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, ${P_{n}}$ , and ${P_{n}^{'}}$ .

Let ${P_{n}}$ be any logical inductor over PA.

We consider an infinite sequence of sentences $ϕ_{n} :\Leftrightarrow" P_{n} (ϕ_{n}) < 1 / 2 "$ .

Let $P_{n} (ϕ_{n})$ be computable in $f (n)$ time.

We construct ${P_{n}^{'}}$ 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, ${P_{n}^{'}}$ is a logical inductor.

However, since the truth value of $ϕ_{n}$ is computable in $f (n)$ time, if the difference between $P_{n}^{'} (ϕ_{n})$ and the indicator of $ϕ_{n}$ did not converge to 0, a trader running in time polynomial in f(n) can easily exploit ${P_{n}^{'}}$ . Thus,

$P_{n} (ϕ_{n}) ≂_{n} 1 / 2,$

and

$P_{n}^{'} (ϕ_{n}) ≂_{n} T h m_{P A} (ϕ_{n}) .$

Now, consider the market

${P_{n}^{''} = \frac{P_{n} + P_{n}^{'}}{2}} .$

Observe that

$P_{n}^{''} (ϕ_{n}) ≂_{n} \frac{1 + 2 \cdot T h m_{P A} (ϕ_{n})}{4},$

$| P_{n}^{''} (ϕ_{n}) - 1 / 2 | ≂_{n} 1 / 4.$

Now, consider the trader who exploits ${P_{n}^{''}}$ by repeatedly either buying a share of $ϕ_{n}$ 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 ${P_{n}^{''}}$ is not a logical inductor.