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.
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.
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 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,