Satan, Saint Peter and Saint Petersburg

Abstract

We examine a distinctive kind of problem for decision theory, involving what we call discontinuity at infinity. Roughly, it arises when an infinite sequence of choices, each apparently sanctioned by plausible principles, converges to a ‘limit choice’ whose utility is much lower than the limit approached by the utilities of the choices in the sequence. We give examples of this phenomenon, focusing on Arntzenius et al.’s Satan’s apple, and give a general characterization of it. In these examples, repeated dominance reasoning (a paradigm of rationality) apparently gives rise to a situation closely analogous to having intransitive preferences (a paradigm of irrationality). Indeed, the agents in these examples are vulnerable to a money pump set-up despite having preferences that exhibit no obvious defect of rationality. We explore several putative solutions to such problems, particularly those that appeal to binding and to deliberative dynamics. We consider the prospects for these solutions, concluding that if they fail, the examples show that money pump arguments are invalid.

Appendices

Appendix 1: Continuity and transitivity

In this section, our main objective is to state and prove an interesting partial converse to Theorem 4.1. To state this converse, we need to define two notions of limit.

First, let $U$ be a real-valued function on a topological space X, not necessarily continuous. Then the limit of $U(x')$ as $x' \rightarrow x$ (or for short, the limit of $U$ at $x$) is defined to be the unique real number $r$ with the following property: for any open interval ($a, b$) around $r$, there is an open neighborhood V of $x$ such that $U$ maps V $-$ {$x$} into ($a, b$). If no such $r$ exists, the limit of $U$ at $x$ is undefined. The idea here is that we can make $U(x')$ as close as we like to $r$ by making $x'$ sufficiently close to $x$, provided $x \ne x'$. Continuity is definable in terms of limits: $U$ is continuous at $x$ iff the limit of $U$ at $x$ is $U(x)$.

Transitive intransitivity implies that the limit of $U$ at $A$ either does not exist or is less than $U(A)$. When the limit of $U$ at $A$ exists, this necessary condition for transitive intransitivity is almost sufficient. To formulate a sufficient condition, we need one more definition. A topological space X is said to be first countable if for each point $x$, there is a countable family {$V_{i}: i =$ 1, 2, ...} of open neighborhoods of $x$ such that for any open neighborhood $V$ of $x, V_{i} \subseteq V$ for some $i$. Such a family is called a countable base at x. All of the spaces we are concerned with in this paper are first countable. We then have the following result.

Theorem 7.1

Let $U$ be a real-valued function on a first countable topological space X, and for all $x \in$ X, let $U'(x)$ be the limit of $U$ at $x$. If $U'$ is well-defined throughout X (i.e., limits exist everywhere), then the ordering $\le$ generated by $U$ is transfinitely intransitive iff there is an $x \in$ X such that (a) $U(x) < U'(x)$, and (b) every neighborhood of $x$ contains a point $x'$ such that $U(x') < U'(x)$.

Proof

Left-to-right: Let $x < x_{1} < x_{2} < \cdots \rightarrow x$. Then $U(x) < U(x_{1}) < U(x_{2}) < \cdots$. Clearly, the sequence $U(x_{1}), U(x_{2}), \ldots$ converges to $U'(x)$, and this value is clearly greater than any of the $U(x_{i})$, and thus greater than $U(x)$. Thus, (a) holds. For (b), note that any open neighborhood $V$ of $x$ contains some of the points $x_{i}$, and since $U(x_{i}) < U'(x)$, (b) holds.

Right-to-left: Suppose $x$ satisfies (a) and (b), and let {$V_{i}: i =$ 1, 2, ...} be a countable base at $x$. By (b), for each $i$ we can pick $y_{i} \in V_{i}$ such that $U'(x) > U(y_{i})$. Any open neighborhood of X contains some $V_{i}$, and hence contains all but finitely many of the $y_{i}$, so the sequence $y_{1}, y_{2}, \ldots$ converges to $x$. Also, for every $i$, there is a $j > i$ such that $U(y_{i}) < U(y_{j})$. To see this, simply pick an interval ($a, b$) around $U'(x)$ that excludes $U(y_{i})$. By the definition of the limit, there is an open neighborhood $V$ of $x$ such that $U$ maps $V-${$x$} into ($a, b$), and there is some $V_{j} \subseteq V$; this gives us the required $j > i$. Thus, we can pick a subsequence $x_{1}, x_{2}, \ldots$ of $y_{1}, y_{2}, \ldots$ such that $U(x_{1}) < U(x_{2}) < \cdots$. This subsequence clearly converges to $x$, and the sequence $U(x_{1}), U(x_{2}), \ldots$ obviously converges to $U'(x)$. By (a), $U(x) < U'(x)$, and thus, $U(x) < U(x_{i})$ for some $i$. Hence, $U$ is transfinitely intransitive. $\square$

If we drop the condition that $U$ has a well-defined limit everywhere, we are not aware of any simple necessary and sufficient condition for the transfinite intransitivity of the generated ordering $\le$.

There is a second connection between transfinite intransitivity and continuity. Instead of assuming that the preference ordering $\le$ comes from a utility function, let us simply assume that $\le$ is complete, meaning that for all $A, B \in$ X, either $A \le B$ or $B \le A$. There is then a standard definition of continuity for preference orderings. Specifically, for any given point $A$, let U$_{A} =$ {$B$: $A \le B$} and L$_{A} =$ {$C$: $C \le A$}. Then $\le$ is said to be continuous if for every $A$, the sets U$_\mathrm{A}$ and L$_\mathrm{A}$ are topologically closed. (A set $S$ is closed iff whenever a sequence of points in $S$ converges, it converges to a point in $S$.) Now suppose that $\le$ is transfinitely intransitive, and let $A < A_{1} < A_{2} \ldots \rightarrow A$. Then the set {$B: A_{1} \le B$} is not closed, because there is a sequence of points from that set whose limit, $A$, is not in that set. Thus, transfinite intransitivity of $\le$ implies discontinuity of $\le$ in the standard sense, further justifying our talk of “discontinuity at infinity.”

Appendix 2: Extreme equilibria

The purpose of this section is to prove Theorem 5.1. We first prove a result based upon an elementary fact about infinite products.

Lemma

$\hbox {lim}_{n \rightarrow \infty } P\left( {ALL/S_n } \right) = 1 \hbox { iff }\sum \nolimits _{n=1}^\infty {(1-q_n )} <\infty$, and $P$(ALL / $S_{n})$ $=$ 0 for all $n$ iff $\sum _{n=1}^\infty {(1-q_n )} =\infty$.

Proof

If $0 \le u_{n} < 1$, then $\prod _{n=1}^\infty {(1-u_n } )>0 \hbox { iff } \sum _{n=1}^\infty {u_n } <\infty$ (Rudin 1974, p. 322).

Recall that $P(ALL / S_{n})=\prod _{i=n+1}^\infty {q_i }$. If $q_{i}$ = 0 for infinitely many $i$, then $P(ALL / S_{n}) =$ 0 for all $n$ and $\sum _{n=1}^\infty {(1-q_n )} =\infty$. If $q_{i} =$ 0 for finitely many $i$, it follows from (Rudin 1974) that either $P(ALL / S_{n}) =$ 0 and $\sum _{i=n+1}^\infty {(1-q_i )} =\infty$, or $P(ALL / S_{n}) >$ 0 and $\sum _{i=n+1}^\infty {(1-q_i )} <\infty$. In the latter case, $P(ALL / S_{n})$ must converge to 1 as $n \rightarrow \infty$. $\square$

Theorem 7.2

Every equilibrium is an extreme assignment. (For all conditional probabilities $q_{n}=P(S_{n} / S_{1} \cdot {\ldots } \cdot S_{n-1})$, either $q_{n} =$ 0 or $q_{n} =$ 1.)

Proof

Suppose that {$q_{n}$} is an equilibrium probability assignment. Either $\sum _{n=1}^\infty {(1-q_n )} <\infty$ or $\sum _{n=1}^\infty {(1-q_n )} =\infty$. If $\sum _{n=1}^\infty {(1-q_n )} =\infty$, then $P(ALL / S_{n}) = 0$ for all $n$. It follows that for all $n$, the expected utility calculations in 2’ favour Take piece n. The dynamics implies upward pressure ($q_{n}' > q_{n}$) unless $q_{n} =$ 0 or $q_{n} =$ 1. Since we have an equilibrium, it follows that $q_{n} =$ 0 or $q_{n} =$ 1 for all $n$. Hence, we have an extreme assignment.

If instead $\sum _{n=1}^\infty {(1-q_n )} <\infty$, then lim$_{n\rightarrow \infty } P(ALL / S_{n}) =$ 1; also, lim$_{n\rightarrow \infty } q_{n} =$ 1. Combining these facts with 2’, Eve reaches a point $N$ beyond which each $q_{n}$ is close to 1 and $P(ALL / S_{n})$ is so large that her expected utility computations favour Reject piece n. For $n > N$, the dynamics implies downward pressure ($q_{n}' < q_{n}$) unless $q_{n} =$ 1. Since we have an equilibrium, it follows that $q_{n}$ = 1 for all $n > N$. So the ‘tail end’ of the probability assignment, from $N+1$ onwards, consists entirely of 1s.

To finish up, we show that, for this case, each of $q_{1}, {\ldots }, q_{N}$ must also be 0 or 1. If not, let $k$ be the largest index such that 0 $< q_{k} <$ 1. If $q_{n} =$ 0 for some $n > k$, then the dynamics implies upward pressure on $q_{k}$: since $P(ALL / S_{k}) =$ 0, EU$(S_{k}) >$ EU($\sim S_{k}$) and $q_{k}' > q_{k}$, contradicting the fact that we have an equilibrium. If $q_{n} =$ 1 for all $n > k$, then the dynamics implies downward pressure on $q_{k}$: since $P(ALL / S_{k}) =$ 1, EU$(S_{k}) <$ EU($\sim S_{k}$) and $q_{k}' < q_{k}$, again contradicting the fact that we have an equilibrium. Hence, $q_{k} =$ 0 and each of $q_{1}, {\ldots }, q_{N}$ must also be 0 or 1.

Hence, the equilibrium assignments for Eve are exactly the extreme assignments in which all of her conditional probabilities $q_{n}=P(S_{n} / S_{1} \cdot {\ldots } \cdot S_{n-1})$ are either 0 or 1. $\square$