Let F be the SWBI that just says F(S) = number of 0s in S - number of 1s in S. Let M be an arbitrary semicomputable SWBI. Then I don't think M can dominate both F and -F up to an additive constant.
I have a vague, handwavy argument:
Suppose the 'data' is an algorithmically random sequence, which basically gives us a discrete random walk. Now a random walk must return infinitely many times to the 'center' (where F = 0). And in order to dominate both F and -F, when the random walk is far away from the center, M is going to have to be increasing at the rate of 1 per step away from the center. But M can't know exactly when the random walk is going to turn around and head back home, so when it gets home, M is going to have lost about 1. (i.e. M will be about 1 less than it was last time it was at the center.) And this is going to happen infinitely many times.
But it's still possible that some M could dominate all semicomputable SWBIs in the weaker sense of losing less than epsilon per step, for all epsilon > 0.
But it's still possible that some M could dominate all semicomputable SWBIs in the weaker sense of losing less than epsilon per step, for all epsilon > 0.
You may already know that, but for the benefit of onlookers: multiplicative weights over F and -F is computable and kinda sorta dominates both F and -F in the weaker sense that you specified. You can make epsilon arbitrarily close to 0 by choosing a multiplier close to 1, at the cost of also incurring an additive constant loss that goes to infinity as epsilon->0. Multiplicative weights over all l...
This will be unparseable to anyone except maybe ten people here. What the hell, I'm posting it anyway.
Consider game 3 from my earlier post: a possibly uncomputable sequence of bits comes in, you're trying to guess the next bit, the payoff for each correct guess is $1. Obviously any deterministic strategy can be humiliated by a sequence of bits that's chosen to make it always guess wrong. User paulfchristiano suggested that multiplicative weights could yield a randomized strategy that does at most a constant worse than any human in expectation. I noted that, unlike the Solomonoff mixture, multiplicative weights over lower-semicomputable randomized strategies is not itself lower-semicomputable (several pages of algebra omitted), so we don't have a strategy that's optimal within its own class. Then I found some slides by Hutter that include a handy table showing how unusual it is for a class do be dominated by a measure within that same class.
So here's the question: is there an analog of "dominant lower-semicomputable semimeasure" for some class of games other than the log-score game? Desiderata: I guess it must correspond to randomized strategies (because deterministic ones are not enough), it must play at most an additive constant worse than any human in expectation (or maybe a multiplicative constant that can be made arbitrarily close to 1 by parameter choice, as with multiplicative weights), and it must be good within its own class of computability (not just superior to all members of some lower class). As far as I can tell, the literature has no answer to this question, and I already spent about a week on it with no success. Anyone here have more understanding than me?