[EM] The Ultimate Lottery Method!
fsimmons at pcc.edu
fsimmons at pcc.edu
Mon Jan 5 12:00:20 PST 2009
I wonder if the solution to the maximization problem could be done iteratively
analogous to the iterative computation of an eigenvector:
Initialize a column matrix V with 1/n in all n rows, where n is the number of
For each ballot vector b, form another column matrix W(b) whose k_th component is
(1/N)*V[k]*b[k]/(the dot product of b and V)
Update V as the sum of all the W(b) over b.
Repeat until V stabilizes.
> Ballots are ratings with a minimum possible rating of zero.
> Ballots with all zero ratings are thrown out as not valid.
> The lottery probabilities are chosen so as to maximize the product of the
> expected ratings over the ballots.
> This method is (1) monotone, (2) clone free, and (3) gives proportional
> probability to stubborn voters. (4) Each ballot has equal weight in determining
> the winning probabilities. (5) Good opportunities for cooperation are not wasted
> by this method. (6) There is little if any incentive for insincere ratings.
> Let's use the Lagrange multiplier method to find the lottery that maximizes the
> product of the expected ballot ratings:
> Let ProdE represent the product of the expected ratings, and let SumP represent
> the sum of the lottery probabilities. Then a necessary condition for maximality
> of ProdE is the stationarity of the expression
> L = log(ProdE) - Lambda*SumP
> as the lottery probabilities are varied subject to the constraint SumP = 1.
> Setting to zero the partial derivative of L with respect to the lottery
> probability p(k) of the k_th alternative (i.e. candidate number k) verifies
> claim (4) in that each ballot b contributes to p(k) precisely the quantity
> where N = Lambda is the number of ballots, b(k) is ballot b's rating of
> alternative k, and E(b) is ballot b's expected rating
> E(b) = p(1)*b(1) + p(2)*b(2) + ...
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