Raph Frank wrote:
I think there is a slight issue. In PAV, the satisfaction of each
voter is determined by
S(N) = sum(1 + 1/2 + 1/3 + 1/4 + .... + 1/N )
Where N is equal to the number of candidates elected.
An approx function could be created that gives S(N) for non-integer N.
The easiest would be just linear interpolation. However, log(N) is
pretty close, but has a slight offset.
So maybe:
C(N) = log(N) - S(N)
Creal(x) = linear interpolate C(N)
Sreal(x) = log(x) - C(x)
This gives the right answer at the integers and a smooth curve between
them (maybe to much detail here :) ).
Why not have just linear interpolation, like RRV? Say that you've voted
for A and B, and these are elected, and your rating was:
A: 0.9
B: 0.6
Then your satisfication is 0.9 + 0.6/2. Voting approval-style (A: 1, B:
1) would give the familiar 1 + 1/2.
If it's Cardinal-n (limited number of choices), then you would normalize
by maximum (e.g 8 of 10 becomes 0.8) and sum as above.
N would then be set equal to the sum of the ratings divided by the max
rating. E.g. if I rate A as 100 and B as 30, and both are in the
result, then that counts as 1.3 candidates elected, so would take 1.3
terms of the above sum. This would work out as a satisfaction of
around 1.2 with the above formula, which is between a satisfaction of
1.0 for 1 candidate and 1.5 for 2 candidates.
Oh, I see. You're considering a candidate with score 0.3 being not "one
candidate elected at score 0.3", but "a third of a candidate elected".
Let's consider that from basics. In PAV you have a function,
x
f(x) = SUM 1/n
n=1
defined for integers, so that e.g f(2) = 1/1 + 1/2, which fits. Then,
the continuous version would be the integral. The integral of 1/x is
log(x), but that doesn't quite give the same results since 1/1 + 1/2
just sums up two values, whereas the integral takes the entire curve
between 1/1 and 1/2. Thus you'd have to adjust it somehow.
If done correctly, you wouldn't need any sort of interpolation, though.
A simple regression gives f(x) = log(e + 1.773 * (x-1)). The middle
constant there increases very slowly: it's 1.7778 for x=14.
Using 1.773, your 1.3 candidate example would give a satisfaction of
1.17871.
The hard part is finding a way of tallying group support. As mentioned,
first preference wouldn't work, because then you'd get return of Woodall
free-riding: parties would say "just vote for a friend as a write-in, then
the order we give you".
Wouldn't parties want to have as many FPV as possible?
Oops, I must have been thinking about decoy lists. You're right, so the
distortion would be in the other direction. Counting party support by
FPV would discourage personal votes, both in one end (because
independent > A1 > A2 would count for the independent and not at all for
A) and in the other (because A1 > independent > A2 .. would count
completely for A).
In theory, if there was some way to measure vote management, then a
party could be punished for vote management by being assigned a lower
value in a later election, but that is probably to complex.
If there was a method of counting party support, parties with
superproportional support could be deweighted the next time around. I
think it would be too unstable, though.
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