The integral of 1/t is log(t), thus hyperbolically discounted constant amount of torture over infinite time is still infinite. Hence your intuition about the Faustian bargain is consistent with hyperbolic discounting (but not with exponential discounting).
Ah. Very good point. I hadn’t actually checked that hyperbolic discounting would have the property of integrating to a constant. I assumed that since it falls faster than exponential, the integral would be smaller—but it only falls faster initially.
The integral of 1/t is log(t), thus hyperbolically discounted constant amount of torture over infinite time is still infinite. Hence your intuition about the Faustian bargain is consistent with hyperbolic discounting (but not with exponential discounting).
Ah. Very good point. I hadn’t actually checked that hyperbolic discounting would have the property of integrating to a constant. I assumed that since it falls faster than exponential, the integral would be smaller—but it only falls faster initially.