# [EM] The voter median candidate generalized to multidimensional issue spaces.

Forest Simmons fsimmons at pcc.edu
Thu Jan 9 17:46:46 PST 2003

```Example:

4 A>B>C
5 B>C>A
2 C>A>B

We start with a matrix whose rows represent the ballots.  It has four rows
of [3,2,1], five rows of [1,3,2], and two rows of [2,1,3].

The average of all eleven rows is the center of gravity vector
V0=[21,25,20]/11.  We can see from this that B is the Borda winner and C
the Borda loser.

Subtracting this V0 from every row we get a matrix A with four rows of
[12,-3,-9]/11, five rows of [-10,8,2]/11, and two rows of [1,-14,13]/11.

The product of the transpose of A with A is a three by three matrix whose
rows are [98,-52,-46]/11, [-52,68,-16]/11, and [-46,-16,62]/11.

The normalized eigenvectors corresponding to positive eigenvalues are
v1=[.814...,-.462...,-.3518...] and v2=[.06369...,.6731...,-.7367...], so
the issue space is two dimensional.

The vector A*v1 has four entries of approximately 1.3, five of -1.14, and
two of .246, so the median is c1=.246 .

The vector A*v2 has four entries of .489, five of .298, and two of -1.72,
so the median is c2=.298.

Then X=V0+c1*V1+c2*V2 is approximately [2.1,2.4,1.5].

Since 2.4 > 2.1 > 1.5 the winning order is B>A>C, in this case the same as
as the Borda order, rather than the Ranked Pairs order of A>B>C.

Forest

On Wed, 8 Jan 2003, Forest Simmons wrote:

> Suppose that (in a certain election) candidate X is preferred over any
> other candidate Y on any issue Z by some majority (depending on Y and Z).
>
> Such a candidate would seem like a logical choice for winner of the
> election, if there were such a candidate.
>
> How could we locate such a candidate if there happened to be one?
>
> Otherwise, how might one find the candidate closest to this ideal?
>
> Here's an idea along these lines:
>
> Form the matrix whose entry in row i and column j is the rank or rating of
> candidate j by voter i.
>
> Let V0 be the average of all the row vectors of the resulting matrix.
> Subtract V0 from all of the rows of this matrix, and call this new matrix
> A.  Let B be the transpose of A.
>
> Let V1, V2, etc. be normalized eigenvectors of the matrix product B*A, in
> descending order of the magnitudes of the eigenvalues (until the
> eigenvalues are too small to represent anything other than roundoff
> error).
>
> V0 is the center of gravity vector, and the other V's are the
> principal axes of rotation, i.e. those axes (through the center of
> gravity) about which the rigid body (with unit masses positioned by the
> row vectors) can rotate without wobble.
>
> Statistically, the eigenvectors give the uncorrelated directions in voter
> space (the row space of A offset by the cg vector V0). They may be thought
> of as the voter space shadows of the decorrelated issues, in order of
> importance to the voters.
>
> Sort (by numerical value) the components of the vector given by the
> product A*V1 to find the median value c1.
>
> Sort (by numerical value) the components of the vector given by the
> product A*V2 to find the median value c2.
>
> Etc.
>
>
> Let X be the vector V0+c1*V1+c2*V2+...
>
> Then X is the voter space position of the ideal candidate we are looking
> for. In other words, if voter space were identical to issue space, then a
> candidate positioned at X would be unbeaten on any of the (uncorrelated)
> issues by any of the other candidates.
>
> What do we do in the almost sure case that no candidate vector (image in
> voter space) points in the direction of X ?
>
> Here are two suggestions out of many possibilities:
>
> (1) Find the candidate vector (image) that has the largest dot product
> with X.
>
> (2) Look at the components of X, and award the win to the candidate
> corresponding to the largest component.  In other words, let the
> components of X order the candidates from winner to loser.
>
> After experimenting I prefer the second approach for reasons which I will
> explain if this becomes a hot topic.
>
> Remark.  I'm sure that this method doesn't satisfy the Local IIAC, because
> each candidate, no matter how marginal in support contributes something to
> the issue space profile.  In other words, even the losing candidates help
> to plumb the depths of issue space.
>
> Any method that makes intentional, intelligent use of the issue space
> information inherent in the distribution of votes in voter space should
> not be concerned with satisfying any version of the IIAC, because the
> voter response to each candidate helps outline the shape of issue space.
>
> Sometimes I think that the best way to look at Arrow's Paradox is that the
> IIAC was just one of those "good" ideas that didn't pan out. It was based
> on a lack of imagination; nobody thought of using the profiles of the
> losing candidates to help determine the relative positions of stronger
> candidates in issue space.
>
> On the other hand, intelligent issue space methods should automatically
> get rid of clone problems, because clones just give redundant information
> about the shape of issue space.  The singular value decomposition
> decorrelates whatever residual information there might be when the
> redundancy is factored out, just as it finds uncorrelated linear
> combinations of random variables in taxonomy problems associated with
> measuring and recording various dimensions and features of related plants,
> for example.
>
> [In that context the SVD brings out the distinguishing features for
> classification of related species. Similar uses are made in image
> processing. If nostril width is directly proportional to ear lobe
> thickness, the SVD will find that out automatically and count these two
> measurements as only one dimension.]
>
> Thanks for your patience in wading through these rather technical musings.
>
> Forest
>
> ----