[EM] Weighted Webster,contd.

Michael Ossipoff email9648742 at gmail.com
Mon Jul 2 23:38:19 PDT 2012

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

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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