[EM] Free riding

[Kristofer Munsterhjelm]

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.