 A brief note on factoring out certain variables
discussion post by Stuart Armstrong 1164 days ago | Jessica Taylor and Patrick LaVictoire like this | 5 comments

Jessica Taylor and Chris Olah has a post on “Maximizing a quantity while ignoring effect through some channel”. I’ll briefly present a different way of doing this, and compare the two.

Essentially, the AI’s utility is given by a function $$U$$ of a variable $$C$$. The AI’s actions are a random variable $$A$$, but we want to ‘factor out’ another random variable $$B$$.

If we have a probability distribution $$Q$$ over actions, then, given background evidence $$E$$, the standard way to maximise $$U(C)$$ would be to maximise:

• $$\sum_{a,b,c} U(c)P(C=c,B=b,A=a|e) \\ = \sum_{a,b,c} U(c)P(C=c|B=b,A=a,e)P(B=b|A=a,e)Q(A=a|e)$$.

The most obvious idea, for me, is to replace $$P(B=b|A=a,e)$$ with $$P(B=b|e)$$, making $$B$$ artificially independent of $$A$$ and giving the expression:

• $$\sum_{a,b,c} U(c)P(C=c|B=b,A=a,e)P(B=b|e)Q(A=a|e)$$.

If $$B$$ is dependent on $$A$$ - if it isn’t, then factoring it out is not interesting - then $$P(B=b)$$ needs some implicit probability distribution over $$A$$ (which is independent of $$Q$$). So, in essence, this approach relies on two distributions over the possible actions, one that the agent is optimising, the other than is left unoptimised. In terms of Bayes nets, this just seems to be cutting $$B$$ from $$A$$.

Jessica and Chris’s approach also relies on two distributions. But, as far as I understand their approach, the two distributions are taken to be the same, and instead, it is assumed that $$U(C)$$ cannot be improved by changes to the distribution of $$A$$, if one keeps the distribution of $$B$$ constant. This has the feel of being a kind of differential condition - the infinitesimal impact on $$U(C)$$ of changes to $$A$$ but not $$B$$ is non-positive.

I suspect my version might have some odd behaviour (defining the implicit distribution for $$A$$ does not seem necessarily natural), but I’m not sure of the consistency properties of the differential approach.

A very dull coordination game

Suppose that $$A$$ is an integer in the range $$1$$ to $$100$$. Then $$B$$ will simply be set to the value of $$A$$. And the utility is equal to $$A$$ if $$A=B$$ and $$0$$ otherwise.

For the differential approach, all pure distributions $$P(A=a)=1$$ are optimal. For the approach presented here, it depends on the choice of implicit distribution over $$A$$ (and hence over $$B$$). If the distribution is uniform, then $$A=100$$ is the outcome. If there is a default $$A=a$$, then $$A=a$$ is the outcome. A third possibility is that the agent gets to select both the implicit distribution and the true distribution; in that case $$A=100$$ is the outcome.

Mixing in some anti-coordination gets different results (though I’m still not clear if, for the differential method, $$d=d'$$ means $$A=A'$$ or simply that they have the same distribution).  by Jessica Taylor 1163 days ago | Patrick LaVictoire likes this | link I think the main problem with using a pre-specified distribution over actions is that, since it doesn’t reflect the AI’s actual behavior, you can’t say much about $$P(B= b | e)$$ in relation to real life. For example, maybe the implicit policy is to take random actions, which results in humans not pressing the shutdown button; therefore, in real life the AI is confident that the button will not be pressed, so it spends no resources looking at the shutdown button to see if it was pressed. reply  by Patrick LaVictoire 1160 days ago | link Maybe if the pre-specified distribution is a reasonably well-calibrated predictor of the AI (given that distribution)? Like, maybe this is a way that an Oracle AI could help ensure the safety of a somewhat weaker Tool AI. reply  by Stuart Armstrong 1159 days ago | link How about an iterated process? If you have a series of actions and observations $$e_n=a_no_na_{n-1}o_{n-1}\ldots$$, and $$A$$ is the set of possible actions, have the AI pick $$a$$ such that: $$a_{n} = \operatorname{argmax}_{a} \frac{1}{|A|}\sum_{a'\in A,b} U(ao_ne_{n-1}|B=b) P(B=b|a'o_ne_{n-1})$$ (the $$\frac{1}{|A|}$$ is not necessary, just a reminder that we are choosing a uniform distribution over the possible actions). Notice that the AI conditions correctly on all its past actions, but not its current hypothetical. I wonder how this would work if we set $$A={0,1}$$ and had the AI type out messages bit by bit, each time thinking the probability of $$B$$ was independent of its bit-choice? EDIT: What this means is that, as soon as it’s done any action that might cause the shutdown button to be pressed, it will then look at the shutdown button (but maybe doesn’t expect, ahead of time, that it will be looking at the button). reply  by Jessica Taylor 1159 days ago | link I think, to get $$P(B = b)$$, you have to have an implicit policy for the entire rest of the game (not just the next action $$a'$$). I like the idea of using the evidence you have so far to inform the $$P(b)$$, so you spend more effort on looking at the shutdown button if you expect shutdown might be imminent based on your evidence. Of course, you can combine this with the fixed point thing, so the distribution of $$a'$$ is the same as the distribution of $$a$$. My main concern is that this isn’t reflectively stable. If at an early time step the AI has a certain $$P(b)$$ distribution, it may want to modify into an agent that fixes this as the correct $$P(b)$$ rather than changing $$P(b)$$ in response to new evidence; this is because it is modelling $$B$$ as coming independently from $$P(b)$$. reply  by Ryan Carey 259 days ago | link [Note: This comment is three years later than the post] The “obvious idea” here unfortunately seems not to work, because it is vulnerable to so-called “infinite improbability drives”. Suppose $$B$$ is a shutdown button, and $$P(b|e)$$ gives some weight to $$B=pressed$$ and $$B=unpressed$$. Then, the AI will benefit from selecting a Q such that it always chooses an action $$a$$, in which it enters a lottery, and if it does not win, then it the button B is pushed. In this circumstance, $$P(b|e)$$ is unchanged, while both $$P(c|b=pressed,a,e)$$ and $$P(c|b=unpressed,a,e)$$ allocate almost all of the probability to great $$C$$ outcomes. So the approach will create an AI that wants to exploit its ability to determine $$B$$. reply

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