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by Vadim Kosoy 70 days ago | link | parent

Hi Tom!

There is a positive probability that the advisor falls into the trap, but this probability goes to \(0\) as the time discount parameter \(t\) goes to \(\infty\) (which is the limit I study here). This follows from the condition \(\beta(t)=\omega(t^{2/3})\) in the Theorem. To give a simple example, suppose that \(\mathcal{A}=\{0,1,2\}\) and the environment is s.t.:

  • When you take action 0, you fall into a trap and get reward 0 forever.

  • When you take action 1, you get reward 0 for the current round and remain in the same state.

  • When you take action 2, you get reward 1 for the current round (unless you are in the trap) and remain in the same state.

In this case, our advisor would have to take action 0 with probability \(\exp\left(-\omega\left(t^{2/3}\right)\right)\) and action 2 has to be more probable than action 1 by a factor of \(\exp\left(\omega\left(t^{-1/3}\right)\right) \approx 1 + \omega\left(t^{-1/3}\right)\).

by Tom Everitt 69 days ago | link

Hi Vadim!

So basically the advisor will be increasingly careful as the cost of falling into the trap goes to infinity? Makes sense I guess.

What is the incentive for the agent not to always let the advisor choose? Is there always some probability that the advisor saves them from infinite loss, or only in certain situations that can be detected by the agent?


by Vadim Kosoy 69 days ago | link

If the agent always delegates to the advisor, it loses a large fraction of the value. Returning again to the simple example above, the advisor on its own is only guaranteed to get expected utility \(1/2 + \omega(t^{-1/3})\) (because it often takes the suboptimal action 1). On the other hand, for any prior over a countable set of environments that includes this one, the corresponding DIRL agent gets expected utility \(1 - o(1)\) on this environment (because it will learn to only take action 2). You can also add an external penalty for each delegation, adjusting the proof is straightforward.

So, the agent has to exercise judgement about whether to delegate, using its prior + past observations. For example, the policy I construct in Lemma A delegates iff there is no action whose expected loss (according to current beliefs) is less than \(\beta(t)^{-1}t^{-1/3}\).


by Tom Everitt 68 days ago | link

So this requires the agent’s prior to incorporate information about which states are potentially risky?

Because if there is always some probability of there being a risky action (with infinitely negative value), then regardless how small the probability is and how large the penalty is for asking, the agent will always be better off asking.

(Did you see Owain Evans recent paper about trying to teach the agent to detect risky states.)


by Vadim Kosoy 68 days ago | link

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?


by Tom Everitt 59 days ago | link

My confusion is the following:

Premises (*) and inferences (=>):

  • The primary way for the agent to avoid traps is to delegate to a soft-maximiser.

  • Any action with boundedly negative utility, a soft-maximiser will take with positive probability.

  • Actions leading to traps do not have infinitely negative utility.

=> The agent will fall into traps with positive probability.

  • If the agent falls into a trap with positive probability, then it will have linear regret.

=> The agent will have linear regret.

So when you say in the beginning of the post “a Bayesian DIRL agent is guaranteed to attain most of the value”, you must mean that in a different sense than a regret sense?


by Vadim Kosoy 58 days ago | link

Your confusion is because you are thinking about regret in an anytime setting. In an anytime setting, there is a fixed policy \(\pi\), we measure the expected reward of \(\pi\) over a time interval \(t\) and compare it to the optimal expected reward over the same time interval. If \(\pi\) has probability \(p > 0\) to walk into a trap, regret has the linear lower bound \(\Omega(pt)\).

On other hand, I am talking about policies \(\pi_t\) that explicitly depend on the parameter \(t\) (I call this a “metapolicy”). Both the advisor and the agent policies are like that. As \(t\) goes to \(\infty\), the probability \(p(t)\) to walk into a trap goes to \(0\), so \(p(t)t\) is a sublinear function.

A second difference with the usual definition of regret is that I use an infinite sum of rewards with geometric time discount \(e^{-1/t}\) instead of a step function time discount that cuts off at \(t\). However, this second difference is entirely inessential, and all the theorems work about the same with step function time discount.






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