V_V comments on Pascal's Mugging for bounded utility functions - Less Wrong
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Comments (49)
Yes.
It doesn't have to decrease faster than every computable function, only to decrease at least as fast as an exponential function with negative exponent.
Solomonoff induction doesn't try to learn your utility function.
Clearly, if your utility function is super-exponential, then p(X) * U(X) may diverge even if p(X) is light-tailed.