Nash’s 1953 paper covers that, I think. Just about any game theory text should explain. Look in the index for “threat game”. In fact, Googling on the string “Nash bargaining threat game” returns a host of promising-looking links.
Of course, when you go to extend this 2-person result to coalition games, it gets even more complicated. In effect, the Shapley value is a weighted average of values for each possible coalition structure, with the division of spoils and responsibilities within each coalition also being decided by bargaining. The thing is, I don’t see any real justification for the usual convention of giving equal weights to each possible coalition. Some coalitions seem more natural to me than others—one most naturally joins the coalitions with which one communicates best, over which one has the most power to reward and punish, and which has the most power over oneself. But I’m not sure exactly how this fits into the math. Probably a Rubinstein-style answer could be worked out within the general framework of Nash’s program.
Well, sorry. You were right all along and I’m a complete idiot. For some reason my textbook failed to cover that, and I never stumbled on that anywhere else.
Nash’s 1953 paper covers that, I think. Just about any game theory text should explain. Look in the index for “threat game”. In fact, Googling on the string “Nash bargaining threat game” returns a host of promising-looking links.
Of course, when you go to extend this 2-person result to coalition games, it gets even more complicated. In effect, the Shapley value is a weighted average of values for each possible coalition structure, with the division of spoils and responsibilities within each coalition also being decided by bargaining. The thing is, I don’t see any real justification for the usual convention of giving equal weights to each possible coalition. Some coalitions seem more natural to me than others—one most naturally joins the coalitions with which one communicates best, over which one has the most power to reward and punish, and which has the most power over oneself. But I’m not sure exactly how this fits into the math. Probably a Rubinstein-style answer could be worked out within the general framework of Nash’s program.
Well, sorry. You were right all along and I’m a complete idiot. For some reason my textbook failed to cover that, and I never stumbled on that anywhere else.
(reads paper, goes into a corner to think)