Strict fairness of equilibria in asymmetric information economies and mixed markets

Abstract

We investigate the fairness property of equal-division competitive market equilibria (CME) in asymmetric information economies with a space of agents that may contain non-negligible (large) traders. We first propose an extension to our framework of the notion of strict fairness due to Zhou (J Econ Theory 57:158–175, 1992). We prove that once agents are asymmetrically informed, any equal-division CME allocation is strictly fair, but a strictly fair allocation might not be supported by an equilibrium price. Then, we investigate the role of large traders and we provide two sufficient conditions under which, in the case of complete information economies, a redistribution of resources is strictly fair if and only if it results from a competitive mechanism.

Appendix

1.1 The associated atomless economy ${\mathcal {E}}^*$

Given a mixed market ${\mathcal {E}}=\{(T, \mathcal T, \mu ), \mathbb {R}_+^\ell , e, (u_t)_{t \in T}\}$ we construct the associated atomless economy ${\mathcal {E}}^*=\{(T^*, \mathcal T^*, \mu ^*), \mathbb {R}_+^\ell , e, (u_t)_{t \in T^*}\}$, by splitting each atom A of $T_1$ into a continuum of negligible traders identical to A that forms an atomless coalition $A^*$ with the same measure of A. The disjoint union of these atomless coalitions $A^*$ is denoted by $T^*_1$. The measure space of agents $(T^*,{\mathcal {T}}^*,\mu ^*)$ is the direct sum of $(T_0,{\mathcal {T}}_{|_{T_0}},\mu _{|_{T_0}})$ and $T^*_1$ endowed with the Lebesgue measure. The consumption set $\mathbb {R}_+^\ell$ and the initial endowment e are unchanged, while agents’ utility function is such that $u_t=u_A$ for each member t of $A^*$.

For an allocation x of the economy ${\mathcal {E}},$ we define over $T^*$ the allocation $x^*$ of the economy ${\mathcal {E}}^*$ by setting $x^*=x\chi _{T_0}+\sum _{A \in T_1}x(A)\chi _{A^*}$, where $\chi _C$ is the characteristic function of a set C. Reciprocally, given an allocation $x^*$ of ${\mathcal {E}}^*,$ we define the allocation x of ${\mathcal {E}}$ by setting $x=x^*\chi _{T_0}+\sum _{A \in T_1}\bar{x}^*(A^*)\chi _{A}.$

Proof of Proposition 2.4

Let x be an equal-division CME allocation, with price p, and assume to the contrary that it is not strictly fair. Since it is efficient [Proposition 4.3 in Basile et al. (2014)], it follows that in some state $\bar{\omega }$ the set of envious individuals has positive measure. Any envious agent t strictly envies a certain coalition S at x, so that $t \notin S$, $S \subseteq C_t(\bar{\omega })$ and $V_t(\bar{x}(S)| {\mathcal {F}}_t(\bar{\omega }))>V_t( x_t| {\mathcal {F}}_t(\bar{\omega })).$ From $S \subseteq C_t(\bar{\omega })$, it follows that ${\mathcal {F}}_t(\bar{\omega }) = {\mathcal {F}}_s(\bar{\omega })=:{\mathcal {F}}(\bar{\omega })$ and $\mathcal {G}_t(\omega ') = \mathcal {G}_s(\omega ')=:\mathcal {G}(\omega ')$ for all $\omega '\in {\mathcal {F}}(\bar{\omega })$ and for all $s \in S$. Let $y: T \times \varOmega \rightarrow \mathbb {R}_+^\ell$ be such that $y_s(\omega )=\bar{x}(S, \omega )$ if $s \in S$ and $\omega \in {\mathcal {F}}(\bar{\omega })$ and $y_s(\omega )$ equals to a state-independent positive vector $\bar{y}$ otherwise. Thus, $y_s(\cdot )$ is $\mathcal {G}_s$-measurable for all $s \in S \cup \{t\}$. Moreover $V_t(\bar{x}(S)| {\mathcal {F}}_t(\bar{\omega }))>V_t(x_t| {\mathcal {F}}_t(\bar{\omega }))$ implies that $V_t(y_s| {\mathcal {F}}_t(\bar{\omega }))>V_t( x_t| {\mathcal {F}}_t(\bar{\omega }))$ for all $s \in S$, and hence

$$\begin{aligned} 0< & {} \sum _{\omega ' \in {\mathcal {F}}_t(\bar{\omega })} \left[ p(\omega ')\cdot y_s(\omega ')- p(\omega ')\cdot e(\omega ')\right] \\= & {} \sum _{\omega ' \in {\mathcal {F}}(\bar{\omega })} \left[ p(\omega ')\cdot \frac{1}{\mu (S)}\int _S x_s(\omega ')\,\mathrm{d}\mu -p(\omega ')\cdot e(\omega ')\right] \\= & {} \frac{1}{\mu (S)}\int _S\left[ \sum _{\omega ' \in {\mathcal {F}}(\bar{\omega })} \left[ p(\omega ')\cdot x_s(\omega ')- p(\omega ')\cdot e(\omega ') \right] \right] \, \mathrm{d}\mu \leqslant 0, \end{aligned}$$

which is a contradiction.

Now, let x be a strictly envy-free allocation and assume to the contrary that the set of (individually) envious agents at x has positive measure. According to Definition 3.6 of Basile et al. (2014), this means that for some state $\bar{\omega }$ and for any envious agent t in $\bar{\omega }$, the set $S:=\{s\in C_t(\bar{\omega })\,:\,\, \, V_t(x_s|{\mathcal {F}}_t(\bar{\omega }))> V_t(x_t|{\mathcal {F}}_t(\bar{\omega }))\}$ has positive measure. Then, $t \notin S$ and $S \subseteq C_t(\bar{\omega })$, moreover from concavity, $V_t(\bar{x}(S)|{\mathcal {F}}_t(\omega ))>V_t(x_t|{\mathcal {F}}_t(\omega ))$, which contradicts the fact that x is strictly envy-free. $\square$

Proof of Proposition 3.3

Consider a mixed market whose consumption set is $\mathbb {R}_{++}^2,$$T=T_0\cup T_1$ where $T_0=\left[ 0,\frac{1}{2}\right]$ and $T_1=\{A_1,\ldots , A_n\}$ with $\mu (A_i)=\frac{1}{2n}$ for every $i=1,\ldots ,n.$ Note that $\mu (T_1)=\frac{1}{2}.$ The total initial endowment is $e=(1,1)\gg 0.$ Agents’ utility function is given by

$$\begin{aligned} u_t(x,y) = \left\{ \begin{array}{ll} xy &{}\quad {\text {if}}\; t\in T_0\\ x^2y &{}\quad {\text {if}}\; t\in T_1. \end{array} \right. \end{aligned}$$

Consider the following family of feasible allocations

$$\begin{aligned} (x(t),y(t)) = (a,b)\chi _{T_0}+(2-a,2-b)\chi _{T_1} \end{aligned}$$

(1)

with ${b=\frac{(n-1)a+2}{n(2a-1)+1}}$ and ${a\in \left[ \frac{2(3n+1)}{7n+3},1\right) }.$

Fact 1Any (x, y) is strictly envy-free in${\mathcal {E}}.$

For every coalition S, let us denote by $S_0$ and $S_1$, respectively, the sets $S\cap T_0$ and $S\cap T_1.$ Let $\alpha =\frac{\mu (S_0)}{\mu (S)},$ and therefore $1-\alpha = \frac{\mu (S_1)}{\mu (S)}.$ Note that $\alpha \in \left[ 0,\frac{n}{n+1}\right] \cup \{1\}$. Then, the average bundle of (x, y) over S is

$$\begin{aligned} (\bar{x}(S),\bar{y}(S))=\left( 2\alpha (a-1)+2-a, 2\alpha (n+1)\frac{1-a}{n(2a-1)+1}+\frac{(3n+1)a-2n}{n(2a-1)+1}\right) . \end{aligned}$$

For every t in $T_0,$$u_t(x(t),y(t))\geqslant u_t(\bar{x}(S),\bar{y}(S))$ that is

$$\begin{aligned} a\frac{(n-1)a+2}{n(2a-1)+1}\geqslant [2\alpha (a-1)+2-a] \frac{2\alpha (n+1)(1-a)+(3n+1)a-2n}{n(2a-1)+1}, \end{aligned}$$

(2)

because by easy computation (2) is equivalent to $(n+1)\alpha ^2-(2n+1)\alpha +n\geqslant 0,$ which holds for any $\alpha \in [0,\frac{n}{n+1}]\cup \{1\}.$

For every t in $T_1,$$u_t(x(t),y(t))\geqslant u_t(\bar{x}(S),\bar{y}(S))$, that is

$$\begin{aligned} (2-a)^2\frac{(3n+1)a-2n}{n(2a-1)+1}\geqslant [2\alpha (a-1)+2-a]^2 \frac{2\alpha (n+1)(1-a)+(3n+1)a-2n}{n(2a-1)+1},\end{aligned}$$

(3)

because by algebraic computation (3) is equivalent to $4\alpha ^2(n+1)(1-a)^2+2\alpha (1-a)(a(5n+3)-6n-4)+(2-a)(-a(7n+3)+6n+2)\leqslant 0$, which holds, in particular, for any $\alpha \in [0,1].$

Therefore, any (x, y) is strictly envy-free in ${\mathcal {E}}.$

Fact 2The allocation

$$\begin{aligned} (\tilde{x}(t),\tilde{y}(t))= & {} \left( \frac{3n-2+\sqrt{9n^2+16n+8}}{7n+1},\frac{3n-4+\sqrt{9n^2+16n+8}}{5n-1}\right) \chi _{T_0}\\&+\left( \frac{11n+4-\sqrt{9n^2+16n+8}}{7n+1},\frac{7n+2-\sqrt{9n^2+16n+8}}{5n-1}\right) \chi _{T_1}, \end{aligned}$$

belonging to the family of strictly envy-free allocations (1), is efficient.

Assume to the contrary that for some feasible allocation (c, d), $u_t(c(t),d(t))>u_t(\tilde{x}(t), \tilde{y}(t))$ for almost all $t \in T$. From Remark 3.1, also the allocation $(\bar{c},\bar{d}):=(\bar{c}(T_0),\bar{d}(T_0))\chi _{T_0}+(\bar{c}(T_1),\bar{d}(T_1))\chi _{T_1}$ improves upon $(\tilde{x},\tilde{y})$; whereas from the feasibility of $(\bar{c},\bar{d}),$ it follows that $(\bar{c}(T_1),\bar{d}(T_1))=(2-\bar{c}(T_0),2-\bar{d}(T_0)),$ with $\bar{c}(T_0),$$\bar{c}(T_1),$$\bar{d}(T_0)$ and $\bar{d}(T_1)$ in (0, 2). Therefore,

$$\begin{aligned} \left\{ \begin{array}{ll} \bar{c}(T_0)\bar{d}(T_0)>\frac{3n-2+\sqrt{9n^2+16n+8}}{7n+1}\frac{3n-4+\sqrt{9n^2+16n+8}}{5n-1}\\ (2-\bar{c}(T_0))^2(2-\bar{d}(T_0))>\left( \frac{11n+4-\sqrt{9n^2+16n+8}}{7n+1}\right) ^2\frac{7n+2- \sqrt{9n^2+16n+8}}{5n-1}. \end{array} \right. \end{aligned}$$

By the first inequality,

$$\begin{aligned} \bar{d}(T_0)>\frac{2[9n^2-n+8+3(n-1) \sqrt{9n^2+16n+8}]}{(7n+1)(5n-1)\bar{c}(T_0)} \end{aligned}$$

which implies in the second condition

$$\begin{aligned}&(2-\bar{c}(T_0))^2\left[ 2-\frac{2[9n^2-n+8+3(n-1)\sqrt{9n^2+16n+8}]}{(7n+1)(5n-1)\bar{c}(T_0)}\right]> \\&\quad >\left( \frac{11n+4-\sqrt{9n^2+16n+8}}{7n+1}\right) ^2\frac{7n+2-\sqrt{9n^2+16n+8}}{5n-1}, \end{aligned}$$

that is, after algebraic computation,

$$\begin{aligned}&(7n+1)^2\left( \bar{c}(T_0)-\frac{3n-2+\sqrt{9n^2+16n+8}}{7n+1}\right) ^2 \\&\quad \left( \bar{c}(T_0)-\frac{17n-\sqrt{9n^2+16n+8}}{5n-1}\right) >0, \end{aligned}$$

that has no solution in (0, 2). This means that $(\tilde{x},\tilde{y})$ is efficient.

Therefore, $(\tilde{x},\tilde{y})$ is strictly fair but it is not a Walrasian allocation, because the unique equal-division Walrasian allocation is $\left( \frac{6}{7},\frac{6}{5}\right) \chi _{T_0}+\left( \frac{8}{7},\frac{4}{5}\right) \chi _{T_1}.$ This concludes the proof. $\square$

Proof of Proposition 3.4

Consider the same economy described in the proof of Proposition 3.3 and the same family of strictly envy-free allocations (1). For any coalition S with $\mu (S\cap T_0)>0$ and $\mu (S\cap T_1)>0$, $\alpha =\frac{\mu (S\cap T_0)}{\mu (S)}\leqslant \frac{n}{n+1}.$ This restriction does not hold for the coalitions $S^*$ of the associated atomless economy ${\mathcal {E}}^*,$ because $\mu ^*(S^*\cap T^*_1)$ can be any real number in $(0,\frac{1}{2}],$ while $\mu (S\cap T_1)\in [\frac{1}{2n},\frac{1}{2}].$⁷

Let $(x^*,y^*)$ be the associated family of allocations given by

$$\begin{aligned} (x^*(t),y^*(t))=\left\{ \begin{array}{ll} (x(t),y(t)) &{}\quad {\text {if}}\; t\in T_0\\ (x(A),y(A)) &{}\quad {\text {if}}\; t\in A^* \, \hbox {and} \, A\in T_1. \end{array} \right. \end{aligned}$$

Notice that for all t in $T_0$ and for all $S^*$ such that $\frac{\mu ^*(S^*\cap T_0)}{\mu ^*(S)}>\frac{n}{n+1},$

$$\begin{aligned} u_t(x^*(t),y^*(t))< u_t(\bar{x}(S^*{\setminus } \{t\}),\bar{y}(S^*{\setminus } \{t\}))= u_t(\bar{x}(S^*),\bar{y}(S^*)); \end{aligned}$$

meaning that any agent t in $T_0$ is strictly envious, and hence, any allocation of the family $(x^*,y^*)$ is not strictly equitable in ${\mathcal {E}}^*$. This concludes the proof. $\square$

Proof of Lemma 3.5

Since x is strictly envy-free, $u(x(A_1))=u(x(A_2))$ for all $A_1, A_2\in C \cap T_1$. If C contains at least three atoms, by strict quasi-concavity of u,

$$\begin{aligned} x(A_1)=x(A_2)\quad \mathrm{for\,\,all\,\,} A_1, A_2 \in C \cap T_1, \end{aligned}$$

(4)

otherwise a third atom $A_3$ would strictly envy the coalition $\{A_1, A_2\}$. The presence of at least three atoms is needed if $\mu (C{\setminus } T_1)=0$, because according to the notion of strict envy-freeness an individual cannot be member of the coalition she envies. If $\mu (C{\setminus } T_1)>0$, we now prove that $x(t)=\bar{x}(C)$ for almost all $t \in C$, regardless the number of large traders of C. To this end, consider the following sets $B=\{t\in C \,:\, u(\bar{x}(C))>u(x(t))\}$ and $D=\{t\in C \,:\, u(\bar{x}(C))<u(x(t))\}.$ We want to show that $\mu (B)=\mu (D)=0.$

First, assume to the contrary that $\mu (B)>0$ and notice that $\mu (B\cap T_0)=0$, otherwise every agent t in $B \cap T_0$ would strictly envy the coalition $C{\setminus }\{t\}$. Similarly, by Remark 3.1, $\mu (C{\setminus } B)=0$. Hence, $\mu (C)=\mu (B)=\mu (B\cap T_1)$ and from (4) it follows that $u(x(t))=u(\bar{x}(C))$ for almost all $t \in B$, which is impossible, as the set B is defined. Hence, $\mu (B)=0.$

Now, assume to the contrary that $\mu (D)>0$. Define $\alpha =\frac{\mu (D)}{\mu (C)}$ and notice that $\alpha <1,$ that is $\mu (C{\setminus } D)>0$, otherwise Remark 3.1 would induce a contradiction. By the continuity of the utility function, there exist $\varepsilon \in (0,1)$ and a subset E of D with positive measure such that $u(\varepsilon x(t))>u(\bar{x}(C))$ for almost every t in E. Hence, $u(\varepsilon \bar{x}(E))> u(\bar{x}(C))$ and $u(\bar{x}(C {\setminus } E)) \geqslant u(\bar{x}(C)).$

Now, let $\beta =\frac{\mu (E)}{\mu (C)}\in (0,1)$ and notice that $\bar{x}(C)= \beta \bar{x}(E)+ (1-\beta ) \bar{x}(C {\setminus } E)$. Then,

$$\begin{aligned} u(\bar{x}(C))= & {} u\left( \beta (1-\varepsilon )\bar{x}(E) +\beta \varepsilon \bar{x}(E)+(1-\beta )\bar{x}(C {\setminus } E)\right) \\&> u\left( \beta \varepsilon \bar{x}(E)+(1-\beta )\bar{x}(C {\setminus } E) \right) \geqslant u(\bar{x}(C)), \end{aligned}$$

which is a contradiction. Since $\mu (B)=\mu (D)=0,$ then $u(x(t))=u(\bar{x}(C))$ for almost all $t \in C$. Using the strict quasi-concavity, we complete the proof. $\square$

Proof of Theorem 3.6

Let x be a strictly envy-free allocation for the economy ${\mathcal {E}}$. Assume to the contrary that $x^*$ is not strictly envy-free in the associated atomless economy ${\mathcal {E}}^*.$ By Remark 3.1, for any envious agent t there exist a coalition⁸$S^*$ and an allocation $y^*$ such that $u_t(y^*(s))>u_t(x^*(t))$ almost everywhere on $S^*$ and

$$\begin{aligned} \int _{S^*} x^*(s)\mathrm{d}\mu ^*= \int _{S^*} y^*(s) \mathrm{d}\mu ^*. \end{aligned}$$

(5)

Let us consider the statement 1.

By Remark 3.1, $y^*$ can be taken constant on $S^* \cap T^*_1$, i.e., $y^*(s)=\bar{y}$ for all $s\in S^* \cap T_1^*$, while Lemma 3.5 ensures that $x^*(s)=\bar{x}$ for all $s\in S^* \cap T_1^*$. Therefore, from (5) it follows

$$\begin{aligned} \int _{S^*} [y^*(s)-x^*(s)] \mathrm{d}\mu ^*= \int _{S^*\cap T_0} [y^*(s)-x^*(s)] \mathrm{d}\mu ^* + (\bar{y} - \bar{x})\mu ^*(S^* \cap T_1^*)=0. \end{aligned}$$

Notice that $\mu ^*(S^*\cap T^*_1)>0$ otherwise, by taking in ${\mathcal {E}}$, $S=S^*$ and $y=y^*$, we would contradict the strict equitability of x in ${\mathcal {E}}$. Since $T_1$ is countably infinite and ${\mu (T_1) =\sum _{n=1}^{+\infty }\mu (A_n)<+\infty },$ then ${\lim _{n\rightarrow +\infty }\mu (A_n)=0}$, and hence, there exists B belonging to $T_1$ such that $\mu (B)<\mu ^*(S^*\cap T^*_1)$. Then for any $\alpha$, $\alpha \int _{S^*\cap T_0}[y^*(s)-x^*(s)]\mathrm{d}\mu ^*+(\bar{y}-\bar{x})\alpha \mu ^*(S^*\cap T^*_1)=0$, in particular for $\alpha =\frac{\mu (B)}{\mu ^*(S^*\cap T^*_1)}$. By the Lyapunov convexity theorem, there exists R subset of $S^*\cap T_0$ such that $\int _{R}[y^*(s)-x^*(s)]\mathrm{d}\mu +(\bar{y}-\bar{x})\mu (B)=0.$ Without loss of generality, $t\notin R\cup \{B\}$; hence, t strictly envies $R\cup \{B\}$ at x via $z=y^*\chi _{R}+\bar{y}\chi _{B}.$

Let us consider the statement 2.

From Remark 3.2, without loss of generality the set $J:=\{n \in \mathbb {N}:\, \mu ^*(S^*\cap A^*_n)>0, \,\mathrm{with\,}A_n^*\in T_1^*\}$ is finite. For any $n \in J$, let $C^*_n=\{t \in T^*:\, u_t=u_{A^*_n}\}$ be the set of agents identical to the atom $A_n$ and $C^*= \bigcup _{n\in J} C^*_n$. For any $n \in J$, by assumption $\mu (C^*_n\cap T_0)>0$; moreover, Remark 3.1 and Lemma 3.5 imply, respectively, that $y^*$ and $x^*$ can be taken constant on each $C^*_n$. Hence, $y^*(s)=y_n$ and $x^*(s)=x_n$ for almost all $s\in C_n^* \cap S^*$. From (5), it follows

$$\begin{aligned} \int _{S^*{\setminus } C^*} [y^*(s)-x^*(s)] \mathrm{d}\mu ^* +\sum _{n \in J} (y_n - x_n)\mu ^*(C_n^*\cap S^*)=0. \end{aligned}$$

(6)

Let $J_1:=\{n\in J\, :\, \mu ^*(C^*_n\cap S^*)>\mu (C^*_n\cap T_0)\}.$ For any $n \in J_1$, let $\beta _n$ be $\frac{\mu (C^*_n\cap T_0)}{\mu ^*(C^*_n\cap S^*)}$ and $\beta :=\min \{\beta _n, n\in J_1\}$ if $J_1\ne \emptyset$ and $\beta =1$ if $J_1=\emptyset$. Then, from (6) it follows that

$$\begin{aligned} \beta \int _{S^*{\setminus } C^*}[ y^*(s)-x^*(s)]\mathrm{d}\mu ^*+\beta \sum _{n\in J}(y_n- x_n)\mu ^*(C_n^* \cap S^*) =0. \end{aligned}$$

For any $n \in J$, $\beta \mu ^*(C^*_n\cap S^*)\leqslant \mu (C^*_n\cap T_0),$ then there exists $B_n\subseteq C^*_n\cap T_0$ such that $\mu (B_n)=\beta \mu ^*(C^*_n\cap S^*)$. Furthermore, by the Lyapunov convexity theorem there exists $B\subseteq S^*{\setminus } C^*$ such that

$$\begin{aligned} \int _{B}(y^*(s)-x^*(s))\mathrm{d}\mu +\sum _{n\in J} (y_n-x_n) \mu (B_n)=0. \end{aligned}$$

This means that t strictly envies the atomless coalition ${\left( \bigcup _{n\in J} B_n\right) \cup B}$ at x via ${ y=y^*\chi _{B}+\sum _{n\in J}y_n\chi _{B_n}}$, which is a contradiction.

This concludes the proof of the one-to-one correspondence between ${\mathcal {E}}$ and ${\mathcal {E}}^*$ in terms of strictly envy-free allocations. We have now to show the equivalence between the set of equal-division Walrasian allocations and SF. One inclusion has been already shown in Proposition 2.4. To prove the converse, let x be a strictly fair allocation and consider the corresponding allocation $x^*$, which is strictly envy-free in ${\mathcal {E}}^*$, and hence it is strictly fair. Proposition 3.4 of Zhou (1992) ensures that $x^*$ is an equal-division Walrasian allocation of ${\mathcal {E}}^*$. Then, coming back to the original mixed market ${\mathcal {E}}$, the associated allocation x is an equal-division Walrasian allocation. This concludes the proof. $\square$

Proposition 4.1

If $x^*$ is strictly envy-free in ${\mathcal {E}}^*$, the corresponding allocation x is strictly envy-free in ${\mathcal {E}}$.

Proof

Assume by contradiction that x is not strictly equitable in ${\mathcal {E}}$. Thus, for every envious agent t in ${\mathcal {E}}$ there exists a coalition S such that $t \notin S$ and $u_t(\bar{x}(S))>u_t(x(t)).$ Consider the coalition $S^*$ of ${\mathcal {E}}^*$ obtained by splitting each member of $S \cap T_1$. Then, by definition of $x^*$, $u_t\left( \frac{1}{\mu ^*(S^*)}\int _{S^*}x^*(t)\, \mathrm{d}\mu ^*\right) =u_t\left( \bar{x}(S)\right) >u_t(x(t))=u_t(x^*(t)).$ This means that the set of envious agents in ${\mathcal {E}}^*$ has positive measure, which is a contradiction. $\square$