Using Math to deal with Moral Uncertainty

There is a standard result that if a "rational" agent is uncertain about what the outcome of events will be, (i.e. they have to choose between two "lotteries") then they should maximize the expectation of some utility function. Formally, if we define a lottery as $L=\sum_i p_i O_i$ where $\{O_i\}$ are the outcomes and $\{p_i\}$ their associated probabilities, then for any "rational" preference ordering $\preceq$ there is a utility function $u$ such that
$$E\left[u(L)\right]\leq E\left[u(L')\right] \leftrightarrow L \preceq L'$$
Traditionally, this is used when people aren't certain about what the outcomes of their actions will be. However, I recently attended an interesting presentation by Brian Hedden where he discussed using this in cases of normative uncertainty, i.e. in cases when we know what the outcome of our actions will be, but we just don't know what the correct thing to value is.

An analog to equation (1) in this case is to introduce ethical theories $T_1,\dots,T_n$ to which we might subscribe and $u_i(o)$ the value of an outcome $o$ under theory $T_i$ and then ask whether there is a utility function $u$ such that for $M(o) = \sum_i p(T_i)u_i(o)$ we have:
$$M(o)\leq M(o') \leftrightarrow o \preceq o'$$
Brian referred to this "meta-" theory as Maximize InterTheoretical Expectation or MITE. He believes that
There are moral theories which it can be rational to take seriously, such that if you do take them seriously, MITE cannot say anything about what you super-subjectively ought to do, given your normative uncertainty.
I show here that:
  1. Contrary to Brian's argument, a MITE function always exists.
  2. Furthermore, the output of this function is always just a vector of real numbers

Groups

The basis of this post is the fact that we can generalize the above equation (2) to an arbitrary ordered group $G=(\Omega,+,\leq)$. Rather than bore the reader with a recitation of the group axioms, I will just point the reader to Wikipedia and point out that the possibly questionable assumption here is existence of inverses (i.e. the claim that for any lottery $L$ there is a lottery $L'$ such that the agent is indifferent between participating in both lotteries and neither).1

There are probably prettier ways of doing this, but here's a simple way of defining a group which is guaranteed to work. Let's say that:

  • Each theory $T_i$ has some set of possible values $V_i$ and that we can find the (intratheoretic) value of an outcome via $u_i:\mathcal{O}\to V_i$. Crucially, we are not claiming that these values are in any way comparable to each other. ($u_i$ is guaranteed to exist because it could just be the identity function.)
  • $\Omega_i = \mathbb R \times V_i$ is a tuple which joins the probability of an outcome with its value. 
  • $\Omega =\prod_i \Omega_i$ and that $\pi_i:\Omega_i\hookrightarrow \Omega$ is the canonical embedding (i.e. $\pi_i(\omega)$ is zero everywhere except it puts $\omega$ into the $i$th position).
  • $G=(\Omega, +)$ with addition being defined element wise 


Theorem 1: For any partial order $\preceq\in \Omega\times \Omega$, $G$ satisfies (2).

Proof: It's clear that
$$M(o)=\sum_i \pi_i \left(p(T_i), u_i(o)\right)$$
will just embed the information into G, which can easily inherit the order. Of course if we are really dedicated to the notation in (2) we can define $x\cdot y = \pi(x,y)$ and then get
$$M(o)=\sum_i p(T_i) \cdot u_i(o)$$
$\square$

So what?

So far we've managed to show that you can redefine addition to mean whatever you want, and therefore utility functions will basically always exist. But it will turn out that we are actually dealing with some pretty standard groups here.

First, a little commentary on terms. One of the major objections Brian raises is the notion of "options", i.e. the fact that in certain moral theories we have "optional" things and "required" things. For example we might say that donating to charities is optional but not murdering people is required. Furthermore, these types of goods bear a non-Archimedean relationship to each other – that is, no amount of donating to charity can offset a murder.

For any ordered group $G$ there is a chain of subgroups $C_1\subset C_2\subset\dots\subset G$ such that each $C_i$ is "convex". Convex subgroups represents this notion of "optionality": $C_1$ represents all the "optional" things, $C_2$ is everything that is either required or optional, etc. Note that I am not assuming anything new here; it is a standard result that the set of all convex subgroups form a chain in any ordered group (see Glass, Lemma 3.2.1).

Theorem 2: Our above group can be order-embedded into a subset of $\mathbb R ^n$ ordered lexically, i.e. we are just dealing with a set of vectors where each component of the vector is a real number. Furthermore, the number of components in the vector is identical to the number of "degrees" of optionality.
Proof: This is the Hahn embedding theorem. $\square$

Corollary: if (and only if!) none of our theories that we give credence to have "optionality", then we are just dealing with the real numbers.

Example

The above was really abstract, so it's reasonable to ask for an example. But before I do that I would like to give a standard math joke:
(Prof. finishes proving Liouville's theorem that any bounded entire function is constant.)
Student: I'm not sure I really understand. Could you give an example?
Prof.: Sure. 7.
(Prof. goes back to writing on the blackboard.)
The joke here is that $f(x)=7$ is "obviously" a constant function whereas the student somehow wanted a more exotic example. But the professor had just proven that no such examples exist!

So I will give some examples which the astute reader will point out are "obviously" instances of lexically ordered vectors of real numbers. This is because I have just proven that there are no other examples. Hopefully it will still be useful.

First, let's discuss how just satisficing consequentialism by itself is a lexically ordered vector. Consider the decision criterion that $(x_1,x_2)\leq (y_1,y_2)$ if and only if $x_2< y_2$ or both $(x_2 = y_2)$ and $(x_1\leq y_1)$ (i.e. it is lexically ordered from the right). So we could for example represent giving a thousand dollars to charity as $(1000,0)$ and murdering someone as $(0,-10000)$; this gives us our desired result that no amount of donations can offset a murder (i.e. $(x,-10000)\prec(0,0)$ for all $x$). And of course this is a vector of real numbers which is lexically ordered, in accordance with our theorem.

Now let's contrast this with standard utilitarianism, which would say that murdering someone could be offset by donating enough money to charity to prevent someone from dying. Let's call that amount $\$$10,000 (i.e. murdering someone has -10,000 utils). There are no "optional" things in standard utilitarianism, so we can write this as $(0,u)$ where $u$ is the utility of the outcome. In this case we have that $(0,x-10,000)\succ (0,0)$ if $x\geq 10,000$, i.e. donations greater than $\$$10,000 offset a murder.

Now let's ask about the inter-theoretic uncertainty case. We have to choose between either doing nothing or murdering someone and donating $\$$15,000 to charity. We believe in satisficing consequentialism with probability $p$ and in standard utilitarianism with probability $1-p$. Therefore we have
$$\begin{align*}
p(15000,-10000) + (1-p)(0, 5000) & = (15000p,-10000p + 5000(1-p)) \\
& = (15000p,5000-15000p)
\end{align*}
$$ This is strongly preferred to the $(0,0)$ option if $p< 1/3$; if $p=1/3$ exactly then it is weakly preferred.

This isn't the only way we can make inter-theoretic comparisons. I actually don't even think it's the best way. But is one example where we're using a lexically ordered vector of real numbers, and all other examples will be similar.

A Counterexample

It may be useful to construct a decision criterion which can't be represented using a MITE formula. (Obviously, it will have to disobey one of the ordered-group axioms due to theorem 1.)

Here's one example:
Let's say we represent an outcome having deontological value $d$ and utility $u$ as $(d,u)$ and we believe deontology with probability $p$. Then $(d_1,u_1)\preceq (d_2,u_2)$ if and only if $p(u_1\mod d_1)\leq p(u_2\mod d_2)$.
This is not order-preserving because sometimes increasing utility is good but other times increasing utility is bad. So it doesn't make up an ordered group.

Commentary

Brian took as his definition of "rational" the standard von Neumann-Morgenstern axioms. This is of course a perfectly reasonable thing to do in general, but as he points out many individual moral theories fail these axioms. (Insert joke here about utilitarianism being the only "rational" moral system.)

I personally find the idea of optionality pretty stupid and think it causes all sorts of problems even without needing to compare it to other theories. But if you do want to give it some credence, then a MITE formula will work fine for you.

Footnotes

  1. Note that this also requires "modding out" by an indifference relation

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