Now, we have to rescale one of the preference scales: (I think? Having a third, fourth, and fifth preferred choice among three choices doesn’t seem to make any sense.)
(0, 3), (1.2, 2.6), (2, 2)
(3rd, 1st), (2nd, 2nd), (1st,3rd)
And, then we rescale this Ordinally, keeping them in the same preference order:
(x=x/2, y=y-2)
(0, 1), (.6, .6), (1, 0)
(3rd, 1st), (2nd, 2nd), (1st,3rd)
And now the symmetric choice is (.6, .6), which it wasn’t before.
Whereas if we don’t remove the irrelevant alternatives.
Nearly. But I used more than an ordinal symmetry to be able to actually select the points I wanted (ordinally, “choice 3” is the same as “50% between choices 1 and 5″, “50% between choices 2 and 4” and so on; I used the stronger symmetry to be able to choose option 3).
Stuart_Armstrong, I figured I should try and lay out the math without pictures to check for my understanding. Am I doing this correctly?
Here are the choices, which are Ordinally Symmetric:
(0, 3), (1.2, 2.6), (2, 2), (2.6, 1.2), (3, 0).
Here is those choices expressed as a preference order (I.e, 1st is most preferred, 5th is least preferred, units are irrelevant.)
(5th, 1st), (4th, 2nd), (3rd,3rd), (2nd,4th), (1st,5th).
All points are Pareto Optimal, so Symmetry is used to decide that among these:
(0, 3), (1.2, 2.6), (2, 2), (2.6, 1.2), (3, 0).
(5th, 1st), (4th, 2nd), (3rd,3rd), (2nd,4th), (1st,5th).
The choice would be this:
(2,2)
(3rd,3rd)
Now, let’s remove two irrelevant choices:
(0, 3), (1.2, 2.6), (2, 2)
(5th, 1st), (4th, 2nd), (3rd,3rd),
Now, we have to rescale one of the preference scales: (I think? Having a third, fourth, and fifth preferred choice among three choices doesn’t seem to make any sense.)
(0, 3), (1.2, 2.6), (2, 2)
(3rd, 1st), (2nd, 2nd), (1st,3rd)
And, then we rescale this Ordinally, keeping them in the same preference order: (x=x/2, y=y-2)
(0, 1), (.6, .6), (1, 0)
(3rd, 1st), (2nd, 2nd), (1st,3rd)
And now the symmetric choice is (.6, .6), which it wasn’t before.
Whereas if we don’t remove the irrelevant alternatives.
(0, 3), (1.2, 2.6), (2, 2), (2.6, 1.2), (3, 0).
(5th, 1st), (4th, 2nd), (3rd,3rd), (2nd,4th), (1st,5th).
And then we rescale them (x=x/2, y=y-2)
(0, 1), (.6, .6), (1, 0), (1.3, -.8), (1.5, −2).
(5th, 1st), (4th, 2nd), (3rd,3rd), (2nd,4th), (1st,5th).
Now they aren’t Ordinally symmetric. Only removing those points earlier allowed it to be Ordinally symmetric again after the rescale.
Does this correctly capture the point of the above pictures?
Nearly. But I used more than an ordinal symmetry to be able to actually select the points I wanted (ordinally, “choice 3” is the same as “50% between choices 1 and 5″, “50% between choices 2 and 4” and so on; I used the stronger symmetry to be able to choose option 3).