Dynamic Network Formation with Ordered Partitioning and Incomplete Information

Abstract

We introduce a dynamic network formation model with incomplete information and asymmetric players who are partitioned into ordered groups. At each stage, a pair of players is randomly selected to update the link connecting them. The feature of the model is that the cost of a link depends on the distance between the labels of the groups players belong to. As a result, information about the other player’s label is significant in the decision to form a link. Assuming incomplete information, players initially do not know each other’s labels, but learn them upon being connected. Comparing the topologies of stable networks for cases of complete and incomplete information indicates that the presence of the latter can have deep implications. Furthermore, our simulation experiments are focused on three issues: the stabilization time, the heterogeneity index, and the impact of the group size gap.

Appendices

Appendix A. Proof of Theorem 1

Proof

 

• 1. When $E_i>1$ holds for any $i\in N$, as shown in Proposition 1, no pair would form a link in the entire process. Therefore, the empty network is the unique one that emerges and achieves stability.

• 2. We first prove for $G_{C}\subseteq G_{IC}$.

  Let $E_i\le 1$ be true for any $i\in N$ and $g\in G_{C}$. If $g=\varnothing$, there must exist a pair of players $i, j\in N$ such that $f(|\pi (i)-\pi (j)|)>1$. Consider selection path $\gamma _2=((i, j), (i, j))$ under incomplete information. It is evident that link ij will be formed at stage 1 and consequently severed at stage 2, which implies that $g\in G_{IC}$. If $g\ne \varnothing$, consider selection path $\gamma _{t_1}$ such that $g=g_C(\gamma _{t_1})$ and network g emerges for the first time at stage $t_1$. Let construct another selection path ${\bar{\gamma }}_{t_2}$ based on $\gamma _{t_1}$ by removing any pair (i, j) at certain stages if players i and j do not form a link ij when it does not exist or do not sever it when it exists in the network at these stages. It is obvious that $g=g_C({\bar{\gamma }}_{t_2})$ and $t_2\le t_1$. Now we aim to demonstrate that with ${\bar{\gamma }}_{t_2}$ the formation process gives identical link formation and severance results under complete and incomplete information cases.

  Let pair $(i_1, j_1)$ be selected at stage 1 of path ${\bar{\gamma }}_{t_2}$. By construction of ${\bar{\gamma }}_{t_2}$, it is clear that link $i_1j_1$ is built under complete information. Since $E_{i_1}, E_{j_1}\le 1$, $i_1j_1$ is also established under incomplete information, indicating that $g_C({\bar{\gamma }}_{1})=g_{IC}({\bar{\gamma }}_{1})=\{i_1j_1\}$. Suppose $g_C({\bar{\gamma }}_{k})=g_{IC}({\bar{\gamma }}_{k})$ holds true for $k<t_2$. We demonstrate that $g_C({\bar{\gamma }}_{k+1})=g_{IC}({\bar{\gamma }}_{k+1})$. Assume that players $i_{k+1}$ and $j_{k+1}$ are selected at stage $k+1$, and three possible cases are discussed, respectively:

  • (a) $i_{k+1}j_{k+1}\in g_C({\bar{\gamma }}_{k})=g_{IC}({\bar{\gamma }}_{k})$. Then players $i_{k+1}$ and $j_{k+1}$ know each other’s exact group when they choose their actions at stage $k+1$. As a result, each of them will make the same action under complete and incomplete information cases.

  • (b) $i_{k+1}j_{k+1}\notin g_C({\bar{\gamma }}_{k})=g_{IC}({\bar{\gamma }}_{k})$ and players $i_{k+1}$ and $j_{k+1}$ have never been connected before stage $k+1$, i.e., their mutual beliefs remain at the prior. Taking into account selection path ${\bar{\gamma }}_{t_2}$ and assuming that $E_{i_{k+1}}, E_{j_{k+1}}\le 1$, players $i_{k+1}$ and $j_{k+1}$ will form a link under both complete and incomplete information cases.

  • (c) $i_{k+1}j_{k+1}\notin g_C({\bar{\gamma }}_{k})=g_{IC}({\bar{\gamma }}_{k})$ and players $i_{k+1}$ and $j_{k+1}$ have already been connected before stage $k+1$. Then, both players choose their actions knowing the exact payoff. Since such a link will be formed under complete information, it will also be established by these two players under incomplete information case.

  Above all, we conclude that $g=g_C({\bar{\gamma }}_{t_2})=g_{IC}({\bar{\gamma }}_{t_2})$, implying $g\in G_{IC}$.

  To prove for $S_{C}^*\subseteq S_{IC}^*$, we first clarify that any network emerging under complete information can also appear under incomplete information by the analysis given above. Thus, it is sufficient to show that for any $g\in S_C^*$ there exists a selection path under incomplete information after the first appearance of g that makes it stable. We demonstrate it by construction.

  If $g=\varnothing$, consider selection path $\gamma _{\frac{n(n-1)}{2}}$ such that each pair of players is selected twice consecutively. Since $g\in S_C^*$, we know that under incomplete information each pair first forms a link and then severs it at the next stage. Finally, at stage $\frac{n(n-1)}{2}+1$, information is complete and empty network g becomes stable, i.e., $g\in S_{IC}^*$.

  If $g\ne \varnothing$, denote the number of components (including isolated players) in network g by q(g). Under incomplete information let the subsequent selection path after the first appearance of g be such that at first $q(g)(q(g)-1)$ stages, two players from various components are selected twice consecutively, and every two components are involved in the process. As $g\in S_C^*$, even if the two players may form a link at the former stage misled by the prior beliefs, the link will be severed when they are selected at the latter stage. Alternatively, if two players have been connected before, the link would not be even formed at two stages when they are consecutively selected. In either case, players know each other’s group as well as the groups of players in the counter party’s component. Respectively, after $q(g)(q(g)-1)$ stages, information is completely revealed. Again, when $g\in S_C^*$, no link will be modified given any subsequent selection path. Thus, $g\in S_{IC}^*$, which completes the proof. $\square$

Appendix B. Proof of Lemma 2

Suppose $g\in S^*_C$, and it directly follows from the utility function form (no decay of benefit) that g must be minimal. Now we prove the connectivity of network g. Since $b+\frac{a}{4}(m-\frac{1}{m})\le 1$, we obtain that $f(0)=b<1$. Thus for any two players i, j such that $\pi (i)=\pi (j)$, i and j must be connected in g. We assume that there exist two players i and k where $|\pi (i)-\pi (k)|=1$, i and k are not connected in g, then three possible cases should be discussed:

• 1. If $m=3$, then $b+\frac{a}{4}(m-\frac{1}{m})\le 1$ is equivalent to $b+\frac{2a}{3}\le 1$. Moreover, $f(0)=b>0$, so $a<\frac{3}{2}$. Therefore, we obtain

  $$\begin{aligned} f(1)=a+b\le 1+\frac{a}{3}<\frac{3}{2}<M. \end{aligned}$$

  As a result, for player i,

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

  and the same inequality can be obtained for player k.

• 2. If $m=4$, then $b+\frac{a}{4}(m-\frac{1}{m})\le 1$ is equivalent to $b+\frac{15a}{16}\le 1$. Moreover, $f(0)=b>0$, thus, $a<\frac{16}{15}$. So we obtain

  $$\begin{aligned} f(1)=a+b\le 1+\frac{a}{16}<\frac{16}{15}<M. \end{aligned}$$

  As a result, for player i,

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

  and the same inequality can be proved for player k.

• 3. If $m\ge 5$, then with $b+\frac{a}{4}(m-\frac{1}{m})\le 1$ and $a>0, b>0$, we prove that $f(1)=a+b\le 1<M$. As a result, for player i,

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

  and the same inequality can be found for player k.

In any case from the list above, if pair (i, k) is selected, link ik must be established, which contradicts that $g\in S_C^*$. Therefore, any two players from the nearby groups must be connected in g. Hence, g is minimally connected.

Appendix C. Proof of Theorem 3

We consider two cases:

• 1. $\varnothing \in S_C^*$. Then, $f(0)=b>1$; thus, for any player $i\in N$, $E_i>1$. As a result, we obtain that $S_C^*=S_{IC}^*=\{\varnothing \}$.

• 2. $\varnothing \notin S_C^*$. This implies that $f(0)=b\le 1$. From condition (6), we obtain that $b+\frac{a}{4}(m-\frac{1}{m})\le 1$, then from Lemma 2 it follows that g is minimally connected. Moreover, $\max \limits _{i\in N} E_i=b+\frac{a}{4}(m-\frac{1}{m})+\frac{a}{4}(m+\frac{1}{m}-2)<2$. Consider a network $g\in S^{*}_{C}$, such that g emerges and becomes stable after selection path $\gamma$ under complete information. Now we are going to prove that there exists a selection path ${\bar{\gamma }}$, after which network g appears and becomes stable under incomplete information.

  Let $i_f$ be the first neighbor of player $i\in N$ in the selection path $\gamma$ under complete information. We first prove that for any player $i\in N_{\frac{m+1}{2}}$, $E_{i_f}\le 1$. We prove this by contradiction. Suppose there exists player $i\in N_{\frac{m+1}{2}}$ such that $E_{i_f}>1$. Let $c_1=|\pi (i_f)-\frac{m+1}{2}|$. Calculating the expected unit cost $E_{i_f}=b+\frac{a}{4}(m-\frac{1}{m})+\frac{c_{1}^2a}{m}>1$ and combining with condition (6), we will get that $f(c_1)>1$, contradicting that $i_f$ is the first neighbor of player i under complete information. Thus, we prove that $E_{i_f}\le 1$ for any player $i\in N_{\frac{m+1}{2}}$.

  Any link $ij\in g$ corresponds to one of three cases: (i) $E_i, E_j\le 1$, (ii) $E_i>1, E_j\le 1$, or (iii) $E_i>1, E_j>1$. We will construct the selection path ${\bar{\gamma }}$ for the emergence of links corresponding to these three cases following the steps:

• Step 1: We randomly choose a player $d\in N_{\frac{m+1}{2}}$, and let pair $(d, d_f)$ be selected in path ${\bar{\gamma }}$. From the proof given above, we know that $E_{d_f}\le 1$, thus link $dd_f\in g$ will be formed. Let $F=\{i\mid E_i>1, E_{i_f}>1\}$ be the set of players, whose expected costs of forming a link are larger than 1 and for whose first neighbor in path $\gamma$ under complete information, the expected costs of forming a link are also larger than 1. Let $B_k=\{i\in F\mid i_f=k\}$, $k\in N$, be the set of such players from set F, for whom the first neighbor in the formed network under complete information is player k. For any $B_k\ne \varnothing$, let pairs $(d, i^{k}_1), \ldots , (d, i^{k}_{|B_k|}), (d, k)$ be selected in a successive order in path ${\bar{\gamma }}$, where $\{i^{k}_1, \ldots , i^{k}_{|B_k|}\}=B_k$. Because $E_d\le 1$ and $\max \limits _{i\in N} E_i< 2$, all links dh, where $h\in B_k\cup \{k\}$, are formed. Since for each player $h\in B_k\cup \{k\}$, $E_h>1$, i.e., $\frac{a}{4}(m-\frac{1}{m})+b+\frac{(\pi (h)-\pi (d))^2a}{m}>1$, then from (6) we have $f(|\pi (h)-\pi (d)|)>1$. Thus when pairs $(d, i^{k}_1), \ldots , (d, i^{k}_{|B_k|}), (d, k)$ are selected again, all links dh, $h\in B_k\cup \{k\}$ are deleted by player d. Next for each player $i\in F$, pair $(i, i_f)$ is chosen, i.e., $(i_1, (i_1)_{f}), \ldots , (i_{|F|}, (i_{|F|})_f)$ are consequently selected in path ${\bar{\gamma }}$. Each link $ii_f$, where $i\in F$, will be formed since players i and $i_f$ know the labels of each other when they have been connected via player d and $f(|\pi (i)-\pi (i_f)|)\le 1$.

  Remark on Step 1: With respect to the selected path of pairs constructed in Step 1, there are two exceptions: (i) If there exists set $B_k\ne \varnothing$, such that $dk\in g$ or $di\in g$ for some $i\in B_k$, then pair (d, k) (pair (d, i)) is not selected in the second round to save a link dk (link di) at the end of Step 1. (ii) If there exists set $B_k\ne \varnothing$, such that $kd_f\in g$ or $id_f\in g$ for some $i\in B_k$, then pair (d, k) (pair (d, i)) is replaced by pair $(d_f, k)$ (pair $(d_f, i)$) in our selection path. It is also selected once in the first round to keep link $kd_f$ (link $id_f$) at the end of Step 1, but not link dk (link di).

  Moreover, there is a special rule for Step 1: if there exists set $B_k$ satisfying one of the above exceptions (given in Remark on Step 1), then the pairs, in which one of the players belongs to $B_k$, i.e., pairs such that $(d, i^{k}_1), \ldots , (d, i^{k}_{|B_k|}), (d, k)$ should be selected after the pairs described in Step 1, which do not satisfy these exceptions.

  We move on to Step 2 to proceed with construction of path ${\bar{\gamma }}$.

• Step 2: Let $A=\{i, E_i\le 1\mid \text {for any } j\in N_i(g), E_j>1\}$, that is the set of players, whose expected costs are not larger than 1, but all neighbors of these players have expected costs larger than 1. Denote by ${\bar{c}}=\min \limits _{\begin{array}{c} f(c)>1\\ c=1, \ldots , (m-1)/2 \end{array}}c$. Obviously, for any player $i\in A$, $|\pi (i)-\pi (i_f)|<{\bar{c}}$ while $|\pi (i_f)-\frac{m+1}{2}|\ge {\bar{c}}$. As a result, for $i\in A$, either $\frac{m+1}{2}<\pi (i)<\pi (i_f)$, or $\pi (i_f)<\pi (i)<\frac{m+1}{2}$ is satisfied. For any player $k\in N$, denote $A_k=\{i\in A, i_f=k\}$. For a certain set $A_k\ne \varnothing$, without loss of generality, we assume that $\frac{m+1}{2}<\pi (i)<\pi (k)$ for any $i\in A_k$, then some player $a_k\ne d_f$ satisfying $E_{a_k}\le 1$, and such that $\pi (a_k)<\frac{m+1}{2}$ and $|\pi (a_k)-\pi (i)|\ge {\bar{c}}$, can be found. Till now (and after Step 1), the players $a_k$ and $i\in A_k$ have never been chosen. Therefore, when pairs $(a_k, i^{k}_1), \ldots , (a_k, i^{k}_{|A_k|}), (a_k, k)$, where $\{i^{k}_1, \ldots , i^{k}_{|A_k|}\}= A_k$, are selected consequently with two rounds, then all links of type $ia_k$, where $i\in A_k$, as well as of type $ka_k$ will be definitely added in the first round of sequential selection and then be deleted by player $a_k$ in the second round of selection.

  During such a process, players $i\in A_k$ and k recognize the labels of each other. Then, if we select pairs $(i^{k}_1, k), \ldots , (i^{k}_{|A_k|}, k)$ consequently, all these links will be formed. The case when $\pi (i_f)<\pi (i)<\frac{m+1}{2}$ is satisfied for any $i\in A_k$ can be similarly discussed and addressed. For each set $A_k\ne \varnothing$, $k\in N$, the pairs are selected in the same way as we have designed above.

  Remark on Step 2: For the selected pairs in Step 2, there is an exception: if there exists set $A_k\ne \varnothing$ such that $ka_k\in g$, then pair $(a_k, k)$ is not selected in the second round to save a link $ka_k$ in the network after Step 2.

  Moreover, there is a special rule for Step 2: if there exists set $A_k$ satisfying the exception above, then the pairs with the players from $A_k$, that are $(a_k, i^{k}_1), \ldots , (a_k, i^{k}_{|A_k|}), (a_k, k)$, are selected after the pairs from the sets which do not satisfy such an exception.

• Step 3: Denote the network structure formed after Step 2 by ${\bar{g}}$ and let $D=\{ij\in g{\setminus } {\bar{g}}\mid E_i\le 1~ \text {and}~ E_j\le 1\}$. The set $D\ne \varnothing$ since for any player $i\in N_{\frac{m+1}{2}}$ we have $E_{i_f}\le 1$. All pairs (i, j), $ij\in D$ are chosen one by one, then all links belonging to set D can be established since no pair of players has been ever connected and expected costs of both players are not larger than 1.

• Step 4: Denote the network structure formed after Step 3 by $g^{'}$. If $g\setminus g^{'}=\varnothing$, then we finish constructing a selection path ${\bar{\gamma }}$. If $g{\setminus } g^{'}\ne \varnothing$, then we proceed by considering any links $ij\in g{\setminus } g^{'}$ such that $E_i>1$ and $E_j\le 1$. It is easy to check that players i and j have never been connected via other players through Steps 1–3. This indicates that they do not recognize the labels of each other.

  We describe the procedure of further construction of the selection path depending on the neighbors of players i and j in network $g^{'}$. Our steps are described as follows:

  • (a) $n_i(g^{'})>0$ and $n_j(g^{'})>0$. Choose pair (i, j), then link ij is formed.

  • (b) $n_i(g^{'})>0$ and $n_j(g^{'})=0$. Since and link $pq\in g$ such that $E_p\le 1$ and $E_q\le 1$, has been formed at the previous steps, then for any $k\in N_j(g)$, we may only have $E_k>1$, i.e., $j\in A$. As $n_j(g^{'})\ge 1$ for any $j\in A$, hence the situation when $n_j(g^{'})=0$ never happens.

  • (c) $n_i(g^{'})=0$ and $n_j(g^{'})>0$. Choose pair (i, j), and then link ij will be formed.

• Step 5: Denote the network structure formed after Step 4 by $g^{''}$, i.e., $g^{''}=g^{'}\cup \{ij\mid ij\in g, E_i>1, E_j\le 1\}$. In this step we consider the links $ij\in g\setminus g^{'}$, for which $E_i>1$ and $E_j >1$. We can easily show that through Steps 1-4, players i and j have never been connected via other players, thus they do not know the labels of each other.

  Our next steps are described as follows (depending on the neighbors of players i and j in network $g^{''}$):

  • (a) $n_i(g^{''})>0$ and $n_j(g^{''})>0$. Select a pair (i, j), and link ij will be established.

  • (b) $n_i(g^{''})\cdot n_j(g^{''})=0$. We prove that this situation cannot happen. Suppose $n_j(g^{''})=0$, since for any $kl\in g$, such that $E_k>1$ and $E_l\le 1$, we have $kl\in g^{''}$. Thus we may obtain that $E_{j_f}>1$ implying that $j\in F$. Recall that each link $kk_f$, such that $k\in F$, has been already established, contradicting that $n_j(g^{''})=0$.

Finally, the construction of path ${\bar{\gamma }}$ is finished. All links in network g have been built, i.e., g appears after ${\bar{\gamma }}$ under incomplete information. Since g is minimally connected and stable under complete information, thus, g is also stable under incomplete information. We conclude that any $g\in S_{C}^*$ belongs to the set $S_{IC}^*$.

Appendix D. Proof of Proposition 6

• 1. Since $f(0)\le 1$, then $E_i=\frac{M-1}{2M-1}f(0)+\frac{M}{2M-1}f(1)\le 1$ for each player $i\in N$ because $f(1)\le 2-\frac{1}{M}-(1-\frac{1}{M})f(0)$. Thus, for any two players i, j such that $\pi (i)=\pi (j)$, no matter whether they know the labels of each other when they are selected while not connected, link ij must be formed. Thus, in any network $g\in S_{IC}^*$, players in the same group are connected. Next, consider a pair i, k such that $\pi (i)\ne \pi (k)$. Suppose they are not connected in $g\in S_{IC}^*$. When they are selected after the appearance of g, link ik is formed since $E_i=E_k\le 1$ if they do not know the labels of each other, or $\Pi _i(g+ik)-\Pi _i(g)=\Pi _k(g+ik)-\Pi _k(g)=M-f(1)>0$ if they know each other.

• 2. We prove the second item by contradiction. Assume that there exists network $g\in S_{SIC}^{*}$ such that g is not connected and g appears after selection path $\gamma$. Two possible cases should be discussed:

  • (a) There are two players i and j who are not connected in g, and $\pi (i)=\pi (j)$. Without loss of generality, we assume that $\pi (i)=\pi (j)=1$. There are several possible situations for players i and j:

    • i. $n_i(g)=n_j(g)=0$. In this case, players i and j have no idea about the labels of each other, but the belief of at least one of them (say i) has been updated to be higher than the initial one. The belief of player i can be increased only if another player from the same group has been connected with her before. Then, player i is either indirectly connected with such a player or his neighbor in network g due to $f(0)\le 1$, contradicting $n_i(g)=0$.

    • ii. $n_i(g)\ge 1$, $n_j(g)=0$. In this case, players i and j again have no idea about the labels of each other. Moreover, in the component $C_i(g)$, there is at least one player $k\in N_2$. Otherwise, there exists player $l\in N_2$ such that $n_l(g)\ge 1$, and link il will be built if they are selected. It is a contradiction with the fact that g is stable.

      There exists exactly one player, say player q, such that $n_q(g)=0$. If such a player does not exist, then either players i and j know each other (there are only two parts in the network, $C_i(g)$ and isolated j), or link iq is formed when i and q are selected ($n_q(g)\ge 1$). Otherwise, if $\pi (q)=1$, then player i must know the labels of j and q and make the connection when they are selected. If $\pi (q)=2$, then link ij must be created when i and j are selected since $E_i^{\gamma }=\frac{1}{2}\cdot f(0)+\frac{1}{2}\cdot f(1)=1-\frac{1-f(0)}{2\,M}<1$.

  • (b) There are two players i and j which are not connected in g, where $\pi (i)=1, \pi (j)=2$. Following (a) above, player i (j) is connected with each player in the same group with her. Since $f(1)< M$, when i and j are selected, the link ij is formed, contradicting that g is stable.