Ok; I thought of a better way to phrase Stuart’s point. Suppose there are five alternatives, and I rank them 1-2-3-4-5, but you rank them 5-4-3-2-1. If we are equal in power we will compromise on 3. (Well… given some simplifying assumptions, anyway. It’s quite possible that you are almost indifferent between 2 and 3, but I care a lot about that gap. If so, even if we are equal in power I will likely commit a lot more resources to the fight, and drag the compromise up to 2.) But if the available options had been 1, 2, and 3, we would instead have compromised on 2. This demonstrates that removing options changes the outcome.
However, I think there is a problem with carrying the “irrelevant alternatives” axiom into a two-agent problem. If I have A>B>C, then I should choose A whether or not C is an option; fine. But this needn’t be true of problems with multiple agents, because that phrase “we will compromise on” is hiding rather a lot of complexity that doesn’t have anything to do with utility functions, per se. Options 4 and 5 are not, in fact, irrelevant; they are bargaining chips. Removing one side’s bargaining chips breaks the symmetry; it is equivalent to giving the other side more power. Suppose I had left the options as they were, but specified that the agent whose utility is on the y axis suddenly gets a lot more bargaining power; would we then expect the decision to be option 3? Surely not. And this is exactly what is accomplished by asymmetrically removing options.
The problem rises from breaking the game-theoretic symmetry and asserting that only the utility symmetry is important.
The most common formulation of IIA precisely assumes that there is no such thing as “bargaining chips”. So yes, you could rewrite the point of my post as: any symmetric bargaining solution will have bargaining chips.
Ok; I thought of a better way to phrase Stuart’s point. Suppose there are five alternatives, and I rank them 1-2-3-4-5, but you rank them 5-4-3-2-1. If we are equal in power we will compromise on 3. (Well… given some simplifying assumptions, anyway. It’s quite possible that you are almost indifferent between 2 and 3, but I care a lot about that gap. If so, even if we are equal in power I will likely commit a lot more resources to the fight, and drag the compromise up to 2.) But if the available options had been 1, 2, and 3, we would instead have compromised on 2. This demonstrates that removing options changes the outcome.
However, I think there is a problem with carrying the “irrelevant alternatives” axiom into a two-agent problem. If I have A>B>C, then I should choose A whether or not C is an option; fine. But this needn’t be true of problems with multiple agents, because that phrase “we will compromise on” is hiding rather a lot of complexity that doesn’t have anything to do with utility functions, per se. Options 4 and 5 are not, in fact, irrelevant; they are bargaining chips. Removing one side’s bargaining chips breaks the symmetry; it is equivalent to giving the other side more power. Suppose I had left the options as they were, but specified that the agent whose utility is on the y axis suddenly gets a lot more bargaining power; would we then expect the decision to be option 3? Surely not. And this is exactly what is accomplished by asymmetrically removing options.
The problem rises from breaking the game-theoretic symmetry and asserting that only the utility symmetry is important.
The most common formulation of IIA precisely assumes that there is no such thing as “bargaining chips”. So yes, you could rewrite the point of my post as: any symmetric bargaining solution will have bargaining chips.