[Election-Methods] Clarke taxes and group strategies

[Jobst Heitzig]

Hello,

Warren pointed out that my modified taxing scheme involving groups 
cannot possibly solve the group strategy problem completely since one 
Margolis' theorem supposedly showed that such was not possible. I was 
not able unfortunately to find a proof of that "theorem" and can't see 
why it should be true.


However, here's another aspect to Clarke taxes and group strategies. 
Consider again two voters i1,i2 who try to increase each other's net 
outcome by exaggerating their ratings of the alternative winner B over 
the current winner A.

Case 1: Both plan to exaggerate their ratings so much that the winner 
would change from A to B even if one of the two would stick to her 
honest ratings. In this case they face a prisoners' dilemma in which it 
is strictly better to vote honestly than to exaggerate, no matter what 
the other does. Although under certain circumstances, cooperation can 
evolve in a repeated prisoners' dilemma, I doubt that these 
circumstances are given here, so I believe that voters seeking to 
maximize their expected outcome would not cooperate in this situation.

Case 2: In order to avoid the prisoners' dilemma, they plan to 
exaggerate only so much that the winner changes from A to B only if both 
actually exaggerate. Although in this way their gain is much smaller and 
bounded by the total rating difference between A and B, it makes at 
least sure that neither of the two runs the risk of being cheated by the 
other. I guess this case is the more likely one.

But implementing a case 2 group strategy not only depends on very good 
information about the total rating difference between A and B. It also 
becomes more complicated when the cooperating group becomes larger, 
because then they have to take into account that any number of group 
members might decide to cheat the others.

When a randomization scheme is applied, such as Warren's three-group 
scheme or his idea to replace each rating r by some random number 
between 0 and 2r, or my jackknife scheme to make the taxes sum up to 
zero, then such group strategies become even more risky since even in 
the case of perfect information the group cannot be sure that everything 
works as they wish.

So, in all, I am not convinced that group strategies pose a real problem 
in the context of Clarke taxes.

Yours, Jobst


Jobst Heitzig schrieb:
> Hello friends,
> 
> here is another thought about Clarke taxes and similar demand-revealing 
> processes.
> 
> As discussed, these mechanisms make range-voting strategy-free in the 
> sense that no individual voter has an incentive to misrepresent her true 
> ratings. Groups of voters, however, may have incentives to 
> simultaneously exaggerate their ratings in order to elect a different 
> option.
> 
> For example, consider this version of the Clarke taxes: Voter i pays a 
> tax of
> 
>   sum { R(W,k) - R(W(i),k) : k different from i }
> 
> where, as before, R(X,j) is voter j's rating of option X, W is the range 
> voting winner, and W(i) is the range voting winner after removal of 
> ballot i.
> 
> Now suppose a pair of voters (i1, i2) each exaggerate their rating of an 
> option W' different from W by an amount exceeding the total rating 
> difference between W and W':
> 
>   R'(W',i1) := R(W',i1) + epsilon,
>   R'(W',i2) := R(W',i2) + epsilon,
>   epsilon > delta := T(W) - T(W')
> 
> where T(X)=sum{R(X,j):j} is option X's total rating. Then both voters 
> have a net gain (including the utility of the outcome) of
> 
>   epsilon-delta > 0.
> 
> So, already groups of only two voters can easily exploit the mechanism.
> 
> Now, I have the impression that a slight modification of the tax formula 
> may reduce this incentive considerably. Consider this tax:
> 
>   sum { R(W,k) - R(W(i),k)
>         + sum { ( R(W(i),k) - R(W(i,j),k) ) / 2
>               : j different from i and k }
>       : k different from i }
> 
> where W(i,j) is the winner after removal of both i and j. If I'm right, 
> this formula makes it ineffective to misrepresent ratings for both 
> individual voters and pairs of voters.
> 
> Please check this!
> 
> If it works, the thing could be extended to larger groups, perhaps in a 
> way similar to this:
> 
>   sum { R(W,k) - R(W(i),k)
>         + sum { sum { ( R(W(i,D),k) - R(W(i,j,D),k) ) / (2 + |D|)
>                     : D a subset of V\{i,j,k} }
>               : j different from i and k }
>       : k different from i }
> 
> where V is the set of all voters, W(i,D) is the winner after removal of 
> i and all members of D, W(i,j,D) is the winner after removal of i, j, 
> and all members of D, and |D|>=0 is the size of D.
> 
> And if it works, it can probably also used in the zero-sum form 
> suggested by me some days ago.
> 
> Yours, Jobst
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