Q2: No. Counterexample: Suppose there’s one outcome x such that all lotteries are equally good, except for the lottery than puts probability 1 on x, which is worse than the others.
Nice! This, of course, seems like something we should salvage, by e.g. adding an axiom that if A is strictly preferred to B, there should be a lottery strictly between them.
That proposed axiom to add does not work. Consider the function on lotteries over {x,y,z} that gives utility 1 if z is supported, and otherwise gives utility equality to the probability of x. This function is concave but not continuous, satisfies A1-A5 and the extra axiom I just proposed, and cannot be made continuous.
Oh, no, I made a mistake, this counterexample violates A3. However, the proposed fix still doesn’t work, because you just need a function that is decreasing in probability of x, but does not hit 0, and then jumps to 0 when probability of x is 1.
I meant the conclusions to all be adding to the previous one, so this actually also answers the main question I stated, by violating continuity, but not the main question I care about. I will edit the post to say that I actually care about concavity, even without continuity.
I edited the post to remove the continuity assumption from the main conclusion. However, my guess is that if we get a VNM-like result, we will want to add back in another axiom that gives us continuity,
Q2: No. Counterexample: Suppose there’s one outcome x such that all lotteries are equally good, except for the lottery than puts probability 1 on x, which is worse than the others.
Nice! This, of course, seems like something we should salvage, by e.g. adding an axiom that if A is strictly preferred to B, there should be a lottery strictly between them.
That proposed axiom to add does not work. Consider the function on lotteries over {x,y,z} that gives utility 1 if z is supported, and otherwise gives utility equality to the probability of x. This function is concave but not continuous, satisfies A1-A5 and the extra axiom I just proposed, and cannot be made continuous.
Oh, no, I made a mistake, this counterexample violates A3. However, the proposed fix still doesn’t work, because you just need a function that is decreasing in probability of x, but does not hit 0, and then jumps to 0 when probability of x is 1.
Oh, nvm, that is fine, maybe it works.
I think the way I would rule out my counterexample is by strengthening A3 to if A≻B and B≻C then there is p∈(0,1)…
That does not rule out your counterexample. The condition is never met in your counterexample.
Oh, derp. You’re right.
I meant the conclusions to all be adding to the previous one, so this actually also answers the main question I stated, by violating continuity, but not the main question I care about. I will edit the post to say that I actually care about concavity, even without continuity.
I edited the post to remove the continuity assumption from the main conclusion. However, my guess is that if we get a VNM-like result, we will want to add back in another axiom that gives us continuity,