[EM] Joe: Could you explain the question?
Joseph Malkevitch
malkevitch at york.cuny.edu
Tue Jan 23 12:38:58 PST 2007
Dear Mike,
It turns out that Huntington showed that corresponding to the divisor
methods one can think of these methods as "rank index" methods, or
what I like to call table methods. One constructs a table from the
original population where the table involved depends on the method.
(This is the way the Census Bureau describes Huntington-Hill on its
web page.) For Webster the table is constructed by dividing the
populations by a + 1/2 where a = number of states, starting with a =
0 for the first row, a= 1 for the second, etc. . So one obtains rows
for the table by dividing by 1/2, 3/2, 5/2, etc. (In this way of
thinking one can easily give each state one seat, or whatever, at the
start.) However, since all that matters for assigning h seats is to
assign the seats to the entries in the table ordered by size (after
we give each state one seat if required) we can divide instead by 1,
3, 5, etc. rather than the numbers above. This is the way that
Webster (St. Lague) is described in Europe. Viewed in this framework
the presence or lack of the factor 1/e makes no difference. One would
get the same apportionment either way. This means there are are
"equivalence classes" of formulas which for the rank index point of
view one can use and which will give the same apportionment.
That is what I was trying to get at.
Huntington called his method the Method of Equal Proportions which is
a "clever" name but what it really does is optimize some measure of
fairness for transfer between pairs of states as well as some
"axiom," while Webster optimizes a different measure as well as some
axioms. When it comes to bias, there are many different kinds of bias
that one might talk about and differing ways to measure that bias.
Calling something "bias free" does not shed extra light without more
insights.
Regards,
Joe
On Jan 23, 2007, at 8:37 AM, Michael Ossipoff wrote:
>
> Joe--
>
> You asked what is the role of 1/e in the formula for Bias-Free's
> round-off
> point. I tried to answer by saying that ((b^b)/(a^a))(1/e) is just
> the
> result I got when looking for the round-off point that would make
> cycles'
> s/q = 1. That's all I can say about where the formula came from.
>
> What was the meanng of your question? In what sense was "role"
> intended?
>
> Unless I made an error (and I probably didn't make an error), the
> formula
> for Weighted Bias-Free doesn't contain "e". Weighted Bias-Free is
> gotten by
> a/q-1 and b/q-1 by a weight that approximates the frequency density of
> states at a particular q value. I approximate that density function
> by:
>
> B/(q+A), which has the same general behavior as the distribution
> function,
> though of course it will be a rough approximation.
>
> Mike Ossipoff
>
>
> ----
> election-methods mailing list - see http://electorama.com/em for
> list info
------------------------------------------------
Joseph Malkevitch
Department of Mathematics
York College (CUNY)
Jamaica, New York 11451
Phone: 718-262-2551 (Voicemail available)
My new email is:
malkevitch at york.cuny.edu
web page:
http://www.york.cuny.edu/~malk
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