Rationalising humans: another mugging, but not Pascal’s

[Stuart_Armstrong]

A putative new idea for AI control; index here.

In a previous post, I used the $(p,R)$ model to show how a value-learning agent might pull a Pascal’s mugging on humans: it would change the human reward into something much easier to maximise.

But there’s a much easier mugging the agent might want to pull: it might want to make the human fully rational. And this does not depend on low-probability-high reward situations.

Planners and Rewards

In the original mugging, the agent has the ability to let the human continue (event $\neg S$) with ${\pi }_{STA}$, a “standard” human policy according to some criteria, or coercively change their policy (event $S$) into ${\pi }_{MAX}$, another policy.

Thus the actual human policy is ${\pi }_{S/M}$, the policy that follows ${\pi }_{STA}$ given $\neg S$, and ${\pi }_{MAX}$ given $S$.

There are two rewards: $R_{STA}$, the standard human reward according to some criteria, and $R_{S/M}$ a reward that is $R_{STA}$ if the agent chooses $\neg S$ (i.e. chooses not to coerce them), and $R_{MAX}$ if the agent chooses $S$ (coerce them into a new policy), where $R_{MAX}$ is a reward rationally maximised by ${\pi }_{MAX}$.

Thus the agent has two choices: whether to coercively change the human ($S$), and, if so, which $R_{MAX}$ (and hence ${\pi }_{MAX}$) to pick.

There are two planners: $p_r$, the planner under which human behaviour is rational, given $S$, and $p$, a “reasonable” planner that maps $R_{STA}$ to ${\pi }_{S/M}$, the actual human policy. The planer $p$ is assumed to note the fact that action $S$ is overriding human rewards, since ${\pi }_{MAX}$ is—presumably—poor at maximising $R_{STA}$.

There are three compatible pairs: $(p_r,R_{S/M})$, $(p,R_{S/M})$, and $(p,R_{MAX})$. We’ll assume the agent has learnt well, and that the probability of $R_{MAX}$ is some $ϵ<<1$.

The full equation: rationalise the human

Let $V(R,\pi )$ be the expected value, according to the reward $R$, of the human following policy $\pi$. Let $V^{\ast }\left(R\right)$ be the expected value of $R$ if the human follows the optimal $R$-maximising policy.

Then the reward for the agent for not doing the surgery ($\neg S$) is:

• $V(R_{STA},{\pi }_{STA}$. (1)

On the other hand, the reward for doing the surgery is (recall how $R_{S/M}$ splits):

• ${\mathrm{argmax}}_{R_{MAX}}ϵV^{\ast }\left(R_{MAX}\right)+(1-ϵ)V(R_{STA},{\pi }_{MAX})$. (2)

In the previous post, I focused on the $V^{\ast }\left(R_{MAX}\right)$ term, wondering if there could be rewards sufficiently high to overcome the $ϵ$ probability.

But there’s another way (2) could be higher than (1). What if the term $V(R_{STA},{\pi }_{MAX})$ was is quite high? Indeed, if ${\pi }_{STA}$ is not perfectly rational, setting $R_{MAX}=R_{STA}$ will set (2) to be $V^{\ast }\left(R_{STA}\right)$, which is higher than (1).

(In practice, there may be a compromise $R_{MAX}$, chosen so that $V^{\ast }\left(R_{MAX}\right)$ is very high, and $V(R_{STA},{\pi }_{MAX})$ is not too low compared with (1)).

So if the human is not perfectly rational, the agent will always choose to transform them into rational maximisers if it can.

Personal identity and reward

What is `setting $R_{MAX}=R_{STA}$’? That’s essentially agreeing that $R_{STA}$ is the reward to be maximised, but that the human should be surgically changed to be more effective at maximising $R_{STA}$ than it currently is.

Now, if $R_{STA}$ fully captures human preferences, this is fine. But many people value their own identity and absence or coercion, and it’s hard to see how to capture that in reward form.

Is there a way of encoding `don’t force me into becoming a mindless outsourcer’?