[EM] NES--Quick Preliminary Reply

[MIKE OSSIPOFF]

Jobst--

As I say in the subject-line, this is just a quick preliminary reply:

You wrote:

I wonder whether I understand the NES stuff right. Let me consider a
very simple example: 3 options A,B,C, 3 voters, sincere preferences are
complete and strict rankings, base method is plurality.

I reply:

Ok, but remember that no one has proposed NES(Plurality). I propose 
NES(Approval) and               NES(Condorcet(wv)), and 
NES(NES(Condorcet(wv))), NES(NES(Approval)), etc.

So it would be better to test those methods, starting with the simplest, of 
course. Starting preferably with NES(Approval).

You continued:

when (b) are the sincere
preferences, voting (b) is still not an equilibrium in NES(plurality),
but voting (c) is the equilibrium instead!

I reply:

But I never said that NES would violate the Gibbard-Satterthwaite theorem 
and be completely strategy-free. I found that NES seems to share the 
properties of Condorcet(wv) to a large extent.

That means that there's an offensive order-reversal strategy, and defensive 
truncation. It seems to me that defensive equal ranking didn't always work 
in NES(Approval), as it does in Condorcret(wv).

I've recently suggested that maybe nested combinations of NES could be a 
place to look for a method that meets SSFC. For that reason of course I 
thank you for checking NES out.

You continued:

So, when the whole thing should make sense, we would have to consider
NES^2(base method) := NES(NES(base method)) and so on...
Fortunately, for any combination of finitely many options and voters
there is only a finite number of possible voting methods we must take
into account. Hence the sequence NES^k must become stable or periodic
for large k. If it becomes stable, that is, if NES^(k+1)=NES^k for some
k, then it seems that NES^k would be a method in which the equilibrium
result always equals the sincere result.

I reply:

Ok, but remember that I don't expect NES or any other nonprobabilistic 
method to be entirely strategy-free. I don't expect sincere rankings to 
always be a Nash equiilbrium.


Proof: NES^k(equilibrium
votes)=NES^(k+1)(sincere prefs)=NES^k(sincere prefs)? That doesn't mean
that NES^k leaves no incentives to vote strategically at all, but that
at least the situations attained by voting strategically are unstable
and will in the end lead to an equilibrium which gives the same result
as the sincere one. But what if the sequence NES^k does not become
stable but periodic? Is there a similar argument then?

Another question is: When can we be sure that equilibria exist? With
plurality, it seems that the equilibria are exactly those voting
situations where a majority votes for the Condorcet winner. In all other
situations, some majority would have an incentive to vote for the
Condorcet winner instead. Without a Condorcet winner, there would be no
equilibria. Are there base methods of which we know that equilibria must
exist? Does that follow from some well-known criterion for example?

I reply:

NES's rules deal with situations where there's no Nash equilibrium, or in 
which there's more than 1 Nash equilibrium.

When I was looking at NES(Approval), either I didn't run across an example 
in which there were no Nash equiilbria, or I didn't run across an example in 
which there was more than 1 Nash Equilibrium. I'm not sure it was. Probably 
the latter. I'm not saying that that's always so.

Mike Ossipoff

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