DMs open.
seems correct, thanks!
Why do decision-theorists say "pre-commitment" rather than "commitment"?
e.g. "The agent pre-commits to 1 boxing" vs "The agent commits to 1 boxing".
Is this just a lesswrong thing?
Steve Byrnes argument seems convincing.
If there’s 10% chance that the election depends on an event which is 1% quantum-random (e.g. the weather) then the overall event is 0.1% random.
How far back do you think an omniscient-modulo-quantum agent could‘ve predicted the 2024 result?
2020? 2017? 1980?
The natural generalization is then to have one subagent for each time at which the button could first be pressed (including one for “button is never pressed”, i.e. the button is first pressed at ). So subagent maximizes E[ | do( = unpressed), observations], and for all other times subagent T maximizes E[ | do( = unpressed, = pressed), observations]. The same arguments from above then carry over, as do the shortcomings (discussed in the next section).
Can you explain how this relates to Elliot Thornley's proposal? It's pattern matching in my brain but I don't know the technical details.
For the sake of potential readers, a (full) distribution over is some with finite support and , whereas a subdistribution over is some with finite support and . Note that a subdistribution over is equivalent to a full distribution over , where is the disjoint union of with some additional element, so the subdistribution monad can be written .
I am not at all convinced by the interpretation of here as terminating a game with a reward for the adversary or the agent. My interpretation of the distinguished element in is not that it represents a special state in which the game is over, but rather a special state in which there is a contradiction between some of one's assumptions/observations.
Doesn't the Nirvana Trick basically say that these two interpretations are equivalent?
Let be and let be . We can interpret as possibility, as a hypothesis consistent with no observations, and as a hypothesis consistent with all observations.
Alternatively, we can interpret as the free choice made by an adversary, as "the game terminates and our agent receives minimal disutility", and as "the game terminates and our agent receives maximal disutility". These two interpretations are algebraically equivalent, i.e. is a topped and bottomed semilattice.
Unless I'm mistaken, both and demand that the agent may have the hypothesis "I am certain that I will receive minimal disutility", which is necessary for the Nirvana Trick. But also demands that the agent may have the hypothesis "I am certain that I will receive maximal disutility". The first gives bounded infrabayesian monad and the second gives unbounded infrabayesian monad. Note that Diffractor uses in Infra-Miscellanea Section 2.
Okay, mea culpa. You can state the policy clearly like this:
"Suppose that, if you hadn't been told by someone who thinks is exfohazardous, then you wouldn't have known before time . Then you are obligated to not tell anyone before time ."
yep, if that's OP's suggestion then I endorse the policy. (But I think it'd be covered by the more general policy of "Don't share information someone tells you if they wouldn't want you to".) But my impression is that OP is suggesting the stronger policy I described?
“Don't share information that’s exfohazardous on others' models, even if you disagree with those models, except if your knowledge of it isn’t exclusively caused by other alignment researchers telling you of it.”
So if Alice tells me about her alignment research, and Bob thinks that Alice’s alignment research is exfohazardous, then I can’t tell people about Alice’s alignment research?
Unless I’ve misunderstood you, that’s a terrible policy.
Why am I deferring to Bob, who is completely unrelated? Why should I not using my best judgement, which includes the consideration that Bob is worried? What does this look like in practice, given someone people think everything under the sun is exfohazardous?
Of course, if someone tells me some information and asks me not to share it then I won’t — but that’s not a special property of AI xrisk.
if a lab has 100 million AI employs and 1000 human employees then you only need one human employee to spend 1% of their allotted AI headcount on your pet project and you’ll have 1000 AI employees