Comment author: sixes_and_sevens 02 May 2015 11:15:07AM *  1 point [-]

I get that you can do this in principle, but in the specific case of the Allais Paradox (and going off the Wikipedia setup and terminology), if someone prefers options 1B and 2A, what specific sequence of trades do you offer them? It seems like you'd give them 1A, then go 1A -> 1B -> (some transformation of 1B formally equivalent to 2B) -> 2A -> (some transformation of 2A formally equivalent to 1A') -> 1B' ->... in perpetuity, but what are the "(some transformation of [X] formally equivalent to [Y])" in this case?

Comment author: one_forward 02 May 2015 04:21:09PM *  1 point [-]

You can stagger the bets and offer either a 1A -> 1B -> 1A circle or a 2B -> 2A -> 2B circle.

Suppose the bets are implemented in two stages. In stage 1 you have an 89% chance of the independent payoff ($1 million for bets 1A and 1B, nothing for bets 2A and 2B) and an 11% chance of moving to stage 2. In stage 2 you either get $1 million (for bets 1A and 2A) or a 10/11 chance of getting $5 million.

Then suppose someone prefers a 10/11 chance of 5 million (bet 3B) to a sure $1 million (bet 3A), prefers 2A to 2B, and currently has 2B in this staggered form. You do the following:

  1. Trade them 2A for 2B+$1.
  2. Play stage 1. If they don't move on to stage 2, they're down $1 from where they started. If they do move on to stage 2, they now have bet 3A.
  3. Trade them 3B for 3A+$1.
  4. Play stage 2.

The net effect of those trades is that they still played gamble 2B but gave you a dollar or two. If they prefer 3A to 3B and 1B to 1A, you can do the same thing to get them to circle from 1A back to 1A. It's not the infinite cycle of losses you mention, but it is a guaranteed loss.

Comment author: RichardKennaway 02 February 2015 09:43:40PM *  1 point [-]

Actually, I should have been using curly brackets, as when I wrote (0,1) I meant the set with two elements, 0 and 1, which is what I had taken X to be a product of copies of, hence my obtaining 50000 as the expected Manhattan distance between any two members. I'll correct the post to make that clear. I think everything I said would still apply to the continuous case. If it doesn't, that would be better addressed with a separate comment.

Comment author: one_forward 04 February 2015 08:34:14PM 0 points [-]

Yeah, I don't think it makes much difference in high-dimensions. It's just more natural to talk about smoothness in the continuous case.

Comment author: RichardKennaway 27 January 2015 11:36:08PM *  1 point [-]

I think I'm beginning to see the problem for the Bayesian, although I not yet sure what the correct response to it is. I have some more or less rambling thoughts about it.

It appears that the Bayesian is being supposed to start from a flat prior over the space of all possible thetas. This is a very large space (all possible strings of 2^100000 probabilities), almost all of which consists of thetas which are independent of pi. (ETA: Here I mistakenly took X to be a product of two-point sets {0,1}, when in fact it is a product of unit intervals [0,1]. I don't think this makes much difference to the argument though, or if it does, it would be best addressed by letting this one stand as is and discussing that case separately.) When theta is independent of pi, it seems to me that the Bayesian would simply take the average of sampled values of Y as an estimate of P(Y=1), and be very likely to get almost the same value as the frequentist. Indirectly observing a few values of theta (through the observed values of Y) gives no information about any other values of theta, because the prior was flat. This is why the likelihood calculated in the blog post contains almost no information about theta.

Here is what seems to be to be a related problem. You will be presented with a series of some number of booleans, say 100. After each one, you are to guess the next. If your prior is a flat distribution over {0,1}^100, your prediction will be 50% each way at every stage, regardless of what the sequence so far has been, because all continuations are equally likely. It is impossible to learn from such a prior, which has built into it the belief that the past cannot predict the future.

As noted in the blog post, smoothness of theta with respect to e.g. the metric structure of {0,1}^100000 doesn't help, because a sample of only 1000 from this space is overwhelmingly likely to consist of points that are all at a Manhattan distance of about 50000 from each other. No substantial extrapolation of theta is possible from such a sample unless it is smooth at the scale of the whole space.

The flat prior over theta seems to be of a similar nature to the flat prior over sequences. If in this sample of 1000 you noticed that when pi was high, the corresponding value of Y, when sampled, was very likely to be 1, and similarly that when pi was low, Y was usually 0 among those rare times it was sampled, you might find it reasonable to conclude that pi and theta were related and use something like the Horwitz-Thompson estimator. But the flat prior over theta does not allow this inference. However many values of theta you have gained some partial information about by sampling Y, they tell you nothing about any other values.

My guess so far is that that is a problem with the flat prior over theta. The problem for the Bayesian is to come up with a better one that is capable of seeing a dependency between pi and theta.

Is the Robins and Ritov paper the one cited in the blog post, "Toward a Curse of Dimensionality Appropriate (CODA) Asymptotic Theory for Semi-parametric Models"? I looked at that briefly, only enough to see that their example, though somewhat similar, deals with a relatively low dimensional case (5), which in practical terms counts as high dimensional, and what they describe as a "moderate" sample size of 10000. So that's rather different from the present example, and I don't know if anything I just said will be relevant to it.

On reading further in the blog post, I see that a lot of what I said is said more briefly in the comments there, especially comment (4) by Chris Sims:

If theta and pi were independent, we could just throw out the observations where we don’t see Y and use the remaining sample as if there were no “R” variable. So specifying that theta and pi are independent is not a reasonable way to say we have little knowedge. It amounts to saying we are sure the main potential complication in the model is not present, and therefore opens us up to making seriously incorrect inference.

And a flat prior on theta is an assumption that theta and pi are almost certainly independent.

Comment author: one_forward 02 February 2015 07:34:51PM 0 points [-]

A note on notation - [0,1] with square brackets generally refers to the closed interval between 0 and 1. X is a continuous variable, not a boolean one.

Comment author: dankane 16 September 2014 02:56:24PM 2 points [-]

Which is actually one of the annoying things about UDT. Your strategy cannot depend simply on your posterior probability distribution, it has to depend on your prior probability distribution. How you even in practice determine your priors for Newcomb vs. anti-Newcomb is really beyond me.

But in any case, assuming that one is more common, UDT does lose this game.

Comment author: one_forward 16 September 2014 09:48:34PM 2 points [-]

Why does UDT lose this game? If it knows anti-Newcomb is much more likely, it will two-box on Newcomb and do just as well as CDT. If Newcomb is more common, UDT one-boxes and does better than CDT.

Comment author: dankane 16 September 2014 04:51:51PM 1 point [-]

Actually, I take it back. Depending on how you define things, UDT can still lose. Consider the following game:

I will clone you. One of the clones I paint red and the other I paint blue. The red clone I give $1000000 and the blue clone I fine $1000000. UDT clearly gets expectation 0 out of this. SMCDT however can replace its code with the following: If you are painted blue: wipe your hard drive If you are painted red: change your code back to standard SMCDT

Thus, SMCDT never actually has to play blue in this game, while UDT does.

Comment author: one_forward 16 September 2014 09:46:00PM 2 points [-]

You seem to be comparing SMCDT to a UDT agent that can't self-modify (or commit suicide). The self-modifying part is the only reason SMCDT wins here.

The ability to self-modify is clearly beneficial (if you have correct beliefs and act first), but it seems separate from the question of which decision theory to use.

Comment author: dankane 16 September 2014 07:48:42PM 4 points [-]

Actually, here's a better counter-example, one that actually exemplifies some of the claims of CDT optimality. Suppose that the universe consists of a bunch of agents (who do not know each others' identities) playing one-off PDs against each other. Now 99% of these agents are UDT agents and 1% are CDT agents.

The CDT agents defect for the standard reason. The UDT agents reason that my opponent will do the same thing that I do with 99% probability, therefore, I should cooperate.

CDT agents get 99% DC and 1% DD. UDT agents get 99% CC and 1% CD. The CDT agents in this universe do better than the UDT agents, yet they are facing a perfectly symmetrical scenario with no mind reading involved.

Comment author: one_forward 16 September 2014 09:42:06PM 1 point [-]

This is a good example. Thank you. A population of 100% CDT, though, would get 100% DD, which is terrible. It's a point in UDT's favor that "everyone running UDT" leads to a better outcome for everyone than "everyone running CDT."

Comment author: dankane 16 September 2014 02:08:00AM 2 points [-]

Well, if the universe cannot read your source code, both agents are identical and provably optimal. If the universe can read your source code, there are easy scenarios where one or the other does better. For example,

"Here have $1000 if you are a CDT agent" Or "Here have $1000 if you are a UDT agent"

Comment author: one_forward 16 September 2014 03:01:56AM 2 points [-]

Ok, that example does fit my conditions.

What if the universe cannot read your source code, but can simulate you? That is, the universe can predict your choices but it does not know what algorithm produces those choices. This is sufficient for the universe to pose Newcomb's problem, so the two agents are not identical.

The UDT agent can always do at least as well as the CDT agent by making the same choices as a CDT would. It will only give a different output if that would lead to a better result.

Comment author: dankane 16 September 2014 12:50:18AM 2 points [-]

For analogous reasons, a CDT agent would self-modify to do well on all Newcomblike problems that it would face in the future (e.g., it would precommit generally)

I am not convinced that this is the case. A self-modifying CDT agent is not caused to self-modify in favor of precommitment by facing a scenario in which precommitment would have been useful, but instead by evidence that such scenarios will occur in the future (and in fact will occur with greater frequency than scenarios that punish you for such precommitments).

Anyone who can credibly claim to have knowledge of the agent's original decision algorithm (e.g. a copy of the original source) can put the agent into such a situation, and in certain exotic cases this can be used to "blackmail" the agent in such a way that, even if it expects the scenario to happen, it still fails (for the same reason that CDT twoboxes even though it would precommit to oneboxing).

Actually, this seems like a bigger problem with UDT to me than with SMCDT (self-modifying CDT). Either type of program can be punished for being instantiated with the wrong code, but only UDT can be blackmailed into behaving differently by putting it in a Newcomb-like situation.

The story idea you had wouldn't work. Against a SMCDT agent, all that getting the AIs original code would allow people to do is to laugh at it for having been instantiated with code that is punished by the scenario they are putting it in. You manipulate a SMCDT agent by threatening to get ahold of its future code and punishing it for not having self-modified. On the other hand, against a UDT agent you could do stuff. You just have to tell it "we're going to simulate you and if the simulation behaves poorly, we will punish the real you". This causes the actual instantiation to change its behavior if it's a UDT agent but not if it's a CDT agent.

On the other hand, all reasonable self-modifying agents are subject to blackmail. You just have to tell them "every day that you are not running code with property X, I will charge you $1000000".

Comment author: one_forward 16 September 2014 01:44:13AM 2 points [-]

Can you give an example where an agent with a complete and correct understanding of its situation would do better with CDT than with UDT?

An agent does worse by giving in to blackmail only if that makes it more likely to be blackmailed. If a UDT agent knows opponents only blackmail agents that pay up, it won't give in.

If you tell a CDT agent "we're going to simulate you and if the simulation behaves poorly, we will punish the real you," it will ignore that and be punished. If the punishment is sufficiently harsh, the UDT agent that changed its behavior does better than the CDT agent. If the punishment is insufficiently harsh, the UDT agent won't change its behavior.

The only examples I've thought of where CDT does better involve the agent having incorrect beliefs. Things like an agent thinking it faces Newcomb's problem when in fact Omega always puts money in both boxes.

Comment author: Luke_A_Somers 27 August 2014 05:30:40PM 2 points [-]

Once you've asked about decoherence and irreversibility, that immediately raises the question of whether these are what we're aiming at, or something usually very closely related - or indeed whether these are the same thing at all! Suppose we have a quantum computer with three parts, each much larger than the previous.

  • Alice is a simulation of a putatively conscious entity. Suppose that the only reason we'd have not to think it's conscious is what we're about to do to it.
  • Alice's Room is an entropy sink Alice will interact with in the process of its being putatively conscious
  • In order to run Alice and Alice's Room, we also have an entropy sink we use for error correction.

We run Alice and Alice's Room forwards in time for a while, and Alice is doing a bunch of locally-irreversible computations, dumping the resulting entropy into Alice's Room instead of outer space.

At some point, we quantum-randomly either: 1) let Alice's Room shed entropy into outer space, causing the local irreversibility to become permanent, or 2) we time-reverse the dynamics of Alice and Alice's Room until we reach the initial state.

Was Alice conscious in case 1? In case 2? Since the sequence of events in both cases were in fact the same exact sequence of events - not merely identical, but referring to the exact same physically realized sequence of events - up to our quantum coinflip, it's nonsense to say that one was conscious and the other was not.

So yes, consciousness is connected to the arrow of time, but on a local level, not necessarily on the billion-year scale.

This lets us spit out that bullet about the Anti-deSitter space. If you're in an AdS space, you're going to choke on your own waste heat a zillion years before quantum billiards brings you back close to the starting point.

So, I'd say that there's consciousness inside this AdS trap, for a little while, until they die. When quantum billiards has again randomly lowered entropy to the point that a potentially conscious entity might have an entropy sink, then you can again have consciousness.

So, the AdS sphere is 99.999...(insert a lot)..99% not conscious, on account of its being dead, not on account of its being quantum-reversible.

Comment author: one_forward 28 August 2014 05:42:52PM 1 point [-]

wolfgang proposed a similar example on Scott's blog:

I wonder if we can turn this into a real physics problem:

1) Assume a large-scale quantum computer is possible (thinking deep thoughts, but not really self-conscious as long as its evolution is fully unitary).

2) Assume there is a channel which allows enough photons to escape in such a way to enable consciousness.

3) However, at the end of this channel we place a mirror – if it is in the consciousness-OFF position the photons are reflected back into the machine and unitarity is restored, but in the consciousness-ON position the photons escape into the deSitter universe.

4) As you can guess we use a radioactive device to set the mirror into c-ON or c-OFF position with 50% probability.

Will the quantum computer now experience i) a superposition of consciousness and unconsciousness or ii) will it always have a “normal” conscious experience or iii) will it have a conscious experience in 50% of the cases ?

Scott responded that

I tend to gravitate toward an option that’s not any of the three you listed. Namely: the fact that the system is set up in such a way that we could have restored unitarity, seems like a clue that there’s no consciousness there at all—even if, as it turns out, we don’t restore unitarity.

This answer is consistent with my treatment of other, simpler cases. For example, the view I’m exploring doesn’t assert that, if you make a perfect copy of an AI bot, then your act of copying causes the original to be unconscious. Rather, it says that the fact that you could (consistent with the laws of physics) perfectly copy the bot’s state and thereafter predict all its behavior, is an empirical clue that the bot isn’t conscious—even before you make a copy, and even if you never make a copy.

Comment author: polymer 25 August 2014 04:57:00PM *  2 points [-]

Can someone link to a discussion, or answer a small misconception for me?

We know P(A & B) < P(A). So if you add details to a story, it becomes less plausible. Even though people are more likely to believe it.

However, If I do an experiment, and measure something which is implied by A&B, then I would think "A&B becomes more plausible then A", Because A is more vague then A&B.

But this seems to be a contradiction.

I suppose, to me, adding more details to a story makes the story more plausible if those details imply the evidence. Sin(x) is an analytic function. If I know a complex differentiable function has roots on all multiples of pi, Saying the function is satisfied by Sin is more plausible then saying it's satisfied by some analytic function.

I think...I'm screwing up the semantics, since sin is an analytic function. But this seems to me to be missing the point.

I read a technical explanation of a technical explanation, so I know specific theories are better then vague theories (provided the evidence is specific). I guess I'm asking for clarifications on how this is formally consistent with P(A) > P(A&B).

Comment author: one_forward 25 August 2014 06:49:47PM 2 points [-]

A&B cannot be more probable than A, but evidence may support A&B more than it supports A.

For example, suppose you have independent prior probabilities of 1/2 for A and for B. The prior probability of A&B is 1/4. If you are then told "A iff B," the probability for A does not change but the probability of A&B goes up to 1/2.

The reason specific theories are better is not that they are more plausible, but that they contain more useful information.

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