[EM] Markov chain approaches

[bql at bolson.org]

On Sun, 16 May 2004, Jobst Heitzig wrote:

> "Neutrality": Now this is something I did not understand yet.
> Quote:
> "Neutrality requires that if two problems are such that the ranking
> method cannot rank any player [that is, any option! JH] above another,
> then the ranking method should still be unable to rank one player above
> another when considering the two problems together as a single problem.
> And conversely, if two problems are such that in the first one the
> ranking method does not rank one player above another, but in the second
> problem there are two players that are strictly ordered, then it must be
> the case that, when considering the two problems together, there are at
> least two players that are ranked one above the other."
> At least this seems to apply only in rare cases, where the method fails
> to construct any strict ranking at all...

I'd say this means: If you have two elections, A,B,C and D,E,F, if you
merge them into A..F, the relative rankings of A:B:C and D:E:F won't
change in the merged solution vs the separate solutions.

That's what I think they're saying, but I'm not sure why it's really an
interesting property of a voting system.

> Finally, "negative responsiveness to losses" seems to be the crucial
> point implying the fair-bets method: Suppose in some situation, the
> method cannot but rank all options equal. Suppose further that now for
> each option A some non-negative number lambda(A) is chosen and the
> number of voters prefering B to A is multiplied by lambda(A) for each
> A,B. Then the axiom requires that the method must now put A above B if
> and only if lambda(A)<lambda(B).
>
> I don't know what the rationale behind this last axiom is, but perhaps
> someone can find it. However, I have the feeling that it must be somehow
> related to the MinMax rule and to breaking cycles at the weakest link...

It sounds to me like a rule for resolving Condorcet ties I once heard.
This is confusing because it's saying: "if the method ends up in a tie,
the method must resolve ties like this". That's not where I draw the black
box of a voting method for comparison. Tie resolution ought to be totally
internal and we should just be talking about the final output of the
system.

-Brian Olson