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Technical discussion of election methods

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An observation from the Ranked Pairs post

KM
Kristofer Munsterhjelm
Wed, Sep 13, 2023 10:17 AM

I used LIIA to show that ties between no-hopes can be shown to have been
irrelevant to the relative position of the strong candidates.

This got me thinking that there are other implications to LIIA too. Namely:

If X is a set, the method passes the X criterion (ranking members of X
ahead of non-X members), then automatically it must pass independence of
non-X alternatives.

Hence: any method that passes Landau and LIIA is nonmonotone (assuming a
relatively non-contentious additional property). Just like that.

And it also means I can't make a LIIA Resistant/Inner Burial Set method
and hope to retain monotonicity.

However, there's an out: The method can still pass the Y criterion,
where Y is always a subset of X. As long as independence of non-Y is
monotone, we're saved. E.g. it's possible that independence of
non-Copeland alternatives is monotone, thus this impossibility
implication doesn't hold.

But such methods must necessarily spread non-Y X candidates over the
ranking. For instance, suppose that in a particular election, the Landau
set is {A, B, C} and the Copeland set is {A, B}. Then to pass Y but not
X, the outcome could be A>B>D>C.

-km

I used LIIA to show that ties between no-hopes can be shown to have been irrelevant to the relative position of the strong candidates. This got me thinking that there are other implications to LIIA too. Namely: If X is a set, the method passes the X criterion (ranking members of X ahead of non-X members), then automatically it must pass independence of non-X alternatives. Hence: any method that passes Landau and LIIA is nonmonotone (assuming a relatively non-contentious additional property). Just like that. And it also means I can't make a LIIA Resistant/Inner Burial Set method and hope to retain monotonicity. However, there's an out: The method can still pass the Y criterion, where Y is always a subset of X. As long as independence of non-Y is monotone, we're saved. E.g. it's possible that independence of non-Copeland alternatives is monotone, thus this impossibility implication doesn't hold. But such methods must necessarily spread non-Y X candidates over the ranking. For instance, suppose that in a particular election, the Landau set is {A, B, C} and the Copeland set is {A, B}. Then to pass Y but not X, the outcome could be A>B>D>C. -km