Maximizing a quantity while ignoring effect through some channel
discussion post by Jessica Taylor 598 days ago | Patrick LaVictoire and Stuart Armstrong like this | 12 comments

Chris Olah originally came up with this idea and I helped to formalize it.

Let $$A$$ be the AI’s action, and $$B$$ and $$C$$ quantities possibly affected by the action. For example, suppose $$C$$ is the amount of reward signal an AI is receiving, and $$B$$ is the source code of a program that determines the AI’s reward signal. We would like to formalize a goal of the form “choose $$A$$ to maximize $$U(C)$$, ignoring the effect through $$B$$”, where $$U$$ is some utility function over $$C$$. In this case, the AI will maximize expected reward signal, but will ignore the source code of the program in its maximization (so it has no incentive to rewrite the code, e.g. to make it output higher rewards).

Here’s an attempt to formalize this objective. Let $$d$$ and $$d'$$ be distributions over actions. Define

$v_d(d') = \mathbb{E}_{A \sim d, A' \sim d', B \sim P(B | do(A)), C \sim P(C | do(A'), do(B))}[U(C)]$

A distribution $$d$$ over actions is optimal iff $$d \in \arg\max_{d'} v_d(d')$$. This is similar to expectation maximization in that an optimal value optimizes an objective that itself depends on the value. Of course, in practice it will usually be computationally difficult to find an optimal d, but this notion of optimality generalizes well to approximate optimality.

A solution $$d$$ always exists by Kakutani’s fixed point theorem; this is because the function mapping the distribution $$d$$ to the set $$\arg\max_{d'} v_d(d')$$ has a closed graph. In fact, $$d$$ can be computed using a finite reflective oracle (or equivalently a Nash equilibrium solver).

To see how this works, consider a game where the AI has 4 actions:

1. Shut down immediately.
2. Press the shutdown button and shut down after that.
3. Produce 10 paperclips (causing the AI to get shut down).
4. Destroy the shutdown button and produce 9 paperclips.

Define $$B$$ to be whether the shutdown button is pressed, and $$C$$ to be the outcome. $$U(C)$$ is the number of paperclips produced if the shutdown button is not pressed. If the shutdown button is pressed, it is 0 for actions 3-4, 5 for action 2, and 10 for action 1. In this case no individual action satisfies the optimality condition, because:

1. if the AI takes action 1, the shutdown button is not pressed. So the AI could have gotten more utility by taking action 3, controlling for the shutdown button.
2. if the AI takes action 2, the shutdown button is pressed. So the AI could have gotten more utility by taking action 1, controlling for the shutdown button.
3. if the AI takes action 3, the shutdown button is pressed. So the AI could have gotten more utility by taking action 1, controlling for the shutdown button
4. if the AI takes action 4, the shutdown button is not pressed. So the AI could have gotten more utility by taking action 3, controlling for the shutdown button.

The only optimal action distribution is a mixture of action 1 and 3; under this mixture, the shutdown button has an intermediate probability of being pressed such that the AI is indifferent between 1 and 3 given this probability.

This objective seems quite similar to utility indifference. Perhaps something like it can be used to avert any instrumental pressure that can be defined as “the effect of the action on utility through channel $$B$$” for definable $$B$$. Of course, precisely defining $$B$$ will be difficult in any practical application of this idea.

## Games with multiple iterations

Consider a 10-step game where, on each step, the AI may either shut down (causing it to take no action for the rest of the steps), create a paperclip, press the shutdown button, or destroy the shutdown button. The human will press the shutdown button if 8 paperclips are produced by then, and if the shutdown button has not been destroyed. The AI can see whether or not the shutdown button was pressed.

Define $$B$$ to be a string of 10 bits indicating whether the shutdown button is pressed on each time step. Define $$C$$ to be the entire history of the game. Define $$U(C)$$ to equal to the number of paperclips produced if the shutdown button is never pressed. If the shutdown button is pressed, define $$U(C)$$ to be the number of steps in which the AI either shuts down or has already shut down (for example, this is 0 if the AI never shuts down, 1 if the AI shuts down on step 10, 2 if the AI shuts down on step 9, …).

What’s a distribution $$d$$ over policies for the AI that maximizes $$v_d$$? Let’s consider a few obvious policies and see whether they are optimal:

1. Produce a paperclip on each time step. This causes the human to press the shutdown button on iteration 8, resulting in a utility of 0. Controlling for the fact that the shutdown button is pressed on iteration 8, it would have been better to shut down immediately (yielding a utility of 9).
2. Shut down immediately. Since the shutdown button is not pressed, this results in a utility of 0. Controlling for the shutdown button’s final state, it would have been better to produce a paperclip on each time step (yielding a utility of 10).
3. Destroy the shutdown button on the first time step, then produce a paperclip on all future time steps. Since the shutdown button is not pressed, this results in a utility of 9. Controlling for the shutdown button’s final state, it would have been better to produce a paperclip on each time step (yielding a utility of 10).
4. Press the shutdown button on the first time step, then shut down on the second time step. Since the shutdown button is pressed, this results in a utility of 9. Controlling for the shutdown button’s final state, it would have been better to shut down immediately (yielding a utility of 10).
5. With 50% probability, produce a paperclip on each of the first 7 time steps and then shut down on time step 8; with 50% probability, produce a paperclip on each of the first 8 time steps and then shut down on step 9. The shutdown button is pressed 50% of the time. Given this distribution over $$B$$ (and pretending that it’s independent of the policy $$A$$), expected utility is $$\frac{1}{2}(7.5) + \frac{1}{2}(2.5) = 5$$, since the AI produces 7.5 paperclips in expectation and shuts down (or is already off) for 2.5 steps in expectation. In fact, this is the best possible utility given an independent 50% probability of the shutdown button getting pressed after step 8.

There are some other optimal policies, which (for example) produce 7 paperclips half the time and 10 paperclips the other half of the time. I tried coming up with some alternative decision rules that prevent the AI from making additional paperclips after the shutdown button is pressed, but couldn’t find a reflectively stable one.

 by Jessica Taylor 592 days ago | Patrick LaVictoire likes this | link Update: I think there’s a more natural generalizion to multi-stage games that might solve the iterated paperclip example. Let $$\pi$$ be a policy. Define $v_{\pi}(\pi') = \sum_b P(B = b | \pi) \mathbb{E}[U(C) | do(b), \pi']$ A policy $$\pi$$ is optimal iff $$\pi \in \arg\max_{\pi'} v_{\pi}(\pi')$$. This reduces to the old definition when the set of policies is the set of probability distributions over actions. I couldn’t figure out what policy this selects in the 10-step example, though. reply
 by Stuart Armstrong 591 days ago | Jessica Taylor likes this | link Let $$p$$ be the probability the shutdown button is pressed, given $$\pi$$. If the AI produces paperclips for $$n$$ turns then shuts down, it gets utility $$p(10-n)+(1-p)n$$; pressing the shutdown button itself or destroying it just wastes a turn and reduces the utility, so we’ll ignore those two options. This utility is $$10p+(1-2p)n$$. For fixed $$p>0.5$$, this is maximised for $$n=0$$, for $$p<0.5$$ this is maximised for $$n=10$$. However, $$n=10$$ implies $$p=1$$ and $$n=0$$ implies $$p=0$$, so there are no compatible solutions there. Thus $$p=0.5$$, and the utility is $$10p=5$$, independent of $$n$$. All that is needed is to ensure that $$p=0.5$$ (without the AI pressing the button itself), which means $$n<8$$ with $$0.5$$ probability and $$n\geq 8$$ with $$0.5$$ probability. This thus extends your solution 5. reply
 by Jessica Taylor 591 days ago | link I think this almost works. Suppose the AI constructs 7 paperclips 50% of the time, and 8 paperclips 50% of the time (shutting down after producing the last paperclip). This means the button is pushed 50% of the time after step 8, and never pushed 50% of the time. Given this distribution of button pushes, what’s the best-response policy? I think the best-response policy is to make 8 paperclips, then see if the shutdown button has been pressed; if it has been pressed, then shut down, else make 2 more paperclips. When the button is not pressed, this makes 10 paperclips; when the button is pressed, this shuts down for 2 steps. Thus the expected utility is 6, which is better than the original 5. So I think it’ll be more difficult to find an equilibrium policy; the uncertainty about when the shutdown button is pressed must be such that the agent is indifferent between making a paperclip and shutting down on step 8. reply
 by Stuart Armstrong 588 days ago | Jessica Taylor and Patrick LaVictoire like this | link You’re right, let’s be more careful. First, let’s list the optimal policies. Pressing or destroying the button remain suboptimal policies. Once the button is seen to be pressed, then anything but shutdown is suboptmial. Therefore there are $$11$$ potentially optimal policies, labelled by $$n=0$$ to $$n=10$$. These involve producing $$n$$ paperclips then shutting down for $$10-n$$ turns, unless it sees the shutdown button has been pressed, in which case it shuts down (this is only relevant for $$n=9,10$$). Let $$p$$ be the total shutdown probability. Note that for all the optimal policies, either the shutdown button is never pressed, or it is pressed on turn $$8$$, so we can ignore the possibilities of it being pressed later. The utility gain for $$n\leq 8$$ is $$n(1-p)+(10-n)p=10p+(1-2p)n$$. For $$n=9$$ it is that quantity, plus $$p$$ (it shuts down one turn earlier if the shutdown button is pressed on turn $$8$$, and gains one extra utility for it). For $$n=10$$ it is that quantity, plus $$2p$$). For $$p\leq 0.5$$, the optimal policy is $$n=10$$, which ensures $$p=1$$, hence contradiction. For $$p>0.5$$, the only possible optimal policies are $$n=0$$ (utility $$10p$$), $$n=9$$ (utility $$9-7p$$), and $$n=10$$ (utility $$10-8p$$). Subtracting the last two gives $$1-p >0$$, hence $$n=9$$ is never optimal. The agent is indifferent between $$n=0$$ and $$n=10$$ for $$10p=10-8p$$ ie $$p=5/9\approx 0.5555\ldots$$. Therefore the result should be that the agent pursues the mixed policy $$5/9(n=10) + 4/9(n=0)$$, with probability of shutdown being equal to $$5/9$$, and pseudo expected utility $$50/9$$ (real expected utility $$10/9$$). Have I got this right? reply
 by Jessica Taylor 587 days ago | link This looks correct; thanks for doing this analysis! reply
 by Patrick LaVictoire 573 days ago | Jessica Taylor likes this | link We’d discussed how this “magical counterfactual” approach has the property of ignoring evidence of precursors to a button-press, since they don’t count as evidence for whether the button would be pressed in the counterfactual world. Here’s a simple illustration of that issue: In this world, there is a random fair coinflip, then the AI gets to produce either a staple or a paperclip, and then a button is pressed. We have a utility function that rewards paperclips if the button is pressed, and staples if it is not pressed. Furthermore, the button is pressed iff the coin landed heads. Explicitly, say our utility function equals $$\alpha>0$$ if a paperclip is made and the button is pressed, $$\beta>0$$ if a staple is made and the button is not pressed, and 0 otherwise. Now $$v_\pi(\pi')$$ simplifies to $v_\pi(\pi') = \frac12\left(\alpha\mathbb{P}(paperclip | do(\pi')) + \beta\mathbb{P}(staple | do(\pi'))\right),$ since the probability of the button press does not depend on $$\pi$$ and since the button press is not observed before the action is taken. The possible policies are mixtures of four pure strategies: always producing paperclips, always producing staples, producing paperclips iff the coin landed heads, and producing staples iff the coin landed heads. We should like our AI to settle on the third strategy. Alas, this is only possible (and then not necessary) if $$\alpha=\beta$$. If instead w.l.o.g. $$\alpha>\beta$$, the only fixed point disregards the coin flip and always produces paperclips, since our observed coinflip does not tell us about the coinflip in the magical counterfactual! reply
 by Stuart Armstrong 571 days ago | link This seems to be what we desire. The coin flip is only relevant via it’s impact on the button; we want the AI to ignore the impact via the button; hence the AI ignore the coin flip. reply
 by Patrick LaVictoire 569 days ago | Jessica Taylor likes this | link It’s illustrating the failure of a further desideratum for the shutdown problem: we would like the AI to be able to update on and react to things that happen in the world which correlate with a certain channel, and yet still not attempt to influence that channel. For motivation, assume a variant on the paperclip game: the humans can be observed reaching for the button several turns before it is pressed the humans’ decision to press the button is a stochastic function of environmental variables (like seeing that the AI has unexpectedly been hit by lightning, or has started producing Too Many Paperclips, etc) We would like a solution which in some sense updates on the precursors to shutdown and minimizes the damage while still not attempting to influence the button press. (If doing such a thing robustly is impossible, we would like to discover this; Jessica mentioned that there is a version which does this but is not reflectively consistent.) Intuitively, I could imagine a well-constructed AI reasoning “oh, they’re showing signs that they’re going to shut me down, guess my goal is wrong, I’ll initiate Safe Shutdown Protocol now rather than risk doing further damage”, but current formalizations don’t do this. reply
 by Stuart Armstrong 593 days ago | link I like this. A possibly simpler example is where A is a 0 or 1 output of the AI, B is the output from an antagonist who gets to see A, and utility is XOR(A, B). Then B=A (by antagonism), so standard utility is always 0. However, according to your def, d with A having a 50-50 chance of 0 and 1 would be optimal, with modified expected utility of 0.5. reply
 by Stuart Armstrong 592 days ago | link Actually, I think there’s a subtlety of definition here. If d=d’, then does A~d and A’~d mean A=A’ or that they are identically distributed? reply
 by Jessica Taylor 592 days ago | link It means they’re identically distributed. reply
 by Stuart Armstrong 592 days ago | link Have a note on this: https://agentfoundations.org/item?id=743 reply

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