The examples and results in your post are very interesting and surprising. Thanks for writing this.
I’m inclined to reject the dominance axioms you’ve assumed, at least for mixtures of infinitely many lotteries. I think stochastic dominance is a more fundamental axiom, avoids inconsistency and doesn’t give any obviously wrong answers on finite payoff lotteries (even mixtures of infinitely many lotteries or outcomes with intuitively infinite expected value, including St. Petersburg and Pasadena). See Christian Tarsney’s “Exceeding Expectations: Stochastic Dominance as a General Decision Theory” (skip sections 5-7, which are about applying it to models with background uncertainty; the definitions and initial motivation are in section 3, quoted in my reply, and various choice situations are considered in section 9). I think the independence axiom and your dominance axioms restricted to mixtures of finitely many lotteries (almost?) follow from versions of stochastic dominance extended to sequences of lotteries and multiple decisions, based on Dutch book/money pump arguments. See Johan E. Gustafsson’s “The Sequential Dominance Argument for the Independence Axiom of Expected Utility Theory”.
Without good enough ways to distinguish different definite infinite payoffs (i.e. first solving infinite ethics, although not all solutions to infinite ethics would necessarily fit well with stochastic dominance), stochastic dominance can give unintuitive results when comparing lotteries with intuitively different infinite payoffs. It could also allow for incomparability, but a kind of incomparability that doesn’t necessarily break everything (well, not (much) more than infinite ethics already breaks things), since it just means multiple options can be permissible within a given set of options, even if they can’t be treated as if they were equivalent (when mixing with other lotteries, considering background noise, etc.) across all sets of options.
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.
EDIT: p.37-38 in Goodsell, 2023 gives a better proposal, which is to clip/truncate the utilities into the range [−t,t] and compare the expected clipped utilities in the limit as t→∞. This will still suffer from St Petersburg lottery problems, though.
Here’s an order that’s as strong as both expected utility and stochastic dominance, and overall seems promising to me:
tl;dr: For lotteries with finite utility payoffs (but possibly unbounded utility payoffs and infinite expected utility), we can take expectations through any subset with finite and well-defined expected utility, and then compare the resulting lotteries with stochastic dominance. We just need to find any pair of well-behaved “expected utility collapses” for which one lottery stochastically dominates the other. Allowing expected utility collapses over the infinite expected utilities can lead to A<A, so I rule that out.
In practice, you might just take one expectation over everything but the top X% and bottom Y% of each lottery, and compare those lotteries with stochastic dominance, for different values of X and Y. This allows you to focus on the tails of heavy-tailed distributions.
For a lottery X, a utility function U, and a countable (possibly finite and possibly empty) set of mutually exclusive non-empty measurable subsets of the measure space, P={Q1,Q2,…,Qn} (or basically a set of binary random variables whose sum is at most 1) and letting PC=(∪Q∈PQ)C be the complement of their union (so, for their indicator binary random variables, 1PC=1−∑Q∈P1Q), the expected utility collapse of X over P is:
XP=E[U(X)|Q] if Q, for Q∈P, and XP=X|PC, otherwise.
Or, in lottery notation, letting L(c) be the constant lottery with constant value c,
XP=P(PC)X|PC+∑Q∈PP(Q)L(E[U(X)|Q]).
In other words, we replace probability subsets of X with its expected utility over those subsets.
If furthermore, E[U(X)|Q] is well-defined and finite for each Q∈P, we call the expected utility collapse well-behaved.
Then, we define the order over lotteries as follows:
A<B if there exists well-behaved expected utility collapses AP1 and BP2 of A and B respectively such that BP2 strictly stochastically dominates AP1.
If you allow infinite actual utilities (including possibly infinities of different magnitudes), you could add a disjunctive or overriding condition to handle comparisons with those.
Option O first-order stochastically dominates option P if and only if
For any payoff x, the probability that O yields a payoff at least as good as x is equal to or greater than the probability that P yields a payoff at least as good as x, and
For some payoff x, the probability that O yields a payoff at least as good as x is strictly greater than the probability that P yields a payoff at least as good as x.
(...)
Stochastic dominance is a generalization of the familiar statewise dominance relation that holds between O and P whenever O yields at least as good a payoff as P in every possible state, and a strictly better payoff in some state. To illustrate: Suppose that I am going to flip a fair coin, and I offer you a choice of two tickets. The Heads ticket will pay $1 for heads and nothing for tails, while the Tails ticket will pay $2 for tails and nothing for heads. The Tails ticket does not statewise dominate the Heads ticket because, if the coin lands Heads, the Heads ticket yields a better payoff. But the Tails ticket does stochastically dominate the Heads ticket. There are three possible payoffs: winning $0, winning $1, and winning $2. The two tickets offer the same probability of a payoff at least as good as $0, namely 1. And they offer the same probability of an payoff at least as good as $1, namely 0.5. But the Tails ticket offers a greater probability of a payoff at least as good as $2, namely 0.5 rather than 0. Stochastic dominance is generally seen as giving a necessary condition for rational choice:
Stochastic Dominance Requirement (SDR) An option O is rationally permissible in situation S only if it is not stochastically dominated by any other option in S.
This principle is on a strong a priori footing. Various formal arguments can be made in its favor. For instance, if O stochastically dominates P, then O can be made to statewise dominate P by an appropriate permutation of equiprobable states in a sufficiently finegrained partition of the state space (Easwaran, 2014; Bader, 2018).
The examples and results in your post are very interesting and surprising. Thanks for writing this.
I’m inclined to reject the dominance axioms you’ve assumed, at least for mixtures of infinitely many lotteries. I think stochastic dominance is a more fundamental axiom, avoids inconsistency and doesn’t give any obviously wrong answers on finite payoff lotteries (even mixtures of infinitely many lotteries or outcomes with intuitively infinite expected value, including St. Petersburg and Pasadena). See Christian Tarsney’s “Exceeding Expectations: Stochastic Dominance as a General Decision Theory” (skip sections 5-7, which are about applying it to models with background uncertainty; the definitions and initial motivation are in section 3, quoted in my reply, and various choice situations are considered in section 9). I think the independence axiom and your dominance axioms restricted to mixtures of finitely many lotteries (almost?) follow from versions of stochastic dominance extended to sequences of lotteries and multiple decisions, based on Dutch book/money pump arguments. See Johan E. Gustafsson’s “The Sequential Dominance Argument for the Independence Axiom of Expected Utility Theory”.
Without good enough ways to distinguish different definite infinite payoffs (i.e. first solving infinite ethics, although not all solutions to infinite ethics would necessarily fit well with stochastic dominance), stochastic dominance can give unintuitive results when comparing lotteries with intuitively different infinite payoffs. It could also allow for incomparability, but a kind of incomparability that doesn’t necessarily break everything (well, not (much) more than infinite ethics already breaks things), since it just means multiple options can be permissible within a given set of options, even if they can’t be treated as if they were equivalent (when mixing with other lotteries, considering background noise, etc.) across all sets of options.
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.
EDIT: p.37-38 in Goodsell, 2023 gives a better proposal, which is to clip/truncate the utilities into the range [−t,t] and compare the expected clipped utilities in the limit as t→∞. This will still suffer from St Petersburg lottery problems, though.
Here’s an order that’s as strong as both expected utility and stochastic dominance, and overall seems promising to me:
tl;dr: For lotteries with finite utility payoffs (but possibly unbounded utility payoffs and infinite expected utility), we can take expectations through any subset with finite and well-defined expected utility, and then compare the resulting lotteries with stochastic dominance. We just need to find any pair of well-behaved “expected utility collapses” for which one lottery stochastically dominates the other. Allowing expected utility collapses over the infinite expected utilities can lead to A<A, so I rule that out.
In practice, you might just take one expectation over everything but the top X% and bottom Y% of each lottery, and compare those lotteries with stochastic dominance, for different values of X and Y. This allows you to focus on the tails of heavy-tailed distributions.
For a lottery X, a utility function U, and a countable (possibly finite and possibly empty) set of mutually exclusive non-empty measurable subsets of the measure space, P={Q1,Q2,…,Qn} (or basically a set of binary random variables whose sum is at most 1) and letting PC=(∪Q∈PQ)C be the complement of their union (so, for their indicator binary random variables, 1PC=1−∑Q∈P1Q), the expected utility collapse of X over P is:
XP=E[U(X)|Q] if Q, for Q∈P, and XP=X|PC, otherwise.
Or, in lottery notation, letting L(c) be the constant lottery with constant value c,
XP=P(PC)X|PC+∑Q∈PP(Q)L(E[U(X)|Q]).
In other words, we replace probability subsets of X with its expected utility over those subsets.
If furthermore, E[U(X)|Q] is well-defined and finite for each Q∈P, we call the expected utility collapse well-behaved.
Then, we define the order over lotteries as follows:
A<B if there exists well-behaved expected utility collapses AP1 and BP2 of A and B respectively such that BP2 strictly stochastically dominates AP1.
If you allow infinite actual utilities (including possibly infinities of different magnitudes), you could add a disjunctive or overriding condition to handle comparisons with those.
From section 3 of Tarsney’s paper: