Hmm, I do think Intermediate Mixtures is violated, and in a very bad way: it can flip the order.
Consider
A=0, constant
B=∑∞i=112iδ3i=3i with probability 2i for each i=1,2,….
Note that B>A.
Let’s check if B>12A+12B. For a given q sufficiently close to 1 (away from 1⁄2, at least), the integral of the integral of the quantile for 12A+12B will be much further into the series terms of B than the integral of the quantile B, because 12A+12B has to cram the probabilities of B into half as much space. Because the expected value of the terms grow exponentially, halving the probabilities is outweighed by summing the series at a faster rate (with respect to the probabilities). In other words, the quantile integral of 12A+12Bdiverges to infinity faster than B’s.
So, I think it’s actually the case that B<12A+12B, which seems very bad, bad enough to reject this approach.
(I think the order I defined here will be better behaved.)
Hmm, I do think Intermediate Mixtures is violated, and in a very bad way: it can flip the order.
Consider
A=0, constant
B=∑∞i=112iδ3i=3i with probability 2i for each i=1,2,….
Note that B>A.
Let’s check if B>12A+12B. For a given q sufficiently close to 1 (away from 1⁄2, at least), the integral of the integral of the quantile for 12A+12B will be much further into the series terms of B than the integral of the quantile B, because 12A+12B has to cram the probabilities of B into half as much space. Because the expected value of the terms grow exponentially, halving the probabilities is outweighed by summing the series at a faster rate (with respect to the probabilities). In other words, the quantile integral of 12A+12Bdiverges to infinity faster than B’s.
So, I think it’s actually the case that B<12A+12B, which seems very bad, bad enough to reject this approach.
(I think the order I defined here will be better behaved.)