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

stephane.rouillon at sympatico.ca stephane.rouillon at sympatico.ca
Fri Jan 10 07:29:53 PST 2003

```Forest --

I like the mathematical approach...
But maybe I did not follow all.
It seems to me that you will always find the Borda order.
Could you summarize in words why you could not?

Hope I am wrong, show it to me.
Steph.

>
> Date: 2003/01/09 jeu. PM 08:46:46 GMT-05:00
>
> 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
> >
> > ----
> >
> >
>
> ----