The set of Logical Inductors is not Convex
Sam Eisenstat asked the following interesting question: Given two logical inductors over the same deductive process, is every (rational) convex combination of them also a logical inductor? Surprisingly, the answer is no! Here is my counterexample.
We construct two logical inductors over PA, , and .
Let be any logical inductor over PA.
We consider an infinite sequence of sentences .
Let be computable in time.
We construct as in the paper, but instead of using all traders computable in time polynomial in , we use all traders computable in time polynomial in time. Since this also includes all polynomial time traders, is a logical inductor.
However, since the truth value of is computable in time, if the difference between and the indicator of did not converge to 0, a trader running in time polynomial in f(n) can easily exploit . Thus,
Now, consider the market
Now, consider the trader who exploits by repeatedly either buying a share of when the price is near 3⁄4, or selling a share when the price is near 1⁄4, waiting for that sentence to be resolved, and then repeating. Eventually, in each cycle, this trader will make roughly 1⁄4 of a share, because eventually the price will always be close enough to either 1⁄4 or 3⁄4, and all shares that this trader buys will be true, and all shares that this trader sells will be false.
Thus is not a logical inductor.