[EM] How to clone proof your favorite method (whether deterministic or stochastic).

[Forest Simmons]

One little technical detail in step (4) of the method  (see correction in
line below)


On Wed, May 7, 2014 at 3:02 PM, Forest Simmons <fsimmons at pcc.edu> wrote:

> First we need the concept of a covering of a set S of candidates by a
> subset K of S: a subset K of S covers S iff every member of S that is
> beaten (pairwise) by another member of S is also beaten by a member of K.
> A minimal covering is a covering with as few members as possible.
>
> Suppose that your favorite election method FEM is clone dependent.
>
> Let us designate the winner of the clone fixed version of FEM applied to a
> set S of candidates by the notation CloneFix(FEM, S).  To compute this
> winner ...
>
> (1) Among minimal coverings of the set S let K be the one with maximum
> total approval.
>
> (2) If K is a singleton, elect its only member.  Otherwise ...
>
> (3) Let k be the member of K that is elected by (unadorned) FEM restricted
> to K.
>
> (4) Let S' be the set of candidates that have the same pairwise relation
> to the other members of K that k has, i.e. candidate s' is a member of S'
> whenever (for each candidate x in K) candidate s' is beaten by x if and
> only if candidate k is beaten by x.
>

Step (4) should start "Let S' be the subset of S whose members have the
same pairwise relation ..."


> Elect CloneFix(FEM, S') by recursion.
>
> Example: Suppose that FEM is Random Candidate, and that the pairwise
> defeats are given by
>
> A>B>C>A for every member A of the clone set {A1, A2, A3},
>
> and that within the clone set the defeats are  A1>A2>A3>A1.
>
> In this case all three minimal covering sets are of the form  {A, B, C}
> where A is a member of the clone set {A1, A2, A3}.
>
> Suppose that A is the clone with greatest approval.  Then K = {A, B, C},
> and each member of this set has equal likelihood of being candidate k.
>
> If k is B or C, respectively, then B or C is elected.
>
> If k is A, then S' is the clone set {A1, A2, A3}.
>
> In that case, we elect CloneFix(FEM, S') which is A1, A2, or A3 with equal
> likelihood.
>
> In sum the probabilities are 1/9 for each member of {A1, A2. A3},
> and 1/3 for each member of {B,C}.
>
> This distribution of probability respects the clone structure, unlike the
> raw FEM (i.e. random candidate) distribution which would assign a
> probability of 1/5 to each of the five candidates.
>
> Note that in our example the approval scores tuned out to be irrelevant.
>
> Some other way (besides approval scores) of selecting the best minimal
> covering K from among minimal coverings of S could be used.  For example,
> the covering with the least Sum (over the candidates in the covering) of
> the Max (over the candidates in S) pairwise opposition.
>
> If approval scores are used, I suggest singling out the best minimal
> covering as the one with the highest Proportional Approval Voting score
> (PAV score).
>
> If approval scores are not used, and randomization is acceptable, I
> suggest going with the minimal covering that ranks the highest according to
> some random ballot order. (There are various ways of doing this.)
>
> Also note that FEM = Copeland gives the same result as FEM = Random
> Candidate when Copeland ties are broken by random candidate, at least in
> examples where the minimal coverings have three or fewer members, which is
> probably true for almost all public elections (even when the Smith set is
> much larger).
>
> For those that remember it, note the similarity with the "Condorcet
> Lottery."  Only this clone fixed version of random candidate is fully
> monotonic if I am not mistaken.
>
> Does anybody else like this idea?  Does anybody have a FEM method and
> ballot set that they would like to see it applied to?
>
> Forest
>
>
>
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