jsteinhardt comments on A Thought on Pascal's Mugging - Less Wrong

12 Post author: komponisto 10 December 2010 06:08AM

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Comment author: wedrifid 12 December 2010 07:59:06AM 4 points [-]

The way around Pascal's mugging is to have a bounded utility function.

Way around? If my utility function suggests that being mugged by Pascal is the best thing for me to do then I'll be delighted to do it.

Utility functions determine our decisions, not the reverse!

Comment author: jsteinhardt 12 December 2010 10:58:53PM 1 point [-]

A utility function shouldn't suggest anything. It is simply an abstract mathematical function that is guaranteed to exist by the VNM utility theorem. If you're letting an unintuitive mathematical theorem tell you to do things that you don't want to do, then something is wrong.

Again, the problem is there is a namespace collision between the utility function guaranteed by VNM, which we are maximizing the expected value of, and the utility function that we intuitively associate with our preferences, which we (probably) aren't maximizing the expected value of. VNM just says that if you have consistent preferences, then there is some function whose expected value you are maximizing. It doesn't say that this function has anything to do with the degree to which you want various things to happen.

I seem to be having a lot of trouble getting this point across, so let me try to put it another way: Ignore Kolmogorov complexity, priors, etc. for a moment, and if you can, forget about your utility function and just ask yourself what you would want. Now imagine the worst possible thing that could happen (you can even suppose that both time and space are potentially infinite, so infinitely many people being tortured for infinite extents of time is fine). Let us call this thing X. Suppose that you have somehow calculated that, with probability 10^(-100), the mugger will cause X to happen if you don't pay him $5. Would you pay him? If you would pay him, then why?

I am actually quite interested in the answer to this question, because I am having trouble diagnosing the precise source of my disagreement on this issue. And even though I said to forget about utility functions, if you really think that is the answer to the "why" question, feel free to use them in your argument. As I said, at this point I am most interested in determining why we disagree, because previous discussions with other people suggest that there is some hidden inferential distance afoot.

As an aside, if you wouldn't pay him then the definition of utility implies that u($5) > 10^(-100) u(X), which implies that u(X), and therefore the entire utility function, is bounded.

Comment author: komponisto 13 December 2010 12:24:12AM 1 point [-]

Now imagine the worst possible thing that could happen

As was pointed out in the other subthread, you are assuming the conclusion you wish to prove here, viz. that the utility function is (necessarily) bounded.

Comment author: jsteinhardt 13 December 2010 01:39:06AM *  -1 points [-]

Fine, I was slightly sloppy in my original proof (not only in the way you pointed out, but also in keeping track of signs). Here is a rigorous version:

Suppose that there is nothing so bad that you would pay $5 to stop it from happening with probability 10^(-100). Let X be a state of the universe. Then u(-$5) < 10^(-100) u(X), so u(X) > 10^(100) u(-$5). Since u(X) > 10^(100) u(-$5) for all X, u is bounded below.

Similarly, suppose that there is nothing so good that you would pay $5 to have a 10^(-100) chance of it happening. Then u($5) > 10^(100) u(X) for all X, so u(X) < 10^(100) u($5), hence u is also bounded above.

Now I've given proofs that u is bounded both above and below, without looking at argmax u or argmin u (which incidentally probably don't exist even if u is bounded; it is much more likely that u asymptotes out).

My proof is still not entirely rigorous, for instance u(-$5) and u($5) will in general depend on my current level of income / savings. If you really want me to, I can write everything out completely rigorously, but I've been trying to avoid it because I find that diving into unnecessary levels of rigor only obscures the underlying intuition (and I say this as someone who studies math).

Comment author: komponisto 13 December 2010 01:47:03AM 0 points [-]

Suppose that there is nothing so bad that you would pay $5 to stop it from happening with probability 10^(-100)

Again, why assume this?

Comment author: jsteinhardt 13 December 2010 02:31:52AM -1 points [-]

Your question has two possible meanings to me, so I'll try to answer both.

Meaning 1: Why is this a reasonable assumption in the context of the current debate?

Answer: Because if there was something that bad, then you get Pascal's mugged in my hypothetical situation. What I have shown is that either you would give Pascal $5 in that scenario, or your utility function is bounded.

Meaning 2: Why is this a reasonable assumption in general?

Answer: Because things that occur with probability 10^(-100) don't actually happen. Actually, 10^(-100) might be a bit high, but certainly things that occur with probability 10^(-10^(100)) don't actually happen.

Comment author: komponisto 13 December 2010 02:44:14AM 2 points [-]

Because if there was something that bad, then you get Pascal's mugged in my hypothetical situation

You seem not to have understood the post. The worse something is, the more difficult it is for the mugger to make the threat credible. There may be things that are so bad that I (or my hypothetical AI) would pay $5 not to raise their probability to 10^(-100), but such things have prior probabilities that are lower than 10^(-100), and a mugger uttering the threat will not be sufficient evidence to raise the probability to 10^(-100).

Answer: Because things that occur with probability 10^(-100) don't actually happen. Actually, 10^(-100) might be a bit high, but certainly things that occur with probability 10^(-10^(100)) don't actually happen.

We don't need to declare 10^(-100) equal to 0. 10^(-100) is small enough already.

Comment author: jsteinhardt 13 December 2010 02:59:05AM 0 points [-]

I have to admit that I did find the original post somewhat confusing. However, let me try to make sure that I understood it. I would summarize your idea as saying that we should have u(X) = O(1/p(X)), where u is the utility function and p is our posterior estimate of X. Is that correct? Or do you want p to be the prior estimate? Or am I completely wrong?

Comment author: komponisto 13 December 2010 03:11:22AM 1 point [-]

Yes, p should be the prior estimate. The point being that the posterior estimate is not too different from the prior estimate in the "typical" mugging scenario (i.e. someone says "give me $5 or I'll create 3^^^^3 units of disutility" without specifying how in enough detail).

Comment author: jsteinhardt 13 December 2010 06:19:09PM 0 points [-]

So, backing up, let me put forth my biggest objections to your idea, as I see it. I will try to stick to only arguing about this point until we can reach a consensus.

I do not believe there is anything so bad that you would trade $5 to prevent it from happening with probability 10^(-500). If there is, please let me know. If not, then this is a statement that is independent of your original priors, and which implies (as noted before) that your utility function is bounded.

I concede that the condition u(X) = O(1/p(X)) implies that one would be immune to the classical version of the Pascal's mugging problem. What I am trying to say now is that it fails to be immune to other variants of Pascal's mugging that would still be undesirable. While a good decision theory should certainly be immune to [the classical] Pascal's mugging, a failure to be immune to other mugging variants still raises issues.

My claim (which I supported with math above) is that the only way to be immune to all variants of Pascal's mugging is to have a bounded utility function.

My stronger claim, in case you agree with all of the above but think it is irrelevant, is that all humans have a bounded utility function. But let's avoid arguing about this point until we've resolved all of the issues in the preceding paragraphs.

Comment author: jsteinhardt 13 December 2010 03:36:46AM 0 points [-]

Okay, thanks. So it is the prior, not the posterior, which makes more sense (as the posterior will be in general changing while the utility function remains constant).

My objection to this is that, even though you do deal with the "typical" mugging scenario, you run into issues in other scenarios. For instance, suppose that your prior for X is 10^(-1000), and your utility for X is 10^750, which I believe fits your requirements. Now suppose that I manage to argue your posterior up to 10^(-500). Either you can get mugged (for huge amounts of money) in this circumstance, or your utility on X is actually smaller than 10^(500).

Comment author: wedrifid 13 December 2010 02:12:33AM 0 points [-]

As an aside, if you wouldn't pay him then the definition of utility implies that u($5) > 10^(-100) u(X), which implies that u(X), and therefore the entire utility function, is bounded.

This doesn't actually imply that the entire utility function is bounded. It is still possible that u(Y) is infinite, where Y is something that is valued positively.

As an aside we can now consider the possibility of Pascal's Samaritan.

Assume a utility function such that u(Y) is infinite (and neutral with respect to risk). Further assume that you predict that $5 would increase your chance of achieving Y by 1/3^^^3. A Pascal Samaritan can offer to pay you $5 for the opportunity to give you a 90% chance of sending the entire universe into the hell state X. Do you take the $5?

Comment author: jsteinhardt 13 December 2010 02:35:16AM *  0 points [-]

From my reply to komponisto (incidentally, both you and he seem to be making the same objections in parallel, which suggests that I'm not doing a very good job of explaining myself, sorry):

Suppose that there is nothing so bad that you would pay $5 to stop it from happening with probability 10^(-100). Let X be a state of the universe. Then u(-$5) < 10^(-100) u(X), so u(X) > 10^(100) u(-$5). Since u(X) > 10^(100) u(-$5) for all X, u is bounded below.

Similarly, suppose that there is nothing so good that you would pay $5 to have a 10^(-100) chance of it happening. Then u($5) > 10^(100) u(X) for all X, so u(X) < 10^(100) u($5), hence u is also bounded above.

Comment author: wedrifid 13 December 2010 02:00:07AM 0 points [-]

As I said, at this point I am most interested in determining why we disagree,

The meaning of a phrase, primarily. And slightly about the proper use of an abstract concept.

A utility function should be a representation of my values. If my values are such that paying a mugger is the best option then I am glad to pay a mugger.

Suppose that you have somehow calculated that, with probability 10^(-100), the mugger will cause X to happen if you don't pay him $5. Would you pay him? If you would pay him, then why?

If I were to pay him it would be because I happen to value not having a 10^(-100) chance of X happening more than I value $5.

As an aside, if you wouldn't pay him then the definition of utility implies that u($5) > 10^(-100) u(X), which implies that u(X), and therefore the entire utility function, is bounded.

My utility function quite likely is bounded. Not because that is a way around pascal's mugging. Simply because that happens to be what the arbitrary value system represented by this particular bunch of atoms happens to be.

Comment author: jsteinhardt 13 December 2010 02:50:03AM 0 points [-]

Hm...it sounds like we agree on far more than I thought, then.

What I am saying is that my utility function is bounded because it would be ridiculous to be Pascal's mugged, even in the hypothetical universe I created that disobeys komponisto's priors. Put another way, I am simply not willing to seriously consider events at probabilities of, say, 10^(-10^(100)), because such events don't happen. For this same reason, I have a hard time taking anyone seriously who claims to have an unbounded utility function, because they would then care about events that can't happen in a sense at least as strong as the sense that 1 is not equal to 2.

Would you object to anything in the above paragraph? Thanks for bearing with me on this, by the way.

P.S. Am I the only one who is always tempted to write "mugged by Pascal" before realizing that this is comically different from being "Pascal's mugged"?

Comment author: wedrifid 13 December 2010 03:26:32AM 0 points [-]

Put another way, I am simply not willing to seriously consider events at probabilities of, say, 10^(-10^(100)), because such events don't happen.

As far as I know they do happen. To know that such a number cannot represent an altogether esoteric feature of the universe that can nevertheless be the legitimate subject of infinite value I would need to know the smallest number that can be assigned to a quantum state.

(This objection is purely tangential. See below for significant disagreement.)

I have a hard time taking anyone seriously who claims to have an unbounded utility function, because they would then care about events that can't happen in a sense at least as strong as the sense that 1 is not equal to 2.

That isn't true. Someone can assign infinite utility to Australia winning the ashes if that is what they really want. I'd think them rather silly but that is just my subjective evaluation, nothing to do with maths.

Comment author: jsteinhardt 13 December 2010 04:18:59AM *  0 points [-]

To know that such a number cannot represent an altogether esoteric feature of the universe that can nevertheless be the legitimate subject of infinite value I would need to know the smallest number that can be assigned to a quantum state.

I think you are conflating quantum probabilities with Bayesian probabilities here, but I'm not sure. Unless you think this point is worth discussing further I'll move on to your more significant disagreement.

Someone can assign infinite utility to Australia winning the ashes if that is what they really want. I'd think them rather silly but that is just my subjective evaluation, nothing to do with maths.

Hm...I initially wrote a two-paragraph explanation of why you were wrong, then deleted it because I changed my mind. So, I think we are making progress!

I initially thought I accorded disdain to unbounded utility functions for the same reason that I accorded disdain to ridiculous priors. But the difference is that your priors affect your epistemic state, and in the case of beliefs there is only one right answer. On the other hand, there is nothing inherently wrong with being a paperclip maximizer.

I think the actual issue I'm having is that I suspect that most people who claim to have unbounded utility functions would have been unwilling to make the trades implied by this before reading about VNM utility / "Shut up and multiply". So my objection is not that unbounded utility functions are inherently wrong, but that they cannot possibly reflect the preferences of a human.

Comment author: wedrifid 13 December 2010 04:33:00AM 1 point [-]

I think the actual issue I'm having is that I suspect that most people who claim to have unbounded utility functions would have been unwilling to make the trades implied by this before reading about VNM utility / "Shut up and multiply". So my objection is not that unbounded utility functions are inherently wrong, but that they cannot possibly reflect the preferences of a human.

On this I believe we approximately agree.