How to co-exist with nonexistent expectations

Abstract

Dozens of articles have addressed the challenge that gambles having undefined expectation pose for decision theory. This paper makes two contributions. The first is incremental: we evolve Colyvan’s “Relative Expected Utility Theory” into a more viable “conservative extension of expected utility theory” by formulating and defending emendations to a version of this theory that was proposed by Colyvan and Hájek. The second is comparatively more surprising. We show that, so long as one assigns positive probability to the theory that there is a uniform bound on the expected utility of possible gambles (and assuming a uniform bound on the amount of utility that can accrue in a fixed amount of time), standard principles of anthropic reasoning (as formulated by Brandon Carter) place lower and upper bounds on the expected values of gambles advertised as having no expectation–even assuming that with positive probability, all gambles advertised as having infinite expected utility are administered faithfully. Should one accept the uniform bound premises, this reasoning thus dissolves (or nearly dissolves, in some cases) several puzzles in infinite decision theory.

Appendix

In Sect. 5 we saw that ORET subscribers are vulnerable to group Dutch Books and to infinitary Dutch Books. The content of the following theorem is that they needn’t worry about finitary Dutch Books.

Theorem 2

Individual ORET subscribers are not vulnerable to finitary Dutch Books.

Proof

Let $(\Omega , \mu )$ be a probability space and let $({\mathcal F}_n)_{n=0}^\infty$ be a filtration on $\Omega$. That is, for each n, ${\mathcal F}_n$ is a $\sigma$-algebra of measurable subsets of $\Omega$, with ${\mathcal F}_0\subseteq {\mathcal F}_1\subseteq {\mathcal F}_2\subseteq \cdots$. For $n=0,1,2,\ldots$, let $X_n$ be a real valued random variable defined on $\Omega$. We assume that $X_0=0$ a.e. and $E(X_{n+1}-X_n|{\mathcal F}_n) \ge 0$ a.e., $n=0,1,2,\ldots$.

The idea here is that $X_n$ represents the agent’s bankroll at timestep n. Also at time step n (or shortly after), the agent is assumed to learn which cell of ${\mathcal F}_n$ obtains. The agent then consents (between time n and time $n+1$) to a gamble having payoff $X_{n+1}-X_n$. Since the agent knows which cell of ${\mathcal F}_n$ obtains and $E(X_{n+1}-X_n|{\mathcal F}_n) \ge 0$ a.e., the agent is at least indifferent to these gambles (she may view them as favorable), so we can assume that she accepts them.

The challenge for the would-be finitary Dutch Bookie is to construct a random variable T such that the gambles stop at time T (that is, the agent’s final bankroll is $X_T$) and $X_T<0$ a.e. We require that for every n, $\{\omega \in \Omega : T(\omega ) \le n\} \in {\mathcal F}_n$. (T is a stopping time; the bookie is allowed to know no more than the agent–when the gambles have stopped, the agent will know this.) For the Dutch Book to be finitary, one of following two additional conditions must be met (cf. footnote 4).

First condition $T(\omega )\le K$ a.e. for some $K<\infty$.

Note that $E(X_1|{\mathcal F}_0)= E(X_1-X_0|{\mathcal F}_0)\ge 0$ a.e., which implies in turn that $P\big (E(X_1|{\mathcal F}_1)\ge 0\big )>0$.

Next note that $E(X_2|{\mathcal F}_1)\ge E(X_1|{\mathcal F}_1)$ a.e. (both may = $+\infty$), so that

$$\begin{aligned} P\big ( E(X_2|{\mathcal F}_1)\ge E(X_1|{\mathcal F}_1)\ge 0\big ) >0. \end{aligned}$$

This, in turn, implies that¹⁸

$$\begin{aligned} P\big ( E(X_2|{\mathcal F}_2)\ge E(X_1|{\mathcal F}_1)\ge 0\big ) >0. \end{aligned}$$

Having shown that

$$\begin{aligned} P\big ( E(X_n|{\mathcal F}_n)\ge E(X_{n-1}|{\mathcal F}_{n-1})\ge \cdots \ge E(X_1|{\mathcal F}_1)\ge 0\big ) >0, \end{aligned}$$

note that $E(X_{n+1}|{\mathcal F}_{n})\ge E(X_{n}|{\mathcal F}_{n})$ a.e. This implies that

$$\begin{aligned} P\big ( E(X_{n+1}|{\mathcal F}_n)\ge E(X_{n}|{\mathcal F}_{n})\ge E(X_{n-1}|{\mathcal F}_{n-1})\cdots \ge E(X_1|{\mathcal F}_1)\ge 0\big ) >0, \end{aligned}$$

which implies in turn that

$$\begin{aligned} P\big ( E(X_{n+1}|{\mathcal F}_{n+1})\ge E(X_{n}|{\mathcal F}_{n})\ge E(X_{n-1}|{\mathcal F}_{n-1})\cdots \ge E(X_1|{\mathcal F}_1)\ge 0\big ) >0. \end{aligned}$$

By induction, then, one has

$$\begin{aligned} P\big ( E(X_{K}|{\mathcal F}_K)\ge E(X_{K-1}|{\mathcal F}_{K-1})\ge E(X_{K-2}|{\mathcal F}_{K-2})\cdots \ge E(X_1|{\mathcal F}_1)\ge 0\big ) >0. \end{aligned}$$

There is, therefore, a positive measure set $F\in {\mathcal F}_K$ such that

$$\begin{aligned} E(X_{K}|{\mathcal F}_K)\ge E(X_{K-1}|{\mathcal F}_{K-1})\ge E(X_{K-2}|{\mathcal F}_{K-2})\cdots \ge E(X_1|{\mathcal F}_1)\ge 0 \end{aligned}$$

a.e. on F. We may, moreover, assume that $T(\omega )$ is constant on F, say $T(\omega )=n\le K$, $\omega \in F$. We therefore have

$$\begin{aligned} F\subseteq \big \{ \omega : E(X_n|{\mathcal F}_n)(\omega )\ge 0\big \} \cap T^{-1}(n) =F_n\in {\mathcal F}_n . \end{aligned}$$

Now if $X_{T(\omega )} (\omega )<0$ a.e. then $X_n(\omega )<0$ for a.e. $\omega \in F_n$, which implies that $E(X_n|{\mathcal F}_n)(\omega )<0$ for a.e. $\omega \in F_n$, contradicting the definition of $F_n$. So $P\big (X_{T(\omega )}(\omega )\ge 0\big )>0$ and there is no Dutch Book.

Second condition $\sum _{n=0}^\infty E[ (X_{n+1}-X_n)_-\cdot 1_{\{ T>n\}}] <\infty$.

Let $(X^T_n)_{n=0}^\infty$ denote the stopped process, defined by $X^T_n =X_n$ if $n\le T$ and $X^T_n = X^T_T$ when $n>T$. Notice that $E(X^T_{n+1}-X^T_n|{\mathcal F}_n) \ge 0$ a.e., $n=0,1,2,\ldots$ and $X^T_n (\omega )\rightarrow X_{T(\omega )}(\omega )$ a.e. Let now

$$\begin{aligned} M = \sum _{n=0}^\infty (X_{n+1}-X_n)_-\cdot 1_{\{T>n\}} . \end{aligned}$$

Then $E(M) = \sum _{n=0}^\infty E[ (X_{n+1}-X_n)_-\cdot 1_{\{ T>n\}}] <\infty$ by the monotone convergence theorem. Moreover one has

$$\begin{aligned} -M\le X^T_n \le X_{T(\omega )} +M \end{aligned}$$

(1)

a.e. Since $E(X_{T})\ge E(-2M)>-\infty$, either $E(X_{T})=+\infty$, or $E(X_{T})$ is finite. In the latter case, $(X^T_n)$ is uniformly integrable by (1) and so

$$\begin{aligned} E[X_{T}] = \lim _{n\rightarrow \infty } E[X^T_n] \ge E(X^T_0) =0 \end{aligned}$$

by the dominated convergence theorem, since $E(X^T_{n+1})\ge E(X^T_{n})$ in the integrable case. So in either case, $E[X_T] \ge 0$ and there is no Dutch Book. $\square$