Better impossibility result for unbounded utilities

[paulfchristiano]

This post slightly improves on the impossibility results in my last post and A paradox for tiny probabilities and enormous values. This is very similar to those arguments but with minor technical differences. I may refer people here if we’re discussing details of the argument against unbounded utilities, but for more interesting discussion you should check out those other writeups.

Summarizing the key differences from my last post:

• I replace the dominance principle $\forall i:A_i>B_i\Rightarrow \sum p_i{A}_i>\sum p_i{B}_i$ with $\forall i:A_i\ge B_i\Rightarrow \sum p_i{A}_i\ge \sum p_i{B}_i$.

• Assuming unbounded positive and negative utilities, I remove the assumption of Intermediate Mixtures.

Weakening dominance

We’ll represent preferences by a relation $\le$ over probability distributions over some implicit space of outcomes $\Omega$ (and we’ll identify outcomes with the constant probability distribution). Define $A<B$ to mean ($A\le B$ and not $B\le A$).

We’ll show that it’s impossible for $\le$ to satisfy four properties: unbounded utilities, intermediate mixtures, very weak dominance, and transitivity.

The properties

Unbounded utilities: there is an infinite sequence of outcomes $X_1,X_2,X_4,X_8,\dots$ each “more than twice as good”^[1] as the last. More formally, there exists an outcome $X_0$ such that:

• $X_0<X_1$

• $X_{2^k}<\frac{1}{2}{X}_{2^{k+1}}+\frac{1}{2}{X}_0$ for every $k$

That is: $X_1$ is not as good as a $\frac{1}{2}$ chance of $X_2$, which is not as good as a $\frac{1}{4}$ chance of $X_4$, which is not as good as a $\frac{1}{8}$ chance of $X_8$…

Intermediate mixtures: If $A<B$ and $0<p<1$,^[2] then $A<pA+(1-p)B<B$.

That is: a chance of a good outcome is strictly worse than a sure thing.^[3]

Weak lottery-lottery dominance: Let $A_0,A_1,A_2,\dots$, and $B_0,B_1,B_2,\dots$ be sequences of lotteries, and $p_0,p_1,p_2,\dots$ be a sequence of probabilities summing to 1. If $A_i\ge B_i$ for all $i$, then $\sum p_i{A}_i\ge \sum p_i{B}_i$.

Transitivity: if $A\ge B$ and $B\ge C$, then $A\ge C$.^[4]

Inconsistency proof

Consider the lottery $X_{\infty }:=\frac{1}{3}{X}_0+\frac{2}{9}{X}_1+\frac{4}{27}{X}_2+\frac{8}{81}{X}_4+\dots$, where each outcome $X_{2^k}$ is $\frac{2}{3}$ as likely as the one before. Intuitively this sum is “more infinite” than the usual St Petersburg lottery and this will be important for our proof.

By weak dominance we can replace $X_1$ by $X_0$ without making the lottery any better. Thus:

• $X_{\infty }\ge \frac{5}{9}{X}_0+\frac{4}{27}{X}_2+\frac{8}{81}{X}_4+\frac{16}{243}{X}_8+\dots$

We can rewrite this new lottery as a mixture:

• $X_{\infty }\ge \frac{1}{9}{X}_0+\frac{8}{27}(\frac{1}{2}{X}_0+\frac{1}{2}{X}_2)+\frac{16}{81}(\frac{1}{2}{X}_0+\frac{1}{2}{X}_4)+\frac{32}{243}(\frac{1}{2}{X}_0+\frac{1}{2}{X}_8)+\dots$

We can apply bounded utilities to each parenthesized expression. Combining with weak dominance, we obtain:

• $X_{\infty }\ge \frac{1}{9}{X}_0+\frac{8}{27}{X}_1+\frac{16}{81}{X}_2+\frac{32}{243}{X}_4+\dots$

Write $X_{\infty }^2$ for the right hand side of this inequality.

• $X_{\infty }^2:=\frac{1}{9}{X}_0+\frac{8}{27}{X}_1+\frac{16}{81}{X}_2+\frac{32}{243}{X}_4+\dots$

Now note that $X_{\infty }=\frac{1}{4}{X}_0+\frac{3}{4}{X}_{\infty }^2$, despite the fact that $X_{\infty }\ge X_{\infty }^2$. If we knew $X_{\infty }^2>X_0$, then intermediate mixtures would imply that $X_{\infty }^2>X_{\infty }$, contradicting $X_{\infty }\ge X_{\infty }^2$.

So all that remains is to show that $X_{\infty }^2>X_0$. This will involve a bit of annoying arithmetic but it should feel pretty obvious given that all the outcomes in $X_{\infty }^2$ seems much better than $X_0$.

Combining $\frac{1}{2}{X}_0+\frac{1}{2}{X}_2\ge X_1$ with weak dominance and $X_1>X_0$ we get:

• $X_{\infty }^2\ge \frac{13}{27}{X}_0+\frac{1}{27}{X}_1+\frac{7}{81}{X}_2+\frac{32}{243}{X}_4+\frac{64}{729}{X}_8+\dots$

Then we write the right hand side as a mixture and apply unbounded utilities and weak dominance:

• $X_{\infty }^2\ge \frac{1}{27}{X}_1+\frac{14}{81}(\frac{1}{2}{X}_0+\frac{1}{2}{X}_2)+\frac{64}{243}(\frac{1}{2}{X}_0+\frac{1}{2}{X}_4)+\frac{128}{729}(\frac{1}{2}{X}_0+\frac{1}{2}{X}_8)+\dots$

• $X_{\infty }^2\ge \frac{1}{27}{X}_1+\frac{14}{81}{X}_1+\frac{64}{243}{X}_1+\frac{128}{729}{X}_1+\dots$

• $X_{\infty }^2\ge X_1>X_0$

Leveraging negative utilities

To rule out unbounded utilities, we’ve made two substantive consistency assumptions: intermediate mixtures and weak lottery-lottery dominance. The assumption of intermediate mixtures is necessary: we could satisfy the other properties by simply being indifferent between all lotteries with infinite expectations.

But if we can have unboundedly good or unboundedly bad outcomes, then we can obtain a contradiction even without infinite mixtures. That is, we will show that there is no relation $\le$ satisfying transitivity, symmetric unbounded utilities, and weak outcome-lottery dominance.

I think that almost anyone who accepts unbounded utilities (in the informal sense) should accept symmetric unbounded utilities, so I think they probably need to reject weak outcome-lottery dominance (or stop defining preferences over arbitrary probability distributions). I find this pretty damning, but maybe others are more comfortable with it.

The properties

Transitivity. If $A\le B$ and $B\le C$, then $A\le C$.

Symmetric unbounded utilities: There is a pair of outcomes $X_0<X_1$ (i.e. it’s not the case that all pairs of outcomes are either incomparable or equal). Moreover:

• For any^[5] pair of outcomes $X^{+}>X$, there is outcome $X^{-}$ such that $\frac{2}{3}{X}^{+}+\frac{1}{3}{X}^{-}\le X$.

• For any pair of outcomes $X^{-}<X$ there is an outcome $X^{+}$ such that $\frac{2}{3}{X}^{-}+\frac{1}{3}{X}^{+}\ge X$.

In words: no matter how much better $X^{+}$ is than $X$, there’s always an $X^{-}$ that’s 2x “further away” from $X$ on the other side. By that we mean that a $\frac{1}{3}$ risk of moving from $X$ to $X^{-}$ can offset a $\frac{2}{3}$ chance of moving from $X$ to $X^{+}$. And no matter how bad an outcome $X^{-}$ is, there’s always an $X^{+}$ that’s 2x “further away” from $X$ on the other side.

Weak outcome-lottery dominance: Let $X$ be an outcome, let $A_0,A_1,A_2,\dots$ be a sequence of lotteries, and let $p_0,p_1,p_2,\dots$ be a sequence of probabilities summing to 1. If $A_i\ge X$ for all $i$, then $\sum p_i{A}_i\ge X$. Similarly, if $A_i\le X$ for all $i$, then $\sum p_i{A}_i\le X$.

Inconsistency proof

Define a sequence of outcomes $X_0,X_1,X_{-2},X_7,X_{-14},X_{31}\dots$ as follows:

• Pick $X_0<X_1$ arbitrarily.

• Take $X_{-2}$ to be the “reflection” of $X_1$ across $X_0$ as defined in symmetric unbounded utilities.

• Take $X_7$ to be the reflection of $X_{-2}$ across $X_1$.

• Take $X_{-14}$ to be the reflection of $X_7$ across $X_0$.

• Take $X_{31}$ to be the reflection of $X_{-14}$ across $X_1$.

• And so on.

Now define the lottery $X_{\pm \infty }=\frac{1}{2}{X}_1+\frac{1}{4}{X}_{-2}+\frac{1}{8}{X}_7+\frac{1}{16}{X}_{-14}+\frac{1}{32}{X}_{31}+\dots$

We can write $X_{\pm \infty }$ as the mixture:

• $\frac{3}{4}(\frac{2}{3}{X}_1+\frac{1}{3}{X}_{-2})+\frac{3}{16}(\frac{2}{3}{X}_7+\frac{1}{3}{X}_{-14})+\frac{3}{64}(\frac{2}{3}{X}_{31}+\frac{1}{3}{X}_{-62})\dots$

By Unbounded Utilities each of these terms is $\le X_0$. So by weak dominance, $X_{\pm \infty }\le X_0.$

But we can also write $X_{\pm \infty }$ as the mixture:

• $\frac{1}{2}{X}_1+\frac{3}{8}(\frac{2}{3}{X}_{-2}+\frac{1}{3}{X}_7)+\frac{3}{32}(\frac{2}{3}{X}_{-14}+\frac{1}{3}{X}_{31})+\dots$

By Unbounded Utilities each of these terms is $\ge X_1$. So by weak dominance, $X_{\pm \infty }\ge X_1$.

Now we have $X_0\ge X_{\pm \infty }\ge X_1$. By transitivity, $X_0\ge X_1$, contradicting $X_1>X_0$.

1. ^

   Of course the same argument would work if we replaced “good” with “bad.”

2. ^

   Note that we only actually need to apply this principle for $p=\frac{3}{4}$ and $p=\frac{8}{9}$ so a reader squeamish about very small probabilities need not be concerned. By making the argument with different numbers we could probably just fix $p=\frac{1}{2}$.

3. ^

   Despite appearing innocuous, this might be more “controversial” than very weak dominance. Many theories say that if $B$ is infinitely good, then it doesn’t matter whether I achieve $B$ with 100% probability or 50% probability (or 1% probability or 0.000001% probability...) I find this sufficiently unappealing to reject such theories out of hand, but each reader must pick their own poison.

4. ^

   I’m pretty sure this isn’t necessary for the proof, since we only use it as a convenience for short manipulations rather than to establish very long chains. That said, it makes the proof easier and it’s a pretty mild assumption.

5. ^

   We don’t really need this universal quantifier—it would be enough to define a single chain of escalating and alternative outcomes. But the quantitative properties we need out of the chain are somewhat arbitrary, so it seemed clearer to state an axiom capturing why unbounded utilities allow us to construct a particular kind of chain, rather than to directly posit a chain. That said, it results in a stronger assumption—for example it may be that there is an outcome $X_{\infty }$ which can offset any negative outcome, while still having a chain of increasingly large finite outcomes.