by Sam Eisenstat 25 days ago | Abram Demski likes this | link | parent I at first didn’t understand your argument for claim (2), so I wrote an alternate proof that’s a bit more obvious/careful. I now see why it works, but I’ll give my version below for anyone interested. In any case, what you really mean is the probability of deciding a sentence outside of $$\Phi$$ by having it announced by nature; there may be a high probability of sentences being decided indirectly via sentences in $$\Phi$$. Instead of choosing $$\Phi$$ as you describe, pick $$\Phi$$ so that the probability $$\mu(\Phi)$$ of sampling something in $$\Phi$$ is greater than $$1 - \mu(\psi) \cdot \varepsilon / 2$$. Then, the probability of sampling something in $$\Phi - \{\psi\}$$ is at least $$1 - \mu(\psi) \cdot (1 + \varepsilon / 2)$$. Hence, no matter what sentences have been decided already, the probability that repeatedly sampling from $$\mu$$ selects $$\psi$$ before it selects any sentence outside of $$\Phi$$ is at least \begin{align*} \sum_{k = 0}^\infty (1 - \mu(\psi) \cdot (1 + \varepsilon / 2))^k \cdot \mu(\psi) & = \frac{\mu(\psi)}{\mu(\psi) \cdot (1 + \varepsilon / 2)} \\ & > 1 - \varepsilon / 2 \end{align*} as desired. Furthermore, this argument makes it clear that the probability distribution we converge to depends only on the set of sentences which the environment will eventually assert, not on their ordering! Oh, I didn’t notice that aspect of things. That’s pretty cool.

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