But this is obviously wrong: symmetry implies the correct outcome should be the blue point (0.6, 0.6), not the green (1, 0) which was the outcome before we removed the “irrelevant” extra points. We have derived a contradiction, and IIA must fall.
Isn’t the correct outcome (1.0, 0), not (.6, .6)? If you rescale the utility functions and your decision changes, you aren’t using the utility functions correctly.
That is, as RolfAndreassen points out, you can’t use symmetry to choose the ‘correct’ outcome if you use a symmetry-breaking transformation.
If you rescale the utility functions and your decision changes, you aren’t using the utility functions correctly.
Precisely. So the error arrived when we removed the extra points and IIA implied the decision didn’t change.
That is, as RolfAndreassen points out, you can’t use symmetry to choose the ‘correct’ outcome if you use a symmetry-breaking transformation.
There is no transformation. Scalings are aretefacts of how we represent the utility functions. They have no intrinsic meaning. So I have not been using a “symmetry-braking transformation”—I haven’t used a transformation at all, just a different way of drawing the exact same situation.
Precisely. So the error arrived when we removed the extra points and IIA implied the decision didn’t change.
What? (2,2) is still on the axis of symmetry, regardless of whether or not the point (2.6, 1.2) exists or not, and so if they select that point because of symmetry, they will continue to do so, regardless of the existence or nonexistence of irrelevant alternatives.
Scalings are aretefacts of how we represent the utility functions. They have no intrinsic meaning. So I have not been using a “symmetry-braking transformation”—I haven’t used a transformation at all, just a different way of drawing the exact same situation.
Not in this problem, because you’ve set it up so the axis of symmetry is used for decision-making. If you do the scaling to both the points and the axis of symmetry, then you get the correct answer- (1.0, 0) - because it lies on the line y=2x-2, which is the new axis of symmetry. By only scaling part of the problem, you perform a transformation.
[edit] Thinking about this more, I think I’ve modified my position somewhat: you have two assumptions, first that the outcomes are symmetric and second that there are no canonical choices for utility functions. Those don’t look like they play well together- if outcome (0,3) is symmetric with outcome (3,0), and then by changing your choice of utility function you can make (0,3) symmetric with (2,2), then you have a serious problem. If you fulfill the symmetry assumption by restricting IIA to removing pairs of symmetric points, the crisis is averted.
[edit2] There’s a more general point that should be mentioned- whenever you have a decisionmaker whose decision depends on relative outcome merits, then IIA either breaks or is limited severely. “Irrelevant” needs to be understood not as “outcomes I didn’t choose” but “outcomes which didn’t impact my choice.” When your rule is “pick the second best,” then both the best and second best are relevant alternatives, even though you only picked one. In a bargaining game without a frame, the only way to judge outcomes is by their relative merits- and so you get weak or broken IIA.
But in this particular problem, the symmetry assumption throws a wrench into that general point, because now there is a frame- the symmetric lattice, and the implied axis of symmetry- and so there’s an objective criterion rather than simply relative criteria.
What? (2,2) is still on the axis of symmetry, regardless of whether or not the point (2.6, 1.2) exists or not, and so if they select that point because of symmetry, they will continue to do so, regardless of the existence or nonexistence of irrelevant alternatives.
The axis of symmetry is a property of the figure (in this case, the set of points), not of the axis. In fact, ignore the axis: they don’t exist, only their directions have mathematical meaning (neither their scale nor their points of origin mean anything, because the affine transformations of the utility functions will shift those).
Right- let’s just call the points A, B, C, -B, and -A. C is going to be the middle of that set, even if you remove (B and -B) and/or (A and -A). When you remove (-B and -A), you’ve broken the symmetry of the set, even though the set has new symmetry by virtue of there being a new middle point.
If you’re postulating a (much much) weaker version of IIA saying something like “if you remove symmetric irrelevant points from a symmetric set, the outcome doesn’t change” then you’d be right. But IIA does not require that symmetry be preserved.
Yep, I agree that strong IIA of “if x is chosen from T, and S is a subset of T, then x is chosen from S” doesn’t apply if preferences are based on the relative merits of x rather than the individual merits of x. That statement seems obviously true on its own, and so I think the picture proof of this particular example detracts more than it adds, because there is a natural weak IIA here.
Isn’t the correct outcome (1.0, 0), not (.6, .6)? If you rescale the utility functions and your decision changes, you aren’t using the utility functions correctly.
That is, as RolfAndreassen points out, you can’t use symmetry to choose the ‘correct’ outcome if you use a symmetry-breaking transformation.
Precisely. So the error arrived when we removed the extra points and IIA implied the decision didn’t change.
There is no transformation. Scalings are aretefacts of how we represent the utility functions. They have no intrinsic meaning. So I have not been using a “symmetry-braking transformation”—I haven’t used a transformation at all, just a different way of drawing the exact same situation.
What? (2,2) is still on the axis of symmetry, regardless of whether or not the point (2.6, 1.2) exists or not, and so if they select that point because of symmetry, they will continue to do so, regardless of the existence or nonexistence of irrelevant alternatives.
Not in this problem, because you’ve set it up so the axis of symmetry is used for decision-making. If you do the scaling to both the points and the axis of symmetry, then you get the correct answer- (1.0, 0) - because it lies on the line y=2x-2, which is the new axis of symmetry. By only scaling part of the problem, you perform a transformation.
[edit] Thinking about this more, I think I’ve modified my position somewhat: you have two assumptions, first that the outcomes are symmetric and second that there are no canonical choices for utility functions. Those don’t look like they play well together- if outcome (0,3) is symmetric with outcome (3,0), and then by changing your choice of utility function you can make (0,3) symmetric with (2,2), then you have a serious problem. If you fulfill the symmetry assumption by restricting IIA to removing pairs of symmetric points, the crisis is averted.
[edit2] There’s a more general point that should be mentioned- whenever you have a decisionmaker whose decision depends on relative outcome merits, then IIA either breaks or is limited severely. “Irrelevant” needs to be understood not as “outcomes I didn’t choose” but “outcomes which didn’t impact my choice.” When your rule is “pick the second best,” then both the best and second best are relevant alternatives, even though you only picked one. In a bargaining game without a frame, the only way to judge outcomes is by their relative merits- and so you get weak or broken IIA.
But in this particular problem, the symmetry assumption throws a wrench into that general point, because now there is a frame- the symmetric lattice, and the implied axis of symmetry- and so there’s an objective criterion rather than simply relative criteria.
The axis of symmetry is a property of the figure (in this case, the set of points), not of the axis. In fact, ignore the axis: they don’t exist, only their directions have mathematical meaning (neither their scale nor their points of origin mean anything, because the affine transformations of the utility functions will shift those).
Right- let’s just call the points A, B, C, -B, and -A. C is going to be the middle of that set, even if you remove (B and -B) and/or (A and -A). When you remove (-B and -A), you’ve broken the symmetry of the set, even though the set has new symmetry by virtue of there being a new middle point.
If you’re postulating a (much much) weaker version of IIA saying something like “if you remove symmetric irrelevant points from a symmetric set, the outcome doesn’t change” then you’d be right. But IIA does not require that symmetry be preserved.
Yep, I agree that strong IIA of “if x is chosen from T, and S is a subset of T, then x is chosen from S” doesn’t apply if preferences are based on the relative merits of x rather than the individual merits of x. That statement seems obviously true on its own, and so I think the picture proof of this particular example detracts more than it adds, because there is a natural weak IIA here.