# A simple example of non-VNM total orders

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)

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## 2 thoughts on “A simple example of non-VNM total orders”

1. Veky

Regarding 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. :-)

1. david Post author

I 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!