[EM] Yee diagrams and Condorcet

[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