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, \(\{\mathbb{P}_n\}\), and \(\{\mathbb{P}^\prime_n\}\).
Let \(\{\mathbb{P}_n\}\) be any logical inductor over PA.
We consider an infinite sequence of sentences \(\phi_n:\Leftrightarrow"\mathbb{P}_n(\phi_n)<1/2"\).
Let \(\mathbb{P}_n(\phi_n)\) be computable in \(f(n)\) time.
We construct \(\{\mathbb{P}^\prime_n\}\) as in the paper, but instead of using all traders computable in time polynomial in \(n\), we use all traders computable in time polynomial in \(f(n)\) time. Since this also includes all polynomial time traders, \(\{\mathbb{P}^\prime_n\}\) is a logical inductor.
However, since the truth value of \(\phi_n\) is computable in \(f(n)\) time, if the difference between \(\mathbb{P}^\prime_n(\phi_n)\) and the indicator of \(\phi_n\) did not converge to 0, a trader running in time polynomial in f(n) can easily exploit \(\{\mathbb{P}^\prime_n\}\). Thus,
\[\mathbb{P}_n(\phi_n)\eqsim_n 1/2,\]
and
\[\mathbb{P}^\prime_n(\phi_n)\eqsim_n Thm_{PA}(\phi_n).\]
Now, consider the market
\[\left\{\mathbb{P}^{\prime\prime}_n=\frac{\mathbb{P}_n+\mathbb{P}^\prime_n}{2}\right\}.\]
Observe that
\[\mathbb{P}^{\prime\prime}_n(\phi_n)\eqsim_n \frac{1+2\cdot Thm_{PA}(\phi_n)}{4},\]
so
\[\mathbb{P}^{\prime\prime}_n(\phi_n)1/2\eqsim_n 1/4.\]
Now, consider the trader who exploits \(\{\mathbb{P}^{\prime\prime}_n\}\) by repeatedly either buying a share of \(\phi_n\) 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 \(\{\mathbb{P}^{\prime\prime}_n\}\) is not a logical inductor.
