[EM] An incentive to take positions a direct democracy would choose (was Re: Arrow's theorem and cardinal voting systems)

[Kristofer Munsterhjelm]

On 12/01/2020 21.59, Steve Eppley wrote:
> On 1/11/2020 2:42 PM, Kristofer Munsterhjelm wrote:
>> I don't think it would in every such scenario. Consider this election pair:
>>
>> Before cloning:
>>
>> 110: A
>> 100: X>A
>> 100: Y>A
>>
>> X and Y are eliminated and then A wins.
>>
>> After cloning:
>>
>> 110: A2>A1
>> 100: X>A1>A2
>> 100: Y>A1>A2
>>
>> First A1 is eliminated, and then X and Y are eliminated, and then A2
>> wins. But A1 is the CW and beats A2 pairwise 200-110.
>>
>> If the q-preferring majority ranks A1 and A2 low enough, then IRV may
>> exclude A1 before it gets to determine who should win of A1 and A2. It's
>> the usual center squeeze.
>>
>> Does that make the clone criterion more suited to your purposes, or
>> would it have to be stronger? I suppose the clone criterion is a sort of
>> local optimum criterion (if Alice exists, then Bob can copy all of
>> Alice's positions except the one a majority dislikes, and overtake
>> Alice), while your non-rigorous criterion is a global optimum criterion.
>>
>> (In passing, I think I see that LIAA + clone independence implies this
>> clone criterion, as well.)
> 
> You're right that Instant Runoff fails "clone A1 should win."
> 
> I don't know whether its satisfaction implies satisfaction of "the
> incentive to take positions the voters would choose."  My election
> method analysis skills are very rusty.
> 
> I don't recall LIAA.  I assume you mean LIIA (Local Independence of
> Irrelevant Alternatives, promoted by Peyton Young).

Yes, that should have been LIIA.

> There appears to be a flaw in that clone criterion.  Suppose 3
> clones majority cycle: Bob > Alice > Charlie > Bob.  The premise of
> the "clone A1 should win" criterion could hold: In the "original"
> scenario where Bob doesn't run, Alice wins.  We don't have enough
> information to show that Bob will win if Bob runs too.  Alice could
> still win if the Bob>Alice majority is the smallest of the three
> cyclic majorities. (When I described my thinking about the incentive
> in MAM, I wrote: "The larger the majority who prefer q over p, the
> larger the majority who would tend to rank Bob over Alice.")  But
> that clone failure isn't necessarily a failure of the voting method
> to create the strong incentive.  My hunch is that typically,
> candidates like Alice won't be able to rely on a Bob>Alice majority
> being the smallest in a cycle, when taking positions on issues.  The
> chance that Bob>Alice won't be smallest in a cycle is a risk to be
> avoided, all else being equal.

I'm not entirely sure what you mean. Do you mean that even if "A1 should
win" happens to be necessary, it isn't sufficient; or that even if it
happens to be sufficient, it isn't necessary?

I *think* you're saying it's not sufficient, because there could already
be a clone of Alice, and then when Bob enters, he could have the
smallest of the cyclic majorities and create a cycle, and he won't win.

More generally, we can say that he creates a three-cycle and the cloning
comes out so that, according to the cycle-resolution mechanism of the
method in question, he doesn't win even though he's in the Smith set.
But then it would seem that no matter what method you have, it's
possible to construct the cloning so that the right clone loses.

If that's right, then there has to be some kind of additional structure
that makes it possible for the method to distinguish the right clone
from the other clones. In the original example, that is that Bob copies
all of Alice's positions except the disliked one, where he does better
according to a majority. For the three clones to create a cycle, there
has to be some set of properties so that a majority prefers Bob's to
Alice's, Alice's to Charlie's, and Charlie's to Bob's. But then,
wouldn't Bob have to differ from Alice by more than one property?

> Thanks for spending time on this.  I hope you can continue.
> 
> --Steve
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