Normalisation procedures: if they are ‘structural’ (not caring about details like the names of the theories or outcomes), then the two theories are symmetric, so they must be normalised in the same way. WLOG, as follows:
Then letting q = (1-p) the aggregate preferences T are given by:
T(A) = 2p, T(B) = q, T(C) = p, T(D) = q
So:
if p > 2⁄3, the aggregate chooses A and C
if 1⁄3 < p < 2⁄3, the aggregate chooses A and D
if p < 1⁄3, the aggregate chooses B and D
The advantage of this simple set-up is that I didn’t have to make any assumptions about the normalisation procedure beyond that it is structural. If the bargaining outcome agrees with this we may need to look at more complicated cases; if it disagrees we have discovered something already.
For the bargaining outcome, I’ll assume we’re looking for a Nash Bargaining Solution (as suggested in another comment thread).
The defection point has expected utility 3p/2 for Theory I and expected utility 3q/2 for Theory II (using the same notation as I did in this comment).
I don’t see immediately how to calculate the NBS from this.
Then Theory I has expected utility 1, and Theory 2 has expected utility 1⁄2.
Assume (x,y) is the solution point, where x represents probability of voting for A (over B), and y represents probability of voting for C (over D). I claim without proof that the NBS has x=1 … seems hard for this not to be the case, but would be good to check it carefully.
Then the utility of Theory 1 for the point (1, y) = 1 + y/2, and utility of Theory 2 = 1 - y. To maximise the product of the benefits over the defection point we want to maximise y/2*(1/2 - y). This corresponds to maximising y/2 - y^2. Taking the derivative this happens when y = 1⁄4.
Note that the normalisation procedure leads to being on the fence between C and D at p = 2⁄3.
If I’m correct in my ad-hoc approach to calculating the NBS when p = 2⁄3, then this is firmly in the territory which prefers D to C. Therefore the parliamentary model gives different solutions to any normalisation procedure.
I propose to work through a simple example to check whether it aligns with the methods which normalise preferences and sum even in a simple case.
Setup:
Theory I, with credence p, and and Theory II with credence 1-p.
We will face a decision either between A and B (with probability 50%), or between C and D (with probability 50%).
Theory I prefers A to B and prefers C to D, but cares twice as much about the difference between A and B as that between C and D.
Theory II prefers B to A and prefers D to C, but cares twice as much about the difference between D and C as that between B and A.
Questions: What will the bargaining outcome be? What will normalisation procedures do?
Normalisation procedures: if they are ‘structural’ (not caring about details like the names of the theories or outcomes), then the two theories are symmetric, so they must be normalised in the same way. WLOG, as follows:
T1(A) = 2, T1(B) = 0, T1(C) = 1, T1(D) = 0 T2(A) = 0, T2(B) = 1, T2(C) = 0, T2(D) = 2
Then letting q = (1-p) the aggregate preferences T are given by:
T(A) = 2p, T(B) = q, T(C) = p, T(D) = q
So:
if p > 2⁄3, the aggregate chooses A and C
if 1⁄3 < p < 2⁄3, the aggregate chooses A and D
if p < 1⁄3, the aggregate chooses B and D
The advantage of this simple set-up is that I didn’t have to make any assumptions about the normalisation procedure beyond that it is structural. If the bargaining outcome agrees with this we may need to look at more complicated cases; if it disagrees we have discovered something already.
For the bargaining outcome, I’ll assume we’re looking for a Nash Bargaining Solution (as suggested in another comment thread).
The defection point has expected utility 3p/2 for Theory I and expected utility 3q/2 for Theory II (using the same notation as I did in this comment).
I don’t see immediately how to calculate the NBS from this.
Assume p = 2⁄3.
Then Theory I has expected utility 1, and Theory 2 has expected utility 1⁄2.
Assume (x,y) is the solution point, where x represents probability of voting for A (over B), and y represents probability of voting for C (over D). I claim without proof that the NBS has x=1 … seems hard for this not to be the case, but would be good to check it carefully.
Then the utility of Theory 1 for the point (1, y) = 1 + y/2, and utility of Theory 2 = 1 - y. To maximise the product of the benefits over the defection point we want to maximise y/2*(1/2 - y). This corresponds to maximising y/2 - y^2. Taking the derivative this happens when y = 1⁄4.
Note that the normalisation procedure leads to being on the fence between C and D at p = 2⁄3.
If I’m correct in my ad-hoc approach to calculating the NBS when p = 2⁄3, then this is firmly in the territory which prefers D to C. Therefore the parliamentary model gives different solutions to any normalisation procedure.