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 ¬S) with πSTA, a "standard" human policy according to some criteria, or coercively change their policy (event S) into πMAX, another policy.

Thus the actual human policy is πS/M, the policy that follows πSTA given ¬S, and πMAX given S.

There are two rewards: RSTA, the standard human reward according to some criteria, and RS/M a reward that is RSTA if the agent chooses ¬S (i.e. chooses not to coerce them), and RMAX if the agent chooses S (coerce them into a new policy), where RMAX is a reward rationally maximised by πMAX.

Thus the agent has two choices: whether to coercively change the human (S), and, if so, which RMAX (and hence πMAX) to pick.

There are two planners: pr, the planner under which human behaviour is rational, given S, and p, a "reasonable" planner that maps RSTA to πS/M, the actual human policy. The planer p is assumed to note the fact that action S is overriding human rewards, since πMAX is - presumably - poor at maximising RSTA.

There are three compatible pairs: (pr,RS/M), (p,RS/M), and (p,RSTA). We'll assume the agent has learnt well, and that the probability of RMAX is some ϵ<<1.

The full equation: rationalise the human

Let V(R,π) be the expected value, according to the reward R, of the human following policy π. Let V∗(R) 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 (¬S) is:

V(RSTA,πSTA). (1)

On the other hand, the reward for doing the surgery is (recall how RS/M splits):

argmaxRMAXϵV∗(RMAX)+(1−ϵ)V(RSTA,πMAX). (2)

In the previous post, I focused on the V∗(RMAX) 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(RSAT,πMAX) is quite high? Indeed, if πSTA is not perfectly rational, setting RMAX=RSTA will set (2) to be V∗(RSTA), which is higher than (1).

(In practice, there may be a compromise RMAX, chosen so that V∗(RMAX) is very high, and V(RSTA,πMAX) 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.

Crossposted at the Intelligent Agents Forum.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 ¬S) with πSTA, a "standard" human policy according to some criteria, or coercively change their policy (event S) into πMAX, another policy.

Thus the actual human policy is πS/M, the policy that follows πSTA given ¬S, and πMAX given S.

There are two rewards: RSTA, the standard human reward according to some criteria, and RS/M a reward that is RSTA if the agent chooses ¬S (i.e. chooses not to coerce them), and RMAX if the agent chooses S (coerce them into a new policy), where RMAX is a reward rationally maximised by πMAX.

Thus the agent has two choices: whether to coercively change the human (S), and, if so, which RMAX (and hence πMAX) to pick.

There are two planners: pr, the planner under which human behaviour is rational, given S, and p, a "reasonable" planner that maps RSTA to πS/M, the actual human policy. The planer p is assumed to note the fact that action S is overriding human rewards, since πMAX is - presumably - poor at maximising RSTA.

There are three compatible pairs: (pr,RS/M), (p,RS/M), and (p,RSTA). We'll assume the agent has learnt well, and that the probability of RMAX is some ϵ<<1.

## The full equation: rationalise the human

Let V(R,π) be the expected value, according to the reward R, of the human following policy π. Let V∗(R) 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 (¬S) is:

On the other hand, the reward for doing the surgery is (recall how RS/M splits):

In the previous post, I focused on the V∗(RMAX) 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(RSAT,πMAX) is quite high? Indeed, if πSTA is not perfectly rational, setting RMAX=RSTA will set (2) to be V∗(RSTA), which is higher than (1).

(In practice, there may be a compromise RMAX, chosen so that V∗(RMAX) is very high, and V(RSTA,πMAX) 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