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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