Decision making in phantom spaces

Abstract

This paper introduces a new model of decision making under uncertainty. Aiming to provide a more realistic depiction of decision making, it generalizes the von Neumann–Morgenstern theory by including additional tiers of uncertainty. In this model, beliefs about the probabilities of events are ambiguous and their consequential utilities are vague; both are naturally formulated in the phantom space using phantom numbers. The degree of uncertainty, determined by the decision maker’s beliefs, is distinguished from the attitude toward uncertainty, which is drawn from her preferences. Decision making under ambiguity is a particular case of our model in which probabilities are ambiguous, but resulting utilities of events are knowable.

Appendix

1.1 Phantom spaces: special attributes

The special properties of the phantom framework are explored in this section. For this purpose, we use the following definition. A function $f$ taking phantom arguments is said to have a realization form if it can be written as

$$\begin{aligned} f\left( a+\wp \,b \right) := f_{ {\mathrm{re }}} \left( a \right) + \wp \,\left( \widehat{f}\left( a+b \right) - f_{ {\mathrm{re }}} \left( a \right) \right) , \end{aligned}$$

where $f_{ {\mathrm{re }}}$ and $\widehat{f}$ are real functions, called the real component and the reduction of  $f$, respectively.

Recall that the division of a nonnegative real number by a greater or equal positive number lies in the interval $\left[ 0,1 \right]$. The phantom analog of this definition, with respect to $\bar{\varLambda }$, is as follows.

Claim

Suppose $z_2 \ \underline{\gg } \ z_1$ are, respectively, pseudo-positive and pseudo-nonnegative, then $\frac{z_1}{z_2} \in \bar{\varLambda }$.

Proof

Write the fraction $\frac{z_1}{z_2} = \frac{a_1}{a_2} + \wp \,\left( \frac{ \widehat{z_1}}{\widehat{z_2}} - \frac{a_1}{a_2} \right)$, then $0 \le \frac{a_1}{a_2} \le 1$ and $0 \le \frac{\widehat{z_1}}{\widehat{z_2}} \le 1$. Since $\frac{\widehat{z_1}}{\widehat{z_2}} = \widehat{(\frac{{z_1}}{{z_2}} )}$, then $- \frac{a_1}{a_2} \le \widehat{(\frac{{z_1}}{{z_2}} )}- \frac{a_1}{a_2} \le 1- \frac{a_1}{a_2}$, implying that $\frac{z_1}{z_2} \in \bar{\varLambda }$. $\square$

Later, when using phantom affinity and convexity, we also need the following.

Claim

Let $z_1,z_2,z_3 \in \mathbb {PH}$, then there exists $\lambda \in \mathbb {PH}$ such that $z_2 = \lambda z_1 + \left( 1-\lambda \right) z_3$. If $z_1 \ne z_3$ and at least one of them is not a zero divisor, then $\lambda$ is unique.

Proof

Let $z_i = a_i + \wp \,b_i$, $i = 1,2,3$, and let $\lambda = \alpha + \wp \,\beta$. Expanding the product $\lambda z_1 + \left( 1-\lambda \right) z_3$, we have

$$\begin{aligned} \big (\alpha a_1 + \left( 1-\alpha \right) a_3 \big )+ \wp \,\big (\alpha \left( b_1 - b_3 \right) + \beta \left( a_1 + b_1 - a_3 -b_3 \right) + b_3\big ). \end{aligned}$$

This shows that each term is linear in $\alpha$ and in $\beta$, and that there exists $\lambda$ that satisfies the requirement. By the same token, when $z_1 \ne z_3$, i.e., $a_1 \ne a_3$, then $\alpha$ is unique, and if $z_1$ or $z_2$ are not zero divisors, then $\beta$ is also unique. $\square$

The following definition is analogous to the known definition of affinity over the real numbers.

Definition 10

A phantom-valued function $f: \mathbb {PH}\rightarrow \mathbb {PH}$ is affine if it is of the form

$$\begin{aligned} f \left( \alpha _1 z_1, \alpha _2 z_2, \ldots , \alpha _n z_n \right) = \alpha _1 z_1+ \alpha _2 z_2+ \cdots + \alpha _n z_n + \beta , \end{aligned}$$

for some $\alpha _1, \ldots , \alpha _n,\beta \in \mathbb {PH}.$

The class of affine functions has subclasses of functions $f: \mathbb {PH}\rightarrow {\mathbb {R}}$ and $f: {\mathbb {R}}\rightarrow {\mathbb {R}}$, which satisfy the following lemma.

Lemma 1

(Phantom affinity) If the function $f: \mathbb {PH}\rightarrow \mathbb {PH}$ is affine, then for any $\lambda \in \varLambda$

$$\begin{aligned} f(\lambda z_1 + (1-\lambda )z_2) = \lambda f(z_1) + (1-\lambda ) f(z_2). \end{aligned}$$

Proof

By affinity, $f\left( \lambda z_1 + (1-\lambda )z_2 \right) = \lambda \left( \alpha z_1 + \beta \right) + (1-\lambda )\left( \alpha z_2+ \beta \right) = \lambda f(z_1) + (1-\lambda ) f(z_2).$ $\square$

Lemma 2

Let $f: \mathbb {PH}\rightarrow {{\mathbb {R}}}$ be a continuous function.⁴⁶ For any real $\alpha \in \left[ 0,1 \right]$, there exists $\lambda \in \bar{\varLambda }$ such that

$$\begin{aligned} \alpha f(z_1) + (1-\alpha ) f(z_2) = f\left( \lambda z_1 + (1-\lambda )z_2 \right) . \end{aligned}$$

(2)

Conversely, if $f$ is monotonic, for any $\lambda \in \bar{\varLambda }$, there exists a real $\alpha \in [0,1]$ for which Eq. (2) holds.

Proof

In the view of $f$ as a function $f: {\mathbb {R}}^2 \rightarrow {\mathbb {R}}$, the proof is obtained by the continuity of $f$ over ${\mathbb {R}}^2$. $\square$

In order to compare phantom numbers, one needs to equip $\mathbb {PH}$ with a weak order $\succsim$, which is assumed to be given with the phantom structure. In our case, this order corresponds to the DM’s preferences and, by Theorem 1, has an order-preserving function ${\mathrm{\Gamma }}$.

Assumption 1

The order $\succsim$ is taken to be compatible with the familiar order of ${\mathbb {R}}$. That is, if $z_1 \succsim z_2$ for $z_1,z_2 \in {\mathbb {R}}$, then $z_1 \ge z_2$.

Definition 11

A function $f: \mathbb {PH}\rightarrow \mathbb {PH}$ is (i) Monotonically increasing if $f(z_1) \succsim f(z_2)$ for all $z_1 \succsim z_2$, and monotonically decreasing if $f(z_1) \precsim f(z_2)$; (ii) Strictly increasing if $f(z_1) \succ f(z_2)$ for all $z_1 \succ z_2$, and strictly decreasing if $f(z_1) \prec f(z_2)$.

In the special case when $f:\mathbb {PH}\rightarrow {\mathbb {R}}$, the inequality between the values of functions is given by the standard order of ${\mathbb {R}}$. This leads us to define the convexity of the phantom-valued function.

Definition 12

A continuous function $f: \mathbb {PH}\rightarrow \mathbb {PH}$ is convex on a set $X \subseteq \mathbb {PH}$ if for any nonzero divisor $z_1,z_2 \in X$ and any $\lambda \in \varLambda$: $f \left( \lambda z_1 + \left( 1-\lambda \right) z_2 \right) \precsim \lambda f \left( z_1 \right) + \left( 1-\lambda \right) f\left( z_2 \right) .$ $f$ is said to be concave on $X$ if $f \left( \lambda z_1 + \left( 1-\lambda \right) z_2 \right) \succsim \lambda f \left( z_1 \right) + \left( 1-\lambda \right) f\left( z_2 \right) .$

When the target of $f$ is restricted to ${{\mathbb {R}}}$, the weight $\lambda$ belongs to the real interval $[0,1]$. Definition 12 leads to a generalization of the Jensen inequality of functions having a finite support.

Lemma 3

Let $f: \mathbb {PH}\rightarrow \mathbb {PH}$ be a continuous convex function on $X \subseteq \mathbb {PH}$, and let $\lambda _i \in \varLambda$, $i = 1,2,\ldots ,n$, be phantom weights such that $\sum \lambda _i =1$. Then, $f \left( {\sum \lambda _i z_i} \right) \precsim {\sum \lambda _i f(z_i)},$ assuming $z_i$ is a nonzero divisor for every $i$. In the case where $f: \mathbb {PH}\rightarrow {\mathbb {R}}$ and $\lambda _i \in [0,1]$ are real weights, $f \left( {\sum \lambda _i z_i)} \right) \le {\sum \lambda _i f(z_i)}$, assuming each $z_i$ is a nonzero divisor for every $i$.

Proof

We prove the lemma by induction on $n$. Assume that $f$ has a finite support, when $n=2$, the convexity of $f$ implies $f \left( \lambda _1 z_1 + \lambda _2 z_2 \right) \precsim \lambda _1 f \left( z_1 \right) + \lambda _2 f \left( z_2 \right) .$ For $n+1$, write $\sum _{i=1}^{n+1} \lambda _i = \mu + \lambda _{n+1}$. Then, $\sum _{i=1}^{n+1} \lambda _i z_i = \mu (\sum _{i=1}^n \frac{\lambda _i}{\mu } z_i) + \lambda _{n+1}z_{n+1}$, and by convexity

$$\begin{aligned} f \bigg ( \lambda _{n+1}z_{n+1} + \mu \sum _{i=1}^n \frac{\lambda _i}{\mu } z_i \bigg ) \precsim \lambda _{n+1}f(z_{n+1}) + \mu f\bigg ( \sum _{i=1}^n \frac{\lambda _i}{\mu } z_i \bigg ). \end{aligned}$$

The induction step shows that the right-hand side is

$$\begin{aligned} \lambda _{n+1}f(z_{n+1}) + \mu \sum _{i=1}^n \frac{\lambda _i}{\mu } f( z_i) = \sum _{i=1}^{n+1} {\lambda _i} f( z_i). \end{aligned}$$

$\square$

1.2 Properties of preferences

This section studies additional properties of the preference relation $\succsim$ over phantom lotteries in $\mathcal {L}$, defined in Axioms 1–6.

Lemma 4

(Intermediacy) If $\mathcal P\succ \mathcal Q$ and $\alpha \in \varLambda$, then $\mathcal P\succsim \alpha \mathcal P+ (1 - \alpha ) \mathcal Q\ \succsim \mathcal Q$. When $\alpha \ne 0$ and $\alpha \ne 1$, the relations are strict.

Proof

Since $\mathcal P\succ \mathcal Q$, Independence (Axiom 4) implies that $\mathcal P= \alpha \mathcal P+ (1 - \alpha ) \mathcal P\succsim \alpha \mathcal P+ (1 - \alpha ) \mathcal Q\succsim \alpha \mathcal Q+(1 - \alpha ) \mathcal Q= \mathcal Q$. $\square$

Lemma 5

(Mixture monotonicity) Suppose that $\mathcal P\succ \mathcal Q$ and $\alpha , \beta \in [0,1]$, then

$$\begin{aligned} \beta \mathcal P+ (1 - \beta ) \mathcal Q\succ \alpha \mathcal P+ (1 - \alpha ) \mathcal Q\ \Leftrightarrow \ 1 > \beta > \alpha >0. \end{aligned}$$

Proof

$(\Leftarrow )$: Assume $\beta > \alpha$. Define $\gamma \in [0,1]$ to be $\gamma = \frac{ \beta - \alpha }{1 - \alpha }$ and write $\beta \mathcal P+ (1 - \beta ) \mathcal Q= \gamma \mathcal P+ (1 - \gamma ) \left( \alpha \mathcal P+ (1 - \alpha ) \mathcal Q \right) .$ Since $\mathcal P\succ \mathcal Q$ and $\alpha \in [0,1] \subseteq \varLambda$, Lemma 4 implies that $\gamma \mathcal P+ (1 - \gamma ) \left( \alpha \mathcal P+ (1 - \alpha ) \mathcal Q \right) \succ \alpha \mathcal P+ (1 - \alpha ) \mathcal Q$. We conclude that $\beta \mathcal P+ (1 - \beta ) \mathcal Q\succ \alpha \mathcal P+ (1 - \alpha ) \mathcal Q$.

$(\Rightarrow ):$ Assume $\beta \le \alpha$ and $\beta \mathcal P+ (1 - \beta ) \mathcal Q\succ \alpha \mathcal P+ (1 - \alpha ) \mathcal Q$. Interchanging $\alpha$ and $\beta$ in the previous argument, one gets $\alpha \mathcal P+ (1 - \alpha ) \mathcal Q\succsim \beta \mathcal P+ (1 - \beta ) \mathcal Q$—a contradiction. $\square$

Lemma 6

(Unique solvability) For $\mathcal P\succsim \mathcal Q\succsim \mathcal R$, there exists a unique $\alpha ^*\in [0,1]$ such that $\mathcal Q\backsim \alpha ^*\mathcal P+ (1 - \alpha ^*) \mathcal R$.

Proof

When $\mathcal P\backsim \mathcal Q$ ($\mathcal Q\backsim \mathcal R$), take $\alpha ^*= 1$ ($\alpha ^*= 0).$ Otherwise, consider the sets    ${\mathcal {H}}_{A} = \left\{ \alpha \in [0,1] \ \big | \ \mathcal Q\succ \alpha \mathcal P+ (1 - \alpha ) \mathcal R \right\}$    and ${\mathcal {H}}_{B} = \left\{ \alpha \in [0,1] \ \big | \ \alpha \mathcal P+ (1 - \alpha ) \mathcal R\succ \mathcal Q \right\} .$ By Archimedean (Axiom 3), these are disjoint nonempty open sets, which do not cover $[0,1]$. Thus, there exists $\alpha ^*\in [0,1]$ for which $\mathcal Q\backsim \alpha ^*\mathcal P+ (1 - \alpha ^*) \mathcal R$, which is unique by mixture monotonicity (Lemma 5). $\square$

Theorem 5

Let $x_1,x_2 \in \mathbb {PH}$ be consequences in $\mathcal {X}$.⁴⁷ Then, there exists a continuous real order-preserving function (phantom-valued representation) ${\mathrm{\Gamma }}: \mathbb {PH}\rightarrow {\mathbb {R}}$, such that

$$\begin{aligned} x_1 \succsim x_2 \ \Leftrightarrow \ {\mathrm{\Gamma }}\left( x_1 \right) \ge {\mathrm{\Gamma }}\left( x_2 \right) , \qquad \forall x_1, x_2 \in \mathcal {X}. \end{aligned}$$

The function ${\mathrm{\Gamma }}$ is unique up to a strictly increasing transformation.

Proof

The preference $\succsim$ on $\mathcal {X}$ is induced by the order on $\mathcal {L}$. To see this, one can construct a constant lottery for each element in $\mathcal {X}$. Archimedean (Axiom 3) provides the convexity of $\mathcal {X}$, implying that $\mathcal {X}$ is connected. Let $\mathcal {Y}$ be the set of the phantom consequence in $\mathcal {X}$. Since $\mathcal {Y}$ is a connected subset of the topological space $\mathbb {PH}$, realized as the Cartesian product ${{\mathbb {R}}}\times {{\mathbb {R}}}$, and since the rational numbers are dense in ${{\mathbb {R}}}$, then $\mathcal {Y}$ contains a countable dense subset. That is, $\mathcal {Y}$ is a separable. Thus, by Debreu (1954) Theorem I, there is a function ${\mathrm{\Gamma }}: \mathcal {Y}\rightarrow {\mathbb {R}}$ that preserves $\succsim$.⁴⁸ Since $\succsim$ is assumed to coincide with the order of ${\mathbb {R}}$, any strictly increasing transformation of the value function ${\mathrm{\Gamma }}$ preserves $\succsim$.⁴⁹ $\square$

Proposition 10

Suppose $\mathcal P$ and $\mathcal R$ are phantom lotteries whose outcomes $X$ and $Z$ are independent and ${\mathrm{E }}_{\mathcal R} \left( Z \right) = 0$. Let $\mathcal Q_i$, $i = 1,2$, be the phantom lotteries having the outcomes $X + \lambda _i Z$, where $\lambda _1, \lambda _2 \in \mathbb {PH}$. Then, $\mathcal Q_1$ is strictly more uncertain than $\mathcal Q_2$ for every pseudo-positive $\lambda _1 \gg \lambda _2$.

Proof

Let $\mu =\frac{\lambda _2}{\lambda _1}$ and write $X + \lambda _2 Z = \mu (X + \lambda _1 Z) + (1-\mu ) X$, which by Claim 11.1 is in $\varLambda$. Then,⁵⁰

$$\begin{aligned} {\mathrm{U }}\left( X + \lambda _2 Z \right) >_{\mathrm{\Gamma }}\mu {\mathrm{U }}\left( X + \lambda _1 Z \right) + (1 -\mu ){\mathrm{U }}\left( X \right) , \end{aligned}$$

(3)

since ${\mathrm{U }}: \mathcal {X}\rightarrow \mathbb {PH}$ is strictly concave. Taking expectations implies ${\mathrm{E }}_{\mathcal Q_2} \left( {\mathrm{U }}\left( X + \lambda _2 Z \right) \right) >_{\mathrm{\Gamma }}\mu {\mathrm{E }}_{\mathcal Q_1} \left( {\mathrm{U }}\left( X + \lambda _1 Z \right) \right) + (1 -\mu ){\mathrm{E }}_{\mathcal P} \left( {\mathrm{U }}\left( X \right) \right)$. Since $\mathcal Q_1$, whose outcome is $X + \lambda _1 Z$, is strictly more uncertain than $\mathcal P$, we have ${\mathrm{E }}_{\mathcal P} \left( {\mathrm{U }}\left( X \right) \right) >_{\mathrm{\Gamma }}{\mathrm{E }}_{\mathcal Q_1} \left( {\mathrm{U }}\left( X + \lambda _1 Z \right) \right)$. Together with Eq. (3), this yields, ${\mathrm{E }}_{\mathcal Q_2} \left( {\mathrm{U }}\left( X + \lambda _2 Z \right) \right) >_{\mathrm{\Gamma }}{\mathrm{E }}_{\mathcal Q_1} \left( {\mathrm{U }}\left( X + \lambda _1 Z \right) \right)$. Then, Theorem 3 asserts that $\mathcal Q_1$ is strictly more uncertain than $\mathcal Q_2$. $\square$

Corollary 2

The phantom lottery $\mathcal Q'$ with outcomes $\lambda Y$ is strictly more uncertain than lottery $\mathcal Q$ with outcomes $Y$ for any $\lambda \gg 1$.

Proof

The proof is obtained immediately by applying Proposition 10 to a constant phantom lottery $\mathcal P$ with outcome 0 and to phantom lottery $\mathcal R$ whose outcome is given by $Y - {\mathrm{E }}_{\mathcal Q} \left( Y \right)$. $\square$

1.3 Proofs

Proof of Theorem 1

Recall that ${\mathrm{C }}: \mathcal {L}\rightarrow {\mathbb {R}}$ is given by ${\mathrm{C }}: \mathcal P\mapsto {\mathrm{\Gamma }}\left( \sum {\mathrm{U }}\left( x \right) \mathcal {P}\left( x \right) \right)$, i.e., it is a composition of the utility function ${\mathrm{U }}: \mathcal {X}\rightarrow \mathbb {PH}$ and the value function ${\mathrm{\Gamma }}: \mathbb {PH}\rightarrow {\mathbb {R}}$.

$(\Leftarrow ):$ The following two-step proof first proves the assertion for a closed and bounded (by a worst and a best consequence) subset of lotteries $\mathcal {L}$ and then extends to the entire $\mathcal {L}$.⁵¹

I) Assume $x^*$ and $x_*$, where $x^* \succsim x_*$ are a best and a worst consequences, respectively. Note that if $x^* \thicksim x_*$, then $\mathcal P\backsim \mathcal Q$ for any $\mathcal P, \mathcal Q\in \mathcal {L}$, and the proof is accomplished by taking a constant value function.

We first show, inductively, that each (not necessarily constant) phantom lottery $\mathcal P\in \mathcal {L}$ is equivalent to a compound lottery constructed of $x^*$ and $x_*$. Suppose $\mathcal P: x_i \mapsto \mathfrak {p}_i$, $i =1,\ldots ,n$, then for the sequence of constant lotteries $\mathcal Q_i^{\mathrm{C }}: x_i \mapsto 1$, we have

$$\begin{aligned} \mathcal P\backsim \mathfrak {p}_1 \mathcal Q_1^{\mathrm{C }}+ \mathfrak {p}_2 \mathcal Q^{\mathrm{C }}_2 + \cdots + \mathfrak {p}_n \mathcal Q^{\mathrm{C }}_n, \qquad \mathfrak {p}_i \in \varLambda , \quad \sum _i \mathfrak {p}_i =1. \end{aligned}$$

By Archimedean (Axiom 3), $\mathcal Q^{\mathrm{C }}_i \backsim \alpha _i x^* + (1-\alpha _i) x_*$ for some $\alpha _i \in \varLambda$. Therefore,

$$\begin{aligned} \mathcal P\backsim \sum _i \mathfrak {p}_i( \alpha _i x^* + (1-\alpha _i) x_*) = x^* \sum _i \mathfrak {p}_i \alpha _i + x_*\bigg (1 - \sum _i \mathfrak {p}_i \alpha _i\bigg ). \end{aligned}$$

Recall that $\mathfrak {p}_i= \mathcal P(x_i)$ and set each phantom value $\alpha _i ={\mathrm{U }}\left( x_i \right)$, thus

$$\begin{aligned} \mathcal P\backsim x^* \bigg (\sum _{x\in \mathcal {X}} {\mathrm{U }}\left( x \right) \mathcal {P}\left( x \right) \bigg ) + x_* \bigg ( 1 - \sum _{x\in \mathcal {X}} {\mathrm{U }}\left( x \right) \mathcal {P}\left( x \right) \bigg ), \end{aligned}$$

i.e., ${\mathrm{U }}\left( x \right) \mathcal {P}\left( x \right) \in \varLambda$ for any $x \in \mathcal {X}$.

On the other hand, by Unique Solvability (Lemma 6), there is a unique (real) $\alpha ^*\in [0,1]$ such that

$$\begin{aligned} \mathcal P\backsim x^*\alpha ^*+ x_*\left( 1 - \alpha ^* \right) . \end{aligned}$$

By Theorem 5, we have ${\mathrm{\Gamma }}\left( \alpha ^* \right) = \alpha ^*= {\mathrm{\Gamma }}\left( \sum {\mathrm{U }}\left( x \right) \mathcal {P}\left( x \right) \right) \in [0,1]$. Hence,

$$\begin{aligned} \mathcal P\backsim x^*{\mathrm{\Gamma }}\bigg (\sum _{x\in \mathcal {X}} {\mathrm{U }}\left( x \right) \mathcal {P}\left( x \right) \bigg ) + x_*\bigg ( 1 - {\mathrm{\Gamma }}\bigg (\sum _{x\in \mathcal {X}} {\mathrm{U }}\left( x \right) \mathcal {P}\left( x \right) \bigg )\bigg ). \end{aligned}$$

(4)

To complete this part of the proof, suppose $\mathcal P\succsim \mathcal Q$ are two lotteries in $\mathcal {L}$. By the above considerations (see Eq. (4)), for $\mathcal Q$, we have

$$\begin{aligned} \mathcal Q\backsim x^*{\mathrm{\Gamma }}\bigg (\sum _{x\in \mathcal {X}} {\mathrm{U }}\left( x \right) \mathcal {Q}\left( x \right) \bigg ) + x_*\bigg ( 1 - {\mathrm{\Gamma }}\bigg (\sum _{x\in \mathcal {X}} {\mathrm{U }}\left( x \right) \mathcal {Q}\left( x \right) \bigg )\bigg ). \end{aligned}$$

Then, Lemma 5 (Mixture monotonicity) completes the proof of this part:

$$\begin{aligned} \mathcal P\succsim \mathcal Q\ \Leftrightarrow \ {\mathrm{\Gamma }}\bigg (\sum _{x\in \mathcal {X}} {\mathrm{U }}\left( x \right) \mathcal {P}\left( x \right) \bigg ) \ge {\mathrm{\Gamma }}\bigg (\sum _{x\in \mathcal {X}} {\mathrm{U }}\left( x \right) \mathcal {Q}\left( x \right) \bigg ). \end{aligned}$$

II) Next, we extend the result of part I to the case where there is no best or worst lottery. If there is no best or worst lottery, we take an arbitrary pair of constant phantom lotteries $\mathcal P_0 \succ \mathcal Q_0$ in $\mathcal {L}$ and define the set of lotteries $\mathcal {T}_0 = \left\{ {\mathcal R} \in \mathcal {L}\big | \mathcal P_0 \succsim \mathcal R\succsim \mathcal Q_0 \right\}$. If $\mathcal P_0 \backsim \mathcal Q_0$, for all $\mathcal P_0,\mathcal Q_0 \in \mathcal {L}$, then the conclusion holds trivially. Otherwise, following the reasoning above, we can obtain a functional representation ${\mathrm{C }}_0$ for all the lotteries in $\mathcal {T}_0$.

Next, consider lotteries $\mathcal P_1 \succ \mathcal P_0$ and $\mathcal Q_0 \succ \mathcal Q_1$, i.e., they are outside $\mathcal {T}_0$ and, as before, define the set of lotteries $\mathcal {T}_1 = \left\{ \mathcal R\in \mathcal {L}\big | \mathcal P_1 \succsim \mathcal R\succsim \mathcal Q_1 \right\}$. Clearly, $\mathcal {T}_0 \subset \mathcal {T}_1$, and therefore, the choice function ${\mathrm{C }}_0$ on $\mathcal {T}_0$ agrees with the choice function ${\mathrm{C }}_1$, representing the preference on $\mathcal {T}_1$. That is, ${\mathrm{C }}_0(\mathcal R) = {\mathrm{C }}_1(\mathcal R)$ for every $\mathcal R\in \mathcal {T}_1 \cap \mathcal {T}_0$. Proceeding inductively, we obtain a sequence $\mathcal {T}_0 \subset \mathcal {T}_1 \subset \mathcal {T}_2 \subset \cdots \subset \mathcal {L}$ of sets of lotteries, each having a preference representative ${\mathrm{C }}_i$ with ${\mathrm{C }}_i(\mathcal R) = {\mathrm{C }}_{i+1}(\mathcal R) = {\mathrm{C }}_{i+2}(\mathcal R)= \cdots$ for all $\mathcal R\in \mathcal {T}_{i}$, and thereby construct a choice function ${\mathrm{C }}$ whose domain is $\mathcal {L}$.

$(\Rightarrow )$: Suppose that ${\mathrm{C }}: \mathcal {L}\rightarrow {\mathbb {R}}$ is a choice function representing $\succsim$. We need to prove that Axioms 1–6 are satisfied. By the continuity of ${\mathrm{C }}$ (since ${\mathrm{U }}$ and ${\mathrm{\Gamma }}$ are continuous), for any phantom lottery $\mathcal P\in \mathcal {L}$, one can find a constant lottery $\mathcal P^{\mathrm{C }}$ such that ${\mathrm{C }}\left( \mathcal P^{{\mathrm{C }}} \right) = {\mathrm{C }}\left( \mathcal P \right)$. Therefore, one may consider only constant lotteries, which are identified with consequences $x\in \mathcal {X}$, and ${\mathrm{C }}$ can be viewed as a function sending $\mathcal {X}$ to ${\mathbb {R}}$.

Completeness (Axiom 1) and Transitivity (Axiom 2) are immediately obtained by the fact that any function ${\mathrm{\Gamma }}: \mathbb {PH}\rightarrow {\mathbb {R}}$ induces a weak order on $\mathbb {PH}$ and that $\mathcal {X}$ is contained in the domain of ${\mathrm{C }}$. Archimedean (Axiom 3): The representation of $\mathcal P\succsim \mathcal Q\succsim \mathcal R$ implies ${\mathrm{C }}\left( \mathcal P \right) \ge {\mathrm{C }}\left( \mathcal Q \right) \ge {\mathrm{C }}\left( \mathcal R \right)$, which are reals. Thus, there is $\lambda \in (0, 1)$ for which $\lambda {\mathrm{C }}\left( \mathcal P \right) + (1-\lambda ){\mathrm{C }}\left( \mathcal R \right) \ge {\mathrm{C }}\left( \mathcal Q \right)$. By Lemma 2, there is $\alpha \in \varLambda$ such that $\lambda {\mathrm{C }}\left( \mathcal P \right) + (1-\lambda ){\mathrm{C }}\left( \mathcal R \right) = {\mathrm{C }}\left( \alpha \mathcal P+ (1-\alpha )\mathcal R \right)$, hence ${\mathrm{C }}\left( \alpha \mathcal P+ (1-\alpha )\mathcal R \right) \ge {\mathrm{C }}\left( \mathcal Q \right)$, which by representation (converse direction) yields $\alpha \mathcal P+ (1-\alpha )\mathcal R\succsim \mathcal Q$. The same argument shows that $\mathcal Q\succsim \beta \mathcal P+ (1-\beta )\mathcal R$ for some $\beta \in \varLambda$. Independence (Axiom 4) is then obtained by the Archimedean (Axiom 3), the continuity of ${\mathrm{C }}$, and its monotonicity. Consequence monotonicity (Axiom 5) is obtained since ${\mathrm{\Gamma }}$ is monotonic function, i.e., monotonic in phantom numbers. Uniformity (Axiom 6) is obtained immediately, since one can find a constant phantom lottery $\mathcal P^{{\mathrm{C }}}$ such that ${\mathrm{C }}\left( \mathcal P^{{\mathrm{C }}} \right) = {\mathrm{\Gamma }}\left( {\mathrm{U }}\left( \mathcal P^{{\mathrm{C }}} \right) \right) = {\mathrm{\Gamma }}\left( \sum {\mathrm{U }}\left( x \right) \mathcal P\left( x \right) \right) = {\mathrm{C }}\left( \mathcal P \right)$. $\square$

Proof of Proposition 1

Since, by Theorem 5, ${\mathrm{\Gamma }}$ is invariant to strictly increasing transformations, then ${\mathrm{C }}= {\mathrm{\Gamma }}\circ {\mathrm{U }}$ is also invariant to strictly increasing transformations. $\square$

Proof of Proposition 2

The proof is straightforward: ${\mathrm{C }}$ is a composition of two order-preserving functions ${\mathrm{\Gamma }}$ and ${\mathrm{U }}$. $\square$

Proof of Proposition 3

The proof is derived directly from the concavity of the representation functions (Definition 12) and the convexity of preferences, together with the representation theorems (Theorems 5 and 1). $\square$

Proof of Proposition 4

We prove for risk aversion. The proof for risk loving and risk neutrality is obtained by similar considerations. Write $(i)$ ${\mathrm{E }}\left( \mathcal P \right) \succsim \mathcal P$ $\ \Leftrightarrow \$ $(ii)$ ${\mathrm{U }}$ is concave $\ \Leftrightarrow \$ $(iii)$ ${\mathrm{U }}\left( {\mathrm{E }}_{\mathcal {P}} \left( x \right) \right) \ge _{\mathrm{\Gamma }}{\mathrm{E }}_{\mathcal {P}} \left( {\mathrm{U }}\left( x \right) \right)$. $(i) \Rightarrow (ii)$: Assume that $\succsim$ exhibits risk aversion, represented by ${\mathrm{U }}$. That is, ${\mathrm{E }}_{\mathcal {P}} \left( x \right) = \sum x\mathcal {P}\left( x \right) \succsim \mathcal P$. The latter can be written as a compound phantom lottery $\bigoplus _i \alpha _i \mathcal P_i^{\mathrm{C }}$ of constant lotteries (cf. Remark 1) $\mathcal P_1^{\mathrm{C }}, \ldots , \mathcal P_n^{\mathrm{C }}$, each identified with a consequence $x_i \in \mathcal {X}$. By Theorem 1, ${\mathrm{U }}\left( {\mathrm{E }}_{\mathcal {P}} \left( x \right) \right) \ge _{\mathrm{\Gamma }}\sum _{i=1}^n \alpha _i {\mathrm{U }}\left( x_i \right)$, where $\alpha _i = \mathcal {P}\left( x_i \right)$, and thus, by Lemma 3, ${\mathrm{U }}$ is concave. $(ii) \Rightarrow (iii)$: Directly by Lemma 3. $(iii) \Rightarrow (i)$: If ${\mathrm{U }}$ is concave, then ${\mathrm{U }}\left( {\mathrm{E }}_{\mathcal {P}} \left( x \right) \right) \ge _{\mathrm{\Gamma }}{\mathrm{E }}_{\mathcal {P}} \left( {\mathrm{U }}\left( x \right) \right)$ by Lemma 3; thus, ${\mathrm{E }}_{\mathcal P} \left( x \right) \succsim \mathcal P$. $\square$

Proof of Proposition 5

The second equality is obtained by applying proposition 1.18 of Izhakian and Izhakian (2009) to ${\mathrm{U }}\left( x \right) = -\frac{e^{-\gamma x}}{\gamma }$. The Arrow-Pratt measure $A\left( x \right) =-\frac{{\mathrm{U }}''\left( x \right) }{{\mathrm{U }}'\left( x \right) }$, computed using phantom derivatives, implies that $A\left( x \right) =\gamma$, cf. Izhakian and Izhakian (2009, section 1.7). $\square$

Proof of Proposition 6

Suppose $\gamma = \frac{m}{n}$. Then, $\frac{x^{1-\gamma }}{1-\gamma }$ can be written ${\mathrm{U }}\left( x \right) = \frac{n}{n-m}\root n \of {x^{m-n}}$. The second equality is then obtained by the realization property for exponents and roots of phantoms, cf. Izhakian and Izhakian (2009, section 1.4). The phantom derivatives imply $A\left( x \right) =-\frac{{\mathrm{U }}''\left( x \right) }{{\mathrm{U }}'\left( x \right) }x=\gamma$, cf. Izhakian and Izhakian (2009, section 1.7). $\square$

Proof of Proposition 7

We prove the case (i); the other two cases are proved by the same considerations. Suppose ${\mathrm{\Gamma }}$ is concave, then by Lemma 3, ${\mathrm{\Gamma }}\left( \check{x} \right) \ge \frac{{\mathrm{\Gamma }}\left( x \right) +{\mathrm{\Gamma }}\left( \overline{x} \right) }{2}$, which by Theorem 1 implies $\check{x} \succsim \frac{1}{2} x \oplus \frac{1}{2} \overline{x}$. Conversely, suppose that $\succsim$ exhibits phantom aversion, then $\check{x} \succsim \frac{1}{2} x \oplus \frac{1}{2} \overline{x}$, which by Theorem 1 implies ${\mathrm{\Gamma }}\left( \check{x} \right) \ge \frac{{\mathrm{\Gamma }}\left( x \right) +{\mathrm{\Gamma }}\left( \overline{x} \right) }{2}$, i.e., ${\mathrm{\Gamma }}$ is concave. $\square$

Proof of Theorem 2

Recall that since DMs are assumed to have identical risk attitudes, phantom lotteries can be represented by their expected utilities $x \in \mathbb {PH}$. In particular, $\check{x}$ can be referred to as a constant real lottery in $\mathcal {L}$ and $\frac{1}{2}x \oplus \frac{1}{2}\overline{x}$ as a phantom lottery in $\mathcal {L}$. Recall also that ${\mathrm{\Gamma }}: \mathbb {PH}\rightarrow {\mathbb {R}}$ is assumed to be monotonically increasing with respect to an order that coincides with the order on ${{\mathbb {R}}}$. Therefore, ${\mathrm{\Gamma }}^{-1}$, is well defined on the restriction ${\mathrm{\Gamma }}|_{{\mathbb {R}}}$ of ${\mathrm{\Gamma }}$ to ${{\mathbb {R}}}$. ($\Rightarrow$): Suppose that DM $A$ is at least as phantom averse as DM $B$ and assume $x,y,z \in \mathbb {PH}$ such that ${\mathrm{\Gamma }}_B\left( x \right) < {\mathrm{\Gamma }}_B\left( y \right) < {\mathrm{\Gamma }}_B\left( z \right)$. Let $\lambda \in (0,1)$ satisfy ${\mathrm{\Gamma }}_B\left( y \right) = \lambda {\mathrm{\Gamma }}_B\left( x \right) + (1-\lambda ){\mathrm{\Gamma }}_B\left( z \right)$. One have to show that ${\mathrm{\Gamma }}_A\left( y \right) \ge \lambda {\mathrm{\Gamma }}_A\left( x \right) + (1-\lambda ){\mathrm{\Gamma }}_A\left( z \right)$. Suppose by way of negation that ${\mathrm{\Gamma }}_A\left( y \right) < \lambda {\mathrm{\Gamma }}_A\left( x \right) + (1-\lambda ){\mathrm{\Gamma }}_A\left( z \right)$. Then, by monotonicity and continuity, there is $y' \succ y$ close enough to $y$ for which $y' \prec _A \lambda x \oplus (1-\lambda )z$ and $y' \succ _B \lambda x \oplus (1-\lambda )z$—a contradiction. Therefore, $y' \succsim _A \lambda x \oplus (1-\lambda )z$, and by Theorem 1 ${\mathrm{\Gamma }}_A\left( y \right) \ge \lambda {\mathrm{\Gamma }}_A\left( x \right) + (1-\lambda ){\mathrm{\Gamma }}_A\left( z \right)$. Then, for $g = {\mathrm{\Gamma }}_A|_{{\mathbb {R}}}\circ {\mathrm{\Gamma }}_B^{-1}|_{{\mathbb {R}}}$, it follows that $g\left( {\mathrm{\Gamma }}_B\left( y \right) \right) \ge \lambda g\left( {\mathrm{\Gamma }}_B\left( x \right) \right) + (1-\lambda )g\left( {\mathrm{\Gamma }}_B\left( z \right) \right)$. That is, $g$ is concave. ($\Leftarrow$): Assume that ${\mathrm{\Gamma }}_A$ is a concave transformation of ${\mathrm{\Gamma }}_B$. Then, by definition, $\frac{{\mathrm{\Gamma }}_B\left( x \right) +{\mathrm{\Gamma }}_B\left( \overline{x} \right) }{2}={\mathrm{\Gamma }}_B\left( {\mathrm{RCE }}_B(\mathcal P) \right) ,$ i.e., ${\mathrm{RCE }}_B(\mathcal P) = {\mathrm{\Gamma }}_{B}^{-1} \left( \frac{{\mathrm{\Gamma }}_B\left( x \right) +{\mathrm{\Gamma }}_B\left( \overline{x} \right) }{2} \right)$. Let $g: {{\mathbb {R}}}\rightarrow {{\mathbb {R}}}$ be defined by $g = {\mathrm{\Gamma }}_A|_{{\mathbb {R}}}\circ {\mathrm{\Gamma }}_B^{-1}|_{{\mathbb {R}}}$. Since $g$ is concave, by the Jensen inequality,

$$\begin{aligned} {\mathrm{\Gamma }}_A\left( {\mathrm{RCE }}_B(\mathcal P) \right)&= {\mathrm{\Gamma }}_A\left( {\mathrm{\Gamma }}_B^{-1} \left( \frac{{\mathrm{\Gamma }}_B\left( x \right) +{\mathrm{\Gamma }}_B\left( \overline{x} \right) }{2} \right) \right) = g\left( \frac{{\mathrm{\Gamma }}_B\left( x \right) +{\mathrm{\Gamma }}_B\left( \overline{x} \right) }{2} \right) \\&\ge \frac{g\left( {\mathrm{\Gamma }}_B\left( x \right) \right) +g\left( {\mathrm{\Gamma }}_B\left( \overline{x} \right) \right) }{2} = \frac{{\mathrm{\Gamma }}_A\left( x \right) +{\mathrm{\Gamma }}_A\left( \overline{x} \right) }{2} \\&= {\mathrm{\Gamma }}_A\left( {\mathrm{RCE }}_A(\mathcal P) \right) . \end{aligned}$$

The monotonicity of ${\mathrm{\Gamma }}_A$ implies that ${\mathrm{RCE }}_B \left( \mathcal P \right) \ge {\mathrm{RCE }}_A \left( \mathcal P \right)$. Now, suppose a constant lottery $\mathcal Q^{\mathrm{C }}$ satisfying $\mathcal P\succsim _A \mathcal Q^{\mathrm{C }}$, then ${\mathrm{RCE }}_A\left( \mathcal P \right) \succsim _A {\mathrm{RCE }}_A\left( \mathcal Q^{\mathrm{C }} \right)$. This implies ${\mathrm{RCE }}_A\left( \mathcal P \right) \ge {\mathrm{RCE }}_A\left( \mathcal Q^{\mathrm{C }} \right)$, since $\succsim _A$ is compatible with the order of ${{\mathbb {R}}}$. But ${\mathrm{RCE }}_B\left( \mathcal P \right) \ge {\mathrm{RCE }}_A\left( \mathcal P \right)$ and ${\mathrm{RCE }}_A\left( \mathcal Q^{\mathrm{C }} \right) = {\mathrm{RCE }}_B\left( \mathcal Q^{\mathrm{C }} \right)$ which, by Theorem 5, implies that ${\mathrm{RCE }}_B\left( \mathcal P \right) \succsim _B {\mathrm{RCE }}_B\left( \mathcal Q^{\mathrm{C }} \right)$ and by transitivity $\mathcal P\succsim _B \mathcal Q^{\mathrm{C }}$. $\square$

Proof of Corollary 1

Let $g: {\mathbb {R}}\rightarrow {\mathbb {R}}$ be a twice-differentiable function such that ${\mathrm{\Gamma }}_A = g \circ {\mathrm{\Gamma }}_B$. Thus, $\partial _{\wp \,}{{\mathrm{\Gamma }}_A} = g'({\mathrm{\Gamma }}_B) \partial _{\wp \,}{{\mathrm{\Gamma }}_B}$ and $\partial _{\wp \,\wp \,}{{\mathrm{\Gamma }}_A} = g''({\mathrm{\Gamma }}_B) (\partial _{\wp \,}{{\mathrm{\Gamma }}_B})^2 + g'({\mathrm{\Gamma }}_B) \partial _{\wp \,\wp \,}{{\mathrm{\Gamma }}_B}$. Since ${\mathrm{\Gamma }}_A$ and ${\mathrm{\Gamma }}_B$ are strictly increasing, $g'({\mathrm{\Gamma }}_B)=\frac{\partial _{\wp \,}{{\mathrm{\Gamma }}_A}}{\partial _{\wp \,}{{\mathrm{\Gamma }}_B}}>0$. Putting all together:

$$\begin{aligned} \vartheta _A = - \frac{ \partial _{\wp \,\wp \,}{{\mathrm{\Gamma }}_A}}{\partial _{\wp \,}{{\mathrm{\Gamma }}_A}} = - \frac{ g''({\mathrm{\Gamma }}_B )}{g'({\mathrm{\Gamma }}_B )}{\partial _{\wp \,}{{\mathrm{\Gamma }}_B}} - \frac{ \partial _{\wp \,\wp \,}{{\mathrm{\Gamma }}_B}}{\partial _{\wp \,}{{\mathrm{\Gamma }}_B}} = - \frac{ g''({\mathrm{\Gamma }}_B )}{g'({\mathrm{\Gamma }}_B )}{\partial _{\wp \,}{{\mathrm{\Gamma }}_B}} + \vartheta _B. \end{aligned}$$

Recall that, by consequence monotonicity (Axiom 5), $\partial _{\wp \,}{{\mathrm{\Gamma }}_B} \ge 0$. Therefore, $\vartheta _A\left( x \right) \ge \vartheta _B\left( x \right)$ iff $g''({\mathrm{\Gamma }}_B)/g'({\mathrm{\Gamma }}_B) < 0$, i.e., $g$ is concave. The proof is then completed by Theorem 2. $\square$

Proof of Proposition 8

We prove the case of pessimism; the cases of optimism and apathy are proved similarly. Take $x = a + b \wp \,\in \mathbb {PH}$ with $\check{x} = a + \frac{1}{2}b \in {\mathbb {R}}$. Then, by Theorem 1, $\check{x} \succ x$ iff ${\mathrm{\Gamma }}\left( a + \frac{1}{2}b \right) > {\mathrm{\Gamma }}\left( a + b \wp \, \right)$. Since ${\mathrm{\Gamma }}$ is assumed to be differentiable,

$$\begin{aligned} {\mathrm{\Gamma }}\left( a + \frac{1}{2}b \right) \backsimeq {\mathrm{\Gamma }}\left( a \right) + \partial _{r \,}{{\mathrm{\Gamma }}\left( a \right) }\frac{1}{2}b > {\mathrm{\Gamma }}\left( a \right) + \partial _{\wp \,}{{\mathrm{\Gamma }}\left( a \right) }b \backsimeq {\mathrm{\Gamma }}\left( a + b \wp \, \right) , \end{aligned}$$

which implies that if $b > 0$, then $\theta = \frac{\partial _{r \,}{{\mathrm{\Gamma }}}}{\partial _{\wp \,}{{\mathrm{\Gamma }}}} >2$, and if $b < 0$, then $\theta = \frac{ \partial _{r \,}{{\mathrm{\Gamma }}}}{\partial _{\wp \,}{{\mathrm{\Gamma }}}}<2$. $\square$

Proof of Proposition 9

Proposition 8 is a special case of Proposition 7, and the proof is derived from the latter. $\square$

Proof of Theorem 3

$(\Rightarrow )$: Assume $\mathcal P$ is more uncertain than $\mathcal Q$. Let $x=X-{\mathrm{E }}_{\mathcal P} \left( X \right)$ and $y=Y-{\mathrm{E }}_{\mathcal Q} \left( Y \right)$ be random variables. By Definition 6, $x =_dy+z$ for some phantom lottery $\mathcal R\in \mathcal {L}$ with ${\mathrm{E }}_{\mathcal R} \left( z \right) =0$. Considering a DM’s preference $\succsim$, characterized by ${\mathrm{U }}:\mathcal {X}\rightarrow \mathbb {PH}$ and ${\mathrm{\Gamma }}:\mathbb {PH}\rightarrow {\mathbb {R}}$, we have

$$\begin{aligned} {\mathrm{E }}_{\mathcal {P}} \left( {\mathrm{U }}\left( x \right) \right) =_{\mathrm{\Gamma }}{\mathrm{E }}_{\mathcal Q} \left( {\mathrm{E }}_{\mathcal R} \left( {\mathrm{U }}\left( y+z \right) \right) \right) . \end{aligned}$$

Risk aversion implies a concavity of ${\mathrm{U }}$ and thus, by Lemma 3,

$$\begin{aligned} {\mathrm{E }}_{\mathcal R} \left( {\mathrm{U }}\left( y+z \right) \right) \le _{\mathrm{\Gamma }}{\mathrm{U }}\left( {\mathrm{E }}_{\mathcal R} \left( y+z \right) \right) = {\mathrm{U }}\left( y \right) . \end{aligned}$$

Taking expectations implies ${\mathrm{E }}_{\mathcal P} \left( {\mathrm{U }}\left( x \right) \right) \le _{\mathrm{\Gamma }}{\mathrm{E }}_{\mathcal Q} \left( {\mathrm{U }}\left( y \right) \right)$. Applying the same arguments with respect to ${\mathrm{\Gamma }}$ provides ${\mathrm{\Gamma }}\left( {\mathrm{E }}_{\mathcal P} \left( {\mathrm{U }}\left( x \right) \right) \right) \le {\mathrm{\Gamma }}\left( {\mathrm{E }}_{\mathcal Q} \left( {\mathrm{U }}\left( y \right) \right) \right)$. Hence, $\mathcal P\precsim \mathcal Q$.

$(\Leftarrow )$: On the contrary, assume that $\mathcal P\succsim \mathcal Q$, but $\mathcal P$ is more uncertain than $\mathcal Q$, i.e., $x =_dy+z$. However, by the arguments of the first part of the proof, if the DM is uncertainty averse, then ${\mathrm{E }}_{\mathcal P} \left( {\mathrm{U }}\left( x \right) \right) \le _{\mathrm{\Gamma }}{\mathrm{E }}_{\mathcal Q} \left( {\mathrm{U }}\left( y \right) \right)$, and therefore $\mathcal P\precsim \mathcal Q$, which is a contradiction. $\square$

Proof of Theorem 4

Recall that a phantom variance is pseudo-positive (Izhakian and Izhakian 2009, Lemma 1.10). ($\Rightarrow$): Suppose $\mathcal P$ is strictly more uncertain than $\mathcal Q$. Then, $X -{\mathrm{E }}_{\mathcal P} \left( X \right) =_dY - {\mathrm{E }}_{\mathcal Q} \left( Y \right) + Z$, for some phantom lottery $\mathcal R$ with ${\mathrm{E }}_{\mathcal R} \left( Z \right) = 0$. Since ${\mathrm{E }}_{\mathcal P} \left( X \right) = {\mathrm{E }}_{\mathcal Q} \left( Y \right)$, then $X =_dY + Z$. Because $Y$ and $Z$ are independent, $\sigma _{\mathcal P}^2 = \sigma _{\mathcal Q}^2 + \sigma _{\mathcal R}^2$, yielding $\sigma _{\mathcal P}^2 \gg \sigma _{\mathcal Q}^2$, since $\sigma _{\mathcal R}^2$ is pseudo-positive.

($\Leftarrow$): The phantom lottery $\mathcal Q'$ having the outcomes $Y' = \lambda \left( Y-{\mathrm{E }}_{\mathcal Q} \left( Y \right) \right)$ is symmetric with mean 0 and variance $\lambda ^2 \sigma _{\mathcal Q}^2$. Let $\lambda =\frac{\sigma _{\mathcal P}^2}{\sigma _{\mathcal Q}^2}$, which is pseudo-positive, and $\lambda \gg 1$ by hypothesis. Clearly, $\sigma _{\mathcal P}^2 = \lambda \sigma _{\mathcal Q}^2$, and therefore, the distribution of $\lambda \left( Y -{\mathrm{E }}_{\mathcal Q} \left( Y \right) \right)$ is identical to the distribution of $X-{\mathrm{E }}_{\mathcal P} \left( X \right)$. By Corollary 2, $\lambda \left( Y-{\mathrm{E }}_{\mathcal Q} \left( Y \right) \right)$ is strictly more uncertain than $Y-{\mathrm{E }}_{\mathcal Q} \left( Y \right)$, and hence $X-{\mathrm{E }}_{\mathcal P} \left( X \right)$ is strictly more uncertain than $Y-{\mathrm{E }}_{\mathcal Q} \left( Y \right)$. Since expectations play no role here, $\mathcal P$ is strictly more uncertain than $\mathcal Q$. $\square$

Proof of Claim 1

Pessimism implies

, but for the LHS, we have

, cf. Table (1). Thus, ${\mathrm{E }}_{\mathcal P_{R}} \left( X \right) \succsim {\mathrm{E }}_{\mathcal P_{B}} \left( X \right)$, which by Axiom 6 implies ${\mathcal P_{R}} \succsim {\mathcal P_{B} }$. By the same consideration, starting with

, one can see that $\mathcal P_{BY} \succsim \mathcal P_{RY}$. Therefore, Independence (Axiom 4) is satisfied. $\square$