If we define A<B whenever B stochastically dominates A, then I think that you have Dominance (since mixtures preserve stochastic dominance) but not Unbounded Utilities (since it’s impossible for a smaller chance of a good outcome to dominate a higher chance of a less-good outcome), right?
If you’re defining the order purely based on stochastic dominance (and no stronger), then ya, I think you’ll have Dominance but not Unbounded Utilities for the reasons you give.
However, I think stochastic dominance is consistent with Unbounded Utilities in general and the expected utility order when choosing between bounded lotteries, since the order based on expected utilities is well-defined and stronger than the stochastic dominance one over bounded lotteries. That is, if A strictly (or weakly) stochastically dominates B, and both are bounded lotteries, then A has a higher expected utility than B (or, at least as high, respectively). So, you could use an order that’s at least as strong as using expected utilities, when they’re well-defined, and also generally at least as strong as stochastic dominance, but that doesn’t imply Dominance for mixtures of infinitely many lotteries. Your specific example with X∞ would prove that Dominance does not hold.
Also, if you’re extending expected utility anyway, you’d probably want to go with something stronger than stochastic dominance, something that also implies sequential dominance or some kind of independence.
If we define A<B whenever B stochastically dominates A, then I think that you have Dominance (since mixtures preserve stochastic dominance) but not Unbounded Utilities (since it’s impossible for a smaller chance of a good outcome to dominate a higher chance of a less-good outcome), right?
If you’re defining the order purely based on stochastic dominance (and no stronger), then ya, I think you’ll have Dominance but not Unbounded Utilities for the reasons you give.
However, I think stochastic dominance is consistent with Unbounded Utilities in general and the expected utility order when choosing between bounded lotteries, since the order based on expected utilities is well-defined and stronger than the stochastic dominance one over bounded lotteries. That is, if A strictly (or weakly) stochastically dominates B, and both are bounded lotteries, then A has a higher expected utility than B (or, at least as high, respectively). So, you could use an order that’s at least as strong as using expected utilities, when they’re well-defined, and also generally at least as strong as stochastic dominance, but that doesn’t imply Dominance for mixtures of infinitely many lotteries. Your specific example with X∞ would prove that Dominance does not hold.
Also, if you’re extending expected utility anyway, you’d probably want to go with something stronger than stochastic dominance, something that also implies sequential dominance or some kind of independence.