Desirability foundations of robust rational decision making

Abstract

Recent work has formally linked the traditional axiomatisation of incomplete preferences à la Anscombe-Aumann with the theory of desirability developed in the context of imprecise probability, by showing in particular that they are the very same theory. The equivalence has been established under the constraint that the set of possible prizes is finite. In this paper, we relax such a constraint, thus de facto creating one of the most general theories of rationality and decision making available today. We provide the theory with a sound interpretation and with basic notions, and results, for the separation of beliefs and values, and for the case of complete preferences. Moreover, we discuss the role of conglomerability for the presented theory, arguing that it should be a rationality requirement under very broad conditions.

Appendices

A Decomposition and completeness of preferences

We now turn to the formalisms of desirability and coherent lower previsions to briefly show how they can represent and generalise well-known concepts in rational preferences. The following discussion is a summary of the one we presented in Zaffalon and Miranda (2017, Section 5), with some additional remarks and proofs for the case of infinitely many values.

1.1 A.1 In terms of sets of desirable gambles

We begin by considering the case where our assessments are modelled by sets of desirable gambles.

1.1.1 A.1.1 State independence

State independence is the condition that allows us to have separate models for beliefs and values.

We showed in Zaffalon and Miranda (2017, Section 5.3) that the traditional notion of state independence in the literature of preferences is equivalent to the notion of ‘strong independence’ in imprecise probability. In the case of totally ordered Archimedean preferences (i.e., precise probability and utility models), this means that probability and utility are stochastically independent models, when utility is mathematically represented through a probability function (like we do in this paper, see Sect. 3.4).

Such a notion of independence extends over multiple probability-utility pairs, and even to sets of desirable gambles. However, it ceases to exist when both $\varOmega$ and $\mathcal {X}$ are infinite (Miranda and Zaffalon 2015a, Example 1).⁹ This holds also for most other notions of independence while, interestingly, that need not be the case for notions of ‘irrelevance’: these are asymmetric notions where mutual (symmetric) independence is replaced by independence in only one direction. In the case of this paper, for instance, this means assuming that values are independent of states but not vice versa; we say that states are irrelevant to values.

This makes of irrelevance a very natural candidate for a notion of state independence that exists no matter the cardinalities of the spaces involved. But there is more to it, as it is arguable that irrelevance qualifies itself as the right notion to use in the present context, unlike independence. The reason is that while it makes sense that states may be irrelevant to values, the opposite seems to be questionable: in fact, as we have already remarked, it appears meaningless to update a model on elements of $\mathcal {X}$, which do not occur; as a consequence the irrelevance of values to states appears meaningless too.

Let us then consider the following definition of irrelevance for desirability:

Definition 29

($\varOmega$-$\mathcal {X}$ irrelevant product for gambles) A coherent set of gambles $\mathcal {R}$ on $\varOmega \times \mathcal {X}$ is called an $\varOmega$-$\mathcal {X}$ irrelevant product of its marginal sets of gambles $\mathcal {R}_{\varOmega }',\mathcal {R}_{\mathcal {X}}'$ if it includes the set

$$\begin{aligned} \mathcal {R}|\varOmega :=\{h\in \mathcal {L}(\varOmega \times \mathcal {X}): (\forall \omega \in \varOmega )\ h(\omega ,\cdot )\in \mathcal {R}_{\mathcal {X}}'\cup \{0\}\}{\setminus }\{0\}. \end{aligned}$$

We have already considered these products, in another context and for lower previsions, in Miranda and Zaffalon (2015a) (see also the work by de Cooman and Miranda 2009; de Bock and de Cooman 2014, 2015b).

The rationale of the definition is the following. First, the requirement that $h(\omega ,\cdot )\in \mathcal {R}_\mathcal {X}'\cup \{0\}$ is there so that the inferences conditional on $\{\omega \}$ encompassed by $\mathcal {R}|\varOmega$ should yield the marginal set $\mathcal {R}_\mathcal {X}'$. This is just a way to formally state that $\omega$ is irrelevant to $\mathcal {X}$. The same is repeated for every $\omega$, so that $\mathcal {R}|\varOmega$ can be regarded as being born out of aggregating all the irrelevant conditional sets.¹⁰ Finally, that $\mathcal {R}$ contains $\mathcal {R}|\varOmega$ is imposed to make sure that $\mathcal {R}$ is a model coherent with the irrelevance of beliefs on values.

If $\mathcal {R}$ satisfies Definition 29 we say it models state-independent preferences. The least-committal among these models is given by¹¹

Proposition 5

Given two marginal coherent sets of desirable gambles $\mathcal {R}_{\varOmega },\mathcal {R}_{\mathcal {X}}$ defined on the subsets of $\mathcal {L}(\varOmega \times \mathcal {X})$ given by the $\varOmega$-measurable and $\mathcal {X}$-measurable gambles, respectively, their smallest $\varOmega$-$\mathcal {X}$ irrelevant product is given by the marginal extension of $\mathcal {R}_{\varOmega }$ and $\mathcal {R}|{\{\omega \}}:=\mathcal {R}_{\mathcal {X}}$ for all $\omega \in \varOmega$. Namely,

$$\begin{aligned} \hat{\mathcal {R}}:=\{g+h: g\in \mathcal {R}_{\varOmega }\cup \{0\}, h\in \mathcal {R}|\varOmega \cup \{0\}\}{\setminus }\{0\}. \end{aligned}$$

(12)

Definition 29 and Eq. (12) represent our main proposal to model state-independent preferences.

1.1.2 A.1.2 State dependence

We say that a set of desirable gambles $\mathcal {R}$ models state-dependent preferences when it does not satisfy the conditions in Definition 29, i.e., when it does not include the set $\mathcal {R}|\varOmega$ derived from its marginal $\mathcal {R}_\mathcal {X}$ and the assumption of $\varOmega$-$\mathcal {X}$ irrelevance.

1.1.3 A.1.3 Completeness

So far we have discussed the general case of coherent sets of gambles $\mathcal {R}$ without discussing in particular the so-called complete or maximal coherent sets for which it holds that $f\notin \mathcal {R}\Rightarrow -f\in \mathcal {R}$ for all $f\in \mathcal {L},f\ne 0$. Given the equivalence between coherent sets of gambles in $\mathcal {L}(\varOmega \times \mathcal {X})$ and coherent preferences in $\mathcal {H}\times \mathcal {H}$, discussing the case of maximal sets amounts to discussing the case of complete preferences.

Definition 30

(Completeness of beliefs and values for gambles) A coherent set of desirable gambles $\mathcal {R}\subseteq \mathcal {L}(\varOmega \times \mathcal {X})$ is said to represent complete beliefs if $\mathcal {R}_{\varOmega }$ is maximal. It is said to represent complete values if $\mathcal {R}_{\mathcal {X}}$ is maximal. Finally, if $\mathcal {R}$ is maximal, then it is said to represent complete preferences.

The rationale of this definition is straightforward: when we say, for instance, that $\mathcal {R}$ represents complete beliefs, we mean that there is never indecision between two options that are concerned only with $\varOmega$. The situation is analogous in the case of complete values. Finally, in the case of complete preferences, the definition is just the direct application of the definition of maximality.

1.2 A.2 In terms of lower previsions

We now consider the case where preferences are modelled by means of a coherent lower prevision ${\underline{P}}$ on $\mathcal {L}(\varOmega \times \mathcal {X})$.

1.2.1 A.2.1 State independent preferences

We proceed as in the case of desirability by focusing on irrelevance of states to values.

Definition 31

($\varOmega$-$\mathcal {X}$ irrelevant product for lower previsions) Given marginal coherent lower previsions ${\underline{P}}_{\varOmega }',{\underline{P}}_{\mathcal {X}}'$, let

$$\begin{aligned} {\underline{P}}(f|\{\omega \}):={\underline{P}}_\mathcal {X}'(f(\omega ,\cdot )), \end{aligned}$$

for all $f\in \mathcal {L}(\varOmega \times \mathcal {X})$, and

$$\begin{aligned} {\underline{P}}(\cdot |\varOmega ):=\sum _{\omega \in \varOmega }I_{\{\omega \}}{\underline{P}}(\cdot |\{\omega \}). \end{aligned}$$

Then a coherent lower prevision ${\underline{P}}$ on $\mathcal {L}(\varOmega \times \mathcal {X})$ is called an $\varOmega$-$\mathcal {X}$ irrelevant product of ${\underline{P}}_{\varOmega }',{\underline{P}}_{\mathcal {X}}'$ if

$$\begin{aligned} {\underline{P}}\ge {\underline{P}}_\varOmega '({\underline{P}}(\cdot |\varOmega )). \end{aligned}$$

In this case we also say that ${\underline{P}}$ models state-independent preferences.

Here ${\underline{P}}(\cdot |\varOmega )$ plays the role that $\mathcal {R}|\varOmega$ took in Section A.1, that is, conglomerating all the conditional information. The concatenation ${\underline{P}}_\varOmega '({\underline{P}}(\cdot |\varOmega ))$ is a marginal extension: in particular, it is the least-committal coherent $\varOmega$-conglomerable model built out of the given marginals and the assessment of irrelevance that defines ${\underline{P}}(\cdot |\varOmega )$ through ${\underline{P}}_\mathcal {X}'$. Every coherent lower prevision that dominates ${\underline{P}}_\varOmega '({\underline{P}}(\cdot |\varOmega ))$ is compatible with the irrelevance assessment but is also more informative than that:

Proposition 6

Given two marginal coherent lower previsions ${\underline{P}}_{\varOmega }',{\underline{P}}_{\mathcal {X}}'$, their smallest $\varOmega$-$\mathcal {X}$ irrelevant product is given by the marginal extension

$$\begin{aligned} {\underline{P}}_\varOmega '({\underline{P}}(\cdot |\varOmega )). \end{aligned}$$

We refer to the work of Walley (1991, Section 6.7), Miranda and de Cooman (2007), de Cooman and Miranda (2009) and de Cooman and Hermans (2008) for additional information on the marginal extension and its use in a number of different contexts.

1.2.2 A.2.2 State dependent preferences

As in the case of desirable gambles, we define state dependence as the lack of state independence: this means that if we consider a coherent lower prevision ${\underline{P}}$ with marginals ${\underline{P}}_{\varOmega }',{\underline{P}}_{\mathcal {X}}'$, it is said to model state-dependent preferences when it does not dominate the concatenation ${\underline{P}}_{\varOmega }'({\underline{P}}(\cdot |\varOmega ))$, where ${\underline{P}}(\cdot |\varOmega )$ is derived from ${\underline{P}}_{\mathcal {X}}'$ by an assumption of epistemic irrelevance.

1.2.3 A.2.3 Completeness

The definition of completeness for lower previsions is a rephrasing of that for desirable gambles:

Definition 32

(Completeness of beliefs and values for lower previsions) A coherent lower prevision ${\underline{P}}$ on $\mathcal {L}(\varOmega \times \mathcal {X})$ is said to represent complete beliefs if its marginal ${\underline{P}}_{\varOmega }$ is linear. It is said to represent complete values if its marginal ${\underline{P}}_{\mathcal {X}}$ is linear. Finally, if ${\underline{P}}$ is linear, then it is said to represent complete preferences.

There are a number of equivalent ways in which we can characterise the linearity of previsions in terms of the corresponding set of desirable gambles:

Proposition 7

Let ${\underline{P}}$ be a coherent lower prevision on $\mathcal {L}$ and $\mathcal {R}$ its corresponding coherent set of strictly desirable gambles. The following are equivalent:

• (i) ${\underline{P}}$ is a linear prevision.

• (ii) If $f\notin \mathcal {R}$ then $\varepsilon -f\in \mathcal {R}$ for all $\varepsilon >0$.

• (iii) $\mathcal {R}$ is negatively additive, meaning that $f,g\notin \mathcal {R}\Rightarrow (\forall \varepsilon >0)\ f+g-\varepsilon \notin \mathcal {R}$.

We can then focus on any of those formulations to immediately deduce characterisations for all the cases of completeness in preferences:

Proposition 8

Consider a coherent preference relation $\succ$ on gambles represented by a coherent lower prevision ${\underline{P}}$ on $\mathcal {L}(\varOmega \times \mathcal {X})$, whose corresponding coherent set of gambles is denoted by $\mathcal {R}$. Let $\mathcal {R}_\varOmega ,\mathcal {R}_\mathcal {X}$ denote its marginals. Then:

• (i) ${\underline{P}}$ represents complete beliefs $\Leftrightarrow$ $\mathcal {R}_\varOmega$ is negatively additive.

• (ii) ${\underline{P}}$ represents complete values $\Leftrightarrow$ $\mathcal {R}_\mathcal {X}$ is negatively additive.

• (iii) ${\underline{P}}$ represents complete preferences $\Leftrightarrow$ $\mathcal {R}$ is negatively additive.

B Proofs

Proof of Proposition 1

Using the mixture-independence axiom A4 together with the assumption,

$$\begin{aligned} p\succ q\Leftrightarrow \alpha p+(1-\alpha )r\succ \alpha q+(1-\alpha )r=\alpha p+(1-\alpha )s\Leftrightarrow r\succ s. \end{aligned}$$

Proof of Proposition 2

By $p\gneq q$ we deduce that $0\lneq p-q\le 1$. Then $(p-q)\in \mathcal {H}$ and $(p-q)\succ 0$ by A1. And since

$$\begin{aligned} \frac{1}{2}(p-q)=\frac{1}{2}[(p-q)-0], \end{aligned}$$

we deduce by Proposition 1 that $p\succ q$.

Proof of Lemma 1

Consider a set of desirable gambles $\mathcal {D}\subseteq \mathcal {L}_1(\varOmega \times \mathcal {X})$ and define the linear relation $\succ \subseteq \mathcal {H}\times \mathcal {H}$ by $p\succ q$ if and only if $p-q\in \mathcal {D}$.

Observe also that for any $f\in \mathcal {D}$ we can always find $p,q\in \mathcal {H}$ such that $f=p-q$: it is sufficient to let $p:=\max (f,0)$ and $q:=\min (0,f)$.

Let us now prove injectivity. Assume by contradiction that there are sets $\mathcal {D}_1\ne \mathcal {D}_2$ in $\mathcal {L}_1$ such that the respective linear relations are equal: $\succ _1=\succ _2$. Since the sets are different, then we can assume without loss of generality that there is $f_1\in \mathcal {D}_1{\setminus }\mathcal {D}_2$. At the same time for all $p,q\in \mathcal {H}$ with $p-q=f_1$, we have that $p\succ _1q$ and $p\succ _2 q$. This implies that there is $f_2\in \mathcal {D}_2$ such that $f_2=p-q$. But then $f_1=f_2$, a contradiction.

Let us focus on surjectivity. Consider a linear relation $\succ \subseteq \mathcal {H}\times \mathcal {H}$ and let $\mathcal {D}':=\{f\in \mathcal {L}_1(\varOmega \times \mathcal {X}):(\exists p,q\in \mathcal {H})\ f=p-q,p\succ q\}$. The relation it induces is given by $p\succ 'q\Leftrightarrow p-q\in \mathcal {D}'$. Now, if $p\succ q$, then $p-q\in \mathcal {D}'$, whence $p\succ 'q$. Conversely, if $p\succ 'q$, then $p-q\in \mathcal {D}'$; as a consequence, there are $p',q'\in \mathcal {H}$, $p'\succ q'$ such that $p-q=p'-q'$. But relation $\succ$ is linear, whence $p\succ q$. It follows that $\succ =\succ '$.

Proof of Theorem 1

Consider a set of desirable gambles $\mathcal {D}\subseteq \mathcal {L}_1$ and its corresponding linear relation $\succ$. They are related through

$$\begin{aligned}&\mathcal {D}=\{f\in \mathcal {L}_1(\varOmega \times \mathcal {X}):(\exists p,q\in \mathcal {H})\ f=p-q,p\succ q\}, \\&p\succ q\Leftrightarrow p-q\in \mathcal {D}. \end{aligned}$$

($\hbox {A1}\Leftrightarrow \hbox {D1}'$):

If $p\succ 0$ for all $p\in \mathcal {H},p\ne 0$, then $p\in \mathcal {D}$; and since $\mathcal {L}_1^+=\mathcal {H}{\setminus }\{0\}$, then we have the thesis. Conversely, if $\mathcal {L}_1^+\subseteq \mathcal {D}$, then $p\in \mathcal {D}$ for all $p\in \mathcal {H}{\setminus }\{0\}$; as a consequence $p\succ 0$ for all $p\in \mathcal {H},p\ne 0$.

($\hbox {A2}\Leftrightarrow \hbox {D2}'$):

By contradiction, if $p\succ p$ for some $p\in \mathcal {H}$, then $0\in \mathcal {D}$; and if $0\in \mathcal {D}$, then for some $p\in \mathcal {D}$, $p\succ p$.

($\hbox {A3}\Leftrightarrow \hbox {D3}'$):

Assume that the relation is transitive and take $f,g\in \mathcal {D}$ such that $f=p-q,g=q-r$ for some $p,q,r\in \mathcal {H}$. Then we have that $p\succ q$ and $q\succ r$; by transitivity we conclude that $p\succ r$, whence $f+g\in \mathcal {D}$.

Conversely, assume that $\mathcal {D}$ is additive and take $p\succ q$, $q\succ r$. Letting $f:=p-q,g:=q-r$, we get by additivity that $f+g=p-r\in \mathcal {D}$, so that $p\succ r$.

($\hbox {A4}\Leftrightarrow \hbox {D4'}$):

Assume that $\succ$ satisfies mixture independence and take $f\in \mathcal {D}$ and $\lambda >0$ such that $\lambda f\in \mathcal {L}_1$. Then there are $p\succ q$ such that $f=p-q$. Moreover, there must be $r,s\in \mathcal {H}$ such that $\lambda f=r-s$ given that $\lambda f\le 1$ (it is enough as usual to take the positive and negative parts of $\lambda f$, respectively). Using Proposition 1, we get that $r\succ s$, whence $\lambda f\in \mathcal {D}$.

Conversely, assume that $f\in \mathcal {D}$ implies that $\lambda f\in \mathcal {D}$ for all $\lambda >0$ such that $\lambda f\in \mathcal {L}_1$. Taking $p\succ q$, we have then that $\lambda p+(1-\lambda )r\succ \lambda q+(1-\lambda )r$ for all $p,q,r\in \mathcal {H}$ and $\lambda \in (0,1]$. On the other hand, if $\lambda p+(1-\lambda )r\succ \lambda q+(1-\lambda )r$ for all $p,q,r\in \mathcal {H}$ and $\lambda \in (0,1]$, then trivially $p\succ q$.

($\hbox {A0}\Leftrightarrow \hbox {D0'}$):

Assume that for all $p\succ q$, with $p\gneq q$ not holding, there is $\alpha \in (0,1)$ such that $\alpha p\succ q$. We want to show that for all $f\in \mathcal {D}{\setminus }\mathcal {L}_1^+$ and all $g\in \mathcal {L}_1$ such that $\max (0,-f)\le g\le \min (1,1-f)$, there is $\alpha \in (0,1)$ such that $\alpha f-(1-\alpha )g\in \mathcal {D}$.

Let $q:=g$ and $p:=f+q$, so that $f=p-q$. Let us show that $q\in \mathcal {H}$; it is enough to prove that $\max (0,-f)\le \min (1,1-f)$. In case $f(\omega ,x)\ge 0$, the inequality reduces itself to $0\le 1-f(\omega ,x)$, which is true since $f(\omega ,x)\le 1$. In case $f(\omega ,x)<0$, the inequality becomes $-f(\omega ,x)\le 1$, which is again true since $f(\omega ,x)\ge -1$.

Let us similarly show that $p=f+q\in \mathcal {H}$ too. We know that $f+\max (0,-f)\le p\le f+\min (1,1-f)$. Again, if $f(\omega ,x)\ge 0$, we see that $0\le f(\omega ,x)\le p(\omega ,x)\le 1$; if $f(\omega ,x)<0$, we have that $0\le p(\omega ,x)\le f+1\le 1$.

At this point we can apply the hypothesis, recalling that $f\notin \mathcal {L}_1^+$ and hence that $p\gneq q$ does not hold, and so we deduce that there is $\alpha \in (0,1)$ such that $\alpha p-q=\alpha (p-q)-(1-\alpha )q=\alpha f-(1-\alpha )q=\alpha f-(1-\alpha )g\in \mathcal {D}$.

Conversely, suppose that for all $f\in \mathcal {D}{\setminus }\mathcal {L}_1^+$ and all $g\in \mathcal {L}_1$ such that $\max (0,-f)\le g\le \min (1,1-f)$, there is $\alpha \in (0,1)$ such that $\alpha f-(1-\alpha )g\in \mathcal {D}$. Let us show that $p\succ q$, with $p\gneq q$ not holding, implies that there is $\alpha \in (0,1)$ such that $\alpha p\succ q$.

Let $f:=p-q$, from which it follows that $f\notin \mathcal {L}_1^+$, and $g:=q$. Observe that if $\max (0,-f)\le g\le \min (1,1-f)$ holds, then from the assumption the thesis follows immediately. In other words, we have to show that $\max (0,q-p)\le q\le \min (1,q+(1-p))$. This is immediate since on the left-hand side we decrease q by some non-negative amount while on the right-hand side we are increasing it by a non-negative amount.

Proof of Lemma 2

That D3$''$ implies D3$'$ is trivial. Therefore let us assume that $\mathcal {D}$ satisfies D3$'$ and prove that D3$''$ holds.

Consider $f,g\in \mathcal {D}$ such that $f+g\in \mathcal {L}_1$. By D4$'$ we have that $\frac{1}{2}f,\frac{1}{2}g\in \mathcal {D}$. We can always write $f=p-q,g=r-s$ for some $p,q,r,s\in \mathcal {H}$. It follows that

$$\begin{aligned} \left( \frac{1}{2}p+\frac{1}{2}r\right) -\left( \frac{1}{2}q+\frac{1}{2}r\right) \in \mathcal {D},\\ \left( \frac{1}{2}q+\frac{1}{2}r\right) -\left( \frac{1}{2}q+\frac{1}{2}s\right) \in \mathcal {D}. \end{aligned}$$

Using D3$'$ we obtain that $(\frac{1}{2}p+\frac{1}{2}r)-(\frac{1}{2}q+\frac{1}{2}s)=\frac{1}{2}f+\frac{1}{2}g\in \mathcal {D}$. Using D4$'$ again, we have the thesis.

Proof of Theorem 2

Let us prove the five points of the theorem.

• 1. Assume that $\mathcal {D}$ satisfies D1$'$–D4$'$ and show that $\mathcal {R}_\mathcal {D}$ is coherent.

  • D1. Consider $f\in \mathcal {L}^+$ and let $g:=f/\sup |f|$, so that $g\in \mathcal {L}_1^+$. As a consequence of D1$'$, $g\in \mathcal {D}$, whence $f\in \mathcal {R}_\mathcal {D}$.

  • D2. Trivial.

  • D3. Take $f,g\in \mathcal {R}_\mathcal {D}$, so that there are $f',g'\in \mathcal {D},\lambda ',\lambda ''>0$ with $f=\lambda 'f',g=\lambda ''g'$. If we let $\lambda :=\lambda '+\lambda ''$, we have that $f/(2\lambda ),g/(2\lambda )\in \mathcal {D}$ by D4$'$ and at the same time $(f+g)/(2\lambda )\in \mathcal {L}_1$, so that $(f+g)/(2\lambda )\in \mathcal {D}$ by D3$''$. As a consequence, $f+g\in \mathcal {R}_\mathcal {D}$.

  • D4. Trivial.

• 2. That $\mathcal {D}_\mathcal {R}$ satisfies D1$'$–D4$'$ once $\mathcal {R}$ is coherent is straightforward.

• 3. We need to show that, under coherence of $\mathcal {R},\mathcal {D}$, it holds that $\mathcal {R}_{\mathcal {D}_{\mathcal {R}}}=\mathcal {R}$ and $\mathcal {D}_{\mathcal {R}_{\mathcal {D}}}=\mathcal {D}$. For the first equality, that $\mathcal {R}_{\mathcal {D}_\mathcal {R}}\supseteq \mathcal {R}$ follows from

  $$\begin{aligned} \mathcal {R}_{\mathcal {D}_\mathcal {R}}=\{\lambda g: \lambda>0,g\in \mathcal {D}_\mathcal {R}\}=\left\{ \lambda \frac{g}{\sup |g|}: \lambda >0,g\in \mathcal {R}\right\} \supseteq \mathcal {R}\end{aligned}$$

  by taking $\lambda :=\sup |g|$.

  To show that $\mathcal {R}_{\mathcal {D}_\mathcal {R}}\subseteq \mathcal {R}$, consider any h in $\mathcal {R}$, implying that $\frac{h}{\sup |h|}\in \mathcal {D}_{\mathcal {R}}$. Note that $\sup |h| > 0$; indeed, if $\sup |h|$ would be 0 then $h=0$. For any given $\lambda >0$ , this implies that $\lambda \frac{h}{\sup |h|}=\lambda ' h$ for $\lambda ':=\frac{\lambda }{\sup |h|}>0$ indeed belongs to $\mathcal {R}$, because this set satisfies D4.

  Let us prove now the equality $\mathcal {D}_{\mathcal {R}_{\mathcal {D}}}=\mathcal {D}$. Take $\mathcal {D}$ satisfying D1$'$–D4$'$. To see that $\mathcal {D}_{\mathcal {R}_{\mathcal {D}}} \subseteq \mathcal {D}$, take $f\in \mathcal {D}_{\mathcal {R}_{\mathcal {D}}}$. Then there is some $g\in \mathcal {R}_\mathcal {D}$ such that $f=\frac{g}{\sup |g|}$. Since $g\in \mathcal {R}_\mathcal {D}$, there is some $h\in \mathcal {D}$ and some $\lambda >0$ such that $g=\lambda h$, whence $f=\frac{\lambda h}{\sup |\lambda h|}=\frac{h}{\sup |h|}$. Since $f\in \mathcal {L}_1$, we deduce from D4$'$ that $f\in \mathcal {D}$.

  Conversely, take $f\in \mathcal {D}$. Then $f\in \mathcal {R}_\mathcal {D}$ and as a consequence $\frac{f}{\sup |f|}\in \mathcal {D}_{\mathcal {R}_\mathcal {D}}$. Since $\mathcal {D}_{\mathcal {R}_\mathcal {D}}$ satisfies D4$'$ and $f\in \mathcal {L}_1$, we deduce that $f\in \mathcal {D}_{\mathcal {R}_\mathcal {D}}$.

• 4. Since $\mathcal {D}\subseteq \mathcal {R}_\mathcal {D}$ and $\mathcal {R}_\mathcal {D}$ is coherent (apply 1), then $\mathcal {D}$ avoids partial loss by (Miranda and Zaffalon 2010, Proposition 3(e)). This implies that the minimal coherent extension of $\mathcal {D}$ exists. Let us call it $\mathcal {E}$. That $\mathcal {R}_\mathcal {D}\subseteq \mathcal {E}$ is trivial. Since $\mathcal {R}_\mathcal {D}$ is coherent, then by definition of $\mathcal {E}$, it must be that $\mathcal {R}_\mathcal {D}\supseteq \mathcal {E}$. As a consequence $\mathcal {R}_\mathcal {D}$ is the natural extension.

  Let us show that $\mathcal {R}_\mathcal {D}\cap \mathcal {L}_1\subseteq \mathcal {D}$ (the converse inclusion is trivial). Consider $f\in \mathcal {R}_\mathcal {D}\cap \mathcal {L}_1$. Then $f=\lambda g$ for some $\lambda >0,g\in \mathcal {D}$. And since $\lambda g\in \mathcal {L}_1$, we get by D4$'$ that $f=\lambda g\in \mathcal {D}$. This shows that $\mathcal {D}$ is coherent relative to $\mathcal {L}_1$.

• 5. ($\Rightarrow$) Showing that $\mathcal {D}_\mathcal {R}$ satisfies D0$'$ is equivalent to showing that the preference relation it induces is weakly Archimedean, thanks to Theorem 1. Therefore consider $g\in \mathcal {D}_\mathcal {R}{\setminus }\mathcal {L}_1^+$ such that $g=p-q$ for some $p,q\in \mathcal {H}$. We want to show that there is $\alpha \in (0,1)$ such that $\alpha p-q\in \mathcal {D}_\mathcal {R}$.

  Thanks to point 4, we have that $g\in \mathcal {R}{\setminus }\mathcal {L}^+$ and hence by strict desirability that there is $\delta >0$ such that $g-\delta \in \mathcal {R}$. Choose $\alpha \in (0,1)$ so that $(1-\alpha )p\le \delta$. Then $\alpha p-q\ge p-q-\delta =g-\delta$, whence $\alpha p-q\in \mathcal {D}_\mathcal {R}$.

  ($\Leftarrow$) Finally, let us assume that $\mathcal {D}_\mathcal {R}$ satisfies D0$'$–D4$'$ and show that for all $h\in \mathcal {R}{\setminus }\mathcal {L}^+$ and $k\in \mathcal {L}$, there is $\alpha \in (0,1)$ such that $\alpha h+(1-\alpha )k\in \mathcal {R}$. If we prove this, then it is enough to take a constant $k<0$ to get that $h-\delta \in \mathcal {R}$, with $\delta :=-\frac{1-\alpha }{\alpha }k>0$. This means that $\mathcal {R}$ would be strictly desirable.

  Now then consider the positive and negative parts of k, namely, $k^+,k^-$, such that $k=k^+-k^-$. Note that for the thesis it is enough to show that there is $\alpha \in (0,1)$ such that $\alpha \frac{h}{2}-(1-\alpha )k^-\in \mathcal {R}$, given that $\alpha \frac{h}{2}+(1-\alpha )k^+$ belongs to $\mathcal {R}$ already, whence by additivity the result follows.

  To that end, choose $\lambda >0$ such that $f:=\frac{h}{2\lambda }$ and $g:=\frac{k^-}{\lambda }$ satisfy the inequalities in D0$'$. Then there is $\alpha \in (0,1)$ such that $\alpha \frac{h}{2\lambda }-(1-\alpha )\frac{k^-}{\lambda }\in \mathcal {D}_\mathcal {R}$; and as a consequence, $\alpha \frac{h}{2}-(1-\alpha )k^-\in \mathcal {R}$.

Proof of Theorem 3

Thanks to Theorems 1 and 2, we have an equivalence between coherent preference relations $\succ$ and sets of desirable gambles coherent relative to $\mathcal {L}_1$; the latter are in a one-to-one correspondence with their natural extensions, from which we deduce the first part of the theorem. The second part is granted again by Theorems 1 and 2.

Proof of Proposition 3

Assume first of all that $\succ$ is negatively transitive, and let $f,g\notin \mathcal {R}$. Then $f\nsucc 0$ and $g\nsucc 0$, from which it follows that $0\nsucc -g$. Applying the negative transitivity of $\succ$, we deduce that $f\nsucc -g$, or, equivalently, that $f+g \nsucc 0$, whence $f+g\notin \mathcal {R}$.

Conversely, assume that Eq. (7) holds and let $p,q,r\in \mathcal {H}$ satisfy $p\nsucc q$ and $q\nsucc r$. Then $p-q\notin \mathcal {R}$ and $q-r\notin \mathcal {R}$, whence $p-r\notin \mathcal {R}$, and using the correspondence between $\succ$ and $\mathcal {R}$ we conclude that $p\nsucc r$.

Proof of Proposition 4

From (10) and (11), we see that simply defining

$$\begin{aligned} p_2:=\psi (p_1) \end{aligned}$$

(13)

for all horse lotteries $p_1\in \mathcal {H}_z$, we obtain a correspondence between $\mathcal {H}$ and $\mathcal {H}_z$ made of elements that are equally valuable for us: their respective probability currencies are just the same. Moreover, the correspondence is one-to-one as it follows from Zaffalon and Miranda (2017, Remark 5 in p. 1098) (note also that $\psi$ is a linear operator).

It is now a simple step to prove the two inclusions.

$\mathcal {R}_2\subseteq \mathcal {R}_1$:

If $\lambda (p_2-q_2)\in \mathcal {R}_2$ for some $p_2,q_2\in \mathcal {H}$ and $\lambda >0$, then also $\lambda \psi (p_1-q_1)\in \mathcal {R}_1$, where $p_1,q_1$ are obtained through (13). This follows from the identity $p_2-q_2=\psi (p_1-q_1)$, as obtained from linearity of $\psi$, and the fact that the lotteries defined through (13) are equally valuable.

$\mathcal {R}_1\subseteq \mathcal {R}_2$:

Similarly, consider any $\lambda \psi (p_1-q_1)\in \mathcal {R}_1$ and let $p_2:=\psi (p_1),q_2:=\psi (q_1)$, so that $p_2,q_2\in \mathcal {H}$. It follows that $\lambda (p_2-q_2)\in \mathcal {R}_2$ for the same reasons given in the proof of the converse inclusion.

Proof of Proposition 5

It follows from (Miranda et al. 2012, Proposition 29) that $\hat{\mathcal {R}}$ is the smallest conglomerable and coherent set of gambles that includes $\mathcal {R}_\varOmega$ and $\mathcal {R}|\{\omega \}$ (for all $\omega \in \varOmega$). It includes $\mathcal {R}|\varOmega$ by construction. We are left to show that $\hat{\mathcal {R}}$ induces both $\mathcal {R}_{\varOmega }$ and $\mathcal {R}_{\mathcal {X}}$.

We begin by proving that the $\varOmega$-marginal of $\hat{\mathcal {R}}$ is $\mathcal {R}_{\varOmega }$. Consider an $\varOmega$-measurable gamble $f\in \hat{\mathcal {R}}$. Then there are $g\in \mathcal {R}_{\varOmega }$ and $h\in \mathcal {R}|\varOmega$ such that $f\ge g+h$. For any $\omega \in \varOmega$ , it holds that

$$\begin{aligned} 0 \le \sup _{x\in \mathcal {X}} h(\omega ,x) \le \sup _{x\in \mathcal {X}} (f(\omega ,x)-g(\omega ,x))=f'(\omega )-g'(\omega ), \end{aligned}$$

where in last equality we are using that both f, g are $\varOmega$-measurable, and are denoting by $f',g'$ their equivalent representations as gambles on $\varOmega$.

Thus, $f\ge g$, and since $\mathcal {R}_{\varOmega }$ is a coherent set of gambles we conclude that also $f\in \mathcal {R}_{\varOmega }$. The converse inclusion follows from Eq. (12).

Next, consider an $\mathcal {X}$-measurable gamble $f\in \hat{\mathcal {R}}$. Then there are $g\in \mathcal {R}_{\varOmega }$ and $h\in \mathcal {R}|\varOmega$ such that $f\ge g+h$. If $g\ne 0$, then there exists $\omega \in \varOmega$ such that $g'(\omega )> 0$. Then,

$$\begin{aligned} fI_{\{\omega \}}\ge g'(\omega )+hI_{\{\omega \}}\ge hI_{\{\omega \}}, \end{aligned}$$

and since $h(\omega ,\cdot )\in \mathcal {R}_{\mathcal {X}}\cup \{0\}$, either $h(\omega ,\cdot )\in \mathcal {R}_{\mathcal {X}}$, in which case $f(\omega ,\cdot )\in \mathcal {R}_{\mathcal {X}}$, or $h(\omega ,\cdot )=0$, whence $f(\omega ,\cdot )\ge g'(\omega )>0$, meaning that also $f(\omega ,\cdot )\in \mathcal {R}_{\mathcal {X}}$. Finally, the proof when $g=0$ is similar (in that case we can pick any $\omega \in \varOmega$ such that $hI_{\{\omega \}}\ne 0$, since one is bound to exist). Again, the converse inclusion follows from Eq. (12).

Proof of Proposition 6

The result follows immediately from Definition 31, taking into account that ${\underline{P}}_\varOmega '({\underline{P}}(\cdot |\varOmega ))$ is a coherent lower prevision by Walley (1991, Section 6.7.2).

Proof of Proposition 7

We make a circular proof.

$(i)\Rightarrow (ii)$:

It $f\notin \mathcal {R}$, it follows from Eq. (3) that ${\underline{P}}(f)={\overline{P}}(f)\le 0$, whence ${\underline{P}}(-f)=-{\overline{P}}(f)\ge 0$, and therefore ${\underline{P}}(\varepsilon -f)>0$ for every $\varepsilon >0$. Thus, Eq. (3) implies that $\varepsilon -f\in \mathcal {R}$.

$(ii)\Rightarrow (iii)$:

Assume ex-absurdo the existence of $f,g\notin \mathcal {R}$ such that $f+g-\varepsilon \in \mathcal {R}$. It follows from (ii) that $\frac{\varepsilon }{4}-f,\frac{\varepsilon }{4}-g\in \mathcal {R}$, and applying D3 we conclude that $-\frac{\varepsilon }{2}\in \mathcal {R}$, a contradiction with D2.

$(iii)\Rightarrow (i)$:

If ${\underline{P}}$ is not linear, we can find a gamble f such that ${\underline{P}}(f)<{\overline{P}}(f)$. The conjugacy of the lower and upper previsions and Eq. (1) implies that for every $\varepsilon >0$ the gambles $f-{\underline{P}}(f)-\varepsilon$ and ${\overline{P}}(f)-f-\varepsilon$ do not belong to $\mathcal {R}$, and (iii) implies then that for every $\delta >0$ the gamble $(f-{\underline{P}}(f)-\varepsilon )+({\overline{P}}(f)-f-\varepsilon )-\delta$ does not belong to $\mathcal {R}$. But for $\varepsilon ,\delta$ satisfying $2\varepsilon +\delta <{\overline{P}}(f)-{\underline{P}}(f)$ this sum is a non-negative gamble. This is a contradiction with D1.

Proof of Proposition 8

This is an immediate consequence of Proposition 7.