A seemingly trivial result, that I haven't seen posted anywhere in this form, that I could find. It simply shows that we expect evidence to increase the posterior probability of the true hypothesis.

Let H be the true hypothesis/model/environment/distribution, and ~H its negation. Let e be evidence we receive, taking values e₁, e₂, ... e_n. Let p_i=P(e=e_i|H) and q_i=P(E=e_i|~H).

The expected posterior weighting of H, P(e|H), is Σp_ip_i while the expected posterior weighting of ~H, P(e|~H), is Σq_ip_i. Then since the p_i and q_i both sum to 1, Cauchy–Schwarz implies that

• E(P(e|H)) ≥ E(P(e|~H)).

Thus, in expectation, the probability of the evidence given the true hypothesis, is higher than or equal to the probability of the evidence given its negation.

This, however, doesn't mean that the Bayes factor - P(e|H)/P(e|~H) - must have expectation greater than one, since ratios of expectation are not the same as expectations of ratio. The Bayes factor given e=e_i is (p_i/q_i). Thus the expected Bayes factor is Σ(p_i/q_i)p_i. The negative logarithm is a convex function; hence by Jensen's inequality, -log[E(P(e|H)/P(e|~H))] ≤ -E[log(P(e|H)/P(e|~H))]. That last expectation is Σ(log(p_i/q_i))p_i. This is the Kullback–Leibler divergence of P(e|~H) from P(e|H), and hence is non-negative. Thus log[E(P(e|H)/P(e|~H))] ≥ 0, and hence

• E(P(e|H)/P(e|~H)) ≥ 1.

Thus, in expectation, the Bayes factor, for the true hypothesis versus its negation, is greater than or equal to one.

Note that this is not true for the inverse. Indeed E(P(e|~H)/P(e|H)) = Σ(q_i/p_i)p_i = Σq_i = 1.

In the preceding proofs, ~H played no specific role, and hence

• For all K,    E(P(e|H)) ≥ E(P(e|K))    and    E(P(e|H)/P(e|K)) ≥ 1    (and E(P(e|K)/P(e|H)) = 1).

Thus, in expectation, the probability of the true hypothesis versus anything, is greater or equal in both absolute value and ratio.

Now we can turn to the posterior probability P(H|e). For e=e_i, this is P(H)*P(e=e_i|H)/P(e=e_i). We can compute the expectation of P(e|H)/P(e) as above, using the non-negative Kullback–Leibler divergence of P(e) from P(e|H), and thus showing it has an expectation greater than or equal to 1. Hence:

• E(P(H|e)) ≥ P(H).

Thus, in expectation, the posterior probability of the true hypothesis is greater than or equal to its prior probability.

In a previous post, I looked at some of the properties of using the median rather than the mean.

Inspired by Househalter's comment, it seems we might be able to take a compromise between median and mean. It seems to me that simply taking the mean of the lower quartile, median, and upper quartile would also have the nice features I described, and would likely be closer to the mean.

Furthermore, there's no reason to stop there. We can take the mean of the n-1 n-quantiles.

Two questions:

1. As n increases, does this quantity tend to the mean if it exists? (I suspect yes).
2. For some distributions (eg Cauchy distribution) this quantity will tend to a limit as n increases, even if there is no mean. Is this an effective way of extending means to distributions that don't possess them?

Note the unlike the median approach, for large enough n, this maximiser will pay Pascal's mugger.

tl;dr A median maximiser will expect to win. A mean maximiser will win in expectation. As we face repeated problems of similar magnitude, both types take on the advantage of the other. However, the median maximiser will turn down Pascal's muggings, and can say sensible things about distributions without means.

Prompted by some questions from Kaj Sotala, I've been thinking about whether we should use the median rather than the mean when comparing the utility of actions and policies. To justify this, see the next two sections: why the median is like the mean, and why the median is not like the mean.

Why the median is like the mean

The main theoretic justifications for the use of expected utility - hence of means - are the von Neumann Morgenstern axioms. Using the median obeys the completeness and transitivity axioms, but not the continuity and independence ones.

It does obey weaker forms of continuity; but in a sense, this doesn't matter. You can avoid all these issues by making a single 'ultra-choice'. Simply list all the possible policies you could follow, compute their median return, and choose the one with the best median return. Since you're making a single choice, independence doesn't apply.

So you've picked the policy π_m with the highest median value - note that to do this, you need only know an ordinal ranking of worlds, not their cardinal values. In what way is this like maximising expected utility? Essentially, the more options and choices you have - or could hypothetically have - the closer this policy must be to expected utility maximalisation.

Assume u is a utility function compatible with your ordinal ranking of the worlds. Then π_u = 'maximise the expectation of u' is also a policy choice. If we choose π_m, we get a distribution d_mu of possible values of u. Then E(u|π_m) is within the absolute deviation (using d_mu) of the median value of d_mu. This absolute deviation always exists for any distribution with an expectation, and is itself bounded by the standard deviation, if it exists.

Thus maximising the median is like maximising the mean, with an error depending on the standard deviation. You can see it as a risk averse utility maximising policy (I know, I know - risk aversion is supposed to go in defining the utility, not in maximising it. Read on!). And as we face more and more choices, the standard deviation will tend to fall relative to the mean, and the median will cluster closer and closer to the mean.

For instance, suppose we consider the choice of whether to buckle our seatbelt or not. Assume we don't want to die in a car accident that a seatbelt could prevent; assume further that the cost of buckling a seatbelt is trivial but real. To simplify, suppose we have an independent 1/Ω chance of death every time we're in a car, and that a seatbelt could prevent this, for some large Ω. Furthermore, we will be in a car a total of ρΩ, for ρ < 0.5. Now, it seems, the median recommends a ridiculous policy: never wear seatbelts. Then you pay no cost ever, and your chance of dying is less than 50%, so this has the top median.

And that is indeed a ridiculous result. But it's only possible because we look at seatbelts in isolation. Every day, we face choices that have small chances of killing us. We could look when crossing the street; smoke or not smoke cigarettes; choose not to walk close to the edge of tall buildings; choose not to provoke co-workers to fights; not run around blindfolded. I'm deliberately including 'stupid things no-one sensible would ever do', because they are choices, even if they are obvious ones. Let's gratuitously assume that all these choices also have a 1/Ω chance of killing you. When you collect together all the possible choices (obvious or not) that you make in your life, this will be ρ'Ω choice, for ρ' likely quite a lot bigger than 1.

Assume that avoiding these choices has a trivial cost, incommensurable with dying (ie no matter how many times you have to buckle your seatbelt, it still better than a fatal accident). Now median-maximisation will recommend taking safety precautions for roughly (ρ'-0.5)Ω of these choices. This means that the decision of a median maximiser will be close to those of a utility maximiser - they take almost the same precautions - though the outcomes are still pretty far apart: the median maximiser accepts a 49.99999...% chance of death.

But now add serious injury to the mix (still assume the costs are incommensurable). This has a rather larger probability, and the median maximiser will now only accept a 49.99999...% chance of serious injury. Or add light injury - now they only accept a 49.99999...% chance of light injury. If light injuries are additive - two injuries are worse than one - then the median maximiser becomes even more reluctant to take risks. We can now relax the assumption of incommensurablility as well; the set of policies and assessments becomes even more complicated, and the median maximiser moves closer to the mean maximiser.

The same phenomena tends to happen when we add lotteries of decisions, chained decisions (decisions that depend on other decisions), and so on. Existential risks are interesting examples: from the selfish point of view, existential risks are just other things that can kills us - and not the most unlikely ones, either. So the median maximiser will be willing to pay a trivial cost to avoid an xrisk. Will a large group of median maximisers be willing to collectively pay a large cost to avoid an xrisk? That gets into superrationality, which I haven't considered yet in this context.

But let's turn back to the mystical utility function that we are trying to maximise. It's obvious that humans don't actually maximise a utility function; but according to the axioms, we should do so. Since we should, people on this list tend to often assume that we actually have one, skipping over the process of constructing it. But how would that process go? Let's assume we've managed to make our preferences transitive, already a major good achievement. How should we go about making them independent as well? We can do so as we go along. But if we do it ahead of time, chances are that we will be comparing hypothetical situations ("Do I like chocolate twice as much as sex? What would I think of a 50% chance of chocolate vs guaranteed sex? Well, it depends on the situation...") and thus construct a utility function. This is where we have to make decisions about very obscure and unintuitive hypothetical tradeoffs, and find a way to fold all our risk aversion/risk love into the utility.

When median maximising, we do exactly the same thing, except we constrain ourselves to choices that are actually likely to happen to us. We don't need a full ranking of all possible lotteries and choices; we just need enough to decide in the situations we are likely to face. You could consider this a form of moral learning (or preference learning). From our choices in different situations (real or possible), we decide what our preferences are in these situations, and this determines our preferences overall.

Why the median is not like the mean

Ok, so the previous paragraph argues that median maximising, if you have enough choices, functions like a clunky version of expected utility maximising. So what's the point?

The point is those situations that are not faced sufficiently often, or that have extreme characteristics. A median maximiser will reject Pascal's mugging, for instance, without any need for extra machinery (though they will accept Pascal's muggings if they face enough independent muggings, which is what we want - for stupidly large values of "enough"). They cope fine with distributions that have no means - such as the Cauchy distribution or a utility version of the St Petersburg paradox. They don't fall into paradox when facing choices with infinite (but ordered) rewards.

In a sense, median maximalisation is like expected utility maximalisation for common choices, but is different for exceptionally unlikely or high impact choices. Or, from the opposite perspective, expected utility maximising gives high probability of good outcomes for common choices, but not for exceptionally unlikely or high impact choices.

Another feature of the general idea (which might be seen as either a plus or a minus) is that it can get around some issues with total utilitarianism and similar ethical systems (such as the repugnant conclusion). What do I mean by this? Well, because the idea is that only choices that we actually expect to make matter, we can say, for instance, that we'd prefer a small ultra happy population to a huge barely-happy one. And if this is the only choice we make, we need not fear any paradoxes: we might get hypothetical paradoxes, just not actual ones. I won't put too much insistence on this point, I just thought it was an interesting observation.

For lack of a Cardinal...

Now, the main issue is that we might feel that there are certain rare choices that are just really bad or really good. And we might come to this conclusion by rational reasoning, rather than by experience, so this will not show up in the median. In these cases, it feels like we might want to force some kind of artificial cardinal order on the worlds, to make the median maximiser realise that certain rare events must be considered beyond their simple ordinal ranking.

In this case, maybe we could artificially add some hypothetical choices to our system, making us address these questions more than we actually would, and thus drawing them closer to the mean maximising situation. But there may be other, better ways of doing this.

Anyway, that's my first pass at constructing a median maximising system. Comments and critics welcome!

EDIT: We can use the absolute deviation (technically, the mean absolute deviation around the mean) to bound the distance between median and mean. This itself is bounded by the standard deviation, if it exists.

A putative new idea for AI control; index here.

There's a result that's almost a theorem, which is that an agent that is an expected utility maximiser, is an agent that is stable under self-modification (or the creation of successor sub-agents).

Of course, this needs to be for "reasonable" utility, where no other agent cares about the internal structure of the agent (just its decisions), where the agent is not under any "social" pressure to make itself into something different, where the boundedness of the agent itself doesn't affect its motivations, and where issues of "self-trust" and acausal trade don't affect it in relevant ways, etc...

So quite a lot of caveats, but the result is somewhat stronger in the opposite direction: an agent that is not an expected utility maximiser is under pressure to self-modify itself into one that is. Or, more correctly, into an agent that is isomorphic with an expected utility maximiser (an important distinction).

What is this "pressure" agent are "under"? The known result is that if an agent obeys four simple axioms, then its behaviour must be isomorphic with an expected utility maximiser. If we assume the Completeness axiom (trivial) and Continuity (subtle), then violations of Transitivity or Independence correspond to situations where the agent has been money pumped - lost resources or power for no gain at all. The more likely the agent is to face these situations, the more pressure they're under to behave as an expected utility maximiser, or simply lose out.

Unbounded agents

I have two models for how idealised agents could deal with this sort of pressure. The first, post-hoc, is the unlosing agent I described here. The agent follows whatever preferences it had, but kept track of its past decisions, and whenever it was in a position to violate transitivity or independence in a way that it would suffer from, it makes another decision instead.

Another, pre-hoc, way of dealing with this is to make an "ultra choice" and choose between not decisions, but all possible input output maps (equivalently, between all possible decision algorithms), looking to the expected consequences of each one. This reduces the choices to a single choice, where issues of transitivity or independence need not necessarily apply.

Bounded agents

Actual agents will be bounded, unlikely to be able to store and consult their entire history when making every single decision, and unable to look at the whole future of their interactions to make a good ultra choice. So how would they behave?

This is not determined directly by their preferences, but by some sort of meta-preferences. Would they make an approximate ultra-choice? Or maybe build up a history of decisions, and then simplify it (when it gets to large to easily consult) into a compatible utility function? This is also determined by their interactions, as well - an agent that makes a single decision has no pressure to be an expected utility maximiser, one that makes trillions of related decisions has a lot of pressure.

It's also notable that different types of boundedness (storage space, computing power, time horizons, etc...) have different consequences for unstable agents, and would converge to different stable preference systems.

Investigation needed

So what is the point of this post? It isn't presenting new results; it's more an attempt to launch a new sub-field of investigation. We know that many preferences are unstable, and that the agent is likely to make them stable over time, either through self-modification, subagents, or some other method. There are also suggestions for preferences that are known to be unstable, but have advantages (such as resistance to Pascal Muggings) that standard maximalisation does not.

Therefore, instead of saying "that agent design can never be stable", we should be saying "what kind of stable design would that agent converge to?", "does that convergent stable design still have the desirable properties we want?" and "could we get that stable design directly?".

The first two things I found in this area were that traditional satisficers could converge to vastly different types of behaviour in an essentially unconstrained way, and that a quasi-expected utility maximiser of utility u might converge to an expected utility maximiser, but it might not be u that it maximises.

In fact, we need not look only at violations of the axioms of expected utility; they are but one possible reason for decision behaviour instability. Here are some that spring to mind:

1. Non-independence and non-transitivity (as above).
2. Boundedness of abilities.
3. Adversaries and social pressure.
4. Evolution (survival cost to following “odd” utilities (eg time-dependent preference)).
5. Unstable decision theories (such as CDT).

Now, some categories (such as "Adversaries and social pressure") may not possess a tidy stable solution, but it is still worth asking what setups are more stable than others, and what the convergence rules are expected to be.

I use the phrase 'clever argument' deliberately: I have reached a conclusion that contradicts the usual wisdom around here, and want to check that I didn't make an elementary mistake somewhere.

Consider a lottery ticket that costs $100 for a one-in-ten-thousand chance of winning a million dollars, expected value, $100. I can take this deal or leave it, and of course a realistic ticket actually costs 100+epsilon where epsilon covers the arranger's profit, which is a bad deal.

But now consider this deal in terms of time. Suppose I've got a well-paid job in which it takes me an hour to earn that $100. Suppose further that I work 40 hours a week, 50 weeks a year, and that my living expenses are a modest $40k a year, making my yearly savings $160k. Then, with 4% interest on my $160k yearly, it would take me about 5.5 years to accumulate that million dollars, or 11000 hours. Also note that with these assumptions, once I have my million I don't need to work any more.

It seems to me that, given the assumptions above, I could view the lottery deal as paying one hour of my life for a one-in-ten-thousand chance to win 11000 hours, expected value, 1.1 hours. (Note that leisure hours when young are probably worth more, since you'll be in better health to enjoy it; but this is not necessary to the argument.)

Of course it is possible to adjust the numbers. For example, I could scrimp and save during my working years, and make my living expenses only 20k; in that case it would take me less than 5 years to accumulate the million, and the ticket goes back to being a bad deal. Alternatively, if I spend more than 40k a year, it takes longer to accumulate the million; in this case my standard of living drops when I retire to live off my 4% interest, but the lottery ticket becomes increasingly attractive in terms of hours of life.

I think, and I could be mistaken, that the reason this works is that the rate at which I'm indifferent between money and time changes with my stock of money. Since I work for 8 hours a day at $100 an hour, we can reasonably conclude that I'm *roughly* indifferent between an hour and $100 at my current wealth. But I'm obviously not indifferent to the point that I'd work 24 hours a day for $2400, nor 0 hours a day for $0. Further, once I have accumulated my million dollars (or more generally, enough money to live off the interest), my indifference level becomes much higher - you'd have to offer me way more money per hour to get me to work. Notice that in this case I'm postulating a very sharp dropoff, in that I'm happy to work for $100 an hour until the moment my savings account hits seven digits, and then I am no longer willing to work at all; it seems possible that the argument no longer works if you allow a more gradual change in indifference, but on the other hand "save to X dollars and then retire" doesn't seem like a psychologically unrealistic plan either.

Am I making any obvious mistakes? Of course it may well be the case that the actual lottery tickets for sale in the real world do not match the wages-and-savings situations of real people in such a way that they have positive expected value; that's an empirical question. But it does seem in-principle possible for an epsilon chance at one-over-epsilon dollars paid out right away to be of positive expected value after converting to expected hours of life, even though it's neutral in expected dollars. Am I mistaken?

Edit: Wei Dai found the problem. Briefly, the 100 dollars added to my savings would cut more than 1.1 hours off the time I had to work at the end of the 5.5 years.