I'd written an expression for the expected number of seats for a state whose quotient, resulting from division by the final divisor, is between a and b.
I didn't have an expression for such a state's expected population. I'm going to suggest a way of getting that quantity, though I'm not claiming that it's the most efficient or convenient way.This is just the first way that occurs to me at the moment. Find the inverse of S(x), S(x) is the interpolating exponential function that I spoke of in my other post about this, so that inverse will be a logarithmic function. Population, expressed in the unit that I spoke of, as a function of state number. I'll call that P(s), where s is the state number and P is the population. If S(x) is A*e^(-k*x), where A and k are constants, then x = (-1/k)*ln(s/A). Since I'm calling x "P(s) now, then P(s) = (-1/k)*ln(s/A). If P(s) is summed between the limits of S(a) and S(b), and the result divided by (S(b)-S(a)), that (at least it seems to me tonight, as my first impression) is the expectation for the populaton of a state whose population is somewhere between x=a and x=b. Then divide the expected number of seats for that state (for which an expression was written in my previous post about this), by that expected population, to get the expected s/p of a state whose population is somewhere between a and b. Set that equal to 1, and solve for R. That gives the roundoff point for the interval between a and b. That expression, in terms of a and b, when found, is what defines a divisor method, such as this Weighted Webster divisor method. It differs from Hill, Webster, etc., only in that it takes more work to calculate the roundoff point, R. Hill requires a little more calculation than Webster. Weighted Webster requires more than that--starting with finding the constants, A and k, for the interpolating function, S(x). When Warren and I discussed this subject around the beginning of 2007, on EM, I posted a definition of Weighted Webster then too. I don't know how much it resembles what I've posted here. (I haven't been able to find it yet, in the archives). I've only just begun to take a look at this problem, just before it was necessary to leave the computer earlier this evening. And now I've posted what has just occurred to me as a possible solution. Of course this is tentative, because it's only a first look at the problem. It's late, and I should get off the computer, but I just wanted to post this sketch of Weighted Webster. WW is unbiased, if the interpolatng function S(x) is assumed accurate in the interval from a to b, for which the rounding point R, is being calculated. Of course the whole process, including finding the interpolation function's constants, k and A, would have to be done separately for each interval between two integers, on the population number line. Mike Ossipoff
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