AI ALIGNMENT FORUM
AF

Stuart, what's your view on the problem I described in Is the potential astronomical waste in our universe too small to care about? Translated to this setting, the problem is that if you do a normalisation when you're uncertain about the size of the universe (i.e., $E_{π_{r d}}$ is computed under this uncertainty), and then later find out the actual size of the universe (or just gets some information that shifts your expectation of the size of the universe or of how many lives or observer-moments it can support), you'll end up putting almost all of your efforts into Total Utilitarianism (if the shift is towards the universe being bigger) or almost none of your efforts into it (if the shift is in the opposite direction).

[-]Stuart_Armstrong6y10

Hum... It seems that we can stratify here. Let $X$ represent the values of a collection of variables that we are uncertain about, and that we are stratifying on.

When we compute the normalising factor for utility $U$ under two policies $π$ and $π^{'}$ , we normally do it as:

$U \to U / N_{U}$ , with $N_{U} = \sum_{x} P (X = x) (E_{π, X = x} U - E_{π^{'}, X = x} U)$ .

And then we replace $U$ with $U / N_{U}$ .

Instead we might normalise the utility $U$ separately for each value of $x$ :

Conditional on $X = x$ , then $U \to U / N_{U, x}$ , with $N_{U, x} = E_{π, X = x} U - E_{π^{'}, X = x} U$ .

The problem is that, since we're dividing by the $N$ , the expectation of $U / N_{U, x}$ is not the same $U / N_{U}$ .

Is there an obvious improvement on this?

Note that here, total utilitarianism get less weight in large universes, and more in small ones.

I'll think more...

Moderation Log

Random dictator policy

These problems can be solved by switching instead to the random dictator policy as the default, rather than a random policy.

Assume we are hesitating between utility functions $U_{1}$ , $U_{2}$ , ... $U_{n}$ , with $π_{i}^{*}$ the optimal policy for utility $U_{i}$ . Then the random dictator policy is just $π_{r d}$ which picks a $π_{i}^{*}$ at random and then follows that. So

$π_{r d} = \frac{1}{n} π_{1}^{*} + \frac{1}{n} π_{2}^{*} + \dots + \frac{1}{n} π_{n}^{*}$ .

Normalising to the random dictator policy

This $π_{r d}$ is an excellent candidate for replacing the random policy in the normalisation. It is well defined, it would never choose options that all utilities object to, and it doesn't care about how options are labelled or about how to count them.

Therefore we can present the random dictator normalisation: if you are hesitating between utility functions $U_{1}$ , $U_{2}$ , ... $U_{n}$ , then normalise each one to ${ˆ U}_{i}$ as follows:

${ˆ U}_{i} = \frac{U_{i}}{E_{π_{i}^{*}} [U_{i}] - E_{π_{r d}} [U_{i}]}$ ,

where $E_{π_{i}^{*}} [U_{i}]$ is the expected utility of $U_{i}$ given optimal policy, and $E_{π_{r d}} [U_{i}]$ is its expected utility given the random dictator policy.

Our overall utility to maximise then becomes:

$U = \frac{1}{n} ({ˆ U}_{1} + {ˆ U}_{2} + \dots + {ˆ U}_{n})$ .

Note that that normalisation has a singularity when $E_{π_{i}^{*}} [U_{i}] = E_{π_{r d}} [U_{i}]$ . But realise what that means: it means that the random dictator policy is optimal for $U_{i}$ . That means that every single $π_{j}^{*}$ is optimal for $U_{i}$ . So, though the explosion in the normalisation means that we must pick an optimal policy for $U_{i}$ , this set is actually quite large, and we can use the normalisations of the other $U_{j}$ to pick from among it (so maximising $U_{i}$ becomes a lexicographic preference for us).

Normalising a distribution over utilities

Now suppose that there is a distribution over the utilities - we're not equally sure of each $U_{i}$ , instead we assign a probability $p_{i}$ to them. Then the random dictator policy is defined quite obviously as:

$π_{r d} = p_{1} π_{1}^{*} + p_{2} π_{2}^{*} + \dots + p_{n} π_{n}^{*}$ .

And the normalisation can proceed as before, generating the ${ˆ U}_{i}$ , and maximising the normalised sum:

$U = p_{1} {ˆ U}_{1} + p_{2} {ˆ U}_{2} + \dots + p_{n} {ˆ U}_{n}$ .

Properties

The random dictator normalisation has all the good properties of the mean-max normalisation in this post, namely that the utility is continuous in the data and that it respects indistinguishable choices. It is also invariant under cloning (ie adding another option that is completely equivalent to one of the options already there), which the mean-max normalisation does not.

But note that, unlike all the normalisations in that post, it is not a case of normalising each $U_{i}$ without looking at the other $U_{j}$ , and only then combining them. Each normalisation of $U_{i}$ takes the other $U_{j}$ into account, because of the definition of the random dictator policy.

Problems? Double counting, or the rich get richer

Suppose we are hesitating between utilities $U_{1}$ (with $9 / 10$ probability) and $U_{2}$ (with $1 / 10$ ) probability.

Then $π_{r d} = (9 / 10) π_{1}^{*} + (1 / 10) π_{1}^{*}$ is the random dictator policy, and is likely to be closer to optimal for $U_{1}$ than for $U_{2}$ .

Because of this, we expect $U_{1}$ to get "boosted" more by the normalisation process than $U_{2}$ does (since the normalisation is the inverse of the difference between $π_{r d}$ and the optimal policies).

But then when we take the weighted sum, this advantage is compounded, because the boosted ${ˆ U}_{1}$ is weighted $9 / 10$ versus $1 / 10$ for the relatively unboosted ${ˆ U}_{2}$ . It seems that the weight of $U_{1}$ thus gets double-counted.

A similar phenomena happens when we are equally indifferent between utilities $U_{1}$ , $U_{2}$ , ... $U_{10}$ , if the $U_{1}$ , ... $U_{9}$ all roughly agree with each other while $U_{10}$ is completely different: the similarity of the first nine utilities seems to give them a double boost effect.

There are some obvious ways to fix this (maybe use $\sqrt{p_{i}}$ rather than $p_{i}$ ), but they all have problems with continuity, either when $p_{i} \to 0$ , or when $U_{i} \to U_{j}$ .

I'm not sure how much of a problem this is.