A general equilibrium evolutionary model with two groups of agents, generating fashion cycle dynamics

Abstract

We propose a discrete-time exchange economy evolutionary model with two groups of agents. In our setting the definition of equilibrium depends also on agents’ population shares, which affect the market clearing conditions. We prove that, despite such difference with the classical Walrasian framework, for all economies and population shares there exists at least one equilibrium, and we show that for all population shares, generically in the set of the economies, equilibria are finite and regular. We then introduce the dynamic law governing the evolution of the population shares, and we investigate the existence and the stability of the resulting stationary equilibria. We assume that the reproduction level of a group is related to its attractiveness degree, which depends on the social visibility level, determined by the consumption choices of the agents in that group. The attractiveness of a group is described via a generic bell-shaped map, increasing for low visibility levels, but decreasing when the visibility of the group exceeds a given threshold value, due to a congestion effect. The model is able to reproduce the recurrent dynamic behavior typical of the fashion cycle, presenting booms and busts in the agents’ consumption choices, and in the groups’ attractiveness and population shares.

Proof of the analytical results

Proof of Proposition 1

Since the arguments we shall employ are independent of the considered time period, in order not to overburden notation, we will omit the subscript t, as well as the stars, which will refer to the Pareto optimal allocation¹⁶ only.

We define $\theta =\left( p_{x},p_{y},\left( x_{i},y_{i}\right) _{i\in \{\alpha ,\beta \}}\right) \in \varTheta =(0,+\infty )^{6}$ and, according to Definition 2, we say that $\theta \in \varTheta$ is a market equilibrium given $E\in {\mathcal {E}}$ and $a\in (0,1)$ if, for every $i\in \{\alpha ,\beta \}$, $\left( x_{i},y_{i}\right)$ solves problem (4) at $\left( p_x,p_y,E\right)$ and $\left( x_{i},y_{i}\right) _{i\in \{\alpha ,\beta \}}$ satisfies market clearing conditions (3) at $\left( E,a\right)$.

We denote by $\varTheta (E,a)$ the set of market equilibria for E and a, and we introduce

$$\begin{aligned} \varTheta _\mathrm {n}(E,a)=\left\{ \theta \in \varTheta (E,a):p_x=1\right\} , \end{aligned}$$

that is, the set of normalized¹⁷ market equilibria for E and a. We will show that $\varTheta _{\mathrm {n}}(E,a)\ne \emptyset$, for every $(E,a)\in {\mathcal {E}}\times (0,1)$.

As one Walras’ law holds true in our model, just one market clearing condition in Definition 2 is significant. To fix ideas, we will focus on the equation for commodity y, i.e., on

$$\begin{aligned} a\, y_{\alpha }+(1-a)\, y_{\beta }=w_{y}. \end{aligned}$$

The extended system for our economy with two groups of agents, whose numerosity may not coincide, reads as

$$\begin{aligned} \left\{ \begin{array}{l} \frac{\partial u_{\alpha }}{\partial x_{\alpha }}(x_{\alpha },y_{\alpha })-\lambda _{\alpha } p_x=0\\ \frac{\partial u_{\alpha }}{\partial y_{\alpha }}(x_{\alpha },y_{\alpha })-\lambda _{\alpha } p_y=0\\ -p_{x}\, x_{\alpha }-p_{y}\, y_{\alpha }+ p_{x}\, w_x+p_{y}\, w_y=0\\ \frac{\partial u_{\beta }}{\partial x_{\beta }}(x_{\beta },y_{\beta })-\lambda _{\beta } p_x=0\\ \frac{\partial u_{\beta }}{\partial y_{\beta }}(x_{\beta },y_{\beta })-\lambda _{\beta } p_y=0\\ -p_{x}\, x_{\beta }-p_{y}\, y_{\beta }+ p_{x}\, w_x+p_{y}\, w_y=0\\ a\, y_{\alpha }+(1-a)\, y_{\beta }-w_{y}=0\\ p_x-1=0 \end{array} \right. \end{aligned}$$

(19)

Since we are going to study market equilibria in terms of first-order conditions associated with households’ maximization problems and (significant) market clearing conditions, we define

$$\begin{aligned} \xi =\left( p_{x},p_{y},\left( x_{i},y_{i},\lambda _i\right) _{i\in \{\alpha ,\beta \}}\right) \in (0,+\infty )^{8}, \end{aligned}$$

and the function

$$\begin{aligned} {\mathcal {F}}:(0,+\infty )^{8}\times {\mathcal {E}}\times (0,1)\rightarrow {\mathbb {R}}^{8},\quad {\mathcal {F}}(\xi ,E,a)=\text{ lhs } \text{ of } (19). \end{aligned}$$

(20)

Given $(E,a)\in {\mathcal {E}}\times (0,1)$, it is immediate to prove that if $\theta =\bigl (p_{x},p_{y},\left( x_{i},y_{i}\right) _{i\in \{\alpha ,\beta \}}\bigr )$ belongs to $\varTheta _{\mathrm {n}}(E,a)$, then there exists the vector $(\lambda _{\alpha },\lambda _{\beta })\in (0,+\infty )^{2}$ such that $\xi =\left( p_{x},p_{y},\left( x_{i},y_{i},\lambda _i\right) _{i\in \{\alpha ,\beta \}}\right) \in (0,+\infty )^{8}$ solves the system ${\mathcal {F}}(\xi ,E,a)=0$. Vice versa, if $\xi =\left( p_{x},p_{y},\left( x_{i},y_{i},\lambda _i\right) _{i\in \{\alpha ,\beta \}}\right) \in (0,+\infty )^{8}$ solves ${\mathcal {F}}\left( \xi ,E,a\right) =0$, then $\left( p_{x},p_{y},\left( x_{i},y_{i}\right) _{i\in \{\alpha ,\beta \}}\right) \in \varTheta _{\mathrm {n}}(E)$.

Let us then show that for all $(E,a)\in {\mathcal {E}}\times (0,1)$ there exists $\xi \in (0,+\infty )^{8}$ which solves system ${\mathcal {F}}(\xi ,E,a)=0$. Fixing $(E,a)\in {\mathcal {E}}\times (0,1)$, we define

$$\begin{aligned} F:(0,+\infty )^{8}\rightarrow {\mathbb {R}}^{8},\quad F(\xi )={\mathcal {F}}(\xi ,E,a). \end{aligned}$$

Let us also introduce the homotopy

$$\begin{aligned} H:(0,+\infty )^{8}\times [0,1]\rightarrow {\mathbb {R}}^{8},\quad H(\xi ,\tau )=\text{ lhs } \text{ of } (21), \end{aligned}$$

with

$$\begin{aligned} \left\{ \begin{array}{l} \frac{\partial u_{\alpha }}{\partial x_{\alpha }}(x_{\alpha },y_{\alpha })-\lambda _{\alpha } p_x=0\\ \frac{\partial u_{\alpha }}{\partial y_{\alpha }}(x_{\alpha },y_{\alpha })-\lambda _{\alpha } p_y=0\\ -p_{x}\, x_{\alpha }-p_{y}\, y_{\alpha }+ p_{x}\,((1-\tau )w_x+\tau x_{\alpha }^{*})+p_{y}\,((1-\tau )\, w_y+\tau y_{\alpha }^{*})=0\\ \frac{\partial u_{\beta }}{\partial x_{\beta }}(x_{\beta },y_{\beta })-\lambda _{\beta } p_x=0\\ \frac{\partial u_{\beta }}{\partial y_{\beta }}(x_{\beta },y_{\beta })-\lambda _{\beta } p_y=0\\ -p_{x}\, x_{\beta }-p_{y}\, y_{\beta }+ p_{x}\,((1-\tau )w_x+\tau x_{\beta }^{*})+p_{y}\,((1-\tau )\, w_y+\tau y_{\beta }^{*})=0\\ a\, y_{\alpha }+(1-a)\, y_{\beta }-\left( (1-\tau )w_y+\tau (a y_{\alpha }^{*}+(1-a) y_{\beta }^{*})\right) =0\\ p_x-1=0, \end{array} \right. \end{aligned}$$

(21)

where $\left( x_{i}^{*},y_{i}^{*}\right) _{i\in \{\alpha ,\beta \}}\in (0,+\infty )^{4}$ is an instantaneous Pareto optimal allocation, whose existence can be proven as in Section 8.5 in Villanacci et al. (2002). In particular, denoting by r the vector of the total resources associated with E, we have $r=(w_x,w_y)\in (0,+\infty )^{2}$ and, denoting by $\underline{U}^r$ the set of utility level vectors attainable with resources r in correspondence to $E\in {\mathcal {E}}$ and $a\in (0,1)$, we have

$$\begin{aligned} \underline{U}^r=\left\{ (\underline{u}_{\alpha },\underline{u}_{\beta })\in {\mathbb {R}}^2: \exists (x_{\alpha },x_{\beta },y_{\alpha },y_{\beta })\in (0,+\infty )^{4} \text{ s.t. } \begin{array}{ll} a\, j_{\alpha }+(1-a)\, j_{\beta }=w_{j}, \text{ for } j\in \{x,y\}, \\ \!\! \text{ and } u_i(x_i,y_i)-\underline{u}_i=0, \text{ for } i\in \{\alpha ,\beta \} \end{array} \right\} . \end{aligned}$$

We notice that $H(\xi ,0)=F(\xi ),\,\forall \xi \in (0,+\infty )^{8}$. Setting

$$\begin{aligned} G:(0,+\infty )^{8}\rightarrow {\mathbb {R}}^{8},\quad G(\xi )=H(\xi ,1), \end{aligned}$$

it holds that $F,\,H$ and G are continuous functions. If we prove that, for some ${\widehat{\xi }}\in (0,+\infty )^{8}$,

$$\begin{aligned}&G^{-1}(0)=\{{\widehat{\xi }}\} \text{ and } G \text{ is } {\mathcal {C}}^1 \text{ in } \text{ an } \text{ open } \text{ neighborhood } \text{ of } {\widehat{\xi }}, \end{aligned}$$

(22)

$$\begin{aligned}&D_\xi G({\widehat{\xi }}) \text{ is } \text{ not } \text{ singular }, \end{aligned}$$

(23)

$$\begin{aligned}&H^{-1}(0) \text{ is } \text{ compact }, \end{aligned}$$

(24)

then Theorem 1 can be applied with the identifications $M=(0,+\infty )^{8},\,N={\mathbb {R}}^{8},\,y=0\in {\mathbb R^8},\,\varPhi =F,\,\varGamma =G,\,\varPsi =H$, to get $F^{-1}(0)\ne \emptyset$, so that a market equilibrium exists in correspondence to the fixed $(E,a)\in {\mathcal {E}}\times (0,1)$.

The proof of conditions (22), (23) and (24) follows by standard arguments and it is omitted. $\square$

Proof of Proposition 2

Since the arguments we shall employ are independent of the considered time period, like in the proof of Proposition 1, in order not to overburden notation, we will omit the subscript t.

Recalling the definition of the map ${\mathcal {F}}$ in (20), it is evident that ${\mathcal {F}}$ is continuous. In order to apply Theorem 2 with the identifications $f={\mathcal {F}},\,\mathbb R^n={\mathbb {R}}^8,\,O=(0,+\infty )^8,\,B={\mathcal {E}}\times (0,1),\,{\mathscr {B}}={\mathscr {E}}\times {\mathbb {R}}$, with ${\mathscr {E}}$ as in (1), we have to check that ${\mathcal {F}}\in \mathcal {C}^1((0,+\infty )^8\times {\mathcal {E}}\times (0,1),{\mathbb {R}}^8)$ according to the definition in (5), i.e., that

$$\begin{aligned} d{\mathcal {F}}:\left( (0,+\infty )^8\times \mathcal E\times (0,1)\right) \times \left( {\mathbb {R}}^8\times \mathscr {E}\times {\mathbb {R}}\right) \rightarrow {\mathbb {R}}^8 \end{aligned}$$

is well defined and continuous. Considering any $(\xi ,E,a)\in (0,+\infty )^8\times {\mathcal {E}}\times (0,1)$ and $(\nu ,\eta ,\kappa )\in {\mathbb {R}}^8\times {\mathscr {E}}\times {\mathbb {R}}$, it holds indeed that the limit

$$\begin{aligned} \hbox {d}{\mathcal {F}}(\xi ,E,a)=\lim _{\varepsilon \rightarrow 0}\frac{{\mathcal {F}}(\xi +\varepsilon \nu ,E+\varepsilon \eta ,a+\varepsilon \kappa )-{\mathcal {F}}(\xi ,E,a)}{\varepsilon } \end{aligned}$$

exists, and it is easy to check that the map $\hbox {d}{\mathcal {F}}$ is continuous.

Next, we want to show that for every $a\in (0,1)$ the set

$$\begin{aligned} {\mathcal {D}}(a)=\left\{ E\in {\mathcal {E}}: {\mathcal {F}}\left( \xi ,E,a\right) =0\Rightarrow \det D_{\xi }{\mathcal {F}}\left( \xi ,E,a\right) \ne 0 \right\} \end{aligned}$$

is an open and full measure subset of ${\mathcal {E}}$. Namely, applying Theorem 2, we obtain the smooth dependence of the market equilibria associated with all population shares $a\in (0,1)$ and to any economy $E\in {\mathcal {D}}(a)$ on the elements $(E,a)\in {\mathcal {E}}\times (0,1)$.

Let us then sketch the proof that, for every $a\in (0,1)$, ${\mathcal {D}}(a)$ is an open and full measure subset of ${\mathcal {E}}$.

The openness of ${\mathcal {D}}(a)$ follows by the closedness of the complement set ${\mathcal {E}}\setminus {\mathcal {D}}(a)$, due to the continuity of the involved functions. In order to show that ${\mathcal {D}}(a)$ is a full measure subset of ${\mathcal {E}}$, it suffices to show that, for every $(u_{\alpha },u_{\beta },a)\in \mathcal U^2\times (0,1)$ the set

$$\begin{aligned} {\mathcal {D}}(u_{\alpha },u_{\beta },a)=\left\{ (w_x,w_y)\in (0,+\infty )^2: {\mathcal {F}}\left( \xi ,u_{\alpha },u_{\beta },w_x,w_y,a\right) =0\Rightarrow \det D_{\xi }{\mathcal {F}}\left( \xi ,E,a\right) \ne 0 \right\} \end{aligned}$$

is a full measure subset of $(0,+\infty )^2$. Since it can be proven that 0 is a regular value for the map

$$\begin{aligned} \widetilde{{\mathcal {F}}}:(0,+\infty )^8\times (0,+\infty )^2\rightarrow {\mathbb {R}}^8,\quad \widetilde{{\mathcal {F}}}(\xi ,w_x,w_y)={\mathcal {F}}(\xi ,u_{\alpha },u_{\beta },w_x,w_y,a), \end{aligned}$$

then, by a transversality result (cf. Theorem 6.3.294 in Villanacci et al. 2002), there exists a full measure subset $\widetilde{{\mathcal {D}}}(u_{\alpha },u_{\beta },a)$ of $(0,+\infty )^2$ such that, for all $(w_x,w_y)\in \widetilde{{\mathcal {D}}}(u_{\alpha },u_{\beta },a)$, 0 is a regular value for the map $\widetilde{{\mathcal {F}}}(\cdot ,w_x,w_y)$. As $\widetilde{{\mathcal {D}}}(u_{\alpha },u_{\beta },a)\subseteq {\mathcal {D}}(u_{\alpha },u_{\beta },a)$, it follows that ${\mathcal {D}}(u_{\alpha },u_{\beta },a)$ is a full measure subset of $(0,+\infty )^2$. Consequently, for every $a\in (0,1)$, ${\mathcal {D}}(a)$ is a full measure subset of ${\mathcal {E}}$.

The finiteness of the number of market equilibria associated with an element (a, E), with $a\in (0,1)$ and $E\in {\mathcal {D}}(a)$, comes now from the fact that, as it is easy to check, the projection

$$\begin{aligned} \pi :{\mathcal {F}}^{-1}(0)\rightarrow {\mathcal {E}}\times (0,1),\quad \pi \left( \xi ,E,a\right) =(E,a), \end{aligned}$$

is proper, i.e., the inverse image through $\pi$ of a compact subset of ${\mathcal {E}}\times (0,1)$ is compact, too.

The proof is complete. $\square$

Proof of Proposition 3

By Theorem 1 we know that, for any $E\in {\mathcal {E}},\,t\in {\mathbb {N}}$ and $a_t\in (0,1)$, there exists at least a market equilibrium at time t. Let us then check that the equilibrium is unique, omitting as usual the dependence on t, in order not to overburden notation.

Recalling Definition 2 and the formulation of the market clearing conditions with shares in (3), we have to prove that if the gross substitute property holds for all the individual demand functions there exists one solution $({\widehat{p}}_x,\widehat{p}_y)\in (0,+\infty )^2$ to the system

$$\begin{aligned} a\, x^{*}_{\alpha }(p_x,p_y){+}(1{-}a)\, x^{*}_{\beta }(p_x,p_y)-w_{x}{=}0{=}a\, y^{*}_{\alpha }(p_x{,}p_y){+}(1{-}a)\, y^{*}_{\beta }(p_x{,}p_y){-}w_{y}.\nonumber \\ \end{aligned}$$

(25)

Since we can normalize one price, to show the uniqueness of the solution, we may focus on the price vectors $(1,p)\in (0,+\infty )^2$, with $p=p_y/p_x$. Assuming that $a\, x^{*}_{\alpha }(1,\widehat{p})+(1-a)\, x^{*}_{\beta }(1,{\widehat{p}})-w_{x}=0=a\, y^{*}_{\alpha }(1,{\widehat{p}})+(1-a)\, y^{*}_{\beta }(1,\widehat{p})-w_{y}$, let us prove that no $(1,{\widetilde{p}})$, with $\widetilde{p}\ne {\widehat{p}}$, may solve (25). Namely, if ${\widetilde{p}}< {\widehat{p}}$, by the gross substitute property it would hold that $x^{*}_i(1,{\widetilde{p}})<x^{*}_i(1,\widehat{p}),\,i\in \{\alpha ,\beta \}$, and thus, since $a\in (0,1)$, we would have

$$\begin{aligned} a\, x^{*}_{\alpha }(1,{\widetilde{p}})+(1-a)\, x^{*}_{\beta }(1,{\widetilde{p}})-w_{x}<a\, x^{*}_{\alpha }(1,{\widehat{p}})+(1-a)\, x^{*}_{\beta }(1,{\widehat{p}})-w_{x}=0, \end{aligned}$$

so that $(1,{\widetilde{p}})$ would not be a solution to (25). Similarly, if ${\widetilde{p}}>{\widehat{p}}$, by the gross substitute property it would hold that $x^{*}_i(1,\widetilde{p})>x^{*}_i(1,{\widehat{p}}),\,i\in \{\alpha ,\beta \}$, and thus, since $a\in (0,1)$, it would follow that

$$\begin{aligned} a\, x^{*}_{\alpha }(1,{\widetilde{p}})+(1-a)\, x^{*}_{\beta }(1,{\widetilde{p}})-w_{x}>a\, x^{*}_{\alpha }(1,{\widehat{p}})+(1-a)\, x^{*}_{\beta }(1,{\widehat{p}})-w_{x}=0, \end{aligned}$$

so that $(1,{\widetilde{p}})$ would not solve (25).

This completes the proof. $\square$

Proof of Proposition 4

The conclusion immediately follows by observing that the solutions to the fixed-point equation $g(a)=a$, with g as in (10), are given by $a=0,\,a=1$, as well as by all solutions to the equation $V_{\alpha }(a)=V_{\beta }(a)$, if any, and by all the solutions to the equation $(V_{\alpha }(a)+V_{\beta }(a))/2={\overline{V}}$, if any. Namely, the nontrivial equilibria for Equation (9) are found as solutions to the equation ${\mathscr {A}}_{\alpha }=f(\sigma \,d^2(\overline{V},V_{\alpha }))=f(\sigma \,d^2(\overline{V},V_{\beta }))={\mathscr {A}}_{\beta }$. Since f is strictly decreasing, all solutions have to satisfy $d(\overline{V},V_{\alpha })=d({\overline{V}},V_{\beta })$, i.e., $V_{\alpha }=V_{\beta }$ or $(V_{\alpha }+V_{\beta })/2={\overline{V}}$, as desired. This concludes the proof. $\square$

Proof of Proposition 5

Since $g'(0)=1/(\exp \left( \mu \left( {\mathscr {A}}_{\beta }(0)-{\mathscr {A}}_{\alpha }(0)\right) \right) )$, the condition $g'(0)>-1$ is always fulfilled, while $g'(0)<1$ is fulfilled for ${\mathscr {A}}_{\beta }(0)>{\mathscr {A}}_{\alpha }(0)$. Independently of f, exploiting the strictly decreasing behavior of ${\mathscr {A}}_{i}$ with respect to $d^2({\overline{V}},V_i)$, we then find $(V_{\alpha }(0)-{\overline{V}})^2>(V_{\beta }(0)-{\overline{V}})^2$, as desired.

As concerns the local stability of $a=1$ for map g, we find that $g'(1)= \exp \left( \mu \left( {\mathscr {A}}_{\beta }(1)-{\mathscr {A}}_{\alpha }(1)\right) \right)$. Hence, the condition $g'(1)>-1$ is always fulfilled, while $g'(1)<1$ is fulfilled for ${\mathscr {A}}_{\beta }(1)<{\mathscr {A}}_{\alpha }(1)$. Independently of f, exploiting again the strictly decreasing behavior of ${\mathscr {A}}_{i}$ with respect to $d^2({\overline{V}},V_i)$, we find $(V_{\alpha }(1)-{\overline{V}})^2<(V_{\beta }(1)-{\overline{V}})^2$.

In regard to $a={\hat{a}}$, recalling that $V_{\alpha }({\hat{a}})=V_{\beta }({\hat{a}})$, we find that $g'({\hat{a}})=1-2\mu \sigma {\hat{a}}(1-{\hat{a}})\frac{\partial {\mathscr {A}}_{\beta }(a)}{\partial (\sigma d_{\beta }^2)}\vert _{a={\hat{a}}}(V_{\beta }({\hat{a}})-\overline{V})(V_{\beta }'({\hat{a}})-V_{\alpha }'({\hat{a}}))$. Since $\frac{\partial {\mathscr {A}}_{\beta }(a)}{\partial (\sigma d_{\beta }^2)}\vert _{a={\hat{a}}}$ is negative, then $a={\hat{a}}\in (0,1)$ is unstable for map g for all positive values of $\mu$ if $(V_{\beta }({\hat{a}})-{\overline{V}})(V_{\beta }'({\hat{a}})-V_{\alpha }'({\hat{a}}))>0$. If instead $(V_{\beta }({\hat{a}})-{\overline{V}})(V_{\beta }'({\hat{a}})-V_{\alpha }'({\hat{a}}))<0$, then $g'({\hat{a}})<1$ is always fulfilled, while, if ${\mathscr {A}}_{\beta }$ does not depend on $\mu$, $g'({\hat{a}})>-1$ is fulfilled for $\mu <{\hat{\mu }}$, with ${\hat{\mu }}$ as in (11), obtaining the desired conclusion about the local stability of $a={\hat{a}}\in (0,1)$ for map g, too. The condition for the flip bifurcation follows by setting $g'({\hat{a}})=-1$.

Finally, in regard to $a={\tilde{a}}$, recalling that $V_{\beta }(\tilde{a})-{\overline{V}}={\overline{V}}-V_{\alpha }({\tilde{a}})$, we find that $g'({\tilde{a}})=1-2\mu \sigma {\tilde{a}}(1-\tilde{a})\frac{\partial {\mathscr {A}}_{\beta }(a)}{\partial (\sigma d_{\beta }^2)}\vert _{a={\tilde{a}}}({\overline{V}}-V_{\alpha }(\tilde{a}))(V_{\alpha }' ({\tilde{a}})+V_{\beta }'({\tilde{a}}))$. Since $\frac{\partial {\mathscr {A}}_{\beta }(a)}{\partial (\sigma d_{\beta }^2)}\vert _{a={\tilde{a}}}$ is negative, then $a=\tilde{a}\in (0,1)$ is unstable for map g for all positive values of $\mu$ if $({\overline{V}}-V_{\alpha }({\tilde{a}}))(V_{\alpha }'(\tilde{a})+V_{\beta }'({\tilde{a}}))>0$. If instead $(\overline{V}-V_{\alpha }({\tilde{a}}))(V_{\alpha }'({\tilde{a}})+V_{\beta }'(\tilde{a}))<0$, then $g'({\tilde{a}})<1$ is always fulfilled, while, if ${\mathscr {A}}_{\beta }$ does not depend on $\mu$, $g'({\tilde{a}})>-1$ is fulfilled for $\mu <{\tilde{\mu }}$, with ${\tilde{\mu }}$ as in (12). The condition for the flip bifurcation follows by setting $g'(\tilde{a})=-1$. This concludes the proof. $\square$