Impact of Utilities on the Structures of Stable Networks with Ordered Group Partitioning

Abstract

This paper considers a society partitioned into ordered groups and examines the stable networks that players form. Four utility functions combining benefits and costs from network connections are introduced. The utilities significantly influence players’ incentives in the formation of links and, consequently, network structure. Costs are affected by a given partition in one of two different ways: (i) a link’s cost depends only on the distance between the types of groups players belong to (the larger the distance, the larger the cost), and (ii) cost is not affected only by distance but also by the composition of a player’s neighborhood (the more members of a group the player has in her neighborhood, the less the average cost of a link is within this group). We observe that a player may prefer linking with players in other groups with a higher average link cost and reject linking with players in her own group when the second type of costs is applied. This never occurs with the first type of costs. We examine when specific network structures (i.e., empty network, complete network, minimal and minimally connected network, inner star and inner complete network) are pairwise stable with different utility functions. Stable networks regarding a special class of partitions with a unique large group and many individual players are also examined.

Appendices

Appendix 1: Proof of Proposition 1

Proof

First, we prove Item 1. Suppose a nonempty network g is stable but not minimal, then there exist players i and j such that $ij\in g$ and $i{\mathop {\longleftrightarrow }\limits ^{g-ij}}j$. As

$$\begin{aligned} {\Pi }_i(g)- {\Pi }_i(g-ij)={\Pi }_j(g)- {\Pi }_j(g-ij)=-f(|\pi (i)-\pi (j)|)<0 \end{aligned}$$

contradicting the stability of network g, therefore, any nonempty stable network is minimal.

Second, we prove Item 2. Since for any two players i and j, we have

$$\begin{aligned} {\Pi }_i(\varnothing )- {\Pi }_i(\{ij\})={\Pi }_j(\varnothing )- {\Pi }_j(\{ij\})=f(|\pi (i)-\pi (j)|)-1>0, \end{aligned}$$

and the empty network is stable when $f(0)>1$.

Next, we prove the uniqueness of a stable network. Suppose a nonempty network g is stable. When $f(0)>1$, network g is such that for any player $i\in N(g)$, $n_i(g)\ge 2$. Otherwise, if there exists a player $i\in N(g)$ such that $n_i(g)=1$, and $ij\in g$, then

$$\begin{aligned} {\Pi }_j(g)- {\Pi }_j(g-ij)&=1+ [n_j^{\pi (i)}(g-ij)-n_j^{\pi (i)}(g)]f(|\pi (i)-\pi (j)|)\\&=1-f(|\pi (i)-\pi (j)|)<0, \end{aligned}$$

which contradicts the stability of network g.

Therefore, the only possible case is that $n_i(g)\ge 2$ for any $i\in N(g)$. Let p and q be the players such that $d_{pq}(g)=\max \limits _{i{\mathop {\longleftrightarrow }\limits ^{g}}j}d_{ij}(g)$. If $d_{pq}(g)=1$, then from the fact that $n_p(g)\ge 2$ and $n_q(g)\ge 2$, there exist players $i\ne q$, $j\ne p$ such that $pi\in g, qj\in g$, and it follows that $i=j$. Otherwise, if $i\ne j$ and $ij\notin g$, then $d_{ij}(g)\ge 2$ contradicts $\max \limits _{i{\mathop {\longleftrightarrow }\limits ^{g}}j}d_{ij}(g)=1$; but if $i\ne j$ and $ij\in g$, then g is not minimal. It contradicts Item 1. If $i=j$, then players p, q and i compose a cycle in network g, which makes g not minimal. Thus, $d_{pq}(g)\ge 2$, then there exists player h such that $ph\in g$ and $h{\mathop {\longleftrightarrow }\limits ^{g}}q$, and from $n_p(g)\ge 2$, there exists player $r\ne h$ such that $pr\in g$. One of two options must be true: $d_{rq}(g-rp)=\infty$ or $d_{rq}(g-rp)\ne \infty$. The first option implies that $d_{rq}(g)=d_{pq}(g)+1$, contradicting that $d_{pq}(g)$ is the maximal distance. The second option contradicts the fact that network g is minimal.

Therefore, when $f(0)>1$, the empty network is a unique stable network with utility function (3). $\square$

Appendix 2: Proof of Proposition 2

Proof

First, we prove Item 1 by contradiction. Suppose a nonempty network g is stable but not minimal, then there must exist at least one link $ij\in g$ such that $i{\mathop {\longleftrightarrow }\limits ^{g-ij}}j$. For players i and j, there are two possible cases:

• 1. $\pi (i)=\pi (j)$, i.e., players i and j are in the same group, then for player i we have

  $$\begin{aligned} {\Pi }_i(g-ij)-{\Pi }_i(g)&=|\{k\mid i{\mathop {\longleftrightarrow }\limits ^{g-ij}}k \}|-\sum _{\begin{array}{c} ik\in g-ij \end{array}}c_{ik}^2(g-ij)-\Big [ |\{k\mid i{\mathop {\longleftrightarrow }\limits ^{g}}k \}|-\sum _{\begin{array}{c} ik\in g \end{array}}c_{ik}^2(g)\Big ]\\&=f(0)+\sum \limits _{k\ne \pi (i)} n_i^k(g)\Big [ \frac{n_i^k(g)-1}{n_i(g-ij)}-\frac{n_i^k(g)-1}{n_i(g)} \Big ]\\&=f(0)+\sum \limits _{k\ne \pi (i)} n_i^k(g)\Big [ \frac{n_i^k(g)-1}{n_i(g)-1}-\frac{n_i^k(g)-1}{n_i(g)} \Big ]\ge f(0)>0, \end{aligned}$$

  which implies that g is not stable since player i strictly gains by severing a link $ij\in g$, which is a contradiction.

• 2. $|\pi (i)- \pi (j)|=l>0$, i.e., players i and j are in different groups. For player i, we have

  $$\begin{aligned}&{\Pi }_i(g-ij)-{\Pi }_i(g)\\&\quad =|\{k\mid i{\mathop {\longleftrightarrow }\limits ^{g-ij}}k \}|-\sum _{\begin{array}{c} ik\in g-ij \end{array}}c_{ik}^2(g-ij)-\Big [ |\{k\mid i{\mathop {\longleftrightarrow }\limits ^{g}}k \}|-\sum _{\begin{array}{c} ik\in g \end{array}}c_{ik}^2(g)\Big ]\\&\quad =f(l)+\sum \limits _{\begin{array}{c} k\ne \pi (i)\\ k\ne \pi (j) \end{array}}n_i^k(g)\Big (\frac{n_i^k(g)-1}{n_i(g)-1}-\frac{n_i^k(g)-1}{n_i(g)}\Big )+\Big ( n_i^{\pi (j)}(g)-1\Big ) \frac{n_i^{\pi (j)}(g)-2n_i(g)}{(n_i(g)-1)n_i(g)}. \end{aligned}$$

  We notice that

  $$\begin{aligned} \sum \limits _{\begin{array}{c} k\ne \pi (i)\\ k\ne \pi (j) \end{array}}n_i^k(g)\Big (\frac{n_i^k(g)-1}{n_i(g)-1}-\frac{n_i^k(g)-1}{n_i(g)}\Big )\ge 0. \end{aligned}$$

  Then we consider a term

  $$\begin{aligned} \Big ( n_i^{\pi (j)}(g)-1\Big ) \frac{n_i^{\pi (j)}(g)-2n_i(g)}{(n_i(g)-1)n_i(g)}, \end{aligned}$$

  which is a decreasing function of $n_i^{\pi (j)}(g)$ for a given $n_i(g)$. The relation $n_i^{\pi (j)}(g)\le n_i(g)$ follows from the definitions of $n_i^{\pi (j)}(g)$ and $n_i(g)$. Hence, the minimal value of the term is obtained at $n_i^{\pi (j)}(g)=n_i(g)$, and we get

  $$\begin{aligned} \min \limits _{n_i^{\pi (j)}(g)} \Big [ \big ( n_i^{\pi (j)}(g)-1\big )\cdot \frac{n_i^{\pi (j)}(g)-2n_i(g)}{(n_i(g)-1)n_i(g)}\Big ]= \big ( n_i(g)-1\big )\cdot \frac{n_i(g)-2n_i(g)}{(n_i(g)-1)n_i(g)}=-1. \end{aligned}$$

  As a result,

  $$\begin{aligned} {\Pi }_i(g-ij)-{\Pi }_i(g)\ge f(l)-1>0, \end{aligned}$$

  which implies that g is not stable since player i strictly gains by deleting a link $ij\in g$, which is also a contradiction. Therefore, when $f(1)>1$, and the utility function is given by (4), any nonempty stable network must be minimal.

To complete the proof, we turn to Item 2. We can easily prove that the empty network is stable since $f(x)>1$ for $x\ge 0$. Now we prove the uniqueness of a stable network and suppose there exists a nonempty network g which is stable. From Item 1, it follows that g must be minimal. Similar to the latter part of Proposition 1 proof, we can state that there exists a player $j\in N$ such that $n_j(g)=1$. Suppose that $ij\in g$ and consider two cases:

• 1. If $\pi (i)=\pi (j)$, then

  $$\begin{aligned}&{\Pi }_i(g)- {\Pi }_i(g-ij)\\&\quad =1-f(0)+\sum \limits _{\begin{array}{c} k\ne \pi (i) \end{array}}n_i^k(g)\Big (\frac{n_i^k(g)-1}{n_i(g)}-\frac{n_i^k(g)-1}{n_i(g)-1}\Big )\le 1-f(0)<0, \end{aligned}$$

  which contradicts that network g is stable.

• 2. If $|\pi (i)-\pi (j)|=l>0$, then

  $$\begin{aligned} {\Pi }_i(g)- {\Pi }_i(g-ij)&= 1-f(l)+\sum \limits _{\begin{array}{c} k\ne \pi (i)\\ k\ne \pi (j) \end{array}}n_i^k(g)\Big (\frac{n_i^k(g)-1}{n_i(g)}-\frac{n_i^k(g)-1}{n_i(g)-1}\Big )\\&\quad +\frac{-(n_i^{\pi (j)}(g))^2+(2n_i(g)+1)n_i^{\pi (j)}(g)-2n_i(g)}{(n_i(g)-1)\cdot n_i(g)}\\&\le 1-f(l)+\frac{-(n_i^{\pi (j)}(g))^2+(2n_i(g)+1)n_i^{\pi (j)}(g)-2n_i(g)}{(n_i(g)-1)\cdot n_i(g)}. \end{aligned}$$

  When $n_i^{\pi (j)}(g)\le n_i(g)+\frac{1}{2}$, the term

  $$\begin{aligned} -(n_i^{\pi (j)}(g))^2+(2n_i(g)+1)n_i^{\pi (j)}(g)-2n_i(g) \end{aligned}$$

  increases as $n_i^{\pi (j)}(g)$ increases. Since $n_i(g)\ge n_i^{\pi (j)}(g)$, as a result,

  $$\begin{aligned} \frac{-(n_i^{\pi (j)}(g))^2+(2n_i(g)+1)n_i^{\pi (j)}(g)-2n_i(g)}{(n_i(g)-1)\cdot n_i(g)}\le 1. \end{aligned}$$

  Then

  $$\begin{aligned} 1-f(l)+\frac{-(n_i^{\pi (j)}(g))^2+(2n_i(g)+1)n_i^{\pi (j)}(g)-2n_i(g)}{(n_i(g)-1)\cdot n_i(g)}\le 2-f(l)<0, \end{aligned}$$

  which contradicts that g is a stable network.

$\square$

Appendix 3: Proof of Proposition 3

Proof

Since for any pair of players i and j such that $\pi (i)=\pi (j)$,

$$\begin{aligned} {\Pi }_i(\{ij\})-{\Pi }_i(\varnothing )={\Pi }_j(\{ij\})-{\Pi }_j(\varnothing )=1-f(0)>0, \end{aligned}$$

then the empty network cannot be stable when $f(0)<1$. In Proposition 1, it is proved that if the network is nonempty and stable, then it is minimal.

Now suppose nonempty network g is minimal but not connected stable network. Then there exist players i and j such that $d_{ij}(g)=\infty$. If $\pi (i)=\pi (j)$, then for player i,

$$\begin{aligned} {\Pi }_i(g+ij)- {\Pi }_i(g)=|\{k\mid j{\mathop {\longleftrightarrow }\limits ^{g}}k \}|+1-f(0)>0, \end{aligned}$$

and for player j,

$$\begin{aligned} {\Pi }_j(g+ij)- {\Pi }_j(g)=|\{k\mid i{\mathop {\longleftrightarrow }\limits ^{g}}k \}|+1-f(0)>0, \end{aligned}$$

which contradict an assumption that g is stable. Therefore, if $\pi (i)=\pi (j)$, then $i{\mathop {\longleftrightarrow }\limits ^{g}}j$, i.e., each player has an access to all members of the group she belongs to, owing to the fact of which, if $|\pi (i)-\pi (j)|=1$, a lower bound of the number of players that can be reached by of player i (j) linking with j (i) is $\eta _{\pi (j)}$ ($\eta _{\pi (i)}$). Then for player i,

$$\begin{aligned} {\Pi }_i(g+ij)- {\Pi }_i(g)=|\{k\mid j{\mathop {\longleftrightarrow }\limits ^{g}}k \}|+1-f(1)\ge \eta _{\pi (j)}-f(1)>0, \end{aligned}$$

and for player j,

$$\begin{aligned} {\Pi }_j(g+ij)- {\Pi }_j(g)=|\{k\mid i{\mathop {\longleftrightarrow }\limits ^{g}}k \}|+1-f(1)\ge \eta _{\pi (i)}-f(1)>0, \end{aligned}$$

which also contradicts an assumption that g is stable. As a result, for any pair of players $i\in N_l$, $j\in N_{l+1}$, $0\le l\le m-1$, i and j are connected in network g. This finishes the proof that if network g is stable, then it must be minimally connected. $\square$

Appendix 4: Proof of Proposition 4

Proof

The empty network is not stable which immediately follows from $f(0)<\frac{2}{n-1}\le 1$. Let g be a stable network that must be minimal which is proved in Proposition 2. Suppose g is not connected, so there exist players i and j such that $d_{ij}(g)=\infty$.

If $\pi (i)=\pi (j)$, i.e., players i and j are in the same group, then we consider the possible cases of the neighbor sizes:

• 1. Let $n_i(g)=n_j(g)=0$, i.e., players i and j be both isolated in network g, then

  $$\begin{aligned} {\Pi }_i(g+ij)-{\Pi }_i(g)={\Pi }_j(g+ij)- {\Pi }_j(g)=1-f(0)>0, \end{aligned}$$

  which contradicts that network g is stable.

• 2. Let $n_i(g)=0$, $n_j(g)\ge 1$, i.e., one be isolated and the other be not. Then for player j, we obtain

  $$\begin{aligned} {\Pi }_j(g+ij)- {\Pi }_j(g)&= 1-f(0)-\sum \limits _{\begin{array}{c} k\ne \pi (j) \end{array}}\Big [ n_j^k(g+ij)\Big ( f(|k-\pi (j)|)\\&\quad - \frac{n_j^k(g+ij)-1}{n_j(g+ij)} \Big ) -n_j^k(g)\Big ( f(|k-\pi (j)|) -\frac{n_j^k(g)-1}{n_j(g)} \Big ) \Big ]\\&= 1-f(0)+\sum \limits _{\begin{array}{c} k\ne \pi (j) \end{array}}n_j^k(g)\Big (\frac{n_j^k(g)-1}{n_j(g+ij)}- \frac{n_j^k(g)-1}{n_j(g+ij)-1} \Big )\\&= 1-f(0)+\sum \limits _{\begin{array}{c} k\ne \pi (j) \end{array}}n_j^k(g) \frac{1- n_j^k(g) }{\Big ( n_j(g+ij)-1\Big )n_j(g+ij)}. \end{aligned}$$

  Consider the multiplier

  $$\begin{aligned} \frac{1- n_j^k(g) }{\Big ( n_j(g+ij)-1\Big )n_j(g+ij)} \end{aligned}$$

  from the above sum. Obviously, when $n_j(g+ij)$ is fixed, it is a decreasing function of $n^{k}_j(g)$. Since $n_j^k(g)\le n_j(g+ij)-1$, the minimal value of this function is obtained at $n_j^{k}(g)=n_j(g+ij)-1$. Then

  $$\begin{aligned} \sum \limits _{\begin{array}{c} k\ne \pi (j) \end{array}}n_j^k(g) \frac{1- n_j^k(g) }{\Big ( n_j(g+ij)-1\Big )n_j(g+ij)}\ge \frac{2-n_j(g+ij)}{\Big ( n_j(g+ij)-1\Big )n_j(g+ij)} \sum \limits _{\begin{array}{c} k\ne \pi (j) \end{array}}n_j^k(g). \end{aligned}$$

  The expression in the RHS of the above inequality is nonpositive and reaches the minimal value when $\sum \nolimits _{\begin{array}{c} k\ne \pi (j) \end{array}}n_j^k(g)$ reaches the maximal value, that is, $n_j(g+ij)-1$. As a result,

  $$\begin{aligned} \sum \limits _{\begin{array}{c} k\ne \pi (j) \end{array}}n_j^k(g) \frac{1- n_j^k(g) }{\Big ( n_j(g+ij)-1\Big )n_j(g+ij)}\ge \frac{2-n_j(g+ij)}{n_j(g+ij)}. \end{aligned}$$

  Therefore,

  $$\begin{aligned} {\Pi }_j(g+ij)- {\Pi }_j(g)\ge \frac{2}{n_j(g+ij)}-f(0)\ge \frac{2}{n-1}-f(0)>0, \end{aligned}$$

  and for player i,

  $$\begin{aligned} {\Pi }_i(g+ij)- {\Pi }_i(g)>1-f(0)>0. \end{aligned}$$

  The above two inequalities contradict the stability of network g.

• 3. Let $n_i(g)\ge 1$ and $n_j(g)\ge 1$, i.e., both players i and j be not isolated in network g. Then we consider one of them, say, player i,

  $$\begin{aligned}&{\Pi }_i(g+ij)-{\Pi }_i(g)\\&\quad =|\{k\mid j{\mathop {\longleftrightarrow }\limits ^{g}}k \}|+1-f(0)+\sum \limits _{\begin{array}{c} k\ne \pi (i) \end{array}}n_i^k(g)\Big (\frac{n_i^k(g)-1}{n_i(g+ij)}- \frac{n_i^k(g)-1}{n_i(g+ij)-1} \Big ) \\&\quad \ge 2-f(0)+\sum \limits _{\begin{array}{c} k\ne \pi (i) \end{array}}n_i^k(g)\Big (\frac{n_i^k(g)-1}{n_i(g+ij)}- \frac{n_i^k(g)-1}{n_i(g+ij)-1} \Big ) \\&\quad \ge 2-f(0)+ \frac{2-n_i(g+ij)}{n_i(g+ij)}\ge 1-f(0)+\frac{2}{n-2}>0. \end{aligned}$$

  The same conclusion can be obtained for player j, which implies that network g is not stable.

From the analysis of the case $\pi (i)=\pi (j)$ given above, we conclude that in stable network g, any player has access to all members of the group she belongs to. Then consider the case when $|\pi (i)-\pi (j)|=1$, i.e., players i and j are in the closest groups. It is clear that the lower bound of the number of players, that player i can reach by linking with player j, is $\eta _{\pi (j)}$. Hence, for player i, it is true that

$$\begin{aligned} {\Pi }_i(g+ij)- {\Pi }_i(g)&\ge \eta _{\pi (j)} -f(l)+\sum \limits _{\begin{array}{c} k\ne \pi (i)\\ k\ne \pi (j) \end{array}}n_i^k(g)\Big (\frac{n_i^k(g)-1}{n_i(g+ij)}- \frac{n_i^k(g)-1}{n_i(g+ij)-1} \Big ). \end{aligned}$$

Consider the term

$$\begin{aligned} \frac{n_i^k(g)-1}{n_i(g+ij)}- \frac{n_i^k(g)-1}{n_i(g+ij)-1}, \end{aligned}$$

which is a decreasing function of $n_i^k(g)$ given $n_i(g+ij)$. Since $n_i^k(g)\le n_i(g+ij)-1$, the minimal value of this function is obtained at $n_i^{k}(g)=n_i(g+ij)-1$, then

$$\begin{aligned} \sum \limits _{\begin{array}{c} k\ne \pi (i)\\ k\ne \pi (j) \end{array}}n_i^k(g)\Big (\frac{n_i^k(g)-1}{n_i(g+ij)}- \frac{n_i^k(g)-1}{n_i(g+ij)-1} \Big )\ge \frac{2-n_i(g+ij)}{\Big ( n_i(g+ij)-1\Big )n_i(g+ij)}\sum \limits _{\begin{array}{c} k\ne \pi (i)\\ k\ne \pi (j) \end{array}}n_i^k(g). \end{aligned}$$

The expression in the RHS of the above inequality is nonpositive and reaches its minimal value when $\sum \limits _{\begin{array}{c} k\ne \pi (i)\\ k\ne \pi (j) \end{array}}n_i^k(g)$ reaches the maximal value, that is, $n_i(g+ij)-1$. Therefore,

$$\begin{aligned} {\Pi }_i(g+ij)- {\Pi }_i(g)\ge \eta _{\pi (j)} -f(1)+ \frac{2-n_i(g+ij)}{n_i(g+ij)}\ge \eta _{\pi (j)} -f(1)-1+\frac{2}{n-1}. \end{aligned}$$

Since $f(1)\le {\underline{M}}-1$, then

$$\begin{aligned} \eta _{\pi (j)} -f(1)-1+\frac{2}{n-1}\ge \frac{2}{n-1}>0, \end{aligned}$$

and the same result can be obtained for player j, which contradicts an assumption that g is stable. Thus, we conclude that, for any two players $i\in N_l$, $j\in N_{l+1}$, $0\le l\le m-1$, i and j are connected in network g. Finally, we prove that if network g is stable, then it must be minimally connected. $\square$

Appendix 5: Proof of Theorem 1

Proof

First, we note that the proof for the case when the utility function is defined by formula (3) is similar to that when the utility function is given by formula (4). Therefore, only the proof for function (4) is given below, and the other one is omitted to save the space.

Necessity Let g be a stable network. By Proposition 2, from $f(1)>{\overline{M}}>1$ it follows that g must be minimal. Thus, we need to prove two items for network g: (i) for any pair of players i and j such that $\pi (i)\ne \pi (j)$, i and j are not connected, (ii) for any pair of players i and j such that $\pi (i)=\pi (j)$, i and j are connected.

Suppose there exist players i and j such that $|\pi (i)- \pi (j)|=l>0$ and $ij\in g$. Consider all possible cases:

• 1. Let $|N_i(g)\cap N_{\pi (j)}|=1$. For player i, we have

  $$\begin{aligned} {\Pi }_i(g)- {\Pi }_i(g-ij)&=|\{k\mid k{\mathop {\longleftrightarrow }\limits ^{g-ij}}j \}|+1-f(l)\\&\quad +\sum \limits _{\begin{array}{c} k\ne \pi (i)\\ k\ne \pi (j) \end{array}}n_i^k(g)\Big (\frac{n_i^k(g)-1}{n_i(g)}- \frac{n_i^k(g)-1}{n_i(g)-1} \Big ) \\&\le |\{k\mid k{\mathop {\longleftrightarrow }\limits ^{g-ij}}j \}|+1-f(l). \end{aligned}$$

  Since $f(l)\ge f(1)>{\overline{M}}$, ${\Pi }_i(g)- {\Pi }_i(g-ij)$ may be nonnegative only when $|\{k\mid k{\mathop {\longleftrightarrow }\limits ^{g-ij}}j \}|+1> {\overline{M}}$, which indicates that there exist players p and q such that $\pi (p)\ne \pi (q)$ and $pq\in C_j(g-ij)$. Then the proof for the pair of player p, q can be repeated as for the pair i, j. Since g is minimal, $g \supset C_j(g-ij) \supset C_p(g-pq) \supset \ldots$, i.e., the components become smaller step by step. From the above analysis, we conclude that for any link $pq\in g$ such that $\pi (p)\ne \pi (q)$, it is true that $C_p(g-pq)\ne \varnothing$, and $C_q(g-pq)\ne \varnothing$. Since N is finite and network g is minimal, we finally come to a contradiction.

• 2. Let $|N_i(g)\cap N_{\pi (j)}|\ge 2$, which indicates that there exists at least one player $r\ne j$ such that $\pi (r)=\pi (j)$ and $ir\in g$. For player i,

  $$\begin{aligned}&{\Pi }_i(g)- {\Pi }_i(g-ij)= |\{k\mid k{\mathop {\longleftrightarrow }\limits ^{g-ij}}j \}|+1-f(l)\\&\quad -\sum \limits _{\begin{array}{c} k\ne \pi (i)\\ k\ne \pi (j) \end{array}}\frac{n_i^k(g)(n_i^k(g)-1)}{(n_i(g)-1)n_i(g)}+\frac{(n_i^{\pi (j)}(g)-1)\big (2n_i(g)-n_i^{\pi (j)}(g)\big )}{n_i(g)(n_i(g)-1)}. \end{aligned}$$

  Consider the last term $(n_i^{\pi (j)}(g)-1)\big (2n_i(g)-n_i^{\pi (j)}(g)\big )/\big [n_i(g)(n_i(g)-1)\big ]$. For given $n_i(g)$, it is an increasing function of $n_i^{\pi (j)}(g)$ when $n_i^{\pi (j)}(g)\le n_i(g)$. Then, the maximal value of this term is obtained at $n_i^{\pi (j)}(g)=n_i(g)$. Thus,

  $$\begin{aligned} {\Pi }_i(g)- {\Pi }_i(g-ij)&\le |\{k\mid k{\mathop {\longleftrightarrow }\limits ^{g-ij}}j \}|+2-f(l)+ \sum \limits _{\begin{array}{c} k\ne \pi (i)\\ k\ne \pi (j) \end{array}}\frac{-n_i^k(g)(n_i^k(g)-1)}{(n_i(g)-1)n_i(g)}\\&\le |\{k\mid k{\mathop {\longleftrightarrow }\limits ^{g-ij}}j \}|+2-f(l). \end{aligned}$$

  Obviously, only if term $|\{k\mid k{\mathop {\longleftrightarrow }\limits ^{g-ij}}j \}|+2-f(l)$ is nonnegative, then inequality ${\Pi }_i(g)\ge {\Pi }_i(g-ij)$ may hold. Since $f(l)\ge f(1)>{\overline{M}}$, then $|\{k\mid k{\mathop {\longleftrightarrow }\limits ^{g-ij}}j \}|+2-f(l)$ is nonnegative only when the integer $|\{k\mid k{\mathop {\longleftrightarrow }\limits ^{g-ij}}j \}|+1$ is not less than ${\overline{M}}$, which indicates that there exist players p and q, $\pi (p)\ne \pi (q)$, such that $pq\in C_j(g-ij)$. Then the proof for the pair of players p, q can be repeated as for the pair i, j. The same line of reasonings as in Case 1 leads to a contradiction. In conclusion, in stable network g, for any two players i, j such that $|\pi (i)- \pi (j)|=l>0$, we prove that $ij\notin g$.

Next suppose there exist players i, j such that $\pi (i)=\pi (j)$ and they are not connected in g. Considering players i and j, we obtain the inequalities:

$$\begin{aligned} {\Pi }_i(g+ij)- {\Pi }_i(g)&= |\{k\mid k{\mathop {\longleftrightarrow }\limits ^{g}}j \}|+1-f(0)>0, \\ {\Pi }_j(g+ij)- {\Pi }_j(g)&= |\{k\mid k{\mathop {\longleftrightarrow }\limits ^{g}}i \}|+1-f(0)>0, \end{aligned}$$

which contradict that g is stable. Therefore, for any two players i and j such that $\pi (i)=\pi (j)$, i and j must be connected in network g.

Sufficiency Let g be a partially minimally connected network regarding ${\Delta }$, we check its stability for different cases:

• 1. For any $ij\in g$, we know that $\pi (i)=\pi (j)$, then

  $$\begin{aligned} {\Pi }_i(g)- {\Pi }_i(g-ij)&= |\{k\mid k{\mathop {\longleftrightarrow }\limits ^{g-ij}}j \}|+1-f(0)>0,\\ {\Pi }_j(g)- {\Pi }_j(g-ij)&= |\{k\mid k{\mathop {\longleftrightarrow }\limits ^{g-ij}}i \}|+1-f(0)>0. \end{aligned}$$

• 2. For $ij\notin g$, if $\pi (i)=\pi (j)$, then for player i, it holds that

  $$\begin{aligned} {\Pi }_i(g+ij)- {\Pi }_i(g)=-f(0)<0. \end{aligned}$$

• 3. For $ij\notin g$, if $|\pi (i)-\pi (j)|=l>0$, then for player i, it is true that

  $$\begin{aligned} {\Pi }_i(g+ij)- {\Pi }_i(g)=\eta _{\pi (j)}-f(l)\le {\overline{M}}-f(l)<0. \end{aligned}$$

Therefore, partially minimally connected regarding ${\Delta }$ network g is stable. $\square$

Appendix 6: Proof of Proposition 5

Proof

Before we start the proof, we notice that it is supposed that $f(1)>f(0)+{\overline{M}}\big (1-f(0)\big )$ to have an interval for a decay parameter $\delta$ nonempty.

Suppose g is an inner star network regarding partition ${\Delta }$. To verify its stability, we need to prove two items from Definition 1:

• 1. For any $ij\in g$, we prove that ${\Pi }_i(g)\ge {\Pi }_i(g-ij)$ and ${\Pi }_j(g)\ge {\Pi }_j(g-ij)$. In fact, $\pi (i)=\pi (j)$ for any $ij\in g$. Consider two players $i, j\in N_l$, $l=1, \ldots , m$, and suppose i is in the center of the star, and j is in a noncentral position. For players i and j, the following inequalities hold:

  $$\begin{aligned} {\Pi }_i(g)- {\Pi }_i(g-ij)&=1-f(0)\ge 0,\\ {\Pi }_j(g)- {\Pi }_j(g-ij)&=1+\delta (\eta _{l}-2)- f(0)\ge 0. \end{aligned}$$

• 2. For any $ij\notin g$, we prove that if ${\Pi }_i(g)< {\Pi }_i(g+ij)$, then ${\Pi }_j(g)> {\Pi }_j(g+ij)$. We discuss the following cases:

  • (a) $\pi (i)=\pi (j)$, so both players i and j are in noncentral positions of the common star subnetwork. Players i and j are symmetric, so we just consider one of them, say, player i, and we have

    $$\begin{aligned} {\Pi }_i(g+ij)- {\Pi }_i(g)=1-\delta -f(0)\le \delta -\delta =0. \end{aligned}$$

  • (b) $|\pi (i)-\pi (j)|=l>0$, i.e., players i and j are in different groups. First, suppose that both i and j are in the central positions of the corresponding star subnetworks. We have

    $$\begin{aligned} {\Pi }_i(g+ij)-{\Pi }_i(g)=1+\delta (\eta _{\pi (j)}-1)-f(l). \end{aligned}$$

    Since

    $$\begin{aligned} \delta \le \frac{f(1)-1}{{\overline{M}}-1}\le \frac{f(l)-1}{{\overline{M}}-1}\le \frac{f(l)-1}{\eta _{\pi (j)}-1}, \end{aligned}$$

    then

    $$\begin{aligned} {\Pi }_i(g+ij)- {\Pi }_i(g)=1+\delta (\eta _{\pi (j)}-1)-f(l)\le f(l)-1+1-f(l)=0. \end{aligned}$$

    Similarly, for player j we have

    $$\begin{aligned} {\Pi }_j(g+ij)- {\Pi }_j(g)=1+\delta (\eta _{\pi (i)}-1)-f(l)\le f(l)-1+1-f(l)=0. \end{aligned}$$

    Next, consider player i being in the central position, and j being in a noncentral position. For player i, we have

    $$\begin{aligned} {\Pi }_i(g+ij)- {\Pi }_i(g)=1+\delta +\delta ^2(\eta _{\pi (j)}-2)-f(l)\le 1+\delta (\eta _{\pi (j)}-1)-f(l)\le 0, \end{aligned}$$

    and for player j,

    $$\begin{aligned} {\Pi }_j(g+ij)- {\Pi }_j(g)=1+\delta (\eta _{\pi (i)}-1)-f(l)\le f(l)-1+1-f(l)=0. \end{aligned}$$

    The similar calculations can be done for the case when both players i and j are in noncentral positions of their corresponding star subnetworks.

Finally, we prove that g, the inner star network regarding partition ${\Delta }$, is stable. $\square$

Appendix 7: Proof of Proposition 6

Proof

Let g be the inner complete network regarding ${\Delta }$. First, we prove Item 1.

• 1. For any $ij\in g$, $\pi (i)=\pi (j)$. Since players i and j are symmetric in the network, we consider any player, say player i, and we obtain

  $$\begin{aligned} {\Pi }_i(g)- {\Pi }_i(g-ij)=1-\delta -f(0)\ge 0. \end{aligned}$$

• 2. For any $ij\notin g$, $\pi (i)\ne \pi (j)$, let $l=|\pi (i)-\pi (j)|$. For player i the equality holds:

  $$\begin{aligned} {\Pi }_i(g+ij)- {\Pi }_i(g)=1+\delta (\eta _{\pi (j)}-1)-f(l). \end{aligned}$$

  Since $\delta \le 1-f(0)$, thus

  $$\begin{aligned} \delta (\eta _{\pi (j)}-1)\le (1-f(0))(\eta _{\pi (j)}-1)\le (1-f(0))({\overline{M}}-1). \end{aligned}$$

  As

  $$\begin{aligned} f(1)>{\overline{M}}-f(0)({\overline{M}}-1) \Longleftrightarrow 1-f(0)<\frac{f(1)-1}{{\overline{M}}-1}, \end{aligned}$$

  as a result

  $$\begin{aligned} (1-f(0))({\overline{M}}-1)<f(1)-1\le f(l)-1. \end{aligned}$$

  Then, we obtain the result:

  $$\begin{aligned} {\Pi }_i(g+ij)- {\Pi }_i(g)=1+\delta (\eta _{\pi (j)}-1)-f(l)< f(l)-1+1-f(l)=0. \end{aligned}$$

  The same conclusion can be obtained for player j. Therefore, the inner complete network g regarding ${\Delta }$, is stable when conditions in Item 1 are satisfied.

Second, we prove Item 2.

• 1. For any $ij\in g$, $\pi (i)=\pi (j)$. Since players i and j are symmetric in the network, we consider any player, say i, and get the following:

  $$\begin{aligned} {\Pi }_i(g)- {\Pi }_i(g-ij)=1-\delta -f(0). \end{aligned}$$

  Since

  $$\begin{aligned} \delta \le \frac{f(1)-1}{{\overline{M}}-1}\le \frac{ {\overline{M}}-f(0)({\overline{M}}-1)-1}{{\overline{M}}-1}=1-f(0), \end{aligned}$$

  then

  $$\begin{aligned} {\Pi }_i(g)- {\Pi }_i(g-ij)=1-\delta -f(0)\ge 0. \end{aligned}$$

• 2. For any $ij\notin g$, $\pi (i)\ne \pi (j)$, and let $l=|\pi (i)-\pi (j)|$. For player i we have

  $$\begin{aligned} {\Pi }_i(g+ij)- {\Pi }_i(g)=1+\delta (\eta _{\pi (j)}-1)-f(l). \end{aligned}$$

  Since $\delta \le \frac{f(1)-1}{{\overline{M}}-1}$, then

  $$\begin{aligned} \delta (\eta _{\pi (j)}-1)\le \frac{f(1)-1}{{\overline{M}}-1}(\eta _{\pi (j)}-1)\le f(1)-1. \end{aligned}$$

  We can easily prove that

  $$\begin{aligned} {\Pi }_i(g+ij)- {\Pi }_i(g)=1+\delta (\eta _{\pi (j)}-1)-f(l)\le f(1)-1+1-f(l)\le 0. \end{aligned}$$

  The same result is true for player j. $\square$

Appendix 8: Proof of Proposition 7

Proof

We require $f(m-1)<1$ in order to have interval $\delta \in (0, 1-f(m-1)]$ nonempty. Let g be the complete network. For any $ij\in g$, consider two possible cases:

• 1. If $\pi (i)=\pi (j)$, it is unprofitable for player i to delete the link as

  $$\begin{aligned} {\Pi }_i(g)- {\Pi }_i(g-ij)=1-\delta -f(0)\ge 0. \end{aligned}$$

• 2. If $|\pi (i)-\pi (j)|=l>0$, for player i, we have

  $$\begin{aligned} {\Pi }_i(g)- {\Pi }_i(g-ij)=1-\delta -f(l)\ge 0. \end{aligned}$$

  The same conclusion can be obtained for player j. $\square$

Appendix 9: Proof of Proposition 8

Proof

With $X=1+\frac{(n-{\underline{M}})(1-{\overline{M}})}{(n-1)(n-2)}$, we require $f(m-1)<X+\frac{2{\underline{M}}-2}{n-2}$ and $f(0)<X$ in order to have interval

$$\begin{aligned} \Big (0, \min \Big \{X+\frac{2{\underline{M}}-2}{n-2}-f(m-1), X-f(0)\Big \}\Big ] \end{aligned}$$

nonempty. Let g be the complete network. For any $ij\in g$, consider two possible cases:

• 1. If $\pi (i)=\pi (j)$, for player i, we have

  $$\begin{aligned} {\Pi }_i(g)- {\Pi }_i(g-ij)&=1-\delta -f(0)+\sum \limits _{\begin{array}{c} k\ne \pi (i) \end{array}}\eta _k \frac{1-\eta _k}{(n-1)(n-2)}\\&\ge 1-\delta -f(0)+\frac{(n-{\underline{M}})(1-{\overline{M}})}{(n-1)(n-2)} =X-\delta -f(0)\ge 0. \end{aligned}$$

• 2. If $|\pi (i)-\pi (j)|=l>0$, for player i, we calculate the following difference:

  $$\begin{aligned} {\Pi }_i(g)- {\Pi }_i(g-ij)&=1-\delta -f(l)+\sum \limits _{\begin{array}{c} k\ne \pi (i)\\ k\ne \pi (j) \end{array}}\eta _k \frac{1-\eta _k}{(n-2)}\\&\quad + \sum \limits _{\begin{array}{c} k\ne \pi (i) \end{array}}\eta _k \frac{\eta _k-1}{(n-1)} -\frac{(\eta _{\pi (j)}-1)(\eta _{\pi (j)}-2)}{n-2} \\&=\sum \limits _{\begin{array}{c} k\ne \pi (i) \end{array}}\eta _k \frac{1-\eta _k}{(n-2)(n-1)}+1-\delta -f(l)+\frac{2(\eta _{\pi (j)}-1)}{n-2} \\&\ge \frac{(n-{\underline{M}})(1-{\overline{M}})}{(n-1)(n-2)} +1-\delta -f(l)+\frac{2({\underline{M}}-1)}{n-2}\\&\ge X+\frac{2({\underline{M}}-1)}{n-2}-\delta -f(m-1). \end{aligned}$$

  Since $\delta \le X+\frac{2({\underline{M}}-1)}{n-2}-f(m-1)$, then

  $$\begin{aligned} {\Pi }_i(g)- {\Pi }_i(g-ij)\ge 0. \end{aligned}$$

  The same conclusion can be made for player j. Therefore, the complete network g is stable.

$\square$

Appendix 10: Proof of Corollary 1

Proof

• 1. Armed with Proposition 1, we only need to consider the case when the utility function is defined by (4). Suppose a nonempty network g is stable but not minimal, then there exists at least one link $ij\in g$ such that $i{\mathop {\longleftrightarrow }\limits ^{g-ij}}j$. For a particular partition ${\Delta }\in P^*$, we need to consider two possible cases:

  • (a) Let $\pi (i)=\pi (j)=l^*$, then for player i we obtain

    $$\begin{aligned} {\Pi }_i(g-ij)-{\Pi }_i(g)=f(0)>0, \end{aligned}$$

    which implies that g is not stable since player i strictly gains by severing link $ij\in g$, which leads to a contradiction.

  • (b) Let $|\pi (i)- \pi (j)|=l>0$, i.e., players i and j be in different groups. For player i, we have

    $$\begin{aligned} {\Pi }_i(g-ij)-{\Pi }_i(g)&= f(l)+\sum \limits _{\begin{array}{c} k\ne \pi (i)\\ k\ne \pi (j) \end{array}}n_i^k(g)\Big (\frac{n_i^k(g)-1}{n_i(g)-1}-\frac{n_i^k(g)-1}{n_i(g)}\Big )\\&\quad +\Big ( n_i^{\pi (j)}(g)-1\Big )\cdot \frac{n_i^{\pi (j)}(g)-2n_i(g)}{(n_i(g)-1)n_i(g)}, \end{aligned}$$

    and we discuss all possible situations:

    • i. $\pi (i)=l^*$ and $\pi (j)\ne l^*$, then

      $$\begin{aligned} {\Pi }_i(g-ij)-{\Pi }_i(g)=f(l)>0. \end{aligned}$$

    • ii. $\pi (j)=l^*$ and $\pi (i)\ne l^*$, then

      $$\begin{aligned} {\Pi }_j(g-ij)-{\Pi }_j(g)=f(l)>0. \end{aligned}$$

    • iii. $\pi (i)\ne l^*$ and $\pi (j)\ne l^*$, then

      $$\begin{aligned} {\Pi }_i(g-ij)-{\Pi }_i(g)&=f(l)+\sum \limits _{\begin{array}{c} k\ne \pi (i)\\ k\ne \pi (j) \end{array}}n_i^k(g)\Big (\frac{n_i^k(g)-1}{n_i(g)-1}-\frac{n_i^k(g)-1}{n_i(g)}\Big )\\&\ge f(l)>0. \end{aligned}$$

    The cases i.–iii. imply that g is not stable since player i (or j) strictly gains by deleting a link $ij\in g$, which is also a contradiction.

• 2. Since for any partition ${\Delta }\in P^*$, ${\underline{M}}=1$, then the result can be directly obtained from Proposition 3.

• 3. The empty network is not stable which immediately follows from inequality $f(0)\le f(1)<1$. Let g be a stable network which must be minimal by Item 1. Suppose g is not connected, then there exists a pair of players i and j such that $d_{ij}(g)=\infty$.

  If $\pi (i)=\pi (j)=l^*$, then

  $$\begin{aligned} {\Pi }_i(g+ij)- {\Pi }_i(g)\ge 1-f(0)>0, \end{aligned}$$

  and

  $$\begin{aligned} {\Pi }_j(g+ij)- {\Pi }_j(g)\ge 1-f(0)>0. \end{aligned}$$

  If $|\pi (i)- \pi (j)|=1$, i.e., players i and j are in the nearby groups, and we take into account that $d_{ij}=\infty$. Then there are two cases to discuss:

  • (a) $\pi (i)=l^*$ and $\pi (j)\ne l^*$ (similar to $\pi (i)\ne l^*$ and $\pi (j)= l^*$), and for player i it is true that

    $$\begin{aligned} {\Pi }_i(g+ij)- {\Pi }_i(g)\ge \eta _{\pi (j)} -f(1) =1-f(1)>0, \end{aligned}$$

    and for player j,

    $$\begin{aligned} {\Pi }_j(g+ij)- {\Pi }_j(g)\ge \eta _{\pi (i)} -f(1)=h-f(1)>0. \end{aligned}$$

  • (b) $\pi (i)\ne l^*$ and $\pi (j)\ne l^*$, then

    $$\begin{aligned}&{\Pi }_i(g+ij)- {\Pi }_i(g)\\&\quad \ge \eta _{\pi (j)} -f(1)+\sum \limits _{\begin{array}{c} k\ne \pi (i)\\ k\ne \pi (j) \end{array}}n_i^k(g)\Big (\frac{n_i^k(g)-1}{n_i(g+ij)}- \frac{n_i^k(g)-1}{n_i(g+ij)-1} \Big ) \\&\quad = 1 -f(1)-\frac{n_i^{l^*}(g)\big (n_i^{l^*}(g)-1 \big )}{n_i(g+ij)\big (n_i(g+ij)-1\big )}\ge 1 -f(1)-\frac{n_i^{l^*}(g)\big (n_i^{l^*}(g)-1 \big )}{n_i^{l^*}(g)\big (n_i^{l^*}(g)+1\big )}\\&\quad = \frac{2}{n_i^{l^*}(g)+1}-f(1)\ge \frac{2}{h+1}-f(1)>0, \end{aligned}$$

    and the same result can be also obtained for player j.

The reasonings given above contradict the assumption that g is stable, and this finishes the proof. $\square$