Notation: Throughout I will let $\Delta X$ denote the set of probability distributions over the set $X$.

I have been studying expected utility theory, and especially Savage Acts and Anscombe-Aumann Acts. However I am new to it and am not sure if I have the terminology correct.

Let $S$ be the finite set of possible states, let $Z$ be the set of possible outcomes. An Anscombe-Aumann act is defined as a function $f:S \to \Delta Z$. Let the space of acts be denoted $X$. Preferences are defined over acts.

If preferences have a state independent expected utility representation (SIEU) we have that there is a function $u: Z \to \mathbb{R}$ and a distribution over states $p \in \Delta S$ such that for any two acts $f,g \in X$, $$f \precsim g \iff \sum_{s \in S} p(s) \sum_{z \in Z} u(z) f(s,z) \leq \sum_{s \in S} p(s) \sum_{z \in Z} u(z) g(s,z)$$

Suppose I want to define a lottery $L \in \Delta Z$ in this situation. For simplicity, suppose $L$ is the lottery that assigns probability $1$ to outcome $z^{*}$ and probability $0$ to all other outcomes.

My Question is: Is this lottery (or any lottery for that matter) a compound lottery first over states ($p$) and then over anscombe -aumann acts ($f$)? Or how is a lottery defined in this situation?

Let me know if any clarification is required.


1 Answer 1


You are right, but to make sure that the odds are really exogenous ("objective") you need to make sure that the subjective uncertainty has no bite here. In other words, you need to assume that the objective lottery played after the horse race is independent of the result of the horse race (in Anscombe-Aumann's terminology).

Let's assume that $Z$ is finite, $Z=\{z_1,\cdots,z_n\}$ and that $L$ is the objective lottery that puts weight $l_i$ on the prize $z_i$, with $l_i \geq 0$ and $\sum_{i=1}^{n}{l_i}=1$.

You can identify the lottery $L$ with the act $f:s \rightarrow \Delta Z$ such that $f(s)=L$ for any $s \in S$. That is, for any $s \in S$, $f(s)$ is the objective lottery that puts weight $l_i$ on the prize $z_i$, exactly as $L$ does.


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