You can also think of A5 in terms of its contrapositive: For all A,B∈L, if A≻B, then for all 0<p≤1A≻pA+(1−p)B
This is basically just the strict version of A4. I probably should have written it that way instead. I wanted to use ⪰ instead of ≻, because it is closer to the base definition, but that is not how I was natively thinking about it, and I probably should have written it the way I think about it.
I haven’t actually thought about whether A5 implies A4 though. It is plausible that it does. (together with A1-A3, or some other simple axioms,)
When A≻B, we get A4 from A5, so it suffices to replace A4 with the special case that A∼B. If A∼B, and A,B≻X, a mixture of A and B, then all we need to do is have any Y such that A≻Y≻X, then we can get Y′ between A and X by A3, and then X will also be a mixture of Y′ and B, contradicting A5, since B≻Y′.
A1,A2,A3,A5 do not imply A4 directly, because you can have the function that assigns utility 0 to a fair coin flip between two options, and utility 1 to everything else. However, I suspect when we add the right axiom to imply continuity, I think that will be sufficient to also allow us to remove A4, and only have A5.
[Edit: yeah nevermind I have the inequality backwards]
A5 seems too strong?
Consider lotteries A and B, and a mixture X=pA+(1−p)B in between. Applying A5 twice gives:
If u(X)≥u(A) then u(B)≥u(A)
If u(X)≥u(B) then u(B)≥u(A)
So if u(X)≥u(A) and u(X)≥u(B) then u(A)=u(B)?
Either I’m confused or A5 is a stricter condition than concavity.
You have the inequality backwards. You can’t apply A5 when the mixture is better than the endpoint, only when the mixture is worse than the endpoint.
Got it, thanks!
You can also think of A5 in terms of its contrapositive: For all A,B∈L, if A≻B, then for all 0<p≤1A≻pA+(1−p)B
This is basically just the strict version of A4. I probably should have written it that way instead. I wanted to use ⪰ instead of ≻, because it is closer to the base definition, but that is not how I was natively thinking about it, and I probably should have written it the way I think about it.
I haven’t actually thought about whether A5 implies A4 though. It is plausible that it does. (together with A1-A3, or some other simple axioms,)
When A≻B, we get A4 from A5, so it suffices to replace A4 with the special case that A∼B. If A∼B, and A,B≻X, a mixture of A and B, then all we need to do is have any Y such that A≻Y≻X, then we can get Y′ between A and X by A3, and then X will also be a mixture of Y′ and B, contradicting A5, since B≻Y′.
A1,A2,A3,A5 do not imply A4 directly, because you can have the function that assigns utility 0 to a fair coin flip between two options, and utility 1 to everything else. However, I suspect when we add the right axiom to imply continuity, I think that will be sufficient to also allow us to remove A4, and only have A5.
The way I understand A4 is that it says “if moving by Δ is good, then moving by any fraction λΔ is also good”.
And A5 says “if moving by Δ is good, then moving by any multiple nΔ is also good”, which is much stronger.
Your understanding of A4 is right. In A5, “good” should be replaced with “bad.”
(and everywhere you say “good” and “bad”, they are the non-strict versions of the words)
yep!
Okay, I now think A5 implies: “if moving by Δ is good, then moving by any negative multiple −nΔ is bad”. Which checks out to me re concavity.