On 20/03/2021 19.30, Kristofer Munsterhjelm wrote:
Suppose that, in some election, a method places A ahead of B and C, and
that A beats B pairwise but is beaten by C p.w.
What types of election methods have the property that raising A does not
change the relative order of B and C? I.e. that if the outcome is
A>...>B>...>C>..., then raising A can never change it into A>...>C>...>B>...
Clearly, LIIA implies this criterion ...
Maybe I shouldn't be so quick. Here's the proof I had in mind:
Suppose that the ranking is ...>X>A>{...>B>...>C>...} and the method
passes LIIA. If we raise A so that the outcome changes to
...>A>X>[...>B>...>C>...], then after eliminating every candidate in the
social ordering down to X (after raising) or down to A (before raising),
the election must be the same regardless of whether A was raised or not.
By LIIA, since we've eliminated everybody down to and including X, the
outcome of this reduced election must be {...>B>...>C>...}. Similarly,
because we before the raising eliminated everybody down to and including
A, the outcome of the reduced election must be [...>B>...>C>...]. So the
two post-A orderings are the same, because the election is the same. In
particular, B must beat C in the social ordering both before and after A
is raised, given that both were ranked after A to begin with.
The problem with this proof is that LIIA doesn't preclude that the
before-raising ordering is something like Z>X>A>B>Y, and that it
afterwards turns into Y>A>X>B>Z. In that case, the reduction fails and
so does the proof.
-km
On 20/03/2021 19.30, Kristofer Munsterhjelm wrote:
> Suppose that, in some election, a method places A ahead of B and C, and
> that A beats B pairwise but is beaten by C p.w.
>
> What types of election methods have the property that raising A does not
> change the relative order of B and C? I.e. that if the outcome is
> A>...>B>...>C>..., then raising A can never change it into A>...>C>...>B>...
>
> Clearly, LIIA implies this criterion ...
Maybe I shouldn't be so quick. Here's the proof I had in mind:
Suppose that the ranking is ...>X>A>{...>B>...>C>...} and the method
passes LIIA. If we raise A so that the outcome changes to
...>A>X>[...>B>...>C>...], then after eliminating every candidate in the
social ordering down to X (after raising) or down to A (before raising),
the election must be the same regardless of whether A was raised or not.
By LIIA, since we've eliminated everybody down to and including X, the
outcome of this reduced election must be {...>B>...>C>...}. Similarly,
because we before the raising eliminated everybody down to and including
A, the outcome of the reduced election must be [...>B>...>C>...]. So the
two post-A orderings are the same, because the election is the same. In
particular, B must beat C in the social ordering both before and after A
is raised, given that both were ranked after A to begin with.
The problem with this proof is that LIIA doesn't preclude that the
before-raising ordering is something like Z>X>A>B>Y, and that it
afterwards turns into Y>A>X>B>Z. In that case, the reduction fails and
so does the proof.
-km
