The optimization that I’ve been linking to- take the derivative with respect to exposure, set it equal to 0- is a 1-step optimization problem. That is, the strategy I’m describing as optimal (Shannon’s Demon) is optimal even if there’s only one coin flip, and because of the nature of the setup and the log utility function what’s optimal for one coin flip is optimal for an arbitrary number of coin flips.
The optimization that I’ve been linking to- take the derivative with respect to exposure, set it equal to 0- is a 1-step optimization problem. That is, the strategy I’m describing as optimal (Shannon’s Demon) is optimal even if there’s only one coin flip, and because of the nature of the setup and the log utility function what’s optimal for one coin flip is optimal for an arbitrary number of coin flips.