DanielLC

Expected utility can be expressed as the sum ΣP(Xn)U(Xn). Suppose P(Xn) = 2-n, and U(Xn) = (-2)n/n. Then expected utility = Σ2-n(-2)n/n = Σ(-1)n/n = -1+1/2-1/3+1/4-... = -ln(2). Except there's no obvious order to add it. You could just as well say it's -1+1/2+1/4+1/6+1/8-1/3+1/10+1/12+1/14+1/16-1/5+... = 0. The sum depends on the order you add it. This is known as conditional convergence. This is clearly something we want to avoid. Suppose my priors have an unconditionally convergent expected utility. This would mean that ΣP(Xn)|U(Xn)| converges. Now suppose I observe evidence Y. ΣP(Xn|Y)|U(Xn)| = Σ|U(Xn)|P(Xn∩Y)/P(Y) ≤ Σ|U(Xn)|P(Xn)/P(Y) = 1/P(Y)·ΣP(Xn)|U(Xn)|. As long as P(Y) is nonzero, this must also converge. If my prior expected utility is unconditionally convergent, then given any finite amount of evidence, so is my posterior. This means I only have to come up with a nice prior, and I'll never have to worry about evidence braking expected utility. I suspect that this can be made even more powerful, and given any amount of evidence, finite or otherwise, I will almost surely have an unconditionally convergent posterior. Anyone want to prove it? Now let's look at Pascal's Mugging. The problem here seems to be that someone could very easily give you an arbitrarily powerful threat. However, in order for expected utility to converge unconditionally, either carrying out the threat must get unlikely faster than the disutility increases, or the probability of the threat itself must get unlikely that fast. In other words, either someone threatening 3^^^3 people is so unlikely to carry it out to make it non-threatening, or the threat itself must be so difficult to make that you don't have to worry about it.