# [EM] Yet another weaker IIA criterion

Kristofer Munsterhjelm km_elmet at t-online.de
Sun Feb 4 06:43:34 PST 2018

```On 01/08/2018 10:49 PM, Greg Dennis wrote:
> 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