[EM] Oops! :-) Bias-Free's fallacy
Michael Ossipoff
mikeo2106 at msn.com
Tue Jan 30 08:10:57 PST 2007
I said that Webster is large-biased because, though its step function is
over the 1-seat-per-quota line as much as it is below it, the states in the
low section are more hurt by being lower, because their lower q means that a
given drop in s results in a bigger drop in s/q. What I missed was the fact
that what happens in a state has to be weighted according to its size,
because, since it has fewer people, you're less likely to be in it.
With Webster's step function symmetrically above & below the 1 seat per
quota line, the cycle has one seat per quota. Every cycle has the same
number of seats per person. Your expectation of representation is the same
no matter which cycle you're in.
What brought this to my attention was when I approached BF in a different
way, summing separately the seats and quoas in a cycle, and setting them
equal, and solving for R. R turned out to be (a+b)/2.
I propose Weighted-Webster (WW), in which the integrand is weighted with a
function that approximates the probability density. Just like Weighted
Bias-Free, except that it's Webster now.
I propose that the weighting function be Bexp(-Aq), with A & B positive
constants. But if that results in an antiderivative without an exact
solution, then I'll use B/(q+A).
Mike Ossipoff
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