Suppose that you have inconsistent but monotone credences: if p entails q then P(Q)≥P(p). Level Set Integrals (LSI) provide a way of evaluating expected utilities that escapes Dutch Books and avoids domination failure: if E(f)≥E(g) then g cannot dominate f.
Sadly, by running this simple script, I’ve just discovered that LSI need not escape multi-shot domination failure. Suppose you have these monotonic credences for a coin toss:
Heads: 1/4
Tails: 1/4
Heads or Tails: 1
Neither: 0.
Suppose you’ll first be offered a choice between these two wagers:
- A: $1 if Heads and $1 if Tails
- A′: $3 if Heads and $0 if Tails
and then second you will be offered a choice between these two wagers:
- B: $1 if Heads and $1 if Tails
- B′: $0 if Heads and $3 if Tails.
You will first choose A over A′ and then B over B′. This is true regardless of whether you use the independent multi-shot decision procedure where you ignore previous wagers or the cumulative method where you compare the expected utilities of the sum of all the unresolved wagers. The net result of your choices is getting $2 no matter what. But if you chose A′ over A and B′ over B, you’d have received $3 no matter what.
Stepping back, classical expected utility with consistent credences has the following nice property: When you are offered a sequence of choices between wagers, with the offers not known in advance to you but also not dependent on your choices, and you choose by classical expected utility, you won’t choose a sequence of wagers dominated by another sequence you could have chosen.
Level Set Integrals with in my above case (and I strongly suspect more generally, but I don’t have a theorem yet) do not have this property.
I wonder how much it matters that one does not have this property. The property is one that involves choosing sequentially without knowing what choices are coming in the future, but the choices available in the future are not dependent on the choices already made. This seems a pretty gerrymandered situation.
If you do know what choices will be available in the future, it is easy to avoid being dominated: you figure out what sequence of choices has the overall best Level Set Integral, and you stick to that sequence.
I think I have a sketch of a proof that this issue is going to come up for LSI whenever finite additivity fails. :-(
ReplyDeleteHere is a simple way to visualize 2-option cases. Plot the contours of equal expected utility against first payoff on the X axis and second payoff on the Y axis. A typical contour will be piecewise linear, with all segments having negative slope (or 0 or ∞ in edge cases) and with kinks where it crosses the X axis, the Y axis and the line X=Y.
ReplyDeleteIf there is a ‘concave’ kink point there will be violations of sequential dominance as in the post. (Take the kink point as A and B in the post. Take A’ and B’ as points on either side on the kink point, both just inside the contour through the kink point, but such that their average dominates the kink point. This will be possible because of concavity.)
If the sum of the credences is less than 1 (as in the post), there will be a concave kink point on the X=Y line. If the sum of the credences is greater than 1, there will be concave kink points on the axes (or on one of the axes, in edge cases).
If there are more than 2 options, work with a minimal inconsistent set of options. (i.e. such that the credences for any proper subset add up properly but the credences for the whole set don’t.) Similar concavity-based reasoning will apply (I think), but you will have to get formal.
The property you describe could be called No Regrets. You can be sure that your early choices will not lead you up a blind alley.
Make a plan and stick to it is one standard response to Satan’s Apple (there is no best option, but you can plan to avoid the bad option of taking all the slices) and to similar setups involving ‘discontinuity at infinity’. You may or may not want to endorse this response.
The standard make-a-plan-and-stick case, like Satan's Apple, is where at some or all steps there is rational pressure to break the plan, because there is a better plan available. (E.g., one more slice, and then stop.)
ReplyDeleteThis is not what is going on here. In this case, I am thinking you should look at all the possible plans, of which there are four (A+B, A+B', A'+B and A'+B'). One of these is best according to your prevision: A'+B'. Now comes the first step. At the first step, if you were considering only the utility of the first step alone, you would choose A by your prevision. But you're not considering only the utility of the first step, but of the first and second. And the best two-step plan at this point is A'+B'. So, you go for A', as that's the first step of the still-optimal plan. Then at the next step you go for B' as A'+B' beats A'+B by your prevision. At no point do you have a better plan available.
The case here seems more like the case of a chess player sacrificing a knight in order to take a rook in the next move. Of course you need to stick to the plan: if you sacrifice the knight but don't bother to take the rook afterwards, you're down a knight. But there is no rational pressure to not take the rook in the next move. Similarly, if you go for A', then there is no rational pressure to go for anything other than B' in the next step.
Now, in the case of classical decision theory, when the utilities from the sequential wagers are additive, there is a short-cut: to decide what to do at stage n, you don't need to consider the whole sequence of wagers (past, present and future) to decide which wager to take. You just need to take the present wager. But this is an oversimplification of how real-life choices work. In real life, overall utility isn't just a sum of the values of the experiences resulting from particular wagers, but a non-linear function of the values of individual experiences. This matters for modeling risk (e.g., it's central to Buchak's account of risk -- which account formally is very similar to using LSI with inconsistent credences) as well as goods of variety, once-in-a-lifetime achievements, etc. In all these cases, we need to make plans and stick to them in the face of temptation from short-term gains. But there is nothing particularly paradoxical about sticking to a plan in the face of temptation from short-term gains. That's just the stuff of life.
It's still true that in the case of additive utilities from sequential wagers, the classical decision theorist with consistent credences has an advantage in that they can employ the shortcut. Moreover, they don't need to know what wagers are coming in order to avoid being dominated, as long as the sequence of coming wagers is independent of their choices. But the latter is an idealization that rarely holds in real life.
You are right. Satan’s Apple is relevantly different. There is no best plan, so no solid reason to choose a particular plan. By contrast, in finite cases there is always a best plan (up to ties).
ReplyDeleteI’m not sure that real life is relevant. LSI with inconsistent credences is too complex to apply in any but very simple cases. (e.g. you have inconsistent credences about a coin flip and a die roll. There are 12 basic joint outcomes, so 2^12 = 4096 credences, less the 2 trivial ones. That’s soon going to get out of hand.) It’s interest, as I see it, is theoretical – what justifies standard probabilism? How far can you stretch it? From this point of view, sequential violation of dominance counts against LSI with inconsistent credences.
In Buchak’s approach, the contours, in the representation I outlined above, are always consistently flat (representing risk neutrality), convex (risk appetite) or concave (risk aversion). For LSI with inconsistent credences, the contours are never entirely convex or concave (e.g. in your example, the kinks on the axes are convex and the kink on X=Y is concave.) So LSI with inconsistent credences cannot represent consistent risk appetite or aversion. This does not by itself make LSI unreasonable, but it raises the question of what attitudes and preferences it could represent.
Here a puzzle about inconsistent credences. How do you treat joint events? For example, you have inconsistent credences about each of two coin flips that you take to be independent. What should be your credences in joint events e.g. (HH or TT)? Note that there are 4 basic outcomes, so 2^4 = 16 credences to be assigned (less the 2 trivial ones). Four of them come from the credences for individual flips. But what about the others? Note that you can’t assume the rules of standard probability theory.