The continuity hypothesis really is an unimportant “technical assumption.” The only kind of thing it rules out are lexicographical preferences, like if you maximize X, but use Y as a tie-breaker.
Specifically, it follows from independence that if AP; the only thing the continuity axiom requires is that at P there is no preference between B and the mixture; there is no tie-breaker. (Without the continuity axiom, it may well be that P is 0 or 1.)
This is still true if you only have preferences involving p a rational number: the above is a Dedekind cut. If you restrict p to some smaller set that isn’t dense, it’s probably bad, but then I’d say you aren’t taking probability seriously.
Correct, by definition, if you have a dense set (which by default we treat the probability space as) and we map it into another space than either that space is also dense, in which case the converging sequences will have limits or it will not be dense (in which case continuity fails). In the former case, continuity reduces to point-wise continuity.
Note, setting the limit to “no preference” does not resolve the discontinuity. But by intermediate value, there will exist at least one such point in any continuous approximation of the discontinuous function.
Note, setting the limit to “no preference” does not resolve the discontinuity. But by intermediate value, there will exist at least one such point in any continuous approximation of the discontinuous function.
What is the discontinuous function? the function that assigns a preference to a dilemma? (particularly, mixed dilemmas parameterized by probabilities) With discrete range, that can never be continuous.
I think you are complaining about the name “continuity axiom”; I am not the right target of that complaint! I don’t know why it’s called that, but I suspect you have jumped from the name to false beliefs about the axiom system.
There is another continuous function, which is the assignment of utilities to lotteries. But I think this is continuous (to the extent that it can be defined) without invoking the continuity axiom. It is more the inverse map, from utilities to indifference-classes of lotteries, that risks not being continuous. I would complain more that this map is not well-defined, but there may be a way of arranging something like indifference-classes to have a finer topology than the order topology (eg, the left-limit topology, or the discrete topology).
I was talking about utility functions, but I can see your point about generalizing the result to the mapping from arbitrary dilemmas to preferences. Realize though, that preference space isn’t discrete.
You can describe it as the function from a mixed dilemma to the joint relation space for < and =. Which you can treat as a somewhat more complex version of the ordinals (certainly you can construct a map to a dense version of the ordinals if you have at least 2 dilemmas and dense probability space). That gives you a notion of the preference space where a calculus concept of continuity does apply (as the continuity axiom is a variation on the intermediate value theorem for this space which implies typical continuity).
From this perspective, the point I’m making about continuous approximations should make more sense.
The function whose continuity is at issue is the function from real numbers to lotteries that mixes A and B. C is being used to build open sets in the space of lotteries of the form of all lotteries better (or worse) than C, whose preimage in the real numbers must be open, rather than half-open.
We are talking about the same thing here just at different levels of generality. The function you describe is the same as the one I’m describing, except on a much narrower domain (only a single binary lottery between A and B). Then you project the range to just a question about C.
In the specific function you are talking about, you must hold that this is true for all A, B, and C to get continuity. In the function I describe, the A, B, and C are generalized out, so the continuity property is equivalent to the continuity of the function.
Note, setting the limit to “no preference” does not resolve the discontinuity. But by intermediate value, there will exist at least one such point in any continuous approximation of the discontinuous function.
I meant that setting the limit to no preference for a given C doesn’t equate to a globally continuous function. But that when you adjust your preferences function to approximate the discontinuous function by a continuous one, the result will contain (at least one) no preference point between any two A < B.
Now perhaps there is a result which says that if you take the limit as you set all discontinuous C to no preference, that the resulting function is complete, consistent, transitive, and continuous, but I wouldn’t take that to be automatic.
Consider, for example, a step discontinuity, where an entire swatch of pA + (1-p)B are stuck on the same set of < and = mappings and then there is a sharp jump to a very large set of < and = mappings at a critical p’. If you map the ordinals to the real line, this is analogous to a y-coordinate jump. To remove this discontinuity you would need to do more than split the preferences at p’ around no preference, because all this does is add a single point to the mix. To fully resolve it, you need to add an entire continuous curve, which means a process of selecting new A, B, and C, and showing that the transfinite limit always converges to a valid result.
The continuity hypothesis really is an unimportant “technical assumption.” The only kind of thing it rules out are lexicographical preferences, like if you maximize X, but use Y as a tie-breaker.
Specifically, it follows from independence that if AP; the only thing the continuity axiom requires is that at P there is no preference between B and the mixture; there is no tie-breaker. (Without the continuity axiom, it may well be that P is 0 or 1.)
This is still true if you only have preferences involving p a rational number: the above is a Dedekind cut. If you restrict p to some smaller set that isn’t dense, it’s probably bad, but then I’d say you aren’t taking probability seriously.
Correct, by definition, if you have a dense set (which by default we treat the probability space as) and we map it into another space than either that space is also dense, in which case the converging sequences will have limits or it will not be dense (in which case continuity fails). In the former case, continuity reduces to point-wise continuity.
Note, setting the limit to “no preference” does not resolve the discontinuity. But by intermediate value, there will exist at least one such point in any continuous approximation of the discontinuous function.
What is the discontinuous function? the function that assigns a preference to a dilemma? (particularly, mixed dilemmas parameterized by probabilities) With discrete range, that can never be continuous.
I think you are complaining about the name “continuity axiom”; I am not the right target of that complaint! I don’t know why it’s called that, but I suspect you have jumped from the name to false beliefs about the axiom system.
There is another continuous function, which is the assignment of utilities to lotteries. But I think this is continuous (to the extent that it can be defined) without invoking the continuity axiom. It is more the inverse map, from utilities to indifference-classes of lotteries, that risks not being continuous. I would complain more that this map is not well-defined, but there may be a way of arranging something like indifference-classes to have a finer topology than the order topology (eg, the left-limit topology, or the discrete topology).
I was talking about utility functions, but I can see your point about generalizing the result to the mapping from arbitrary dilemmas to preferences. Realize though, that preference space isn’t discrete.
You can describe it as the function from a mixed dilemma to the joint relation space for < and =. Which you can treat as a somewhat more complex version of the ordinals (certainly you can construct a map to a dense version of the ordinals if you have at least 2 dilemmas and dense probability space). That gives you a notion of the preference space where a calculus concept of continuity does apply (as the continuity axiom is a variation on the intermediate value theorem for this space which implies typical continuity).
From this perspective, the point I’m making about continuous approximations should make more sense.
The function whose continuity is at issue is the function from real numbers to lotteries that mixes A and B. C is being used to build open sets in the space of lotteries of the form of all lotteries better (or worse) than C, whose preimage in the real numbers must be open, rather than half-open.
We are talking about the same thing here just at different levels of generality. The function you describe is the same as the one I’m describing, except on a much narrower domain (only a single binary lottery between A and B). Then you project the range to just a question about C.
In the specific function you are talking about, you must hold that this is true for all A, B, and C to get continuity. In the function I describe, the A, B, and C are generalized out, so the continuity property is equivalent to the continuity of the function.
So what did you mean by
I meant that setting the limit to no preference for a given C doesn’t equate to a globally continuous function. But that when you adjust your preferences function to approximate the discontinuous function by a continuous one, the result will contain (at least one) no preference point between any two A < B.
Now perhaps there is a result which says that if you take the limit as you set all discontinuous C to no preference, that the resulting function is complete, consistent, transitive, and continuous, but I wouldn’t take that to be automatic.
Consider, for example, a step discontinuity, where an entire swatch of pA + (1-p)B are stuck on the same set of < and = mappings and then there is a sharp jump to a very large set of < and = mappings at a critical p’. If you map the ordinals to the real line, this is analogous to a y-coordinate jump. To remove this discontinuity you would need to do more than split the preferences at p’ around no preference, because all this does is add a single point to the mix. To fully resolve it, you need to add an entire continuous curve, which means a process of selecting new A, B, and C, and showing that the transfinite limit always converges to a valid result.
Thanks, that’s very useful to know. Do you have a link to the proof?