who belongs to the wheel $W\left( m_{2}+1\right)$ is best-off among the agents in $N\left( i^{\prime },g^{\prime \prime \prime }\right)$ so by Lemma 2 a wheel of dimension $\left\| N\left( i^{\prime },g^{\prime \prime \prime }\right) \right\|$ forms. Repeat until an agent gets to observe $n^{*}$ (which completes the proof) or no such i, that is linked to the wheel and observes fewer than $n^{*}$ agents overall, exists anymore. Call this new network $\check{g}$ and the wheel dimension $\check{m}=\left\| N\left( i,g^{\prime \prime \prime }\right) \right\| <n^{*}$. We now enlarge the wheel further so as to partition the agents in $\hat{P}$ into those who belong to the wheel and observe fewer than $n^{*}$ players and those who observe more than $n^{*}$ players. So we eliminate all the agents observing fewer than $n^{*}$ agents who do not belong to the wheel. In $\check{g}$ if agent j does not belong to the $W\left( \check{m}\right)$ then either $\left\| N\left( j,\check{g}\right) \right\| \ge n^{*}+1$ or $\left\| N\left( j,\check{g}\right) \right\| \le m_{2}$ by definition of $m_{2}$. So $\left\| N\left( j,\check{g}\right) \right\| \le \check{m}$ whenever $\left\| N\left( j,\check{g}\right) \right\| \le n^{*}-1$ for all $j\in \hat{P}$. Take agent $h_{1}$ who does not belong to the $W\left( \check{m}\right)$ such that $\left\| N\left( h_{1},\check{g}\right) \right\| \le n^{*}-1$. As $\left\| N\left( h_{1},\check{g}\right) \right\| <\check{m}$ and $\check{m}\le n^{*}-1$ then $h_{1}$ has a better response of cutting all his links and linking to the $W\left( \check{m}\right)$ (from outside). So he does this and observes $\check{m}+1$ agents. If $\check{m}+1=n^{*}$ then the proof is complete. Otherwise take $h_{2}$ among the agents not observed by $h_{1}$ in the new network such that $h_{2}$ observes fewer than $n^{*}$. By the same reasoning applied to $h_{1}$, agent $h_{2}$ improves his payoff by cutting all his links and linking to $h_{1}$. Repeat this argument until either some agents observe $n^{*}$ or there exist no agents in the network who do not belong to the wheel and observe fewer than $n^{*}$ players. If some agents observe $n^{*}$ then the proof is complete. Otherwise call $\check{h}$ the last agent who moved and note that he is best-off among all those he observes as he observes the most and still observes fewer than $n^{*}$ players with one link only. So by Lemma 2 a wheel forms among the agents observed by $\check{h}$. Call this new network $\hat{g}$ and the (new) wheel dimension $\hat{m}\le n^{*}-1$. In $\hat{g}$ all the agents in $\hat{P}$ either observe more than $n^{*}$ agents or belong to the wheel $W\left( \hat{m}\right)$ and thus observe $\hat{m}\le n^{*}-1$. Take $j_{0}=\arg \min _{h\in \left\{ \ell \in \hat{P}:\left\| N\left( \ell ,\hat{g}\right) \right\| >n^{*}\right\} }\left\| N\left( h,\hat{g}\right) \right\|$. Let $m_{0}=\left\| N\left( j_{0},\hat{g}\right) \right\|$. Assume first $u\left( \hat{m},1\right) <u\left( m_{0},1\right)$ then $i_{1}$ who belongs to the wheel $W\left( \hat{m}\right)$ has a better response of cutting his only link and linking to $j_{0}$. By so doing $i_{1}$ breaks the wheel. As in $\hat{g}$ everyone who observed fewer than $n^{*}$ agents belonged to the wheel $W\left( \hat{m}\right)$ now there exists no agent $\ell \in \hat{P}$ such that $\left\| N\left( \ell ,\hat{g}\right) \right\| <n^{*}$. So by Lemma 4 one agent observing $n^{*}$ agents surely emerges. Assume instead $u(\hat{m},1)\ge u(m_{0},1)$ then $j_{0}$ cuts all his links and links to the wheel. By definition of $m_{0}$ for all $j^{\prime }\in \hat{P}$ with $m^{\prime }\ge n^{*}$ where $m^{\prime }=\left\| N\left( j^{\prime },\hat{g}\right) \right\|$, we have $m^{\prime }\ge m_{0}$ and hence $u \left( \hat{m},1\right) \ge u\left( m^{\prime },1\right)$. So all $j^{\prime }\in \hat{P}$ with $\left\| N\left( j^{\prime },\hat{g}\right) \right\| \ge n^{*}$ have the same better response and link to the wheel. As $\hat{m}\le n^{*}-1$ and $||\hat{P}||>n^{*}$ the wheel enlarges to $n^{*}-\hat{m}-1$ (peripheral) agents and an agent observing $n^{*}$ players will emerge.

• 5. Now in the network at least one agent observes $n^{*}$ players. In finite time all the agents in the network observing $n^{*}$ players belong to a $SW\left( n^{*},m\right)$.

  • (a) If a player observes $n^{*}$ agents then this player belongs to a $W\left( n^{*}\right) \$ in finite time.

Proof

By assumption there exists a player, say $i_{1}^{*}$, who observes $n^{*}$ agents, that is $\left\| N\left( i_{1}^{*},g\right) \right\| =n^{*}$. Consider $i_{2}^{*}\in \arg \max _{\ell \in N\left( i_{1}^{*},g\right) }d\left( i_{1}^{*},\ell ;g\right)$, i.e., the an agent farthest away from $i_{1}^{*}$. We already know that $i_{1}^{*}$ does not use any of $i_{2}^{*}$’s links to observe anyone else. Agent $i_{2}^{*}$ improves his payoff (in fact it is his best response) by cutting all his links and linking to $i_{1}^{*}$ directly. So he does this. Call this new network $g^{\prime }$. Now $i_{2}^{*}$ enjoys the maximum payoff attainable: he observes $n^{*}$ agents with one link. Notice that $i_{2}^{*}$ is best-off in $N\left( i_{2}^{*},g^{\prime }\right)$ and $\left\| N\left( i_{2}^{*},g^{\prime }\right) \right\| =n^{*}$ so by Lemma 2 a wheel $W\left( n^{*}\right)$ forms. Replicate this argument until there exist no agents observing $n^{*}$ players who do not belong to a wheel $W\left( n^{*}\right)$.

• (b) If all the agents who observe $n^{*}$ players belong to a wheel $W\left( n^{*}\right)$ and there exists an agent who does not belong to a $W\left( n^{*}\right)$ then a starred wheel of dimension $n^{*}$ emerges in finite time.

Proof

Call i the agent who does not belong to a $W\left( n^{*}\right)$. Let $m_{i}=\left\| N\left( i,g\right) \right\|$. Note that $m_{i}\ne n^{*}$. We claim that agent i has a better response of cutting all his links and linking to the wheel from outside. Assume first that either $m_{i}>n^{*}$ or $m_{i}<n^{*}-1$ then $\pi _{i}\left( g\right) \le u(n^{*}+1,1)$. So i’s better response is to cut all his links and to link to a wheel of dimension $n^{*}$ forming a starred wheel of the same dimension. Assume instead that $m_{i}=n^{*}-1$. There we consider two cases. First suppose that $u(n^{*}-1)\le u(n^{*}+1)$ then any agent observing $n^{*}-1$ has a better response of linking to a $W\left( n^{*}\right)$ forming a starred wheel of the same dimension. Assume now that $u(n^{*}-1)>u(n^{*}+1)$. If there exists another agent j who does not observe $n^{*}$ then j’s better response is to cut all the links and link to i getting to observe $n^{*}$ agents. Then a new wheel will emerge by part a) of this Step. If no such agent j exists then there exists an integer d such that $P=\left( d+1\right) n^{*}-1$ which is the integer problem we excluded.

• 6. In finite time the better-reply dynamics converges to a constellation of h starred wheels of dimension $n^{*}$, $h=1,\ldots ,\left\lfloor \frac{P}{n^{*}}\right\rfloor$.

Proof

By Proposition 2 a constellation of starred wheels of dimension $n^{*}$ is a PSSE set which is absorbing for the better-reply dynamics (3).

This concludes the proof of Proposition 3.