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

[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.