[EM] Help me understand some notation

[Forest Simmons]

De-cloned Borda would go particularly well with de-cloned Copeland ... but
it is not necessary because one of the greatest advantages of de-cloning
Copeland is that it makes ties statistically rare: unlike standard Copeland
whose scores take on only integer values, de-cloned Copeland scores are
almost never integers.

And it would be a shame to help propose a clone dependent election method,
especially if there is a sufficiently simple clone free version.

Here's my first attempt at a greatly simplified version without sacrificing
its integrity:

Elect the candidate X that maximizes the number of its first place votes
minus the number of candidates that head-to-head defeat X, weighted by the
average of their first place votes.

In other words, the winner is the candidate X that maximizes the difference
between its own first place vote count and the total of the  first place
counts of the candidates that defeat X head-to-head.

An equivalent formulation that is more likely to yield a positive
difference is ...

Elect the candidate X that minimizes the score S(X), defined as the total
first place vote count of the candidates that pairwise defeat X, minus X's
own first place count.

The first formulation of this method most clearly shows its comparison with
the standard version of Copeland, but the last formulation is easiest to
work with.

Now let's prove monotonicity:

1. Raising X neither increases the number of candidates defeating X, nor
any of their first place votes. So raising X cannot increase S(X), unless
the first place votes of X decrease, which is not possible from raising X.

But could raising X decrease S(Y)?
If so it would either (1) have to decrease the number of candidates that
defeat Y, or (2)lower their total number of first place votes, or (3)
increase the number of first place votes of Y.

Let's look at the first possibility ... could raising X make Y defeat or
tie a candidate Z that defeated it before. Certainly not if Z=X. And if Z
!= X, then the relative pairwise count between Z and Y is not affected.

Saving the most difficult possibility for last, we consider (3) ... could
raising X, increase the first place count of Y. No, it could only decrease
Y's first place votes.

Lastly, (3) how about lowering the total number of first place votes among
the candidates that pairwise defeat Y?

If X replaces Z as first place on some ballot, then X's first place count
increases, (which decreases S(X)) while Z's first place count decreases,
possibly decreasing S(Y) by the same amount as S(X). This is no problem if
X defeats Y (because that would increase S(Y)), or if Z does not defeat Y,
because that would have no effect on S(Y).

The only problem is if S(Y) decreases by more than S(X) does.

The problematic case is where Z defeats Y (and Y defeats X). If Y defeats
X, then X contributes nothing to the total of first place votes of the
candidates defeating Y, so has no direct effect on S(Y) . But indirectly
decreasing  Z's first place votes by one, decreases S(Y) by one unit. This
decrease maintains the difference between S(X) and S(Y).

So the difference S(X)-S(Y) either does not change with the raising of X,
or decreases.

This fact shows that raising X does not change the winner from X to Y.

The method satisfies mono-raise!

(But check me on this!)



El jue., 20 de ene. de 2022 4:22 a. m., Kristofer Munsterhjelm <
km_elmet at t-online.de> escribió:

> On 20.01.2022 12:26, Daniel Carrera wrote:
>
> > Ha! So I want to look for methods similar to Smith,X. Hmm... let me try:
> >
> > 1) Every candidate has a 1-1 pairwise match against every other
> candidate.
> >
> > 2) The candidates with the most won matches are "finalists".
> >
> > 3) The finalist with the greatest margin of victory against any other
> > candidate is elected.
> >
> > So step (2) basically gives the Copeland set. The whole method should be
> > "Copeland,X" where "X" is the method "elect the candidate with the
> > largest victory". Since X must be a monotonic method, would it follow
> > that Copeland,X is monotonic too? Conversely, if step (3) had said "...
> > against any other *finalist*" that would have created "Copeland//X" and
> > it would probably be non-monotonic. Does that sound right?
> >
> > This is interesting because something like Smith,X or Copeland,X as in
> > my example allows you to consider methods X that you would normally have
> > considered too simple to be interesting. This might be a good way to
> > design good methods (e.g. Smith-efficient and monotone) without them
> > being very complicated.
>
> Yes. I forgot to mention two things, though:
>
> - Borda is particularly nice in that Copeland//Borda is monotone. This
> happens because you can infer the Borda score from the pairwise matrix,
> and the remaining pairwise matrix entries don't change when you remove
> candidates. So although as a rule X//Y is nonmonotone, some monotone
> pairwise-based set restrictions of Borda (in particular Smith and
> Copeland) will be monotone. However, Borda is unusually susceptible to
> strategy, which is a problem, and neither Borda nor Copeland are
> cloneproof.
>
> - X//Y methods give up monotonicity, but they get something in return:
> if the X set itself is independent of candidates not in it (i.e. you
> can't make the X set smaller or larger by introducing candidates not in
> it), then X//Y is independent of X-dominated candidates. E.g. in
> Smith//Plurality it doesn't matter if there's vote-splitting among the
> losers as long as there's no vote-splitting in the Smith set itself. But
> in Smith,Plurality, losers outside the Smith set may draw votes away
> from the Smith set, making someone else inside it win.
>
> (Note that Copeland fails the "independence of candidates not in it"
> criterion. So Copeland//Borda doesn't get you independence of
> Copeland-dominated candidates -- but it does get you independence of
> Smith-dominated candidates.)
>
> -km
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>
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