MichaelStJules comments on Impossibility results for unbounded utilities

[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.