A new approach to the rational expectations equilibrium: existence, optimality and incentive compatibility

Abstract

Rational expectations equilibrium seeks a proper treatment of behavior under private information by assuming that the information revealed by prices is taken into account by consumers in their decisions. Typically agents are supposed to maximize a conditional expectation of state-dependent utility function and to consume the same bundles in indistiguishable states [see Allen (Econometrica 49(5):1173–1199, 1981), Radner (Econometrica 47(3):655–678, 1979)]. A problem with this model is that a rational expectations equilibrium may not exist even under very restrictive assumptions, may not be efficient, may not be incentive compatible, and may not be implementable as a perfect Bayesian equilibrium (Glycopantis et al. in Econ Theory 26(4):765–791, 2005). We introduce a notion of rational expectations equilibrium with two main features: agents may consume different bundles in indistinguishable states and ambiguity is allowed in individuals’ preferences. We show that such an equilibrium exists universally and not only generically without freezing a particular preferences representation. Moreover, if we particularize the preferences to a specific form of the maxmin expected utility model introduced in Gilboa and Schmeidler (J Math Econ 18(2):141–153, 1989), then we are able to prove efficiency and incentive compatibility. These properties do not hold for the traditional (Bayesian) Rational Expectation Equilibrium.

Appendix

1.1 Proofs of Section 3

In order to prove the existence theorem, we show below that the set of V-REE allocations and a fortiori of maxmin REE allocations contains all the selections from the Walrasian equilibrium correspondence of the associated family of complete information economies.

Lemma 8.1

If (p, x) is an ex post Walrasian equilibrium, then (p, x) is a V-REE, and in particular it is a maxmin REE.

Proof

Let (p, x) an ex post Walrasian equilibrium, we want to show that (p, x) is a V-REE. First, notice that x is feasible in the economy ${{\mathcal {E}}}$ because so is x(s) in the economy ${{\mathcal {E}}}( s )$ for each s, and p is a price function because for any $s\in S$, $p(s) >0$. Consider the algebra generated by p denoted by $\sigma ({p})$, and for each agent i let $\mathcal {G}^p_i=\mathcal {F}_i \vee \sigma ( {p})$. We show that ( p, x) is a V-REE for ${{\mathcal {E}}}$. Clearly, for all $i\in I$ and $s\in S,$$p( s )\cdot x_i( s )\le p( s )\cdot e_i( s )$, hence ${x}_i\in B_i( s , {p})$. It remains to prove that ${x}_i$ maximizes $V_i(\cdot | {\mathcal {G}}^p_i(s))$ on $B_i(s, p)$. Assume, on the contrary, that there exists an alternative allocation y such that for some agent i and some state s,

$$\begin{aligned}&V_i(y_i| {\mathcal {G}}^p_i(s))> V_i(x_i|{\mathcal {G}}^p_i(s)) \,\,{{\mathrm{and}}}\,\,y_i\in B_i ( s , {p})\,\,{{\mathrm{that}}\,{\mathrm{is}}}\nonumber \\&\quad {p}( s ^\prime )\cdot y_i( s ^\prime )\le p( s ^\prime )\cdot e_i( s ^\prime )\quad {{\mathrm{for}}\,\,{\mathrm{all}}\,\,} s ^\prime \in {\mathcal {G}}^p_i( s ). \end{aligned}$$

(13)

From (A2) it follows that there exists a state ${\bar{s}} \in {\mathcal {G}}^p_i( s )$ such that

$$\begin{aligned} u_i({\bar{s}} , y_i({\bar{s}} ))>u_i({\bar{s}} , x_i({\bar{s}} )). \end{aligned}$$

Since $(p({\bar{s}} ), x({\bar{s}} ))$ is a Walrasian equilibrium for ${{\mathcal {E}}}({\bar{s}} )$, it follows that, ${p}({\bar{s}} )\cdot y_i({\bar{s}} )> p({\bar{s}} )\cdot e_i({\bar{s}} )$, which clearly contradicts (13). $\square$

Proof of Theorem 3.2

Since S is finite, there is a finite number of complete information economies ${{\mathcal {E}}}(s)=\{I, {\mathbb {R}}_+^\ell , (u_i(s), e_i(s))_{i \in I}\}$, where for any $i \in I$ and any $s \in S$, $u_i(s):=u_i(s, \cdot ): {\mathbb {R}}_+^\ell \rightarrow {\mathbb {R}}$ is continuous, monotone, quasi-concave; and $e_i(s)\gg 0$. For any $s\in S$, let $W({{\mathcal {E}}}(s))$ the set of Walrasian equilibrium allocations for the economy ${{\mathcal {E}}}(s)$. The above assumptions ensure that for any $s\in S$, $W({{\mathcal {E}}}(s))\ne \emptyset$ and hence the set $W=\left\{ x:I \times S \rightarrow {\mathbb {R}}_+^\ell \,| \,\,x(s)\in W({{\mathcal {E}}}( s )) {\,\, {\mathrm{for}}\,\, {\mathrm{all}}\,\,} s \in S \right\}$ is non empty. An element of W is an ex post Walrasian equilibrium allocation and from Lemma 8.1 it is a V-REE.$\square$

Lemma 8.1 states that any ex post Walrasian equilibrium is a V-REE. The converse is not true (see Example 3.3) unless the strict version of (A2) holds, as stated in Proposition 3.4.

Proof of Proposition 3.4

One inclusion is shown in Lemma 8.1. In order to prove the converse, let (p, x) a V-REE and consider for any agent $i \in I$ the algebra ${\mathcal {G}}^p_i={\mathcal {F}}_i \vee \sigma (p)$. Condition $(A2^*)$ implies that for any $s\in S$ the equilibrium price p(s) is positive in ${{\mathcal {E}}}(s)$, i.e., $p(s)> 0$ for any $s \in S$. Clearly, feasibility and budget constrains hold. Assume to the contrary that for some state s the pair (x(s), p(s)) is not a Walrasian equilibrium for the complete information economy ${{\mathcal {E}}}(s)$. This means that there exists an alternative bundle $y\in {\mathbb {R}}_+^\ell$ such that for some agent j

$$\begin{aligned}&(i)\quad u_j(s, y)>u_j(s, x_j(s))\\&(ii)\quad p(s)\cdot y\le p(s)\cdot e_i(s). \end{aligned}$$

If ${\mathcal {G}}^p_j(s)=\{s\}$, from (A1) we get the contradiction. If ${\mathcal {G}}^p_j(s){\setminus } \{s\}\ne \emptyset$, let $z_j(s)=y$ and $z_j(s^\prime )=x_j(s^\prime )$ for any $s^\prime \in {\mathcal {G}}^p_j(s){\setminus } \{s\}$. Condition $(A2^*)$ implies that $V_j(z_j|{\mathcal {G}}^p_j(s))> V_j(x_j|{\mathcal {G}}^p_j(s))$, and hence there must exist $\bar{s}\in {\mathcal {G}}^p_j(s)$ such that

$$\begin{aligned} p(\bar{s})\cdot z_j(\bar{s})> p(\bar{s})\cdot e_j(\bar{s}). \end{aligned}$$

This is impossible by the definition of $z_j$. $\square$

1.2 Proofs of Section 4

Proof of Proposition 4.3

Assume to the contrary that there exist an agent $i \in I$ and two states $a, b \in S$ such that $a \in \mathcal {G}^p_i(b)$ and $x_i(a)\ne x_i(b)$. Consider $z_i( s )= \alpha x_i(a)+ (1-\alpha ) x_i(b)$ for all $s \in \mathcal {G}^p_i(b)$, where $\alpha \in (0,1)$, and notice that $z_i$ is constant in the event $\mathcal {G}^p_i(b)$. Moreover,

$$\begin{aligned} \underline{u}_i^{REE}(b, z_i)= \min _{ s \in \mathcal {G}^p_i(b)} u_i( s ,z_i( s ))= \min _{ s \in \mathcal {G}^p_i(b)} u_i( s , \alpha x_i(a)+ (1-\alpha ) x_i(b)). \end{aligned}$$

Since $u_i(\cdot ,y)$ is $\mathcal {G}^p_i$-measurable for all $y \in {\mathbb {R}}_+^\ell$, from strict quasi-concavity of $u_i$ it follows that

$$\begin{aligned} \underline{u}_i^{REE}(b, z_i)= & {} u_i(b, \alpha x_i(a)+ (1-\alpha ) x_i(b))> \min \{u_i(b, x_i(a)), u_i(b, x_i(b))\}\\= & {} \min \{ u_i(a, x_i(a)), u_i(b, x_i(b))\}\ge \min _{ s \in {\mathcal {G}}^p_i(b)} u_i( s , x_i( s ))\\= & {} \underline{u}_i^{REE}(b, x_i). \end{aligned}$$

Since (p, x) is a maxmin rational expectations equilibrium it follows that $z_i \notin B_i (b, p)$, that is, there exists a state $s _i\in \mathcal {G}^p_i(b)$ such that

$$\begin{aligned}&p( s _i) \cdot z_i( s _i)> p( s _i)\cdot e_i( s _i)\,\,\Rightarrow \,\,\alpha p( s _i)\cdot x_i(a)\\&\quad + (1-\alpha ) p( s _i) \cdot x_i(b) > p( s _i) \cdot e_i( s _i). \end{aligned}$$

Moreover, since $p(\cdot )$ and $e_i(\cdot )$ are $\mathcal {G}^p_i$-measurable and $p( s )\cdot x_i( s )\le p( s )\cdot e_i( s )$ for all $s \in S$, it follows that $p( s _i) \cdot e_i( s _i)> p( s _i)\cdot e_i( s _i)$, which is a contradiction.$\square$

Proof of Proposition 4.5

Clearly if $\sigma (e_i) \subseteq {\mathcal {G}}^p_i$ and $y_i(\cdot )$ is ${\mathcal {G}}^p_i$-measurable, then $p(s)\cdot y_i(s)\le p(s)\cdot e_i(s)$ is equivalent to $p(s') \cdot y_i(s')\le p(s') \cdot e_i(s')$ for all $s' \in {\mathcal {G}}^p_i(s)$. Thus, all we need to show is that the maxmin utility and the (Bayesian) interim expected utility coincide. Since for all $i\in I$, $\sigma (u_i)\subseteq \mathcal {F}_i$ and $\mathcal {F}_i \subseteq \mathcal {G}^p_i$, then $\sigma (u_i)\subseteq {\mathcal {G}}^p_i$.

Moreover, since for each $i \in I$, $x_i(\cdot )$ is $\mathcal {G}^p_i$-measurable it follows that for all $i \in I$ and $s \in S$, both maxmin and interim utility function are equal to the ex post utility function. That is,

$$\begin{aligned} \underline{u}^{REE}_i( s , x_i)= \min _{ s ^\prime \in \mathcal {G}^p_i( s )} u_i( s ^\prime , x_i( s ^\prime ))=u_i( s , x_i( s )) \end{aligned}$$

(14)

and

$$\begin{aligned} E_{\pi }(u_i(\cdot , x(\cdot ))|{\mathcal {G}}^p_i(s))= \sum _{ s ^\prime \in \mathcal {G}^p_i( s )} u_i( s ^\prime , x_i( s ^\prime ))\frac{\pi _i( s ^\prime )}{\pi _i\left( \mathcal {G}^p_i( s )\right) }=u_i( s , x_i( s )). \end{aligned}$$

(15)

From (14) and (15) it follows that for all i and s, $\underline{u}^{REE}_i( s , x_i)=E_{\pi }(u_i(\cdot , x(\cdot ))|{\mathcal {G}}^p_i(s))$. Therefore, we can conclude that if (p, x) is a traditional REE, then (p, x) is a MREE; the converse is also true if $x_i(\cdot )$ is $\mathcal {G}^p_i$-measurable for all $i\in I$. $\square$

From Lemma 8.1 it follows that any ex post Walrasian equilibrium is a maxmin REE. The converse is not true (see Example 3.3) unless agents’ utility functions and initial endowments are private information measurable. The next lemma, which is useful for the proof of Proposition 4.7, holds true for the general MEU formulation (5) provided that for any agent i and state s, the set ${\mathcal {M}}_i^s$ contains only positive priors (see Sect. 8.5).

Lemma 8.2

If $(u_i, e_i)\subseteq {\mathcal {F}}_i$ for all $i\in I$, then any ex post Walrasian equilibrium is a maxmin REE and vice versa.

Proof

One inclusion is shown in Lemma 8.1 and no private information measurability assumption is needed. In order to prove the converse, let (p, x) a maxmin REE and consider for any agent $i \in I$ the algebra ${\mathcal {G}}^p_i={\mathcal {F}}_i \vee \sigma (p)$. The monotonicity assumption on agents’ utility function ensures that for any $s\in S$ the equilibrium price p(s) is positive in ${{\mathcal {E}}}(s)$, i.e., $p(s)> 0$ for any $s \in S$. Clearly, feasibility and the budget constraint hold. Assume to the contrary that for some state s the pair (x(s), p(s)) is not a Walrasian equilibrium for the complete information economy ${{\mathcal {E}}}(s)$. This means that there exist an agent j and an alternative bundle $y\in {\mathbb {R}}_+^\ell$ such that

$$\begin{aligned}&(i)\quad u_j(s, y)>u_j(s, x_j(s)),\\&(ii)\quad p(s)\cdot y\le p(s)\cdot e_i(s). \end{aligned}$$

Let $z_j(s^\prime )=y$ for any $s^\prime \in {\mathcal {G}}^p_j(s)$, and notice that since $u_j(\cdot , z)$ is ${\mathcal {F}}_j$-measurable and a fortiori ${\mathcal {G}}^p_j$-measurable, from (i) it follows that $\underline{u}_j^{REE}(s, z_j)=u_j(s,y)>u_j(s,x_j(s))\ge \underline{u}_i^{REE}(s, x_i).$ Recall that (p, x) is a maxmin rational expectations equilibrium, thus there exists $\bar{s}\in {\mathcal {G}}^p_j(s)$ such that $p(\bar{s})\cdot z_j(\bar{s})> p(\bar{s})\cdot e_j(\bar{s}).$ Since $e_j(\cdot )$ and $p(\cdot )$ are ${\mathcal {G}}^p_j$-measurable, it follows that $p(s)\cdot y> p(s)\cdot e_i(s),$ which contradicts (ii) above. $\square$

Remark 8.3

Proposition 4.7 states that if in addition $u_i(s, \cdot )$ is strict quasi-concave for all $i\in I$ and $s \in S$, the ex post Walrasian equilibria coincide also with the (traditional) rational expectations equilibrium (see also Einy et al. 2000b and De Simone and Tarantino 2010).

Proof of Proposition 4.7

The equivalence between (1) and (2) is obtained by combining Propositions 4.3 and 4.5 (see Remark 4.6). The equivalence between (1) and (3) is instead stated in Lemma 8.2. $\square$

Proof of Proposition 4.10

For each $s \in S$, let

$$\begin{aligned} H(s)=\{h\in \{1,\ldots , \ell \}\,:\,\,p^h(s)=0\}, \end{aligned}$$

and let

$$\begin{aligned} {\bar{S}}=\{s\in S\,:\,\,H(s)\ne \emptyset \}. \end{aligned}$$

Since (p, x) is a maxmin REE, we consider the information generated by the equilibrium price, that is the algebra $\sigma (p)$. Clearly, $H(\cdot )$ is $\sigma (p)$-measurable,²⁰ because $p(s_1)=p(s_2)$ whenever $\sigma (p)(s_1)=\sigma (p)(s_2)$. Moreover, since for any $i \in I$, $\sigma (p) \subseteq$$\mathcal {G}^p_i=\mathcal {F}_i\vee \sigma (p)$, it follows that for all $i \in I$

$$\begin{aligned} H(\cdot )\,\,\,{{\mathrm{is}}\,\,}\mathcal {G}^p_i{\text {-}}{{\mathrm{measurable}}}. \end{aligned}$$

(16)

Now, assume to the contrary that ${\bar{S}}$ is non empty and let ${\bar{s}}\in {\bar{S}}$. Hence, $H({\bar{s}})\ne \emptyset$, i.e., there exists at least a “free” good h such that $p^h({\bar{s}})=0$. Let $i \in I$ be the agent such that $u_i(s, \cdot )$ is strictly monotone for any $s \in S$; and define the following allocation:

$$\begin{aligned} z_i^h(s) = \left\{ \begin{array}{ll} x_i^h(s)+K &{} \quad \text {if }s \in \mathcal {G}^p_i({\bar{s}})\text { and }h \in H(s)\\ x_i^h(s) &{} \quad \text {otherwise,} \end{array} \right. \end{aligned}$$

where $K>0$.

Notice that for any $s \in \mathcal {G}^p_i({\bar{s}})$, since $H(s)=H({\bar{s}})\ne \emptyset$ [see (16)], from the strict monotonicity it follows that $u_i(s, z_i(s))>u_i(s, x_i(s))$ for all $s \in {\mathcal {G}}^p_i({\bar{s}})$, and hence

$$\begin{aligned} \underline{u}_i^{REE}({\bar{s}}, z_i)>\underline{u}_i^{REE}({\bar{s}}, x_i). \end{aligned}$$

Since (p, x) is a maxmin REE, $z_i \notin B_i({\bar{s}}, p)$, that is there exists a state $s_i \in \mathcal {G}^p_i({\bar{s}})$ such that

$$\begin{aligned} p(s_i)\cdot [z_i(s_i)-e_i(s_i)]>0. \end{aligned}$$

From (16), it follows that $H(s_i)=H(\bar{s})\ne \emptyset$, and therefore

$$\begin{aligned}&0<p(s_i)\cdot [z_i(s_i)-e_i(s_i)]\\&\quad =\sum _{h\in H(s_i)}p^h(s_i) [x_i^h(s_i)+K-e^h_i(s_i)]+ \sum _{h\notin H(s_i)}p^h(s_i) [x_i^h(s_i)-e^h_i(s_i)]\\&\quad =0+ \sum _{h\notin H(s_i)}p^h(s_i) [x_i^h(s_i)-e^h_i(s_i)]\\&\quad =\sum _{h\in H(s_i)}p^h(s_i) [x_i^h(s_i)-e^h_i(s_i)]+ \sum _{h\notin H(s_i)}p^h(s_i) [x_i^h(s_i)-e^h_i(s_i)]\\&\quad =p(s_i)\cdot [x_i(s_i)-e_i(s_i)]\le 0. \end{aligned}$$

This is a contradiction, hence $p(s)\gg 0$ for each $s \in S$. $\square$

Proof of Proposition 4.11

Let (p, x) be a maxmin rational expectations equilibrium and define for each agent $i \in I$ and state $s \in S$ the following set:

$$\begin{aligned} M_i( s )=\left\{ s ^\prime \in {\mathcal {G}}^p_i( s ):\,\, \underline{u}_i^{REE}( s , x_i)=u_i( s ^\prime , x_i( s ^\prime ))\right\} . \end{aligned}$$

Clearly, since S is finite, for all $i\in I$ and $s \in S$, the set $M_i( s )$ is nonempty, i.e., $M_i( s )\ne \emptyset$. Moreover, if $s ^\prime \in \mathcal {G}^p_i( s ){\setminus } M_i( s )$ it means that $\underline{u}_i^{REE}( s , x_i)<u_i( s ^\prime , x_i( s ^\prime ))$. Thus, we want to show that for all $i\in I$ and $s \in S$, $M_i( s )={\mathcal {G}}^p_i( s )$.

Assume to the contrary that there exist an agent $j\in I$ and a state ${\bar{s}} \in S$ such that $\mathcal {G}^p_j({\bar{s}} ){\setminus } M_j({\bar{s}} )\ne \emptyset$. Notice that

$$\begin{aligned} \underline{u}_j^{REE}({\bar{s}} , x_j)<u_j( s , x_j( s ))\quad {{\mathrm{for}}\,\,{{\mathrm{any}}}\,\,} s \in \mathcal {G}^p_j({\bar{s}} ){\setminus } M_j({\bar{s}} ). \end{aligned}$$

Fix $s ^\prime \in \mathcal {G}^p_j({\bar{s}} ){\setminus } M_j({\bar{s}} )$ and define the following allocation

$$\begin{aligned} y_j( s ) = \left\{ \begin{array}{ll} x_j( s ) &{} \text {if } s \in \mathcal {G}^p_j({\bar{s}} ){\setminus } M_j({\bar{s}} )\\ x_j( s ^\prime ) &{} \text {if } s \in M_j({\bar{s}} ). \end{array} \right. \end{aligned}$$

Since the utility functions are assumed to be private information measurable, it follows that $u_j( s , y_j( s ))>\underline{u}^{REE}_j({\bar{s}} , x_j)$ for any $s \in \mathcal G_j({\bar{s}} )$, and hence $\underline{u}^{REE}_j({\bar{s}} , y_j)>\underline{u}^{REE}_j({\bar{s}} , x_j)$. Recall that (p, x) is a maxmin REE, therefore there exists $s \in \mathcal {G}^p_j(\bar{s} )$ such that $p( s )\cdot y_j( s )>p( s )\cdot e_j( s )$. If $s \in M_j({\bar{s}} )$, then $p( s )\cdot x_j( s ^\prime )>p( s )\cdot e_j( s )$. Since $p(\cdot )$ and $e_j(\cdot )$ are both ${\mathcal {G}}^p_j$-measurable, it follows that $p( s ^\prime )=p( s )$ and $e_j( s ^\prime )=e_j( s )$. This implies that $p( s ^\prime )\cdot x_j( s ^\prime )>p( s ^\prime )\cdot e_j( s ^\prime )$, which is clearly a contradiction. On the other hand, if $s \in G_j({\bar{s}} ){\setminus } M_j({\bar{s}} )$, we have that $p( s )\cdot x_j( s )>p( s )\cdot e_j( s )$ which is a contradiction as well. Therefore, for each $i \in I$ and $s \in S$, $M_i( s )={\mathcal {G}}^p_i( s )$. $\square$

1.3 Proofs of Section 5

Proof of Proposition 5.3

Let x be a maxmin Pareto optimal allocation with respect to the information structure $\Pi$ and assume to the contrary that there exists a feasible allocation y such that $u_i({ s }, y_i( s )) \ge u_i({ s }, x_i( s ))$ for all $i\in I$ and all $s\in S$ with at least one strict inequality.

Let $j\in I$ and $\bar{s}\in S$ such that $u_j({ \bar{s} }, y_j( \bar{s} )) >u_j({ \bar{s}}, x_j( \bar{s} ))$. Hence, $y_j( \bar{s} )>0$ and if $u_j({ \bar{s} }, t )= u_j({ \bar{s} }, 0)$ for any $t \in \partial {\mathbb {R}}_+^\ell$, then $y_j( \bar{s} )\gg 0$. Thanks to continuity of $u_j(\bar{s}, \cdot )$ there exists $\epsilon \in (0,1)$ for which $u_j({ \bar{s} }, \epsilon y_j( \bar{s} )) >u_j({ \bar{s}}, x_j( \bar{s} ))$. Consider the feasible allocation z given by $z_i(s)=y_i(s)$ for any $i\in I$ and $s\in S{\setminus }\{\bar{s}\}$; while in $\bar{s}$

$$\begin{aligned} z_i(\bar{s}) = \left\{ \begin{array}{ll} \epsilon y_j(\bar{s}) &{} \quad \text {if }i=j\\ y_i(\bar{s}) + \frac{1-\epsilon }{n-1}y_j(\bar{s}) &{} \quad \text {otherwise.} \end{array} \right. \end{aligned}$$

From the strict monotonicity it follows that the feasible allocation z is such that

$$\begin{aligned} u_i(s, z_i(s))\ge & {} u_i(s,x_i(s)) \quad {{\mathrm{for}}\,\,{{\mathrm{any}}}\,\,} i\in I\,\,{{\mathrm{and}}}\,\,s\in S,\\ u_i(\bar{s}, z_i(\bar{s}))> & {} u_i(\bar{s},x_i(\bar{s})) \quad {{\mathrm{for}}\,\,{{\mathrm{any}}}\,\,} i\in I. \end{aligned}$$

The same happens if $u_j({ \bar{s} }, t )= u_j({ \bar{s} }, 0)$ for any $t \in \partial {\mathbb {R}}_+^\ell$ because $y_j( \bar{s} )\gg 0$ and $u_i({ \bar{s} }, \cdot )$ is monotone.

Let $k\in I$ be such that $\Pi _k(\bar{s})=\{\bar{s}\}$, then

$$\begin{aligned}&\underline{u}^{\Pi _i}_i(s, z_i)\ge \underline{u}^{\Pi _i}_i(s,x_i) \quad {{\mathrm{for}}\,\,{{\mathrm{any}}}\,\,} i\in I\,\,{{\mathrm{and}}}\,\,s\in S,\\&\underline{u}_k^{\Pi _k}(\bar{s}, z_k)=u_k(\bar{s}, z_k(\bar{s}))> u_k(\bar{s}, x_k(\bar{s}))= \underline{u}_k^{\Pi _k}(\bar{s}, x_k). \end{aligned}$$

Therefore, x is not maxmin efficient with respect to the information structure $\Pi$, which is a contradiction. We now show that the converse may not be true.²¹ To this end consider an asymmetric information economy with two agents $I=\{1,2\}$, two goods and two states $S=\{a,b\}$. The primitives are as follows:

$$\begin{aligned} \begin{array}{ll} \Pi _{1}=\{\{a\},\{b\}\} &{} \Pi _{2}=\{\{a,b\}\} \\ e_{1}(a)=(1,2) &{} e_{2}(a)=(1,1) \\ e_1(b)=(2,1) &{} e_2(b)=(1,1) \\ u_{i}(a, x,y)= \sqrt{xy} &{} u_i(b,x,y)=xy\end{array} \end{aligned}$$

Notice that since the first agent is fully informed, the information structure $\Pi$ satisfies the assumption that for any state s there exists an agent i such that $\Pi _i(s)=\{s\}$. The following feasible allocation

$$\begin{aligned} (x_i(a), y_i(a))=\left( 1, \frac{3}{2}\right) \quad (x_i(b), y_i(b))=\left( \frac{3}{2},1\right) \quad {{\mathrm{for}}\,\,{\mathrm{any}}\,\,}i\in I \end{aligned}$$

is ex post efficient. Indeed assume to the contrary the existence of an alternative feasible allocation (t, z) such that $t_i(s)z_i(s)\ge \frac{3}{2}$ for all $i \in I$ and $s\in S$, with at least one strict inequality.

Without loss of generality let $t_1(a)z_1(a)> \frac{3}{2}$, which means that²²$z_1(a)>\frac{3}{2t_1(a)}$. This together with feasibility imply that

$$\begin{aligned} (2-t_1(a))\left( 3-\frac{3}{2t_1(a)}\right) > (2-t_1(a))\left( 3-z_1(a)\right) =t_2(a)z_2(a)\ge \frac{3}{2}, \end{aligned}$$

which causes the contradiction $(t_1(a) -1)^2<0$. Hence, (x, y) is ex post Pareto optimal. We now show that it is not maxmin efficient with respect to the information structure $\Pi$. To this end consider the following feasible allocation

$$\begin{aligned} (t_i(a), z_i(a))= & {} (x_i(a), y_i(a))\quad {{\mathrm{for}}\,\,{{\mathrm{any}}}\,\,}i \in I,\\ (t_1(b), z_1(b))= & {} \left( \frac{7}{4},1\right) \\ (t_2(b), z_2(b))= & {} \left( \frac{5}{4},1\right) , \end{aligned}$$

and notice that

$$\begin{aligned} \underline{u}_1^{\Pi _1}(a,t_1,z_1)= & {} \underline{u}_1^{\Pi _1}(a,x_1,y_1)\\ \underline{u}_1^{\Pi _1}(b,t_1,z_1)= & {} \frac{7}{4}>\frac{3}{2}=\underline{u}_1^{\Pi _1}(b,x_1,y_1)\\ \underline{u}_2^{\Pi _2}(a,t_2,z_2)= & {} \underline{u}_2^{\Pi _2}(b,t_2,z_2)= \min \left\{ \sqrt{\frac{3}{2}}, \frac{5}{4}\right\} =\sqrt{\frac{3}{2}}\\= & {} \min \left\{ \sqrt{\frac{3}{2}}, \frac{3}{2}\right\} =\underline{u}_2^{\Pi _2}(b,x_2,y_2)=\underline{u}_2^{\Pi _2}(a,x_2,y_2). \end{aligned}$$

Thus, the allocation (x, y) is ex post efficient but not maxmin Pareto optimal with respect to the information structure $\Pi$. $\square$

The next example shows that the assumption that for any state s there exists an agent $i\in I$ such that $\Pi _i(s)=\{s\}$ is crucial in the proof of Proposition 5.3.

Example 8.4

Consider an asymmetric information economy with two agents $I=\{1,2\}$, two goods and three states $S=\{a,b,c\}$, whose primitives are given as follows:

$$\begin{aligned} \begin{array}{ll} \Pi _{1}=\{\{a\},\{b,c\}\} &{} \Pi _{2}=\{\{a,b\}, \{c\}\} \\ e_{1}(a)=(4,4) &{} e_2(a)=(0,0) \\ e_{1}(b)=(2,2) &{} e_2(b)=(2,2) \\ e_{1}(c)=(0,0) &{} e_2(c)=(4,4), \end{array} \end{aligned}$$

$u_{i}(\cdot , x,y)=xy$ for any $i \in I$. Notice that $\{b\}\ne \Pi _i(b)$ for any $i \in I$. The following feasible allocation $(x_i(s),y_i(s))=e_i(s)$ for any i and any $s\ne b$; $(x_{1}(b),y_{1}(b))=(1,3)$ and $(x_{2}(b),y_{2}(b))=(3,1)$ is not ex post efficient since it is blocked by the initial endowment, but it is maxmin Pareto optimal with respect to the information structure $(\Pi _1, \Pi _2)$. Indeed, assume by the way of contradiction the existence of an alternative feasible allocation (t, z) such that

$$\begin{aligned}&(i)\quad t_1(a)z_1(a))\ge 16\\&(ii)\quad \min \{t_1(b)z_1(b),t_1(c)z_1(c)\}\ge \min \{3,0\}=0\\&(iii)\quad \min \{t_2(a)z_2(a),t_2(b)z_2(b)\}\ge \min \{0,3\}=0\\&(iv)\quad t_2(c)z_2(c)\ge 16, \end{aligned}$$

with at least one strict inequality. If one of (i) and (iii) is strict, then $(4- t_1(a))\left( 4-\frac{16}{t_1(a)}\right) >0$ or equivalently that $(t_1(a)-4)^2<0$ which is a contradiction. Similarly if one of (ii) and (iv) is strict.

Proof of Theorem 5.4

Let (p, x) be a maxmin rational expectations equilibrium.

I CASE:  If $\sigma (u_i, e_i)\subseteq {\mathcal {F}}_i$ for each $i \in I$, Lemma 8.2 ensures that x is an ex post Walrasian equilibrium allocation and therefore it is ex post efficient. We now show that it is also maxmin Pareto optimal. To this end, assume to the contrary that there exists an alternative feasible allocation y such that $\underline{u}^{REE}_i({ s }, y_i) \ge \underline{u}^{REE}_i({ s }, x_i)$ for all $i\in I$ and all $s\in S$, with at least one strict inequality. Proposition 4.11 implies that for any agent $i\in I$ and any state $s\in S$

$$\begin{aligned} u_i(s, y_i(s))\ge \underline{u}_i^{REE} (s, y_i)\ge \underline{u}_i^{REE} (s, x_i)=u_i(s, x_i(s)), \end{aligned}$$

with at least one strict inequality. This means that x is not ex post efficient which is a contradiction.

II CASE:   Assume that p is fully revealing. Clearly since ${\mathcal {G}}^p_i(s)=\{s\}$ for all i and s, maxmin Pareto optimality with respect to the information structure ${\mathcal {G}}^p$ coincides with the ex post efficiency. We have already observed that in this case a maxmin REE is an ex post Walrasian equilibrium and hence it is both ex post and maxmin efficient.

Example 8.5 and Remark 5.11 show that if none of the above conditions is satisfied, a maxmin REE may not be maxmin efficient. $\square$

Proof of Proposition 5.8

Let x be a weak maxmin efficient allocation and assume, on the contrary, that there exists an alternative feasible allocation y such that $u_i({ s }, y_i( s )) > u_i({ s }, x_i( s ))$ for all $i\in I$ and all $s\in S$. Thus, for each agent $i\in I$ whatever her information partition is $\Pi _i$, it follows that $\underline{u}_i^{\Pi _i}( s , y_i)>\underline{u}_i^{\Pi _i}( s , x_i)$ for each state s. Hence, we get a contradiction since x is weak maxmin Pareto optimal. In order to show that the converse may not be true, consider an economy with two agents, three states of nature, $S =\{a,b,c\}$, and two goods, such that

$$\begin{aligned} u_i(a, x_i,y_i)= & {} \sqrt{x_i y_i} \quad u_i(b, x_i,y_i)=x_i y_i\quad u_i(c, x_i,y_i)= x_i^2 y_i \quad {{\mathrm{for}}\,\,{{\mathrm{all}}}\,\,}i=1,2.\\ e_1(a)= & {} (2,1)\quad e_2(a)=e_1(b)=e_2(b)=e_1(c)=e_2(c)=(1,2)\\ \Pi _1= & {} \{\{a,c\},\{b\}\}\quad \Pi _2=\{\{a\},\{ b,c\}\}. \end{aligned}$$

Consider the following feasible allocation:

$$\begin{aligned} (x_1(a), y_1(a))= & {} \left( 3, \frac{1}{3}\right) \quad (x_2(a), y_2(a))= \left( 0, \frac{8}{3}\right) \\ (x_1(b), y_1(b))= & {} \left( 1,2\right) \quad (x_2(b), y_2(b))= \left( 1,2\right) ,\\ (x_1(c), y_1(c))= & {} \left( 2,1\right) \quad (x_2(c), y_2(c))= \left( 0,3\right) . \end{aligned}$$

Notice that it is weak ex post efficient. Indeed, if on the contrary there exists (t, z) such that

$$\begin{aligned} u_i(s, t_i(s),z_i(s))>u_i(s, x_i(s),y_i(s))\quad {{\mathrm{for}}\,\,{{\mathrm{all}}}\,\,}i \in I\,\,{{\mathrm{and}}\,\,{{\mathrm{all}}}\,\,}s \in S, \end{aligned}$$

in particular,

$$\begin{aligned} \left\{ \begin{array}{l} t_1(b)z_1(b)> 2\\ t_2(b)z_2(b)> 2\\ t_1(b)+t_2(b)=2\\ z_1(b)+z_2(b)=4, \end{array} \right. \end{aligned}$$

then²³

$$\begin{aligned} \left\{ \begin{array}{l} z_1(b)> \frac{2}{t_1(b)}\\ (2-t_1(b))(2t_1(b) -1)>t_1(b). \end{array} \right. \end{aligned}$$

This implies that $(t_1(b)-1)^2<0$, which is impossible. Thus, the above allocation is weak ex post Pareto optimal, but it is not weak maxmin efficient with respect to the information structure $\Pi$, since it is (maxmin) blocked by the following feasible allocation:

$$\begin{aligned} (t_1(a), z_1(a))= & {} \left( \frac{5}{4},\frac{5}{2}\right) \quad (t_2(a), z_2(a))= \left( \frac{7}{4},\frac{1}{2}\right) \\ (t_1(b), z_1(b))= & {} \left( 1,\frac{8}{3}\right) \quad (t_2(b), z_2(b))= \left( 1,\frac{4}{3}\right) \\ (t_1(c), z_1(c))= & {} \left( \frac{3}{4}, {2}\right) \quad (t_2(c), z_2(c))= \left( \frac{5}{4}, {2}\right) . \end{aligned}$$

Indeed,

$$\begin{aligned} \underline{u}_1^{\Pi _1}(a,t_1, z_1)= & {} \underline{u}_1^{\Pi _1}(c,t_1, z_1)=\min \left\{ \sqrt{\frac{25}{8}}, \frac{9}{8}\right\} =\frac{9}{8}\\> & {} 1=\min \{1,4\}=\underline{u}_1^{\Pi _1}(c, x_1,y_1)=\underline{u}_1^{\Pi _1}(a, x_1,y_1)\\ \underline{u}_1^{\Pi _1}(b,t_1, z_1)= & {} {u}_1(b,t_1(b), z_1(b))=\frac{8}{3}\\> & {} 2= {u}_1(b, x_1(b),y_1(b))=\underline{u}_1^{\Pi _1}(b,x_1, y_1)\\ \underline{u}_2^{\Pi _2}(a,t_2, z_2)= & {} {u}_2(a,t_2(a), z_2(a))=\sqrt{\frac{7}{8}}\\> & {} 0= {u}_2(a, x_2(a),y_2(a))=\underline{u}_2^{\Pi _2}(a,x_2, y_2)\\ \underline{u}_2^{\Pi _2}(b,t_2, z_2)= & {} \underline{u}_2^{\Pi _2}(c,t_2, z_2)= \min \left\{ \frac{4}{3}, \frac{25}{8}\right\} = \frac{4}{3}\\> & {} 0=\min \{2, 0\}=\underline{u}_2^{\Pi _2}(c,x_2, y_2)=\underline{u}_2^{\Pi _2}(b,x_2, y_2). \end{aligned}$$

$\square$

Proof of Theorem 5.9

Clearly in the first two cases the result easily follows from Theorem 5.4 and from the observation that any allocation maxmin efficient with respect to $\Pi$ is weak maxmin Pareto optimal with respect to $\Pi$.

Let (p, x) be a maxmin rational expectations equilibrium, and assume to the contrary that there exists an alternative feasible allocation y such that $\underline{u}^{REE}_i({ s }, y_i) > \underline{u}^{REE}_i({ s }, x_i)$ for all $i\in I$ and all $s \in S$.

III CASE: there exists a state of nature ${\bar{s}} \in S$, such that $\{{\bar{s}} \}={\mathcal {G}}^p_i({\bar{s}} )$ for all $i\in I$.

Since for each $i \in I$, $\{{\bar{s}} \}={\mathcal {G}}^p_i({\bar{s}} )$; it follows that $\underline{u}^{REE}_i({{\bar{s}} }, y_i)=u_i({\bar{s}} , y_i({\bar{s}} )) > u_i({\bar{s}} , x_i({\bar{s}} ))= \underline{u}^{REE}_i({{\bar{s}} }, x_i)$ for all $i\in I$. Hence, since (p, x) is a MREE, for each agent i there exists at least one state $s _i\in {\mathcal {G}}^p_i({\bar{s}} )=\{{\bar{s}} \}$ (that is $s _i={\bar{s}}$ for all $i\in I$) such that $p({\bar{s}} )\cdot y_i({\bar{s}} )> p({\bar{s}} )\cdot e_i({\bar{s}} )$. Therefore,

$$\begin{aligned} \sum _{i \in I}p({\bar{s}} )[y_i({\bar{s}} )-e_i({\bar{s}} )]>0, \end{aligned}$$

which contradicts the feasibility of y.

IV CASE: $n-1$ agents are fully informed.

Since (p, x) is a MREE and y is preferred by anyone to x, it follows that for any state $s\in S$ and any agent $i\in I$ there exists at least one state $s _i\in {\mathcal {G}}^p_i( s )$ such that $p( s _i)\cdot y_i( s _i)> p( s _i)\cdot e_i( s _i)$. Let j be the unique not fully informed agent, and consider the state $s _j$ for which $p( s _j)\cdot y_j( s _j)> p( s _j)\cdot e_j( s _j)$. Since each agent $i \ne j$ is fully informed, it follows that ${\mathcal {G}}^p_i( s _j)=\{ s _j\}$ for all $i \ne j$. Thus,

$$\begin{aligned} p( s _j)\cdot y_i( s _j)> p( s _j)\cdot e_i( s _j)\quad {{\mathrm{for}}\,\,{{\mathrm{all}}}\,\,} i \in I. \end{aligned}$$

Hence,

$$\begin{aligned} \sum _{i \in I} p( s _j)\cdot y_i( s _j)> \sum _{i \in I}p( s _j)\cdot e_i( s _j), \end{aligned}$$

which is a contradiction.

Example 8.5 below shows that if no condition of Theorem 5.9 is satisfied, then a maxmin REE may not be weak maxmin efficient (and a fortiori it may not be maxmin Pareto optimal). $\square$

Example 8.5

Consider an asymmetric information economy with three states of nature, $S =\{a, b, c\}$, two goods, $\ell =2$ (the first good is considered as numeraire) and three agents, $I=\{1,2,3\}$ whose characteristics are given as follows:

$$\begin{aligned} \left. \begin{array}{lll} e_1(a)=e_1(b)=(2,1)&{} e_1(c)=(3,1)&{} {\mathcal {F}}_1=\{\{a,b\};\{c\}\} \\ e_2(a)=e_2(c)=(1,2)&{} e_2(b)=(2,2)&{} {\mathcal {F}}_2=\{\{a,c\};\{b\}\} \\ e_3(b)=e_3(c)=(2,1)&{} e_3(a)=(3,1)&{} {\mathcal {F}}_3=\{\{a\};\{b,c\}\}.\\ u_1(a,x,y)= \sqrt{xy} &{} u_1(b,x,y)=log(xy)&{} u_1(c,x,y)=\sqrt{xy},\\ u_2(a,x,y)= log(xy) &{} u_2(b,x,y)=\sqrt{xy}&{} u_2(c,x,y)=\sqrt{xy},\\ u_3(a,x,y)= \sqrt{xy} &{} u_3(b,x,y)=\sqrt{xy}&{} u_3(c,x,y)=log(xy). \end{array} \right. \end{aligned}$$

Consider the following maxmin rational expectations equilibrium

$$\begin{aligned} \left. \begin{array}{llll} (p(a), q(a))=\left( 1,\frac{3}{2}\right) &{}(x_1(a), y_1(a))=\left( \frac{7}{4}, \frac{7}{6}\right) &{}(x_2(a), y_2(a))=\left( 2,\frac{4}{3}\right) &{}(x_3(a), y_3(a))=\left( \frac{9}{4}, \frac{3}{2}\right) \\ (p(b), q(b))=\left( 1,\frac{3}{2}\right) &{}(x_1(b), y_1(b))=\left( \frac{7}{4}, \frac{7}{6} \right) &{}(x_2(b), y_2(b))=\left( \frac{5}{2}, \frac{5}{3}\right) &{}(x_3(b), y_3(b))=\left( \frac{7}{4}, \frac{7}{6}\right) \\ (p(c), q(c))=\left( 1,\frac{3}{2}\right) &{}(x_1(c), y_1(c))=\left( \frac{9}{4}, \frac{3}{2}\right) &{}(x_2(c), y_2(c))=\left( 2, \frac{4}{3}\right) &{}(x_3(c), y_3(c))=\left( \frac{7}{4}, \frac{7}{6}\right) ,\\ \end{array} \right. \end{aligned}$$

and notice that it is a non revealing equilibrium, since $(p(a), q(a))= (p(b), q(b))= (p(c),q(c))$ and hence $\sigma (p,q)=\{\{a,b,c\}\}$, that is ${\mathcal {G}}^p_i={\mathcal {F}}_i$ for any $i\in I$. Moreover, notice that no condition of Theorems 5.4 and 5.9 is satisfied. We now show that the equilibrium allocation is not weak maxmin Pareto optimal with respect to the information structure ${\mathcal {G}}^p=({\mathcal {G}}^p_{i})_{i\in I}$ and a fortiori it is neither maxmin efficient. Indeed, consider the following feasible allocation

$$\begin{aligned} (t_1(a), z_1(a))= & {} \left( \frac{20}{12}, \frac{13}{12}\right) \quad (t_2(a), z_2(a))= \left( \frac{25}{12}, \frac{16}{12}\right) \quad (t_3(a), z_3(a))= \left( \frac{27}{12}, \frac{19}{12}\right) \\ (t_1(b), z_1(b))= & {} \left( \frac{22}{12}, \frac{14}{12}\right) \quad (t_2(b), z_2(b))= \left( \frac{30}{12}, \frac{21}{12}\right) \quad (t_3(b), z_3(b))= \left( \frac{20}{12}, \frac{13}{12}\right) \\ (t_1(c), z_1(c))= & {} \left( \frac{28}{12}, \frac{18}{12}\right) \quad (t_2(c), z_2(c))= \left( \frac{23}{12}, \frac{15}{12}\right) \quad (t_3(c), z_3(c))= \left( \frac{21}{12}, \frac{15}{12}\right) , \end{aligned}$$

and notice that,

$$\begin{aligned} \underline{u}_1^{REE}(a,t_1, z_1)= & {} \underline{u}_1^{REE}(b,t_1, z_1) = \min \left\{ \sqrt{\frac{260}{144}}, log \frac{308}{144}\right\} = log \frac{308}{144}> log \frac{49}{24}\\= & {} \min \left\{ \sqrt{\frac{49}{24}}, log \frac{49}{24}\right\} = \underline{u}_1^{REE}(a, x_1,y_1)=\underline{u}_1^{REE}(b, x_1,y_1),\\ \underline{u}_1^{REE}(c,t_1, z_1)= & {} {u}_1 (c,t_1(c), z_1(c)) = \sqrt{\frac{504}{144}}> \sqrt{ \frac{27}{8}}\\= & {} {u}_1 (c, x_1(c),y_1(c))=\underline{u}_1^{REE}(c, x_1,y_1),\\ \underline{u}_2^{REE}(a,t_2, z_2)= & {} \underline{u}_2^{REE}(c,t_2, z_2) = \min \left\{ log \frac{400}{144}, \sqrt{\frac{345}{144}} \right\} = log \frac{400}{144}> log \frac{8}{3}\\= & {} \min \left\{ log \frac{8}{3}, \sqrt{\frac{8}{3}}\right\} = \underline{u}_2^{REE}(a, x_2,y_2)=\underline{u}_2^{REE}(c, x_2,y_2),\\ \underline{u}_2^{REE}(b,t_2, z_2)= & {} {u}_2 (b,t_2(b), z_2(b)) = \sqrt{\frac{630}{144}}> \sqrt{ \frac{25}{6}}= {u}_2 (b, x_2(b),y_2(b))\\= & {} \underline{u}_2^{REE}(b, x_2,y_2),\\ \underline{u}_3^{REE}(a,t_3, z_3)= & {} {u}_3 (a,t_3(a), z_3(a)) = \sqrt{\frac{513}{144}}> \sqrt{ \frac{27}{8}}= {u}_3 (a, x_3(a),y_3(a))\\= & {} \underline{u}_3^{REE}(a, x_3,y_3),\\ \underline{u}_3^{REE}(b,t_3, z_3)= & {} \underline{u}_3^{REE}(c,t_3, z_3) = \min \left\{ \sqrt{\frac{260}{144}}, log \frac{315}{144}\right\} = log \frac{315}{144} > log \frac{49}{24}\\= & {} \min \left\{ \sqrt{\frac{49}{24}}, log \frac{49}{24}\right\} = \underline{u}_3^{REE}(b, x_3,y_3)=\underline{u}_3^{REE}(c, x_3,y_3). \end{aligned}$$

Hence, the equilibrium allocation (x, y) is not weak maxmin Pareto optimal with respect to the information structure ${\mathcal {G}}^p=({\mathcal {G}}^p_{i})_{i\in I}$.

The following example shows that if there exists a state that everybody may distinguish (see condition (iii) of Theorem 5.9) then according to Theorem 5.9, a maxmin REE allocation is weak maxmin efficient with respect to the information structure $({\mathcal {G}}^p_i)_{i\in I}$, but it is not maxmin Pareto optimal.

Example 8.6

Consider an asymmetric information economy with five states of nature, $S =\{a, b, c, d, f\}$, two goods and two agents, $I=\{1,2\}$ whose characteristics are given as follows:

$$\begin{aligned} \left. \begin{array}{l@{\quad }l@{\quad }l} e_1(a)=e_1(b)=(1,2)&{} e_1(c)=e_1(d)=e_1(f)=(2,1)&{} {\mathcal {F}}_1=\{\{a,b\};\{c,d\};\{f\}\} \\ e_2(a)=e_2(c)=e_2(d)=e_2(f)=(2,1)&{} e_2(b)=(1,2)&{} {\mathcal {F}}_2=\{\{a,c\};\{b\};\{d,f\}\}.\\ u_i(a,x,y)= u_i(c,x,y)=\sqrt{xy} &{} u_i(b,x,y)=u_i(d,x,y)=log(xy)&{} u_i(f,x,y)={xy}. \end{array} \right. \end{aligned}$$

Consider the following maxmin rational expectations equilibrium

$$\begin{aligned} \left. \begin{array}{lll} (p(a), q(a))=\left( 1,1\right) &{}(x_1(a), y_1(a))=\left( \frac{3}{2}, \frac{3}{2}\right) &{}(x_2(a), y_2(a))=\left( \frac{3}{2},\frac{3}{2}\right) \\ (p(b), q(b))=\left( 1,\frac{1}{2}\right) &{}(x_1(b), y_1(b))=\left( 1,2 \right) &{}(x_2(b), y_2(b))=\left( 1,2\right) \\ (p(c), q(c))=\left( 1, {2}\right) &{}(x_1(c), y_1(c))=\left( 2,1\right) &{}(x_2(c), y_2(c))=\left( 2, 1\right) \\ (p(d), q(d))=\left( 1,2\right) &{}(x_1(d), y_1(d))=\left( 2,1\right) &{}(x_2(d), y_2(d))=\left( 2,1\right) \\ (p(f), q(f))=\left( 1,2\right) &{}(x_1(f), y_1(f))=\left( 2,1\right) &{}(x_2(f), y_2(f))=\left( 2,1\right) , \end{array} \right. \end{aligned}$$

and notice that $\sigma (p,q)=\{\{a\},\{b\},\{c,d,f\}\}$ and hence, ${\mathcal {G}}^p_1=\{\{a\},\{b\},\{c,d\},\{f\}\}$ and ${\mathcal {G}}^p_2=\{\{a\},\{b\},\{c\},\{d,f\}\}$.

For any $i\in I$ the equilibrium allocation $(x_i,y_i)$ is ${\mathcal {G}}^p_i$-measurable but not ${\mathcal {F}}_i$-measurable. Moreover notice that the utility functions are not ${\mathcal {F}}_i$-measurable neither ${\mathcal {G}}^p_i$-measurable, the equilibrium price is not fully revealing, and no agent is fully informed. On the other hand, there exists a state s such that ${\mathcal {G}}^p_i(s)=\{s\}$ for any agent i, for example states a and b, but such a condition does not hold for the initial information structure $({\mathcal {F}}_i)_{i \in I}$. Thus, only condition (iii) of Theorem 5.9 is satisfied. From this it follows that the equilibrium allocation (x, y) is weak efficient with respect to the information structure $({\mathcal {G}}^p_i)_{i \in I}$. We now show that x is not maxmin efficient with respect to the information structure $({\mathcal {G}}^p_i)_{i \in I}$. To this end, consider the following feasible allocation

$$\begin{aligned} (t_i(s), z_i(s))= & {} (x_i(s), y_i(s))\quad {{\mathrm{for}} \,\,{{\mathrm{any}}}\,\,}i=\{1,2\} \,\,{{\mathrm{and}}\,\,{{\mathrm{any}}}\,\,}s\in \{a,b,d\}\\ (t_1(c), z_1(c))= & {} \left( \frac{3}{2}, 1\right) \quad (t_2(c), z_2(c))= \left( \frac{5}{2},1\right) \\ (t_1(f), z_1(f))= & {} \left( \frac{5}{2}, 1\right) \quad (t_2(f), z_2(f))= \left( \frac{3}{2},1\right) , \end{aligned}$$

and notice that,

$$\begin{aligned} \underline{u}_i^{REE}(s,t_i, z_i)= & {} \underline{u}_i^{REE}(s, x_i, y_i)\quad {{\mathrm{for}}\,\,{{\mathrm{any}}}\,\,}i\in \{1,2\}\,\,{{\mathrm{and}}\,\,{{\mathrm{any}}}\,\,}s\in \{a,b\}\\ \underline{u}_1^{REE}(c,t_1, z_1)= & {} \underline{u}_1^{REE}(d,t_1, z_1)=\min \left\{ \sqrt{\frac{3}{2}}, log 2\right\} = log2\\= & {} \min \{\sqrt{2}, log2\}=\underline{u}_1^{REE}(d,x_1, y_1)=\underline{u}_1^{REE}(c,x_1, y_1)\\ \underline{u}_2^{REE}(c,t_2, z_2)= & {} u_2(c, t_2(c), z_2(c))=\sqrt{\frac{5}{2}}\\> & {} \sqrt{2}= u_2(c,x_2(c), y_2(c))= \underline{u}_2^{REE}(c,x_2, y_2)\\ \underline{u}_1^{REE}(f,t_1, z_1)= & {} u_1(f, t_1(f), z_1(f))= {\frac{5}{2}}\\> & {} {2}= u_1(f,x_1(f), y_1(f))= \underline{u}_1^{REE}(f,x_1, y_1)\\ \underline{u}_2^{REE}(d,t_2, z_2)= & {} \underline{u}_2^{REE}(f,t_2, z_2)=\min \left\{ log 2, \frac{3}{2}\right\} =log2\\= & {} \min \{log2,2\}=\underline{u}_2^{REE}(f,x_2, y_2)=\underline{u}_2^{REE}(d,x_2, y_2). \end{aligned}$$

Hence, the equilibrium allocation is not maxmin Pareto optimal with respect to the information structure ${\mathcal {G}}^p=({\mathcal {G}}^p_{i})_{i\in I}$.

The next example shows that if all agents except one are fully informed (i.e., condition (iv) of Theorem 5.9 holds), then a maxmin REE allocation is weak maxmin efficient with respect to the information structure $({\mathcal {G}}^p_i)_{i\in I}$ but it may not be maxmin Pareto optimal.

Example 8.7

Consider an asymmetric information economy with two states of nature, $S =\{a, b\}$, two goods and three agents, $I=\{1,2,3\}$ whose characteristics are given as follows:

$$\begin{aligned} \left. \begin{array}{ll} e_1(a)=e_1(b)=\left( \frac{1}{3},\frac{1}{3}\right) &{} {\mathcal {F}}_1=\{\{a\};\{b\}\} \\ e_2(a)=e_2(b)=\left( \frac{1}{3},\frac{1}{3}\right) &{} {\mathcal {F}}_2=\{\{a\};\{b\}\}.\\ e_3(a)=e_3(b)=\left( \frac{1}{3},\frac{1}{3}\right) &{} {\mathcal {F}}_3=\{\{a,b\}\}.\\ u_i(a,x,y)= \sqrt{xy} &{} u_i(b,x,y)=xy\,\,{{\mathrm{for}}\,\,{\mathrm{all}}\,\,}i\in I. \end{array} \right. \end{aligned}$$

Notice that for any $i \in I$$e_i(\cdot )$ is ${\mathcal {F}}_i$-measurable, while $u_i$ is not. Two agents are fully informed. The initial endowment is a non-revealing maxmin rational expectations equilibrium and there does not exist a state s such that ${\mathcal {G}}^p_i(s)=\{s\}$ for any i, neither ${\mathcal {F}}_i(s)=\{s\}$ for any i. Thus, only condition (iv) of Theorem 5.9 is satisfied. From this it follows that the equilibrium allocation e is weak maximin efficient with respect to the information structure $({\mathcal {G}}^p_i)_{i \in I}$, and since it is a non-revealing maxmin REE it is also weak maximin efficient with respect to the information structure $({\mathcal {F}}_i)_{i \in I}$ (because ${\mathcal {G}}^p_i={\mathcal {F}}_i$ for any $i\in I$). We now show that e is not maxmin efficient with respect to the information structure $({\mathcal {G}}^p_i)_{i \in I}$ and hence neither with respect to $({\mathcal {F}}_i)_{i \in I}$. To this end, consider the following feasible allocation

$$\begin{aligned} (t_i(a),z_i(a))= & {} \left( \frac{5}{12},\frac{5}{12}\right) \quad {{\mathrm{for}}\,\,{\mathrm{any}}\,\,}i\in \{1,2\}, \\ (t_3(a),z_3(a))= & {} \left( \frac{1}{6},\frac{1}{6}\right) , \\ (t_i(b),z_i(b))= & {} \left( \frac{1}{3},\frac{1}{3}\right) \quad {{\mathrm{for}}\,\,{{\mathrm{any}}}\,\,}i\in \{1,2,3\}. \end{aligned}$$

Notice that,

$$\begin{aligned} \underline{u}_i^{REE}(a,t_i, z_i)= & {} \frac{5}{12}>\frac{1}{3}=\underline{u}_i^{REE}(a,x_i, y_i)\quad {{\mathrm{for}}\,\,{{\mathrm{any}}}\,\,}i\in \{1,2\}\\ \underline{u}_i^{REE}(b,t_i, z_i)= & {} \underline{u}_i^{REE}(b, x_i, y_i)\quad {{\mathrm{for}}\,\,{{\mathrm{any}}}\,\,}i\in \{1,2\}\\ \underline{u}_3^{REE}(a,t_3, z_3)= & {} \underline{u}_3^{REE}(b,t_3, z_3)=\min \left\{ \frac{1}{6}, \frac{1}{9}\right\} =\frac{1}{9}\\= & {} \min \left\{ \frac{1}{3}, \frac{1}{9}\right\} =\underline{u}_3^{REE}(a,x_3, y_3)=\underline{u}_3^{REE}(b,x_3, y_3). \end{aligned}$$

Hence, the equilibrium allocation e is not maxmin Pareto optimal with respect to the information structure ${\mathcal {G}}^p=({\mathcal {G}}^p_{i})_{i\in I}$ neither with respect to ${\mathcal {F}}=({\mathcal {F}}_{i})_{i\in I}$.

Proof of Theorem 5.14

Let (p, x) be a maxmin rational expectations equilibrium and assume to the contrary that there exists an alternative feasible allocation y such that

$$\begin{aligned} \underline{u}_i(s, y_i)>\underline{u}_i (s, x_i)\quad \mathrm{{{ for}}\,\,{{\mathrm{all}}}\,\,}i\in I\,\,{{\mathrm{and}}\,\,}s\in S. \end{aligned}$$

(17)

(a) CASE:    If there exists a state of nature ${\bar{s}} \in S$, such that $\{{\bar{s}} \}={\mathcal {F}}_i({\bar{s}} )$ for all $i\in I$, then in particular from (17) it follows that for all $i\in I$

$$\begin{aligned} \underline{u}_i^{REE}(\bar{s}, y_i)=u_i(\bar{s}, y_i(\bar{s}))= \underline{u}_i (\bar{s}, y_i)>\underline{u}_i (\bar{s}, x_i)=u_i(\bar{s}, x_i(\bar{s}))= \underline{u}_i^{REE}(\bar{s}, x_i). \end{aligned}$$

Thus, since (p, x) is a maxmin rational expectations equilibrium for all $i\in I$ there exists a state $s_i\in {\mathcal {G}}^p_i(\bar{s})=\{\bar{s}\}$ (i.e., $s_i=\bar{s}$ for all $i\in I$) such that $p(s_i)\cdot y_i(s_i)>p(s_i)\cdot e_i(s_i)$, that is

$$\begin{aligned} p(\bar{s})\cdot y_i(\bar{s})>p(\bar{s})\cdot e_i(\bar{s})\quad \mathrm{{{ for}}\,\,{{\mathrm{all}}}\,\,}i\in I. \end{aligned}$$

Hence,

$$\begin{aligned} p(\bar{s}) \cdot \sum _{i\in I} [y_i(\bar{s})-e_i(\bar{s})]>0, \end{aligned}$$

which contradicts the feasibility of the allocation y. Thus, x is weak maxmin efficient with respect to the information structure ${\mathcal {F}}$. Moreover, notice that if there is a state of nature ${\bar{s}}$ such that ${\mathcal {F}}_i({\bar{s}})=\{{\bar{s}}\}$ for all $i \in I$, then a fortiori ${\mathcal {G}}^p_i({\bar{s}})=\{{\bar{s}}\}$ for all $i \in I$. This means that condition (iii) of Theorem 5.9 is satisfied and hence x is maxmin Pareto optimal also with respect to the information structure ${\mathcal {G}}^p$.

(b) CASE:    If the $n-1$ agents are fully informed, condition (iv) of Theorem 5.9 holds and hence x is weak maxmin efficient with respect to the information structure ${\mathcal {G}}^p$. We want to show that x is maxmin Pareto optimal also with respect to the information structure ${\mathcal {F}}$. To this end, assume without loss of generality that 1 is the unique non fully informed agent and let s be a state of nature. From (17) it follows in particular that there exists $\bar{s} \in {\mathcal {F}}_1(s)$ such that

$$\begin{aligned} \underline{u}_1^{REE}(\bar{s}, y_1)\ge \underline{u}_1 (\bar{s}, y_1)>\underline{u}_1 (\bar{s}, x_1)= \underline{u}_1^{REE}(\bar{s}, x_1). \end{aligned}$$

Since (p, x) is a maxmin REE, there exists a state $s^\prime \in {\mathcal {G}}^p_1(\bar{s})$ such that

$$\begin{aligned} p(s^\prime )\cdot y_1(s^\prime )> p(s^\prime )\cdot e_1(s^\prime ). \end{aligned}$$

(18)

Any agent $i\ne 1$ is fully informed, then (17) implies that

$$\begin{aligned} \underline{u}_i^{REE}(s^\prime , y_i)= u_i(s^\prime , y_i(s^\prime ))=\underline{u}_i (s^\prime , y_i)>\underline{u}_i (s^\prime , x_i)= u_i(s^\prime , x_i(s^\prime ))= \underline{u}_i^{REE}(s^\prime , x_i). \end{aligned}$$

Thus, for all $i\ne 1$ there exists a state $s_i\in {\mathcal {G}}^p_i(s^\prime )=\{s^\prime \}$ (i.e., $s_i=s^\prime$ for all $i\ne 1$, because they are all fully informed) such that $p(s_i)\cdot y_i(s_i)>p(s_i)\cdot e_i(s_i)$, that is

$$\begin{aligned} p(s^\prime )\cdot y_i(s^\prime )>p(s^\prime )\cdot e_i(s^\prime )\quad {{\mathrm{for}}\,\,{{\mathrm{all}}}\,\,}i\ne 1. \end{aligned}$$

(19)

Hence, from (18) and (19) it follows that

$$\begin{aligned} p(s^\prime ) \cdot \sum _{i\in I} [y_i(s^\prime )-e_i(s^\prime )]>0, \end{aligned}$$

which contradicts the feasibility of the allocation y.

We now show that if none of the above conditions is satisfied, then a maxmin REE may not be weak maxmin efficient with respect to the information structure ${\mathcal {F}}=({\mathcal {F}}_i)_{i \in I}$ and a fortiori may not be maxmin Pareto optimal. To this end, consider an asymmetric information economy with two states of nature, $S =\{a, b\}$, two goods, $\ell =2$ (the first good is considered as numeraire) and three agents, $I=\{1,2,3\}$ whose characteristics are given as follows:

$$\begin{aligned} \left. \begin{array}{lll} e_1(a)=(2,1)&{} e_1(b)=(1,2)&{} {\mathcal {F}}_1=\{\{a\};\{b\}\} \\ e_2(a)=(1,2)&{} e_2(b)=(1,2)&{} {\mathcal {F}}_2=\{\{a,b\}\} \\ e_3(a)=(2,1)&{} e_3(b)=(2,1)&{} {\mathcal {F}}_3=\{\{a,b\}\}.\\ u_1(s,x,y)= x^2y, &{} u_2(s,x,y)= \sqrt{xy}, &{} u_3(s,x,y)=xy, \quad {{\mathrm{for}}\,\,{\mathrm{any}}\,\,}s \in S. \end{array} \right. \end{aligned}$$

Notice that agents’ initial endowments and utility functions are private information measurable. Consider the following fully revealing maxmin rational expectations equilibrium

$$\begin{aligned}&(p(a), q(a))=\left( 1,1\right) \quad (p(b), q(b))=\left( 1,\frac{11}{17}\right) \\&\left. \begin{array}{lll} (x_1(a), y_1(a))=\left( 2,1 \right) &{}(x_2(a), y_2(a))=\left( \frac{3}{2},\frac{3}{2}\right) &{}(x_3(a), y_3(a))=\left( \frac{3}{2}, \frac{3}{2}\right) \\ \\ (x_1(b), y_1(b))=\left( \frac{26}{17}, \frac{13}{11} \right) &{}(x_2(b), y_2(b))=\left( \frac{39}{34}, \frac{39}{22}\right) &{}(x_3(b), y_3(b))=\left( \frac{45}{34}, \frac{45}{22}\right) .\\ \end{array} \right. \end{aligned}$$

The above fully revealing maxmin REE is maxmin efficient (and a fortiori weak maxmin Pareto optimal) with respect to the information structure ${\mathcal {G}}^p=({\mathcal {G}}^p_i)_{i \in I}$ (see Theorem 5.4). Of course it is also ex post efficient since it coincides with an ex post Walrasian equilibrium. On the other hand, we now show that it is not weak maxmin efficient (and a fortiori it is not maxmin Pareto optimal) with respect to the initial private information structure ${\mathcal {F}}=({\mathcal {F}}_i)_{i \in I}.$ To this end, consider the following feasible allocation (t, z)

$$\begin{aligned} \left. \begin{array}{ll} (t_1(a), z_1(a))= \left( \frac{33}{16}, 1\right) &{} (t_1(b), z_1(b))= \left( \frac{105}{68}, \frac{13}{11}\right) \\ \\ (t_2(a), z_2(a))= \left( \frac{22}{16}, \frac{3}{2}\right) &{} (t_2(b), z_2(b))= \left( \frac{79}{68}, \frac{39}{22}\right) \\ \\ (t_3(a), z_3(a))= \left( \frac{25}{16}, \frac{3}{2}\right) &{} (t_3(b), z_3(b))= \left( \frac{88}{68}, \frac{45}{22}\right) , \end{array} \right. \end{aligned}$$

and notice that,

$$\begin{aligned} \underline{u}_1 (a,t_1, z_1)= & {} u_1(a, t_1(a), z_1(a))= \left( \frac{33}{16}\right) ^2>4\\= & {} u_1(a, x_1(a), y_1(a))=\underline{u}_1(a,x_1, y_1)\\ \underline{u}_1 (b,t_1, z_1)= & {} u_1(b, t_1(b), z_1(b))= \left( \frac{105}{68}\right) ^2 \frac{13}{11}>\left( \frac{26}{17}\right) ^2 \frac{13}{11} \\= & {} u_1(b, x_1(b), y_1(b))=\underline{u}_1(b,x_1, y_1)\\ \underline{u}_2 (a,t_2, z_2)= & {} \underline{u}_2 (b,t_2, z_2)=\min \left\{ \sqrt{\frac{22}{16}\frac{3}{2}},\sqrt{\frac{79}{68}\frac{39}{22}} \right\} = \sqrt{\frac{79}{68}\frac{39}{22}}> \sqrt{\frac{39}{34}\frac{39}{22}} \\= & {} \min \left\{ {\frac{3}{2}},\sqrt{\frac{39}{34}\frac{39}{22}}\right\} = \underline{u}_2 (b,x_2, y_2)=\underline{u}_2 (a,x_2, y_2)\\ \underline{u}_3(a,t_3, z_3)= & {} \underline{u}_3 (b,t_3, z_3)=\min \left\{ \frac{25}{16} \frac{3}{2}, \frac{88}{68} \frac{45}{22} \right\} = \frac{25}{16} \frac{3}{2}> \frac{9}{4}\\= & {} \min \left\{ \frac{9}{4}, \frac{45}{34} \frac{45}{22} \right\} = \underline{u}_3(b,x_3, y_3)=\underline{u}_3 (a,x_3, y_3). \end{aligned}$$

Hence, the equilibrium allocation (x, y) is not weak maxmin Pareto optimal with respect to the information structure ${\mathcal {F}}=({\mathcal {F}}_{i})_{i\in I}$. $\square$

1.4 Proofs of Section 6

Before proving Proposition 6.6 the following lemma is needed.

Lemma 8.8

Condition (iii) and $(*)$ in the Definition 6.4, imply that for all $i \in C$,

$$\begin{aligned} u_i(a, x_i(a))=\min _{ s \in {\Pi _i}(a)} u_i( s , x_i( s ))=\underline{u}^{\Pi _i}_i(a, x_i), \end{aligned}$$

and

$$\begin{aligned} u_i(a, x_i(a))< u_i( s , x_i( s ))\quad for\,\,all\,\, s \in \Pi _i(a){\setminus }\{a\}. \end{aligned}$$

Proof

Assume, on the contrary, there exist an agent $i \in C$ and a state $s _1 \in {\Pi _i}(a){\setminus } \{a\}$ such that $\underline{u}^{\Pi _i}_i(a, x_i )=\min _{ s \in {\Pi _i}(a)} u_i( s , x_i( s ))=u_i( s _1, x_i( s _1))$.

Notice that

$$\begin{aligned} \underline{u}^{\Pi _i}_i(a, y_i )=\min \{u_i(a, e_i(a)+x_i(b)-e_i(b));\min _{ s \in {\Pi _i}(a){\setminus } \{a\}} u_i( s , x_i( s ))\}. \end{aligned}$$

If, $u_i(a, e_i(a)+x_i(b)-e_i(b))= u_i(a, y_i(a))=\underline{u}^{\Pi _i}_i(a, y_i )$, then in particular $u_i(a, y_i(a))\le u_i( s _1, x_i( s _1))=\underline{u}^{\Pi _i}_i(a, x_i )$. This contradicts (iii) in Definition 6.4. On the other hand, if there exists $s _2 \in {\Pi _i}(a){\setminus } \{a\}$ such that $u_i( s _2, x_i( s _2))= \underline{u}^{\Pi _i}_i(a, y_i)$, then in particular $\underline{u}^{\Pi _i}_i(a, y_i)=u_i( s _2, x_i( s _2))\le u_i( s _1, x_i( s _1))= \underline{u}^{\Pi _i}_i(a, x_i)$. This again contradicts (iii) in Definition 6.4. Thus, for each member i of C, there does not exist a state $s \in \Pi _i(a){\setminus }\{a\}$ such that $\underline{u}^{\Pi _i}_i(a, x_i)=u_i( s , x_i( s ))$. This means that

$$\begin{aligned} u_i(a, x_i(a))=\min _{ s \in {\Pi _i}(a)} u_i( s , x_i( s ))=\underline{u}^{\Pi _i}_i(a, x_i), \end{aligned}$$

and

$$\begin{aligned} u_i(a, x_i(a))< u_i( s , x_i( s ))\quad {{\mathrm{for}}\,\,{{\mathrm{all}}}}\,\, s \in \Pi _i(a){\setminus }\{a\}. \end{aligned}$$

$\square$

Proof of Proposition 6.6

Let x be a CIC with respect to the information structure $\Pi$ and assume to the contrary that there exist a coalition C and two states a and b such that

$$\begin{aligned}&(i)\quad {\Pi _i}(a)={\Pi _i}(b)\qquad {{{{\mathrm{for}}~{\mathrm{all}}}}}\,\, i \notin C,\\&(ii)\quad e_i(a)+x_i(b)-e_i(b) \in {\mathbb {R}}_+^\ell \qquad {{{{\mathrm{for}}~{\mathrm{all}}}}}\,\,i \in C,\,\,{\mathrm{and}}\\&(iii)\quad \underline{u}^{\Pi _i}_i(a, y_i)>\underline{u}^{\Pi _i}_i(a, x_i )\quad {{{{\mathrm{for}}~{{\mathrm{all}}}}}}\,\,i \in C, \end{aligned}$$

where for all $i \in C$,

$$\begin{aligned} y_i( s ) = \left\{ \begin{array}{ll} e_i(a)+x_i(b)-e_i(b) &{} \quad \text {if } s =a\\ x_i( s ) &{} \quad \text {otherwise}. \end{array} \right. \end{aligned}$$

Notice that from (iii) and Lemma 8.8 it follows that for all $i \in C$,

$$\begin{aligned} u_i(a, e_i(a)+x_i(b)-e_i(b))= u_i(a, y_i(a))\ge \underline{u}^{\Pi _i}_i(a, y_i )> \underline{u}^{\Pi _i}_{i}(a, x_i)=u_i(a, x_i(a)). \end{aligned}$$

Hence, x is not CIC with respect to the information structure $\Pi$, which is a contradiction. For the converse, we construct the following counterexample. Consider the economy, described in Example 6.2, with two agents, three states of nature, denoted by a, b and c, and one good per state denoted by x. Assume that

$$\begin{aligned} \left. \begin{array}{lll} u_1(\cdot , x_1)=\sqrt{x_1}; &{} e_1(a,b,c)=(20,20,0); &{} \mathcal {F}_1=\{\{a,b\};\{c\}\}. \\ u_2(\cdot , x_2)=\sqrt{x_2}; &{} e_2(a,b,c)=(20,0,20);&{} \mathcal {F}_2=\{\{a,c\};\{b\}\}. \end{array} \right. \end{aligned}$$

Consider the allocation

$$\begin{aligned} x_1(a,b,c)= & {} (20,10,10)\\ x_2(a,b,c)= & {} (20,10,10). \end{aligned}$$

We have already noticed that such an allocation is not Krasa-Yannelis incentive compatible with respect to the initial private information structure ${\mathcal {F}}=({\mathcal {F}}_1, {\mathcal {F}}_2)$ (see Example 6.2), but it is maxmin CIC with respect to ${\mathcal {F}}$ (see Remark 6.5). $\square$

Proof of Theorem 6.7

Let (p, x) be a maxmin rational expectations equilibrium. Since agents take into account the information generated by the equilibrium price p, the private information of each individual i is given by $\mathcal {G}^p_i=\mathcal {F}_i \vee \sigma (p)$. Thus, for each agent $i \in I$, $\Pi _i= \mathcal {G}_i$ and $\underline{u}_i^{\Pi _i}= \underline{u}_i^{REE}$. Assume to the contrary that (p, x) is not maxmin CIC. This means that there exist a coalition C and two states $a, b\in S$ such that

$$\begin{aligned}&(i)\quad \mathcal {G}^p_i(a)= \mathcal {G}^p_i(b)\qquad {{\mathrm{for}}\,\,{{\mathrm{all}}}\,\,} i \notin C,\\&(ii)\quad e_i(a)+x_i(b)-e_i(b) \in {\mathbb {R}}_+^\ell \qquad {{\mathrm{for}}\,\,{\mathrm{all}}\,}\,i \in C,\,\,{\mathrm{and}}\\&(iii)\quad \underline{u}^{REE}_i(a, y_i )>\underline{u}^{REE}_i(a, x_i )\quad {{\mathrm{for}}\,\,{{\mathrm{all}}}\,}\,i \in C, \end{aligned}$$

where for all $i \in C$,

$$\begin{aligned} y_i( s ) = \left\{ \begin{array}{ll} e_i(a)+x_i(b)-e_i(b) &{} \quad \text {if } s =a\\ x_i( s ) &{} \quad \text {otherwise}. \end{array} \right. \end{aligned}$$

Notice that condition (i) implies that $p(a)=p(b)$, meaning that the equilibrium price is partially revealing.²⁴ Clearly, if p is fully revealing, since for any $i \in I$, ${\mathcal {G}}^p_{i}={\mathcal {F}}$, then there does not exist a coalition C and two states a and b such that ${\mathcal {G}}^p_{i}(a)={\mathcal {G}}^p_{i}(b)$ for all $i \notin C$. Therefore, any fully revealing MREE is maxmin coalitional incentive compatible. On the other hand, since (p, x) is a maxmin rational expectations equilibrium, it follows from (iii) that for all $i\in C$ there exists a state $s _i \in \mathcal {G}^p_i (a)$ such that

$$\begin{aligned} p( s _i)\cdot y_i( s _i)> p( s _i) \cdot e_i( s _i)\ge p( s _i) \cdot x_i( s _i). \end{aligned}$$

By the definition of $y_i$, it follows that for all $i \in C$, $s _i =a$, that is $p(a)\cdot y_i(a)> p(a) \cdot e_i(a)$, and hence $p(a)\cdot [x_i(b)-e_i(b)]>0$. Furthermore, since $p(a)=p(b)$ it follows that $p(b)\cdot x_i(b)> p(b) \cdot e_i(b)$. This contradicts the fact that (p, x) is a maxmin rational expectations equilibrium. $\square$

Proof of Proposition 6.11

Let (p, x) be a maxmin REE and assume to the contrary that there exist a coalition C and two states $a, b\in S$ such that

$$\begin{aligned}&(I)\quad \mathcal {F}_i(a)=\mathcal {F}_i(b)\qquad {{\mathrm{for}}\,\,{{\mathrm{all}}}\,\,} i \notin C,\\&(II)\quad u_i(a, x_i(a))=u_i(a, x_i(b))\quad {{\mathrm{for}}\,\,{\mathrm{all}}\,\,}i \notin C,\\&(III)\quad e_i(a)+x_i(b)-e_i(b) \in {\mathbb {R}}_+^\ell \qquad {{\mathrm{for}}\,\,{{\mathrm{all}}}\,}\,i \in C,\,\,{\mathrm{and}}\\&(IV)\quad \underline{u}_i(a, y_i )>\underline{u}_i(a, x_i )\quad {{\mathrm{for}}\,\,{{\mathrm{all}}}\,}\,i \in C, \end{aligned}$$

where for all $i \in C$,

$$\begin{aligned} y_i( s ) = \left\{ \begin{array}{ll} e_i(a)+x_i(b)-e_i(b) &{} \quad \text {if } s =a\\ x_i( s ) &{} \quad \text {otherwise}. \end{array} \right. \end{aligned}$$

If (p, x) is a non revealing MREE, then the proposition holds true with no additional assumptions on utility functions (see Remark 6.8).

I CASE:   Assume that $\sigma (u_i)\subseteq {\mathcal {F}}_i$ for any $i \in I$. Observe that if p is partially revealing and ${\mathcal {G}}^p_i(a){\setminus } \{a\}\ne \emptyset$ for some agent i in C, then the allocation x is (private) maxmin coalitional incentive compatible and hence weak (private) maxmin CIC. Indeed, from Lemma 8.8 and condition (IV), it follows that

$$\begin{aligned} \underline{u}^{REE}_i(a, x_i)=\underline{u}_i(a, x_i)=u_i(a, x_i(a))< u_i( s , x_i( s ))\quad {{\mathrm{for}}\,\,{\mathrm{all}}\,\,} s \in {\mathcal {F}}_i(a){\setminus }\{a\}. \end{aligned}$$

In particular the above inequality holds for all $s \in \mathcal G_i(a){\setminus }\{a\}$, and this contradicts Proposition 4.11. Moreover, if for some agent $i \notin C$, ${\mathcal {G}}^p_i(a)= {\mathcal {G}}^p_i(b)$, then it follows that $p(a)=p(b)$, and hence p is partially revealing. However, even if utility functions are not private information measurable, we can conclude that x is (private) maxmin coalitional incentive compatible and hence weak (private) maxmin CIC. In fact, from (IV) and Lemma 8.8, it follows that for all $i \in C$,

$$\begin{aligned} \underline{u}_i^{REE}(a, y_i)\ge \underline{u}_i(a, y_i)>\underline{u}_i(a, x_i)={u}_i(a, x_i(a))=\underline{u}_i^{REE}(a, x_i). \end{aligned}$$

Therefore, since (p, x) is a maxmin REE, from the definition of the allocation y, it follows that for each $i \in C$, $p(a)\cdot y_i(a)>p(a)\cdot e_i(a)$, and hence $p(a)\cdot x_i(b)>p(a)\cdot e_i(b)$, which is a contradiction because $p(a)=p(b)$.

Thus, let us assume that ${\mathcal {G}}^p_i(a)=\{a\}$ for all $i \in C$ and ${\mathcal {G}}^p_i(a)\ne {\mathcal {G}}^p_i(b)$ for any $i \notin C$. Again from (IV) and Lemma 8.8, it follows that for all $i \in C$,

$$\begin{aligned} \underline{u}_i^{REE}(a, y_i)\ge \underline{u}_i(a, y_i)>\underline{u}_i(a, x_i)={u}_i(a, x_i(a))=\underline{u}_i^{REE}(a, x_i), \end{aligned}$$

while from (II) it follows that for all $i \notin C$,

$$\begin{aligned} \underline{u}_i^{REE}(a, y_i)= & {} \min \left\{ \min _{ s \in {\mathcal {G}}^p_i(a){\setminus } \{a\}} u_i( s , x_i( s )), u_i(a, y_i(a))\right\} \\= & {} \min \left\{ \min _{ s \in {\mathcal {G}}^p_i(a){\setminus } \{a\}} u_i( s , x_i( s )), u_i(a, x_i(b))\right\} \\= & {} \min \left\{ \min _{ s \in {\mathcal {G}}^p_i(a){\setminus } \{a\}} u_i( s , x_i( s )), u_i(a, x_i(a))\right\} \\= & {} \underline{u}_i^{REE}(a, x_i). \end{aligned}$$

Moreover, y is feasible. Indeed, for each state $s \ne a$, y is feasible because so is x. On the other hand, if $s =a$, then

$$\begin{aligned} \sum _{i \in I} y_i(a)=\sum _{i \in I} e_i(a) + \sum _{i \in I} x_i(b)- \sum _{i \in I} e_i(b) = \sum _{i \in I} e_i(a). \end{aligned}$$

Hence, there exists a feasible allocation y such that

$$\begin{aligned} \underline{u}_i^{REE}( s , y_i)\ge \underline{u}_i^{REE}( s , x_i)\quad {{\mathrm{for}}\,\,{\mathrm{all}}\,\,}i \in I\,\,{{\mathrm{and}}\,\,{\mathrm{all}}\,\,} s \in S , \end{aligned}$$

with a strict inequality for each $i \in C$ in state a. Since x is a maxmin REE and ${\mathcal {G}}^p_i(a)=\{a\}$ for all $i \in C$, it follows that

$$\begin{aligned} p(a)\cdot y_i(a)> p(a) \cdot e_i(a)\quad {{\mathrm{for}}\,\, {\mathrm{any}}} \,\,i \in C. \end{aligned}$$

Moreover, since y is feasible, there exists at least one agent $j\notin C$ such that

$$\begin{aligned} p(a )\cdot y_j(a)<p(a )\cdot e_j(a ). \end{aligned}$$

Notice that

$$\begin{aligned} p( s )\cdot y_j(a)<p( s )\cdot e_j( s )\qquad {{\mathrm{for}}\,\,{\mathrm{all}}\,\,} s \in {\mathcal {G}}^p_j(a ), \end{aligned}$$

(20)

because $p(\cdot )$ and $e_j(\cdot )$ are $\mathcal G_j$-measurable. Define the allocation²⁵$z_j$ as follows:

$$\begin{aligned} z_j( s )=y_j(a)+ \frac{\mathbf {1} p( s )\cdot [e_j( s )-y_j(a)]}{\sum _{h=1}^\ell p^h( s )}\qquad {{\mathrm{for}} \,\,{{\mathrm{any}}}\,\,} s \in {\mathcal {G}}^p_j(a), \end{aligned}$$

where ${\mathbf {1}}$ is the vector with $\ell$ components each of them equal to one, i.e., ${\mathbf {1}}=(1,\ldots ,1)$. Notice that $z_j(\cdot )$ is constant in the event ${\mathcal {G}}^p_j (a)$; for any $s \in {\mathcal {G}}^p_j(a)$, $z_j(s)\gg y_j(a)$ and $p( s )\cdot z_j( s )=p( s )\cdot e_j( s )$. Therefore, since (p, x) is a maxmin REE and $u_j(\cdot , x)$ is ${\mathcal {F}}_j$-measurable, from the monotonicity of $u_j(a, \cdot )$, it follows that

$$\begin{aligned} \underline{u}_j^{REE}(a, x_j)\ge \underline{u}_j^{REE}(a, z_j)= & {} u_j(a, z_j(a))>u_j(a, y_j(a))\ge \underline{u}_j^{REE}(a , y_j)\\= & {} \underline{u}_j^{REE}(a, x_j), \end{aligned}$$

a contradiction.

II CASE:   Assume now that the equilibrium price p is fully revealing; hence $\mathcal {G}^p_i(a)=\{a\}$ for any $i \in I$. From (IV) and Lemma 8.8 it follows that for all $i \in C$,

$$\begin{aligned} \underline{u}_i^{REE}(a, y_i)\ge \underline{u}_i(a, y_i)>\underline{u}_i(a, x_i)={u}_i(a, x_i(a))=\underline{u}_i^{REE}(a, x_i), \end{aligned}$$

and hence

$$\begin{aligned} p(a)\cdot y_i(a)> p(a) \cdot e_i(a)\quad {{\mathrm{for}}\,\, {{\mathrm{any}}}} \,\,i \in C. \end{aligned}$$

while from (II) it follows that for all $i \notin C$,

$$\begin{aligned} \underline{u}_i^{REE}(a, y_i)=u_i(a, x_i(b))= u_i(a, x_i(a))=\underline{u}_i^{REE}(a, x_i). \end{aligned}$$

Since, we have already observed that y is feasible, we conclude that for some agent $j \notin C$,

$$\begin{aligned} p(a )\cdot y_j(a)<p(a )\cdot e_j(a ). \end{aligned}$$

Define the following bundle²⁶

$$\begin{aligned} z_j( a )=y_j(a)+ \frac{\mathbf {1}p( a )\cdot [e_j( a)-y_j(a)]}{\sum _{h=1}^\ell p^h( a)}\gg y_j(a), \end{aligned}$$

where ${\mathbf {1}}$ is the vector with $\ell$ components each of them equal to one, i.e., ${\mathbf {1}}=(1,\ldots ,1)$. Notice that $p( a )\cdot z_j( a )=p( a)\cdot e_j( a )$ and

$$\begin{aligned} \underline{u}_j^{REE}(a, z_j)= u_j(a, z_j(a))>u_j(a, y_j(a))=\underline{u}_j^{REE}(a , y_j)= \underline{u}_j^{REE}(a, x_j), \end{aligned}$$

contradicts the fact that x is a maxmin REE allocation. $\square$

1.5 Counterexamples for a general set of priors

As we commented above, Propositions 4.3, 4.7, 4.11, Theorems 5.4, 5.9, and Lemma 8.2 are valid for the general MEU models, provided that all priors are strictly positive. In this section, we give counterexamples to these results if some priors are not strictly positive.

Consider the following asymmetric information economy:

$$\begin{aligned} \left. \begin{array}{lll} I=\{1,2,3\} &{} S=\{a,b,c,d\} &{} \ell =2 \\ {\mathcal {F}}_1=\{\{a,b,c\},\{d\}\} &{} {\mathcal {F}}_2=\{\{a,b,c,d\}\} &{} {\mathcal {F}}_3=\{\{a\},\{b\},\{c\},\{d\}\} \\ e_1(s)=(1,3)\,\,{{\mathrm{for}} \,\,{{\mathrm{all}}}}\,\,s \in \{a,b,c\} &{} e_1(d)=(2,2) &{} e_2(s)=(2,1)\,\,\mathrm{for \,\,all}\,\,s \in S\\ e_3(a)=(1,4)&{}e_3(b)=(2,6)&{} e_3(c)=(0,2)\\ e_3(d)=(1,7) &{} u_i(s,x,y)=\sqrt{xy}&{} \forall i \,\, {\mathrm{and }}\,\,\forall s \in S. \end{array} \right. \end{aligned}$$

Notice that $\sigma (u_i, e_i)\subseteq {\mathcal {F}}_i$ and $u_i(s, \cdot )$ is concave. For any $i\in I$ and any $F\in {\mathcal {F}}_i$, let ${\mathcal {C}}_i^F$ be the set of all priors with support contained in F. Clearly if $F=\{s\}$ then ${\mathcal {C}}_i^{\{s\}}$ consists of only one measure assigning one to $\{s\}$. Let ${\mathcal {M}}_1^F=\{\alpha :S\rightarrow [0,1]\,:\, \alpha (a)+\alpha (b)=1\}$ if $F=\{a,b,c\}$ and ${\mathcal {M}}_2^F=\{\alpha :S\rightarrow [0,1]\,:\, \alpha (a)+\alpha (b)+\alpha (d)=1\}$ for $F=S$. ${\mathcal {M}}^F_1$ and ${\mathcal {M}}^F_2$ are proper subsets of ${\mathcal {C}}^F_1$ and ${\mathcal {C}}^F_2$ and they do not contain only positive priors.

Consider the following allocation $\{(x_i^*(s),y_i^*(s))\}_{i\in I, s \in S}$

$$\begin{aligned} \left. \begin{array}{lll} (x_1^*(a), y^*_1(a))=\left( \frac{5}{4}, \frac{5}{2}\right) &{} (x^*_2(a), y^*_2(a))=\left( \frac{5}{4}, \frac{5}{2}\right) &{}(x^*_3(a), y^*_3(a))=\left( \frac{3}{2}, 3\right) \\ &{}&{}\\ (x^*_1(b), y^*_1(b))=\left( \frac{5}{4}, \frac{5}{2}\right) &{} (x^*_2(b), y^*_2(b))=\left( \frac{5}{4}, \frac{5}{2}\right) &{}(x^*_3(b), y^*_3(b))=\left( \frac{5}{2}, 5\right) \\ &{}&{}\\ (x^*_1(c), y^*_1(c))=\left( \frac{9}{4}, \frac{1}{2}\right) &{} (x^*_2(c), y^*_2(c))=\left( \frac{1}{4},\frac{9}{2}\right) &{}(x^*_3(c), y^*_3(c))=\left( \frac{1}{2},1\right) \\ &{}&{}\\ (x^*_1(d), y_1^*(d))=\left( \frac{3}{2},3\right) &{} (x^*_2(d), y^*_2(d))=\left( \frac{5}{4}, \frac{5}{2}\right) &{}(x^*_3(d), y^*_3(d))=\left( \frac{9}{4}, \frac{9}{2}\right) \end{array} \right. \end{aligned}$$

and the following price $(p(s), q(s))=\left( 1, \frac{1}{2}\right)$ for all $s \in S$. Thus, (p, q) is non revealing and hence ${\mathcal {G}}^p_i={\mathcal {F}}_i$ for all i.

We now show that the allocation above is a MREE where agents’ preferences are represented by (the general) maxmin expected utility (5). Indeed, $\{(x_i^*(s),y_i^*(s))\}_{i\in I, s \in S}$ is feasible and it satisfies the budget constraints. Moreover it maximizes the MEU subject to the budget constraint. Indeed, assume to the contrary that

• I case ($i=1$ and $s \in \{a,b,c\}$)

  there exists a random bundle $(x_1(s), y_1(s))$ such that

  $$\begin{aligned} \inf _{\alpha \in {\mathcal {M}}_1^F} \sum _{s' \in \{a,b,c\}} \sqrt{x_1(s')y_1(s')}\alpha (s')>\inf _{\alpha \in {\mathcal {M}}_1^F} \sum _{s' \in \{a,b,c\}} \sqrt{x^*_1(s')y^*_1(s')}\alpha (s') \end{aligned}$$

  and $x_1(s)+ \frac{1}{2} y_1(s)\le 1 + \frac{3}{2}$ for any $s\in \{a,b,c\}$. Since for all $\alpha \in {\mathcal {M}}_1^F$, $\alpha (c)=0$ and there exists $\beta \in {\mathcal {M}}_1^F$ such that $\beta (a)=1$ and $\beta (b)=\beta (c)=0$, it follows in particular that $\sqrt{x_1(a)y_1(a)}> \sqrt{ \frac{25}{8} }$ and $x_1(a) + \frac{1}{2}y_1(a)\le \frac{5}{2}.$ Thus, $\frac{1}{2}(5-y_1(a))y_1(a)>\frac{25}{8}$, i.e., $\left( y_1(a)- \frac{5}{2}\right) ^2<0$, a contradiction.

• II case ($i=1$ and $s=d$)

  there exists a random bundle $(x_1(d), y_1(d))$ such that $\sqrt{x_1(d)y_1(d)} >\sqrt{\frac{9}{2}}$ and $x_1(d)+ \frac{1}{2} y_1(d)\le 3.$ This implies that $(3-\frac{1}{2}y_1(d))y_1(d)>\frac{9}{2}$, i.e., $\left( y_1(a)- 3\right) ^2<0$, a contradiction.

• III case ($i=2$ and $s \in S$)

  there exists a random bundle $(x_2(s), y_2(s))$ such that $\inf _{\alpha \in {\mathcal {M}}_2^F} \sum _{s' \in S} \sqrt{x_2(s')y_2(s')}\alpha (s')>\inf _{\alpha \in {\mathcal {M}}_2^F} \sum _{s' \in S} \sqrt{x^*_2(s')y^*_2(s')}\alpha (s')$, and $x_2(s)+ \frac{1}{2} y_2(s)\le 2 + \frac{1}{2}$ for all $s \in S.$ Since for all $\alpha \in {\mathcal {M}}_2^F$, $\alpha (c)=0$ and there exists $\beta \in {\mathcal {M}}_2^F$ such that $\beta (a)=1$ and $\beta (b)=\beta (c)=\beta (d)=0$, it follows in particular that $\sqrt{x_2(a)y_2(a)}> \sqrt{ \frac{25}{8} }$ and $x_2(a) + \frac{1}{2}y_2(a)\le \frac{5}{2}.$ As in the first case, this implies a contradiction.

• IV case ($i=3$ and $s=a$)

  there exists a random bundle $(x_3(a), y_3(a))$ such that $\sqrt{x_3(a)y_3(a)}> \sqrt{\frac{9}{2}}$ and $x_3(a)+ \frac{1}{2} y_3(a)\le 3.$ As in the second case, this implies a contradiction.

• V case ($i=3$ and $s=b$)

  there exists a random bundle $(x_3(b), y_3(b))$ such that $\sqrt{x_3(b)y_3(b)}> \sqrt{\frac{25}{2}}$ and $x_3(b)+ \frac{1}{2} y_3(b)\le 5.$ This implies that $(5-\frac{1}{2}y_3(b))y_3(b)>\frac{25}{2}$, i.e., $\left( y_3(b)- 5\right) ^2<0$, a contradiction.

• VI case ($i=3$ and $s=c$)

  there exists a random bundle $(x_3(c), y_3(c))$ such that $\sqrt{x_3(c)y_3(c)}> \sqrt{\frac{1}{2}}$ and $x_3(c)+ \frac{1}{2} y_3(c)\le 1.$ This implies that $\left( 1-\frac{1}{2}y_3(c)\right) y_3(c)>\frac{1}{2}$, i.e., $\left( y_3(c)- 1\right) ^2<0$, a contradiction.

• VII case ($i=3$ and $s=d$)

  there exists a random bundle $(x_3(d), y_3(d))$ such that $\sqrt{x_3(d)y_3(d)}> \sqrt{\frac{81}{8}}$ and $x_3(d)+ \frac{1}{2} y_3(d)\le \frac{9}{2}.$ This implies that $\left( \frac{9}{2}-\frac{1}{2}y_3(d)\right) y_3(d)>\frac{81}{8}$, i.e., $\left( y_3(d)- \frac{9}{2}\right) ^2<0$, a contradiction.

Notice that

• the allocation $(x_i^*(\cdot ), y_i^*(\cdot ))$ is not ${\mathcal {G}}^p_i$-measurable. Thus, this is a counterexample to Proposition 4.3 for the general MEU case if the set of priors contains priors that are not strictly positive.²⁷

• agents’ utilities are not constant in the event ${\mathcal {G}}^p_i(s)$. Thus, this is a counterexample to Proposition 4.11 for the general MEU case if the set of priors contains priors that are not strictly positive.

• the allocation $(x_i^*(\cdot ), y_i^*(\cdot ))$ is not ex post efficient, since it is blocked by

  $$\begin{aligned} (t_i(s), z_i(s))= & {} (x_i^*(s), y_i^*(s)) \,\,\forall i\in I\,\,{{\mathrm{if}} \,\,}s\ne c,\,\,{{\mathrm{and}}}\\ (t_i(c), z_i(c))= & {} \left( \frac{5}{4}, \frac{5}{2}\right) \,\,\forall i\in \{1,2\}\\ (t_3(c), z_3(c))= & {} (x_3^*(c), y_3^*(c)) = \left( \frac{1}{2}, 1\right) . \end{aligned}$$

  Indeed (t, z) is feasible, and $u_i(s, t,z)=u_i(s,x^*,y^*)$ for all $i \in I$ if $s\ne c$, and

  $$\begin{aligned} u_1(t_1(c), z_1(c))= & {} \sqrt{\frac{25}{8}}> \sqrt{\frac{9}{8}}=u_1(x_1^*(c), y_1^*(c))\\ u_2(t_2(c), z_2(c))= & {} \sqrt{\frac{25}{8}}> \sqrt{\frac{9}{8}} =u_2(x_2^*(c), y_2^*(c))\\ u_3(t_3(c), z_3(c))= & {} u_3(x_3^*(c), y_3^*(c)) \end{aligned}$$

  Thus, this is a counterexample to Theorem 5.4 for the general MEU case if the set of priors contains priors that are not strictly positive.

• the allocation $(x_i^*(\cdot ), y_i^*(\cdot )$ is not maxmin efficient, since it is blocked by

  $$\begin{aligned} (t_i(s), z_i(s))= & {} (x_i^*(s), y_i^*(s)) \,\,\forall i\in I\,\,{{\mathrm{if}} \,\,}s\ne c,\,\,{{\mathrm{and}}}\\ (t_i(c), z_i(c))= & {} (0,0) \,\,\forall i\in \{1,2\}\\ (t_3(c), z_3(c))= & {} (3,6). \end{aligned}$$

  Indeed (t, z) is feasible, and $u_i(s, t,z)=u_i(s,x^*,y^*)$ for all $i \in I$ if $s\ne c$, and

  $$\begin{aligned} \inf _{\alpha \in {\mathcal {M}}_1^c} \sum _{s' \in \{a,b,c\}} \sqrt{t_1(s')z_1(s')}\alpha (s')= & {} \sqrt{\frac{25}{8}}=\inf _{\alpha \in {\mathcal {M}}_1^c} \sum _{s' \in \{a,b,c\}} \sqrt{x^*_1(s')y^*_1(s')}\alpha (s')\\ \inf _{\alpha \in {\mathcal {M}}_2^c} \sum _{s' \in S} \sqrt{t_2(s')z_2(s')}\alpha (s')= & {} \sqrt{\frac{25}{8}}=\inf _{\alpha \in {\mathcal {M}}_2^c} \sum _{s' \in S} \sqrt{x^*_2(s')y^*_2(s')}\alpha (s') \\ u_3(t_3(c), z_3(c))= & {} \sqrt{18}> \sqrt{\frac{1}{2}}=u_3(x_3^*(c), y_3^*(c)) \end{aligned}$$

  Thus, this is a counterexample to Theorem 5.9 for the general MEU case if the set of priors contains measures that are not strictly positive.

• the allocation $(x_i^*(\cdot ), y_i^*(\cdot ))$ is not an ex post Walrasian equilibrium allocation. Indeed consider for example agent 2 in state c and the bundle $\left( \frac{5}{4}, \frac{5}{2}\right)$ which is such that

  $$\begin{aligned} \sqrt{\frac{25}{8}}> \sqrt{\frac{9}{8}}\quad {{\mathrm{and}}}\quad \frac{5}{4}+\frac{5}{4}=2 + \frac{1}{2}. \end{aligned}$$

  Thus, this is a counterexample to Proposition 4.7 and Lemma 8.2 for the general MEU case if the set of priors contains priors that are not strictly positive.