Comment author: Nisan 15 January 2013 07:20:28AM 0 points [-]

This example doesn't satisfy the hypotheses of the theorem because you wouldn't want to optimize for v1 if your water was held fixed. Presumably, if you have 3 units of water and no food, you'd prefer 3 units of food to a 50% chance of 7 units of food, even though the latter leads to a higher expectation of v1.

Comment author: jacobt 15 January 2013 07:22:11AM 0 points [-]

You would if you could survive for v1*v2 days.

Comment author: DaFranker 14 January 2013 07:57:45PM *  0 points [-]

For example, you might want you and everyone else to both be happy, and happiness of one without the other would be much less valuable.

Now you've got me curious. I don't see what selections of values representative of the agent they're trying to model could possibly desire non-Pareto-optimal scenarios. The given example (quoted), for one, is something I'd represent like this:

Let x = my happiness, y = happiness of everyone else

To model the fact that each is worthless without the other, let:

v1 = min(x, 10y)
v2 = min(y, 10x)

Choice A: Gain 10 x, 0 y
Choice B: Gain 0 x, 10 y
Choice C: Gain 2 x, 2 y

It seems very obvious that the sole Pareto-optimal choice is the only desirable policy. Utility is four for choice C, and zero for A and B.

This may reduce to exactly what AlexMennen said, too, I guess. I have never encountered any intuition or decision problem that couldn't at-least-in-principle resolve to a utility function with perfect modeling accuracy given enough time and computational resources.

Comment author: jacobt 14 January 2013 10:40:04PM *  1 point [-]

I do think that everything should reduce to a single utility function. That said, this utility function is not necessarily a convex combination of separate values, such as "my happiness", "everyone else's happiness", etc. It could contain more complex values such as your v1 and v2, which depend on both x and y.

In your example, let's add a choice D: 50% of the time it's A, 50% of the time it's B. In terms of individual happiness, this is Pareto superior to C. It is Pareto inferior for v1 and v2, though.

EDIT: For an example of what I'm criticizing: Nisan claims that this theorem presents a difficulty for avoiding the repugnant conclusion if your desiderata are total and average happiness. If v1 = total happiness and v2 = average happiness, and Pareto optimality is desirable, then it follows that utility is a*v1 + b*v2. From this utility function, some degenerate behavior (blissful solipsist or repugnant conclusion) follows. However, there is nothing that says that Pareto optimality in v1 and v2 is desirable. You might pick a non-linear utility function of total and average happiness, for example atan(average happiness) + atan(total happiness). Such a utility function will sometimes pick policies that are Pareto inferior with respect to v1 and v2.

Comment author: AlexMennen 14 January 2013 07:33:02PM 3 points [-]

Given some value v1 that you are risk averse with respect to, you can find some value v1' that your utility is linear with. For example, if with other values fixed, utility = log(v1), then v1':=log(v1). Then just use v1' in place of v1 in your optimization. You are right that it doesn't make sense to maximize the expected value of a function that you don't care about the expected value of, but if you are VNM-rational, then given an ordinal utility function (for which the expected value is meaningless), you can find a cardinal utility function (which you do want to maximize the expected value of) with the same relative preference ordering.

Comment author: jacobt 14 January 2013 10:35:12PM 2 points [-]

I didn't say anything about risk aversion. This is about utility functions that depend on multiple different "values" in some non-convex way. You can observe that, in my original example, if you have no water, then utility (days survived) is linear with respect to food.

Comment author: DaFranker 14 January 2013 02:49:14PM *  2 points [-]

Wha...?

I believe your Game is badly-formed. This doesn't sound at all like how Games should be modeled. Here, you don't have two agents each trying to maximize something that they value of their own, so you can't use those tricks.

As a result, apparently you're not properly representing utility in this model. You're implicitly assuming the thing to be maximized is health and life duration, without modeling it at all. With the model you make, there are only two values, food and water. The agent does not care about survival with only those two Vs. So for this agent, yes, picking one of the "1000" options really truly spectacularly trivially is better. The agent just doesn't represent your own preferences properly, that's all.

If your agent cares at all about survival, there should be a value for survival in there too, probably conditionally dependent on how much water and food is obtained. Better yet, you seem to be implying that the amount of food and water obtained isn't really important, only surviving longer is - strike out the food and water values, only keep a "days survived" value dependent upon food and water obtained, and then form the Game properly.

Comment author: jacobt 14 January 2013 06:55:14PM 0 points [-]

I think we agree. I am just pointing out that Pareto optimality is undesirable for some selections of "values". For example, you might want you and everyone else to both be happy, and happiness of one without the other would be much less valuable.

I'm not sure how you would go about deciding if Pareto optimality is desirable, now that the theorem proves that it is desirable iff you maximize some convex combination of the values.

Comment author: jacobt 14 January 2013 06:35:57AM *  1 point [-]

I think that, depending on what the v's are, choosing a Pareto optimum is actually quite undesirable.

For example, let v1 be min(1000, how much food you have), and let v2 be min(1000, how much water you have). Suppose you can survive for days equal to a soft minimum of v1 and v2 (for example, 0.001 v1 + 0.001 v2 + min(v1, v2)). All else being equal, more v1 is good and more v2 is good. But maximizing a convex combination of v1 and v2 can lead to avoidable dehydration or starvation. Suppose you assign weights to v1 and v2, and are offered either 1000 of the more valued resource, or 100 of each. Then you will pick the 1000 of the one resource, causing starvation or dehydration after 1 day when you could have lasted over 100. If which resource is chosen is selected randomly, then any convex optimizer will die early at least half the time.

A non-convex aggregate utility function, for example the number of days survived (0.001 v1 + 0.001 v2 + min(v1, v2)), is much more sensible. However, it will not select Pareto optima. It will always select the 100 of each option; always selecting 1000 of one leads to greater expected v1 and expected v2 (500 for each).

Comment author: Vladimir_Nesov 23 September 2012 06:49:28PM 4 points [-]

It's often pointless to argue about probabilities, and sometimes no assignment of probability makes sense, so I was careful to phrase the thought experiment as a decision problem. Which decision (strategy) is the right one?

Comment author: jacobt 23 September 2012 07:12:31PM 1 point [-]

Actually you're right, I misread the problem at first. I thought that you had observed yourself not dying 1000 times (rather than observing "heads" 1000 times), in which case you should keep playing.

Applying my style of analyzing anthropic problems to this one: Suppose we have 1,000,000 * 2^1000 players. Half flip heads initially, half flip tails. About 1,000,000 will get heads 1,000 times. Of them, 500,000 will have flipped heads initially. So, your conclusion is correct.

Comment author: jacobt 23 September 2012 06:37:52PM *  1 point [-]

I think you're wrong. Suppose 1,000,000 people play this game. Each of them flips the coin 1000 times. We would expect about 500,000 to survive, and all of them would have flipped heads initially. Therefore, P(I flipped heads initially | I haven't died yet after flipping 1000 coins) ~= 1.

This is actually quite similar to the Sleeping Beauty problem. You have a higher chance of surviving (analogous to waking up more times) if the original coin was heads. So, just as the fact that you woke up is evidence that you were scheduled to wake up more times in the Sleeping Beauty problem, the fact that you survive is evidence that you were "scheduled to survive" more in this problem.

On the other hand, each "heads" you observe doesn't distinguish the hypothetical where the original coin was "heads" from one where it was "tails".

This is the same incorrect logic that leads people to say that you "don't learn anything" between falling asleep and waking up in the Sleeping Beauty problem.

I believe the only coherent definition of Bayesian probability in anthropic problems is that P(H | O) = the proportion of observers who have observed O, in a very large universe (where the experiment will be repeated many times), for whom H is true. This definition naturally leads to both 2/3 probability in the Sleeping Beauty problem and "anthropic evidence" in this problem. It is also implied by the many-worlds interpretation in the case of quantum coins, since then all those observers really do exist.

Comment author: jacobt 20 July 2012 07:28:53AM 13 points [-]

I vote for range voting. It has the lowest Bayesian regret (best expected social utility). It's also extremely simple. Though it's not exactly the most unbiased source, rangevoting.org has lots of information about range voting in comparison to other methods.

Comment author: jacobt 06 June 2012 09:22:42PM *  2 points [-]

For aliens with a halting oracle:

Suppose the aliens have this machine that may or may not be a halting oracle. We give them a few Turing machine programs and they decide which ones halt and which ones don't. Then we run the programs. Sure enough, none of the ones they say run forever halt, and some of them they say don't run forever will halt at some point. Suppose we repeat this process a few times with different programs.

Now what method should we use to predict the point at which new programs halt? The best strategy seems to be to ask the aliens which ones halt, give the non-halting ones a 0 probability of halting at every step, and give the other ones some small nonzero probability of halting on every step. Of course this strategy only works if the aliens actually have a halting oracle so it its predictions should be linearly combined with a fallback strategy.

I think that Solomonoff induction will find this strategy, because the hypothesis that the aliens have a true halting oracle is formalizable. Here's how: we learn a function from aliens' answers -> distribution over when the programs halt. We can use the strategy of predicting that the ones that the aliens say don't halt don't halt, and using some fallback mechanism for predicting the ones that do halt. This strategy is computable so Solomonoff induction will find it.

For Berry's paradox:

This is a problem with every formalizable probability distribution. You can always define a sequence that says: see what bit the predictor predicts with a higher probability, then output the opposite. Luckily Solomonoff induction does the best it can by having the estimates converge to 50%. I don't see what a useful solution to this problem would even look like; it seems best to just treat the uncomputable sequence as subjectively random.

Comment author: jacobt 02 June 2012 12:57:37AM 1 point [-]

I found this post interesting but somewhat confusing. You start by talking about UDT in order to talk about importance. But really the only connection from UDT to importance is the utility function, so you might as well start with that. And then you ignore utility functions in the rest of your post when you talk about Schmidhuber's theory.

It just has a utility function which specifies what actions it should take in all of the possible worlds it finds itself in.

Not quite. The utility function doesn't specify what action to take, it specifies what worlds are desirable. UDT also requires a prior over worlds and a specification of how the agent interacts with the world (like the Python programs here). The combination of this prior and the expected value computations that UDT does would constitute "beliefs".

Informally, your decision policy tells you what options or actions to pay most attention to, or what possibilities are most important.

I don't see how this is. Your decision policy tells you what to do once you already know what you can do. If you're using "important" to mean "valuable" just say that instead.

I do like the idea of modelling the mind as an approximate compression engine. This is great for reducing some thought processes to algorithms. For example I think property dualism can be thought of as a way to compress the fact that I am me rather than some other person, or at least make explicit the fact that this must be compressed.

Schmidhuber's theory is interesting but incomplete. You can create whatever compression problem you want through a template, e.g. a pseudorandom sequence you can only compress by guessing the seed. Yet repetitions of the same problem template are not necessarily interesting. It seems that some bits are more important than other bits; physicists are very interested in compressing the number of spacial dimensions in the universe even though this quantity can be specified in a few bits. I don't know any formal approaches to quantifying the importance of compressing different things.

I wrote a paper on this subject (compression as it relates to theory of the mind). I also wrote this LessWrong post about using compression to learn values.

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