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Harsanyi power solutions for graph-restricted games

Harsanyi power solutions for graph-restricted games,10.1007/s00182-009-0220-3,International Journal of Game Theory,René van den Brink,Gerard van der L

Harsanyi power solutions for graph-restricted games   (Citations: 4)
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We consider cooperative transferable utility games, or simply TU-games, with limited communication structure in which players can cooperate if and only if they are connected in the communication graph. Solutions for such graph games can be obtained by applying standard solutions to a modified or restricted game that takes account of the cooperation restrictions. We discuss Harsanyi solutions which distribute dividends such that the dividend shares of players in a coalition are based on power measures for nodes in corresponding communication graphs. We provide axiomatic characterizations of the Harsanyi power solutions on the class of cycle-free graph games and on the class of all graph games. Special attention is given to the Harsanyi degree solution which equals the Shapley value on the class of complete graph games and equals the position value on the class of cycle-free graph games. The Myerson value is the Harsanyi power solution that is based on the equal power measure. Finally, various applications are discussed.
Journal: International Journal of Game Theory - INT J GAME THEORY , vol. 40, no. 1, pp. 87-110, 2011
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    • ...This expression is used to show that the average tree solution with respect to T is a Harsanyi solution (see van den Brink et al. 2010b) if and only if T is a subset of the set introduced in Herings et al. (2010)...
    • ...solutions, introduced by Vasil’ev (1982) for TU-games and studied by van den Brink etal.(2010b)forcommunicationsituations...

    Richard Baronet al. Average tree solutions and the distribution of Harsanyi dividends

    • ...The graph (N,L) induces 3 trees, one for each of the 3 dierent nodes of N, T(1) = {(1,2),(2,3)}, T(2) = {(2,1),(2,3)}, and T(3) = {(3,2),(2,1)}...
    • ...The graph (N,L) induces 3 trees, one for each of the 3 dierent nodes of N, T(1) = {(1,2),(2,3)}, T(2) = {(2,1),(2,3)}, and T(3) = {(3,2),(2,1)}...
    • ...The graph (N,L) induces 3 trees, one for each of the 3 dierent nodes of N, T(1) = {(1,2),(2,3)}, T(2) = {(2,1),(2,3)}, and T(3) = {(3,2),(2,1)}...

    P. Jean Jacques Heringset al. The average tree solution for cycle-free graph games

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