[EM] Geometric Condorcet cycle example, improved
fsimmons at pcc.edu
fsimmons at pcc.edu
Wed Dec 30 14:31:30 PST 2009
Actually it's the symmetry property of metrics (d(p,q)=d(q,p)), not the
triangle inequality, that guarantees the absence of a Condorcet cycle in the
triangle case:
Suppose that A, B, and C are the only candidates, and that voters are
concentrated very near their favorites.
Assume that preferences are determined by distances, i.e. a voter prefers a
nearer candidate over a more distant one.
Without loss in generality, assume that d(A,B) is the greatest of the three
distances.
Then we have the preference profile:
x: A>C>B
y:B>C>A
z: C >????
Suppose, by way of contradiction, that we have a Condorcet cycle. The without
loss in generality suppose that the cycle is A beats B beats C beats A.
The first step in this beat path ( A beats B) implies (x+z)>y, while the
second step (B beats C) implies y>(x+z).
We got into this contradiction by assuming the existence of a cycle. So that
assumption is untenable.
We have made tacit use of the symmetry property of distance by assuming that the
longest side of the triangle was agreed upon by both A and B, i.e. A thought B
was the most distant candidate, and B considered A to be the most distant
candidate.
To show that this property is essential, suppose that we use taxicab "distance"
is a grid of one way streets, and that A, B, and C are located three of the
corners [say (0,0), (1,0), and (0,1)] of a block about which the one way
streets have a counterclockwise orientation.
Then (by taxi) it is quicker to go from A to B than from A to C, from B to C
than from B to A, and from C to A than from C to B. The preference profile has
to be of the form...
x: A>B>C
y: B>C>A
z: C>A>B
which leads to a Condorcet cycle whenever no candidate has a majority.
More information about the Election-Methods
mailing list