# Better impossibility result for unbounded utilities

This post slightly improves on the impossibility results in [my last post](https://www.lesswrong.com/posts/hbmsW2k9DxED5Z4eJ) and [*A paradox for tiny probabilities and enormous values*](https://globalprioritiesinstitute.org/wp-content/uploads/Beckstead-Thomas-A-Paradox-for-Tiny-Probabilities-and-Enormous-Values-Version-2.pdf). 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 $\forall i : A_i \geq B_i \Rightarrow \sum p_i A_i \geq \sum p_i B_i$.
*   Assuming unbounded positive *and* negative utilities, I remove the assumption of Intermediate Mixtures.

Weakening dominance
===================

We'll represent preferences by a relation $\leq$ 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 ($A \leq B$ and not $B \leq A$).

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

The properties
--------------

**Unbounded utilities**: there is an infinite sequence of outcomes $X_1, X_2, X_4, X_8, \ldots$ each "more than twice as good"[^bq8hk6ft2bd] as the last. More formally, there exists an outcome $X_0$ such that:

*   $X_0 < X_1$
*   $X_{2^k} < \frac 12 X_{2^{k+1}} + \frac 12 X_0$ for every $k$

That is: $X_1$ is not as good as a $\frac 12$ chance of $X_2$, which is not as good as a $\frac 14$ chance of $X_4$, which is not as good as a $\frac 18$ chance of $X_8$...

**Intermediate mixtures**: If $A < B$ and $0 < p < 1$,[^ymwdst1iwxi] then $A < p A + (1-p) B < B$.

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

**Weak lottery-lottery dominance**: Let $A_0, A_1, A_2, \ldots$, and $B_0, B_1, B_2, \ldots$ be sequences of lotteries, and $p_0, p_1, p_2, \ldots$ be a sequence of probabilities summing to 1. If $A_i \geq B_i$ for all $i$, then $\sum p_i A_i \geq \sum p_i B_i$.

**Transitivity**: if $A \geq B$ and $B \geq C$, then $A \geq C$.[^4rg3dygiurn]

Inconsistency proof
-------------------

Consider the lottery $X_{\infty} := \frac 13  X_0 + \frac 29 X_{1} + \frac 4{27} X_{2} + \frac 8{81} X_{4} + \ldots$, where each outcome $X_{2^k}$ is $\frac 23$ 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:

*   $X_{\infty} \geq \frac 59 X_0 + \frac 4{27} X_2 + \frac 8{81} X_4 + \frac {16}{243} X_8 + \ldots$

We can rewrite this new lottery as a mixture:

*   $X_{\infty} \geq \frac 19 X_0 + \frac 8{27} \left( \frac 12 X_0 + \frac 12 X_2\right) + \frac {16}{81} \left( \frac 12 X_0 + \frac 12 X_4 \right) + \frac {32}{243} \left(\frac 12 X_0 + \frac 12 X_8\right) + \ldots$

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

*   $X_{\infty} \geq \frac 19 X_0 + \frac 8{27} X_1 + \frac {16}{81} X_2 + \frac {32}{243}X_4 + \ldots$

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

*   $X^2_{\infty} := \frac 19 X_0 + \frac 8{27} X_1 + \frac {16}{81} X_2 + \frac {32}{243}X_4 + \ldots$

Now note that $X_{\infty} = \frac 14 X_0 + \frac 34 X^2_{\infty}$, despite the fact that $X_{\infty} \geq X_{\infty}^2$. $$If we knew $X_{\infty}^2 > X_0$, then intermediate mixtures would imply that $X^2_{\infty} > X_{\infty}$, contradicting $X_{\infty} \geq X_{\infty}^2$.

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 $X^2_{\infty}$ seems much better than $X_0$.

Combining $\frac 12 X_0 + \frac 12 X_2 \geq X_1$ with weak dominance and $X_1 > X_0$ we get:

*   $X_{\infty}^2 \geq \frac {13}{27} X_0 + \frac 1{27} X_1 + \frac 7{81} X_2 + \frac {32}{243} X_4 + \frac {64}{729} X_8 + \ldots$

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

*   $X^2_{\infty} \geq \frac 1{27} X_1 + \frac {14}{81} \left( \frac 12 X_0 + \frac 12 X_2\right) + \frac {64}{243} \left(\frac 12 X_0 + \frac 12 X_4\right) + \frac {128}{729} \left( \frac 12 X_0 + \frac 12 X_8\right) + \ldots$
*   $X^2_{\infty} \geq \frac 1{27} X_1 + \frac {14}{81} X_1 + \frac {64}{243} X_1 + \frac {128}{729} X_1 + \ldots$
*   $X^2_{\infty} \geq X_1 > X_0$

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 $\leq$ 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 $A \leq B$ and $B \leq C$, then $A \leq C$.

**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[^7yotq9qa93] pair of outcomes  $X^{+} > X$, $$there is outcome $X^{-}$ such that $\frac 23 X^+ + \frac 13 X^{-} \leq X$.
*   For any pair of outcomes $X^- < X$ there is an outcome $X^{+}$ such that $\frac 23 X^- + \frac 13 X^{+} \geq X$.

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 $\frac 13$ risk of moving from $X$ to  $X^-$ can offset a $\frac 23$ 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 $A_0, A_1, A_2, \ldots$ be a sequence of lotteries, and let $p_0, p_1, p_2, \ldots$ be a sequence of probabilities summing to 1. If $A_i \geq X$ for all $i$, then $\sum p_i A_i \geq X$. Similarly, if $A_i \leq X$ for all $i$, then $\sum p_i A_i \leq X$. 

### Inconsistency proof

Define a sequence of outcomes $X_0, X_1, X_{-2}, X_7, X_{-14}, X_{31} \ldots$ 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 $X_{\pm \infty} = \frac 12 X_{1} + \frac 14 X_{-2} + \frac 18 X_{7} + \frac 1{16} X_{-14} + \frac 1{32} X_{31} + \ldots$

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

*   $$$\frac 34 \left( \frac 23 X_{1} + \frac 13 X_{-2}\right) + \frac 3{16} \left(\frac 23 X_{7} + \frac 13 X_{-14}\right) + \frac 3{64} \left(\frac 23 X_{31} + \frac 13 X_{-62}\right) \ldots$

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

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

*   $\frac 12 X_{1} + \frac 38 \left (\frac 23 X_{-2} + \frac 13 X_{7}\right) + \frac 3{32} \left(\frac 23 X_{-14} + \frac 13 X_{31}\right) + \ldots$

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

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

[^bq8hk6ft2bd]: Of course the same argument would work if we replaced "good" with "bad." 

[^ymwdst1iwxi]: Note that we only actually need to apply this principle for \(p = \frac 34\) and \(p = \frac 89\) so a reader squeamish about very small probabilities need not be concerned. By making the argument with different numbers we could probably just fix \(p = \frac 12\). 

[^pvewkqaqsv]: 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. 

[^4rg3dygiurn]: 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. 

[^7yotq9qa93]: 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.