Anyway, the above decision process satisfies ⇐ independence (but not < independence).
To see that the decision process satisfies ⇐ independence, note that f(pa+(1-p)b)=max(f(a),f(b)). So if f(a) ⇐ f(b), then f(pa+(1-p)c)=max(f(a),f(c)) ⇐ max(f(b),f(c)) = f(pb+(1-p)c)).
Yes, the two versions of independence are equivalent given the other axioms. If you don’t have continuity to make them equivalent, I think the natural thing to do is to ask for both types of independence.
(The intuition is: independence feels like it should demand all of these things. Normally it’s not stated like that because it’s clunky to add extra statements when one is enough.)
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).
Take a reasoner who can make pre-commitments (or a UDT/TDT type). This reasoner, in effect, only has to make a single decision for all time.
Let A, B, C… be pure outcomes, a, b, c,… be lotteries. Then define the following pseudo-utility function f:
f(a) = 1 if the outcome A appears with non-zero probability in a, f(a) = 0 otherwise. The decision maker will use f to rank options.
This clearly satisfies completeness and transitivity (because it uses a numerical scale). And then… It gets tricky. I’ve seen independence written both in a < form and a ⇐ form (see http://en.wikipedia.org/wiki/Expected_utility_hypothesis vs http://en.wikipedia.org/wiki/Von_Neumann%E2%80%93Morgenstern_utility_theorem ). I have a strong hunch that the two versions are equivalent, given the other axioms.
Anyway, the above decision process satisfies ⇐ independence (but not < independence).
To see that the decision process satisfies ⇐ independence, note that f(pa+(1-p)b)=max(f(a),f(b)). So if f(a) ⇐ f(b), then f(pa+(1-p)c)=max(f(a),f(c)) ⇐ max(f(b),f(c)) = f(pb+(1-p)c)).
Yes, the two versions of independence are equivalent given the other axioms. If you don’t have continuity to make them equivalent, I think the natural thing to do is to ask for both types of independence.
(The intuition is: independence feels like it should demand all of these things. Normally it’s not stated like that because it’s clunky to add extra statements when one is enough.)
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.