Aggregation of Bayesian preferences: unanimity vs monotonicity

Abstract

This article reconsiders the issue of Bayesian aggregation by pointing at a conflict that may arise between two logically independent dominance criteria, Pareto dominance and statewise dominance, that are commonly imposed on social preferences. We propose a weaker dominance axiom that restricts statewise dominance to Pareto dominant alternatives and, symmetrically, Pareto dominance to statewise dominant alternatives. The associated aggregation rule is a convex combination of two components, the first being a weighted sum of the individuals’ subjective expected utility (SEU) functional, the second being a social SEU functional, with associated social utility function and social belief. Such representation establishes the existence of a trade off between adherence to the Pareto principle and compliance with statewise dominance. We then investigate what are the consequences of adding to our assumptions either of the two dominance criteria in their full force and obtain that each of them is equivalent to discarding the other, unless there is essentially a common prior probability across individuals.

Appendix: proofs of all results

1.1 Four useful lemmata

Lemma 1

Consider an integer $P\ge 1$, and a mapping $\varphi _p:\mathcal {F}\rightarrow \mathbb {R}$ for any $p\in [0\ldots P]$. Suppose that each $\varphi _p$ is mixture affine; that is, for any $f,g\in \mathcal {F}$ and any $\alpha \in [0,1]$, $\varphi _p(\alpha f+(1-\alpha )g)=\alpha \varphi _p( f)+(1-\alpha )\varphi _p(g)$. Moreover, suppose that the $\varphi _p$ are related by a Pareto condition; that is, for any $f,g\in \mathcal {F}$, if $\varphi _p(f)\ge \varphi _p(g)$ for any $p\in [1\ldots P]$, then $\varphi _0(f)\ge \varphi _0(g)$.

Then, there exist non-negative numbers $\alpha _1,\ldots ,\alpha _P$ and a real number $\beta$ such that, for any $f\in \mathcal {F}$, we have

$$\begin{aligned} \varphi _0(f)\ =\ \sum _{p=1}^P\ \alpha _p\cdot \varphi _p(f)\ +\ \beta . \end{aligned}$$

(5)

Proof

Define a mapping $\Phi : \mathcal {F} \rightarrow \mathbb {R}^{P+1}$ by, for any $f \in \mathcal {F}$, $\Phi (f)=(\varphi _0(f),\ldots ,\varphi _P(f))$. Let $K\subseteq \mathbb {R}^{P+1}$ denote the range of $\Phi$. Since $\varphi _p$ is mixture affine for any $p\in [0\ldots P]$, the subset K must be convex. Moreover, since the $\varphi _p$ are related by a Pareto condition, we can invoke Proposition 1 in De Meyer and Mongin (1995) and finally obtain $\alpha _1,\ldots ,\alpha _P$ and $\beta$ as in Eq. (5). $\square$

Lemma 2

Let the structure $\{\gamma ,(\alpha _i)_{i\in N},u_0,\lambda _0\}$ provide a representation of $\succsim _0$ as in Theorem 1 with $\gamma >0$. Then, for any $i\in N$, i is null if and only if $\alpha _i=0$.

Proof

First, suppose that i is null. Then, there exist $f,g\in \mathcal {F}$ such that: (a) $f\succ _ig$, (b) $f\sim _jg$ for any $j\in N{\setminus } \{i\}$, (c) $f(s)\sim _0g(s)$ for any $s\in S$, and yet (d) $f\sim _0g$. Since the structure $\{\gamma ,(\alpha _i)_{i\in N},u_0,\lambda _0\}$ provide a representation of $\succsim _0$ as in Theorem 1, and by Conditions (b), (c) and (d),

$$\begin{aligned} 0\ =\ V_0(f)-V_0(g)\ =\ \gamma \cdot \alpha _i\cdot [V_i(f)-V_i(g)] \end{aligned}$$

By Condition (a), we have $V_i(f)-V_i(g)>0$. Since $\gamma >0$ by assumption, we must have $\alpha _i=0$. Now, suppose that $\alpha _i=0$. By Risk Diversity, there are $l,m\in Y$ such that $l\succ _im$ and $l\sim _jm$ for any $j\in N{\setminus }\{i\}$. Let $s\in S$ be a state such that $\lambda _i(s)>0$. Define $f,g\in \mathcal {F}$ by $g(s')=m$ for any $s'\in S$, and $f(s')=l$ if $s'=s$ and $f(s')=m$ otherwise. By construction, we have $f\sim _jg$ for any $j\in N{\setminus }\{i\}$, and $f\sim _ig$. Moreover, since $\gamma >0$ and $\alpha _i=0$, we have $u_0=\sum _{j\ne i}\alpha _ju_j$ by (Th1.1). So $u_0(f(s))=u_0(g(s))$ and, therefore, $f(s)\sim _0g(s)$ for any $s\in S$. But then we must also have $\mathbb {E}_{\lambda _0}(u_0\circ f)=\mathbb {E}_{\lambda _0}(u_0\circ g)$. Finally, by (Th1.2), we also have for any $h\in \mathcal {F}$

$$\begin{aligned} V_0(h)\ =\ \gamma \ \cdot \sum _{j\in N{\setminus }\{i\}}\alpha _j\cdot V_j(h)\ +\ (1-\gamma )\cdot \mathbb {E}_{\lambda _0}(u_0\circ h). \end{aligned}$$

Therefore, $V_0(f)=V_0(g)$, and $f\sim _0g$. This shows that i is null. $\square$

Lemma 3

Assume Risk Diversity. For any function $\psi :N\times S\rightarrow \mathbb {R}$ such that, for any $f\in \mathcal {F}$,

$$\begin{aligned} \sum _{i\in N,s\in S} \psi (i,s) \cdot u_i(f(s))=0, \end{aligned}$$

(6)

we have $\psi (i,s)=0$ for any $i\in N$ and $s\in S$.

Proof

Fix $i\in N$ and $s\in S$. We construct $f,g\in \mathcal {F}$ such that $f(t)\sim _j g(t)$ for any $(j,t)\in N\times S$ with $(j,t)\ne (i,s)$, and $f(s)\succ _i g(s)$. By Risk Diversity, there exist $l,m \in Y$ such that $l\succ _i m$ and $l\sim _{j}m$. for any $j\in N{\setminus }\{i\}$. Let $g \in \mathcal {F}$ be such that $g(t)=m$ for any $t\in S$. Let $f \in \mathcal {F}$ be such that $f(t)=l$ if $t=s$ and $f(t)=m$ otherwise. These two acts satisfy the conditions above. Then, by Eq. (6), we have

$$\begin{aligned} 0= & {} \sum _{j\in N,t\in S} \psi (j,t)\cdot u_j(f(t))\ \\&-\ \sum _{j\in N,t\in S} \psi (j,t)\cdot u_j(g(t))=\psi (i,s)\cdot [u_i(f(s))-u_i(g(s))]. \end{aligned}$$

But by construction we have $u_i(f(s))>u_i(g(s))$. So it must be that $\psi (i,s)=0$. This holds for any $i\in N$ and $s\in S$. $\square$

We say that $\succsim _0$ satisfies STP if for any $E\subseteq S$ and any $f,g,h,k\in \mathcal {F}$, we have $f_Eh\succsim _0g_Eh$ iff $f_Ek\succsim _0g_Ek$.

Lemma 4

If $\succsim _0$ satisfies completeness, transitivity, STP and State Independence, then it satisfies Statewise Dominance.

Proof

Suppose that $\succsim _0$ satisfies STP and State Independence. We have:

• Claim 1: For any $s\in S$, $f,g\in \mathcal {F}$, and $l,m\in Y$, if $l_sf\succ _0 m_sf$, then $l_tg\succ _0 m_tg$ for any nonnull state $t\in S$.

Indeed, suppose $l_sf\succ _0 m_sf$ for some $s\in S$, $f\in \mathcal {F}$ and $l,m\in Y$. Let $t\in S$ be a nonnull state. Then, by State Independence, $l_tf\succ _0 m_tf$. Finally, by STP, we obtain $l_tg\succ _0 m_tg$. $\square$

• Claim 2: For any $s\in S$, $f\in \mathcal {F}$, and $l,m\in Y$, if $l\succsim _0 m$, then $l_sf\succsim _0 m_sf$.

Let $S^*:=\{s_1,\ldots s_N\}$ be the set of nonnull states in S. Consider $s\in S$, $f\in \mathcal {F}$, and $l,m\in Y$ such that $l_sf\succ _0 m_sf$. We will show that $l\succ _0m$. By Claim 1, we obtain:

$$\begin{aligned} l_tg\succ _0 m_tg\quad \text {for any}\quad t\in S^*\quad \text {and any}\quad g\in \mathcal {F}. \end{aligned}$$

(7)

In particular, for $t=s_1$ and $g=l$, we get $l\succ _0 m_{s_1}l$.

Now, note that $m_{s_1}l=l_{s_2}(m_{s_1}l)$. By applying (7) to $t=s_2$ and $g=m_{s_1}l$, we get $m_{s_1}l=l_{s_2}(m_{s_1}l)\succ _0m_{s_2}(m_{s_1}l)=m_{\{s_1,s_2\}}l$. By the conclusion of the previous paragraph, $l\succ _0 m_{\{s_1,s_2\}}l$.

By repeating iteratively this process, we obtain $l\succ _0 m_{S^*}l$. Now, let $S{\setminus } S^*=\{t_1,\ldots , t_M\}$. Note that $m_{S^*}l$ and $m_{S^*\cup \{t_1\}}l$ are equal to each other except on the null state $t_1$. So $l\succ _0m_{S^*}l\sim _0m_{S^*\cup \{t_1\}}l$. Moreover, $m_{S^*\cup \{t_1\}}l$ and $m_{S^*\cup \{t_1,t_2\}}l$ are equal to each other except on the null state $t_2$. So $m_{S^*\cup \{t_1\}}l\sim _0m_{S^*\cup \{t_1,t_2\}}l$ and, therefore, $l\succ _0m_{S^*\cup \{t_1,t_2\}}l$. By repeating iteratively this process, we finally obtain $l\succ _0 m_Sl=m$, as desired. $\square$

• Claim 3:$\succsim _0$ satisfies Statewise Dominance.

Let $S=\{s_1,\ldots ,s_N\}$. Let $f,g\in \mathcal {F}$ be such that $f(s)\succsim _0g(s)$ for any $s\in S$. Define $f_0=f$ and, for any $n\in [1\ldots N]$, $f_n=g(s_n)_{s_n}f_{n-1}$. For any $n\in [1\ldots N]$, we clearly have $f_{n-1}(s_n)=f(s_n)\succsim _0 g(s_n)$ so, by Claim 2, $f_{n-1}(s_n)_{s_n}f_{n-1}\succsim _0g(s_n)_{s_n}f_{n-1}$; that is, $f_{n-1}\succsim _0f_n$. Then, $f_0\succsim _0 f_N$; that is, $f\succsim _0g$. $\square$

Claim 3 completes the proof of the lemma. $\square$

1.2 Proof of Theorem 1

The necessity of Weak Order, Continuity and Independence is straightforward. So we only prove the necessity of Weak Dominance, Boundedness and Non-triviality. To do so, assume a representation as in Theorem 1. Since $u_0$ is normalized, we have $u_0(l_1)>u_0(l_0)$. Moreover, by confronting (Th1.1) and (Th1.2), we see that $V_0(l')=u_0(l')$ for any $l'\in Y$. Thus, $V_0(l_1)>V_0(l_0)$. But since $V_0$ represents social preferences, we finally obtain $l_1\succ _0 l_0$. To show Weak Dominance, assume that $f, g\in \mathcal {F}$ satisfy $f \succsim _i g$ for any $i\in N$ and $f(s) \succsim _0 g(s)$ for any $s\in S$. Then, we have $\mathbb {E}_{\lambda _i}(u_i\circ f)\ge \mathbb {E}_{\lambda _i}(u_i\circ g)$ for any $i\in N$ and $V_0(f(s))\ge V_0(g(s))$ for any $s\in S$. Since we have proved that $u_0$ is the restriction of $V_0$ to lotteries, we have $u_0(f(s))\ge u_0(g(s))$ and, therefore, $\mathbb {E}_{\lambda _0}(u_0\circ f)\ge \mathbb {E}_{\lambda _0}(u_0\circ g)$. By Eq. (2), this is sufficient to obtain $V_0(f)\ge V_0(g)$ and, finally, $f\succsim _0g$. Hence Weak Dominance. Finally, to show Boundedness, consider an arbitrary act $f\in \mathcal {F}$. If $\gamma =0$, then $\succsim _0$ satisfies Satatewise Dominance. Moreover, the range of f is finite. So there exist $l,m\in Y$ such that $l\succsim _0f(s)\succsim _0m$. Then, by Statewise Dominance, we have $l\succsim _0f\succsim _0m$. If $\gamma >0$, then by (Th1.1) we have $u_0=\sum _{i\in N}\alpha _iu_i$. Since the range of f is finite, we can apply c-Agreement and obtain $l,m\in Y$ such that $l\succsim _if(s)\succsim _im$ for any $s\in S$ and $i\in N$. Since individual preferences satisfy Statewise Dominance, we obtain $l\succsim _if\succsim _im$ for any $i\in N$. Moreover, we have $u_i(l)\ge u_i(f(s))\ge u_i(m)$ for any $s\in S$ and $i\in N$. Since $u_0=\sum _{i\in N}\alpha _iu_i$, we obtain $u_0(l)\ge u_0(f(s))\ge u_0(m)$ for any $s\in S$. Therefore, $l\succsim _0f(s)\succsim _0m$ for any $s\in S$. Then, we can apply Weak Dominance since we have already proved that it is necessary, and get $l\succsim _0f\succsim _0m$.

From now on, we assume that social preferences satisfy Weak Order, Continuity, Independence, Boundedness, Weak Dominance and Non-triviality. Observe that Weak order, Continuity and Independence imply that the restriction of $\succsim _0$ to lotteries satisfies the von Neumann and Morgenstern axioms. Then, by their theorem, there exists a mixture affine function $u_0:Y\rightarrow \mathbb {R}$ that provides a representation for the restriction of $\succsim _0$ to lotteries. Moreover, by Non-triviality, we must have $u_0(l_1)>u_0(l_0)$. By applying positive affine transformations if necessary, we may suppose without loss of generality that $u_0(l_1)=1$ and $u_0(l_0)=0$. Thus, $u_0$ is normalized.

Now, to construct a functional $V_0$ providing a representation for $\succsim _0$, we first show that each act $f \in \mathcal {F}$ has a certainty equivalent for $\succsim _0$. That is, for any $f \in \mathcal {F}$, there exists $l \in Y$ such that $f \sim _0 l$. Fix $f\in \mathcal {F}$. By Boundedness, there are $l,m\in Y$ such that $l\succsim _0f\succsim _0m$. Then, the construction of a certainty equivalent follows from standard arguments, which we only briefly sketch. First, if $f\sim _0 l$ or $f\sim _0 m$, we are done. So we assume without loss of generality that $l\succ _0 f\succ _0m$. Then, the sets

$$\begin{aligned} \{ \alpha \in [0,1],\quad \alpha l + (1- \alpha )m \succ _0 f \} \quad \text {and}\quad \{ \alpha \in [0,1],\quad f\succ _0\alpha l + (1- \alpha )m\} \end{aligned}$$

are of the form $(\alpha ,1]$ and $[0,\alpha )$ respectively, for some $\alpha \in (0,1)$. Then, the lottery $p\in Y$ defined by $p=\alpha l+(1-\alpha )m$ satisfies $f\sim _0p$.

Next, define the function $V_0: \mathcal {F} \rightarrow \mathbb {R}$ by, for any $f\in \mathcal {F}$:

$$\begin{aligned} V_0(f) = u_0(l) \end{aligned}$$

(8)

where l is a certainty equivalent associated to f. Clearly, $V_0(f)$ is well-defined, independent of the choice of a specific certainty equivalent since $u_0$ represents the restriction of $\succsim _0$ to Y. Moreover, $V_0$ represents social preferences; that is, for any $f,g\in \mathcal {F}$, $f \succsim _0 g$ if and only if $V_0(f) \ge V_0(g)$. Last, it is simple to use the axiom of Independence to see that $V_0$ is mixture-linear on $\mathcal {F}$.

Note that the functional $V_i$ is mixture affine for any $i\in N$. So is the functional $f\rightarrow u_0(f(s))$ for any $s\in S$. Let k be the cardinalty of S. Since Weak Dominance holds, Lemma 1 provides non-negative numbers $(\sigma _j)_{j=1}^{n+k} \in \mathbb {R}^{n+k}$ and a real number $\mu \in \mathbb {R}$ such that, for any $f\in \mathcal {F}$,

$$\begin{aligned} V_0(f)= \sum _{i\in N}\sigma _i \cdot V_i(f) +\sum _{s\in S} \sigma _s\cdot u_0(f(s))+ \mu . \end{aligned}$$

By assumption, we have $V_i(l_1)=u_i(l_1)=1$ and $V_i(l_0)=u_i(l_0)=0$ for any $i\in N$. Meanwhile, by construction, we have $V_0(l_1)=u_0(l_1)=1$ and $V_0(l_0)=u_0(l_0)=0$. Then, it must necessarily be that $\sigma :=\sum _{j \in N \cup S} \sigma _j =1$ and $\mu =0$. Next, define $\gamma = \sum _{i\in N}\sigma _i$. There are now different cases:

• Case 1: $\gamma =0$. Then, set $\lambda _0(s)=\sigma _s$ for $s\in S$. This defines a probability measure $\lambda _0$ on S, and we obtain $V_0(f)=\mathbb {E}_{\lambda _0}(u_0\circ f)$. Hence the representation of Theorem 1.

• Case 2: $\gamma =1$. Then, set $\alpha _i = \sigma _i$ for $i\in N$. These numbers $\alpha _i$ are non-negative and sum up to 1. We obtain $V_0(f)=\sum _{i\in N}\alpha _i\cdot \mathbb {E}_{\lambda _i}(u_i\circ f)$. Hence the representation of Theorem 1.

• Case 3: $\gamma \in (0,1)$. Then, set $\lambda _0(s)=\sigma _s/(1-\gamma )$ for $s\in S$ and $\alpha _i = \sigma _i/\gamma$ for $i\in N$ to obtain the representation of Theorem 1.

Thus, in all three cases, we have

$$\begin{aligned} V_0(h) = \gamma \cdot \sum _{i\in N}\alpha _i\cdot \mathbb {E}_{\lambda _i}(u_i\circ h)+(1-\gamma )\cdot \mathbb {E}_{\lambda _0}(u_0\circ h). \end{aligned}$$

Given that $u_0$ is the restriction of $V_0$ to lotteries, applying the latter equation to the case of a constant act establishes (Th1.2).

1.3 Proof of Theorem 2

The representation in Theorem 2 is a particular case of the representation from Theorem 1. Hence, it implies all of Weak Order, Continuity, Independence, Boundedness, and Non-triviality. Moreover, since $\lambda _0$ to be a convex combination of $(\lambda _i)_{i\in N}$, there exists a collection of weakly positive weights $(\beta _i)_{i\in N}$ summing up to 1 such that $\lambda _0=\sum _{i\in N}\beta _i\cdot \lambda _i$. Then, with the notations form Theorem 1, for every $f\in \mathcal {F}$, we have

$$\begin{aligned} V_0(f)=\gamma \cdot \sum _{i\in N}\alpha _i\cdot \mathbb {E}_{\lambda _i}(u_i\circ f)+(1-\gamma )\cdot \sum _{i\in N}\beta _i\cdot \mathbb {E}_{\lambda _i}(u_0\circ f). \end{aligned}$$

Hence, Belief-Adjusted Weak Dominance easily follows from the representation.

As for the sufficiency of the axioms, we proceed exactly as in the proof of Theorem 1 and obtain a normalized and mixture affine function $u_0$ from Y to $\mathbb {R}$ and a functional $V_0$ from $\mathcal {F}$ to $\mathbb {R}$ extending $u_0$ and providing a representation of $\succsim _0$. Moreover, by Belief-Adjusted Weak Dominance and Lemma 1, we obtain two collections of non-negative numbers $(\sigma _i)_{i\in N}$ and $(\tau _i)_{i\in N}$ summing up to 1 such that, for every $f\in \mathcal {F}$,

$$\begin{aligned} V_0(f)=\sum _{i\in N}\sigma _i\cdot \mathbb {E}_{\lambda _i}(u_i\circ f)+\sum _{i\in N}\tau _i\cdot \mathbb {E}_{\lambda _i}(u_0\circ f). \end{aligned}$$

Now, let $\sigma =\sum _{i\in N}\sigma _i$. We split the proof in three different cases:

• Case 1: $\sigma =0$. Then, define $\lambda _0=\sum _{i\in N}\tau _i\cdot \lambda _i$, a convex combination of $(\lambda _i)_{i\in N}$. The structure $\{0,(\alpha _i)_{i\in N},u_0,\lambda _0\}$ provides a representation of $\succsim _0$, with a totally arbitrary family $(\alpha _i)_{i\in N}$.

• Case 2: $\sigma =1$. Then define $\lambda _0$ as an arbitrary convex combination of $(\lambda _i)_{i\in N}$. The structure $\{1,(\sigma _i)_{i\in N},u_0,\lambda _0\}$ provides a representation of $\succsim _0$.

• Case 3: $\sigma \in (0,1)$. Then, define $\alpha _i=\sigma _i/\sigma$ for evey $i\in N$ and $\lambda _0=\sum _{i\in N}\tau _i/(1-\sigma )\cdot \lambda _i$, a convex combination of $(\lambda _i)_{i\in N}$. The structure $\{\sigma ,(\alpha _i)_{i\in N},u_0,\lambda _0\}$ provides a representation of $\succsim _0$.

1.4 Proof of Proposition 1

• Claim 1: If $\succsim _0$ satisfies Pareto Dominance and if $\gamma \in (0,1)$, then the social prior is essentially a common prior, and $u_0$ is an adapted convex combination of $\{u_i,\, i\in N\}$.

Since $\gamma >0$, we already know from (Th1.1) that $u_0$ is a convex combination of $\{u_i,\, i\in N\}$ with coefficients given by $\{\alpha _i,\, i\in N\}$. By Lemma 2, this convex combination is adapted. Moreover, note that $V_i$ is mixture affine for any $i\in N_*$. Since Pareto Dominance holds, Lemma 1 provides non-negative numbers $\beta _i$, $i\in N$, and $\mu \in \mathbb {R}$ such that, for any $f\in \mathcal {F}$,

$$\begin{aligned} V_0(f)\ =\ \sum _{i\in N}\beta _i\cdot V_i(f)\ + \ \mu . \end{aligned}$$

If we apply this to $f=l_1,l_0$, we obtain $\mu =0$ and $\sum _{i\in N}\beta _i=1$ since $u_0$ is normalized and since it is the restriction of $V_0$ to lotteries. But since $\gamma >0$, by (Th1.1) we have $u_0\ =\ \sum _{i\in N}\alpha _i\cdot u_i$. So we must have

$$\begin{aligned} \sum _{i\in N}\alpha _i\cdot u_i\ =\ u_0\ =\ \sum _{i\in N}\beta _i\cdot u_i \end{aligned}$$

Fix $i\in N$ and let $l,m\in Y$ be as in Risk Diversity. Then, since $l\sim _jm$ for any $j\in N{\setminus }\{i\}$

$$\begin{aligned} 0= & {} \sum _{j\in N}(\alpha _j -\beta _j)\cdot u_j(l)\ -\ \sum _{i\in N}(\alpha _j-\beta _j)\cdot u_j(m) \\= & {} (\alpha _i -\beta _i)\cdot (u_i(l)-u_i(m)) \end{aligned}$$

Since by construction $l\succ _i m$, we must have $\alpha _i=\beta _i$. Therefore, for any $f\in \mathcal {F}$, we obtain

$$\begin{aligned} V_0(f)\ =\ \sum _{i\in N}\alpha _i\cdot V_i(f)\ \end{aligned}$$

(9)

On the one hand, by Eq. (2) in (Th1.2), we have for any act $f\in \mathcal {F}$,

$$\begin{aligned} V_0(f)=\sum _{i\in N,s\in S}[\gamma \cdot \alpha _i\cdot \lambda _i(s)+(1-\gamma )\cdot \alpha _i\cdot \lambda _0(s)]\cdot u_i(f(s)). \end{aligned}$$

(10)

On the other hand, by Eq. (9)

$$\begin{aligned} V_0(f)=\sum _{i\in N,s\in S}\alpha _i\cdot \lambda _i(s)\cdot u_i(f(s)). \end{aligned}$$

(11)

Then, define $\psi (i,s)=[\gamma \alpha _i\lambda _i(s)+(1-\gamma )\alpha _i\lambda _0(s)]-\alpha _i\lambda _i(s)$ for any $i\in N$ and $s\in S$. By confronting Eqs. (10) and (11) and applying Lemma 3, we obtain:

$$\begin{aligned} \gamma \cdot \alpha _i\cdot \lambda _i(s)+(1-\gamma )\cdot \alpha _i\cdot \lambda _0(s)=\alpha _i\cdot \lambda _i(s). \end{aligned}$$

(12)

Rearranging Eq. (12), and using $\gamma <1$, we obtain that $\lambda _i=\lambda _0$ for any $i\in N$ such that $\alpha _i>0$. But by Lemma 2, $i\in N$ is non-null if and only if $\alpha _i>0$. So for any non-null individual $i\in N$, we have $\lambda _i=\lambda _0$. Hence, the social prior is essentially a common prior. $\square$

• Claim 2: If $\succsim _0$ satisfies Pareto Dominance and if $\gamma =0$, then the social prior is essentially a common prior, and $u_0$ is an adapted convex combination of $\{u_i,\, i\in N\}$

Note that $V_i$ is mixture affine for any $i\in N_*$. Since Pareto Dominance holds, Lemma 1 provides non-negative numbers $\beta _i$, $i\in N$, and $\mu \in \mathbb {R}$ such that, for any $f\in \mathcal {F}$,

$$\begin{aligned} V_0(f)\ =\ \sum _{i\in N}\beta _i\cdot V_i(f)\ + \ \mu . \end{aligned}$$

If we apply this to $f=l_1,l_0$, we obtain $\mu =0$ and $\sum _{i\in N}\beta _i=1$ since $u_0$ is normalized and since it is the restriction of $V_0$ to lotteries. Thus, we obtain, for any $f\in \mathcal {F}$,

$$\begin{aligned} V_0(f)\ =\ \sum _{i\in N,s\in S}\beta _i\cdot \lambda _i(s)\cdot u_i(f(s)). \end{aligned}$$

(13)

In particular, we have $u_0=\sum _{i\in N}\beta _i\cdot u_i$. This shows that $u_0$ is a convex combination of individual utilities. Moreover, Eq. (13) provides another representation of $\succsim _0$ as in Theorem 1. By applying Lemma 2 to this other representation, $i\in N$ is non-null if and only if $\beta _i>0$. So the convex combination is adapted. Since $\gamma =0$, Eq. (2) in (Th1.2) gives, for any $f\in \mathcal {F}$,

$$\begin{aligned} V_0(f)\ =\ \sum _{i\in N,s\in S}\beta _i\cdot \lambda _0(s)\cdot u_i(f(s)). \end{aligned}$$

(14)

Then, define $\psi (i,s)=\beta _i\lambda _i(s)-\beta _i\lambda _0(s)$ for any $i\in N$ and $s\in S$. By confronting Eqs. (13) and (14) and applying Lemma 3, we obtain for any $i\in N$ and $s\in S$:

$$\begin{aligned} \beta _i\cdot \lambda _i(s)\ =\ \beta _i\cdot \lambda _0(s). \end{aligned}$$

(15)

So if $\beta _i>0$, then $\lambda _i=\lambda _0$. Then, if $i\in N$ is non-null, we have $\beta _i>0$ and therefore $\lambda _i=\lambda _0$. The social prior is essentially a common prior. $\square$

Combining Claims 1 and 2, we obtain that if Pareto Dominance holds, then either $\gamma =1$, or the social prior is essentially a common prior, both when $\gamma >0$ (Claim 1) and when $\gamma =0$ (Claim 2). Moreover, $u_0$ is always an adapted convex combination of $\{u_i,\, i\in N\}$. Reciprocally, assume that $\gamma =1$, or the social prior is essentially a common prior and $u_0$ is an adapted convex combination of $\{u_i,\, i\in N\}$. We show Pareto Dominance. If $\gamma =1$, this is trivial given Eq. (2). So we assume that $\gamma <1$.

• Case 1: $\gamma >0$. Then, by (Th1.2) we have $u_0=\sum _{i\in N}\alpha _iu_i$. Let I stand for the set of $i\in N$ that are non-null. Then, by Lemma 2, $I=\{i\in N,\, \alpha _i>0\}$. Since the social prior is essentially a common prior, we have that if $i\in I$, then $\lambda _i=\lambda _0$. Then, Eq. (2) gives for any $f\in \mathcal {F}$,

  $$\begin{aligned} V_0(f)= & {} \gamma \cdot \sum _{i\in N}\alpha _i\cdot \mathbb {E}_{\lambda _i}(u_i\circ f)+ (1-\gamma )\cdot \sum _{i\in N}\alpha _i\cdot \mathbb {E}_{\lambda _0}(u_i\circ f)\\= & {} \gamma \cdot \sum _{i\in N}\alpha _i\cdot \mathbb {E}_{\lambda _i}(u_i\circ f)+ (1-\gamma )\cdot \sum _{i\in I}\alpha _i\cdot \mathbb {E}_{\lambda _i}(u_i\circ f)\\= & {} \gamma \cdot \sum _{i\in N}\alpha _i\cdot \mathbb {E}_{\lambda _i}(u_i\circ f)+ (1-\gamma )\cdot \sum _{i\in N}\alpha _i\cdot \mathbb {E}_{\lambda _i}(u_i\circ f)\\= & {} \sum _{i\in N}\alpha _i\cdot \mathbb {E}_{\lambda _i}(u_i\circ f). \end{aligned}$$

  Then, it is straightforward to see that Pareto Dominance holds.

• Case 2: $\gamma =0$. Let I stand for the set of $i\in N$ that are non-null. By assumption, there exist non-negative coefficients $(\beta _1,\ldots ,\beta _n)$ summing to 1 such that $u_0=\sum _{i\in N}\beta _iu_i$ and $I=\{i\in N,\, \beta _i>0\}$. Proceeding as in Case 1, we obtain $V_0(f)=\sum _{i\in N}\beta _i\cdot \mathbb {E}_{\lambda _i}(u_i\circ f)$ for any $f\in \mathcal {F}$. Hence Pareto Dominance. This completes the proof of (P1.1).

• Claim 3: If $\succsim _0$ satisfies Statewise Dominance and if $\gamma >0$, then there is essentially a common prior.

For any $s\in S$, the mapping $f\rightarrow u_0(f(s))$ is mixture affine; so is $V_0$. Since Statewise Dominance holds, Lemma 1 gives the existence of non-negative numbers $\mu _0(s)$, $s\in S$, and $\mu \in \mathbb {R}$ such that, for any $f\in \mathcal {F}$,

$$\begin{aligned} V_0(f)\ =\ \sum _{s\in S}\mu _0(s)\cdot u_0(f(s))\ + \ \mu . \end{aligned}$$

If we apply this to $f=l_1,l_0$, we obtain $\mu =0$ and $\sum _{s\in S}\mu _0(s)=1$ since $u_0$ is normalized and since it is the restriction of $V_0$ to lotteries. Thus, $\mu _0$ defines a probability measure on S. Moreover, since $\gamma >0$, by (Th1.1) we have $u_0=\sum _{i\in N}\alpha _iu_i$. Then, for any $f\in \mathcal {F}$,

$$\begin{aligned} V_0(f)\ =\ \sum _{i\in N,s\in S}\alpha _i\cdot \mu _0(s)\cdot u_i(f(s)). \end{aligned}$$

(16)

On the other hand, by Eq. (2), we have for any $f\in \mathcal {F}$,

$$\begin{aligned} V_0(f)=\sum _{i\in N,s\in S}[\gamma \cdot \alpha _i\cdot \lambda _i(s)+(1-\gamma )\cdot \alpha _i\cdot \lambda _0(s)]\cdot u_i(f(s)). \end{aligned}$$

(17)

Then, define $\psi (i,s)=[\gamma \alpha _i\lambda _i(s)+(1-\gamma )\alpha _i\lambda _0(s)]-\alpha _i\mu _0(s)$ for any $i\in N$ and $s\in S$. By confronting Eqs. (16) and (17) and applying Lemma 3, we obtain:

$$\begin{aligned} \gamma \cdot \alpha _i\cdot \lambda _i(s)+(1-\gamma )\cdot \alpha _i\cdot \lambda _0(s)=\alpha _i\cdot \mu _0(s). \end{aligned}$$

(18)

Therefore, for any $i\in N$ such that $\alpha _i>0$, we have $\gamma \cdot \lambda _i+(1-\gamma )\cdot \lambda _0=\mu _0$. Since $\gamma >0$, this shows that $\lambda _i$ is independent of i provided i satisfies $\alpha _i>0$. Let $\lambda$ denote this common measure. To conclude, if i is non-null, then by Lemma 2, $\alpha _i>0$ and therefore $\lambda _i=\lambda$. So there is essentially a common prior. $\square$

Claim 3 shows if Statewise Dominance holds, then either $\gamma =0$, or there is essentially a common prior. Reciprocally, if $\gamma =0$, it is straightfroward to show Statewise Dominance given Eq. (2). So we assume that $\gamma >0$, and that there is essentially a common prior. By (Th1.1), we have $u_0=\sum _{i\in N}\alpha _iu_i$. Let $\lambda$ be the measure on S such that, for any non-null $i\in N$, $\lambda _i=\lambda$. Note also that, by Lemma 2, $i\in N$ is non-null if and only if $\alpha _i>0$. Let $I=\{i\in N,\, \alpha _i>0\}$. For any $f\in \mathcal {F}$, we have:

$$\begin{aligned} V_0(f)= & {} \gamma \cdot \sum _{i\in I}\alpha _i\cdot \mathbb {E}_{\lambda }(u_i\circ f)+ (1-\gamma )\cdot \mathbb {E}_{\lambda _0}(u_0\circ f)\\= & {} \gamma \cdot \mathbb {E}_{\lambda }\left( \sum _{i\in I}\alpha _i\cdot u_i\circ f\right) + (1-\gamma )\cdot \mathbb {E}_{\lambda _0}(u_0\circ f)\\= & {} \gamma \cdot \mathbb {E}_{\lambda }(u_0\circ f)+ (1-\gamma )\cdot \mathbb {E}_{\lambda _0}(u_0\circ f) \end{aligned}$$

Then, $\succsim _0$ has an SEU representation where a utility function is given by $u_0$ and the prior is given by $\gamma \cdot \lambda +(1-\gamma )\cdot \lambda _0$. It becomes straightforward to see that Statewise Dominance holds.

• Claim 4: Assume that $\succsim _0$ has an SEU representation. Then $\succsim _0$ satisfies State Independence.

Let $\mu$ be the prior on S and v be a utility function on Y providing an SEU representation. Let $s\in S$, $f\in \mathcal {F}$, and $l,m\in Y$ be such that $l_sf\succ _0 m_sf$. Then, by the assumed SEU representation, we have $\mu (s)\cdot \{v(l)-v(m)\}>0$. So it must be the case that $v(l)-v(m)>0$. Consider now any nonnull state $t\in S$. We have $\mu (t)>0$ and therefore $\mu (t)\cdot \{v(l)-v(m)\}>0$. But then still by the assumed SEU representation we obtain $l_tf\succ _0 m_tf$. $\square$

Finally, we show that Statewise Dominance and State Independence are equivalent. First, assume State Independence. Given the representation (2), it is clear that $\succsim _0$ satisfies STP. Then, Lemma 4 implies that $\succsim _0$ satisfies Statewise Dominance. Now, assume Statewise Dominance. By (P1.2), either $\gamma =0$, or there is essentially a common prior denote by $\lambda$. In the first case, $\succsim _0$ has an SEU representation with respect to $u_0$ and $\lambda _0$. In the second case, as shown above, it has an SEU representation with respect to $u_0$ and $\gamma \cdot \lambda +(1-\gamma )\cdot \lambda _0$. In both cases, it has an SEU representation. Then, by Claim 4, it satisfies State Independence.

1.5 Proof of Proposition 2

Let $V_0$ and $W_0$ be the functionals defined respectively by the structures $\{\gamma ,(\alpha _i)_{i\in N},u_0,\lambda _0\}$ and $\{\delta ,(\alpha _i)_{i\in N},u_0,\lambda _0\}$ as in Eq. (2). First, we show the equivalence between (P2.1) and (P2.2).

Assume that $\succsim _0$ is more prone to Pareto Dominance that $\succsim _0'$. Then, $f\succsim _ig$ for all $i\in N$ and $f\succsim _0'g$ imply $f\succsim _0g$ for any $f,g\in \mathcal {F}$. Since the functionals $V_i$, for $i\in N$, $V_0$ and $W_0$ are all mixture affine, Lemma 1 and normalization provide non-negative numbers $\beta _i$, for all $i\in N$, and $\beta _0$ summing to 1 such that for any $f\in \mathcal {F}$,

$$\begin{aligned} V_0(f)\ =\ \sum _{i\in N}\beta _i\cdot V_i(f)\ +\ \beta _0\cdot W_0(f). \end{aligned}$$

(19)

Moreover, note if $\delta =0$, then we have $\gamma \ge \delta$. So we can suppose that $\delta >0$. Then, $u_0=\sum _{i\in N}\alpha _iu_i$. Therefore, for any $f\in \mathcal {F}$,

$$\begin{aligned} W_0(f) = \delta \cdot \sum _{i\in N}\alpha _i\cdot \mathbb {E}_{\lambda _i}(u_i\circ f)+ (1-\delta )\cdot \sum _{i\in N} \alpha _i\cdot \mathbb {E}_{\lambda _0}(u_i\circ h). \end{aligned}$$

(20)

Combining Eqs. (19) and (20), we obtain for any $f\in \mathcal {F}$,

$$\begin{aligned} V_0(f) = \sum _{i\in N,s\in S}[\beta _i\lambda _i(s)+\beta _0\delta \alpha _i\lambda _i(s)+\beta _0(1-\delta )\alpha _i\lambda _0(s)]\cdot u_i(f(s)). \end{aligned}$$

(21)

On the other hand, by Eq. (2) and $u_0=\sum _{i\in N}\alpha _iu_i$, we also have for any $f\in \mathcal {F}$,

$$\begin{aligned} V_0(f) = \sum _{i\in N,s\in S}[\gamma \alpha _i\lambda _i(s)+(1-\gamma )\alpha _i\lambda _0(s)]\cdot u_i(f(s)). \end{aligned}$$

(22)

Then, define $\psi (i,s)=[\beta _i\lambda _i(s)+\beta _0\delta \alpha _i\lambda _i(s)+\beta _0(1-\delta )\alpha _i\lambda _0(s)]-[\gamma \alpha _i\lambda _i(s)+(1-\gamma )\alpha _i\lambda _0(s)]$ for any $i\in N$ and $s\in S$. By confronting Eqs. (21) and (22) and applying Lemma 3, we obtain for any $i\in N$ and $s\in S$,

$$\begin{aligned} \beta _i\lambda _i(s)+\beta _0\delta \alpha _i\lambda _i(s)+\beta _0(1-\delta )\alpha _i\lambda _0(s)=\gamma \alpha _i\lambda _i(s)+(1-\gamma )\alpha _i\lambda _0(s). \end{aligned}$$

(23)

By summing these equalities for $s\in S$, we obtain $\beta _i=(1-\beta _0)\alpha _i$. Reinjecting this in Eq. (23), we obtain for any $i\in N$ such that $\alpha _i>0$ and $s\in S$,

$$\begin{aligned} (1-\beta _0)\lambda _i(s)+\beta _0\delta \lambda _i(s)+\beta _0(1-\delta )\lambda _0(s)=\gamma \lambda _i(s)+(1-\gamma )\lambda _0(s). \end{aligned}$$

(24)

Now, suppose that $1-\beta _0+\beta _0\delta -\gamma \ne 0$. Then, for any non-null individual $i\in N$, we have $\alpha _i>0$ by Lemma 2. Then, since $1-\beta _0+\beta _0\delta -\gamma \ne 0$, Eq. (23) shows that $\lambda _i$ is independent of i. In other words, there is essentially a common prior, which contradicts our assumptions. Therefore, we must have $1-\gamma =\beta _0(1-\delta )$. Then, $1-\gamma \le 1-\delta$ and finally $\gamma \ge \delta$.

Assume now that $\gamma \ge \delta$. If $\delta =1$, then $\gamma =\delta =1$. Furthermore, we have $V_0=W_0$. So the two binary relations must agree with each other. Thus, $\succsim _0$ is more prone to Pareto Dominance that $\succsim _0'$. If $\delta <1$, then let $\beta _0$ be such that $1-\gamma =\beta _0(1-\delta )$, and for all $i\in N$ set $\beta _i=\alpha _i(1-\beta _0)$. It is then easy to see that Eq. (19) holds. Hence, $\succsim _0$ is more prone to Pareto Dominance that $\succsim _0'$.

Now, we show the equivalence between (P2.2) and (P2.3). Assume that $\succsim _0$ is less prone to Statewis Dominance that $\succsim _0'$. Then, since the utility function $u_0$ is the same for $\succsim _0$ and $\succsim _0'$, we have: $f(s)\succsim _0g(s)$ for all $i\in N$ and $f\succsim _0g$ imply $f\succsim _0'g$ for any $f,g\in \mathcal {F}$. Since the functionals $f\rightarrow u_0(f(s))$, for $s\in S$, $V_0$ and $W_0$ are all mixture affine, Lemma 1 and normalization provide non-negative numbers $\mu (s)$, for all $s\in S$, and $\mu _0$ summing to 1 such that for any $f\in \mathcal {F}$,

$$\begin{aligned} W_0(f)\ =\ \sum _{s\in S}\mu (s)\cdot u_0(f(s))\ +\ \mu _0\cdot V_0(f). \end{aligned}$$

(25)

Moreover, note if $\delta =0$, then we have $\gamma \ge \delta$. So we can suppose that $\delta >0$. Then, $u_0=\sum _{i\in N}\alpha _iu_i$. Therefore, for any $f\in \mathcal {F}$,

$$\begin{aligned} V_0(f) = \gamma \cdot \sum _{i\in N}\alpha _i\cdot \mathbb {E}_{\lambda _i}(u_i\circ f)+ (1-\gamma )\cdot \sum _{i\in N} \alpha _i\cdot \mathbb {E}_{\lambda _0}(u_i\circ h). \end{aligned}$$

(26)

Combining Equations (25) and (26), we obtain for any $f\in \mathcal {F}$,

$$\begin{aligned} W_0(f) = \sum _{i\in N,s\in S}[\alpha _i\mu (s)+\mu _0\gamma \alpha _i\lambda _i(s)+\mu _0(1-\gamma )\alpha _i\lambda _0(s)]\cdot u_i(f(s)). \end{aligned}$$

(27)

On the other hand, by Eq. (2) and $u_0=\sum _{i\in N}\alpha _iu_i$, we also have for any $f\in \mathcal {F}$,

$$\begin{aligned} W_0(f) = \sum _{i\in N,s\in S}[\delta \alpha _i\lambda _i(s)+(1-\delta )\alpha _i\lambda _0(s)]\cdot u_i(f(s)). \end{aligned}$$

(28)

Then, define $\psi (i,s)=[\alpha _i\mu (s)+\mu _0\gamma \alpha _i\lambda _i(s)+\mu _0(1-\gamma )\alpha _i\lambda _0(s)]-[\delta \alpha _i\lambda _i(s)+(1-\delta )\alpha _i\lambda _0(s)]$ for any $i\in N$ and $s\in S$. By confronting Eqs. (27) and (28) and applying Lemma 3, we obtain for any $i\in N$ and $s\in S$,

$$\begin{aligned} \alpha _i\mu (s)+\mu _0\gamma \alpha _i\lambda _i(s)+\mu _0(1-\gamma )\alpha _i\lambda _0(s)\ =\ \delta \alpha _i\lambda _i(s)+(1-\delta )\alpha _i\lambda _0(s). \end{aligned}$$

(29)

Suppose that $\delta \ne \mu _0\gamma$. For any non-null individual $i\in N$, we have $\alpha _i>0$ by Lemma 2 and, therefore,

$$\begin{aligned} \mu +\mu _0\gamma \lambda _i+\mu _0(1-\gamma )\lambda _0\ =\ \delta \lambda _i+(1-\delta )\lambda _0. \end{aligned}$$

(30)

Equation (30) shows that $\lambda _i$ is independent of i. In other words, there is essentially a common prior, which contradicts our assumptions. Therefore, we must have $\delta =\mu _0\gamma \le \gamma$. Now, suppose $\delta \le \gamma$. If $\gamma =0$, then $\delta =\gamma =0$. Furthermore, we have $V_0=W_0$. So the two binary relations must agree with each other. Thus, $\succsim _0$ is less prone to Statewise Dominance that $\succsim _0'$. If $\gamma >0$, then let $\mu _0$ be such that $\delta =\mu _0\gamma$. For any $s\in S$, set $\mu (s)=(1-\mu _0)\lambda _0(s)$. It is then easy to see that Eq. (25) holds. Hence, $\succsim _0$ is less prone to Statewise Dominance that $\succsim _0'$.

1.6 Proof of Proposition 3

Suppose first $\gamma =0$. Then, by Eq. (2), $\succsim _0$ must satisfy Statewise Dominance. But then, by Proposition 1, either $\delta =0$, or there is essentially a common prior. The latter possibility is excluded by assumption. So we must have $\delta =0$, and therefore $\delta =\gamma$. Then, still by Eq. (2), the two representations of $\succsim _0$ reduce to SEU representations: the one is given by $u_0$ and $\lambda _0$, the other one is given by $v_0$ and $\mu _0$. By the uniqueness part of the Anscombe and Aumann (1963) theorem, we obtain $\mu _0=\lambda _0$, and $v_0$ is a positive affine transformation of $u_0$. However, the two utility functions are normalized. Hence $v_0=u_0$.

Suppose now $\gamma =1$. Then, by Eq. (2), $\succsim _0$ must satisfy Pareto Dominance. But then, by Proposition 1, either $\delta =1$, or the social prior is essentially a common prior. The latter possibility is excluded by assumption. So we must have $\delta =1$, and therefore $\delta =\gamma$. Consider now the restriction of $\succsim _0$ to constant acts; that is, to lotteries in Y. By Theorem 1, each of $u_0$ and $v_0$ provides a representation for this restriction. By the uniqueness part of the von Neumann and Morgenstern (1944) theorem, $u_0$ and $v_0$ must be positive affine transformation of each other. But since they are normalized, they are in fact equal to each other. Then, since $\gamma >0$ and $\delta >0$, we have $u_0=\sum _{i\in N}\alpha _iu_i$ and $v_0=\sum _{i\in N}\beta _iu_i$. Therefore, $\sum _{i\in N}\alpha _iu_i=\sum _{i\in N}\beta _iu_i$. Since Risk Diversity holds, we must have $\beta _i=\alpha _i$ for any $i\in N$ (Proceed as in Claim 1).

Finally, suppose $\gamma \in (0,1)$. Then, by the previous paragraphs, it must be the case that $\delta \in (0,1)$. Consider now the restriction of $\succsim _0$ to constant acts; that is, to lotteries in Y. By Theorem 1, each of $u_0$ and $v_0$ provides a representation for this restriction. By the uniqueness part of the von Neumann and Morgenstern (1944) theorem, $u_0$ and $v_0$ must be positive affine transformation of each other. But since they are normalized, they are in fact equal to each other. Let $V_0$ and $W_0$ be the functionals defined respectively by $\{\gamma ,(\alpha _i)_{i\in N},u_0,\lambda _0\}$ and $\{\delta ,(\beta _i)_{i\in N},v_0,\mu _0\}$ as in Eq. (2). For any $l\in Y$, we have by Theorem 1$V_0(l)=u_0(l)=v_0(l)=W_0(l)$. Moreover, for each $f\in \mathcal {F}$, there exists $l\in Y$ such that $f\sim _0l$ (See proof of Theorem 1). Then, $V_0(f)=V_0(l)=W_0(l)=W_0(f)$. On the other hand, the functionals $V_0$ and $W_0$ can be written in the following way: for any $f\in \mathcal {F}$

$$\begin{aligned} V_0(f)=\sum _{i\in N,s\in S}[\gamma \cdot \alpha _i\cdot \lambda _i(s)+(1-\gamma )\cdot \alpha _i\cdot \lambda _0(s)]\cdot u_i(f(s)), \end{aligned}$$

(31)

and

$$\begin{aligned} W_0(f)=\sum _{i\in N,s\in S}[\delta \cdot \beta _i\cdot \lambda _i(s)+(1-\delta )\cdot \beta _i\cdot \mu _0(s)]\cdot u_i(f(s)). \end{aligned}$$

(32)

Then, define $\psi (i,s)=[\gamma \alpha _i\lambda _i(s)+(1-\gamma )\alpha _i\lambda _0(s)]-[\delta \beta _i\lambda _i(s)+(1-\delta )\beta _i\mu _0(s)]$ for any $i\in N$ and $s\in S$. By confronting Eqs. (31) and (32) and applying Lemma 3, we obtain:

$$\begin{aligned} \gamma \cdot \alpha _i\cdot \lambda _i(s)+(1-\gamma )\cdot \alpha _i\cdot \lambda _0(s)=\delta \cdot \beta _i\cdot \lambda _i(s)+(1-\delta )\cdot \beta _i\cdot \mu _0(s). \end{aligned}$$

(33)

By summing these equalities on S, we obtain $\beta _i=\alpha _i$ for any $i\in N$. Moreover, since there is essentially no common prior, there exist two non-null $i,j\in N$ such that $\lambda _i\ne \lambda _j$. Let $s\in S$ be such that $\lambda _i(s)\ne \lambda _j(s)$. Moreover, by Lemma 2, we have $\alpha _i>0$ and $\alpha _j>0$. By applying twice Eq. (33) and substracting, we get

$$\begin{aligned} \gamma \cdot (\lambda _i(s)-\lambda _j(s))\ =\ \delta \cdot (\lambda _i(s)-\lambda _j(s)). \end{aligned}$$

Since $\lambda _i(s)\ne \lambda _j(s)$, it must be that $\delta =\gamma$. Finally, applying again Eq. (33) to any $s\in S$ and some non-null individual $i\in N$ gives $\mu _0=\lambda _0$.

1.7 Proof of Proposition 4

Assume that $\succsim _0$ satisfies Weak Order, Continuity, Independence, Boundedness, Weak Dominance and Non-triviality. Then, by Theorem 1, there exists a structure $\{\gamma ,(\alpha _i)_{i\in N},u_0,\lambda _0\}$ providing a representation of $\succsim _0$ as in Theorem 1. Let $V_0$ be the functional this structure defines according to Eq. (2).

Moreover, let I stand for the set of $i\in N$ that are non-null. Suppose momentarily that $\gamma >0$. Then, by Lemma 2, $I=\{i\in N,\, \alpha _i>0\}$. If $i\in I$, then $\lambda _i=\lambda$. So for any $f\in \mathcal {F}$

$$\begin{aligned} V_0(f)= & {} \gamma \cdot \sum _{i\in I}\alpha _i\cdot \mathbb {E}_{\lambda }(u_i\circ f)+ (1-\gamma )\cdot \mathbb {E}_{\lambda _0}(u_0\circ f)\\= & {} \gamma \cdot \mathbb {E}_{\lambda }\left( \sum _{i\in I}\alpha _i\cdot u_i\circ f\right) + (1-\gamma )\cdot \mathbb {E}_{\lambda _0}(u_0\circ f)\\= & {} \gamma \cdot \mathbb {E}_{\lambda }(u_0\circ f)+ (1-\gamma )\cdot \mathbb {E}_{\lambda _0}(u_0\circ f) \end{aligned}$$

Note that the latter equality also holds if $\gamma =0$. So it holds for any $\gamma \in [0,1]$. This shows that $\succsim _0$ has a SEU representation where the normalized utility function is given by $u_0$ and the probability measure is given by $\gamma \lambda +(1-\gamma )\lambda _0$, thereby showing Eq. (4) as well as (P4.1) and (P4.3). Moreover, (P4.2) follows from (Th1.1). Finally, the uniqueness of $\mu _0$ follows from the Anscombe and Aumann (1963) theorem, which also provides the uniquenes of $v_0$ up to positive affine transformation. But since $v_0$ is normalized, it is in fact unique.

Now, suppose that $\succsim _0$ has an SEU representation with respect to a normalized and mixture affine function $v_0:Y\rightarrow \mathbb {R}$ and probability measure $\mu _0$ on S. Then, it is easy to see that it satisfies all of Weak Order, Continuity, Independence, and Non-triviality. Moreover, $\succsim _0$ satisfies Statewise Dominance, and therefore satisfies Weak Dominance. Finally, since $\succsim _0$ satisfies Statewise Dominance, it also satisfies Boundedness.