## Friday, April 14, 2023

### Independence axiom

Here is an argument for the von Neumann – Morgenstern axiom of independence.

Consider these axioms for a preference structure on lotteries.

1. If L ≺ M and K ≺ N, then pL + (1−p)K ≾ pM + (1−p)N.

2. If M dominates L, then pL + (1−p)N ≾ pM + (1−p)N.

3. If A ≾ pM + (1−p)N for all N′ dominating N, then A ≾ pM + (1−p)N.

4. If L ≺ M, there is an M′ that dominates L but is such that M′ ≺ M.

5. If M dominates L, then L ≺ M.

6. Transitivity and completeness.

7. 0 ⋅ L + 1 ⋅ N ∼ 1 ⋅ N + 0 ⋅ L ∼ N.

Now suppose that L ≺ M and 0 < p < 1. Let M dominate L but be such that M′ ≺ M by (4). Let N dominate N. Then pL + (1−p)N ≾ pM′ + (1−p)N by (2) and pM′ + (1−p)N ≾ pM + (1−p)N by (1). This is true for all N that dominates N, so pL + (1−p)N ≾ pM + (1−p)N by (3).

Now suppose that L ≾ M. Let M dominate M. Then L ≺ M by (5). By the above pL + (1−p)N ≾ pM′ + (1−p)N. This is true for all M′ dominating M, so by (3) we have pL + (1−p)N ≾ pM + (1−p)N. Hence we have independence for 0 < p < 1. And by (7) we get it for p = 0 and p = 1.

Enough mathematics. Now some philosophy. Can we say something in favor of the axioms? I think so. Axioms (5)–(7) are pretty standard fare. Axioms (3) and (4) are something like continuity axioms for the space of values. (I think axiom (4) actually follows from the other axioms.)

Axioms (1) and (2) are basically variants on independence. That’s where most of the philosophical work happens.

Axiom (2) is pretty plausible: it is a weak domination principle.

That leaves Axiom (1). I am thinking of it as a no-regret posit. For suppose the antecedent of (1) is true but the consequent is false, so by completeness pM + (1−p)N ≺ pL + (1−p)K. Suppose you chose pL + (1−p)K over pM + (1−p)N. Now imagine that the lottery is run in a step-wise fashion. First a coin that has probability p of heads is tossed to decide if the first (heads) or second (tails) option in the two complex lotteries is materialized, and then later M, N, L, K are resolved. If the coin is heads, then you now know you’re going to get L. But L ≺ M, so you regret your choice: it would have been much nicer to have gone for pM + (1−p)N. If the coin is tails, then you’re going to get K. But K ≺ N, so you regret your choice, too: it would have been much nicer to have gone for pM + (1−p)N. So you regret your initial choice no matter how the coin flip goes.

Moreover, if there are regrets, there is money to be made. Your opponent can offer to switch you to pM + (1−p)N for a small fee. And you’ll do it. So you have made a choice such that you will pay to undo it. That’s not rational.

So, we have good reason to accept Axiom (1).

This is a fairly convincing argument to me. A pity that the conclusion—the independence axiom—is false.

IanS said...

What do you mean by calling the independence axiom ‘false’? That you reject it as a necessary condition for rational choice? Or something stronger?

If you reject standard EU maximization, it’s hard to avoid ‘regrets’ (inconsistency in sequences). For example, there is a strong intuition that EU approaches are not reasonable for your cut-off St Petersburg. But if you repeated this setup often enough, the laws of large numbers would cut in, and the EU approach would be uncontroversial. (For a cut-off at the nth flip, ‘often enough’ would be order of magnitude greater that 2^n.) So your feeling about the whole sequence would differ from your feeling about the parts.

This sort of inconsistency is also a standard complaint against Buchak’s risk-weighted utility. I suspect that if you want to reject the standard approach, you have to bite the bullet on regrets. Also, that this could be defensible in some cases. (For example, One-off setups, few repetitions, extreme variance, sequence known in advance.)

Alexander R Pruss said...

By false, I mean that it is false that it is a constraint on rationality.

No disagreement with the rest of your, as usual, astute comment.