davidad comments on Impossibility results for unbounded utilities

[davidad]

I think this order does satisfy Homogeneous Mixtures, but not Intermediate Mixtures. Homogeneous Mixtures is a theorem if you model lotteries as measures, because it’s asking that your preference ordering respect a straight-up equality of measures (which it must if it’s reflexive).

Intermediate Mixtures and Weak Dominance are asking that your preference ordering be willing to strictly order mixtures if it would strictly order their components in a certain way, and the ordering you’ve proposed preserves sanity by sometimes refusing to rank pathological mixtures.

[MichaelStJules]

Hmm, I do think Intermediate Mixtures is violated, and in a very bad way: it can flip the order.

Consider

1. $A=0$, constant

2. $B={\sum }_{i=1}^{\infty }\frac{1}{2^i}{\delta }_{3^i}=3^i$ with probability $2^i$ for each $i=1,2,\dots$.

Note that $B>A$.

Let’s check if $B>\frac{1}{2}A+\frac{1}{2}B$. For a given $q$ sufficiently close to 1 (away from ¹⁄₂, at least), the integral of the integral of the quantile for $\frac{1}{2}A+\frac{1}{2}B$ will be much further into the series terms of $B$ than the integral of the quantile $B$, because $\frac{1}{2}A+\frac{1}{2}B$ 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 $\frac{1}{2}A+\frac{1}{2}B$diverges to infinity faster than $B$’s.

So, I think it’s actually the case that $B<\frac{1}{2}A+\frac{1}{2}B$, which seems very bad, bad enough to reject this approach.

(I think the order I defined here will be better behaved.)

[MichaelStJules]

Which equality (not preference equivalence) of measures are you talking about for Homogenous Mixtures?

The order doesn’t satisfy Homogenous Mixtures, but maybe it also doesn’t satisfy Intermediate Mixtures. For Homogenous Mixtures, using lotteries over actual utilities, i.e. the payoff/​outcome is its utility, for $i=0,1,2,\dots$,

1. $X=0$

2. $A_i=\frac{1}{2}{\delta }_{2^i}+\frac{1}{2}{\delta }_{-2^i}=2^i\text{or}-2^i,$ each with probability ¹⁄₂. This is equivalent to $X=0$ since we’re ranking based on expected utility when both lotteries are bounded.

3. $p_i=\frac{1}{2^{i+1}}$

It doesn’t order ${\sum }_{i=1}^{\infty }{p}_i{A}_i$ vs $X$, because fixing $q$ and taking $p\to 0$, the integral for ${\sum }_i{p}_i{A}_i\ge 0$ diverges to $-\infty$, and the integral for $0\ge {\sum }_i{p}_i{A}_i$ diverges to $-\infty$. Note that

${\sum }_{i=1}^{\infty }{p}_i{A}_i=\frac{1}{2}{A}_{+}+\frac{1}{2}{A}_{-}$, where

$A_{+}={\sum }_{i=0}^{\infty }\frac{1}{2^{i+1}}{\delta }_{2^i},A_{-}={\sum }_{i=0}^{\infty }\frac{1}{2^{i+1}}{\delta }_{-2^i}$,

so a mixture of two lotteries, one diverging to $+\infty$, and the other diverging to $-\infty$.

On the other hand, if you require $q=1-p$, then the integral will actually be 0 for each $p$ for this particular lottery, since the lottery is symmetric around $p=\frac{1}{2}$, so you do get equivalence. I suspect we can come up with another counterexample by messing around with how fast each tail is approached and get the liminf of the integral to be positive, and so rank ${\sum }_i{p}_i{A}_i>0$. Maybe instead defining $A_i$ this way would work:

$A_i=(1-\frac{1}{1+2^i}){\delta }_{2^i}+\frac{1}{1+2^i}{\delta }_{-4^i}$.

The idea is that the negative terms get much less far for the same $p$, because far more of the weight is in much lower probability events. The integral for ${\sum }_i{p}_i{A}_i\ge 0$ is roughly counting the number of positive terms whose probability is above $cp$ and subtracting the number of negative terms whose probability is above $cp$, for some constant $c$ (I can’t be bothered to figure out exactly which $c$). This should go to $+\infty$ as $p\to 0$.

I think you can make things worse, too, again with $q=1-p$. You can choose $A_i<0$ for each $i$, but have ${\sum }_i{p}_i{A}_i>0$ by replacing the $-4^i$ with $-4^i-1$. I think we can even get the gap between $A_i$ and 0 to diverge as $i\to \infty$, with something like $-4^i-i(1+2^i)$ or even $-4^i-3^i$ instead of $-4^i$. If we allow $p$ and $q$ to vary independently again, ${\sum }_i{p}_i{A}_i$ and 0 are just incomparable, which seems okay.

Unfortunately, the quantile function is a pain to work with, so checking whether Intermediate Mixtures holds seems tricky.