I don’t understand why people insist on unboundedness.
Possibly because there do appear to be potential solutions to Pascal’s mugging that do not require bounding your utility function.
Example:
I claim that in general, I find it reasonable to submit and give the Pascal’s mugger the money, and I am being Pascal’s mugged and considering giving a Pascal’s mugger money.
I also consider: What is the chance that a future mugger will make a Pascal’s mugging with a higher level of super exponentiation and that I won’t be able to pay?
And I claim that the answer appears to be: terribly unlikely, but considering the risks of failing at a higher level of super exponentiation, likely enough that I shouldn’t submit to the current Pascal’s mugger.
Except, that’s ALSO true for the next Pascal’s mugger.
So despite believing in Pascal’s mugging, I act exactly as if I don’t, and claiming that I ‘believe’ in Pascal’s mugging, doesn’t actually pay rent (for the muggers)
End of Example.
Since there exist examples like this and others that appear to solve Pascal’s mugging without requiring a bounded utility function, a lot of people wouldn’t want to accept a utility bound just because of the mugging.
I’m afraid that this kind of reasoning cannot avoid the real underlying problem, namely that Solomonoff expectation values of unbounded utility functions tend to divergence since utility grows as BB(n) and probability falls only as 2^{-n}.
What if the utility function is bound but and the bound itself is expandable without limit in at least some cases?
For instance, take a hypothetical utility function, Coinflipper bot.
Coinflipper bot has utility equal to the number of fair coins it has flipped.
Coinflipper bot has a utility bound equal to the 2^(greatest number of consecutive heads on fair coins it has flipped+1)
For instance, a particular example of Coinflipper bot might have flipped 512 fair coins and it’s current record is 10 consecutive heads on fair coins, so it’s utility is 512 and it’s utility bound is 2^(10+1) or 2048.
On the other hand, a different instance of Coinflipper bot might have flipped 2 fair coins, gotten 2 tails, and have a utility of 2 and a utility bound of 2^(0+1)=2.
How would the math work out in that kind of situation?
What if, depending on other circumstances(say the flip of a fair coin), your utility function can take values in either a finite(if heads) or infinite(if tails) interval?
Would that entire situation be bounded, unbounded, neither, or is my previous question ill posed?
Possibly because there do appear to be potential solutions to Pascal’s mugging that do not require bounding your utility function.
Example:
I claim that in general, I find it reasonable to submit and give the Pascal’s mugger the money, and I am being Pascal’s mugged and considering giving a Pascal’s mugger money.
I also consider: What is the chance that a future mugger will make a Pascal’s mugging with a higher level of super exponentiation and that I won’t be able to pay?
And I claim that the answer appears to be: terribly unlikely, but considering the risks of failing at a higher level of super exponentiation, likely enough that I shouldn’t submit to the current Pascal’s mugger.
Except, that’s ALSO true for the next Pascal’s mugger.
So despite believing in Pascal’s mugging, I act exactly as if I don’t, and claiming that I ‘believe’ in Pascal’s mugging, doesn’t actually pay rent (for the muggers)
End of Example.
Since there exist examples like this and others that appear to solve Pascal’s mugging without requiring a bounded utility function, a lot of people wouldn’t want to accept a utility bound just because of the mugging.
I’m afraid that this kind of reasoning cannot avoid the real underlying problem, namely that Solomonoff expectation values of unbounded utility functions tend to divergence since utility grows as BB(n) and probability falls only as 2^{-n}.
What if the utility function is bound but and the bound itself is expandable without limit in at least some cases?
For instance, take a hypothetical utility function, Coinflipper bot.
Coinflipper bot has utility equal to the number of fair coins it has flipped.
Coinflipper bot has a utility bound equal to the 2^(greatest number of consecutive heads on fair coins it has flipped+1)
For instance, a particular example of Coinflipper bot might have flipped 512 fair coins and it’s current record is 10 consecutive heads on fair coins, so it’s utility is 512 and it’s utility bound is 2^(10+1) or 2048.
On the other hand, a different instance of Coinflipper bot might have flipped 2 fair coins, gotten 2 tails, and have a utility of 2 and a utility bound of 2^(0+1)=2.
How would the math work out in that kind of situation?
I don’t understand what you mean by “utility bound”. A bounded utility function is just a function which takes values in a finite interval.
Let me try rephrasing this a bit.
What if, depending on other circumstances(say the flip of a fair coin), your utility function can take values in either a finite(if heads) or infinite(if tails) interval?
Would that entire situation be bounded, unbounded, neither, or is my previous question ill posed?