Discussion:
[EM] Yet another weaker IIA criterion
Greg Dennis
2018-01-08 21:49:01 UTC
Permalink
To anyone's knowledge, has the following weaker IIA variant ever been
named/defined before?

If W is elected, then if a candidate is added that finishes behind W (lower
in the social ordering than W), the winner is still W.

The idea is a new candidate can't "drag the winner down," so to speak, but
can overtake them and cause neither to win. If it hasn't already been
named, I'm tempted to call this "Independence of Weaker Alternatives" but
open to other suggestions.

Greg
Kristofer Munsterhjelm
2018-02-04 14:43:34 UTC
Permalink
Post by Greg Dennis
To anyone's knowledge, has the following weaker IIA variant ever been
named/defined before?
If W is elected, then if a candidate is added that finishes behind W
(lower in the social ordering than W), the winner is still W.
The idea is a new candidate can't "drag the winner down," so to speak,
but can overtake them and cause neither to win. If it hasn't already
been named, I'm tempted to call this "Independence of Weaker
Alternatives" but open to other suggestions.
I'm not aware of any previous mention of this variant of IIA. I'm also
unaware of any method that passes it -- short of perhaps
Approval/Range/MJ under the independent evaluation assumptions that are
required to have them pass ordinary IIA.

At first one would think that it would be met by any method that works
by sequentially eliminating losers, and by any method that's equivalent
to its loser-elimination modification. The tempting proof is something
along the lines of "if W is the original winner and is ranked ahead of X
when X is admitted, then X must be eliminated before W, thus X can't
affect whether or not W wins".

However, that is false: X can affect the order of elimination, either
protecting someone who would lose right away, or exposing someone who
wouldn't were X not present. See e.g. this IRV example:

3: A>B>C
3: B>C>A
2: C>A>B

First C is eliminated, then B is eliminated, so the outcome is A>B>C.

Now add candidate D to get:

3:A>B>C>D
1:B>C>A>D
2:C>A>B>D
2:D>B>C>A

First B is eliminated (because D is hiding some B first preferences),
then D is eliminated, then A is eliminated, and C wins. So the outcome
is C>A>D>B.

D, who was added, finishes behind A (the original winner), yet changed
the winner from A to C; so IRV fails this IIA variant. And because IRV
is a loser elimination method, that means that the class of candidate
elimination methods can't all pass this IIA variant.

For an elimination method to pass the variant, it would suffice to have
an additional property that adding some candidate X can't affect the
elimination order of candidates who are eliminated before X. But that
seems to be a very hard criterion to satisfy.

For the same reason, methods that pass LIIA may not necessarily pass
this variant. If such a method also meets the property that adding some
candidate X can't alter the social order after X, then it also passes
variant IIA if it passes LIIA. But I suspect that the "can't alter
social order after X" property plus LIIA is too tall an order for a
ranked method, because of the Condorcet paradox.
----
Election-Methods mailing list - see http://electorama.com/em for list info
Loading...