Hmm, apparently that wasn’t as clearly expressed as I thought. Let’s try that again. I said that a predictor P and utility function U are vulnerable to Pascal’s mugging if
exists function X of type number => evidence-set
such that X(a) differs from X(b) only in that one number appearing literally, and
forall u exists n such that P(X(n))U(X(n)) > u
The last line is the delta-epsilon definition for limits diverging to infinity. It could be equivalently written as
lim[n->inf] P(X(n))U(X(n)) = inf
If that limit diverges to infinity, then you could scale the probability down arbitrarily far and the mugger will just give you a bigger n. But if it doesn’t diverge that way, then there’s a maximum amount of expected utility the mugger can offer you just by increasing n, and the only way to get around it would be to offer more evidence that wasn’t in X(n).
(Note that while the limit can’t diverge to infinity, it is not required to converge. For example, the Pebblesorter utility function, U(n pebbles) = if(isprime(n)) 1 else 0, does not converge when combined with the null predictor P(X)=0.5.)
(The reductio you gave in the other reply does not apply, because the high-utility statement you gave is not parameterized, so it can’t diverge.)
Hmm, apparently that wasn’t as clearly expressed as I thought. Let’s try that again. I said that a predictor P and utility function U are vulnerable to Pascal’s mugging if
The last line is the delta-epsilon definition for limits diverging to infinity. It could be equivalently written as
If that limit diverges to infinity, then you could scale the probability down arbitrarily far and the mugger will just give you a bigger n. But if it doesn’t diverge that way, then there’s a maximum amount of expected utility the mugger can offer you just by increasing n, and the only way to get around it would be to offer more evidence that wasn’t in X(n).
(Note that while the limit can’t diverge to infinity, it is not required to converge. For example, the Pebblesorter utility function, U(n pebbles) = if(isprime(n)) 1 else 0, does not converge when combined with the null predictor P(X)=0.5.)
(The reductio you gave in the other reply does not apply, because the high-utility statement you gave is not parameterized, so it can’t diverge.)