[EM] Condorcet Flavored PR Methods

[Forest Simmons]

A few more thoughts on this subject:

(1)  If we interpret the set difference AB-CD as being just A because A is
the only member of AB that is not equivalent to a member of CD, then we
naturally assume that CD-AB would be have to be empty, since each member
of CD is equivalent to some member of AB.

And what is the median of the empty set?

The way around this is to think of sets having several equivalent members
as multi-sets.  So if a=b=c, and d=e=f, then we interpret the difference

{a,b,d,e,f}-{c,e} is the equivalent to the multiset difference

{a,a,d,d,d}-{a,d}, which is equivalent to  {a,d,d}, as well as

{b,e,f}, {c,d,f}, etc.

That solves the problem, but the increase in complexity might not be worth
it.

(2) A simpler way around this problem is to stick with the familiar
interpretation of set differences, and use the max and min of the
symmetric difference as M and m, respectively.  The more I think of it the
more I like it. [The symmetric difference of two sets is the set
difference of their union and their intersection.]

(3) PAV does not satisfy reverse symmetry, essentially because it uses
only the k's and not the j's in the computation below.  If symmetry is not
valued in Condorcet flavored methods, we can also leave out the j's. In
that case candidates ranked equal to M should receive full weight, rather
than half.

(4) If we leave out the j's, and use the M of paragraph (2) above, then
the sequential version becomes very simple.  It gives results similar to
STV because, like STV, it is sequential, and it gives all of the decision
making weight to the high ranking candidates.

(5) The integral I mentioned for use with fractional k+j sums is easy to
do in closed form for half-integral values:

                     For each whole number n

        The integral (for x from 0 to 1) of (1-x^(n+.5))/(1-x)

          has the numerical value given by the finite sum

          -ln(4) + 1/.5 + 1/1.5 + 1/2.5 + ... + 1/(n+.5).

Forest


On Thu, 7 Nov 2002, Forest Simmons wrote:

> Now I'll try to tackle the second question:
>
> On Thu, 7 Nov 2002, Adam Tarr wrote in part:
>
> > - I will admit this is the first election method I've dealt with where I
> > have trouble manipulating small examples.  Here's a very small example that
> > was giving me trouble: say we are electing two candidates out of four.
> > My
> > ballot is: A(>B=C=D).  The pairwise matrix will be 6x6 (with 6 empty
> > slots).  With respect to my ballot, every comparison is equivalent to one
> > of the following:
> >
> > AB vs. CD
> > (k1 + j2 = 1 + 1 = 2?  k2 + j1 = 0 + .5 = .5?)
> >
> > or
> >
> > AB vs. BD
> > (k1 + j2 = .5 + 1 = 1.5?  k2 + j1 = 0 + 1 = 1?)
>
> On this one I get M=A and m=D, so k1=.5, k2=0, and j2=.5+.5=1, as you have
> noted, but I get j1=.5, since B=D=m, instead of the j1=1 that you got.
>
>
> >
> > or
> >
> > AB vs. AC
> >
> > (k1 + j2 = k2 + j1 = 1.5 + .5 = 2)
> >
> > Do the summations I wrote make sense?  The results (except the trivial last
> > one) seem a bit odd.  I was toying with this example to try and measure how
> > equivalent your Condorcet-PR method is to PAV in situations where the
> > voters vote in an approval-like fashion.
>
>
> It does seem odd that AB should beat CD by a larger margin than AB's win
> over BD when B and D are ranked equally.
>
> Perhaps we should interpret the difference AB-CD as just A when B is equal
> in rank to a member of CD.
>
> In other words, if we have two sets H and K, then we should interpret H-K
> as consisting of those members of H that are not equivalent to any member
> of K.
>
> [Two candidates are "equivalent" on a ballot if they have equal rank.]
>
> If we do this, then AB beats CD by the same margin as in the AB vs. BD
> contest, which makes more sense.
>
> Thanks for pointing that out.
>
> Forest
>

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