It occurred to me yesterday that there is an extremely common and natural example of a total ordering (modulo indifference. i.e. it’s really a total pre-ordering) of distributions which doesn’t satisfy The VNM Axioms.

(Note: We are talking about distributions rather than random variables in this, so we will follow the VNM notation of using pA + (1-p)B to mean “choose A with probability p or B with probability (1 – p)”. This has the effect of a similar combination on the distributions. It’s not addition of random variables).

Specifically, ordering distributions by their lower median value (assuming a total ordering over outcomes). More generally, by any percentile value.

This does not satisfy continuity: To see that it does not satisfy continuity, suppose our outcomes are 1, 2 and 3. Let A, B, C be lotteries choosing each of these with probability 1. Then A < B < C. But the lower median of \(pA + (1-p)B\) is 1 if \(p \geq \frac{1}{2}\) or 3 if \(p < \frac{1}{2}\). The median of \(B\) is always 2, so we are never indifferent between them. You could argue that the problem is that at \(p = \frac{1}{2}\) we should take an averaging of the two values and the median should be 2 there, but that doesn't save you. Instead consider where we have 5 values and A, B, C choose 1, 2 and 5. Then at \(p = \frac{1}{2}\) the median will be 3, so we're still not indifferent to it at any point. It also does not satisfy independence. Here's an example: Suppose we have two possible outcomes, 0 and 1. Consider distributions \(A = [0.35, 0.65]\), \(B = [0.4, 0.6]\), \(C = [0.9, 0.1]\). Let \(p = 0.75\). Then the median of \(A\) and \(B\) are both 1, while the median of \(C\) is 0. Further the median of \(pA + (1-p)C\) is 1, because the probability of it being 0 is \(0.75 * 0.35 + 0.25 * 0.9 \approx 0.49 < 0.5\). The median of \(pB + (1-p)C\) however is 0 because the probability of it being 0 is \(0.525\). So \(A \preceq B\) but \(pA + (1-p)C \succ pB + (1-p)C\) as desired. (Disclaimer: I didn't work this out by inspection, I totally just ran a computer program to find examples for me because I was pretty sure they must exist)

VekyRegarding your computer proof, you might be interested to know that whenever 0<a0<b0<1/2<c0<1 (where a0=P(A=0)…), there is 0<ppB+(1-p)C. :-)

davidPost authorI don’t entirely understand your notation. Do you mean that there is \(p\) the makes the example work whenever \(a_0 = P(A = 0) < b_0 = P(B=0) < \frac{1}{2}\) and \(c_0 = P(C=0) > \frac{1}{2}\)? if so, yes, that is indeed interesting.

I guess that makes sense, because \(P(pA + (1-p)C = 0) = p a_0 + (1-p) c_0\). Increasing \(p \to 1\) this must cross the \(\frac{1}{2}\) boundary at some point, and because \(a_0 < p_0\) this crosses that boundary for \(b_0\) first. Thanks for pointing this out!