I find it quite plausible that would ensure this (60-80% credence?), but it’s not obvious. In particular the way you normally prove that there’s a utility function is that you construct it, and you use the continuity axiom to do this.
Without the continuity axiom, maybe you can prove some representation with something satisfying the axioms for the reals … but it looks hard.
At the very minimum, you need to have a distributivity law for lotteries. If 50%(50%A+50%B)+50%(50%A+50%C) is not defined to be the same thing as 50%A+25%B+25%C, then it’s easy to find counter-examples...
Yes, I think in the classical conception of lotteries these are regarded as the same. You could reject that, but it seems like it would be similar to how some people think that (A, when B was on the table) is a different outcome from (A, when B was not on the table), and so may be assigned a different utility.
However this seems to require a strengthening of the independence axiom, so that the implication goes in the opposite direction in some cases (see axiom 6, page 71).
So if we ditch continuity, do the two independence axioms ensure we have utility functions (possibly non-standard ones)?
I find it quite plausible that would ensure this (60-80% credence?), but it’s not obvious. In particular the way you normally prove that there’s a utility function is that you construct it, and you use the continuity axiom to do this.
Without the continuity axiom, maybe you can prove some representation with something satisfying the axioms for the reals … but it looks hard.
At the very minimum, you need to have a distributivity law for lotteries. If 50%(50%A+50%B)+50%(50%A+50%C) is not defined to be the same thing as 50%A+25%B+25%C, then it’s easy to find counter-examples...
Yes, I think in the classical conception of lotteries these are regarded as the same. You could reject that, but it seems like it would be similar to how some people think that (A, when B was on the table) is a different outcome from (A, when B was not on the table), and so may be assigned a different utility.
I’ll try building counter-examples first, then...
It seems that you do have that result, see here: http://link.springer.com/article/10.1007%2FBF01766393
However this seems to require a strengthening of the independence axiom, so that the implication goes in the opposite direction in some cases (see axiom 6, page 71).
incidentally, < independence isn’t enough on its own either.
Pick a standard expected utility situation, with A<B for two pure outcomes A and B. Then arbitrarily set A=B.
< Independence is of the type “if smeu then blah”. A=B is not a smeu, and will never appear as a blah.
EDIT: this violates transitivity, alas.