Transformative experiences and choosing when you don't know what you value

There is a PDF version of the blogpost here.

There are some things whose value for us we can't know until we have them. Love might be one of them, becoming a parent another. In many such cases, we can't know how much we value that thing until we have it because only by having it can we know what it's like. The nature of the experience of having it---how it feels, its phenomenal character---plays a crucial role in determining the value that we assign to it. What it will feel like to be in love is part of what determines its value for the person who will experience it. The phenomenal quality of the bliss when that love is reciprocated, what it's like for you to feel the pain when it's not---together with the other aspects of being in love, these determine its value for you. And, you might reasonably think, you can't know what it will be like until you experience it. Let's follow Laurie Paul in calling these epistemically transformative experiences. In this post, I'd like to consider Paul's argument in Transformative Experience that they pose a problem for our theories of rational choice. I used to think those theories had the resources to meet this challenge, but now I'm not so hopeful.

Sapphire, uncertain of her utilities, as of so much

To understand Paul's challenge, let me describe it in a particular case. Currently, I'm not a parent. Now, suppose I'm trying to decide whether or not to adopt a child. I know that many factors contribute to the value I assign to becoming a parent, but one of them is what the experience will be like for me. I might enjoy it enormously; I might find it on balance agreeable, but often exhausting and upsetting; or I might hate it. Looking around at my friends and hearing of others' experience of adoption, I come to the conclusion that there are two sorts of person. For one of them, the character of their child doesn't make any difference to their experience of being a parent---they enjoy the experience well enough regardless, and a bit more than not being a parent. For others, it makes an enormous difference---if the child's character is like their own, their experience of parenting is wonderful; if not, it's awful.

To put some numbers on this, suppose you conclude that you are one of these two types of person, but you don't know which, and the two types assign utilities as follows:$$\begin{array}{r|cc} U_1 & \textit{Similar} & \textit{Different} \\ \hline \textit{Child-free} & 0 & 0 \\ \textit{Parent} & 1 & 1 \end{array} \hspace{10mm}  \begin{array}{r|cc} U_2 & \textit{Similar} & \textit{Different} \\ \hline \textit{Child-free} & 0 & 0 \\ \textit{Parent} & 25 & -7 \end{array}$$How should you choose? Paul argues that our theories of rational choice can't help us, since, to apply them, we must know our utilities. To that challenge, I responded in Choosing for Changing Selves that you should simply incorporate your uncertainty about your utilities into your representation of the decision problem. That is, you should not take the states of the world to be simply Similar or Different, since you don't know your utilities for the different options at those states of the world. Rather, you should take them to be Similar & my utilities are given by $U_1$, Similar & my utilities are given by $U_2$, Different & my utilities are given by $U_1$, and Different & my utilities are given by $U_2$, since you do know your utilities at the utilities of the option at these states of the world---the utility of being a parent at Similar & my utilities are given by $U_2$ is 25, for instance. So the decision now looks like this:$$\begin{array}{r|cccc} & \textit{Similar}\ \&\ U_1 & \textit{Different}\ \&\ U_1 & \textit{Similar}\ \&\ U_2 & \textit{Different}\ \&\ U_2 \\ \hline \textit{Child-free} & 0 & 0 & 0 & 0\\ \textit{Parent} & 1 & 1 & 25 & -7 \end{array}$$And you should assign probabilities to these states and then apply your theory of rational choice to this newly-specified decision. I called this the fine-graining response to Paul's challenge.

So, for instance, suppose that you think Similar and Different are equally likely, and $U_1$ and $U_2$ are equally likely, and they're independent of one another; then you should assign credence $\frac{1}{4}$ to each of these states. Suppose that your theory of rational choice tells you to maximise expected utility. Then you should become a parent, because its expected utility is $\frac{1}{4}\times 1 + \frac{1}{4}\times 1 + \frac{1}{4}\times 25 + \frac{1}{4}\times (-7) = 5$, while the expected utility of the alternative is 0. 

Now, notice something about that solution in the case of expected utility theory. Suppose that, whether your utilities are given by $U_1$ or by $U_2$, your expected utility for one option is greater than your expected utility for another. That is, while you don't know what your utilities are, you do know how they'd order those two options. Then, at least assuming that you consider the original states of the world  independent of the facts about your utilities, then the fine-graining strategy I described above will also give higher expected utility to the first option than to the second.*

But I've become convinced over the past few years that expected utility theory isn't the correct theory of rational choice. Rather, it's something closer to Lara Buchak's risk-weighted expected utility theory, which builds on John Quiggin's rank-dependent utility theory. According to this theory, we don't choose between different options by comparing their expected utility, we compare their risk-weighted expected utility instead, which is calculated using our own personal risk function.

A risk function is a function $R : [0, 1] \rightarrow [0, 1]$ that is continuous, strictly increasing, and for which $R(0) = 0$ and $R(1) = 1$. Given a risk function $R$, and an option $o$ such that $U(o, s_1) \leq U(o, s_2) \leq \ldots \leq U(o, s_n)$, and where the probability of state $s_i$ is $p_i$, the risk-weighted expected utility of $o$ is

$REU_R(U(o)) = U(o, s_1) + $

$R(p_2 + \ldots + p_n)(U(o, s_2) - U(o, s_1)) + $

$R(p_3 + \ldots + p_n)(U(o, s_3) - U(o, s_2)) + \ldots +$

$R(p_{n-1} + p_n)(U(o, s_{n-1}) - U(o, s_{n-2})) + $

$R(p_n)(U(o, s_n) - U(o, s_{n-1}))$

One natural family of risk functions is $R_n(p) = p^n$ for $n>0$. If $0 < n < 1$, then $R_n$ is risk-inclined; if $1 < n$, then $R_n$ is risk-seeking; and if $n = 1$, then $R_n$ is risk-netural.

Now suppose we consider my decision whether to become a parent again from the point of view of this theory of rational choice. And let's suppose I'm quite risk-averse, with a risk function $R(p) = p^2$. Then let's begin by looking at the decision from the point of view of $U_1$ and $U_2$ separately. If your utilities are given by $U_1$, then the risk-weighted expected utility of remaining child-free is 0, and of becoming a parent is 1; if $U_2$, then the risk-weighted expected utility of remaining child-free is 0 again, but of becoming a parent is $-7+R(1/2)\times 32 = 1$. So, either way, you'll prefer to become a parent. But now consider the fine-grained version of the decision:$$\begin{array}{r|cccc} & \textit{Similar}\ \&\ U_1 & \textit{Different}\ \&\ U_1 & \textit{Similar}\ \&\ U_2 & \textit{Different}\ \&\ U_2 \\ \hline \textit{Child-free} & 0 & 0 & 0 & 0\\ \textit{Parent} & 1 & 1 & 25 & -7 \end{array}$$Then the risk-weighted expected utility of remaining child-free is still 0, but of becoming a parent is $-7 + R(3/4)\times 8 + R(1/4)\times 24 = -1$. So you prefer remaining child-free.

In general, it turns out, if you use risk-weighted expected utility theory with a risk function that isn't risk-neutral, there will be cases like this, where you are uncertain of your utilities, know that whichever are your utilities they'll prefer one option to another, but when you fine-grain the states of the world and make the decision like that, you'll prefer the second option to the first. 

It seems to me, then, that epistemically transformative experiences do pose a problem for theories of rational choice like Buchak's. It's true that the fine-graining strategy allows us to apply our theory of rational choice even when we are uncertain about our own utilities. But adopting that strategy only pushes the problem elsewhere, for it then turns out that there will be cases in which I know I prefer one option to another---since I know that, whichever of the possible utility functions I actually have, it prefers that option---but I also prefer the other option to the first---since, when I apply the fine-graining strategy to determine which option I prefer, I prefer the other. And this, it seems to me, is an untenable situation.

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* Suppose $U_1, \ldots, U_m$ are the possible utilities you might have, and $q_1, \ldots, q_m$ are the probabilities you might have them; and suppose $s_1, \ldots, s_n$ are the possible states of the world, and $p_1, \ldots, p_n$ are their probabilities; and suppose $U_j(o, s_i)$ is the utility of $o$ at state $s_i$ by the lights of utilities $U_j$. Then, the expected utility of an option $o$ from the point of view of the fine-grained set of states, $s_i\ \&\ U_j$, is $$\sum^m_{j=1} \sum^n_{i=1} q_jp_i U(o, s_i\ \&\ U_j)= \sum^m_{j=1} \sum^n_{i=1} q_jp_i U_j(o, s_i) = \sum^m_{j=1} q_j \sum^n_{i=1} p_i U_j(o, s_i) $$So, if $$\sum^n_{i=1} p_i U_j(o, s_i) < \sum^n_{i=1} p_i U_j(o', s_i) $$for all $j= 1, \ldots, m$, then $$\sum^m_{j=1} \sum^n_{i=1} q_jp_i U(o', s_i\ \&\ U_j) < \sum^m_{j=1} \sum^n_{i=1} q_jp_i U(o, s_i\ \&\ U_j)$$