DavidHolmes comments on Toy model piece #1: Partial preferences revisited

[DavidHolmes]

(actually, my formula doubles the numbers you gave)

Are you sure? Suppose we take $W=W_1\bigsqcup W_2$ with $W_1=\{A,B,C,D\}$, $W_2=\{X,Y,Z\}$, then $n_1=3$, so the values for $W_1$ should be $-3,-1,1,3$ as I gave them. And similarly for $W_2$, giving values $-2,0,2$. Or else I have mis-understood your definition?

I’d simply see that as two separate partial preferences

Just to be clear, by “separate partial preference” you mean a separate preorder, on a set of objects which may or may not have some overlap with the objects we considered so far? Then somehow the work is just postponed to the point where we try to combine partial preferences?

EDIT (in reply to your edit): I guess e.g. keeping conditions 1,2,3 the same and instead minimising
$g\left(G\right)={\sum }_{w\leftarrow w^{\prime }}{\lambda }_{w\leftarrow w^{\prime }}\left(U\right(w^{\prime })-U(w){)}^2,$
where ${\lambda }_{w\leftarrow w^{\prime }}\in R_{>0}$ is proportion to the reciprocal of the strength of the preference? Of course there are lots of variants on this!

[Stuart_Armstrong]

Yep, sorry, I saw −3, −2, −1, etc… and concluded you weren’t doing the 2 jumps; my bad!

Then somehow the work is just postponed to the point where we try to combine partial preferences?

Yes. But unless we have other partial preferences or meta-preferences, then the only resonable way of combining them is just to add them, after weighting.

I like your reciprocal weighting formula. It seems to have good properties.