The Insider-Outsider Model Reexamined
Abstract
:1. Introduction
2. Model Setup
3. Results
- Suppose and , that is there are no links between the two groups. We will show in the following that such a situation cannot happen.First, since g is non-empty, we cannot have simultaneously and . Suppose that , for one . In that case, we know from the point above that is connected.Second, we show in two steps that constitutes a PSS. We begin to show that if , then . To introduce a contradiction, suppose that . In that case there exist two distinct players such that . We have: (i) since , (ii) there is a chain between and which goes through and , (iii) players , , , and belong to . When (i) and (ii) are valid, it is not possible for (iii) to be valid simultaneously by Lemma 1. A contradiction. Thus we can conclude that constitutes a star. In this star, each end player sponsors her link with the center, since . So constitutes a PSS.Third, we show that . Indeed, since , a player strictly gains by forming a link with the center of the PSS formed by the group . It follows that , which contradicts the assumption .
- Suppose or in g. We assume wlog that there are players and such that . Since a player in has formed a link, is connected. We now show that g is connected. To introduce a contradiction suppose g is not connected. In that case there exists a player such that . We have: since and player obtains more resources due to the link with than through her link with . It follows that does not play a strict best response, a contradiction. Thus we conclude that g is connected.
- 1.
- If g contains one inter-groups link: with and , then and , for .
- 2.
- If g contains several inter-groups links between the two groups, then .
- Let , with only one inter-groups link. We assume that this link is formed by with , that is .First, we show that . To introduce a contradiction, we assume that there are such that . There are two cases.(a) Let , with . There are players such that . In that case: (i) we have since , (ii) there exists a chain between players and which contains players and , and (iii) . When (i) and (ii) are valid, it is not possible for (iii) to be valid simultaneously by Lemma 1. A contradiction.(b) Suppose now that if , then either , or . In other words, the player who has formed a link with a player in , is always one of the two players at maximal distance in . In the following we deal with . Note first that since is an end player and , has formed a link with a player, say , who belongs to the chain between and in g. There are two cases.(b1) Suppose player has formed a link in g with a player, say , who belongs to the chain between and . In such a case we have (i) , (ii) there is a chain between players and which contains players and , and (iii) . When (i) and (ii) are valid, it is not possible for (iii) to be valid simultaneously by Lemma 1. A contradiction.(b2) Suppose player has formed no links in g with a player who belongs to the chain between and . In that case, there is a player, say , who has formed the link with and belongs to the chain between and .We show that the link is the only link player has formed in g. Indeed, suppose has formed a link with a player say . Suppose does not belong to the chain between and . We have and by point (a) we know that g is not strict Nash. Suppose now that belongs to the chain between and . We have (i) , (ii) there is a chain between and which contains players and , and (iii) . When (i) and (ii) are valid, it is not possible for (iii) to be valid simultaneously by Lemma 1.Since the link is the only link that player has formed in g, it follows that there is a player who belongs to the chain between and such that . However in this situation player does not play a strict best response, by (R1). Consequently . A contradiction.Second we show that . Using arguments similar to those for it is easy to see that . We now show that . To introduce a contradiction suppose that . We divide the proof into two cases.(a) Suppose there are such that . So there are players such that . In that case: (i) we have since , (ii) there is a chain between and which contains players and , and (iii) we have . When (i) and (ii) are valid, it is not possible for (iii) to be valid simultaneously by Lemma 1.(b) Suppose now that if , then either , or . In the following we deal with . Since is an end player, she has formed a link with a player who belongs to the chain between and . In such a case we have (i) , (ii) there exists a chain between players and which contains players and , and (iii) , and . When (i) and (ii) are valid, it is not possible for (iii) to be valid simultaneously by Lemma 1.The conclusion follows.
- We now consider a minimally connected strict Nash network g in which there are more than one inter-groups link. The arguments which allow to obtain the contradiction in this part are very similar to the arguments used in the previous part. Therefore we briefly show the way to use Lemma 1 in the proof. We know from Lemma 2 that there exists one player who belongs to one group, say , such that is involved in all the inter-groups links.We first deal with players in and show that for all . Let be the set of players such that . By assumption we have . Moreover we know that we have for all .To introduce a contradiction, we suppose that . Since g is connected there are players and a player such that . Let be an end player such that belongs to the chain between and . By construction there is a player such that . Note that we can have and . Moreover, since , we have .Either is an end player, and since , or there exists an end player such that . In the second case by construction and since , we have for one player .If is an end player, then Lemma 1 is contradicted for players and g cannot be a strict Nash network. If is not an end player, then Lemma 1 is contradicted for players and g cannot be a strict Nash network. It follows that . It is worth noting that for all since .We now deal with players in and show that for all . To introduce a contradiction, suppose there is an end player , such that . Since g is connected and there is a player such that . In that case Lemma 1 is contradicted for players , since . Therefore g is not a strict Nash network. The result follows.To conclude, since for all we have .
- 1.
- If there is such that , then g is a D3PSS.
- 2.
- If for each we have , then g is a CIPSS, a NCIPSS or a IMS .
4. Conclusion
References
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Appendix
- The two inter-groups links are sponsored by two players who belong to the same group. Suppose wlog that and . Since g is minimally connected, there are four cases which we analyze in turn.C1. There is a chain between and which contains players and . We know that we have also (i) , . (ii) , . When C1 and (i) are valid, it is not possible for (ii) to be valid simultaneously by Lemma 1.C2. There is a chain between and which does not contain and . In that case, there exists a player such that , otherwise player does not play a strict best response since . Either k is an end player, and since , or there exists an end player such that . Suppose k is an end player. We have (i) , (ii) there is a chain between and k which contains players , (iii) , . When (i) and (ii) are valid, it is not possible for (iii) to be valid simultaneously by Lemma 1. Next suppose the end player is . Then there is a player such that (since ). So we have (i) , (ii) there is a chain between and which contains players , (iii) , . When (i) and (ii) are valid, it is not possible for (iii) to be valid simultaneously by Lemma 1.C3. There is a chain between and which does not contain players and . This case is symmetric to C2.C4. There is a chain between and which does not contain players and . In that case, there exist players such that , otherwise players and do not play a strict best response since . This gives rise to two situations: either both and are end players, or at least one is not an end player. If both and are end players, we have (i) , (ii) there is a chain between and which contains players and , (iii) . When (i) and (ii) are valid, it is not possible for (iii) to be valid simultaneously by Lemma 1.Suppose now wlog that is not an end player. Then there exists an end player . By construction and since , we have also for one player such that . Therefore we have (i) , (ii) there is a chain between and which contains players and , (iii) and . When (i) and (ii) are valid, it is not possible for (iii) to be valid simultaneously by Lemma 1.
- The two inter-groups links are sponsored by players who belong to two distinct groups. Suppose wlog that , . Since g is minimally connected, as before there are four cases which we analyze in turn.C1’. There is a chain between and which contains players and . In that case, there exists a player such that , otherwise player does not play a strict best response since . Either k is an end player, and (since ), or there exists an end player such that . In the second case by construction and since , we have also for one player such that . In the first case, by using Lemma 1 for players , it is easy to check that g cannot be a strict Nash network. In the second case we obtain the same result for players .C2’. There is a chain between and which does not contain and . We have (i) , (ii) there is a chain between and which contains players and , (iii) , and . When (i) and (ii) are valid, it is not possible for (iii) to be valid simultaneously by Lemma 1.C3’. There is a chain between and which does not contain and . In that case, there exist players such that , otherwise players do not play a strict best response since . This gives rise to two situations: (I) both and are end players, and (II) at least one is not an end player.(I) Both and are end players. We have , since . We show that , . Indeed we have (i) , (ii) there is a chain between and which contains players and , (iii) , . By Lemma 1, (i), (ii) and (iii) can be simultaneously satisfied only if and .We now focus on players and to obtain a contradiction when , .
- Suppose . Let . We have (i) , (ii) there is a chain between and which contains players and , (iii) , and . When (i) and (ii) are valid, it is not possible for (iii) to be valid simultaneously by Lemma 1. Let . By the same reasoning we obtain a contradiction.
- Suppose . Let be the sequence of players in the chain between and in g, with and .(a) We first show that , if , then for . By construction we have . Suppose . Then we have: (i) , (ii) there is a chain between and which contains players and , (iii) . By Lemma 1, (i), (ii) and (iii) can not be simultaneously valid if . The result follows.(b) We now show that if , then for . By (a) we have . It follows that we have (i) , (ii) there is a chain between and in g which contains players and , (iii) , . By Lemma 1, (i), (ii) and (iii) can be simultaneously valid only if .It follows that since . A contradiction since we know that .(II) At least one of players or is not an end player. Suppose w.l.o.g. that is not an end player. Then there exists an end player , such that . By construction and since , we have also for one player such that . Therefore we have (i) , (ii) there is a chain between and which contains players and , (iii) , . When (i) and (ii) are valid, it is not possible for (iii) to be valid simultaneously by Lemma 1.C4’. There is a chain between and which does not contain players and . In that case, there exists a player such that , otherwise player does not play a strict best response since . Either k is an end player, and since , or there exists an end player such that . Let k be an end player. We have (i) , (ii) there is a chain between and k which contains players and , (iii) , and . When (i) and (ii) are valid, it is not possible for (iii) to be valid simultaneously by Lemma 1. Next let the end player be . Then there is a player such that (since ). We have (i) , (ii) there is a chain between and which contains players and , (iii) , and . When (i) and (ii) are valid, it is not possible for (iii) to be valid simultaneously by Lemma 1.
- We assume that . First, we characterize the architecture of . Let and be such that . Let . We know by the proof of Proposition 2 1.(a) that if , then either , or . Likewise, we know that and is involved only in two links otherwise there is , with and . Consequently, and we call the player in . By R1, does not form a link with . It follows that player forms a link with . Finally, we have and since for all players who belong to this set we have .Second, we characterize the architecture of . We show that all players has formed a link with . Indeed, we know by Proposition 2 that , that is this subnetwork is a star. We now show that is the center of this star. To introduce a contradiction assume that a player is the center of this star. We have and player has an incentive to replace her link with by a link with in g. A contradiction. As each player is an end player, each player has formed a link with since . The result follows.
- Assume that for each , we have . By using the same arguments as in the previous point, we can show that the subnetwork formed by players of group is a star with as the center and each player forming her link with . Concerning the architecture of the star formed by players of group , we can have two cases: (i) Either is the center of the star, or (ii) a player is the center of the star. In case of (i), all players are end players and have formed a link with . So g is a CIPSS. In case of (ii), all players are end players and have formed a link with . Regarding the link between k and , we can have either , or . If , then g is a NCIPSS, and if , then g is a IMS.
- 1.It is worth noting that decay is just one way of modelling deficiencies in information transmission. Another way is to introduce imperfect reliability of links (see Bala and Goyal, 2000b, for a paper dealing with this).
- 2.Note that Bala and Goyal (2000a) also introduce decay in their basic model. However the authors still assume that players are homogeneous.
- 3.In what follows intermediate costs of intra-groups and inter-groups links has the following interpretation: costs of intra-groups links are such that no player i can have any incentive to form a link with a player j when this link allows i to access only to resources of j; costs of inter-groups links are not so high that this always prevents players to form inter-groups links.
- 4.It is also important from of the point of view of centrality of players in the network. Indeed, while in the first type of architectures, only the players at the center of the stars play a central role in the flow of resources between players, in the second type a third player plays a significant role, since this player allows ressources to flow between the two groups. It follows that in both cases centrality values of the players differ. Note that we refer here to the notion of betweeness centrality, presented by Faust and Wasserman (1994 chapter 5).
- 5.The assumption allows to simplify the proof of Proposition 3.
- 6.Note that for all costs of linking, there always exists a high enough such that if , then no player can have an incentive to form a link with a player she already indirectly accesses in the network.
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Billand, P.; Bravard, C.; Sarangi, S. The Insider-Outsider Model Reexamined. Games 2010, 1, 422-437. https://doi.org/10.3390/g1040422
Billand P, Bravard C, Sarangi S. The Insider-Outsider Model Reexamined. Games. 2010; 1(4):422-437. https://doi.org/10.3390/g1040422
Chicago/Turabian StyleBilland, Pascal, Christophe Bravard, and Sudipta Sarangi. 2010. "The Insider-Outsider Model Reexamined" Games 1, no. 4: 422-437. https://doi.org/10.3390/g1040422
APA StyleBilland, P., Bravard, C., & Sarangi, S. (2010). The Insider-Outsider Model Reexamined. Games, 1(4), 422-437. https://doi.org/10.3390/g1040422