It can make sense to say that a utility function is bounded, but that implies certain other restrictions. For example, bounded utility functions cannot be decomposed into independent (additive or multiplicative, these are the only two options) subcomponents if the number of subcomponents is unknown. Any utility function that is summed or multiplied over an unknown number of independent (e.g.) societies must be unbounded*. Does that mean you believe that utility functions can’t be aggregated over independent societies or that no two societies can contribute independently to the utility function? That latter implies that a utility function cannot be determined without knowing about all societies, which would make the concept useless. Do you believe that utility functions can be aggregated at all beyond the individual level?
Keep in mind that “unbounded” here means “arbitrarily additive”. In the multiplicative case, even if a utility function is always less than 1, if an individual’s utility can be made arbitrarily close to 0, then it’s still unbounded. Such an individual still has enough to gain by betting on a trillion coin tosses.
You mentioned that a utility function should be seen as a proxy to decision making. If decisions can be independent, then their contributions to the definition of a utility function must be independent*. If the utility function is bounded, then the number of independent decisions something can decide between must also be bounded. Maybe that makes sense for individuals since you distinguished a utility function as a summary of “current” decision-making, and any individual is presumably limited in their ability to decide between independent outcomes at any given point in time. Again, though, this causes problems for aggregate utility functions.
Consider the functor F that takes any set of decisions (with inclusion maps between them) to the least-assuming utility function consistent with them. There exists a functor G that takes any utility function to the maximal set of decisions derivable from it. F,G together form a contravariant adjunction between set of decisions and utility functions. F is then left-adjoint to G. Therefore F preserves finite coproducts as finite products. Therefore for any disjoint union of decisions A,B, the least-assuming utility function defined over them exists and is F(A+B)=F(A)*F(B). The proof is nearly identical for covariant adjunctions.
It seems like nonsense to say that utility functions can’t be aggregated. A model of arbitrary decision making shouldn’t suddenly become impossible just because you’re trying to model, say, three individuals rather than one. The aggregate has preferential decision making just like the individual.
I don’t know if my utility function is bounded. My statement was much weaker, that I’m not confident about decision-making in situations involving infinities. You’re right that the problem happens not just for unbounded utilities, but also for arbitrarily fine distinctions between utilities. None of these seem to apply to your original post though, where everything is finite and I can be pretty damn confident.
Algebraic reasoning is independent of the number system used. If you are reasoning about utility functions in the abstract and if your reasoning does not make use of any properties of numbers, then it doesn’t matter what numbers you use. You’re not using any properties of finite numbers to define anything, so the fact of whether or not these numbers are finite is irrelevant.
You’re asking if my utility function is bounded, right? I don’t know. All the intuitions seem unreliable. My original confident answer to you (“second strategy of course”) was from the perspective of an agent for whom your thought experiment is possible, which means it necessarily disagrees with Bob. Didn’t want to make any stronger claim than that.
It can make sense to say that a utility function is bounded, but that implies certain other restrictions. For example, bounded utility functions cannot be decomposed into independent (additive or multiplicative, these are the only two options) subcomponents if the number of subcomponents is unknown. Any utility function that is summed or multiplied over an unknown number of independent (e.g.) societies must be unbounded*. Does that mean you believe that utility functions can’t be aggregated over independent societies or that no two societies can contribute independently to the utility function? That latter implies that a utility function cannot be determined without knowing about all societies, which would make the concept useless. Do you believe that utility functions can be aggregated at all beyond the individual level?
Keep in mind that “unbounded” here means “arbitrarily additive”. In the multiplicative case, even if a utility function is always less than 1, if an individual’s utility can be made arbitrarily close to 0, then it’s still unbounded. Such an individual still has enough to gain by betting on a trillion coin tosses.
You mentioned that a utility function should be seen as a proxy to decision making. If decisions can be independent, then their contributions to the definition of a utility function must be independent*. If the utility function is bounded, then the number of independent decisions something can decide between must also be bounded. Maybe that makes sense for individuals since you distinguished a utility function as a summary of “current” decision-making, and any individual is presumably limited in their ability to decide between independent outcomes at any given point in time. Again, though, this causes problems for aggregate utility functions.
Consider the functor F that takes any set of decisions (with inclusion maps between them) to the least-assuming utility function consistent with them. There exists a functor G that takes any utility function to the maximal set of decisions derivable from it. F,G together form a contravariant adjunction between set of decisions and utility functions. F is then left-adjoint to G. Therefore F preserves finite coproducts as finite products. Therefore for any disjoint union of decisions A,B, the least-assuming utility function defined over them exists and is F(A+B)=F(A)*F(B). The proof is nearly identical for covariant adjunctions.
It seems like nonsense to say that utility functions can’t be aggregated. A model of arbitrary decision making shouldn’t suddenly become impossible just because you’re trying to model, say, three individuals rather than one. The aggregate has preferential decision making just like the individual.
I don’t know if my utility function is bounded. My statement was much weaker, that I’m not confident about decision-making in situations involving infinities. You’re right that the problem happens not just for unbounded utilities, but also for arbitrarily fine distinctions between utilities. None of these seem to apply to your original post though, where everything is finite and I can be pretty damn confident.
Algebraic reasoning is independent of the number system used. If you are reasoning about utility functions in the abstract and if your reasoning does not make use of any properties of numbers, then it doesn’t matter what numbers you use. You’re not using any properties of finite numbers to define anything, so the fact of whether or not these numbers are finite is irrelevant.
The original post doesn’t require arbitrarily fine distinctions, just 2^trillion distinctions. That’s perfectly finite.
Your comment about Bob not assigning a high utility value to anything is equivalent to a comment stating that Bob’s utility function is bounded.
Right, but Bob was based on your claims in this comment about what’s “reasonable” for you. I didn’t claim to agree with Bob.
Fair enough. I have a question then. Do you personally agree with Bob?
You’re asking if my utility function is bounded, right? I don’t know. All the intuitions seem unreliable. My original confident answer to you (“second strategy of course”) was from the perspective of an agent for whom your thought experiment is possible, which means it necessarily disagrees with Bob. Didn’t want to make any stronger claim than that.
I am, and thanks for answering. Keep in mind that there are ways to make your intuition more reliable, if that’s a thing you want.