# Concave Utility Question

This post will just be a concrete math question. I am interested in this question because I have recently come tor reject the independence axiom of VNM, and am thus playing with some weaker versions.

Let $\Omega$ be a finite set of deterministic outcomes. Let $\mathcal{L}$ be the space of all lotteries over these outcomes, and let $\succeq$ be a relation on $\mathcal{L}$. We write $A\sim B$ if A $\succeq$ B and B $\succeq$ A. We write $A\succ B$ if $A\succeq B$ but not $A\sim B$. 

Here are some axioms we can assume about $\succeq$:

A1. For all $A,B\in \mathcal{L}$, either $A\succeq B$ or $B\succeq A$ (or both).

A2. For all $A, B, C \in \mathcal{L}$, if $A\succeq B$, and $B\succeq C$, then $A\succeq C$.

A3. For all $A, B, C \in \mathcal{L}$, if $A\succeq B$, and $B\succeq C$, then there exists a $p\in [0,1]$ such that $B\sim pA+(1-p)C$.

A4. For all $A,B\in \mathcal{L}$, and $p\in [0,1]$ if $A\succeq B$, then $pA+(1-p)B\succeq B$.

A5. For all $A,B\in \mathcal{L}$, and $p\in [0,1]$, if $p>0$ and $B\succeq pA+(1-p)B$, then $B\succeq A$.

Here is one bonus axiom:

B1. For all $A, B, C \in \mathcal{L}$, and $p\in [0,1]$, $A\succeq B$ if and only if $pA+(1-p)C\succeq pB+(1-p)C$.

(Note that B1 is stronger than both A4 and A5)

Finally, here are some conclusions of successively increasing strength:

C1. There exists a function $u:\mathcal{L}\rightarrow[0,1]$ such that $A\succeq B$ if and only if $u(A)\geq u(B)$.

C2. Further, we require $u$ is quasi-concave.

C3. Further, we require $u$ is continuous.

C4. Further, we require $u$ is concave.

C5. Further, we require $u$ is linear.

The standard VNM utility theorem can be thought of as saying A1, A2, A3, and B1 together imply C5.

Here is the main question I am curious about:

Q1: Do A1, A2, A3, A4, and A5 together imply C4? **\[ANSWER: NO\]**

(If no, how can we salvage C4, by adding or changing some axioms?)

Here are some sub-questions that would constitute significant partial progress, and that I think are interesting in their own right:

Q2: Do A1, A2, A3, and A4 together imply C3? **\[ANSWER: NO\]**

Q3: Do C3 and A5 together imply C4? **\[ANSWER: NO\]**

(Feel free to give answers that are only partial progress, and use this space to think out loud or discuss anything else related to weaker versions of VNM.)

EDIT: 

AlexMennen actually resolved the question in the negative as stated, but my curiosity is not resolved, since his argument is violating continuity, and I really care about concavity. My updated main question is now:

Q4: Do A1, A2, A3, A4, and A5 together imply that there exists a concave function $u:\mathcal{L}\rightarrow[0,1]$ such that $A\succeq B$ if and only if $u(A)\geq u(B)$? **\[ANSWER: NO\]**

(i.e. We do not require $u$ to be continuous.)

This modification also implies interest in the subquestion:

**Q5: Do A1, A2, A3, and A4 together imply C2?**

EDIT 2:

Here is another bonus axiom: 

B2.  For all $A,B\in \mathcal{L}$, if $A\succ B$, then there exists some $C\in \mathcal L$ such that $A\succ C\succ B$.

(Really, we don't need to assume $C$ is already in $\mathcal L$. We just need it to be possible to add a $C$, and extend our preferences in a way that satisfies the other axioms, and A3 will imply that such a lottery was already in $\mathcal L$. We might want to replace this with a cleaner axiom later.)

Q6: Do A1, A2, A3, A5, and B2 together imply C4? **\[ANSWER: NO\]**

EDIT 3:

We now have negative answers to everything other than Q5, which I still think is pretty interesting. We could also weaken Q5 to include other axioms, like A5 and B2. Weakening the conclusion doesn't help, since it is easy to get C2 from C1 and A4.

I would still really like some axioms that get us all the way to a concave function, but I doubt there will be any simple ones. Concavity feels like it really needs more structure that does not translate well to a preference relation.