[EM] Webster vs Bias-Free

MIKE OSSIPOFF nkklrp at hotmail.com
Mon Dec 11 14:13:34 PST 2006

```   Let me define a few terms. S(q) is an allocation's function, seats as a
function of quotas. dS is the amount by which S is above the
1-seat-per-quota line (of the graph that I posted about the other day). Yes,
dS should stand for an infinitessimal change in S, but most computers don't
have the letter "delta", and so I'm saying dS for that finite difference. Of
course if ds is negative, then S is below the one-seat-per-quota line.

The horizontal part of the step function below the 1-seat-per-quota line
is the "low section". The high section is similarly defined. A low section
and its subsequent high section is a "cycle" of the step function.

This morning I looked at Webster's justification for proportionality, it's a
sort of blockwise approach that looks at a whole cycle at a time. In each
cycle, in Webster, dS sums to zero. So whether a cycle is in the
high-population or low-population part of the range, the cycles have no
overal net dS--lif they did, then the varying value of q would cause
different ends of the range to have different overall S/q. So, in that rough
sense, Webster is unbiased.

Well, when I looked at that this morning, I said, "You can't say that about
Bias-Free (the method that I defined last night). Its dS doesn't sum to zero
over each cycle. So Webster's rough unbias argument doesn't work for
Bias-Free. So I hurried to the computer to pretty much retract what I'd said
about Bias-Free (BF) being the genuinely unbiased quota & rounding method.

I spoke too soon. Though Webster's rough argument doesn't work for BF, a
finer and more accurate one does. Here's why I said (correcly) last night
that Webster has a liittle bias:

In a partilcular cycle, consider a point on the low-section, and the
corresponding point on the subsequent high-section. They both have the same
dS. But they don't both have the same q. The piont that's on the high
section of course has a larger q. So The point on the low section's negative
dS/q has a greater absolute value than the positive dS/q of the
corresponding point on the high section.

That's true in every cycle. There's a deficit of dS/q, a net negative sum.
And that's more pronounced at the low-population end of the range, because,
there, q is less, and so the dS/q is more negative. So there's less S/q at
the low end of the range.

As I said, the difference is very slight, and Webster's unbias is very
slight. But it's there.

That's why I devised Bias-Free. Sum dS/q over a cycle, set it equal to zero,
and solve for R, the rounding point. In that way, you make dS/q sum to zero
over the cycle. That's Bias-Free. No nonzero sum of dS/q in any cycle. No
S/q deficit in any cycle.

That's a more accurate and detailed unbias than Webster has.

I showed last night that Webster is very close to BF. Hill is significantly
different from BF. Hill's
rounding points are considerably lower.

Or look at the graphs that I described a few days ago. From the above
argument, or from the graphs, Hill gives significantly more seats per quota
to the smallest states. That is not "equal proportions". Yes, Hill advocates
justify it from how it rounds off. But the meaningful consisderation is how
the states' S/q compare. In Hill, the proportions are _not_ equal. The name
Equal Proportions acknowledges equal S/q as the important thing, but Hill
doesn't deliver.

Webster of course is very close to BF. Webster, BF. and Hamilton are all
adequately unbiased. BF & Hamilton are completely unbiased. But Hamilton is
capricious. As you know, states get very annoyed when they lose a seat for
no good reason. For that reason we want BF or Webster. BF would be better,
but Webster is simpler, more naturally obvious, and has precedent.

Mike Ossipoff

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