Money pumping: the axiomatic approach

by Stuart_Armstrong 10y5th Nov 20095 min read93 comments

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This post gets somewhat technical and mathematical, but the point can be summarised as:

  • You are vulnerable to money pumps only to the extent to which you deviate from the von Neumann-Morgenstern axioms of expected utility.

In other words, using alternate decision theories is bad for your wealth.

But what is a money pump? Intuitively it is a series of trades that I propose to you, that end up bringing you back to where you started. All the trades must be indifferent or advantageous to you, so that you will accept them. And if even one of those trades is advantageous, then this is a money pump: I can charge you a tiny amount for that trade, making free money out of you. You are now strictly poorer than if you had not accepted the tradesat all.

A strict money pump happens when every deal is advantageous to you, not simply indifferent. In most situations, there is no difference between a money pump and a strict money pump: I can offer you a tiny trinket at each indifferent deal to make it advantageous, and get these back later. There are odd preference systems out there, though, so the distinction is needed.

The condition "bringing you back to where you started" needs to be examined some more. Thus define:

A strong money pump is a money pump which returns us both to exactly the same situations as when we started: in possession of the same assets and lotteries, with none of them having come due in the meantime.

A weak money pump is a money pump that returns us to the same situation that would have happened if we had never traded at all. Lotteries may have come due in the course of the trades.

What is the difference? Quite simply with a strong money pump, since we return to exactly the same setup, I can then money pump you again, and again, charging you a penny at each round, and draining you of cash until you run out completely or your preferences change. Remember that I charge you that penny only for a deal that is strictly to your advantage, so even with that charge, you are coming out ahead on every deal. It's just at the end of the loop that you're losing out.

For a weak money pump, since we don't return exactly to the same setup, I cannot just money pump you again instantly. I can get cash from you only once for each setup. It might never happen again; or it may happen regularly. But it's not as reusable as a strong money pump.

Now if your preferences are inconsistent, I can certainly money pump you. In his post Zut Allais (French puns remain the property and responsibility of their authors) Eliezer presents a fun version of this, one where subjects prefer A to B and prefer B to A, depending on how they are phrased. Thus I will assume that your preferences are consistent - that you will only invert your preferences under objectively different conditions. I will also assume that you follow the completeness and continuity axioms of the von Neumann-Morgenstern formulation (completeness assumes you are actually capable of deciding between options, while continuity is a technical assumption). Note that the axioms on the Wikipedia article seemed to be incorrect, based on sources such as this book; I have corrected them by replacing >'s with ≥ where appropriate. A lot of the complexities of this post revolve around the difference between these two symbols.

The third von Neumann-Morgenstern axiom is transitivity: that if you like A at least as much as B (designated A ≥ B) and B at least as much as C, then you also like A at least as much as C. Strict transitivity is the weaker statement replacing ≥ with > in the above. This is precisely where the strong money pumps come in:

If your preferences are consistent, complete and continuous, then you are immune to a strong money pump if and only if your preferences are transitive. You are immune to a strict strong money pump if and only if your preferences are strictly transitive.

Proof: A strong money pump is equivalent with a sequence of preferences A ≥ B ≥ ... ≥ S > T ≥ ... ≥ Z ≥ A. Such a sequence can only exist if your preferences are not transitive (note the strict inequality in the middle).

A strict strong money pump is equivalent with a sequence of preferences A > B > ... > Z > A. Such a sequence can only exist if your preferences are not strictly transitive.

The fourth von Neumann-Morgenstern axiom is independence: that if A ≥ B, then for all C, pA + (1-p)C ≥ pB + (1-p)C for all 0 ≤ p ≤ 1. It means essentially that all lotteries can be considered in isolation from each other. This is where the weak money pumps come in:

If your preferences are consistent, complete, continuous and transitive, then you are immune to a weak money pump if and only if your preferences are independent. You are immune to a strict weak money pump if and only if your preferences are not strictly dependent.

The rest of the post is a proof of this, and can be skipped for those unkeen on mathematics.

First of all, we need to explain "strictly dependent" (there is no canonical definition of the term). Given all the other axioms, we can replace independence with the following equivalent axiom:

  • (Independence II) For all A, B, C, D and 0 ≤ p ≤ 1, if pA + (1-p)C > pB + (1-p)D, then A > B or C > D.

The four standard axioms together imply the expected utility hypothesis, and Independence II is a simple consequence of that. Conversely, if we take C and D to be the same lottery, then the axiom states that pA + (1-p)C > pB + (1-p)C implies A > B (since C < C is inconsistent). The contrapositive of this statement is that if A ≤ B, then pA + (1-p)C ≤ pB + (1-p)C. This is just standard independence once more. Thus we can equivalently replace Independence with Independence II.

A decision theory is dependent if it is not independent. Dependency means that there exists p, A, B, C and D such that pA + (1-p)C > pB + (1-p)D while A ≤ B and C ≤ D. A decision theory is strictly dependent if we replace all ≤ in that expression with <. We're now ready to prove the result.

Proof: Assume you have the lottery L, and that there will be a draw to determine one of the random elements. This means L can be written as pA + (1-p)B, where it will end up in as A or B after the draw. Then if I were to trade you L for M = pC + (1-p)D, with M > L, independence implies that A > C or B > D.

Consequently, if you start with lottery L and I trade it for M > L, then for at least one outcome after the random draw, you are left with a lottery strictly better than what would have had if we had not traded. This result continues to be true no matter how often the draws happen, and hence I will not be able to trade you back to your initial situation with an advantageous or indifferent deal. Thus independence implies that you cannot be weakly money pumped with certainty.

Conversely, if your preferences are dependent, then there are p, A, B, C and D such that pA + (1-p)C > pB + (1-p)D and yet A ≤ B and C ≤ D. Then I can weakly money pump you, building on Eliezer's example. Assume you are in possession of pB + (1-p)D, with the first draw being to determine which of B or D you have. Then I can propose a binding contract: I will trade A for B and C for D after that first draw, replacing your current lottery with pA + (1-p)C. This is advantageous to you, so you will accept it. Then, after the first draw, I will propose to trade back B for A or D for C, another advantageous trade that you will accept. Congratulations! You have just been (weakly) money pumped.

In the strict dependency situation, the proof works out exactly the same way, with ≥ replacing > where appropriate, and proving that you cannot be strictly weakly money pumped if your preferences are consistent, transitive, complete, and not strictly dependent.

Note: Of course, if you do not follow independence, you truly do not follow independence. You can play present lotteries off against future lotteries; you can look ahead and see that I will attempt to money-pump you, and compensate for that. You can therefore behave as if you were following independence in these situations, even if you do not. This sort of "arbitraged independence" will be the subject of a future post.

Addendum: Following a discussion with MendelShmiedekamp, I realised that the continuity axiom is not needed for any of the results, as long as one uses independence II instead of independence. Anything that violates independence also violates independence II, so the "if you violate independence, you can be weakly money-pumped" result goes straight through to the non-continuous case.

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