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## Hot answers tagged cooperative-game-theory

6

When preferences are strict, the strong core is unique and the top-trading cycle procedure (TTCP) is the unique mechanism yielding the strong core allocation. Here are the highlights. Lemma 1: Let $x$ be the allocation generated by the Top-Trading Cycle Procedure. Then $x$ is in the strong core. Lemma 2: Let $P$ be the set of players, $\Omega = \{ \omega^{... 4 The Nash bargaining solution DOES maximize the Nash product. You have to separate the playing of the game from the bargaining problem. If the players negotiate a binding agreement they will realize that their maximal total payoff from playing the game is$3$. This can be achieved by playing$(T,L)$or$(T,C)$, or any mixture of those two profiles. The ... 3 The center allocation$z_{i}:=(v(N)/|N|)$for all$i \in N$is in the core of the symmetric game$v$, whenever$z(S) = \sum_{i \in S} z_{i} \ge v(S) $holds for all$S \subseteq N$. Now, if$v(N)/|N| \ge v(S)/|S|$for all$S \subseteq N$, then it holds$z(S) = \sum_{i \in S} z_{i} = |S|\cdot v(N)/|N| \ge v(S)$. Thus,$\mathbf{z} \in C(v)$. We observe that ... 3 It depends on the set of feasible allocations for the coalitions$S$. Suppose for all$S$a best allocation exists (the sum of the individual utilities of the members of$S$is maximal). Then as in usual cooperative games each coalition can be assigned this best utility sum as its value$v(S)$. Let us denote the utility vector each player gets from an ... 2 Chess is EXPTIME-Complete, which makes it significantly harder than NP-Complete problems. Perhaps you are interested in the study of economic networks. Strategic network formation sounds like a good starting point. A lot of the work examines when certain classes of graphs arise under pure strategy Nash equilibria. There are exponentially many pure ... 2 Computing equilibria is an active area of research. Big name complexity theorists like Lance Fortnow are working in this area. When you have continuity, computing core allocations becomes much easier as we can usually describe the core as a linear program. When the feasible allocations are discrete, we run into issues of complexity. Deciding if the core is ... 2 Most of what you write is correct, but the definitions of the$F_i$sets is imprecise. The problem is that in the core$A$and$B$may get goods that do not match their initial endowments. In this case it is not true that the core allocation$x$is Pareto-efficient in the restricted 2-person economy of$A$and$B$, because$x$is not even an allocation in ... 2 Strong Nash equilibrium is different from core mainly because of communication. In a Strong Nash, unlimited private communication is allowed. The core is a concept that is linked to Coalition-proof Nash equilibrium rather than Strong Nash. People can freely communicate but cannot make binding commitment before deciding. In some games, both happen to be the ... 2 If I understand the question correctly, I believe the neatest way to put it is to simply call you solution concept "individual rationality". Individual rationality can easily be seen as a cooperative game theory solution concept in its own right. I am not aware of any other name for a solution concept consisting only of an individual rationality ... 2 We can talk of the core for any arbitrary game. To explain how let me compare it directly to how we define non-cooperative games. In a standard non-cooperative game we define Players,$I$, action sets for each player,$A_i$for each$i\in I$and payoff functions that map any profile of actions to payoffs$u_i: A\rightarrow \mathbb{R}$, where$A=\prod_{i\in I}...

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Finally, I believe it is easy to answer it. Well, in case where some player will not cooperate and the triger strategy is enabled, we have the following soloutions: Soppuse that p2 does not cooperate in the first round so, he will gain a payoof of 1, but in the second round p1 plays $a_2$ and he won't cooperate with p2 in perpetual. Thus, we have that p2 ...

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Define the marginal contribution of $i \in N$ to any $C \subseteq N \setminus \{i\}$ by \begin{align} m_i(C) = v(C \cup \{i\}) - v(C). \end{align} We are going to show C.1 $\Rightarrow$ C.2 $\Rightarrow$ C.3 $\Rightarrow$ C.1. Note that C.1 $\Rightarrow$ C.2 is trivially true. If C.1 holds for all $S,T \subseteq N$, then C.2 must hold for all $S,T \... 1 I suppose it would look something like$\Pi^m$is the profit of a monopolist.$\Pi^c$is the profit under collusion.$\Pi^c=\frac{\Pi^m}{n}\Pi^{comp}$suppose$\Pi^m> \Pi^c> \Pi^{comp} $. The profit a firm gets for deviating through 1 period is$\Pi^d=\Pi^m-\epsilon=\Pi^m$The condition to sustain collusion is$\Pi^d +\delta \Pi^{comp} +\delta ^2 ...

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