Purely subjective extended Bayesian models with Knightian unambiguity

Abstract

This paper provides a model of belief representation in which ambiguity and unambiguity are endogenously distinguished in a purely subjective setting where objects of choices are, as usual, maps from states to consequences. Specifically, I first extend the maxmin expected utility theory and get a representation of beliefs such that the probabilistic beliefs over each ambiguous event are represented by a non-degenerate interval, while the ones over each unambiguous event are represented by a number. I then consider a class of the biseparable preferences. Two representation results are achieved and can be used to identify the unambiguity in the context of the biseparable preferences. Finally a subjective definition of ambiguity is suggested. It provides a choice theoretic foundation for the Knightian distinction between ambiguity and unambiguity.

Appendix

1.1 Proof of Section 3

We first state two facts without proofs.

Fact 1

Axioms 1–3 imply that for every $f\in \mathcal {F}$, there exists $x\in X$ such that $f\sim \bar{x}$.

Fact 2

Axioms 1, 3, and 4 imply that there exist two consequences $x,y$ such that $x\succ y$.

1.1.1 Proof of Theorem 1: (i) implies (ii)

We first show that Axiom 7 (unambiguity independence) implies Certainty independence, which is introduced by Alon and Schmeidler (2014). Therefore, by Theorem 7 in them, the preference relation $\succsim$ is a MEU preference. Then we show that all the probabilities have to agree on the set of unambiguous events.

Recall the Axiom of certainty independence:

Axiom 7 ${}^\prime$ Suppose that a pair of acts $f,g$ and consequences $x,a,w$ satisfy $\langle x;a\rangle \sim ^*\langle a;w\rangle$, and $\langle f(s);g(s)\rangle \sim ^*\langle g(s);w\rangle$ for all $s\in S$. Then $g\sim \bar{a}\Leftrightarrow f\sim \bar{x}$.

Lemma 1

Axiom 7 implies Axiom 7$^{\prime }$.

Proof

Suppose Axiom 7 holds. Let $x=y$ and $z=w$. Then first condition in Axiom 7 becomes $\langle f(s);g(s)\rangle \sim ^*\langle g(s); w\rangle$ for all $s\in S$, and the second condition becomes $\langle x;a\rangle \sim ^*\langle a; w\rangle$, which are the assumption on the Axiom $7'$. Then by (i) in Axiom 7, we have $g\sim \bar{a}\Leftrightarrow f\sim \bar{x}$.$\square$

Proposition 1

(Proposition 12 from Alon and Schmeidler 2014) Suppose $\succsim$ satisfies Axioms 1–5. Then the following is true:

• (1) there exists a continuous function $u:X\rightarrow \mathbb {R}$ such that for all $x,y\in X, x\succsim y\Leftrightarrow u(x)\ge u(y)$. Furthermore, $u$ is unique up to a positive linear transformation.

• (2) $\langle a; b\rangle \sim ^*\langle c;d\rangle$ implies $u(a)-u(b)=u(c)-u(d)$.

• (3) there exists a unique continuous representation function $V_c$ on $\mathcal {F}$ such that $J(\bar{x}=u(x)$ for all $x\in X$.

The lemma below is an analog to Lemma 3.3 from Gilboa and Schmeidler (1989), and follows the Lemma 23 and other results from Alon and Schmeidler (2014). Denoted by $B$ is the set of all real-valued functions on $S$. For any $\gamma \in \mathbb {R}, \bar{gamma}\in B$ is the constant function.

Lemma 2

There exists a functional $I: B\rightarrow \mathbb {R}$ such that

• (i) For any $f\in \mathcal {F}, I(u\circ f)=V_c(f)$.

• (ii) $I$ is monotonic, i.e., for any $a,b\in B, a\ge b\Leftarrow I(a)\ge I(b)$.

• (iii) $I$ is superadditive and homogeneous of degree $1$.

• (iv) $I$ satisfies certainty independence: for any $a\in B$ and constant function $\bar{\gamma }, I(a+\bar{\gamma })=I(a)+\gamma$.

The next lemma will show that the functional $I$ satisfies unambiguity independence: $I$ is additive on an arbitrary act and an unambiguous act. Note that for any $\beta ,\gamma \in \mathbb {R}$ and any $E\in \Sigma , \beta 1_E+\gamma 1_{E^\mathrm{c}}$ is a binary function.

Lemma 3

$I$ satisfies unambiguity independence: for any $E\in \mathcal {U}$, any $a\in B$, and any binary function $\beta 1_E+\gamma 1_{E^\mathrm{c}}, I(a+\beta 1_E+\gamma 1_{E^\mathrm{c}})=I(a)+\beta I(1_E)+\gamma I(1_{E^\mathrm{c}})$.

Proof

Since $I$ satisfies homogeneous of degree one and certainty independence, to show unambiguity independence it suffices to prove that $I(a+1_E)=I(a)+I(1_E)$.

Let $f$ be in $\mathcal {F}$. Let $z.w\in X$ be such that $u\circ (zEw)=1_E$. According to Fact 1, there exists $\xi \in X$ such that $\bar{\xi }\sim zEw$. Therefore $I(1_E)=u(\xi )$.

Let $\eta ,x,y\in X$ be such that $u(\eta )=\frac{1}{2}[u(x)+u(z)]$ and $u(\eta )=\frac{1}{2}[u(y)+u(w)]$. Notice that such construction is always possible: we let $x=w$ and $y=z$, and by continuity of $u$, there is a $\eta$ such that $u(\eta )=\frac{1}{2}[u(w)+u(z)$.

By Fact 1 again, there exists $\psi \in X$ such that $\bar{\psi }\sim xEy$. So $I(u\circ (xEy))=u(\psi )$. Therefore by (ii) of Axiom 7, we have

$$\begin{aligned} u(\eta )=\frac{1}{2}[u(\xi )+u(\psi )] \end{aligned}$$

(8)

Now we construct an act $g$ by,

$$\begin{aligned} u(g(s))=\left\{ \begin{array}{l@{\quad }l} \textstyle \frac{1}{2}[u(f(s))+u(z)], &{} \hbox {if } x\in E, \\ \textstyle \frac{1}{2}[u(f(s))+u(w)], &{} \hbox {if } x\in E^\mathrm{c}.\\ \end{array} \right. \end{aligned}$$

The existence of $g$ is guaranteed by the continuity of $u$. Therefore, $I(u\circ g)=I(\frac{1}{2}(u\circ f+u\circ (zEw)))$. So $f\sim xEy$ and $\bar{\psi }\sim xEy$ imply $f\sim \bar{\psi }$. Hence $I(u\circ f)=u(\psi )$. Axiom 7 implies $g\sim \bar{\eta }$. Thus,

$$\begin{aligned} I\left( \frac{1}{2} u\circ f+\frac{1}{2}u\circ (zEw)\right) =\frac{1}{2}I(u\circ f+1_E)=\frac{1}{2}[u(\xi )+u(\psi )] \end{aligned}$$

That is, $I(u\circ f+1_E)=I(u\circ f)+I(1_E)$.$\square$

Lemma 4

(Theorem 7 of Alon and Schmeidler 2014) $\succsim$ on $\mathcal {F}$ is represented by a MEU functional.

Lemma 5

For every $E\in \mathcal {U}$ and every $P,Q\in K$, we have $P(E)=Q(E)$.

Proof

Let $E$ be in $\mathcal {U}$. By lemma 4, $I(1_E)=\min \{\int 1_E\mathrm {d}P| P\in K\}=\widetilde{P}(E)$ and $I(1_{E^\mathrm{c}})=\min \{\int 1_{E^\mathrm{c}}\mathrm {d}P|P\in K\}=\widehat{P}(E^\mathrm{c})$. So for every $Q\in K$, we must have $Q(E)\ge \widetilde{P}(E)$ and $Q(E^\mathrm{c})\ge \widehat{P}(E^\mathrm{c})$. Hence $\widetilde{P}(E)\le Q(E)\le \widehat{P}(E)$. Since $I$ is unambiguity independent, $I(1)=I(1_E)+I(1_{E^\mathrm{c}})$. That is $\widetilde{P}(E)+\widehat{P}(E^\mathrm{c})=1$. So $\widetilde{P}(E)=\widehat{P}(E)$. Thus, for any $P,Q\in K$, we must have $P(E)=Q(E)$ for all $E\in \mathcal {U}$.$\square$

1.1.2 Proof of Theorem 1: (ii) implies (i)

We first define a functional $I: B\rightarrow \mathbb {R}$ by $I(b)=\min _{P\in K}\{\int _S bdP\}$. For any $E\in \mathcal {U}$, and any $P,Q\in K, P(E)=Q(E)$. We denote $P^*(E)$ the probability of $E$ on $K$. Hence the preference relation is represented by: for any $f\in \mathcal {F}, V_c(f)=I(u\circ f)$, and for any $xEy\in \mathcal {F}^u, V_c(xEy)=u(x)P^*(E)+u(y)P^*(E^\mathrm{c})$.

By Theorem 7 of Alon and Schmeidler (2014), (ii) implies Axioms 1,2,3,4,5,6, and 8. For the rest, we only need to show that Axiom 7 (Unambiguity independence) holds. If $E\in \mathcal {U}$ and $b\in B$, then $I(b+1_E)=I(b)+I(1_E)$. If $\alpha >0$, then $I(\alpha b)=\alpha I(b)$. Let $E\in \mathcal {U}$, and $f,g\in \mathcal {F}$, and $x,y,z,w,a,b,c\in X$ satisfy the following conditions:

• (a) $\langle f(s),g(s)\rangle \sim ^*\langle g(s),z\rangle$ for all $s\in E$ and $\langle f(s),g(s)\rangle \sim ^*\langle g(s),w\rangle$ for all $s\in E^\mathrm{c}$;

• (b) $\langle x,a\rangle \sim ^*\langle a,z\rangle$ and $\langle y,a\rangle \sim ^*\langle a,w\rangle$

According to the representation, condition (a) implies that

$$\begin{aligned} u(g(s))=\left\{ \begin{array}{l@{\quad }l} \textstyle \frac{1}{2}[u(f(s)+u(z)], &{} \hbox {if }s\in E; \\ \textstyle \frac{1}{2}[u(f(s)+u(w)], &{} \hbox {if }s\in E^\mathrm{c}. \\ \end{array} \right. \end{aligned}$$

And condition (b) implies that

$$\begin{aligned} u(a)=\frac{1}{2}[u(x)+u(z)]\quad \text { and }\quad u(a)=\frac{1}{2}[u(y)+u(w)] \end{aligned}$$

Therefore, by condition (a) we have:

$$\begin{aligned} V_c(g)&= I(u\circ g)\\&= I\left( \frac{1}{2}[u\circ f+u(z)]1_E+\frac{1}{2}[u\circ f+ u\circ (w)]1_{E^\mathrm{c}}\right. \\&= I\left( \frac{1}{2} u\circ f+\frac{1}{2}[u(z)1_E+u(w)1_{E^\mathrm{c}}]\right) \\&= \left. \frac{1}{2}I(u\circ f)+\frac{1}{2}[u(z)I(1_E)+u(w)I(1_{E^\mathrm{c}})]\right) \\&= \frac{1}{2}V_c(f)+\frac{1}{2}\left[ u(z)P^*(E)+u(w)P^*(E^\mathrm{c})\right] \end{aligned}$$

Also by condition (b) we have

$$\begin{aligned} U(a)=\frac{1}{2}\left[ u(x)P^*(E)+u(y)P^*(E^\mathrm{c})]+\frac{1}{2}[u(z)P^*(E)+u(w)P^*(E^\mathrm{c})\right] \end{aligned}$$

Hence $g\sim \bar{a}$ iff

$$\begin{aligned} \frac{1}{2}V_c(f)+\frac{1}{2}\left[ u(z)P^*(E)+u(w)P^*(E^\mathrm{c})\right]&= \frac{1}{2}\left[ u(x)P^*(E)+u(y)P^*(E^\mathrm{c})\right] \\&\quad +\frac{1}{2}\left[ u(z)P^*(E)+u(w)P^*(E^\mathrm{c})\right] \end{aligned}$$

That is $V_c(f)=u(x)P^*(E)+u(y)P^*(E^\mathrm{c})$. Therefore $f\sim xEy$.

Now we assume that $b\sim zEw$ and $c\sim xEy$. Then we have

$$\begin{aligned} u(b)=u(z)P^*(E)+u(w)P(E^\mathrm{c})\quad \text { and }\quad u(c)=u(x)P^*(E)+u(y)P(E^*) \end{aligned}$$

By condition (b), $\frac{1}{2}[u(b)+u(c)]=u(a)$. Hence $\langle c;a\rangle \sim ^*\langle a;b\rangle$. Therefore, Axiom 7 (unambiguity independence) is satisfied.

1.2 Proof of Section 4

We begin by listing a theorem of Ghirardato and Marinacci (2001) a là Alon and Schmeidler (2014).

Theorem 5

Suppose that structure assumptions on $S$ and $X$ hold. The following statements are equivalent for $\succsim$ on $\mathcal {F}$:

• (i) $\succsim$ satisfies Axioms 1,2,3,4$^*$,4$^{**}$, and $5^*$.

• (ii) $\succsim$ is a biseparable preference with representation as in (5).

1.2.1 Proof of Theorem 2: (i) $\Rightarrow$ (ii)

By Theorem 5, $\succsim$ is a biseparable preference. We only need to show for each $E\in \mathcal {U}, \rho (E)+\rho (E^\mathrm{c})=1$. Let $x,y,z$ be consequences such that $x\succ y\succ z$. By Axiom 9 and Theorem 2,

$$\begin{aligned} u(\Psi _E(\Psi _E(x,z),\Psi _E(y,y)))=u(\Psi _E(\Psi _E(x,y),\Psi _E(z,y))) \end{aligned}$$

(9)

Note that $\Psi _E(y,y)=y$. Since $\succsim$ satisfies monotonicity, we have $\Psi _E(x,y)\succ \Psi _E(z,y)$. Therefore (5) implies:

$$\begin{aligned}&u\big (\Psi _E(\Psi _E(x,y),\Psi _E(z,y))\big )\\&\quad = u(\Psi _E(x,y))\rho (E)+u(\Psi _E(z,y))(1-\rho (E)) \\&\quad = \big [u(x)\rho (E)+u(y)(1-\rho (E))\big ] \rho (E)+\big [u(z)(1-\rho (E^\mathrm{c}))+u(y)\rho (E^\mathrm{c})\big ] (1-\rho (E))\\&\quad = u(x)\rho (E)^2+u(y)\big (\rho (E)+\rho (E^\mathrm{c})-\rho (E)^2-\rho (E)\rho (E^\mathrm{c})\big )\\&\qquad +u(z)(1-\rho (E))(1-\rho (E^\mathrm{c})) \end{aligned}$$

Now consider the left of (9). There are two cases need to be discussed: $\Psi _E(x,z)\succsim y$ and $y\succsim \Psi _E(x,z)$.

Case 1 $\Psi _E(x,y)\succsim y$.

Since $\succsim$ is a biseparable preference, by (5) we have:

$$\begin{aligned} u\big (\Psi _E(\Psi _E(x,z),y)\big )&= u(\Psi _E(x,z))\rho (E)+u(y)(1-\rho (E)) \nonumber \\&= u(x)\rho (E)^2+u(y)(1-\rho (E))+u(z)\rho (E)(1-\rho (E)) \end{aligned}$$

(10)

Because of (9), we get:

$$\begin{aligned}&u(x)\rho (E)^2+u(y)(1-\rho (E))+u(z)\rho (E)(1-\rho (E))\\&\quad = u(x)\rho (E)^2+u(y)\big (\rho (E)+\rho (E^\mathrm{c})\\&\qquad -\rho (E)^2-\rho (E)\rho (E^\mathrm{c})\big )+u(z)(1-\rho (E))(1-\rho (E^\mathrm{c})), \end{aligned}$$

which yields $\big (u(y)-u(z)\big )\big (1-\rho (E)\big )\big (1-\rho (E)-\rho (E^\mathrm{c})\big )=0$. Since $u(y)>u(z)$ and $E$ is essential, we must have $1-\rho (E)-\rho (E^\mathrm{c})=0$. Hence $\rho (E)+\rho (E^\mathrm{c})=1$.

Case 2 $y\succsim \Psi (xEy)$.

Similarly, by (5), we have:

$$\begin{aligned}&u\big (\Psi _E(\Psi _E(x,z),y)\big )= u(\Psi _E(x,z))(1-\rho (E^\mathrm{c}))+u(y)\rho (E^\mathrm{c}) \\&\quad =u(x)\rho (E)(1-\rho (E^\mathrm{c}))+u(y)\rho (E^\mathrm{c})+u(z)\rho (E)(1-\rho (E))(1-\rho (E^\mathrm{c})), \end{aligned}$$

Again, because of (9), we get:

$$\begin{aligned}&u(x)\rho (E)^2+u(y)(1-\rho (E))+u(z)\rho (E)(1-\rho (E))\\&\quad =u(x)\rho (E)(1-\rho (E^\mathrm{c}))+u(y)\rho (E^\mathrm{c})+u(z)\rho (E)(1-\rho (E))(1-\rho (E^\mathrm{c})), \end{aligned}$$

which yields $\big (u(x)-u(y)\big )\big (1-\rho (E)-\rho (E^\mathrm{c})\big )$. Since $u(x)>u(y)$ and $E$ is essential, we must have $1-\rho (E)-\rho (E^\mathrm{c})=0$. Hence $\rho (E)+\rho (E^\mathrm{c})=1$.

1.2.2 Proof of Theorem 2: (ii) $\Rightarrow$ (i)

Because of Theorem 5, we only need to show that Axiom 9 holds. Let $E$ be in $\mathcal {U}$ and $x,x',y,y'$ be in $X$. Since $E$ is unambiguous event, we know that $\rho (E)+\rho (E^\mathrm{c})=1$. So

$$\begin{aligned} u\big (\Psi _E(\Psi _E(x,x'),\Psi _E(y,y'))\big )&= u(\Psi _E(x,x'))\rho (E))+u(\Psi _E(y,y'))\rho (E^\mathrm{c}) \\&= u(x)\rho (E)^2+u(x')\rho (E)\rho (E^\mathrm{c})\\&\quad +\,u(y)\rho (E)\rho (E^\mathrm{c})+u(y')\rho (E^\mathrm{c})^2 \end{aligned}$$

and

$$\begin{aligned} u\big (\Psi _E(\Psi _E(x,y),\Psi _E(x',y'))\big )&= u(\Psi _E(x,y))\rho (E))+u(\Psi _E(x',y'))\rho (E^\mathrm{c}) \\&= u(x)\rho (E)^2+u(x')\rho (E)\rho (E^\mathrm{c})\\&\quad +\,u(y)\rho (E)\rho (E^\mathrm{c})+u(y')\rho (E^\mathrm{c})^2, \end{aligned}$$

Therefore, $u\big (\Psi _E(\Psi _E(x,x'),\Psi _E(y,y'))\big )=u\big (\Psi _E(\Psi _E(x,y),\Psi _E(x',y'))\big )$. Hence $\Psi _E(\Psi _E(x,x'),\Psi _E(y,y'))\sim \Psi _E(\Psi _E(x,y),\Psi _E(x',y'))$ and Axiom 9 holds.

1.2.3 Proof of Theorem 3: (i) $\Rightarrow$ (ii)

Suppose that $\succsim$ satisfies condition (i) in Theorem 3. Note that Axiom 10 implies Axiom 9. We are left to show that for all $f\in \mathcal {F}, V(f)=\int _S u(f(s))\mathrm {d}P$ where $P$ is a probability measure on $\mathcal {U}$. Let $P:\mathcal {U}\rightarrow [0,1]$ be defined by $P(E)=\rho (E)$ for all $E\in \mathcal {U}$. We first show that $P$ is a probability measure on $\mathcal {U}$.

$$\begin{aligned} \Psi _G(x,x,y)&\sim \Psi _G(\Psi _A(z,z'),\Psi _A(w,w'),\Psi _A(w,z'))\nonumber \\&\sim \Psi _G(z,w,w)A\Psi _G(z',w',z') \end{aligned}$$

(11)

Since $u(x)=\rho (A)u(z)+(1-\rho (A))u(z')$, and $u(x)=\rho (A)u(w)+(1-\rho (A))u(w')$ and $u(y)=\rho (A)u(w)+(1-\rho (A))u(z')$, we have

$$\begin{aligned}&u(x)P(E\cup F)+ u(y)(1-P(E\cup F))=[u(z)P(E)+u(w)(1-P(E))] \rho (A)\\&\qquad +\,[u(z')(1-P(F))+u(w')P(F)](1-\rho (A))\\&\quad =[(u(z)-u(w))P(E)+u(w)]\rho (A)+[(u(w')-u(z'))P(F) +u(z')](1-\rho (A)) \end{aligned}$$

According to definition of $u(y)$, we can simplify the above further as

$$\begin{aligned}{}[u(x)-u(y)]P(E\cup F)=[u(z)-u(w)]P(E)\rho (A) +[u(w')-u(z')]P(F)(1-\rho (A)) \end{aligned}$$

Since $x\succ y$ and $u(x)-u(y)=\rho (A)[u(z)-u(w)]=(1-\rho (A))[u(w')-u(z')]$, we can conclude that $P(E\cup F)=P(E)+P(F)$. Hence $P$ is a probability measure on $\mathcal {U}$.

Let $G=\{E_1,\ldots ,E_n\}$ be a ground partition and $f_n=x_1E_1\ldots x_nE_n$. For $n=2, f_2\in \mathcal {F^U}$ and Theorem 2 gives the desired result. For using the mathematical induction, we thus assume the conclusion is true for all $n-1$ ground partition. Then it suffices to show that for $n$ ground partition

$$\begin{aligned} V(f_n)=\sum ^n_{i=1}u(x_i)P(E_i) \end{aligned}$$

where $P:\mathcal {U}\rightarrow [0,1]$ is such that $P(E)=\rho (E)$ for all $E\in \mathcal {U}$.

If $x_i\sim x_j$ for some $1\le i<j\le n$, then the desired result easily follows from $E_i\cup E_j\in \mathcal {U}$ by $\lambda$-system assumption of $\mathcal {U}$, and the hypothesis of the induction. Let $G'=\{E_1,\ldots ,E_{n-2},E_{n-1}\cup E_n\}$ and $G''=\{E_1\cup E_2, E_3,\ldots ,E_n\}$. So $G'$ and $G''$ are $n-1$ ground partition since $\mathcal {U}$ is a $\lambda$-system. Without loss of generality, we assume that $x_1\succ \ldots x_n$.

Below we have three cases to discuss:

Case 1 There is some $x,y\in X$ and $E\in \Sigma$ such that $xEx_2\sim x_1$ and $x_{n-1}Ey\sim x_n$.

It follows from Axiom 10 and monotonicity that

$$\begin{aligned} f_n&\sim \Psi _G(x_1,\ldots ,x_n)\\&\sim \Psi _G(\Psi _E(x,x_2),\Psi _E(x_2,x_2),\ldots ,\Psi _E(x_{n-1},x{n-1}),\Psi _E(x_{n-1},u)\\&\sim \Psi _G(x,x_2,\ldots ,x_{n-1},x_{n-1})E\Psi _G(x_2,x_2,\ldots ,x_{n-1},y)\\&\sim \Psi _{G'}(x,x_2,\ldots ,x_{n-1})E\Psi _{G''}(x_2,\ldots ,x_{n-1},y) \end{aligned}$$

So $f_n\sim \Psi _E(\Psi _{G'}(x,x_2,\ldots ,x_{n-1}),\Psi _{G''}(x_2,\ldots ,x_{n-1},y))$. Since $xEx-2\sim x_1$ and $x_{n-1}Ey\sim x_n$, Theorem 2 implies that

$$\begin{aligned}&u(x_1)=u(x)\rho (E)+u(x_2)(1-\rho (E))\quad \text { and }\quad u(x_n)=u(x_{n-1})\rho (E)\\&\qquad +\,u(y)(1-\rho (E)) \end{aligned}$$

Thus by the definition of $u(f_n)$ and the hypothesis of induction, we get

$$\begin{aligned} V(f_n)&= u(\Psi _E(\Psi _{G'}(x,x_2,\ldots ,x_{n-1}),\Psi _{G''}(x_2,\ldots ,x_{n-1},y)) \\&= \rho (E)u(\Psi _{G'}(x,x_2,\ldots ,x_{n-1}))+(1-\rho (E))u(\Psi _{G''}(x_2,\ldots ,x_{n-1},y))\\&= \rho (E)\left( \displaystyle \sum _{i=2}^{n-1}u(x_i)P(E_i)+u(x)P(E_1)+u(x_{n-1})P(E_{n-1})\right) \\&+(1-\rho (E))\left( \displaystyle \sum _{i=2}^{n-1}u(x_i)P(E_i)+u(x_2)P(E_i)+u(y)P(E_n)\right) \\&= \displaystyle \sum _{i=2}^{n-1}u(x_i)P(E_i)+P(E_1)(\rho (E)u(x)+(1-\rho (E))u(x_2))\\&\quad +\,P(E_n)(\rho (E)u(x_{n-1})+(1-\rho (E))u(y))\\&= \displaystyle \sum _{i=1}^n u(x_i)P(E_i) \end{aligned}$$

Case 2 Either there is some $x\in X$ and non-universal $E\in \Sigma$ such that $xEx_2\sim x_1$ or there is some $y\in X$ and non-null $F\in \Sigma$ such that $x_{n-1}Fy\sim x_n$.

A similar proof as in Case 1 applies when $xEx_2\sim x_1$ for some $x\in X$ and non-universal $E\in \Sigma$. Now suppose there exist $y$ and non-null $F$ satisfying $x_{n-1}Fy\sim x_n$. First notice that such $F$ is neither non-null nor universal. Then let $z\sim x_1Fx_2$ for some $z\in X$. So $x_1\succ z\succ x_2$. Thus by Axiom 10,

$$\begin{aligned} zE_1x_2E_2\ldots x_nE_n&\sim \Psi _F(x_1,x_2)E_1\Psi _F(x_2,x_2)E_2\ldots \Psi _F(x_n,x_n)E_n \\&\sim \Psi _G(x_1,\ldots ,x_n)E\Psi _G(x_2,x_2,\ldots ,x_n)\\&\sim \Psi _G(x_1,x_2,\ldots ,x_n)F\Psi _{G''}(x_2,x_3,\ldots ,x_n) \end{aligned}$$

Since $f_n\sim \Psi _G(x_1,\ldots ,x_n)$, Theorem 2 implies

$$\begin{aligned} u(\Psi _G(z,x_2,\ldots ,x_n))=\rho (F)u(\Psi _G(x_1,\ldots ,x_n))+(1-\rho (F))u(\Psi _{G''}(x_2,\ldots ,x_n)) \end{aligned}$$

Note that by case 1,

$$\begin{aligned} u(\Psi _G(z,x_2,\ldots ,x_n))=P(E_1)u(z)+\displaystyle \sum ^n_{i=2}u(x_i)P(E_i) \end{aligned}$$

Also by Theorem 2, $u(z)=\rho (F)u(x_1)+(1-\rho (F))u(x_2))$. Therefore, we rearrange the above equation and derive:

$$\begin{aligned}&\rho (F)V(f_n)\\&\quad = P(E_1)u(z)+\displaystyle \sum ^n_{i=2}u(x_i)P(E_i) -(1-\rho (F))\left\{ P(E_1)u(x_2)+\sum ^n_{i=2}u(x_i)P(E_i)\right\} \\&\quad = \rho (F)\displaystyle \sum ^n_{i=1}u(x_i)P(E_i) \end{aligned}$$

Therefore we obtain the desired result.

Case 3 There is no $x,y\in X$, non-universal $E\in \Sigma$, and non-null $F\in \Sigma$ such that $xEx_2\sim x_1$ or $x_{n-1}Fy\sim x_n$.

Now let $z\sim x_1Ex_2$ for some $E\in \Sigma$ which is neither null nor universal. Then $x_1\succ z\succ x_2$. Applying the result in Case 2, we can get the desired result.

1.2.4 Proof of Theorem 3: (ii) $\Rightarrow$ (i)

Suppose Condition (ii) holds. Then $\succsim$ satisfies Axiom 1–5$^*$ is straightforward. The detailed proofs are referred to Alon and Schmeidler (2014). We are only left to show that Axiom 10 holds. Let $\Omega =\{A_1,\ldots ,A_n\}$ be a ground partition. Then $x_1A_1\ldots x_nA_n$ and $y_1A_1\ldots x_nA_n$ have SEU representation where the probability is consistent with capacity on $\Omega$. So $V(x_1A_1\ldots x_nA_n)=\sum _iu(x_i)\rho (A_i)$ and $V(y_1A_1\ldots y_nA_n)=\sum _iu(y_i)\rho (A_i)$. Therefore,

$$\begin{aligned}&V(\Psi _\Omega (x_1,\ldots ,x_n)E\Psi _\Omega (y_1,\ldots ,y_n))\\&\quad =u(\Psi _\Omega (x_1,\ldots ,x_n))\rho (E)+ u(\Psi _\Omega (y_1,\ldots ,y_n)(1-\rho (E))\\&\quad =\sum _i u(x_i)\rho (A_i)\rho (E)+\sum _iu(y_i)\rho (A_i)(1-\rho (E))\\&\quad = \sum _i V(x_iEy_i)\rho (A_i)\\&\quad = V(\Psi _E(x_1,y_1)A_1\ldots \Psi _E(x_n,y_n)A_n) \end{aligned}$$

Therefore, Axiom 10 holds.

1.2.5 Proof of Corollary 1

Condition (ii) implies Condition (i) is similar to the proof of Theorem 3. We only prove that Condition (i) implies (ii). Since MEU preference is also a biseparable preference, capacity $\rho$, defined by $\rho (E)=\min _{P\in K}P(E)$ for all $E\in \mathcal {U}$, is complementarily additive on each $E\in \mathcal {U}$. That is, $\min _{P\in K}P(E)+\min _{P\in K}P(E^\mathrm{c})=1$. Hence for all $E\in \mathcal {U}$ and all $P,Q\in K, P(E)=Q(E)$.

1.2.6 Proof of Corollary 2

We already know that $\succsim$ satisfies 1,2,3,$4^*,4^{**}$, and 5.1 iff $\succsim$ has a CEU representation. Since CEU preference is also a biseparable preference, by Theorem 3, it has a SEU representation on $\mathcal {F^U}$.

1.3 Proof of Section 5

1.3.1 Proof of Theorem 4

Let the consequences $x,x',y,y',z,z'$ satify the condition of Definition 3. Suppose that $xAz'\sim zAy'$ and $zAx'\sim yAz'$. Since preference is biseparable, by representation we have

$$\begin{aligned} u(x)\rho (A)+u(z')(1-\rho (A))&= u(z)(1-\rho (A^\mathrm{c}))+u(y')\rho (A^\mathrm{c})\\ u(z)(1-\rho (A^\mathrm{c}))+u(x')\rho (A^\mathrm{c})&= u(y)\rho (A)+u(z')(1-\rho (A)) \end{aligned}$$

Adding them together yields

$$\begin{aligned} u(x)\rho (A)+u(x')\rho (A^\mathrm{c})=u(y)\rho (A)+u(y')\rho (A^\mathrm{c}) \end{aligned}$$

(12)

Since event $A$ is unambiguous, by definition $xAx'\sim yAy'$, which implies

$$\begin{aligned} u(x)\rho (A)+u(x')(1-\rho (A))=u(y)\rho (A)+u(y')(1-\rho (A)) \end{aligned}$$

Hence, $[u(x')-u(y')(1-\rho (A)-\rho (A^\mathrm{c})=0$. Since $x'\succ y'$ by assumption, we must have $1-\rho (A)-\rho (A^\mathrm{c})=0$. Thus $\rho$ is complementarily additive on event $A$.