Uncertainty and binary stochastic choice

Abstract

Experimental evidence suggests that decision-making has a stochastic element and is better described through choice probabilities than preference relations. Binary choice probabilities admit a strong utility representation if there exists a utility function u such that the probability of choosing a over b is a strictly increasing function of the utility difference $u(a) -u(b)$. Debreu (Econometrica 26(3):440–444, 1958) obtained a simple set of sufficient conditions for the existence of a strong utility representation when alternatives are drawn from a suitably rich domain. Dagsvik (Math Soc Sci 55:341–370, 2008) specialised Debreu’s result to the domain of lotteries (risky prospects) and provided axiomatic foundations for a strong utility representation in which the underlying utility function conforms to expected utility. This paper considers general mixture set domains. These include the domain of lotteries, but also the domain of Anscombe–Aumann acts: uncertain prospects in the form of state-contingent lotteries. For the risky domain, we show that one of Dagsvik’s axioms can be weakened. For the uncertain domain, we provide axiomatic foundations for a strong utility representation in which the utility function represents invariant biseparable preferences (Ghirardato et al. in J Econ Theory 118:133–173, 2004). The latter is a wide class that includes subjective expected utility, Choquet expected utility and maxmin expected utility preferences. We prove a specialised strong utility representation theorem for each of these special cases.

Appendices

Appendix 1: Proofs

1.1 Proof of Theorem 1

In what follows we will make frequent use of the following well-known result:

Lemma 3

(Davidson and Marschak 1959) Let P be a BCPF. Then P satisfies Strong Stochastic Transitivity (Axiom 1) iff it satisfies the the following condition:

$$\begin{aligned} P(a,b) \ge \frac{1}{2}~~~~\Rightarrow ~~~~P\left( a,c\right) \ge P\left( b,c\right) \end{aligned}$$

(20)

for any $a,b,c\in A$.

Property (20) is called weak substitutability. If we replace “$\Rightarrow$” with “$\Leftrightarrow$” in (20), we obtain the substitutability condition (Tversky and Russo 1969). Weak substitutability is obviously implied by substitutability, but the converse is false (Fishburn 1973).

The next result sets the foundation for our proof of Theorem 1.

Lemma 4

Let $M\subseteq A$ and let P satisfy Axioms 1–3. Suppose further that $\succsim ^{P}$ has an M-linear representation $u:A\rightarrow {\mathbb {R}}$ and let $\varSigma =u\left( A\right)$. Then there exists a function $\pi :\varSigma \times \varSigma \rightarrow [0,1]$ such that

$$\begin{aligned} P(a,b) =\pi \left( u(a) ,u(b) \right) \end{aligned}$$

(21)

for any $a,b\in A$. Moreover, $\pi$ satisfies the following conditions for any $\lambda \in [0,1]$ and any $x,y,x^{\prime },y^{\prime } ,z\in \varSigma _{u}$:

• (i) $\pi \left( x,y\right) =\frac{1}{2}$ iff $x=y$.

• (ii) $\pi \left( x,y\right) +\pi \left( y,x\right) =1$.

• (iii) $\pi \left( x,y\right) =\pi \left( x^{\prime } ,y^{\prime }\right)$ implies $\pi \left( x\lambda z,y\lambda z\right) =\pi \left( x^{\prime } \lambda z,y^{\prime } \lambda z\right)$.

• (iv) $\pi$ is continuous in each argument.

• (v) $\pi$ is strictly increasing (respectively, strictly decreasing) in its first (respectively, second) argument.

Proof

Since u is M-linear, $\varSigma$ is an interval in ${\mathbb {R}}$ (Lemma 2). The weak substitutability condition—which is implied by Axiom 1 (Lemma 3)—together with (3) gives

$$\begin{aligned} u(a)\! =\!u(b) ~~~\Leftrightarrow ~~~P(a,b) \!=\! \frac{1}{2}~~~\Rightarrow ~~~P\left( a,c\right) \!=\!P\left( b,c\right) ~~~\Leftrightarrow ~~~P\left( c,a\right) \!=\!P\left( c,b\right) \end{aligned}$$

for any $a,b,c\in A$. It follows that (21) determines a well-defined function $\pi :\varSigma \times \varSigma \rightarrow [0,1]$. It remains to show that $\pi$ satisfies properties (i)–(v).

Property (i) is immediate from the definition of u. Property (ii) follows from (3).

Strong M-Independence of P and the M-linearity of u imply

$$\begin{aligned} \pi \left( x,y\right) =\pi \left( x^{\prime } ,y^{\prime } \right) ~~~\Rightarrow ~~~\pi \left( x\lambda z,y\lambda z\right) =\pi \left( x^{\prime } \lambda z,y^{\prime } \lambda z\right) \end{aligned}$$

for any $x,y,x^{\prime } ,y^{\prime } \in \varSigma$ and any $z\in u\left( M\right)$. Since $u\left( M\right) =u\left( A\right) =\varSigma$, property (iii) holds.

To establish property (iv), note that $\pi$ inherits solvability with respect to its second argument: for any $x,y,z\in \varSigma$ and any $q\in \left[ 0,1\right]$, if $\pi \left( z,x\right) \ge q\ge \pi \left( z,y\right)$ then $\pi \left( z,w\right) =q$ for some $w\in \varSigma$. The weak substitutability property, together with (ii), implies that $\pi$ is also non-increasing in its second argument. It follows that $\pi$ is continuous in its second argument. Continuity in its first argument therefore follows from (ii).

As already noted, Strong Stochastic Transitivity (Axiom 1) implies that $\pi$ is non-increasing in its second argument. Suppose, contrary to what we need to show, that $\pi \left( x,y^{0}\right) =\pi \left( x,y^{1}\right)$ for some $y^{1}>y^{0}$. We can exclude the possibility that $x\in \left[ y^{0},y^{1}\right]$ since (i) and weak monotonicity would then imply $\pi \left( x,y^{0}\right) <\pi \left( x,y^{1}\right)$. Hence, either $x>y^{1}>y^{0}$ or $y^{1}>y^{0}>x$. We only consider the former case, since the latter may be handled similarly. Thus:

$$\begin{aligned} \pi \left( x,y^{0}\right) =\ \pi \left( x,y^{1}\right) \ >\ \frac{1}{2} \end{aligned}$$

(22)

Let $\lambda \in \left( 0,1\right)$ satisfy $y^{1}=y^{0}\lambda x$ and define $y^{n}=y^{n-1}\lambda x=y^{0}\lambda ^{n}x$ for each $n\in \left\{ 2,3,\ldots \right\}$. Since $\pi \left( x,y^{0}\right) =\pi \left( x,y^{1}\right)$, property (iii) gives $\pi \left( x,y^{0}\right) =\pi \left( x,y^{n}\right)$ for all n. By continuity

$$\begin{aligned} \lim _{n\rightarrow \infty } \ \pi \left( x,y^{n}\right) =\ \pi \left( x,x\right) =\ \frac{1}{2} \end{aligned}$$

which contradicts (22). This proves that $\pi$ is strictly decreasing in its second argument. Using (ii), we deduce that it is also strictly increasing in its first argument. $\square$

Corollary 2

Let $M\subseteq A$ and let P satisfy Axioms 1–3. If there is an M-linear utility representation for $\succsim ^{P}$, then P satisfies the substitutability property; that is,

$$\begin{aligned} P(a,b) \ge \frac{1}{2}~~~~\text {iff}~~~~P\left( a,c\right) \ge P\left( b,c\right) \end{aligned}$$

(23)

for any $a,b,c\in A$.

Proof

Strong Stochastic Transitivity (Axiom 1) implies the “only if” part of (23)—recall Lemma 3. The “if” part follows from property (v) in Lemma 4. $\square$

Before proving Theorem 1, let us outline the main obstacle to be overcome. Suppose P satisfies Axioms 1–2 and Strong M-Independence, and suppose further that u is an M-linear representation for $\succsim ^{P}$. Consider the weak order $\ge ^{*}$ on $\varSigma \times \varSigma$ represented by the function $\pi$ defined in Lemma 4:

$$\begin{aligned} \left( x,y\right) \ge ^{*} \left( x^{\prime } ,y^{\prime } \right) \ \ \ \Leftrightarrow \ \ \ \pi \left( x,y\right) \ge \pi \left( x^{\prime } ,y^{\prime } \right) \text {.} \end{aligned}$$

If we can show that $\ge ^{*}$ has a linear representation, then properties (i) and (v) in Lemma 1 imply that $\pi \left( x,y\right) =F\left( x-y\right)$ for some strictly increasing function F. This gives the desired strong utility representation for P. Establishing the existence of a linear representation for $\ge ^{*}$ is complicated by the fact that property (iii) is weaker than the usual (Anscombe–Aumann) independence condition on $\ge ^{*}$. Property (iii) ensures only the Certainty independence (or C-independence) of $\ge ^{*}$.²⁵ We must therefore use the other properties of $\pi$ to make up for this deficiency.

Proof of Theorem 1 Suppose $u:A\rightarrow {\mathbb {R}}$ is an M-linear representation for $\succsim ^{P}$ with $u\left( A\right) =u\left( M\right) =\varSigma$. Then $\varSigma$ is an interval (Lemma 2). The result is trivial if $\varSigma$ is a singleton so suppose otherwise. It is without loss of generality to assume that 0 is contained in the interior of $\varSigma$—if not, apply a suitable affine transformation to u.

Let $\pi$ be defined as in Lemma 4 and let

$$\begin{aligned} \varLambda =\left\{ \left( x,y\right) \in \varSigma \times \varSigma \ |\ x=y\right\} \text {.} \end{aligned}$$

Figure 1 illustrates. Lemma 4 implies that $\varLambda$ is the $\pi$-contour along which $\pi =\frac{1}{2}$. Lemma 4 further implies that $\pi$ is strictly increasing as we move southwards or eastwards in Fig. 1. We need to show that all the other contours of $\pi$ are parallel to $\varLambda$. Property (ii) ensures that these contours are symmetric about $\varLambda$ so it suffices to consider the contours with $\pi \left( x,y\right) <\frac{1}{2}$ (i.e. those lying above $\varLambda$ in Fig. 1).

Suppose, contrary to what we seek to show, that there exist $x,y,x^{\prime } ,y^{\prime } \in \varSigma$ with

$$\begin{aligned} x-y=x^{\prime } -y^{\prime } <0 \end{aligned}$$

but

$$\begin{aligned} \pi \left( x,y\right) >\pi \left( x^{\prime } ,y^{\prime } \right) . \end{aligned}$$

See Fig. 1 (which assumes $x^{\prime } >x$).

Fig. 1

Illustrating points (x, y) and $(x',y')$ in the proof of Theorem 1

Fig. 2

Constructing point $(x',y'')$ in the proof of Theorem 1

By properties (i), (iv) and (v) of $\pi$, there exists some $y^{\prime \prime } \in \varSigma$ such that $x^{\prime }<y^{\prime \prime } <y^{\prime }$ and $\pi \left( x,y\right) =\pi \left( x^{\prime } ,y^{\prime \prime } \right)$. Since $\left( x^{\prime },y^{\prime \prime } \right)$ lies below the line segment joining $\left( x,y\right)$ to $\left( x^{\prime } ,y^{\prime } \right)$, we must have

$$\begin{aligned} \left( x^{\prime } ,y^{\prime \prime } \right) =\ \lambda \left( x,y\right) +\left( 1-\lambda \right) \left( z,z\right) =\ \left( x\lambda z,y\lambda z\right) \end{aligned}$$

(24)

for some $z\in {\mathbb {R}}$ and some $\lambda \in \left( 0,1\right)$. Figure 2 illustrates (again for the case $x<x^{\prime }$). We may further assume that $z\in \varSigma$; otherwise, we can use property (iii) to contract $\left( x,y\right)$ and $\left( x^{\prime },y^{\prime \prime } \right)$ towards $\left( 0,0\right)$ until this condition is satisfied.²⁶

From (24) and the fact that

$$\begin{aligned} \frac{1}{2}>\pi \left( x,y\right) =\pi \left( x^{\prime } ,y^{\prime \prime } \right) =\left( x\lambda z,y\lambda z\right) \text {,} \end{aligned}$$

repeated applications of property (iii) give

$$\begin{aligned} \frac{1}{2}>\pi \left( x,y\right) =\pi \left( x\lambda ^{n}z,y\lambda ^{n}z\right) \ \ \ \ \ \text {for}\, \text {each}~\, n\in \left\{ 1,2,\ldots \right\} \end{aligned}$$

(25)

Since $\left( x\lambda ^{n}z,y\lambda ^{n}z\right) \rightarrow \left( z,z\right)$ as $n\rightarrow \infty$ and $\pi \left( z,z\right) =\frac{1}{2}$, we deduce a contradiction to the continuity property (iv). The contradiction is not quite immediate, as (iv) does not assert the joint continuity of $\pi$. However, for the scenario depicted in Fig. 2 we obtain the required contradiction by considering the sequence $\left\{ \left( x\lambda ^{n}z,\ z\right) \right\} _{n=1}^{\infty }$, which also converges to $\left( z,z\right)$ and satisfies

$$\begin{aligned} \pi \left( x\lambda ^{n}z,z\right) \ \le \ \pi \left( x\lambda ^{n}z,y\lambda ^{n}z\right) \end{aligned}$$

for all n by property (v). For the alternative scenario in which $x>x^{\prime }$, we may consider the sequence $\left\{ \left( z,y\lambda ^{n}z\right) \right\} _{n=1}^{\infty }$ instead.

This contradiction establishes that $\pi \left( x,y\right) =\pi \left( x^{\prime } ,y^{\prime } \right)$ whenever $x-y=x^{\prime }-y^{\prime }$. Since $\pi \left( x,y\right)$ is strictly increasing in $x-y$ by property (v), it follows that $\pi \left( x,y\right) =F\left( x-y\right)$ for some strictly increasing function F. $\square$

1.2 Proof of Corollary 1

The proof of Theorem 1 establishes that $P(a,b) =F\left( u(a) -u(b) \right)$ for some strictly increasing F—see the discussion following the proof of Lemma 4. Property (iv) in Lemma 4 ensures that F is continuous. We can obviously construct a continuous extension of F to I that is non-decreasing everywhere, satisfies $F(x) +F\left( -x\right) =1$ for all $x\in I$, and whose limit is 1 approaching the right-hand end of I. This will be the distribution function for a random variable with the required properties.

1.3 Proof of Proposition 1

Suppose P satisfies Axioms 1, 4 and Strong Independence. By Theorem 1, it suffices to prove that $\succsim ^{P}$ has a mixture linear (i.e. A-linear) representation. We do so by showing that it satisfies the conditions of Theorem 1 in Fishburn (Fisburn 1982, Chapter 2).

It is obvious that $\succsim ^{P}$ is complete. It is transitive by virtue of the Strong Stochastic Transitivity of P (Axiom 1). Setting $c=d$ in the definition of Strong A-Independence, we see that $\succsim ^{P}$ inherits the following independence property: for all $a,b,e\in A$ and all $\lambda \in \left( 0,1\right)$

$$\begin{aligned} a\succsim ^{P}b~~~\Rightarrow ~~~a\lambda e\succsim ^{P}b\lambda e \end{aligned}$$

(26)

The proof is completed by showing that

$$\begin{aligned} \left\{ \lambda \in \left[ 0,1\right] \ \left| \ a\lambda b\succsim ^{P}c\right. \right\} \end{aligned}$$

and

$$\begin{aligned} \left\{ \lambda \in \left[ 0,1\right] \ \left| \ c\succsim ^{P}a\lambda b\right. \right\} \end{aligned}$$

are closed for any $a,b,c\in A$.

Fix some $a,b,c\in A$ and consider the set

$$\begin{aligned} \left\{ \lambda \in \left[ 0,1\right] \ \left| \ c\succsim ^{P}a\lambda b\right. \right\} \end{aligned}$$

(27)

It is without loss of generality to assume $a\succsim ^{P}b$. Then $a\lambda b\succsim ^{P}b$ for any $\lambda \in \left( 0,1\right)$ by (26). Hence, applying (26) once more, we have:

$$\begin{aligned} a\lambda b\succsim ^{P}b\mu \left( a\lambda b\right) \end{aligned}$$

for any $\mu \in \left( 0,1\right)$. Since $b\mu \left( a\lambda b\right) =a \left[ \lambda \left( 1-\mu \right) \right] b$, it follows that if $\lambda$ is in (27) then so is any $\lambda ^{\prime } <\lambda$. It remains to exclude the possibility that (27) is of the form $\left[ 0,\zeta \right)$ for some $\zeta >0$.

Let $\left\{ \lambda _{m}\right\} _{m=1}^{\infty } \subseteq \left[ 0,\zeta \right)$ be a convergent sequence with limit $\zeta$. Suppose $a\lambda _{m}b\succsim ^{P}c$ for each m, while $c\succ ^{P}a\zeta b$. Then

$$\begin{aligned} P\left( c,a\zeta b\right) \ >\ \frac{1}{2}\ \ge \ P\left( c,a\lambda _{m}b\right) \end{aligned}$$

for each m. Let

$$\begin{aligned} \rho \in \left( P\left( c,a\zeta b\right) ,\frac{1}{2}\right) \text {.} \end{aligned}$$

Given some $m\in \left\{ 1,2,\ldots \right\}$, Mixture Solvability (Axiom 4) ensures that there exists $\mu \in \left( 0,1\right)$ such that

$$\begin{aligned} \rho =\ P\left( c,\left( a\zeta b\right) \mu \left( a\lambda _{m}b\right) \right) =\ P\left( c,a\left[ \mu \zeta +\left( 1-\mu \right) \lambda _{m} \right] b\right) \text {.} \end{aligned}$$

Since $\mu \zeta +\left( 1-\mu \right) \lambda _{m}\in \left[ 0,\zeta \right)$ and $\rho >\frac{1}{2}$ we have the desired contradiction.

The same argument (mutatis mutandis) shows that the set

$$\begin{aligned} \left\{ \lambda \in \left[ 0,1\right] \ \left| \ a\lambda b\succsim ^{P}c\right. \right\} \end{aligned}$$

is closed. This completes the proof of Proposition 1.

1.4 Proof of Proposition 2

Lemma 5

Suppose that P satisfies Axioms 1, 4 and 7. Then for every $a\in A$ there exists some ${\overline{a}}\in {\overline{A}}$ such that $a\sim ^{P}{\overline{a}}$.

Proof

Given Strong Stochastic Transitivity (Axiom 1), the binary relation $\succsim ^{P}$ weakly orders the elements of

$$\begin{aligned} \left\{ a(s) \in {\overline{A}}\ |\ s\in {\mathcal {S}}\right\} \text { .} \end{aligned}$$

Suppose that $s^{\prime } ,s^{\prime \prime } \in {\mathcal {S}}$ are such that $a\left( s^{\prime \prime } \right) \succsim ^{P}a(s) \succsim ^{P}a\left( s^{\prime } \right)$ for all $s\in {\mathcal {S}}$. Then Stochastic Monotonicity (Axiom 7) implies $a\left( s^{\prime \prime } \right) \succsim ^{P}a\succsim ^{P}a\left( s^{\prime } \right)$. That is:

$$\begin{aligned} P\left( a,a\left( s^{\prime } \right) \right) \ge \frac{1}{2}\ge P\left( a,a\left( s^{\prime \prime } \right) \right) \text {.} \end{aligned}$$

Hence, Mixture Solvability (Axiom 4) ensures that

$$\begin{aligned} a\sim ^{P}a\left( s^{\prime \prime } \right) \lambda a\left( s^{\prime } \right) \end{aligned}$$

for some $\lambda \in \left[ 0,1\right]$. $\square$

When P satisfies Axioms 1, 4 and 7, we may therefore associate each non-constant act $a\in A\diagdown {\overline{A}}$ with a specific (but arbitrarily chosen) certainty equivalent, denoted by ${\overline{a}}$. For every constant act $a\in {\overline{A}}$, we define ${\overline{a}}=a$. Hence, ${\overline{a}}\in {\overline{A}}$ and $a\sim ^{P}{\overline{a}}$ for every $a\in A$.

Proof of Proposition 2

The restriction of P to ${\overline{A}}\times {\overline{A}}$ satisfies Strong Stochastic Transitivity, Mixture Solvability and Strong ${\overline{A}}$-Independence. By Proposition 1, there exists a mixture linear function $v: {\overline{A}}\rightarrow {\mathbb {R}}$ such that

$$\begin{aligned} P(a,b) \ge P\left( c,d\right) ~~~\Leftrightarrow ~~~v(a) -v(b) \ge v\left( c\right) -v(d) \ \ \ \ \text {for}\, \text {all}\, a,b,c,d\in {\overline{A}} \qquad \end{aligned}$$

(28)

In particular, v represents $\succsim ^{P}$ on ${\overline{A}}$. Let u extend v to A by using Lemma 5 to define $u(a) =v\left( {\overline{a}}\right)$ for all $a\in A\diagdown {\overline{A}}$. Then u represents $\succsim ^{P}$ by construction. We next show that u is ${\overline{A}}$-linear.

Suppose $a\in A, x\in {\overline{A}}$ and $\lambda \in \left( 0,1\right)$. Since $P\left( a,{\overline{a}}\right) =\frac{1}{2}$, Strong ${\overline{A}}$ -Independence implies that $P\left( a\lambda x,{\overline{a}}\lambda x\right) = \frac{1}{2}$. Thus:

$$\begin{aligned} u\left( a\lambda x\right) =\ u\left( {\overline{a}}\lambda x\right) =\ v\left( {\overline{a}}\lambda x\right) =\ \lambda v\left( {\overline{a}} \right) +\left( 1-\lambda \right) v(x) =\ \lambda u(a) +\left( 1-\lambda \right) u(x) \text {,} \end{aligned}$$

where the penultimate equality uses the fact that $v:{\overline{A}}\rightarrow {\mathbb {R}}$ is mixture linear. This proves that u is ${\overline{A}}$ -linear.

Applying Corollary 2 (with $M={\overline{A}}$), and making use of (3), it follows that

$$\begin{aligned} P(a,b) \ge P\left( c,d\right) ~~~\Leftrightarrow ~~~P\left( {\overline{a}},{\overline{b}}\right) \ge P\left( {\overline{c}},{\overline{d}} \right) \ \ \ \ \text {for all}\,a,b,c,d\in A \end{aligned}$$

(29)

Combining (28) and (29), we see that u is a strong utility for P.

It remains to show that u is of the invariant biseparable class.

Observe that v and u have the same range by Lemma 5. Call this common range $\varSigma$. Since v is mixture linear, $\varSigma$ is an interval in ${\mathbb {R}}$ (Lemma 2). Using Stochastic Monotonicity (Axiom 7), it is easy to see that there exists a well-defined function $h:\varSigma ^{S}\rightarrow \varSigma$ such that

$$\begin{aligned} u(a) =\ h\left( v\left( a\left( 1\right) \right) ,\ldots ,v\left( a\left( S\right) \right) \right) \end{aligned}$$

(30)

for all $a\in A$. That is, the utility of any act depends only on the v-utility obtained in each state and is uniquely determined by these v-utilities. We use h to denote the function that converts vectors of state-(v-)utilities into the (u-)utility of the corresponding act.

Let C denote the set of constant vectors in $\varSigma ^{S}$, and identify elements of C with elements of $\varSigma$ in the obvious fashion. We now observe that h has the following properties: (a) $h\left( c\right) =c$ for all $c\in \varSigma$; (b) h is non-decreasing in each argument by virtue of Stochastic Monotonicity; and (c) h is C-linear by virtue of the ${\overline{A}}$-linearity of u. In the language of Ghirardato et al. (2005), h is normalised, monotonic and constant affine. If P is non-trivial, then $\varSigma$ is a non-degenerate interval and it follows (ibid., p. 132) that h has a unique extension $h:{\mathbb {R}} ^{S}\rightarrow {\mathbb {R}}$ satisfying constant linearity: $h\left( ax+b\right) =ah(x) +b$ for all $x\in {\mathbb {R}}^{S}$ and all $a,b\in {\mathbb {R}}$ with $a\ge 0$. The desired result is now immediate from Ghirardato, Maccheroni and Marinacci (2004, Lemma 1 and Theorem 11). $\square$

1.5 Proof of Proposition 3

Let v and u be defined as in the proof of Proposition 2 and let h: $\varSigma ^{S}\rightarrow \varSigma$ be the function defined by (30). Non-triviality of P implies that $\varSigma$ is a non-degenerate interval (Lemma 2), which we assume, without loss of generality, to contain $\left[ -1,1\right]$.

We first show that Strong A-Independence of P implies A-linearity of u. Let $a,b\in A$ and $\lambda \in \left( 0,1\right)$. Since $P\left( a, {\overline{a}}\right) =\frac{1}{2}$ we deduce that $P\left( a\lambda b, {\overline{a}}\lambda b\right) =\frac{1}{2}$ from Strong A-Independence. Thus, $u\left( a\lambda b\right) =u\left( {\overline{a}}\lambda b\right)$. Using the ${\overline{A}}$-linearity of u, we have:

$$\begin{aligned} u\left( a\lambda b\right)= & {} \ u\left( {\overline{a}}\lambda b\right) \ \\= & {} \ \lambda u\left( {\overline{a}}\right) +\left( 1-\lambda \right) u(b) \\= & {} \ \lambda u(a) +\left( 1-\lambda \right) u(b) \end{aligned}$$

Hence, u is A-linear.²⁷

Next, we show that h is mixture linear. Let $a,b\in A$ and $\lambda \in \left[ 0,1\right]$. If $x_{s}=v\left( a(s) \right)$ and $y_{s}=v\left( b(s) \right)$, then $u(a) =h(x)$ and $u(b) =h\left( y\right)$. Using the ${\overline{A}}$-linearity of v and the A-linearity of u we have:

$$\begin{aligned} h\left( \lambda x+\left( 1-\lambda \right) y\right)= & {} \ u\left( a\lambda b\right) \\= & {} \ \lambda u(a) +\left( 1-\lambda \right) u(b) \\= & {} \ \lambda h(x) +\left( 1-\lambda \right) h\left( y\right) \text {.} \end{aligned}$$

Hence, h is mixture linear. It has a unique linear extension h: ${\mathbb {R}} ^{S}\rightarrow {\mathbb {R}}$ by standard arguments.²⁸ Stochastic Monotonicity (Axiom 7) implies that the coefficients of this linear function are non-negative. Since u is non-constant, at least one coefficient is strictly positive so the vector of coefficients can be positively scaled to lie in $\varTheta$. $\square$

1.6 Proof of Proposition 4

Once again, let v and u be defined as in the proof of Proposition 2 and let $h:\varSigma ^{S}\rightarrow \varSigma$ be the function defined by (30).

From Stochastic Uncertainty Aversion (Axiom 8) and the ${\overline{A}}$-linearity of v, we see that $h:\varSigma ^{S}\rightarrow \varSigma$ satisfies—in addition to the properties noted in the proof of Proposition 2—the following: for any $x,y\in \varSigma ^{S}$,

$$\begin{aligned} h(x) =h\left( y\right) \ \ \ \Rightarrow \ \ \ h\left( \lambda x+\left( 1-\lambda \right) y\right) \ge h\left( y\right) \end{aligned}$$

(31)

Its unique constant-linear extension h: ${\mathbb {R}}^{S}\rightarrow {\mathbb {R}}$ is therefore superadditive: that is,

$$\begin{aligned} h\left( x+y\right) \ \ge \ h(x) +h\left( y\right) \end{aligned}$$

(32)

for any $x,y\in {\mathbb {R}}^{S}$. To see why, let $x,y\in {\mathbb {R}}^{S}$, ${\overline{x}}=h(x)$, ${\overline{y}}=h\left( y\right)$ and $\varepsilon ={\overline{x}}-{\overline{y}}$. Then constant linearity and (31) give

$$\begin{aligned} \frac{1}{2}h\left( x+y\right) +\frac{1}{2}\varepsilon= & {} \ h\left( \frac{1}{2} \left( x+y\right) +\frac{1}{2}\varepsilon \right) \\\ge & {} \ \frac{1}{2} h(x) +\frac{1}{2}h\left( y+\varepsilon \right) \\= & {} \frac{1}{2} \left[ h(x) +h\left( y\right) \right] +\frac{1}{2}\varepsilon \end{aligned}$$

which implies (32).

The proof is completed by applying the following well-known result (see, for example, Lemma 3.5 in Gilboa and Schmeidler 1989):

Lemma 6

Let $h:{\mathbb {R}}^{S}\rightarrow {\mathbb {R}}$ be a superadditive, constant-linear function that is non-decreasing in each argument. Then there exists a closed and convex set ${\mathcal {P}}\subseteq \varTheta$ such that

$$\begin{aligned} h(x) =\ \min _{\theta \in {\mathcal {P}}}\sum _{s=1}^{S}\theta (s) x_{s} \end{aligned}$$

for every $x\in {\mathbb {R}}^{S}$.

1.7 Proof of Proposition 5

As usual, we let v and u be defined as in the proof of Proposition 2 and let $h:\varSigma ^{S}\rightarrow \varSigma$ be the function defined by (30). Since $\varSigma$ is a non-degenerate interval we may assume, without loss of generality, that it contains $\left[ -1,1\right]$.

Definition 6

Vectors $x,y\in \varSigma ^{S}$ are comonotonic if there do not exist $s,s^{\prime } \in {\mathcal {S}}$ with $x_{s}>x_{s^{\prime } }$ and $y_{s^{\prime } }>y_{s}$.

If P satisfies Stochastic Comonotonic Independence (Axiom 9), then $h:\varSigma ^{S}\rightarrow \varSigma$ satisfies

$$\begin{aligned} h(x)>h\left( y\right) ~~~\Rightarrow ~~~h\left( \lambda x+\left( 1-\lambda \right) z\right) >h\left( \lambda y+\left( 1-\lambda \right) z\right) \end{aligned}$$

(33)

for any pairwise comonotonic $x,y,z\in \varSigma ^{S}$ and any $\lambda \in \left( 0,1\right)$. The desired representation is now an immediate implication of the following result:²⁹

Lemma 7

(Schmeidler 1986, Corollary) Let $\varSigma \subseteq {\mathbb {R}}$ be a non-degenerate interval containing $\left[ -1,1\right]$ and let $h:\varSigma ^{S}\rightarrow \varSigma$ be a normalised, monotonic function that satisfies (33) for any for any pairwise comonotonic $x,y,z\in \varSigma ^{S}$ and any $\lambda \in \left( 0,1\right)$. Then there exists a capacity $\omega$ such that

$$\begin{aligned} h(x) =\ \int x\ d\omega \end{aligned}$$

(34)

for all x.³⁰

Appendix 2: Strengthening Theorem 2

The following strengthens Theorem 2 by replacing the Quadruple Condition (Axiom 5) with Strong Stochastic Transitivity (Axiom 1).

Theorem 3

Let $A=\varDelta ^{n}$. If P satisfies Axioms 1, 2, 6 and Strong Independence, then P has a strong utility that is mixture linear.

We use a similar argument to that for Proposition 1, except we can no longer appeal to Mixture Solvability to establish the continuity of $\succsim ^{P}$. Instead, we show that $\succsim ^{P}$ possesses a linear representation by an alternative route.

Proof of Theorem 3

As in the proof of Proposition 1, we may assume that $\succsim ^{P}$ is a weak order satisfying

$$\begin{aligned} a\succsim ^{P}b~~~\Rightarrow ~~~a\lambda c\succsim ^{P}b\lambda c \end{aligned}$$

(35)

for all $a,b,c\in A$ and all $\lambda \in \left( 0,1\right)$. Since the result is obvious if $\succsim ^{P}$ is trivial (i.e. $a\sim ^{P}b$ for all $a,b\in A$), let us assume otherwise. From Axiom 6, we also know that $\succsim ^{P}$ satisfies the following Archimedean property: for any $a,b,c\in A$

$$\begin{aligned} a\succ ^{P}b\succ ^{P}c~~~\Rightarrow ~~~a\lambda c\succ ^{P}b\succ ^{P}a\mu c \end{aligned}$$

(36)

for some $\lambda ,\mu \in \left( 0,1\right)$. It therefore suffices (Fisburn 1982, Chapter 2) to establish that

$$\begin{aligned} a\succ ^{P}b~~~\Rightarrow ~~~a\lambda c\succ ^{P}b\lambda c \end{aligned}$$

(37)

for all $a,b,c\in A$ and all $\lambda \in \left( 0,1\right)$.

Lemma 8

Condition (37) holds on the interior of A (that is, for any $a,b,c\in A\cap {\mathbb {R}}_{++}^{n}$).

Proof

Suppose $a,b,c\in A\cap {\mathbb {R}}_{++}^{n}$ with $a\succ ^{P}b$ and $a\lambda c\sim ^{P}b\lambda c$.³¹ That is,

$$\begin{aligned} P(a,b) >\frac{1}{2}~~~\text {and}~~~P\left( a\lambda c,b\lambda c\right) =\frac{1}{2} \end{aligned}$$

(38)

We claim that

$$\begin{aligned} P(a,b) \ge P\left( d,e\right) \ge \frac{1}{2}~~~\Rightarrow ~~~P\left( d\lambda c,e\lambda c\right) =\frac{1}{2} \end{aligned}$$

(39)

for any $d,e\in A$. To see why, observe that Strong Independence and

$$\begin{aligned} P\left( d,e\right) \ge \frac{1}{2}=P\left( c,c\right) \end{aligned}$$

give

$$\begin{aligned} P\left( d\lambda c,e\lambda c\right) \ge \frac{1}{2}\text {.} \end{aligned}$$

The reverse inequality follows by applying Strong Independence to $P(a,b) \ge P\left( d,e\right)$ and using (38).

We next show that $d=a\mu b$ and $e=b$ satisfy the antecedent in (39) for any $\mu \in \left[ 0,1\right]$. The cases $\mu \in \left\{ 0,1\right\}$ are trivial so we focus on $\mu \in \left( 0,1\right)$.

Since

$$\begin{aligned} P(a,b) >\frac{1}{2}=P\left( b,b\right) =P\left( a,a\right) \end{aligned}$$

we have

$$\begin{aligned} P\left( a\mu b,b\right) \ge \frac{1}{2} \end{aligned}$$

(40)

and

$$\begin{aligned} P\left( a,a\mu b\right) =P\left( a,b\left( 1-\mu \right) a\right) \ge \frac{ 1}{2} \end{aligned}$$

by Strong Independence. From the latter inequality and weak substitutability (which is implied by Strong Stochastic Transitivity—Lemma 3), we have

$$\begin{aligned} P(a,b) \ge P\left( a\mu b,b\right) \end{aligned}$$

(41)

Thus, from (39)–(41):

$$\begin{aligned} P\left( \left( a\mu b\right) \lambda c,b\lambda c\right) =\frac{1}{2} \end{aligned}$$

(42)

for any $\mu \in \left[ 0,1\right]$.

From (42) we obtain a linear segment of strictly positive length,³² parallel to the line joining a and b, which forms part of an indifference curve for $\succsim ^{P}$. Since a and b are in the interior of the simplex, we can use the existence of this linear segment, together with transitivity of $\succsim ^{P}$ and the independence property (35), to deduce that the line segment joining a and b must also be part of an indifference curve. Figure 3 illustrates the required construction (for the case $n=3$). By moving point z along the segment from x to y, we sweep out an indifference curve joining a to b.³³ This is the desired contradiction, since $a\succ ^{P}b$. $\square$

Since $A\cap {\mathbb {R}}_{++}^{n}$ is a mixture set (under the usual mixing operation for ${\mathbb {R}}^{n}$), it follows that $\succsim ^{P}$ possesses a linear representation on $A\cap {\mathbb {R}}_{++}^{n}$. Let u be such a representation.

Fig. 3

Illustrating the construction in the proof of Lemma 8

Observe that Axiom 6 is not required for the proof of Lemma 8—Strong Independence does all the work. We now use Axiom 6 to extend the linear representation to the boundary of the simplex.

First, u has a unique mixture linear extension to A. We use u to denote the extended function, as no confusion should arise. Let a be a boundary point of A. There are two possibilities: either (i) $u(a) =u(b)$ for some $b\in A\cap {\mathbb {R}}_{++}^{n}$ or else (ii) the face of the simplex containing a is part of a contour of u.

For case (i) we show that $a\sim ^{P}b$. (It follows that $a\sim ^{P}b$ for any boundary point a and any interior point b on the same utility contour as a.) If, for example, $a\succ ^{P}b$, then we can find some $c\in A\cap {\mathbb {R}}_{++}^{n}$ with $u\left( c\right) <u(b) =u(a)$. Hence $a\succ ^{P}b\succ ^{P}c$, but $u\left( a\lambda c\right) <u(b)$ for all $\lambda \in \left( 0,1\right)$ by the mixture linearity of u. Since $a\lambda c\in A\cap {\mathbb {R}}_{++}^{n}$ for any $\lambda \in \left( 0,1\right)$, we have $b\succ ^{P}a\lambda c$ for all $\lambda \in \left( 0,1\right)$, which contradicts the Archimedean property (36). Assuming $b\succ ^{P}a$ leads similarly to contradiction.

Now consider case (ii). Thus, u is constant on the face of the simplex containing a. Suppose, in particular, that $u(a) >u(b)$ for all $b\in A\cap {\mathbb {R}}_{++}^{n}$. The alternative scenario, in which $u(a) <u(b)$ for all $b\in A\cap {\mathbb {R}}_{++}^{n}$, may be handled similarly. We will show that $a\sim ^{P}b$ for any b on the same face as a, and $a\succ ^{P}b$ for any $b\in A$ not contained in the face containing a. Combined with case (i), this shows that $\succsim ^{P}$ is represented by u on all of A.

Suppose $b\in A$ with $u(a) =u(b)$, so b lies on the same face of the simplex as a. If $a\succ ^{P}b$ then $a\succ ^{P}b\succ ^{P}c$ for any $c\in A\cap {\mathbb {R}}_{++}^{n}$. Since $b\succ ^{P}a\lambda c\in A\cap {\mathbb {R}}_{++}^{n}$ for any $\lambda \in \left( 0,1\right)$, we deduce a contradiction to (36). Assuming $b\succ ^{P}a$ leads similarly to contradiction.

Finally, let $b\in A$ with $u(a) >u(b)$. Assume, contrary to what we seek to show, that $b\succsim ^{P}a$. Since $u(a) >u(b)$ we may choose some $c,d\in A\cap {\mathbb {R}} _{++}^{n}$ with

$$\begin{aligned} u(a)>u(d)>u\left( c\right) >u(b) \text { .} \end{aligned}$$

Therefore, $d\succ ^{P}c\succ ^{P}b\succsim ^{P}a$ and

$$\begin{aligned} u\left( d\lambda a\right) =\lambda u(d) +\left( 1-\lambda \right) u(a)>u(d) >u\left( c\right) \end{aligned}$$

It follows that $d\lambda a\succ ^{P}c$ for all $\lambda \in \left( 0,1\right)$. This contradicts the Archimedean property (36) and completes the proof of Theorem 3. $\square$