I meant to make this information available to the whole EM list. ---------- Forwarded message ---------- Date: Tue, 3 Dec 2002 14:40:05 -0800 (PST) From: Forest Simmons <[EMAIL PROTECTED]> To: Elisabeth Varin/Stephane Rouillon <[EMAIL PROTECTED]> Subject: Re: [EM] The Copeland/Borda wv hybrid
Rob liked his method because it beat Black (Borda Completed Condorcet) in expected utility, and always seemed to pick the CW when there was one. But I showed by counter-example that Rob's method didn't always pick a member of the Smith set for the ranked ballots. My fix was designed to maintain the spirit of Rob's method along with high expected utility and reverse symmetry criterion satisfaction, while ensuring that the winner would be a member of the Smith set. Because of the high expected utility I believe that this method must have problems similar to those inherited by Black from Borda, namely clone problems, as well as strategic incentives for insincere ranking. However, the Stochastic Matrix idea that I used to "fix" Rob's method has many variations, so perhaps you might find a good way of adapting it ... a way that doesn't suffer from the above mentioned problems. In a nutshell here's the stochastic matrix idea: A stochastic matrix is a matrix of non-negative numbers such that each of its rows is normalized to have a sum of one. For each such matrix P there is at least one normalized vector of non-negative numbers v such that vP=v . Usually such a vector is interpreted as an equilibrium vector in a discrete dynamical system called a Markov Process. Suppose that P is a stochastic matrix such that each off diagonal entry Pij (in row i and column j) is greater than zero if and only if candidate j is preferred over candidate i on more ballots than candidate i is preferred over candidate j. Theorem: Let v be an equilibrium vector satisfying vP=v . Then every non-zero entry in v corresponds to a member of the Smith set. So every such matrix P gives rise to a Condorcet Method. Notice that there is a lot of freedom in the choice of the entries Pij, so there are lots of possible Condorcet Methods here. Generally, you can use the diagonal elements of P to give the method your preferred flavor. If you like Borda, put Borda counts down the main diagonal before normalizing the row sums. If you like Approval, put approval counts down the main diagonal before normalizing the row sums. The off-diagonal positive elements could all be ones (before the normalization) or they could be proportional to the corresponding margins or winning votes. Many other possibilities exist. Any such method can be modified in a natural way to satisfy reverse symmetry by constructing the matrix Q for the order reversed ballots, while letting w be the appropriate normalized equilibrium vector satisfying wQ=w. Then (as a corollary to the above theorem) the positive entries of w correspond to members of the reverse Smith set for the ballots. To satisfy reverse symmetry, take as the winner for the modified method the candidate with the largest positive entry in the difference v-w of the two vectors. [If v-w has no positive entry, then no element is negative either since the sum of its elements must be zero, since it is the difference of two sum normalized vectors. In this unlikely case, all of the candidates are tied. Note that if there is both a Condorcet Winner and a Condorcet Loser the vector v-w will have exactly two non-zero elements, a 1 and a -1 for the winner and loser respectively. For example, let P be the matrix obtained by first putting one in all of the off diagonal entries Pij for which candidate j beats candidate i. Then along the main diagonal let Pjj be the percentage of the electorate that approves of candidate j. Finally, normalize the row sums. For Q first use ones for the off diagonal entries Qij for which candidate i beats candidate j. Then along the main diagonal let Qjj be the percentage of the ballots that list candidate j below the approval cutoff. Finally, normalize the row sums. To find v start with a normalized vector of ones. Multiply it by P and normalize the result. Multiply that result by P and normalize the result. Repeat until no change in the tenth decimal place. [The repeated normalization is not really necessary, but should be done occasionally to counter round off error.] Find w analogously (using Q instead of P). Suppose that the largest member of v-w is the fifth entry. Then candidate number five is the winner of this Approval Flavored Condorcet Method. Notice that the method is completely deterministic, even though it makes use of stochastic matrices, which can be given a probabilistic interpretation if anybody is interested. I hope that this at least partially answers your question. Forest On Tue, 3 Dec 2002, Elisabeth Varin/Stephane Rouillon wrote: > The one you think is the best. > I would like to see what are the advantages... > > Forest Simmons a �crit : > > > Which would you like summarized? [Rob's original or my fix.] > > > > On Mon, 2 Dec 2002, Elisabeth Varin/Stephane Rouillon wrote: > > > > > Could you summarize the method? > > > > > > Steph. > > > > > > Forest Simmons a �crit : > > > > > > > Rob, remember that method that you proposed that turned out to be a kind > > > > of Borda/Copeland hybrid that usually picked the CW, but not always? > > > > > > > > I have a way to fix it. > > > > ---- For more information about this list (subscribe, unsubscribe, FAQ, etc), please see http://www.eskimo.com/~robla/em
