[EM] Invariance to affine transformations, keeping cardinal honest

[Kristofer Munsterhjelm]

Good news - I'm almost done with my non-EM big project, and so I should 
be able to create some strategy simulations soon to investigate which 
Friendly variants are good :-)

But here's a thought that I've talked about earlier, that came up in a 
private mail response.

Suppose that von Neumann-Morgenstern utilities are the best we can do as 
far as interpersonal comparisons go.

Then we can imagine that every cardinal method has an idealized 
(Platonic, so to speak) form that operates not on ballots with discrete 
ratings, but on real ratings. E.g. the idealized version of Range 
involves giving each candidate a rating on [0...1), and the candidate 
with the highest mean wins.

Then vNM utilities are unique up to some affine scaling. Let u(v, X) be 
voter v's utility should X be elected, on an objective scale that we 
can't access due to interpersonal comparison problems. Then if v is 
asked to rate X, he would rate X as r(v, X) = a_v * u(v, X) + b_v, where 
a_v and b_v are the positive scaling constants for that voter.

For something like Range, suppose that when v goes to vote, v chooses 
some a and b so that every rating fits on the scale, then submits that vote.

For an idealized method, we can then propose the following invariance 
property:

- Affine invariance: The outcome of the election should not depend on 
the particular a and b constants chosen by the voter.

Methods that pass affine invariance plus anonymity and symmetry satisfy 
a type of one-man-one-vote because expanding your scale is equivalent to 
making the a constant larger. Hence such a method, if it accepts 
ratings, doesn't have to limit itself to a bounded scale, though the 
user interface is probably more pleasant if it does.

Furthermore, these methods have another property that there exists one 
honest ballot, which solves rb-j's objection of how much a voter should 
vote his second favorite if he's honest. Every such method that I know 
of fails IIA, but that's better than the sort of "I pass IIA but the 
outcome still depends on who's in the running" thing that Range does.

What methods pass this criterion?

Well, ranked methods clearly do. No amount of positive affine scaling 
can turn r(v, X) > r(v, Y) into r(v, X) < r(v, Y).

In addition, there's this "double normalized Range": the voter has to 
fill in a ballot where at least one candidate is rated max and at least 
one is rated min. The rules are otherwise as Range.

Each forced rating gets rid of one degree of freedom, so forcing two 
candidates fixes both a and b. The intermediate candidates' ratings can 
be found as follows:

Suppose Best is the best candidate, and Worst is the worst, and the 
scale is from 1 being best to 0 being worst. Then for any other 
candidate X, rate X at p if you're indifferent between getting Best with 
probability p and Worst with probability 1-p, and getting X for sure.

Incentivizing honesty here would be hard, though.

(Perhaps a Ranked Pairs-like algorithm is possible for triples rather 
than pairs.)


I also had an idea that perhaps we could find the best VSE affine 
invariant method by something like my "optimum strategy" search, but 
over the space of cardinal ballots instead of ordinal ones, for a small 
number of candidates. In particular, the results for three candidates 
could be interesting.

We couldn't use the idealized method because there are infinite reals on 
a bounded interval. But the observation about double normalized Range 
above suggests that we could replace the ballot format for three 
candidates with:
	(Who's min), (Who's max), (the p value for the third candidate),

or really, one of:
	(A:  X, B:  0, C: 10)
	(A:  X, B: 10, C:  0)
	(A:  0, B:  X, C: 10)
	(A: 10, B:  X, C:  0)
	(A:  0, B: 10, C:  X)
	(A: 10, B:  0, C:  X)

for a scale of 0..10, where X is an integer between 0 and 10 inclusive. 
This gives each voter 60 different ballots to choose between (contrasted 
with 6 per voter for ordinal).

So for e.g. 10 voters, 3 candidates, there are about (60+10-1) choose 10 
= 3e11 elections. There are 3003 ordinal ones. Perhaps a rating scale of 
0..10 is too high for an integer programming search. The trick would be 
to find a quantization level that's high enough that the search is 
feasible, yet low enough that it's possible to interpolate to an 
idealized method, somewhat like I did with fpA-fpC.


Finally, I'd say it is perhaps possible to go beyond vNM, but I'm not 
entirely sure how to do it, much less how to do it in a 
strategy-resistant manner. On the other hand, there may be limitations 
beyond vNM itself: e.g. it might be hard to tell who would be a better 
winner between Stalin and Ivan the Terrible, because they're both so 
thoroughly awful as to saturate the scale.