MichaelStJules comments on Maximising expected utility follows from self-evident premises?

[MichaelStJules]

If your utility function can take arbitrarily large but finite values, then you can design a prospect/​lottery with infinitely many possible outcomes and infinite expected value, like the St Petersburg paradox. Then you can treat such a prospect/​lottery as if it has infinite actual value, and demonstrate violations of Continuity the same way you would with an outcome with infinite value. This is assuming Continuity applies to arbitrary prospects/​lotteries, including with infinitely many possible outcomes, not just finitely many possible outcomes per prospect/​lottery.

(Infinitary versions of Independence and the Sure-Thing Principle also rule out “unbounded” utility functions. See Russell & Isaacs, 2020.)

[Karthik Tadepalli]

Yes, continuity doesn’t rule out St Petersburg paradoxes. But i don’t see how unbounded utility leads to a contradiction. can you demonstrate it?

[MichaelStJules]

Assume your utility function $u$ is unbounded from above. Pick outcomes $x_1,x_2,...$ such that $u\left(x_n\right)\ge 2^n$. Let your lottery $X$ be $x_n$ with probability $1/2^n$. Note that ${\sum }_{n=1}^{\infty }1/2^n=1$, so the probabilities sum to 1.

Then this lottery has infinite expected utility:

$$E\left[u\right(X\left)\right]=\sum _{n=1}^{\infty }\frac{1}{2^n}u\left(x_n\right)\ge \sum _{n=1}^{\infty }\frac{1}{2^n}{2}^n=\sum _{n=1}^{\infty }1=\infty .$$

Now, consider any two other lotteries $A$ and $B$ with finite expected utility, such that $A\prec B\prec X$. There’s no way to mix $A$ and $X$ probabilistically to be equivalent to $B$, because

$$E\left[u\right(pA+(1-p)X\left)\right]=pE\left[u\right(A\left)\right]+(1-p)E\left[u\right(X\left)\right]=pE\left[u\right(A\left)\right]+\infty =\infty >E\left[u\right(B\left)\right],$$

whenever $p<1$. For $p=1$, $E[pA+(1-p\left)X\right]=E\left[u\right(A\left)\right]<E\left[u\right(B\left)\right]$.

So Continuity is violated.

[Vasco Grilo🔸]

Thanks, Michael! Nitpick, E((X)) in the 3rd line from the bottom should be E(u(X)).

[MichaelStJules]

Thanks, fixed!

[Karthik Tadepalli]

Got it, yes I agree now.