I think the the Nash bargaining solution should be pretty good if there are only two members of the parliament, but it’s not clear how to scale up to a larger parliament.
For the NBS with more than two agents, you just maximize the product of everyone’s gain in utility over the disagreement point. For Kalai-Smodorinsky, you continue to equate the ratios of gains, i.e. picking the point on the Pareto frontier on the line between the disagreement point and vector of ideal utilities.
Agents could be given more bargaining power by giving them different exponents in the Nash product.
Giving them different exponents in the Nash product has some appeal, except that it does seem like NBS without modification is correct in the two-delegate case (where the weight assigned to the different theories is captured properly by the fact that the defection point is more closely aligned with the view of the theory with more weight). If we don’t think that’s right in the two-delegate case we should have some account of why not.
The issue is when we should tilt outcomes in favor of higher credence theories. Starting from a credence-weighted mixture, I agree theories should have equal bargaining power. Starting from a more neutral disagreement point, like the status quo actions of a typical person, higher credence should entail more power / votes / delegates.
On a quick example, equal bargaining from a credence-weighted mixture tends to favor the lower credence theory compared to weighted bargaining from an equal status quo. If the total feasible set of utilities is {(x,y) | x^2 + y^2 ≤ 1; x,y ≥ 0}, then the NBS starting from (0.9, 0.1) is about (0.95, 0.28) and the NBS starting from (0,0) with theory 1 having nine delegates (i.e. an exponent of nine in the Nash product) and theory 2 having one delegate is (0.98, 0.16).
If the credence-weighted mixture were on the Pareto frontier, both approaches are equivalent.
Update: I now believe I was over-simplifying things. For two delegates I think is correct, but in the parliamentary model that corresponds to giving the theories equal credence. As credences vary so do the number of delegates. Maximising the Nash product over all delegates is equivalent to maximising a product where they have different exponents (exponents in proportion to the number of delegates).
I think the the Nash bargaining solution should be pretty good if there are only two members of the parliament, but it’s not clear how to scale up to a larger parliament.
For the NBS with more than two agents, you just maximize the product of everyone’s gain in utility over the disagreement point. For Kalai-Smodorinsky, you continue to equate the ratios of gains, i.e. picking the point on the Pareto frontier on the line between the disagreement point and vector of ideal utilities.
Agents could be given more bargaining power by giving them different exponents in the Nash product.
Giving them different exponents in the Nash product has some appeal, except that it does seem like NBS without modification is correct in the two-delegate case (where the weight assigned to the different theories is captured properly by the fact that the defection point is more closely aligned with the view of the theory with more weight). If we don’t think that’s right in the two-delegate case we should have some account of why not.
The issue is when we should tilt outcomes in favor of higher credence theories. Starting from a credence-weighted mixture, I agree theories should have equal bargaining power. Starting from a more neutral disagreement point, like the status quo actions of a typical person, higher credence should entail more power / votes / delegates.
On a quick example, equal bargaining from a credence-weighted mixture tends to favor the lower credence theory compared to weighted bargaining from an equal status quo. If the total feasible set of utilities is {(x,y) | x^2 + y^2 ≤ 1; x,y ≥ 0}, then the NBS starting from (0.9, 0.1) is about (0.95, 0.28) and the NBS starting from (0,0) with theory 1 having nine delegates (i.e. an exponent of nine in the Nash product) and theory 2 having one delegate is (0.98, 0.16).
If the credence-weighted mixture were on the Pareto frontier, both approaches are equivalent.
Update: I now believe I was over-simplifying things. For two delegates I think is correct, but in the parliamentary model that corresponds to giving the theories equal credence. As credences vary so do the number of delegates. Maximising the Nash product over all delegates is equivalent to maximising a product where they have different exponents (exponents in proportion to the number of delegates).