[Election-Methods] Clarke taxes and group strategies
Jobst Heitzig
heitzig-j at web.de
Tue Jul 29 15:35:48 PDT 2008
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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