[EM] Cardinal systems: A generalization of STAR

[Kristofer Munsterhjelm]

Oops, I made some mistakes in my previous post... let's try to correct them.

On 19.10.2021 14:10, Kristofer Munsterhjelm wrote:

> Now, there may still be an incentive to strategically misreport. In
> particular, it doesn't solve the three-candidate Burr dilemma. Forest
> suggested that Euclidean normalized cumulative voting would fix the
> problem because the optima would not be found at the corners of the
> polytope, i.e the optimal strategic vote doesn't rate everybody either
> max or min.

It should solve the Burr dilemma with honest voting, now that I think
about it: if the candidates are Excellent, Good, and Bad; and a majority
ranks Bad least, then any normalization that sets the lowest ranked
candidate to 0 will ensure that Bad is evicted from the potential winner
set; and then the Excellent vs Good contest is decided by Condorcet.

So this suggests the following vote guide for STAR with three
candidates: "Give your favorite 100% strength, your loathed candidate 0%
strength, and the in-between candidate a percentage X% so that you're
indifferent between having that candidate elected for sure, and an X%
chance of your favorite winning, (100-X)% chance of your despised
candidate winning".

Perhaps a little too verbose :-)

> But I'm not sure just how to do the normalization. Standard L2
> normalization involves subtracting the mean and dividing by the standard
> deviation; but AFAIK, quadratic voting proposals constrain all weights
> to be positive.
> 
> The former gives a constrained setup like:
>     ||v||_2 = 1
>     pv_A + (1-p)v_C = v_B
>     v_A >= v_B >= v_C
> for an A>B>C voter with probability p on the indifference lottery, and
> output rating values given by the vector v; but this still has free
> variables.
> 
> The latter gives:
>     ||v||_2 = 1
>     v_B = pv_A
>     v_A >= v_B >= v_C nonnegative

That's not just nonnegative weights, but with the additional constraint
that the despised candidate has weight 0, which I forgot to mention.
This idea comes from that if you have to pay for quadratic voting
tokens, then you probably wouldn't spend any on the candidate you like
the least; at least not in the context of a type-two method.

I know it's still somewhat shaky a justification.

-km