On the regularity of smooth production economies with externalities: competitive equilibrium à la Nash

Abstract

We consider a general equilibrium model of a private ownership economy with consumption and production externalities. The choices of all agents (households and firms) may affect utility functions and production technologies. The allocation of a competitive equilibrium is a Nash equilibrium. We provide an example showing that, under standard assumptions, competitive equilibria are indeterminate in an open set of the household’s endowments. Next, we consider a new version of this model, with firms’ endowments in the spirit of Geanakoplos et al. (J Math Econ 19:113–151, 1990). In our model, firms’ endowments impact the technologies of the other firms. We then prove that, generically in the space of endowments of households and firms, each economy has a finite number of competitive equilibria and each competitive equilibrium is locally a differentiable map of the fundamental parameters.

Appendix

In this section, we prove all the lemmas stated in Sect. 4. The following notation helps in writing the proofs,

$$\begin{aligned} t_{j}(y_{j} - \eta _{j},y_{-j},x)=(t_{j} \circ g_{j})(y_{j},y_{-j} ,x,\eta _{j}) \end{aligned}$$

(15)

where the mapping $g_{j}$ is defined by $g_{j}(y_{j},y_{-j} ,x,\eta _{j}):=(y_{j} - \eta _{j},y_{-j} ,x)$.

Proof of Lemma 11

In order to prove this lemma, one must pay attention to Assumption 1.2 and Lemma 4.

First, notice that for any given endowment $\eta _{j}$, the function given in (15) satisfies all the assumptions given in Assumption 1, except Assumption 1.2. However, the total production plan $y_{j}=\eta _{j}$ acts for $(t_{j} \circ g_{j})(y_{j},y_{-j} ,x,\eta _{j})$ as $y_{j}=0$ acts for $t_{j}(y_{j}, y_{-j},x)$. Indeed, Assumption 1.2 implies that $(t_{j} \circ g_{j})(\eta _{j},y_{-j} ,x,\eta _{j})=0$, and then using standard maximization arguments, one gets $p \cdot y^{*}_{j} \ge p \cdot \eta _{j}$ if $y^{*}_{j}$ solves problem (14). Consequently, the individual wealth $p \cdot (e_{ h} + \displaystyle \sum \nolimits _{j\in {\mathcal {J}}}s_{jh} y^{*}_{j})$ of household h is greater than $p \cdot (e_{ h} + \displaystyle \sum \nolimits _{j\in {\mathcal {J}}}s_{jh} \eta _{j} )$ which is strictly positive since $(e,\eta ) \gg 0$.

Second, one needs to establish the result analogous to Lemma 4 for the functions $(t_{j} \circ g_{j})_{j \in {\mathcal {J}}}$ and the vector $r=\displaystyle \sum \nolimits _{h\in {\mathcal {H}}}e_{h}$. For this purpose, the set K(r) must be replaced with the set $\left( K(r+\displaystyle \sum \nolimits _{j\in {\mathcal {J}}} \eta _{j}) \right) + {\widetilde{\eta }}$, where ${\widetilde{\eta }}:=(0,\eta )$ and the set $K(r+\displaystyle \sum \nolimits _{j\in {\mathcal {J}}} \eta _{j})$ is provided by Lemma 4 for the functions $(t_{j})_{j \in {\mathcal {J}}}$ and the vector $r+\displaystyle \sum \nolimits _{j\in {\mathcal {J}}} \eta _{j}$. The proof of this lemma is then established by adapting the proof of Theorem 8. $\square$

Proof of Lemma 13

The main difficulty of adapting classical arguments arises from the presence of consumption externalities in the preferences. More precisely, one must pay attention to equilibrium consumption allocations that may converge on the boundary of the positive orthant. However, Assumptions 5.1 and 5.4 imply that the limit point of the household h’s equilibrium consumptions is strictly positive, even when the limit point of the equilibrium consumption of everybody else is on the boundary. Since the same argument applies for every household h, one concludes that the limit point of the equilibrium consumption allocations has to be strictly positive. A detailed proof of this lemma is given in del Mercato and Platino (2015b). $\square$

Proof of Lemma 14

We show that for each $( \xi ^{*},e^{*}, \eta ^{*}) \in F^{-1}(0)$, the Jacobian matrix $D_{\xi ,e,\eta } F( \xi ^{*},e^{*}, \eta ^{*})$ has full row rank. It is enough to prove that $\Delta D_{\xi ,e,\eta }F(\xi ^{*},e^{*}, \eta ^{*})=0$ implies $\Delta =0$, where

$$\begin{aligned}\Delta :=((\Delta x_{h},\Delta \lambda _{h})_{h \in {\mathcal {H}}},(\Delta y_{j},\Delta \alpha _{j})_{j \in {\mathcal {J}}},\Delta p^{\backslash }) \in {\mathbb {R}}^{H(C+1)}\times {\mathbb {R}}^{J(C+1)} \times {\mathbb {R}}^{C-1} \end{aligned}$$

The system $\Delta D_{\xi ,e,\eta }F( \xi ^{*},e^{*}, \eta ^{*}) =0$ is written in detail below.¹⁸

We remind that $(t_{j} \circ g_{j})$ is defined in (15).

$$\begin{aligned} \left\{ \begin{array}{l} (1)\; {\displaystyle \sum _{h \in {\mathcal {H}}}}\Delta x_{h} D_{x_{k}x_{h}}^{2}u_{h}(x_{h}^ {*},x_{-h}^{*},y^{*})-\Delta \lambda _{k}p^ {*} - {\displaystyle \sum _{j\in {\mathcal {J}}}}\alpha _{j}^{*}\Delta y_{j}D_{x_{k}y_{j}}^{2} (t_{j} \circ g_{j})(y_{j}^{*},y_{-j}^{*},x^{*},\eta _{j}^{*}) +\\ - \displaystyle \sum _{j \in {\mathcal {J}}} \Delta \alpha _{j} D_{x_{k}} (t_{j} \circ g_{j})(y_{j}^{*},y_{-j}^{*},x^{*},\eta _{j}^{*}) +\Delta p^{\backslash } \left[ I_{C-1}|0\right] =0, \; \forall \; k \in {\mathcal {H}}\\ (2)\; - \Delta x_{h} \cdot p^{*} =0, \; \forall \; h \in {\mathcal {H}} \\ (3)\; {\displaystyle \sum _{h \in {\mathcal {H}}}}\Delta x_{h} D_{y_{f}x_{h}}^{2}u_{h}(x_{h}^ {*},x_{-h}^{*},y^{*})+\displaystyle \sum _{h \in {\mathcal {H}}}\Delta \lambda _{h}s_{fh}p^ {*} - {\displaystyle \sum _{j\in {\mathcal {J}}}} \alpha _{j}^{*}\Delta y_{j}D_{y_{f}y_{j}}^{2} (t_{j} \circ g_{j})(y_{j}^{*},y_{-j}^{*},x^{*},\eta _{j}^{*}) +\\ - \displaystyle \sum _{j \in {\mathcal {J}}} \Delta \alpha _{j} D_{y_{f}} (t_{j} \circ g_{j})(y_{j}^{*},y_{-j}^{*},x^{*},\eta _{j}^{*}) - \Delta p^{\backslash } \left[ I_{C-1}|0\right] =0, \; \forall \; f \in {\mathcal {J}}\\ (4)\; - \Delta y_{j}\cdot D_{y_{j}} (t_{j} \circ g_{j})(y_{j}^{*},y_{-j}^{*},x^{*},\eta _{j}^{*}) =0, \; \forall \; j \in {\mathcal {J}} \\ (5)\; \Delta \lambda _{h}p^{*} - \Delta p^{\backslash } \left[ I_{C-1}|0\right] =0, \; \forall \; h \in {\mathcal {H}}\\ (6)\; -{\displaystyle \sum _{h \in {\mathcal {H}}}}\lambda _{h}^{*}\Delta x_{h}^{\backslash }- {\displaystyle \sum _{h \in {\mathcal {H}}}}\Delta \lambda _{h}\left( x_{h}^{*\backslash }-e_{h}^ {*\backslash }-\displaystyle \sum _{j\in {\mathcal {J}}}s_{jh}y_{j}^{*\backslash }\right) +\displaystyle \sum _{j\in {\mathcal {J}}}\Delta y_{j}^{ \backslash }=0\\ (7)\; - \alpha _{j}^{*} \Delta y_{j} D^{2}_{\eta _{j} y_{j}} (t_{j} \circ g_{j})(y_{j}^{*},y_{-j}^{*},x^{*},\eta _{j}^{*}) - \Delta \alpha _{j} D_{\eta _{j}}(t_{j} \circ g_{j})(y_{j}^{*},y_{-j}^{*},x^{*} , \eta _{j}^{*}) = 0,\; \forall \; j \in {\mathcal {J}} \\ \end{array} \right. \end{aligned}$$

Using the definition of $(t_{j} \circ g_{j})$, we have that

$$\begin{aligned} D_{\eta _{j}} (t_{j} \circ g_{j})(y_{j}^{*},y_{-j}^{*},x^{*} , \eta _{j}^{*}) = - D_{y_{j}} (t_{j} \circ g_{j})(y_{j}^{*},y_{-j}^{*},x^{*} , \eta _{j}^{*}) \end{aligned}$$

and

$$\begin{aligned} D^{2}_{\eta _{j}y_{j}} (t_{j} \circ g_{j})(y_{j}^{*},y_{-j}^{*},x^{*} , \eta _{j}^{*}) = - D^{2}_{y_{j}} (t_{j} \circ g_{j})(y_{j}^{*},y_{-j}^{*},x^{*} , \eta _{j}^{*}) \end{aligned}$$

Therefore, for every $j \in {\mathcal {J}}$, equation (7) becomes

$$\begin{aligned} \alpha _{j}^{*} \Delta y_{j} D^{2}_{y_{j}} (t_{j} \circ g_{j})(y_{j}^{*},y_{-j}^{*},x^{*} , \eta _{j}^{*}) + \Delta \alpha _{j} D_{y_{j}} (t_{j} \circ g_{j})(y_{j}^{*},y_{-j}^{*},x^{*} , \eta _{j}^{*}) = 0\nonumber \\ \end{aligned}$$

(16)

Multiplying the equation above by $\Delta y_{j}$ and using equation (4), since $\alpha _{j}^{*}>0$ one gets

$$\begin{aligned} \Delta y_{j} D^{2}_{y_{j}} (t_{j} \circ g_{j})(y_{j}^{*},y_{-j}^{*},x^{*} , \eta _{j}^{*}) (\Delta y_{j} ) =0 \end{aligned}$$

Then, equation (4) and Assumption 1.3 imply that $\Delta y_{j}=0$.

Therefore, Eq. (16) becomes

$$\begin{aligned} \Delta \alpha _{j} D_{y_{j}} (t_{j} \circ g_{j})(y_{j}^{*},y_{-j}^{*},x^{*} , \eta _{j}^{*}) = 0 \end{aligned}$$

and then $\Delta \alpha _{j}=0$ by Assumption 1.4. Thus, we get $(\Delta y_{j}, \Delta \alpha _{j}) = 0$ for every $j \in {\mathcal {J}}$.

Since $p^{*C}=1$, from equation (5) one gets $\Delta \lambda _{h}=0$ for all $h \in {\mathcal {H}}$, and then $\Delta p^{\backslash }=0$. Thus, the above system becomes

$$\begin{aligned} \left\{ \begin{array}{ll} (1)\;&{} {\displaystyle \sum _{h \in {\mathcal {H}}}}\Delta x_{h} D_{x_{k}x_{h}}^{2}u_{h}(x_{h}^{*},x_{-h}^{*},y^{*}) =0, \; \forall \; k \in {\mathcal {H}}\\ (2)\;&{}- \Delta x_{h} \cdot p^{*} =0, \; \forall \; h \in {\mathcal {H}} \\ (3)\;&{}\displaystyle \sum _{h \in {\mathcal {H}}}\Delta x_{h} D_{y_{f}x_{h}}^{2}u_{h}(x_{h}^ {*},x_{-h}^{*},y^{*})=0, \; \forall \; f \in {\mathcal {J}}\\ (6)\;&{}-{\displaystyle \sum _{h \in {\mathcal {H}}}}\lambda _{h}^{*}\Delta x_{h}^{\backslash }=0 \end{array} \right. \end{aligned}$$

(17)

Since $F^{h.1}(\xi ^{*},e^{*}, \eta ^{*})=0$, equation (2) in system (17) implies that for every $h \in {\mathcal {H}}$, $\Delta x_{h} \in \mathop {\mathrm{Ker}}D_{x_{h}}u_{h}( x_{h}^{*},x_{-h}^{*},y^{*})$. Now, for every $h \in {\mathcal {H}}$ define $v_{h}:=\lambda _{h}^{*}\Delta x_{h}$. The vector $(x_{h}^{*},v_{h})_{h \in {\mathcal {H}}}$ satisfies the following conditions.

$$\begin{aligned} (v_{h})_{h \in {\mathcal {H}}} \in {\displaystyle \prod _{h\in {\mathcal {H}}}} \mathop {\mathrm{Ker}}D_{x_{h}}u_{h}( x_{h}^{*},x_{-h}^{*},y^{*}) \quad \text {and}\quad {\displaystyle \sum _{h \in {\mathcal {H}}}} v_{h}=0 \end{aligned}$$

(18)

where the last equality comes from equation (6) in system (17). Multiplying both sides of equation (1) in system (17) by $v_{k}$, and using the definition of $v_{h}$, one gets ${\displaystyle { \sum \nolimits _{h \in {\mathcal {H}}}}\frac{v_{h}}{\lambda _{h}^{*}} } D_{x_{k}x_{h}}^{2}u_{h}(x_{h}^ {*},x_{-h}^{*},y^{*})(v_{k})=0$ for every $k \in {\mathcal {H}}$. Summing up $k \in {\mathcal {H}}$, we obtain ${\displaystyle { \sum \nolimits _{h \in {\mathcal {H}}}}\frac{v_{h}}{\lambda _{h}^{*}} }\displaystyle { \sum \nolimits _{k \in {\mathcal {H}}}} D_{x_{k}x_{h}}^{2}u_{h}(x_{h}^ {*},x_{-h}^{*},y^{*})(v_{k})=0$.

Note that the gradients $(D_{x_{h}}u_{h}( x_{h}^{*},x_{-h}^{*},y^{*}))_{h \in {\mathcal {H}}}$ are positively proportional because $F^{h.1}(\xi ^{*},e^{*}, \eta ^{*})=0$ for every $h \in {\mathcal {H}}$. Therefore, from (18) all the conditions of Assumption 10 are satisfied, and then, $v_{h}=0$ for each $h \in {\mathcal {H}}$ since $\lambda _{h}^{*}>0$. Thus, we get $\Delta x_{h}=0$ for all $h \in {\mathcal {H}}$. Consequently, one has $\Delta =0$ which completes the proof. $\square$