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Optimal sharing with an infinite number of commodities in the presence of optimistic and pessimistic agents

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Abstract

We prove that under mild conditions individually rational Pareto optima will exist even in the presence of non-convex preferences. We consider decision-makers (DMs) dealing with a countable flow of pay-offs or choosing among financial assets whose outcomes depend on the realization of a countable set of states of the world. Our conditions for the existence of Pareto optima can be interpreted as a requirement of impatience in the first context and of some pessimism or not unrealistic optimism in the second context. A non-existence example is provided when, in the second context, some DM is too optimistic. We furthermore show that at an individually rational Pareto optimum at most one strictly optimistic DM will avoid ruin at each state or date. Considering a risky context, this entails that even if risk averters will share risk in a comonotonic way as usual, at most one classical strong risk lover will avoid ruin at each state or date. Finally, some examples illustrate circumstances when a risk averter could take advantage of sharing risk with a risk lover rather than with a risk averter.

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Notes

  1. It is important to remark that, in this work, as in Brown and Lewis (1981), the term impatience is simply a specialization of the broader concept of myopia. Some authors such as Chateauneuf and Rébillé (2004) propose alternative definitions for these two concepts.

  2. More precisely, Theorem 4.1 of this work shows sufficient conditions for the existence of extreme \(\alpha \)-core allocations, a notion that in our framework coincides with the definition of IRPO allocations when preferences are complete.

  3. One will find in “Appendix 1” some brief recalls concerning \(\ell ^\infty \) the Banach space of real bounded sequences, and also the weak* topology and the Mackey topology, which are, respectively, the coarsest and the finest topology on \(\ell ^\infty \) for which the dual is \(\ell ^1\) the Banach space of absolutely convergent real sequences.

  4. In this work, they refer to the concept of upper semi-myopia as strong myopia.

  5. In Sect. 4, one will find an example of preferences which are weak* sequentially upper semi-continuous but not weak* upper semi-continuous. This illustrates that our result strengthens Aliprantis and Burkinshaw’s one at least in the \(\ell ^\infty \) case.

  6. “Appendix 2” makes precise that sequential weak* convergence on \(\ell ^\infty \) is as tractable as the standard convergence in the Euclidean space \(\mathbb {R}^m\) (\(m\in \mathbb {N}\)).

  7. Under uncertainty \(\vartheta (A)\) is the subjective evaluation of the likelihood of the event A, while when valuing flow of payments, \(\vartheta (A)\) is the weight given by the DM to the time period A.

  8. Indeed such a DM has non-convex preferences, but actually concave preferences.

  9. While corollary 2 concerns “pathological” situations resulting from the presence of (reasonably) strictly optimistic agents, the reader will find in Dana (2004) a thorough study of welfare implications resulting from the effects of pessimism when only pessimistic agents are considered.

  10. For sake of simplicity, we will assume that \(P(\{n\}) > 0\)  \(\forall \, n \in \mathbb {N}\).

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Correspondence to Jean-Marc Bonnisseau.

Additional information

We would like to thank participants of SAET 2011 at Ancão, of the \(28^{\circ }\) Colóquio Brasileiro de Matemática, IMPA, at Rio de Janeiro (2011), of the 2011 Conference on Theoretical Economics at Kansas University, of the Manchester workshop in Economic Theory (2012), of SWET (2014) in honour of Bernard Cornet and of EWGET (2015) at Naples and Paulo K. Monteiro for helpful discussions and suggestions. Chateauneuf thanks IMPA for the generous financial support from the “Brazilian-French Network in Mathematics” and from the “Ciências sem Fronteiras” fellowship. Novinski gratefully acknowledges the financial support from the “Brazilian-French Network in Mathematics” and CERMSEM of the University of Paris 1 for its hospitality. The valuable detailed suggestions of anonymous referees are gratefully acknowledged.

Appendices

Appendix 1: The space \(\ell ^\infty \)

\(\mathbb {N}\) denotes the set of natural numbers \(\{1,2,3,\ldots \}\).

The space \(\ell ^\infty \) is the Banach space of real bounded sequences equipped with the norm defined by \(||x|| = \sup \nolimits _t |x_t|\). The space \(\ell ^1\) is the Banach space set of absolutely convergent real sequences equipped with the norm defined by \(||x||_1 = \sum \nolimits _{t=1}^\infty |x_t|\).

When \(\ell ^\infty \) is endowed with the norm topology, the dual is denoted by \((\ell ^\infty )^*\). A coarser topology is the Mackey topology, defined as the finest topology on \(\ell ^\infty \) for which the dual is \(\ell ^1\). Now, \((x^n)\) converges to x in this topology if and only if, for any weakly compact subset A of \(\ell ^1\), \(\langle x^n, y\rangle \rightarrow \langle x,y\rangle \) uniformly on \(y \in A\).

The weak* topology is defined as the coarsest topology on \(\ell ^\infty \) for which the dual is \(\ell ^1\).

By definition a sequence \((x^n)\) converges to x in the weak\(^*\) topology, denoted \(x^n \xrightarrow {w^*} x\), if we have \((x^n)\) in \(\ell ^{\infty }\), \(x \in \ell ^{\infty }\) and \(\left\langle x^n,y\right\rangle \rightarrow \left\langle x,y\right\rangle \) for any \(y \in \ell ^1\). Furthermore (see for instance Brézis 2011) \(x^n \xrightarrow {w^*} x\) implies (\(\left||x^n\right||\)) is bounded.

It turns out that, in the particular case of \(\ell ^{\infty }\), the simple following tractable property—which looks like the standard convergence in the Euclidian space \(\mathbb {R}^m\) (\(m \in \mathbb {N}\))—characterizes sequential weak\(^*\) convergence:

Proposition 7

Let (\(x^n\)), \(x^n \in \ell ^{\infty }\) then \(x^n \xrightarrow {w^*} x\) where \(x \in \ell ^{\infty }\)   if and only if \(\lim _{n\rightarrow \infty } x^n(p)=x(p) \in \mathbb {R}\) for all \(p \in \mathbb {N}\) and \((\left||x^n\right||)_n\) is bounded.

Notice that in all this paper when \(x \in \ell ^{\infty }\) we denote indifferently its p-th component as x(p) or \(x_p\).

Proof

Since the necessary condition is immediate, we just prove for sake of completeness the sufficiency property.

So let (\(x^n\)) with \(x^n \in \ell ^{\infty }\)    \(\forall \,n\), be such that \(x^n(p)\rightarrow x(p) \in \mathbb {R}\) \(\, \forall p \in \mathbb {N}\) and \((\left||x^n\right||)\) is bounded.

Let us first prove that \(x \in \ell ^{\infty }\). Let \(K \ge 0\) be such that \(\left||x^n\right|| \le K\)   \(\forall n\). Let \(\varepsilon >0\) be given, \(\lim _{n \rightarrow \infty } x^n(p)=x(p)\), then there exist \(n_0 \in \mathbb {N}\) such that \(n \ge n_0\) implies \(\left| x(p) \right| \le \left| x^n(p)\right| +\varepsilon \) and hence \(\left| x(p) \right| \le K+\varepsilon \) and so \(x \in \ell ^\infty \) since \(\left||x\right|| \le K + \varepsilon \).

It remains to show that for all \(y \in \ell ^1\) we have \(\left\langle x^n,y\right\rangle \rightarrow \left\langle x,y\right\rangle \) i.e. \(S_n \rightarrow S \) where \(S_n = \sum ^{\infty }_{p=1} x^n(p)y(p)\) and \(S=\sum _{p=1}^{\infty } x(p)y(p)\). Let \(m \in \mathbb {N}\), then:

$$\begin{aligned} \left| S_n -S \right| \le \sum _{p=1}^{m}\left| x^n(p)-x(p)\right| \cdot \left| y(p)\right| +\sum _{p=m+1}^{\infty }\left| x^n(p)-x(p)\right| \cdot \left| y(p)\right| . \end{aligned}$$

Hence,

$$\begin{aligned} \left| S_n -S \right| \le \left||y\right||_1\cdot \left( \sum _{p=1}^{m}\left| x^n(p)-x(p)\right| \right) +(K+\left||x\right||)\cdot \sum _{p=m+1}^{\infty }\left| y(p)\right| . \end{aligned}$$

Since \(y \in \ell ^1\), there exist \(m_0\) such that \(\sum _{p=m_0+1}^{\infty }\left| y(p)\right| \le \varepsilon \). For such \(m_0\), \(x^n(p) \rightarrow x(p)\) for all \(1 \le p \le m_0\). Then there is \(n_0(\varepsilon )\) such that \(n \ge n_0(\varepsilon )\) implies \(\sum _{p=1}^{m_0}\left| x^n(p)-x(p)\right| \le \varepsilon \).

Therefore \(\forall \varepsilon >0\), \(\exists n_0(\varepsilon )\) such that \(\left| S_n -S \right| \le (\left||y\right||_1+K+\left||x\right||)\varepsilon \). Hence \(\left\langle x^n,y\right\rangle \rightarrow \left\langle x,y\right\rangle \), which completes the proof. \(\square \)

Appendix 2: Mackey and weak* sequential convergence coincide on \(\ell ^\infty \)

For sake of completeness, we additionally give a direct and elementary simple proof of Proposition 8.

Proposition 8

On \(\ell ^\infty \) the weak* \(\sigma (\ell ^\infty , \ell ^1)\) sequential convergence and the Mackey \(\tau (\ell ^\infty , \ell ^1)\) sequential convergence coincide.

Proof

Since \(\sigma (\ell ^\infty ,\ell ^1) \subset \tau (\ell ^\infty , \ell ^1)\) we just need to prove that for \((x^n)\), \(x^n \in \ell ^\infty \), \(x^n \overset{\omega ^*}{\longrightarrow } x\) implies \(x^n \overset{\tau }{\longrightarrow }x\).

From Proposition 3.13 in Brézis (2011), it comes that \(x^n \overset{\mathrm{\omega ^*}}{\longrightarrow } x\) if and only if the sequence \((x^n)\) is bounded and \(\forall \,i \in \mathbb {N}\)  \(\lim \limits _n x^n(i) = x(i)\), therefore we may assume without loss of generality that \((x^n)\) is a sequence in the closed unit ball \(\overline{B}(0,1)\) of \(\ell ^\infty \) and that x belongs to \(\overline{B}(0,1)\). Recall that on \(\overline{B}(0,1)\) the metric \(d(x,y) = \sum \nolimits _{i=0}^{\infty } \dfrac{1}{2^i}\,|x(i)-y(i)|\) defines the weak star topology (see e.g. Exercise 4 p.136 in Conway 1994).

Let us recall furthermore that a basis of neighbourhoods of 0 for the Mackey topology on \(\ell ^\infty \) is given by the sets: \(\vartheta \big ((a_i),0\big ) = \left\{ z \in \ell ^\infty , |z(i)| \le \dfrac{1}{a_i}\,\, \forall \, i \in \mathbb {N}\right\} \) where \((a_i)\) is a sequence of real numbers strictly decreasing towards 0 (see e.g. Brown and Lewis (1981) p.34). Therefore we need to show, letting \(z^n = |x^n-x|\) that \(z^n \overset{\omega ^*}{\longrightarrow } 0\) in \(\overline{B}(0,1)\) implies \(z^n \overset{\tau }{\longrightarrow } 0\).

Let \(\vartheta \big ((a_i),0\big )\) be a neighbourhood of 0 for the Mackey topology \(\tau \). From \(\dfrac{1}{a_i}\,\uparrow \, +\infty \) there exists \(i_0 \in \mathbb {N}^*\) such that \(i \ge i_0 \Rightarrow 2 \le \dfrac{1}{a_i}\) and hence \(|z^n(i)| \le \dfrac{1}{a_i}\,\forall \, n\) since \(|z^n(i)| \le 2\)  \(\forall \,n\), \(\forall \,i\).

So it remains to prove that if \(z^n \overset{\omega ^*}{\longrightarrow } 0\), there exists \(n_0 \in \mathbb {N}\) such that \(n \ge n_0 \Rightarrow z^n \in \vartheta \big ((a_i),0\big )\).

Let \(\varepsilon = \dfrac{1}{2^{i_0-1}\cdot a_1}\) , \(z^n \overset{\omega ^*}{\longrightarrow } 0 \Rightarrow \exists \, n_0(\varepsilon )\) such that \(n \ge n_0(\varepsilon ) \Rightarrow \sum \nolimits _{i=0}^{\infty } \dfrac{1}{2^i}\, |z^n(i)| \le \varepsilon \) hence \(\dfrac{1}{2^i}\, |z^n(i)| \le \dfrac{1}{2^{i_0-1}} \times \dfrac{1}{a_1}\, \forall \,i \le i_0-1\), therefore \(|z^n(i)| \le \dfrac{2^i}{2^{i_0-1}}\cdot \dfrac{1}{a_1}\, \forall \, i \le i_0-1\) so \(|z^n(i)| \le \dfrac{1}{a_i}\, \forall \, i \le i_0-1\) since \(\dfrac{1}{a_i}\) is strictly increasing; from \(|z^n(i)| \le \dfrac{1}{a_i}\, \forall \, i \ge i_0\) as obtained previously, it turns out that \(z^n \in \vartheta \big ((a_i),0\big )\,\forall \, n \ge n_0 = n_0(\varepsilon )\), which completes the proof. \(\square \)

Appendix 3: Strictly optimistic Choquet preferences

Proof of Lemma 4

(b) can be found in Schmeidler (1989)

(c) comes from Proposition 2.

The necessity of (a) is immediate. It remains to prove that under (a) and (c) the preferences are strictly monotone.

So take \(x,x' \in \ell _+^\infty \) with \(x > x'\) and let us show that \(x \succ _i x'\). \(x > x'\) implies there exists \(s_0 \in \mathbb {N}\) s.t. \(x(s_0) > x'(s_0)\). We need to show that

$$\begin{aligned} \int _0^{+\infty } \vartheta (x \ge t)\,{\hbox {d}}t > \int _0^{+\infty } \vartheta (x' \ge t)\,{\hbox {d}}t \end{aligned}$$

or equally \(d > 0\) where

$$\begin{aligned} d = \int _0^{+\infty } \big (\vartheta (x \ge t) - \vartheta (x' \ge t)\big )\,{\hbox {d}}t. \end{aligned}$$

Let \(t_0 = x(s_0)\), therefore \(\{x \ge t_0\} \varsupsetneq \{x' \ge t_0\}\) and (a) implies \(\vartheta (x \ge t_0) > \vartheta (x' \ge t_0)\).

Let \(y \in \ell _+^\infty \), and let us see that \(f(t) = \vartheta (y \ge t)\) is left-continuous.

Take \(A_n = \{s \in \mathbb {N}, y(s) \ge t_n\}\) with \(t_n \in \mathbb {R}^+\), \(t_n \uparrow t\). It is straightforward to see that \(A_n \downarrow A = \{y \ge t\}\), so since from (c) \(\vartheta (A_n) \downarrow \vartheta (A)\), f is actually left-continuous. So since \(t_0 = x(s_0) > 0\), there exists an interval \([t_0-\varepsilon , t_0) \subseteq \mathbb {R}^+\) such that \(\forall \, t \in [t_0-\varepsilon , t_0]\), \(\vartheta (x \ge t) - \vartheta (x' \ge t) > 0\), since indeed \(x \ge x'\) implies \(\vartheta (x \ge u) \ge \vartheta (x' \ge u)\) \(\forall \,u \in \mathbb {R}^+\) one gets \(d > 0\). \(\square \)

Appendix 4: Proof of Proposition 6

Lemma 5

Under the hypotheses of Proposition 6, if \((x_1, x_2)\) is an interior PO allocation, we have:

$$\begin{aligned} \frac{u'_1(x_1(s_1))}{u'_1(x_1(s_2))} = \frac{u'_2(x_2(s_1))}{u'_2(x_2(s_2))} . \end{aligned}$$
(1)

Proof

Let us assume that (1) is not satisfied and then show that this contradicts Pareto optimality. W.l.o.g., assume that

$$\begin{aligned} \frac{u'_2(x_2(s_1))}{u'_2(x_2(s_2))}<\frac{u'_1(x_1(s_1))}{u'_1(x_1(s_2))}. \end{aligned}$$

Define \(\bar{x}_1(s_1)=x(s_1)+h\), \(\bar{x}_1(t)=x_1(s_2)-k\) and \(\bar{x}_2(s_1)=x_2(s_1)-h\), \(\bar{x}_2(s_2)=x_2(s_2)+h\) with \(h>0\), \(k>0\) sufficiently small such that \((\bar{x}_1, \bar{x}_2)\) is an interior feasible allocation.

It is enough to see that for suitable h and k one would get \(\bar{x}_1 {\succ _1} x_1\) and \(\bar{x}_2 {\succ _2} x_2\). Let us define d and \(\delta \) by:

$$\begin{aligned} d= & {} P(s_1)u_1(x_1(s_1)+h)+P(s_2)u_1(x_1(s_2)-k)-P(s_1)u_1(x_1(s_1)) \\&-P(s_2)u_1(x_1(s_2))\\ \delta= & {} P(s_1)u_2(x_2(s_1)-h)+P(s_2)u_2(x_2(s_2)+k)-P(s_1)u_2(x_2(s_1))\\&-P(s_2)u_2(x_2(s_2)) \end{aligned}$$

Thus:

$$\begin{aligned} d= & {} P(s_1)hu'_1(x_1(s_1))(1+{\varepsilon }_1(h))-P(s_2)ku'_1(x_1(s_2))(1+{\varepsilon }_2(k)) \\ \delta= & {} P(s_2)ku'_2(x_2(s_2))(1+\beta _1(k))-P(s_1)hu'_2(x_2(s_1))(1+{\beta }_2(h)) \end{aligned}$$

where \({\varepsilon }_1,{\beta }_2\rightarrow 0\) when \(h\rightarrow 0\) and \({\varepsilon }_2,{\beta }_1\rightarrow 0\) when \(k\rightarrow 0\). So

$$\begin{aligned} d>0 \Leftrightarrow \frac{P(s_1)\cdot h}{P(s_2)\cdot k} > \frac{u'_1(x_1(s_2))}{u'_1(x_1(s_1))} \, \frac{1+{\varepsilon }_2(k)}{1+{\varepsilon }_1(h)} \end{aligned}$$
(2)

and

$$\begin{aligned} \delta>0 \Leftrightarrow \frac{u'_2(x_2(s_2))}{u'_2(x_2(s_1))} \, \frac{1+{\beta }_1(k)}{1+{\beta }_2(h)} > \frac{P(s_1)\cdot h}{P(s_2)\cdot k} \end{aligned}$$
(3)

Since we are assuming that \(\frac{u'_2(x_2(s_2))}{u'_2(x_2(s_1))}>\frac{u'_1(x_1(s_2))}{u'_1(x_1(s_1))}\), it is clear that there exists \(h_0>0\), \(k_0>0\) such that

$$\begin{aligned} \frac{u'_2(x_2(s_2))}{u'_2(x_2(s_1))}> \frac{P(s_1)h_0}{P(s_2)k_0} > \frac{u'_1(x_1(s_2))}{u'_1(x_1(s_1))}. \end{aligned}$$

So, by letting \(h=th_0\), \(k=tk_0\) for \(t>0\) sufficiently small, (2) and (3) are true since indeed \(\frac{1+{\varepsilon }_2(h)}{1+{\varepsilon }_1(h)}\) and \(\frac{1+{\beta }_1(k)}{1+{\beta }_2(k)}\) will converge to 1 when \(t\rightarrow 0\). This completes the proof. \(\square \)

Now we will provide the proof of Proposition 6.

Proof

Recall that, by hypothesis, \(r =\frac{\rho _1}{\rho _2}> a_\mathrm{r}= \frac{\max \{\omega (s_1),\omega (s_2)\}}{\min \{\omega (s_1),\omega (s_2)\}} \ge 1\) (which implies \(\rho _1>\rho _2\)) and \(\rho \) is defined by \(\frac{1}{\rho } = \frac{1}{\rho _1}+ \frac{1}{-\rho _2}\). So, \(\frac{\rho }{\rho _1}=\frac{\rho _2}{\rho _2-\rho _1}<0\) and \(-\frac{\rho }{\rho _2}=-\frac{\rho _1}{\rho _2-\rho _1}>0\).

Our goal is to show that, under the hypotheses, an allocation \((x_1,x_2)\) is an interior PO if and only \(x_1= a + \frac{\rho }{\rho _1}\omega \), \(x_2= -a - \frac{\rho }{\rho _2}\omega \) for \(a \in \mathbb {R}\) such that \(x_1,x_2 \gg 0\).

Let us denote \(\alpha =P(s_1)\) (so, \(1-\alpha = P(s_2)\)), \(\omega _1\) and \(\omega _2\) are the aggregate endowments at state \(s_1\) and state \(s_2\), respectively, and \((x_1(s_1),x_1(s_2))=(y,z)\) (so, \((x_2(s_1),x_2(s_2))=(\omega _1-y,\omega _2-z)\).

Necessity of the condition If \((x_1,x_2)\) is PO, from Lemma 5 we get \(\frac{u_{1}'(y)}{u_{1}'(z)}=\frac{u_{2}'(\omega _1-y)}{u_{2}'(\omega _2-z)}\), which is equivalent to \(\exp (-\rho _1[y-z])=\exp (\rho _2[\omega _1-\omega _2-y+z])\). So,

$$\begin{aligned} y=z+\frac{\rho _2}{\rho _2-\rho _1}[\omega _1-\omega _2]=z+\frac{\rho }{\rho _1}[\omega _1-\omega _2]. \end{aligned}$$
(4)

Defining \(a=z-\frac{\rho }{\rho _1}\omega _2>0\), we obtain \(x_1=a+\frac{\rho }{\rho _1}\omega \). Finally, \(x_2=\omega -x_1=-a-\frac{\rho }{\rho _2}\omega \).

Sufficiency of the condition W.l.o.g. \(\omega _1 \ge \omega _2\). Since \(r>a_\mathrm{r}\) implies \(-\frac{\rho }{\rho _1} \omega _1>-\frac{\rho }{\rho _2} \omega _2>0\), let us pick any \(\bar{a}\) such that

$$\begin{aligned} -\frac{\rho }{\rho _1} \omega _1> \bar{a} >-\frac{\rho }{\rho _2} \omega _2 \end{aligned}$$
(5)

From (5), we have that \(\bar{x}_1:= \bar{a}+\frac{\rho }{\rho _1}\omega \gg 0\) and \(\bar{x}_2:=-\bar{a}-\frac{\rho }{\rho _2} \omega \gg 0\). We will prove that \((\bar{x}_1, \bar{x}_2)\) is a PO allocation. It is enough to show that \(\bar{x}_1\) is a solution of the Problem (E):

$$\begin{aligned} \begin{array}{rlr} \max &{} -\alpha e^{-\rho _1y} -(1-\alpha )e^{-\rho _1z} &{}\\ \mathrm{s.t.} &{} \alpha e^{\rho _2(\omega _1-y)} + (1-\alpha )e^{\rho _2(\omega _2-z)}=\vartheta &{} \quad \quad \mathrm{(E)} \\ &{} 0 \le y \le \omega _1 &{}\\ &{} 0 \le z \le \omega _2 &{} \end{array} \end{aligned}$$

for \(\vartheta =e^{-\rho _2\bar{a}}[\alpha e^{-\rho \omega _1}+(1-\alpha ) e^{-\rho \omega _2}]\).

It is clear that there is a solution \(x^*_1=(y^*,z^*)\) for (E). So, \((x^*_1, \omega -x^*_1)\) is PO. If \(\omega \gg x^*_1 \gg 0\), the associated PO allocation is interior. Thus, from Lemma 5, \(x^*_1=a+\frac{\rho }{\rho _1}\omega \) for some \(a>0\). However, \(\bar{a}\) is the unique such constant for which the first constraint of the Problem (E) holds. Furthermore, it is straightforward to verify that (yz) with \(y \in \{0,\omega _1\}\) or \(z \in \{0, \omega _2\}\) it is not a solution for (E). Therefore, \(x^*_1=\bar{x}_1\), as desired. \(\square \)

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Araujo, A., Bonnisseau, JM., Chateauneuf, A. et al. Optimal sharing with an infinite number of commodities in the presence of optimistic and pessimistic agents. Econ Theory 63, 131–157 (2017). https://doi.org/10.1007/s00199-016-0985-0

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  • Issue Date:

  • DOI: https://doi.org/10.1007/s00199-016-0985-0

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