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Swap Cost Approval

FS
Forest Simmons
Wed, Oct 26, 2022 4:55 PM

Contrary to all expectations it turns out to be easier to define an
objective approval standard in the UD domain than outside it because of the
top/bottom ballot symmetry that Richard Lung has been reminding us of
continually (but without nagging).

Let pt(X) and pb(X) be the respective percentages of the top and bottom
ballot positions occupied by alternative X. [Remember pb is the
abbreviation for lead in the periodic table.]

They can be defined as random ballot rop and bottom (resp) probabilities if
you like, or they can be defined via symmetric completion of the top and
bottom levels of the ballots ... not much difference for practical purposes
... though by rights the latter should be thought of as a decent Q&D
approximation to the former.

The swap cost for the order reversal of X>Y to Y>X of an adjacent pair in a
ranking of the alternatives, is the product pt(X)pb(Y). In other words, the
more popular X, and the more despised Y, the greater the democratic
political cost of sinking X and raising Y in the rankings.

This definition satisfies clone independence in the sense that if X and Y
are replaced with the respective clone sets {X_j} and {Y_k} then the cost
of reversal of X>Y ...

Sum pt(X_j) * Sum pb(Y_k),

turns out to be the same as the sum of the costs of the individual adjacent
pair reversals required to sink all of the clones of X to a level below all
of the clones of Y, one at a time:

Sum pt(X_j)pb(Y_k)

This last sum is over all combinations of j and k. The ordinary algebraic
distributive property for turning products of sums into a sum of products
vouchsafes the equality of the two expressions.

To deal with alternatives ranked equally, just keep in mind that X>Y -> Y>X
can be accomplished in two steps of equal cost, so each of those steps X>Y
-> X=Y, and
X=Y-> Y>X costs half of pt(X)pb(Y).

Now we understand swap cost, we can define the swap cost approval that
ballot B has for candidate X as the sign of the difference between the cost
required to raise X to the top of the ballot strictly above all of the
other alternatives on B, and the cost required to sink X to the bottom of
B, strictly below all of the other alternatives.

So the total approval awarded X by a set of ballots is the number of
ballots for which it is cheaper to raise X to the top than to sink X to the
bottom.

The total disapproval awarded X by a set of ballots is the number of
ballots for which it is cheaper to sink X to the bottom than to raise X to
the top.

If the cost of raising equals the cost of sinking, there is no contribution
to either cost, approval or disapproval.

¿Isn't that "straight out of the book"? (as Paul Erdős used to say)

-Forest

Contrary to all expectations it turns out to be easier to define an objective approval standard in the UD domain than outside it because of the top/bottom ballot symmetry that Richard Lung has been reminding us of continually (but without nagging). Let pt(X) and pb(X) be the respective percentages of the top and bottom ballot positions occupied by alternative X. [Remember pb is the abbreviation for lead in the periodic table.] They can be defined as random ballot rop and bottom (resp) probabilities if you like, or they can be defined via symmetric completion of the top and bottom levels of the ballots ... not much difference for practical purposes ... though by rights the latter should be thought of as a decent Q&D approximation to the former. The swap cost for the order reversal of X>Y to Y>X of an adjacent pair in a ranking of the alternatives, is the product pt(X)pb(Y). In other words, the more popular X, and the more despised Y, the greater the democratic political cost of sinking X and raising Y in the rankings. This definition satisfies clone independence in the sense that if X and Y are replaced with the respective clone sets {X_j} and {Y_k} then the cost of reversal of X>Y ... Sum pt(X_j) * Sum pb(Y_k), turns out to be the same as the sum of the costs of the individual adjacent pair reversals required to sink all of the clones of X to a level below all of the clones of Y, one at a time: Sum pt(X_j)pb(Y_k) This last sum is over all combinations of j and k. The ordinary algebraic distributive property for turning products of sums into a sum of products vouchsafes the equality of the two expressions. To deal with alternatives ranked equally, just keep in mind that X>Y -> Y>X can be accomplished in two steps of equal cost, so each of those steps X>Y -> X=Y, and X=Y-> Y>X costs half of pt(X)pb(Y). Now we understand swap cost, we can define the swap cost approval that ballot B has for candidate X as the sign of the difference between the cost required to raise X to the top of the ballot strictly above all of the other alternatives on B, and the cost required to sink X to the bottom of B, strictly below all of the other alternatives. So the total approval awarded X by a set of ballots is the number of ballots for which it is cheaper to raise X to the top than to sink X to the bottom. The total disapproval awarded X by a set of ballots is the number of ballots for which it is cheaper to sink X to the bottom than to raise X to the top. If the cost of raising equals the cost of sinking, there is no contribution to either cost, approval or disapproval. ¿Isn't that "straight out of the book"? (as Paul Erdős used to say) -Forest