James Payor answers Concave Utility Question

[James Payor]

No on Q4? I think Alex’s counterexample applies to Q4 as well.

(EDIT: Scott points out I’m wrong here, Alex’s counterexample doesn’t apply, and mine violates A5.)

In particular I think A4 and A5 don’t imply anything about the rate of change as we move between lotteries, so we can have movements too sharp to be concave. We only have quasi-concavity.

My version of the counterexample: you have two outcomes $X$ and $\neg X$, we prefer anything with $P\left(X\right)\le \frac{1}{2}$ equally, and we otherwise prefer higher $P\left(X\right)$.

If you give me a corresponding $u\left(p\right)$, it must satisfy $u\left(0\right)=u\left(\frac{1}{2}\right)<u\left(1\right)$, but convexity demands that $u\left(\frac{1}{2}\right)\ge \frac{1}{2}u\left(0\right)+\frac{1}{2}u\left(1\right)$, which in this case means $u\left(0\right)\ge u\left(1\right)$, a contradiction.

[Scott Garrabrant]

Alex’s counterexample as stated is not a counterexample to Q4, since it is in fact concave.

I believe your counterexample violates A5, taking $B=\neg X$, $A=X$, and $p=\frac{1}{2}$.

[James Payor]

Seems right, oops! A5 is here saying that if any part of my $u$ is flat it had better stay flat!

I think I can repair my counterexample but looks like you’ve already found your own.