[EM] Yee diagrams and Condorcet

[Jameson Quinn]

That proof assumes a euclidean distance metric. With a non-Euclidean one,
the "planes" could have kinks in them. I believe I have heard that the
result still holds with, for instance, a city-block metric, but I cannot
intuitively demonstrate it to myself by imagining volumes and planes as in
this proof.

JQ

2011/7/13 <fsimmons at pcc.edu>

> Actually, any centrally symmetric distribution will do, no matter how many
> dimensions.
>
> The property that we need about central symmetry is this: any plane (or
> hyper-plane in higher
> dimensions) that contains the center of symmetry C will have equal numbers
> of voters on each side of
> the plane..
>
> To see how this guarantees a Condorcet winner, let A and B be candidates
> such that A is nearer to the
> center C than B is.  Let pi be the plane (or hyper-plane in dimensions
> greater than three) through C that
> is perpendicular to the line segment AB.
>
> By the symmetry assumption there are just as many voters on one side of the
> plane pi as on the other
> side.
>
> Now move pi parallel to itself until it bisects the line segment AB.
>
> All of the voters that passed through the plane pi during this move went
> from the B side to the A side of
> the plane.  So A beats B pairwise.
>
> Therefore, if there is a unique candidate that is closer to C than any of
> the rest , that candidate will beat
> each of the others pairwise. Otherwise, all of the candidates sharing the
> minimum distance to C will be
> perfectly tied for CW.
>
>
>
> > From: Bob Richard
> > After looking up some old email threads, it now seems to me that
> > I made
> > a significant mistake in the post below. It is true that the
> > model
> > underlying Yee diagrams guarantees that there will always be a
> > Condorcet
> > winner. But apparently that has nothing to do with the two
> > dimensions
> > being orthogonal. It results from the fact that voters are
> > normally
> > distributed on both dimensions.
> >
> > --Bob Richard
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