Rank methods, Participation, Consistency
Richard Moore
rmoore4 at cox.net
Fri Jan 18 18:16:45 PST 2002
Thanks to Mike's PC monotonicity proof, I can see the error in my previous
argument. I didn't claim the argument proved Condorcet methods couldn't be
monotonic, but rather that it made it appear unlikely (at least, for those
methods that only used the Smith set to choose a winner when there's no CW).
However, see below for my comments as to where the argument doesn't hold up
for PC.
I wrote:
> I've read the claim for RP being Monotonic, but never seen proof.
>
> The general problem for Condorcet methods is this: If X is the CW, then
> it is easy to show that any ballot change favoring X won't make X the
> loser. If X is the winner, but not a CW, then it is possible to make
> a ballot change favoring X that breaks up the Smith set. Let's say the
> Smith set consists of W, X, Y, and Z, and that
>
> W beats X
> X beats Z
> Y beats W and X
> Z beats W and Y
>
> Suppose the completion method picks X. If we find some ballots containing
> "W>X" and swap those two candidates on those ballots, and this changes
> things so that X beats W, then W is no longer in the Smith set, and we
> have:
>
> X beats Z
> Y beats X
> Z beats Y
>
> Unless the method picks X again in this new configuration, it is not
> monotonic. Let's suppose it does, so that Monotonicity is not violated
> (yet).
So, supposing PC picked X in both the above cases, that would mean that
"Y beats X" is the smallest defeat in the second election, and in the
first election it would mean that "W beats X" is smaller than "X beats Z",
"Z beats Y", and the larger of "Y beats W" and "Z beats W". So far, no
contradiction.
> Now suppose you started with this:
>
> W beats Y
> X beats W and Z
> Y beats X
> Z beats W and Y
>
> Suppose this configuration has Y as the winner.
That would mean that "W beats Y and "Z beats Y" are both smaller than
"X beats Z", "Y beats X", and the larger of "X beats W" and "Z beats W".
Which, of course, meants that "Z beats Y" is smaller than "Y beats X".
> Now find some ballots
> with "W>Y", and reverse those two. Then, if this causes W to lose to Y,
> you end up with
>
>
> X beats Z
> Y beats X
> Z beats Y
>
>
> which looks very familiar. If the {margins|winning votes} in this
> final Smith set are identical to the ones in the final Smith set of
> the original example, then X is picked again (assuming those numbers
> are the only data used by the completion method). That violates
> Monotonicity.
Now, if X wins the last election, then "Y beats X" is again the
smallest defeat. But as I observed in the third election, "Z beats Y"
is smaller than "Y beats X". So, you can't get the fourth set of
results (with X as the PC winner) from the third set (with Y as the
PC winner, merely by upranking Y over W.
> So, for a Condorcet method to be monotonic, it would be necessary that
> the method is unable to produce identical {margins|winning vote} in
> the final configuration of the two examples given. I suspect that is
> impossible.
Of course, that paragraph is the weak spot in the argument, for now I see
how it is possible for a method to be unable to produce an identical set
of results in the 3-member Smith set from different starting points merely
by upranking one candidate with respect to another (without changing
any other relative rankings). In retrospect, it even seems kind of
obvious.
Anyway, thanks to Mike's insight on PC's monotonicity, I can now
conclude that "monotonicity does not imply consistency", since no
Condorcet method is consistent (correct?). One counterexample being
sufficient for a disproof, I did not look into his CSSD monotonicity
argument.
I don't know what can be said about the reverse, whether consistency
does or doesn't imply monotonicity. It seems possible to me that a
consistent method that is nonmonotone could exist, though I don't
know of any examples.
-- Richard
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