A General Equilibrium Model in Which Consumption Takes Time

Abstract

This chapter examines a general equilibrium competitive economy with many heterogeneous agents. The key feature of the model is that consumption itself takes time so that a typical household is subject to a financial constraint as well as a time constraint. Using the dividend approach proposed by Le-Van and Nguyen (J Math Econ 43:135−152, 2007), it is shown that the economy possesses at least one autarkic Walrasian equilibrium. Sufficient conditions for the uniqueness of the autarkic equilibrium are then derived. Finally, a specific example is provided to illustrate the working of the model, including the derivation of the equilibrium labour allocation and some comparative static results.

Appendices

Appendices

1.1 Appendix 7.1: Proof of Proposition 7.1

• Step 1. Introduce the intermediary economy:

$$ \widehat{\varepsilon}=\left\{{\left({\widehat{X}}^i,{\widehat{u}}^i,{\widehat{e}}^i\right)}_{i\in \mathbf{I}},{\left({\widehat{Y}}_j\right)}_{j\in \mathbf{J}}\right\} $$

where ${\widehat{X}}^i={X}^i\times {\mathrm{\mathbb{R}}}_{+},{\widehat{e}}^i=\left({e}^i,{\delta}^i\right)$ with δⁱ> 0 for all i ∈ I and ${\widehat{Y}}_j=\left({Y}_j,0\right)$ for j ∈ J and the utilities ûⁱ:

$$ {\widehat{u}}^i\left({x}^i,{d}^i\right)=\left\{\begin{array}{lll}{u}^i\left({x}^i\right)& \mathrm{if}& {x}^i\notin {S}^i\\ {}{u}^i\left({x}^i\right)+\mu {d}^i={M}^i+\mu {d}^i& \mathrm{if}& {x}^i\in {S}^i\end{array}\right. $$

where μ > 0 and Mⁱ = max{uⁱ(x): x ∈ Xⁱ}.

By denoting for all i ∈ I and j ∈ J:

$$ {\displaystyle \begin{array}{c}{\widehat{x}}^i=\left({x}^i,{d}^i\right)=\left({c}_1^i,\dots, {c}_J^i,0,-{L}^i,{d}^i\right)\in {\widehat{X}}^i\\ {}{\widehat{e}}^i=\left({e}^i,{\delta}^i\right)=\left(0,\dots, 0,{\overline{K}}^i,0,{\delta}^i\right)\\ {}{\widehat{y}}_j=\left({\tilde{y}}_j,0\right)=\left(0,\dots, {F}_j\left({K}_j,{\tilde{L}}_j\right),\dots, 0,-{K}_j,-{\tilde{L}}_j,0\right)\\ {}\widehat{p}=\left(p,q\right)=\left({p}_1,\dots, {p}_J,r,w,q\right),\end{array}} $$

we can rewrite the definition of quasi-equilibrium with dividends as follows.

A quasi-equilibrium of $\widehat{\varepsilon}$ is a list $\left({\left({\widehat{x}}^{\ast i}\right)}_{i\in \mathbf{I}},{\left({y}_j^{\ast}\right)}_{j\in \mathbf{J}},{\widehat{p}}^{\ast i}\right)\in {\left({\mathrm{\mathbb{R}}}^{J+3}\right)}^I\times {\left({\mathrm{\mathbb{R}}}^{J+3}\right)}^J\times {\mathrm{\mathbb{R}}}^{J+3}$ which satisfies

$$ {\sum}_{i\in \mathbf{I}}{\widehat{x}}^{\ast i}={\sum}_{i\in \mathbf{I}}{\widehat{e}}^{\ast i}+{\sum}_{j\in \mathbf{J}}{\widehat{y}}_j^{\ast } $$

• (a) For each i, one has

$$ {\widehat{p}}^{\ast}\cdot {\widehat{x}}^i={\widehat{p}}^{\ast i}\cdot {\widehat{e}}^i+\sum \limits_{j\in \mathbf{J}}{\theta}_{ij}\left({\widehat{p}}^{\ast}\cdot {\widehat{y}}_j^{\ast}\right) $$

and for each ${\widehat{x}}^i\in {\widehat{X}}^i,$ with ${\widehat{u}}^i\left({\widehat{x}}^i\right)>{\widehat{u}}^i\left({\widehat{x}}^{\ast i}\right),$ it holds

$$ {\widehat{p}}^{\ast}\cdot {\widehat{x}}^i\ge {\widehat{p}}^{\ast i}\cdot {\widehat{e}}^i+\sum \limits_{j\in \mathbf{J}}{\theta}_{ij}\left({\widehat{p}}^{\ast}\cdot {\widehat{y}}_j^{\ast}\right) $$

• (b) For each j ∈ J, one has ${\widehat{y}}_j^{\ast}\in {\widehat{Y}}_j$ and ${\widehat{p}}^{\ast}\cdot {\widehat{y}}_j^{\ast }=\sup \left({\widehat{p}}^{\ast}\cdot {\widehat{Y}}_j\right)={\sup}_{{\widehat{y}}_j\in {\widehat{Y}}_j}\left(\widehat{p}\cdot {\widehat{y}}_j\right)$.

We consider the feasible set  of $\widehat{\varepsilon}:$

$$ {\displaystyle \begin{array}{c}\widehat{A}=\left\{\left({\left({\widehat{x}}^i\right)}_{i\in \mathbf{I}},{\left({\widehat{y}}_j\right)}_{j\in \mathbf{J}}\right):\forall i\in \mathrm{I},{\widehat{x}}^i\in {\widehat{X}}^i;\forall j\in \mathbf{J},{\widehat{y}}_j\in {\widehat{Y}}_j\kern0.5em \mathrm{and}\sum \limits_{i\in \mathbf{I}}{\widehat{x}}^i\right.\\ {}\left.=\sum \limits_{i\in \mathbf{I}}{\widehat{e}}^i+\sum \limits_{j\in \mathbf{J}}{\widehat{y}}_j\right\}\end{array}} $$

We can see that  is compact.

• Step 2. For each i ∈ I, the function ${\widehat{u}}^i$ is strictly quasi-concave, upper semi-continuous and has no satiation point.

Denote ${\mathrm{\mathcal{L}}}_{\alpha}^i=\left\{{x}^i\in {X}^i:{u}^i\left({x}^i\right)\ge \alpha \right\}$; ${\widehat{\mathrm{\mathcal{L}}}}_{\alpha}^i=\left\{\left({x}^i,{d}^i\right)\in {\widehat{X}}^i:{\widehat{u}}^i\left({x}^i,{d}^i\right)\ge \alpha \right\}$. It is obvious that ${\mathrm{\mathcal{L}}}_{\alpha}^i$ and Sⁱ are closed and convex for every α. We will prove that ${\widehat{\mathrm{\mathcal{L}}}}_{\alpha}^i$ is also closed and convex. We have two cases.

Case 1

α < Mⁱ.

• If x ∉ Sⁱ, then ${\widehat{u}}^i\left({x}^i,{d}^i\right)={u}^i\left({x}^i\right).$So $\left({x}^i,{d}^i\right)\in {\mathrm{\mathcal{L}}}_{\alpha}^i\times {\mathrm{\mathbb{R}}}_{+}\iff \left({x}^i,{d}^i\right)\in {\widehat{\mathrm{\mathcal{L}}}}_{\alpha}^i.$

• If x ∈ Sⁱ, then uⁱ(xⁱ) = Mⁱ> α and ${\widehat{u}}^i\left({x}^i,{d}^i\right)={M}^i+{\mu d}^i>\alpha$; it implies $\left({x}^i,{d}^i\right)\in {\mathrm{\mathcal{L}}}_{\alpha}^i\times {\mathrm{\mathbb{R}}}_{+}$and $\left({x}^i,{d}^i\right)\in {\widehat{\mathrm{\mathcal{L}}}}_{\alpha}^i.$

$$ {\widehat{\mathrm{\mathcal{L}}}}_{\alpha}^i={\mathrm{\mathcal{L}}}_{\alpha}^i\times {\mathrm{\mathbb{R}}}_{+}. $$

(Hence)

Case 2

α ≥ Mⁱ. Consider $\left({x}^i,{d}^i\right)\in {\widehat{\mathrm{\mathcal{L}}}}_{\alpha}^i.$

• If xⁱ ∉ Sⁱ, then ${u}^i\left({x}^i\right)={\widehat{u}}^i\left({x}^i,{d}^i\right)\ge \alpha \ge {M}^i$ contradict the definition of Mⁱ.

• Hence xⁱ ∈ Sⁱ. We have ${\widehat{u}}^i\left({x}^i,{d}^i\right)={M}^i+\mu {d}^i\ge \alpha \iff {d}^i\ge \frac{\alpha -{M}^i}{\mu }.$ It implies that ${\widehat{\mathrm{\mathcal{L}}}}_{\alpha}^i={S}^i\times \left\{{d}^i:{d}^i\ge \frac{\alpha -{M}^i}{\mu}\right\}$.

We have proved that ${\widehat{u}}^i$ is upper semi-continuous and quasi-concave for every i ∈ I.

We now prove that ${\widehat{u}}^i$ is strictly quasi-concave. Take $\left({x}^i,{d}^i\right),\left({\tilde{x}}^i,{\tilde{d}}^i\right)\in {X}^i\times {\mathrm{\mathbb{R}}}_{+}$ such that ${\widehat{u}}^i\left({\tilde{x}}^i,{\tilde{d}}^i\right)>{\widehat{u}}^i\left({x}^i,{d}^i\right).$ For any , we will verify that

$$ {\widehat{u}}^i\left(\uplambda {x}^i+\left(1-\uplambda \right){\tilde{x}}^i,\uplambda {d}^i+\left(1-\uplambda \right){\tilde{d}}^i\right)>{\widehat{u}}^i\left({x}^i,{d}^i\right) $$

We have two cases.

Case 1

xⁱ ∈ Sⁱ. We have ${\widehat{u}}^i\left({x}^i,{d}^i\right)={M}^i+{\mu d}^i$.

Since ${\widehat{u}}^i\left({\tilde{x}}^i,{\tilde{d}}^i\right)>{\widehat{u}}^i\left({x}^i,{d}^i\right)$, it implies ${\tilde{x}}^i\in {S}^i$ and ${\tilde{d}}^i>{d}^i$. This follows $\uplambda {x}^i+\left(1-\uplambda \right){\tilde{x}}^i\in {S}^i$, and we have

$$ {\displaystyle \begin{array}{c}{\widehat{u}}^i\left(\uplambda {x}^i+\left(1-\uplambda \right){\tilde{x}}^i,\uplambda {d}^i+\left(1-\uplambda \right){\tilde{d}}^i\right)={M}^i+\mu \left(\uplambda {d}^i+\left(1-\uplambda \right){\tilde{d}}^i\right)\\ {}>{M}^i+\mu \left(\uplambda {d}^i+\left(1-\uplambda \right){d}^i\right)\\ {}={M}^i+{d}^i={\widehat{u}}^i\left({x}^i,{d}^i\right)\end{array}} $$

Case 2

xⁱ ∉ Sⁱ. We have ${\widehat{u}}^i\left({x}^i,{d}^i\right)={u}^i\left({x}^i\right)<{M}^i$ and ${u}^i\left({\tilde{x}}^i\right)>{u}^i\left({x}^i\right).$

If $\uplambda {x}^i+\left(1-\uplambda \right){\tilde{x}}^i\in {S}^i$, then

$$ {\displaystyle \begin{array}{c}{\widehat{u}}^i\left(\uplambda {x}^i+\left(1-\uplambda \right){\tilde{x}}^i,\uplambda {d}^i+\left(1-\uplambda \right){\tilde{d}}^i\right)={M}^i+\mu \left(\uplambda {d}^i+\left(1-\uplambda \right){\tilde{d}}^i\right)\\ {}>{u}^i\left({x}^i\right)={\widehat{u}}^i\left({x}^i,{d}^i\right).\end{array}} $$

If $\uplambda {x}^i+\left(1-\uplambda \right){\tilde{x}}^i\notin {S}^i,$ since uⁱ is concave, we have:

$$ {\displaystyle \begin{array}{c}{\widehat{u}}^i\left(\uplambda {x}^i+\left(1-\uplambda \right){\tilde{x}}^i,\uplambda {d}^i+\left(1-\uplambda \right){\tilde{d}}^i\right)={u}^i\left(\uplambda {x}^i+\left(1-\uplambda \right){\tilde{x}}^i\right)\\ {}\ge \uplambda {u}^i\left({x}^i\right)+\left(1-\uplambda \right){u}^i\left({\tilde{x}}^i\right)\\ {}>{u}^i\left({x}^i\right)={\widehat{u}}^i\left({x}^i,{d}^i\right).\end{array}} $$

We have proved that ûⁱ is strictly quasi-concave. We now prove that ${\widehat{u}}^i$ has no satiation point. Take $\left({x}^i,{d}^i\right)\in {\widehat{X}}^i;$ we will verify that there exists $\left({\tilde{x}}^i,{\tilde{d}}^i\right)\in {\widehat{X}}^i$ such that ${\widehat{u}}^i\left({\tilde{x}}^i,{\tilde{d}}^i\right)>{\widehat{u}}^i\left({x}^i,{d}^i\right).$ We consider two cases:

• xⁱ ∈ Sⁱ. Take ${\tilde{x}}^i={x}^i$ and ${\tilde{d}}^i>{d}^i.$ We have

$$ {\widehat{u}}^i\left({\tilde{x}}^i,{\tilde{d}}^i\right)={M}^i+\mu {\tilde{d}}^i>{M}^i+\mu {d}^i={\widehat{u}}^i\left({x}^i,{d}^i\right). $$

• xⁱ ∉ Sⁱ. Take ${\tilde{x}}^i\in {X}^i$ such that ${u}^i\left({\tilde{x}}^i\right)>{u}^i\left({x}^i\right)$ and ${\tilde{d}}^i={d}^i.$ We have

$$ {\widehat{u}}^i\left({\tilde{x}}^i,{\tilde{d}}^i\right)\ge {u}^i\left({\tilde{x}}^i\right)>{u}^i\left({x}^i\right)={\widehat{u}}^i\left({x}^i,{d}^i\right). $$

We have proved that ${\widehat{u}}^i$ has no satiation point.

• Step 3. We consider a sequence of truncated economies.

Let B(0, n) denote the ball centred at 0 with radius n. Let

$$ {\widehat{X}}_n^i={\widehat{X}}^i\cap B\left(0,n\right),\kern0.5em i\in \mathbf{I}. $$

Let S be the unit sphere of ℝ^J + 4. For every $\left(\widehat{p},s\right)\in S\cap \left({\mathrm{\mathbb{R}}}^{J+3}\times {\mathrm{\mathbb{R}}}_{+}\right),$ define the multivalued mapping

$$ {\xi}_n^i,{Q}_n^i:S\cap \left({\mathrm{\mathbb{R}}}^{J+3}\times {\mathrm{\mathbb{R}}}_{+}\right)\to {\widehat{X}}^i $$

by setting:

$$ {\displaystyle \begin{array}{ll}{\xi}_n^i\left(\widehat{p},s\right)& =\left\{{\widehat{x}}^i\in {\widehat{X}}_n^i:\widehat{p}\cdot {\widehat{x}}^i\le \widehat{p}\cdot {\widehat{e}}^i+\sum \limits_{j\in \mathbf{J}}{\theta}_{ij}{\Pi}_j^n\left(\widehat{p}\right)+s\right\}{Q}_n^i\left(\widehat{p},s\right)\\ {}& =\left\{{\widehat{x}}^i\in {\xi}_n^i\left(\widehat{p},s\right):\mathrm{if}\;{{\widehat{x}}{\prime}}^i\in {\widehat{X}}_n^i\;\mathrm{with}\;{\widehat{u}}_i\left({{\widehat{x}}{\prime}}^i\right)>{\widehat{u}}^i\left({\widehat{x}}^i\right)\;\mathrm{then}\;\widehat{p}\cdot {{\widehat{x}}{\prime}}^i\ge \widehat{p}\cdot {\widehat{e}}^i+\sum \limits_{j\in \mathbf{J}}{\theta}_{ij}{\Pi}_j^n\left(\widehat{p}\right)+s\right\},\end{array}} $$

where ${\Pi}_j^n(p)=\max \left(p\cdot {Y}_j^n\right)$ is profit of firm j in truncated economies.

Define the mapping z _n  : S ∩ (ℝ^J + 3 × ℝ₊) → ℝ^J + 4 by setting

$$ {z}_n\left(\widehat{p},s\right)=\left(\sum \limits_{i\in \mathbf{I}}{Q}_n^i\left(\widehat{p},s\right)-\sum \limits_{i\in \mathbf{I}}{\widehat{e}}^i-\sum \limits_{j\in \mathbf{J}}{\Phi}_j^n\left(\widehat{p}\right)\right)\times \left\{-I\right\} $$

where ${\Phi}_j^n\left(\widehat{p}\right)=\left\{{\widehat{y}}_j\in {\widehat{Y}}_j^n:\widehat{p}\cdot {\widehat{y}}_j=\max \widehat{p}\cdot {\widehat{Y}}_j^n\right\}$.

• Step 4. (Lemma Gale–Nikaido–Debreu) Suppose

• (i) P be a closed nonempty convex cone in the linear space ℝ^land S be the unit sphere in ℝ^l.

• (ii) The multivalued mapping Z from S ∩ P to ℝ^lis upper semi-continuous having nonempty convex compact values.

• (iii) For every p ∈ S ∩ P, ∃ z ∈ Z(p) such that p ⋅ z ≤ 0.

Then there exists $\overline{p}\in S\cap P$ satisfying

$$Z\left(\overline{p}\right)\cap P^0\ne \cancel{0},$$

where P⁰ = {q ∈ ℝ^l : q ⋅ p ≤ 0, ∀p ∈ P}

• Step 5. For each i ∈ I, the mapping ${\xi}_n^i\left(\widehat{p},s\right)$ is upper semi-continuous having nonempty convex compact values.

By choosing ${\widehat{x}}^i=\left(0,\dots, 0,-{\overline{L}}^i,{\delta}^i\right)\in {\widehat{X}}^i$, we can easily see that ${\xi }_n^i\left(\widehat{p},s\right)\ne \cancel{0}$.

If ${\left\{{x}_k^i\right\}}_k\subset {\xi}_n^i\left(\widehat{p},s\right)$ and $\underset{k\to +\infty }{\lim }{x}_k^i={\widehat{x}}^i$, then ${\widehat{x}}^i\in {\widehat{X}}_n^i$ (since ${\widehat{X}}_n^i$ compact) and $\widehat{p}\cdot {\widehat{x}}^i\le \widehat{p}\cdot {\widehat{e}}^i+s.$Hence ${\xi}_n^i\left(\widehat{p},s\right)$ is closed, and since it is a subset of compact set ${\widehat{X}}_n^i$, so ${\xi}_n^i\left(\widehat{p},s\right)$ is compact.

For every ${\widehat{x}}^i,{\widehat{z}}^i\in {\xi}_n^i\left(\widehat{p},s\right)$ and λ ∈ [0, 1], we have

$$ {\displaystyle \begin{array}{c}\widehat{p}\cdot \left[\left(1-\uplambda \right){\widehat{x}}^i+\uplambda {\widehat{z}}^i\right]=\left(1-\uplambda \right)\widehat{p}\cdot {\widehat{x}}^i+\uplambda \widehat{p}\cdot {\widehat{z}}^i\\ {}\le \left(1-\uplambda \right)\left(\widehat{p}\cdot {\widehat{e}}^i+\sum \limits_{j\in \mathbf{J}}{\theta}_{ij}{\Pi}_j^n\left(\widehat{p}\right)+s\right)+\uplambda \left(\widehat{p}\cdot {\widehat{e}}^i+\sum \limits_{j\in \mathbf{J}}{\theta}_{ij}{\Pi}_j^n\left(\widehat{p}\right)+s\right)\\ {}=\widehat{p}\cdot {\widehat{e}}^i+\sum \limits_{j\in \mathbf{J}}{\theta}_{ij}{\Pi}_j^n\left(\widehat{p}\right)+s.\end{array}} $$

Hence $\left(1-\lambda \right){\widehat{x}}^i+\lambda {\widehat{z}}^i\in {\xi }_n^i\left(\widehat{p},s\right)$ which means that ${\xi}_n^i\left(\widehat{p},s\right)$ is convex.

Let $\left\{{\widehat{p}}_k,{s}_k\right\}\subset S\cap \left({\mathrm{\mathbb{R}}}^{J+3}\times {\mathrm{\mathbb{R}}}_{+}\right)$ converge to $\left(\widehat{p},s\right)$ and let $\left\{{\widehat{x}}_k^i\right\}$ be a sequence with ${\widehat{x}}_k^i\in {\xi}_n^i\left({p}_k,{s}_k\right)\forall k.$ Since $\left\{{\widehat{x}}_k^i\right\}\subset {\widehat{X}}_n^i$ and ${\widehat{X}}_n^i$ is compact, there exists subsequence $\left\{{\widehat{x}}_{km}^i\right\}$ converging to ${\widehat{x}}^i\in {\widehat{X}}_n^i.$ We have

$$ {\widehat{p}}_{k_m}\cdot {\widehat{x}}_{km}^i\le {\widehat{p}}_{k_m}\cdot {\widehat{e}}^i+\sum \limits_{j\in \mathbf{J}}{\theta}_{ij}{\Pi}_j^n\left({\widehat{p}}_{k_m}\right)+{s}_{k_m}. $$

Letting m → +∞, we obtain

$$ \widehat{p}\cdot {\widehat{x}}^i\le \widehat{p}\cdot {\widehat{e}}^i+\sum \limits_{j\in \mathbf{J}}{\theta}_{ij}{\Pi}_j^n\left(\widehat{p}\right)+s. $$

This implies that ${\widehat{x}}^i\in {\xi}_n^i\left(\widehat{p},s\right).$ Hence ${\xi}_n^i$ is upper semi-continuous.

• Step 6. For each i ∈ I, the mapping ${Q}_n^i\left(\widehat{p},s\right)$ is upper semi-continuous having nonempty convex compact values.

Since ${\widehat{u}}^i$ is an upper semi-continuous function on ${\xi}_n^i\left(\widehat{p},s\right),$ nonempty compact subset of ℝ^J + 3, then ${\widehat{u}}^i$ has a maximum on ${\xi}_n^i\left(\widehat{p},s\right).$ Let $x\in {\xi}_n^i\left(\widehat{p},s\right)$ and ${\widehat{u}}^i(x)=\max \left\{{\widehat{u}}^i\left({\widehat{x}}^i\right):{\widehat{x}}^i\in {\xi}_n^i\left(\widehat{p},s\right)\right\}$. We will show that $x\in {Q}_n^i\left(\widehat{p},s\right)$.

Indeed, let ${x^{\prime}}^i\in {\widehat{X}}_n^i$ and ${\widehat{u}}^i\left({x^{\prime}}^i\right)>{\widehat{u}}^i(x)$ and then ${x^{\prime}}^i\notin {\xi}_n^i\left(\widehat{p},s\right)$ (by identifying x). Since ${x^{\prime}}^i\notin {\xi}_n^i\left(\widehat{p},s\right)$, we have

$$ \widehat{p}\cdot {x^{\prime}}^i>\widehat{p}\cdot {\widehat{e}}^i+\sum \limits_{j\in \mathbf{J}}{\theta}_{ij}{\Pi}_j^n\left(\widehat{p}\right)+s. $$

Hence $x\in {Q}_n^i\left(\widehat{p},s\right)$ which implies that ${Q}_n^i\left(\widehat{p},s\right)$ is not empty.

For every ${\widehat{x}}^i,{\widehat{z}}^i\in {Q}_n^i\left(\widehat{p},s\right)$ and $\lambda \in \left[0,1\right],$ we will prove that ${w}^i:= \uplambda {\widehat{x}}^i+\left(1-\uplambda \right){\widehat{z}}^i\in {Q}_n^i\left(\widehat{p},s\right).$Indeed, since ${\xi}_n^i\left(\widehat{p},s\right)$ is convex and ${\widehat{x}}^i,{\widehat{z}}^i\in {\xi}_n^i\left(\widehat{p},s\right)$, we have ${w}^i\in {\xi}_n^i\left(\widehat{p},s\right)$. Let ${x^{\prime}}^i\in {\widehat{X}}_n^i$ with ${\widehat{u}}^i\left({x^{\prime}}^i\right)>{\widehat{u}}^i\left({w}^i\right).$ Since ${\widehat{u}}^i$ is strictly quasi-concave,

$$ {\widehat{u}}^i\left({w}^i\right)={\widehat{u}}^i\left(\uplambda {\widehat{x}}^i+\left(1-\uplambda \right){\widehat{z}}^i\right)>\min\;\left({\widehat{u}}^i\left({\widehat{x}}^i\right),{\widehat{u}}^i\left({\widehat{z}}^i\right)\right). $$

It follows that

$$ {\widehat{u}}^i\left(x{\hbox{'}}^i\right)>\min \left({\widehat{u}}^i\left({\widehat{x}}^i\right),{\widehat{u}}^i\left({\widehat{z}}^i\right)\right) $$

since ${\widehat{x}}^i,{\widehat{z}}^i\in {Q}_n^i\left(\widehat{p},s\right);$ we have

$$ \widehat{p}\cdot {x^{\prime}}^i>\widehat{p}\cdot {\widehat{e}}^i+\sum \limits_{j\in \mathbf{J}}{\theta}_{ij}{\Pi}_j^n\left(\widehat{p}\right)+s. $$

Hence ${w}^i\in {Q}_n^i\left(\widehat{p},s\right)$ which implies that ${Q}_n^i\left(\widehat{p},s\right)$ is convex.

Let $\left({\widehat{p}}_k,{s}_k,{\widehat{x}}_k^i\right)\in \mathrm{graph}\;{Q}_n^i$ and assume that $\left({\widehat{p}}_k,{s}_k\right)\to \left(\widehat{p},s\right);{\widehat{x}}_k^i\to {\widehat{x}}^i.$ We will show that $\left(p,s,{\widehat{x}}^i\right)\in \mathrm{graph}{Q}_n^i$.

Since ${\widehat{x}}_k^i\in {Q}_n^i\left({p}_k,{s}_k\right)\subset {\xi}_n^i\left({p}_k,{s}_k\right)$ and ${\xi}_n^i$ is closed, we have ${\widehat{x}}^i\in {\xi}_n^i\left(\widehat{p},s\right).$

Let ${x^{\prime}}^i\in {\widehat{X}}_n^i$ with ${\widehat{u}}^i\left({x^{\prime}}^i\right)>{\widehat{u}}^i\left({\widehat{x}}^i\right).$ By the upper semi-continuity of ${\widehat{u}}^i$, we have the set

$$ E:= \left\{x:{\widehat{u}}^i(x)<{\widehat{u}}^i\left({x^{\prime}}^i\right)\right\} $$

which is open in ℝ^J + 3. Since ${\widehat{x}}^i\in E,$ there exists ε > 0 such that the ball $B\left({\widehat{x}}^i,\varepsilon \right)\subset E.$ On the other hand, since ${\widehat{x}}_k^i\to {\widehat{x}}^i$, with that ε, there exists k₀ such that ${\widehat{x}}_k^i\in B\left({\widehat{x}}^i,\varepsilon \right),\forall k>{k}_0$. Hence ${\widehat{u}}^i\left({\widehat{x}}_k^i\right)<{\widehat{u}}^i\left({x^{\prime}}^i\right)$ for all k large enough.Since ${\widehat{x}}_k^i\in {Q}_n^i\left({p}_k,{s}_k\right),$ we have

$$ {p}_k\cdot {x^{\prime}}^i>{p}_k\cdot {\widehat{e}}^i+\sum \limits_{j\in \mathbf{J}}{\theta}_{ij}{\Pi}_j^n\left({\widehat{p}}_k\right)+{s}_k; $$

letting k → +∞, we obtain

$$ \widehat{p}\cdot {x^{\prime}}^i>\widehat{p}\cdot {\widehat{e}}^i+\sum \limits_{j\in \mathbf{J}}{\theta}_{ij}{\Pi}_j^n\left(\widehat{p}\right)+s. $$

This implies that ${\widehat{x}}^i\in {Q}_n^i\left(\widehat{p},s\right)$. Hence ${Q}_n^i$ is closed. Moreover,

$$ {Q}_n^i\left(\widehat{p},s\right)\subset {\xi}_n^i\left(\widehat{p},s\right)\subset {\widehat{X}}_n^i,\kern0.5em \forall \left(\widehat{p},s\right)\in S\cap \left({\mathrm{\mathbb{R}}}^{J+3}\times {\mathrm{\mathbb{R}}}_{+}\right),n\ge 1 $$

and ${\widehat{X}}_n^i$ is compact; we see that ${Q}_n^i$ is a compact mapping.

It is obvious that S ∩ (ℝ^J + 3 × ℝ₊) is compact. Following with just proven result, ${Q}_n^i$ is closed. Hence ${Q}_n^i$ is upper semi-continuous.

• Step 7. Applying Lemma Gale–Nikaido–Debreu for multivalued mapping z _n .

It is easy to see that the set ℝ^J + 3 × ℝ₊ is a closed nonempty convex cone in ℝ^{J + 4} (which satisfied the condition (i) in Lemma GND).

By the result of step 6, it is easy to see that z _n is upper semi-continuous having nonempty convex compact values (the condition (ii) is satisfied).

For every $\left(\widehat{p},s\right)\in S\cap \left({\mathrm{\mathbb{R}}}^{J+3}\times {\mathrm{\mathbb{R}}}_{+}\right)$, note that $x\in {z}_n\left(\widehat{p},s\right)$ can be written as

$$ x=\left(\sum \limits_{i\in \mathbf{I}}{\widehat{x}}_n^i-\sum \limits_{i\in \mathbf{I}}{\widehat{e}}^i-\sum \limits_{j\in \mathbf{J}}{\widehat{y}}_j^n\right)\times \left(-I\right),\kern0.5em \mathrm{where}\;{\widehat{x}}_n^i\in {Q}_n^i\left(\widehat{p},s\right),{\widehat{y}}_j^n\in {\Phi}_j^n(p). $$

Since ${\widehat{x}}_n^i\in {Q}_n^i\left(\widehat{p},s\right)\subset {\xi}_n^i\left(\widehat{p},s\right)$, we have

$$ {\displaystyle \begin{array}{l}\kern2em \widehat{p}\cdot {\widehat{x}}_n^i\le \widehat{p}\cdot {\widehat{e}}_n^i+p\sum \limits_{j\in \mathbf{J}}{\theta}_{ij}{\Pi}_j^n(p)+s=\widehat{p}\cdot {\widehat{e}}^i+\sum \limits_{j\in \mathbf{J}}{\theta}_{ij}p\cdot {\widehat{y}}_j^n+s\\ {}\Rightarrow \widehat{p}\cdot \sum \limits_{i\in \mathbf{I}}{\widehat{x}}_n^i\le \widehat{p}\cdot \sum \limits_{i\in \mathbf{I}}{\widehat{e}}^i+\sum \limits_{i\in \mathbf{I}}\sum \limits_{j\in \mathbf{J}}{\theta}_{ij}p\cdot {\widehat{y}}_j^n+ Is=\widehat{p}\cdot \sum \limits_{i\in \mathbf{I}}{\widehat{e}}^i+p\cdot \sum \limits_{j\in \mathbf{J}}{\widehat{y}}_j^n+ Is\\ {}\Rightarrow \widehat{p}\cdot \left(\sum \limits_{i\in \mathbf{I}}{\widehat{x}}_n^i-\sum \limits_{i\in \mathbf{I}}{\widehat{e}}^i-\sum \limits_{i\in \mathbf{J}}{\widehat{y}}_j^i\right)- Is\le 0\\ {}\Rightarrow \left(\widehat{p},s\right)x\le 0.\end{array}} $$

Hence $\left(\widehat{p},s\right)x\le 0$ for every $\left(\widehat{p},s\right)\in S\cap {\mathrm{\mathbb{R}}}^{J+3}\times {\mathrm{\mathbb{R}}}_{+}$ and $x\in {z}_n\left(\widehat{p},s\right)$ (the condition (iii) is satisfied).

Let

$$ {\displaystyle \begin{array}{l}P:= {\mathrm{\mathbb{R}}}^{J+3}\times {\mathrm{\mathbb{R}}}_{+}=\left\{a=\left({a}_1,\dots, {a}_6\right):{a}_6\ge 0\right\}\\ {}{P}^0:= \left\{b=\left({b}_1,\dots, {b}_6\right)\in {\mathrm{\mathbb{R}}}^{J+4}:a\cdot b\le 0,\forall a\in P\right\}.\end{array}} $$

Note, for $j=\overline{1,J+4},\kern0.5em {\mathbf{1}}_j$, the vector with 1 in component j and 0 elsewhere. By choosing a =  ± 1 _j , j = 1, …, J + 4 and a = 1_J + 4, since a · b ≤ 0, we obtain

$$ {b}_1=\cdots ={b}_l=0;{b}_{J+4}\le 0. $$

Moreover, $b\in {\mathbf{O}}_{{\mathrm{\mathbb{R}}}^{J+3}}\times {\mathrm{\mathbb{R}}}_{-}$ satisfies a ⋅ b ≤ 0, ∀ a ∈ P. Hence

$$ {\left({\mathrm{\mathbb{R}}}^{J+3}\times {\mathrm{\mathbb{R}}}_{+}\right)}^0={\mathbf{O}}_{{\mathrm{\mathbb{R}}}^{J+3}}\times {\mathrm{\mathbb{R}}}_{-}. $$

Applying the Gale–Nikaido–Debreu Lemma (see Geistdoerfer−Florenzano 1982), we can conclude that there exists $\left({\widehat{p}}_n,{s}_n\right)\in S\cap \left({\mathrm{\mathbb{R}}}^{J+3}\times {\mathrm{\mathbb{R}}}_{+}\right)$ such that

$$z_n\left({\widehat{p}}_n,s_n\right)\cap {\mathbf{O}}_{{\mathrm{ℝ}}^{J+3}}\times {\mathrm{ℝ}}_{-}=\cancel{0}.$$

It follows that there exists ${\widehat{x}}_n\in {\mathrm{\mathbb{R}}}^{J+4}$ such that

$$ {\widehat{x}}_n=\left(\sum \limits_{i\in \mathbf{I}}{\widehat{x}}_n^i-\sum \limits_{i\in \mathbf{I}}{\widehat{e}}^i-\sum \limits_{j\in \mathbf{J}}{\widehat{y}}_j^n\right)\times \left(-{Is}_n\right) $$

(A.7.1)

$$ {\widehat{x}}_n^i\in {Q}_n^i\left({p}_n,{s}_n\right) $$

(A.7.2)

$$ \sum \limits_{i\in \mathbf{I}}{\widehat{x}}_n^i-\sum \limits_{i\in \mathbf{I}}{\widehat{e}}^i-\sum \limits_{j\in \mathbf{J}}{\widehat{y}}_j^n=0 $$

(A.7.3)

From (3), we have $\left({\widehat{x}}_n^i,{\widehat{y}}_j^n\right)\in \widehat{A}.$ Since $\widehat{A}$ is compact, without loss of generality, we may assume that

$$ \left({\widehat{x}}_n^i,{\widehat{y}}_j^n\right)\to \left({\widehat{x}}^{\ast i},{\widehat{y}}_j^{\ast}\right),\kern0.5em i\in \mathbf{I},j\in \mathbf{J}. $$

Since (p _n , s _n ) ∈ S ∩ (ℝ^J + 3 × ℝ₊) and S ∩ (ℝ^J + 3 × ℝ₊) are compact, we can also assume that

$$ \left({\widehat{p}}_n,{s}_n\right)\to \left({\widehat{p}}^{\ast },{s}^{\ast}\right). $$

We will prove the existence of equilibrium with $\left({\left({\widehat{x}}^{\ast i}\right)}_{i\in \mathbf{I}},{\widehat{p}}^{\ast}\right)$ that has been found.

• Step 8. Existence of quasi-equilibrium .

From (3), let n → +∞; we obtain

$$ \sum \limits_{i\in \mathbf{I}}{\widehat{x}}^{\ast i}=\sum \limits_{i\in \mathbf{I}}{\widehat{e}}^i+\sum \limits_{j\in \mathbf{J}}{\widehat{y}}_j^{\ast }; $$

(A.7.4)

hence the condition (a) is satisfied.

From (A.7.2), we have ${\widehat{x}}_n^i\in {\xi}_n^i\left({\widehat{p}}_n,{s}_n\right)$; this implies

$$ {\widehat{p}}_n\cdot {\widehat{x}}_n^i\le {\widehat{p}}_n\cdot {\widehat{e}}^i+\sum \limits_{j\in \mathbf{J}}{\theta}_{ij}{\Pi}_j^n\left({p}_n\right)+{s}_n,\kern0.5em \forall i\in \mathbf{I}. $$

Letting n → +∞, we obtain

$$ {\widehat{p}}^{\ast}\cdot {\widehat{x}}^{\ast i}\le {\widehat{p}}^{\ast}\cdot {\widehat{e}}^i+\sum \limits_{j\in \mathbf{J}}{\theta}_{ij}{\Pi}_j\left({p}^{\ast}\right)+{s}^{\ast },\kern0.5em \forall i\in \mathbf{I}. $$

(A.7.5)

Let ${\widehat{x}}^i\in {\widehat{x}}_n^i$ with ${\widehat{u}}^i\left({\widehat{x}}^i\right)>{\widehat{u}}^i\left({x}_i^{\ast}\right)$. Let $\lambda \in \left]0,1\right].$ Define

$$ {\widehat{x}}_{\uplambda}^i:\left(\uplambda {\widehat{x}}^i\right)+\left(1-\uplambda \right){\widehat{x}}^{\ast i}. $$

Since ${\widehat{u}}^i$ is strictly quasi-concave, we have

$${\widehat{u}}^i\left({\widehat{x}}_{\lambda }^i\right)>{\widehat{u}}^i\left(x_i^{\ast }\right).$$

Since ${\widehat{u}}^i$ is upper semi-continuous and ${\widehat{x}}_n^i\to {\widehat{x}}^{\ast i},$ for all n large enough, we have

$${\widehat{u}}^i\left({\widehat{x}}_{\lambda }^i\right)>{\widehat{u}}^i\left({\widehat{x}}_n^i\right).$$

From (A.7.2), ${\widehat{x}}_n^i\in {Q}_n^i\left({\widehat{p}}_n,{s}_n\right)$, we obtain

$${\widehat{p}}_n\cdot {\widehat{x}}_{\lambda }^i\le {\widehat{p}}_n\cdot {\widehat{e}}^i+\sum _{j\in \mathbf{J}}{\theta }_{\mathit{ij}}{\mathrm{\Pi }}_j^n\left(p_n\right)+s_n$$

$$\iff {\widehat{p}}_n\cdot \left(\lambda {\widehat{x}}^i+\left(1-\lambda \right){\widehat{x}}^{\ast i}\right)\ge {\widehat{p}}_n\cdot {\widehat{e}}^i+\sum _{j\in \mathbf{J}}{\theta }_{\mathit{ij}}{\mathrm{\Pi }}_j^n\left(p_n\right)+s_n.$$

Let n → +∞; we obtain

$${\widehat{p}}^{\ast }\cdot \left(\lambda {\widehat{x}}^i+\left(1-\lambda \right){\widehat{x}}^{\ast i}\right)\ge {\widehat{p}}_n\cdot {\widehat{e}}^i+\sum _{j\in \mathbf{J}}{\theta }_{\mathit{ij}}{\mathrm{\Pi }}_j\left(p^{\ast }\right)+s^{\ast }.$$

(A.7.6)

Let λ → 0; we have

$$ {\widehat{p}}^{\ast}\cdot {\widehat{x}}^{\ast i}\ge {\widehat{p}}^{\ast}\cdot {\widehat{e}}^i+\sum \limits_{j\in \mathbf{J}}{\theta}_{ij}{\Pi}_j\left({p}^{\ast}\right)+{s}^{\ast }. $$

(A.7.7)

Then from (A.7.5) and (A.7.7), follows

$$ {\displaystyle \begin{array}{l}{\widehat{p}}^{\ast}\cdot {\widehat{x}}^{\ast i}\ge {\widehat{p}}^{\ast}\cdot {\widehat{e}}^i+\sum \limits_{j\in \mathbf{J}}{\theta}_{ij}{\Pi}_j\left({p}^{\ast}\right)+{s}^{\ast}\kern0.5em \forall i\in \mathbf{I}\\ {}\begin{array}{cc}\iff {\widehat{p}}^{\ast}\cdot {\widehat{x}}^{\ast i}={\widehat{p}}^{\ast}\cdot {\widehat{e}}^i+\sum \limits_{j\in \mathbf{J}}{\theta}_{ij}{\widehat{p}}^{\ast}\cdot {\widehat{y}}_j^{\ast }+{s}^{\ast }& \forall i\in \mathbf{I}\end{array}.\end{array}} $$

(A.7.8)

Hence

$$ {\displaystyle \begin{array}{c}{\widehat{p}}^{\ast}\cdot \sum \limits_{i\in \mathbf{I}}{\widehat{x}}^{\ast i}={\widehat{p}}^{\ast}\cdot \sum \limits_{i\in \mathbf{I}}{\widehat{e}}^i+\sum \limits_{i\in \mathbf{I}}\sum \limits_{j\in \mathbf{J}}{\theta}_{ij}{\widehat{p}}^{\ast}\cdot {\widehat{y}}_j^{\ast }+{Is}^{\ast}\kern0.5em \forall i\in \mathbf{I}\\ {}\iff {\widehat{p}}^{\ast}\cdot \left(\sum \limits_{i\in \mathbf{I}}{\widehat{x}}^{\ast i}-\sum \limits_{i\in \mathbf{I}}{\widehat{e}}^i-\sum \limits_{i\in \mathbf{I}}{\widehat{y}}_j^{\ast}\right)={Is}^{\ast }.\end{array}} $$

From (A.7.4), follows

$$ {s}^{\ast }=0. $$

Since $\left({\widehat{p}}^{\ast },{s}^{\ast}\right)\in S$, it follows that ${\widehat{p}}^{\ast}\ne 0$

Moreover, by substituting $\lambda =1$ and s^∗ = 0 into (A.7.6), we obtain that

$$ {\widehat{p}}^{\ast}\cdot {\widehat{x}}^i\ge {\widehat{p}}^{\ast}\cdot {\widehat{e}}^i+\sum \limits_{j\in \mathbf{J}}{\theta}_{ij}\sup\;\left({p}^{\ast}\cdot {Y}_j\right) $$

for all ${\widehat{x}}^i\in {\widehat{x}}^i$ with ${\widehat{u}}^i\left({\widehat{x}}^i\right)>{\widehat{u}}^i\left({\widehat{x}}^{\ast i}\right).$ Hence the condition (b) is satisfied.

Thus $\left({\left({\widehat{x}}^{\ast i}\right)}_{i\in \mathbf{I}},{\widehat{p}}^{\ast}\right)$ is a quasi-equilibrium.

1.2 Appendix 7.2: Proof of Proposition 7.2

From Proposition 7.1, there exists a quasi-equilibrium (with dividends) $\left({\left({x}^{\ast i},{d}^{\ast i}\right)}_{i\in \mathbf{I}},{\left({y}_j^{\ast },0\right)}_{j\in \mathbf{J}},\left({p}^{\ast },{q}^{\ast}\right)\right)$ which satisfies

• 1. $\sum \limits_{i\in \mathbf{I}}{x}^{\ast i}=\sum \limits_{i\in \mathbf{I}}{e}^i+\sum \limits_{j\in \mathbf{J}}{y}_j^{\ast };\sum \limits_{i\in \mathbf{I}}{d}^{\ast i}=\sum \limits_{i\in \mathbf{I}}{\delta}^i$

• 2. For any i ∈ I

$$ {p}^{\ast}\cdot {x}^{\ast i}+{q}^{\ast }{d}^{\ast i}={p}^{\ast}\cdot {e}^i+\sum \limits_{j\in \mathbf{J}}{\theta}_{ij}\sup \left({p}^{\ast}\cdot {Y}_j\right)+{q}^{\ast }{\delta}^i. $$

For each (xⁱ, dⁱ) ∈ Xⁱ × ℝ₊, with ${\widehat{u}}^i\left({x}^i,{d}^i\right)>{\widehat{u}}^i\left({x}^{\ast i},{d}^{\ast i}\right),$ it holds

$$ {p}^{\ast}\cdot {x}^i+{q}^{\ast }{d}^i\ge {p}^{\ast}\cdot {e}^i+\sum \limits_{j\in \mathbf{J}}{\theta}_{ij}\sup \left({p}^{\ast}\cdot {Y}_j\right)+{q}^{\ast }{\delta}^i. $$

• 3. For any $j\in \mathbf{J}:{p}^{\ast}\cdotp {y}_j^{\ast }=\sup \left({p}^{\ast}\cdotp {Y}_j\right).$

We will prove that q^∗ = 0 and p^∗ ≠ 0.

If x^∗i ∉ Sⁱ, then there exists xⁱ ∈ Xⁱ : uⁱ(xⁱ) > uⁱ(x^∗i). Let $\left({x}^i,0\right)\in {\widehat{X}}^i:{u}^i\left({x}^i\right)=\widehat{u}\left({x}^i,0\right)>\widehat{u}\left({x}^{\ast i},{d}^{\ast i}\right)=u\left({x}^{\ast i}\right).$ We then have

$$ {\displaystyle \begin{array}{c}{p}^{\ast}\cdot {x}^i\ge {p}^{\ast}\cdot {e}^i+\sum \limits_{j\in \mathbf{J}}{\theta}_{ij}\sup \left({p}^{\ast}\cdot {Y}_j\right)+{q}^{\ast }{\delta}^i\\ {}={p}^{\ast}\cdot {x}^{\ast i}+{q}^{\ast }{d}^{\ast i}.\end{array}} $$

Let x_λ = λxⁱ + (1 − λ)x^∗i for any λ ∈ ]0, 1[. From the concavity of u, we have

$$ u\left({x}_{\uplambda}\right)=u\left(\uplambda {x}^i+\left(1-\uplambda \right){x}^{\ast i}\right)\ge \uplambda u\left({x}^i\right)+\left(1-\uplambda \right)u\left({x}^{\ast i}\right)>u\left({x}^{\ast i}\right). $$

We then have

$$\begin{array}{c}{p}^{\ast }\cdot x_{\lambda }\ge p^{\ast }\cdot x^{\ast i}+q^{\ast }{d}^{\ast i}\\ \iff p^{\ast }\cdot \left(\mathrm{\lambda }{x}^i+\left(1-\mathrm{\lambda }\right)x^{\ast i}\right)\ge p^{\ast }\cdot x^{\ast i}+q^{\ast }{d}^{\ast i}.\end{array}$$

Letting λ converge to 0, we obtain q^∗d^∗i ≤ 0. Hence q^∗d^∗i = 0 for all i ∈ I. Because

∑_i ∈ Id^∗i = ∑_i ∈ Iδⁱ > 0, then q^∗ = 0.

If p^∗ = 0, we then have q^∗δⁱ = q^∗d^∗i = 0, ∀ i ∈ I; it implies q^∗ = 0 contradiction with (p^∗, q^∗) ≠ (0, 0).

Hence we obtain $\left({\left({x}^{\ast i}\right)}_{i\in \mathbf{I}},{\left({y}_j^{\ast}\right)}_{j\in \mathbf{J}},{p}^{\ast}\right)$ is a quasi-equilibrium.

1.3 Appendix 7.3: Proof of Lemma 7.2

Consider the problem of firm j. The FOCs give (for all j ∈ J)

$$ {\displaystyle \begin{array}{l}{p}_j{A}_j\frac{\partial F}{\partial K}\left({K}_j,{L}_j\right)=r\\ {}{p}_j{A}_j\frac{\partial F}{\partial L}\left({K}_j,{L}_j\right)=w.\end{array}} $$

This implies

$$ {p}_j{A}_j\frac{\partial F}{\partial K}\left(\frac{K_j}{L_j},1\right)=r $$

(A.7.9)

$$ {p}_j{A}_j\frac{\partial F}{\partial L}\left(1,\frac{K_j}{L_j}\right)=w. $$

(A.7.10)

Dividing (A.7.9) by (A.7.10), we obtain

$$ \frac{r}{w}=\frac{\frac{\partial F}{\partial K}\left(\frac{K_j}{L_j},1\right)}{\frac{\partial F}{\partial L}\left(1,\frac{L_j}{K_j}\right)},\kern0.5em \forall j\in \mathbf{J}. $$

Hence

$$ \frac{K_1}{L_1}=\frac{K_2}{L_2}=\cdots =\frac{K_J}{L_J}=\Phi \left(\frac{r}{w}\right), $$

and

$$ \frac{K_1+{K}_2+\dots +{K}_J}{L_1+{L}_2+\dots +{L}_J}=\frac{\overline{K}}{L}=\Phi \left(\frac{r}{w}\right). $$

Observe that

$$ {rK}_j+{wL}_j={p}_j{A}_jF\left({K}_j,{L}_j\right),\kern0.5em \forall j\in \mathbf{J} $$

or

$$ {\displaystyle \begin{array}{r}r\frac{K_j}{L_j}+w={p}_j{A}_jF\left(\frac{K_j}{L_j},1\right),\kern0.5em \forall j\in \mathbf{J}\\ {}\Rightarrow \begin{array}{cc}r\frac{\overline{K}}{L}+w={p}_j{A}_jF\left(\frac{\overline{K}}{L},1\right),& \forall j\in \mathbf{J}\end{array}.\end{array}} $$

Hence p₁A₁ = p₂A₂ = … = p _J A _J . Write ς = p₁A₁ = p₂A₂ = … = p _J A _J . We also have

$$ {\displaystyle \begin{array}{l}F\left({K}_1,{L}_1\right)+F\left({K}_2,{L}_2\right)+\cdots +F\left({K}_J,{L}_J\right)\\ {}={L}_1F\left(\frac{K_1}{L_1},1\right)+{L}_2F\left(\frac{K_2}{L_2},1\right)+\cdots +{L}_JF\left(\frac{K_J}{L_J},1\right)\\ {}={L}_1F\left(\frac{\overline{K}}{L},1\right)+{L}_2F\left(\frac{\overline{K}}{L},1\right)+\cdots +{L}_JF\left(\frac{\overline{K}}{L},1\right)\\ {}=\left({L}_1+{L}_2+\cdots +{L}_J\right)F\left(\frac{\overline{K}}{L},1\right)= LF\left(\frac{\overline{K}}{L},1\right)=F\left(\overline{K},L\right)\end{array}} $$

and

$$ {\displaystyle \begin{array}{rl} cr\overline{K}+ wL& ={p}_1{A}_1F\left({K}_1,{L}_1\right)+{p}_2{A}_2F\left({K}_2,{L}_2\right)+\cdots +{p}_J{A}_JF\left({K}_J,{L}_J\right)\\ {}& =\varsigma \left(F\left({K}_1,{L}_1\right)+F\left({K}_2,{L}_2\right)+\cdots +F\left({K}_J,{L}_J\right)\right)\\ {}& =\varsigma F\left(\overline{K},L\right).\end{array}} $$

The budget constraint

$$ {p}_1{c}_1+{p}_2{c}_2+\dots +{p}_J{c}_J\le \varsigma F\left(\overline{K},L\right). $$

Dividing by ς = p₁A₁ = p₂A₂ = … = p _J A _J , we get

$$ \frac{c_1}{A_1}+\frac{c_2}{A_2}+\dots +\frac{c_J}{A_J}\le F\left(\overline{K},L\right). $$

1.4 Appendix 7.4: Proof of Proposition 7.5

1. Consider the problem of consumer

$$ \underset{c_1,{c}_2,L\ge 0}{\max}\ln \left({c}_1\right)+\beta \ln \left({c}_2\right) $$

subject to

$$ \frac{c_1}{A_1}+\frac{c_2}{A_2}={\overline{K}}^{\alpha }{L}^{1-\alpha } $$

(A.7.11)

$$ {a}_1{c}_1+{a}_2{c}_2+L=\overline{L}. $$

(A.7.12)

We have the FOCs (λ₁ > 0, λ₂ > 0, λ₃ > 0)

$$ \frac{1}{c_1}={\uplambda}_2\frac{1}{A_1}+{\uplambda}_3{a}_1 $$

(A.7.13)

$$ \frac{\beta }{c_2}={\uplambda}_2\frac{1}{A_2}+{\uplambda}_3{a}_2 $$

(A.7.14)

$$ {\uplambda}_1={\uplambda}_2\alpha {\overline{K}}^{\alpha -1}{L}^{1-\alpha } $$

(A.7.15)

$$ {\uplambda}_3={\uplambda}_2\left(1-\alpha \right){\overline{K}}^{\alpha }{L}^{-\alpha }. $$

(A.7.16)

From (A.7.13) and (A.7.14), we have

$$ {\displaystyle \begin{array}{l}1={\uplambda}_2\frac{c_1}{A_1}+{\uplambda}_2\left(1-\alpha \right){\overline{K}}^{\alpha }{L}^{-\alpha }{a}_1{c}_1\\ {}\beta ={\uplambda}_2\frac{c_2}{A_2}+{\uplambda}_2\left(1-\alpha \right){\overline{K}}^{\alpha }{L}^{-\alpha }{a}_2{c}_2\end{array}} $$

which implies that

$$ 1+\beta ={\uplambda}_2\left(\frac{c_1}{A_1}+\frac{c_2}{A_2}\right)+{\uplambda}_2\left(1-\alpha \right){\overline{K}}^{\alpha }{L}^{-\alpha}\left({a}_1{c}_1+{a}_2{c}_2\right). $$

Combining with (A.7.11) and (A.7.12), we have

$$ {\displaystyle \begin{array}{c}1+\beta ={\uplambda}_2{\overline{K}}^{\alpha }{L}^{1-\alpha }+{\uplambda}_2\left(1-\alpha \right){\overline{K}}^{\alpha }{L}^{-\alpha}\left(\overline{L}-L\right)\\ {}\iff \frac{1}{\uplambda_2}=\frac{1}{1+\beta }{\overline{K}}^{\alpha }{L}^{-\alpha}\left[\alpha L+\left(1-\alpha \right)\overline{L}\right].\end{array}} $$

(A.7.17)

In the other side, consider (A.7.13) and (A.7.14):

$$ {\displaystyle \begin{array}{c}{c}_1=\frac{A_1}{\uplambda_2+{\uplambda}_3{a}_1{A}_1}=\frac{A_1}{\uplambda_2+{\uplambda}_2\left(1-\alpha \right){\overline{K}}^{\alpha }{L}^{-\alpha }{a}_1{A}_1}\\ {}=\frac{1}{\uplambda_2}\frac{A_1}{1+\left(1-\alpha \right){\overline{K}}^{\alpha }{L}^{-\alpha }{a}_1{A}_1}\end{array}} $$

(A.7.18)

$$ {\displaystyle \begin{array}{c}{c}_2=\frac{\beta {A}_2}{\uplambda_2+{\uplambda}_3{a}_2{A}_2}=\frac{\beta {A}_2}{\uplambda_2+{\uplambda}_2\left(1-\alpha \right){\overline{K}}^{\alpha }{L}^{-\alpha }{a}_2{A}_2}\\ {}=\frac{1}{\uplambda_2}\frac{\beta {A}_2}{1+\left(1-\alpha \right){\overline{K}}^{\alpha }{L}^{-\alpha }{a}_2{A}_2}\end{array}} $$

(A.7.19)

substituting c₁ and c₂ into the equation (A.7.11)

$$ {\displaystyle \begin{array}{l}\kern2.5em \frac{1}{\uplambda_2}\frac{1}{1+\left(1-\alpha \right){\overline{K}}^{\alpha }{L}^{-\alpha }{a}_1{A}_1}+\frac{1}{\uplambda_2}\frac{\beta }{1+\left(1-\alpha \right){\overline{K}}^{\alpha }{L}^{-\alpha }{a}_2{A}_2}\\ {}={\overline{K}}^{\alpha }{L}^{1-\alpha}\iff \frac{1}{1+\beta }{\overline{K}}^{\alpha }{L}^{-\alpha}\left[\alpha L+\left(1-\alpha \right)\overline{L}\right]\\ {}\begin{array}{ll}& \times \left(\frac{1}{1+{\left(1-\alpha \right)}^{\alpha }{L}^{-\alpha }{a}_1{A}_1}+\frac{\beta }{1+\left(1-\alpha \right){\overline{K}}^{\alpha }{L}^{-\alpha }{a}_2{A}_2}\right)\\ {}& ={\overline{K}}^{\alpha }{L}^{1-\alpha}\iff \frac{1}{1+\beta}\left[\frac{1}{1+{\left(1-\alpha \right)}^{\alpha }{L}^{-\alpha }{a}_1{A}_1}+\frac{\beta }{1+\left(1-\alpha \right){\overline{K}}^{\alpha }{L}^{-\alpha }{a}_2{A}_2}\right]\\ {}& =\frac{L}{\alpha L+\left(1-\alpha \right)\overline{L}}.\end{array}\end{array}} $$

We obtain

$$ \frac{1}{1+\beta}\cdot \frac{\beta }{L^{\alpha }+\left(1-\alpha \right){a}_1{A}_1{\overline{K}}^{\alpha }1}+\frac{\beta }{1+\beta}\cdot \frac{1}{L^{\alpha }+\left(1-\alpha \right){a}_2{A}_2{\overline{K}}^{\alpha }}=\frac{L^{1-\alpha }}{\alpha L+\left(1-\alpha \right)\overline{L}}. $$

(A.7.20)

Equation (A.7.20) allows us to identify L^∗. Denote

$$ {\displaystyle \begin{array}{l}{B}_j=\left(1-\alpha \right){a}_j{A}_j{\overline{K}}^{\alpha },j=1,2\\ {}{\beta}_1=\frac{1}{1+\beta };\kern0.5em {\beta}_2=\frac{\beta }{1+\beta }.\end{array}} $$

Consider the function

$$ f\left(L,{B}_1,{B}_2\right)=\frac{\beta_1}{L^{\alpha }+{B}_1}+\frac{\beta_2}{L^{\alpha }+{B}_2}-\frac{L^{1-\alpha }}{\alpha L+\left(1-\alpha \right)\overline{L}}. $$

We have

$$ {\displaystyle \begin{array}{c}f\left(0,{B}_1,{B}_2\right)=\frac{\beta_1}{B_1}+\frac{\beta_2}{B_2}>0\\ {}f\left(\overline{L},{B}_1,{B}_2\right)=\frac{-{\beta}_1}{L^{\alpha }+{B}_1}+\frac{\beta_2}{{\overline{L}}^{\alpha }+{B}_2}-\frac{L^{1-\alpha }}{\alpha \overline{L}+\left(1-\alpha \right)L}\\ {}<\frac{\beta_1}{{\overline{L}}^{\alpha }}+\frac{\beta_2}{{\overline{L}}^{\alpha }}-\frac{1}{\alpha }=0.\end{array}} $$

Equation (A.7.20) has a solution ${L}^{\ast }=L\left({B}_1,{B}_2\right)\in \left(0,\overline{L}\right).$ From Proposition 7.3, ${L}^{\ast}\in \left(0,\overline{L}\right)$ is the unique solution to (A.7.20).

1.5 Appendix 7.5: Proof of Proposition 7.6

• (i) We now claim that L^∗ is decreasing in B _j , for j = 1, 2. By using partial derivatives of f(L(B₁, B₂), B₁, B₂) = 0 with respect to B₁, we have

$$ {\displaystyle \begin{array}{l}\kern2.12em \left[-\frac{{\alpha \beta}_1{L}^{\alpha -1}}{{\left({L}^{\alpha }+{B}_1\right)}^2}-\frac{{\alpha \beta}_2{L}^{\alpha -1}}{{\left({L}^{\alpha }+{B}_2\right)}^2}-\frac{{\left(1-\alpha \right)}^2\overline{L}{L}^{-\alpha }-{\alpha}^2{L}^{1-\alpha }}{{\left(\alpha L+\left(1-\alpha \right)\overline{L}\right)}^2}\right]\frac{\partial L}{\partial {B}_1}\\ {}-\frac{\beta }{{\left({L}^{\alpha }+{B}_1\right)}^2}=0\iff \left[\frac{{\alpha \beta}_1}{{\left({L}^{\alpha }+{B}_1\right)}^2}-\frac{{\alpha \beta}_2}{{\left({L}^{\alpha }+{B}_2\right)}^2}-\frac{{\left(1-\alpha \right)}^2\overline{L}{L}^{1-2\alpha }-{\left(\alpha {L}^{1-\alpha}\right)}^2}{{\left(\alpha L+\left(1-\alpha \right)\overline{L}\right)}^2}\right]\\ {}\times \frac{\partial L}{\partial {B}_1}=\frac{\beta {L}^{1-\alpha }}{{\left({L}^{\alpha }+{B}_1\right)}^2}.\end{array}} $$

(A.7.21)

We will prove that

$$ \frac{{\alpha \beta}_1}{{\left({L}^{\alpha }+{B}_1\right)}^2}-\frac{{\alpha \beta}_2}{{\left({L}^{\alpha }+{B}_2\right)}^2}\ge \frac{{\left(\alpha {L}^{1-\alpha}\right)}^2}{{\left(\alpha L+\left(1-\alpha \right)\overline{L}\right)}^2}. $$

(A.7.22)

Indeed

$$ {\displaystyle \begin{array}{c}{\left(\frac{\alpha {L}^{1-\alpha }}{\alpha L+\left(1-\alpha \right)\overline{L}}\right)}^2={\left(\frac{{\alpha \beta}_1}{L^{\alpha }+{B}_1}+\frac{{\alpha \beta}_2}{L^{\alpha }+{B}_2}\right)}^2\\ {}\le \alpha \left(\frac{{\alpha \beta}_1}{{\left({L}^{\alpha }+{B}_1\right)}^2}+\frac{{\alpha \beta}_2}{{\left({L}^{\alpha }+{B}_2\right)}^2}\right)\\ {}\le \frac{{\alpha \beta}_1}{{\left({L}^{\alpha }+{B}_1\right)}^2}+\frac{{\alpha \beta}_2}{{\left({L}^{\alpha }+{B}_2\right)}^2}\end{array}} $$

that gives us (A.7.22). Hence $\frac{\partial L}{\partial {B}_1}<0.$ By using the same argument with B₂, we get $\frac{\partial L}{\partial {B}_j}<0$for all j = 1, 2. When a _j increases, B _j increases, and L^∗ decreases.Consider the problem of firm j:

$$ \underset{K_j,{L}_j\ge 0}{\max }{p}_j{A}_j{K}_j^{\alpha }{L}_j^{1-\alpha }-{rK}_j-{wL}_j. $$

The FOC is following:

$$ {\displaystyle \begin{array}{l}\alpha {p}_j{A}_j{K}_j^{\alpha -1}{L}_j^{1-\alpha }=r\\ {}\left(1-\alpha \right){p}_j{A}_i{K}_j^{\alpha }{L}_j^{-\alpha }=w.\end{array}} $$

We obtain

$$ {\displaystyle \begin{array}{l}\frac{r}{w}=\frac{\alpha }{1-\alpha}\frac{L_j}{K_j}\\ {}{\alpha}^{\alpha }{\left(1-\alpha \right)}^{1-\alpha }{p}_j{A}_j={r}^{\alpha }{w}^{1-\alpha }.\end{array}} $$

Hence

$$ {\displaystyle \begin{array}{l}\frac{K_1}{L_1}=\frac{K_2}{L_2}=\frac{K_1+{K}_2}{L_1+{L}_2}=\frac{\overline{K}}{L}=\frac{\alpha }{1-\alpha}\frac{w}{r}\\ {}{p}_1{A}_1={p}_2{A}_2=\frac{r^{\alpha }{w}^{1-\alpha }}{\alpha^{\alpha }{\left(1-\alpha \right)}^{1-\alpha }}.\end{array}} $$

We can represent w and r as functions of L and $\overline{K}$ as follows:

$$ {\displaystyle \begin{array}{c}w={\alpha}^{\alpha }{\left(1-\alpha \right)}^{1-\alpha }{p}_j{A}_j{\left(\frac{w}{r}\right)}^{\alpha }={\alpha}^{\alpha }{\left(1-\alpha \right)}^{1-\alpha }{p}_j{A}_j{\left(\frac{1-\alpha }{\alpha}\frac{\overline{K}}{L}\right)}^{\alpha}\\ {}=\left(1-\alpha \right){p}_j{A}_j{\left(\frac{\overline{K}}{L}\right)}^{\alpha}\end{array}} $$

(A.7.23)

$$ {\displaystyle \begin{array}{c}r=\frac{L}{\overline{K}}\frac{\alpha }{1-\alpha }w=\frac{L}{\overline{K}}\frac{\alpha }{1-\alpha}\left(1-\alpha \right){p}_j{A}_j{\left(\frac{\overline{K}}{L}\right)}^{\alpha}\\ {}=\alpha {p}_j{A}_j{\left(\frac{L}{\overline{K}}\right)}^{1-\alpha }.\end{array}} $$

(A.7.24)

This implies that w^∗ increases and r^∗ decreases when L^∗ decreases.

• (ii) From Eqs. (A.7.18), (A.7.19) and (A.7.20),

$$ {\displaystyle \begin{array}{l}{c}_1^{\ast }=\frac{1}{1+\beta}\frac{A_1{\overline{K}}^{\alpha }{L}^{\ast -\alpha}\left[\alpha {L}^{\ast }+\left(1-\alpha \right)\overline{L}\right]}{1+\left(1-\alpha \right){\overline{K}}^{\alpha }{L}^{\ast -\alpha }{a}_1{A}_1}\\ {}{c}_2^{\ast }=\frac{\beta }{1+\beta}\frac{A_2{\overline{K}}^{\alpha }{L}^{\ast -\alpha}\left[\alpha {L}^{\ast }+\left(1-\alpha \right)\overline{L}\right]}{1+\left(1-\alpha \right){\overline{K}}^{\alpha }{L}^{\ast -\alpha }{a}_2{A}_2}.\end{array}} $$

Market clearing conditions give

$$ {\displaystyle \begin{array}{c}{L}_1^{\ast }=\frac{c_1^{\ast }}{A_1{\overline{K}}^{\alpha }{L}^{\ast -\alpha }}=\frac{1}{1+\beta}\frac{\left[\alpha {L}^{\ast }+\left(1-\alpha \right)\overline{L}\right]}{1+\left(1-\alpha \right){\overline{K}}^{\alpha }{L}^{\ast -\alpha }{a}_1{A}_1}\\ {}{L}_2^{\ast }=\frac{c_1^{\ast }}{A_1{\overline{K}}^{\alpha }{L}^{\ast -\alpha }}=\frac{\beta }{1+\beta}\frac{\left[\alpha {L}^{\ast }+\left(1-\alpha \right)\overline{L}\right]}{1+\left(1-\alpha \right){\overline{K}}^{\alpha }{L}^{\ast -\alpha }{a}_2{A}_2}.\end{array}} $$

If a₁ increases, L^∗ decreases and L^∗ − αa₁ increases, this implies that ${L}_1^{\ast }$ decreases.