Let u be utility function linear in paperclips. Assume the agent has no ability to create or destroy paperclips for the first week; it needs to build up infrastructure and means first. We want it to be maximising u on Monday, -u on Tuesday, and u from Wednesday onwards. How can we accomplish this? And how can we accomplish it without the agent simply turning itself off for Tuesday?
u is a function of paperclips, which is in turn a function of time. So, u(p(t)) is the number of paperclips at time t.
U = integral[some reasonable bounds] {dt p(t) (t in first Tuesday?-1:1)}
So, the AI knows what it wants over all of the future, depending on time. When evaluating future plans for the future, it's able to take that change into account.
Like, it might spend both Monday and Tuesday just building infrastructure. In any case, turning off won't help on Tuesday because it will still know that there were paperclips then - not being on to observe them won't help it.
A putative new idea for AI control; index here.
A simple and easy design for a u-maximising agent that turns into a u-minimising one.
Let X be some boolean random variable outside the agent's control, that will be determined at some future time t (based on a cosmic event, maybe?). Set it up so that P(X=1)=ε, and for a given utility u, consider the utility:
Before t, the expected value of (2/ε)X is 2, so u# = u. Hence the agent is a u-maximiser. After t, the most likely option is X=0, hence a little bit of evidence to that effect is enough to make u# into a u-minimiser.
This isn't perfect corrigibility - the agent would be willing to sacrifice a bit of u-value (before t) in order to maintain its flexibility after t. To combat this effect, we could instead use:
If Ω is large, then the agent is willing to pay very little u-value to maintain flexibility. However, the amount of evidence of X=0 that it needs to become a u-minimiser is equally proportional to Ω, so X better be a clear and convincing event.