MichaelStJules comments on Impossibility results for unbounded utilities

[MichaelStJules]

(Note: I’ve edited this comment a lot since first posting it, mostly small corrections, improving the definition of the order to be better behaved and adding more points.)

I think orders which satisfy (Symmetric) Unbounded Utilities but not (Weak) Dominance or Homogeneous Mixtures can be relatively nice, so we shouldn’t be too upset about giving up (Weak) Dominance and Homogenous Mixtures. Basically no nice order will be sensitive to the kinds of probability rearrangements you’ve done (which we can of course conclude from your results, if we want “nice” to include the other properties).

Here’s such an order, which also extends (is as strong as) expected utility and (I think) stochastic dominance for finite but possibly unbounded utilities and infinite expected utilities:

(EDIT: I think this order is pretty badly behaved based on how badly it violates Intermediate Mixtures, but this other one still seems promising.)

Given a utility function $U$ and lotteries $A$ and $B$, then $A<B$ if

$$\underset{(p,q)\to (0,1),0<p<q<1}{liminf}{\int }_p^q{Q}_{U\left(B\right)}\left(t\right)-Q_{U\left(A\right)}\left(t\right)dt>0,$$

where $Q_{U\left(A\right)}$ and $Q_{U\left(B\right)}$ are the quantile functions of $U\left(A\right)$ and $U\left(B\right)$, respectively. We can also say $A\le B$ when the above holds with $\ge 0$ instead of the strict $>0$.

Basically, take integrals out from the middle of the distributions (based on matching quantiles), and check if one dominates the other in the limit as you cover the whole probability space.

Some properties:

1. The order reduces to comparing expected utilities when the expected utilities of both lotteries are well-defined and they aren’t the same infinity (i.e. not both positive infinity or both negative infinity; in such cases, it can sometimes tell you one still dominates the other). This is because ${lim}_{(p,q)\to (0,1)}{\int }_p^q{Q}_X\left(t\right)dt={\int }_0^1{Q}_X\left(t\right)dt=E\left[X\right]$, whenever the expected value is well-defined.

2. The order is transitive (with bounded actual utilities, including when the expected utilities are infinite or undefined), because lim infs are superadditive, i.e. $liminff+g\ge liminff+liminfg$.

3. The order is not complete. It doesn’t rank $\frac{1}{2}{X}_{\infty }+\frac{1}{2}(1-X_{\infty })$ vs 0, where $1-X_{\infty }$ is just the lottery with the payoffs of $X_{\infty }$ with the negatives of the utilities and an extra utility of 1 added to each.

4. I’m not sure if the independence axiom holds, but I would guess so. It at least holds for lotteries with finite expected utility.

5. It doesn’t satisfy Dominance, Weak Dominance or Homogenous Mixtures because your counterexamples don’t work for it, since it’s not sensitive to probability rearrangements.

[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.

[MichaelStJules]

Here are some ways to get more strict inequalities (less incomparability or equivalence):

1. Require $q=1-p$ to handle some more cases with both positive and negative expected infinities, but I’m not sure that the results would always be intuitive. There might be other relationships between $p$ and $q$ that that depend on the particular lotteries that work better. You could test the lim infs under multiple relationships, $q=1-f\left(p\right)$ for different $f$ from a specific set.

2. Replace the strict inequality condition with

$$\underset{(p,q)\to (0,1),0<p<q<1}{lim}\text{sgn}({\int }_p^q{Q}_{U\left(B\right)}\left(t\right)-Q_{U\left(A\right)}\left(t\right)dt)=1.$$

Equivalently, there are $p_0,q_0,0<p_0<q_0<1$ such that ${\int }_p^q{Q}_{U\left(B\right)}\left(t\right)-Q_{U\left(A\right)}\left(t\right)>0$ for all $p,q,0<p<p_0<q_0<q<1$.

$A<B$ would mean that the integral for $A$ never catches up with that for $B$ in the limit.

[MichaelStJules]

p.37-38 in Goodsell, 2023 gives a better proposal, which is to clip/​truncate the utilities into the range $[-t,t]$ and compare the expected clipped utilities in the limit as $t\to \infty$. This will still suffer from St Petersburg lottery problems, though.