[EM] Candidate Strengths and Significance of Defeats

[Juho]

The S(X) function seems to lead to search of the worst path (of  
length that corresponds to the number of iteration cycles) of defeats  
leading to candidate X. This seems to make losing to bad losers worse  
than losing to some good candidates. But the function seems to get  
less interesting when the cycle length reaches the number of candidates.

Juho Laatu


On Oct 26, 2006, at 21:07 , Simmons, Forest wrote:

> It seems to me that when estimating the strength of  candidate X on  
> the basis of a pairwise comparison with candidate Y, we should take  
> into account the strength of candidate Y ;  if  Y  is a weak  
> candidate, then a large margin of victory by  X  over  Y  may not  
> be as significant as a small margin of victory relative to some  
> other stronger candidate.
>
> How can we implement this concept?
>
> I will start with a margins method because it is simpler to  
> explain, but ultimately I will propose a non-margins version that  
> makes better use of the information in the pairwise matrix.
>
> We need a function  S that assigns to each candidate  X  a  
> strength  S(X).
>
> We would like to have the following equilibrium condition satisfied  
> by this strength function:
>
>                         For each candidate  X,   the equation
>
>                                 S(X) = min over Y  of   S(Y)+m(X,Y)
>
>                         must be satisfied,
>
> where  m(X,Y)  is the margin determined by subtracting the number  
> of ballots that rank Y above X from the number that rank X above Y.
>
> This equilibrium condition may be impossible to satisfy without  
> some form of normalization, and even when possible might be  
> difficule to compute.  So I suggest initializing  S to zero and  
> iterating until some stopping criterion is satisfied.
>
> The winner is the candidate  Z  that maximizes (the last iterate  
> of)  S relative to the other candidates.
>
> So after the first iteration,  S(X) is just X's minimum margin  
> against another candidate, which is the same as the opposite of   
> X's maximum margin of defeat.  If we stopped after one iteration,  
> the winner would be the candidate  Z  with the maximum value of  S 
> (Z) which is the same as the candidate with the minimum value of  -S 
> (Z)  which is the same as the minimum value of her maximum margin  
> of defeat.
>
> In other words, if we stopped after one iteration, the method would  
> yield the MinMax(margins) winner.
>
> So this introductory version is a refinement of  MinMax(margins).
>
> To avoid taxing the patience of the reader I will stop here for now.
>
> Forest
>
> <winmail.dat>
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