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by Vadim Kosoy 5 days ago | link | parent

The only assumptions about the prior are that it is supported on a countable set of hypotheses, and that in each hypothesis the advisor is \(\beta\)-rational (for some fixed \(\beta(t)=\omega(t^{2/3})\)).

There is no such thing as infinitely negative value in this framework. The utility function is bounded because of the geometric time discount (and because the momentary rewards are assumed to be bounded), and in fact I normalize it to lie in \([0,1]\) (see the equation defining \(\mathrm{U}\) in the beginning of the Results section).

Falling into a trap is an event associated with \(\Omega(1)\) loss (i.e. loss that remains constant as \(t\) goes to \(\infty\)). Therefore, we can risk such an event, as long as the probability is \(o(1)\) (i.e. goes to \(0\) as \(t\) goes to \(\infty\)). This means that as \(t\) grows, the agent will spend more rounds delegating to the advisor, but for any given \(t\), it will not delegate on most rounds (even on most of the important rounds, i.e. during the first \(O(t)\)-length “horizon”). In fact, you can see in the proof of Lemma A, that the policy I construct delegates on \(O(t^{2/3})\) rounds.

As a simple example, consider again the toy environment from before. Consider also the environments you get from it by applying a permutation to the set of actions \(\mathcal{A}\). Thus, you get a hypothesis class of 6 environments. Then, the corresponding DIRL agent will spend \(O(t^{2/3})\) rounds delegating, observe which action is chosen by the advisor most frequently, and perform this action forevermore. (The phenomenon that all delegations happen in the beginning is specific to this toy example, because it only has 1 non-trap state.)

If you mean this paper, I saw it?



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