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A dynamic exchange rate model with heterogeneous agents

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Abstract

In this paper, we analyze a heterogeneous agent model in which the fundamental exchange rate is endogenously determined by the real markets. The exchange rate market and the real markets are linked through the balance of payments. We have analytically found that there exists at least a steady state in which the exchange rate is equal to its fundamental value and incomes of both countries are equal to the autonomous components times the multiplier (as in the Income-Expenditure model). This steady state can be unique and unstable when all agents act as contrarians, while when agents act as fundamentalists it is unique but its stability depends on the reactivity of actors of the market. Finally, we show that the (in)stability of the economic system depends on both the reactivity of the markets and that of different types of agents involved. Employing well-know functional forms, we show that the model can replicate some of the statistical features of the true time series of the exchange rate.

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Notes

  1. Our model is close to that developed by Laursen and Metzler in 1950. The Laursen and Metzler model (1950) is static, while its continuous time dynamic version can be found in Gandolfo (1986).

  2. As stated in De Arcangelis and Gandolfo (1997), we do not adhere to the traditional approach to the exchange rate, where the determination of the exchange rate is ’reflected in the balance of payments equation’, which summarizes the foreign exchange market. Moreover, our formulation implies a unitary elasticity.

  3. In line with empirical works, this specification implies that the exchange rate depends only on speculators and not on import/export exchanges.

  4. This is a short-medium-run model so that we believe that the assumption of fixed prices is not so stringent.

  5. Obviously the demand can absorb all the product. However, when there is an excess of demand, we assume that both countries have to ration all components of the demand other than export.

  6. In Westerhoff (2012) that function is linear, while for Naimzada and Pireddu (2014), it is sigmoidal.

  7. This is the only steady state having that property.

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Acknowledgements

We really thank an anonymous referee and the Editor, Uwe Cantner, for suggestions that improved the paper. We also wish to thank Marco Dardi and Frank Westerhoff who discussed with us an early version of the paper, the participants in the workshops held in Scuola Normale di Pisa in December 2013 and in Univerisità Milano-Bicocca in February 2014, and the participants in the conference Computing in Economics and Finance (CEF 2014) held in Oslo in June 2014.

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Appendix: Proof of Proposition 1

Appendix: Proof of Proposition 1

Defining y 1(t) = m p i 1 Y 1(t) and y 2(t) = m p i 2 Y 2(t), we can write system (9) as follows

$$\left\{\begin{array}{ll} y_{1}(t+1)=y_{1}(t)+ mpi_{1} g_{1}\left( A_{1}+\frac{mpc_{1}}{mpi_{1}}y_{1}(t)+\frac{y_{2}(t)}{E(t)}- y_{1}(t)-\frac{1}{mpi_{1}}y_{1}(t)\right)\\ y_{2}(t+1)= y_{2}(t)+mpi_{2} g_{2}\left( A_{2}+\frac{mpc_{2}}{mpi_{2}}y_{2}(t)+ y_{1}(t)E(t)-y_{2}(t)-\frac{1}{mpi_{2}}y_{2}(t)\right)\\ E(t+1)=E(t)+g_{3}\left( \gamma{\sum}_{j=1}^{n} \lambda_{j}\left( k\left( \frac{y_{2}(t)}{y_{1}(t)},E(t)\right)\right) k_{j}\left( \frac{y_{2}(t)}{y_{1}(t)},E(t)\right)\right) \end{array}\right. $$

Defining now

$$ \begin{array}{l} a_{1}=A_{1}>0, \quad a_{2}=A_{2}>0\\ b_{1}= -\frac{mpc_{1}}{mpi_{1}}+1+\frac{1}{mpi_{1}}>1,\quad b_{2}= -\frac{mpc_{2}}{mpi_{2}}+1+\frac{1}{mpi_{2}}>1\\ f_{1}=mpi_{1} g_{1},\quad f_{2}=mpi_{2} g_{2},\quad f_{3}=g_{3} \end{array} $$
(14)

we get the following system

$$ \left\{\begin{array}{ll} y_{1}(t+1)=y_{1}(t)+f_{1}\left( a_{1}-b_{1}y_{1}(t)+\frac{y_{2}(t)}{E(t)}\right)\\ y_{2}(t+1)=y_{2}(t)+ f_{2}\left( a_{2}-b_{2}y_{2}(t)+ y_{1}(t)E(t)\right)\\ E(t+1)=E(t)+f_{3}\left( \gamma{\sum}_{j=1}^{n} \lambda_{j}\left( k\left( \frac{y_{2}(t)}{y_{1}(t)},E(t)\right)\right) k_{j}\left( \frac{y_{2}(t)}{y_{1}(t)},E(t)\right)\right) \end{array}\right. $$
(15)

Denoting by \(\widehat {S}\) the set of steady states of Eq. 15, we have that the function \({\Phi }: \widehat {S}\to S\) mapping \((y_{1},y_{2},E)\in \widehat {S}\) into (y 1/m p i 1,y 2/m p i 2,E) ∈ S is well defined and is a bijection. Moreover, \((y_{1},y_{2},E)\in \widehat {S}\) is asymptotically stable if and only if Φ(y 1,y 2,E) ∈ S is asymptotically stable, and \((y_{1},y_{2},E)\in \widehat {S}\) is unstable if and only if Φ(y 1,y 2,E) ∈ S is unstable. As a consequence, we can prove Proposition 1 about system (9) working with system (15) and properly using the bijection Φ and equalities (14).

Note first that, for every \(s\in \mathbb {R}_{++}\),

$$r_{1}(s)=\frac{a_{1}b_{2}s+a_{2}}{(b_{1}b_{2}-1)s} \quad \text{ and }\quad r_{2}(s)=\frac{a_{2}b_{1}+a_{1}s}{b_{1}b_{2}-1}. $$

Since the system

$$ \left\{\begin{array}{ll} a_{1}-b_{1}y_{1}+\frac{y_{2}}{E}=0\\ a_{2}-b_{2}y_{2}+ y_{1}E=0\\ y_{1},y_{2},E>0 \end{array}\right. $$
(16)

has a set of solutions given by

$$\{(r_{1}(E),r_{2}(E),E)\in\mathbb{R}^{3}_{++}: E>0\}, $$

we have that

$$\widehat{S}=\{(r_{1}(E),r_{2}(E),E)\in\mathbb{R}^{3}_{++}: E\in {\Gamma}\}, $$

where Γ is defined in Eq. 10, and that immediately implies the first statement of Proposition 1. Note also that

$$E^{*}=\frac{a_{2}(b_{1}-1)}{a_{1}(b_{2}-1)}\in {\Gamma}. $$

Then

$$ \left( y^{*}_{1},y^{*}_{2},E^{*}\right)=\left( \frac{a_{1}}{b_{1}-1}, \frac{a_{2}}{b_{2}-1},\frac{a_{2}(b_{1}-1)}{a_{1}(b_{2}-1)}\right), $$
(17)

belongs to \(\widehat {S}\) and since \({\Phi }(y^{*}_{1},y^{*}_{2},E^{*})\) is equal to Eq. 11 and belongs to S, we get the second statement of Proposition 1.

Assume now that, for every \((z_{1},z_{2})\in \mathbb {R}_{++}^{2}\), we have that

$$\text{sign}(k_{1}((z_{1},z_{2})))=\ldots=\text{sign}(k_{n}((z_{1},z_{2}))), $$

and prove that \(\widehat {S}=\{(y^{*}_{1},y^{*}_{2},E^{*})\}\), so that \(S=\{\Phi (y^{*}_{1},y^{*}_{2},E^{*})\}\). Indeed,

$${\sum}_{j=1}^{n} \lambda_{j}\left( k\left( \frac{ r_{2}(E)}{r_{1}(E)},E\right)\right) k_{j}\left( \frac{ r_{2}(E)}{ r_{1}(E)},E\right)=0 $$

if and only if

$$\frac{ r_{2}(E)}{ r_{1}(E)}-E=0, $$

that is,

$$\frac{a_{2}(b_{1}-1)E-a_{1}(b_{2}-1)E^{2}}{a_{2}+a_{1}b_{2} E}=0 $$

and that equation has E has unique solution. That proves the third part of Proposition 1.

Let us move on to prove the last part of the proposition. We analyze then the stability of the state \((y_{1}^{*},y_{2}^{*},E^{*})\) defined in Eq. 17 in order to deduce information about the stability of \({\Phi }(y^{*}_{1},y^{*}_{2},E^{*})\), that is, Eq. 11. Consider the three following functions the domain of which is \(\mathbb {R}_{++}^{3}\),

$$F_{1}(y_{1},y_{2},E)=y_{1}+f_{1}\left( a_{1}-b_{1}y_{1}+\frac{y_{2}}{E}\right) $$
$$F_{2}(y_{1},y_{2},E)=y_{2}+ f_{2}\left( a_{2}-b_{2}y_{2}+ y_{1}E\right) $$
$$F_{3}(y_{1},y_{2},E)=E+ f_{3}\left( \gamma{\sum}_{j=1}^{n} \lambda_{j}\left( k\left( \frac{y_{2}}{y_{1}},E\right)\right) k_{j}\left( \frac{y_{2}}{y_{1}},E\right)\right) $$

Then we have that:

$$\begin{array}{lll} \frac{\partial F_{1}}{\partial y_{1}}\left( y^{*}_{1},y^{*}_{2},E^{*}\right)= & 1-b_{1}f^{\prime}_{1}(0)\\ \frac{\partial F_{1}}{\partial y_{2}}\left( y^{*}_{1},y^{*}_{2},E^{*}\right)= & \frac{a_{1}(b_{2}-1)}{a_{2}(b_{1}-1)} f^{\prime}_{1}(0)\\ \frac{\partial F_{1}}{\partial E}\left( y^{*}_{1},y^{*}_{2},E^{*}\right)= & - \frac{{a^{2}_{1}}(b_{2}-1)}{a_{2}(b_{1}-1)^{2}}f^{\prime}_{1}(0)\\ \frac{\partial F_{2}}{\partial y_{1}}\left( y^{*}_{1},y^{*}_{2},E^{*}\right)= & \frac{a_{2}(b_{1}-1)}{a_{1}(b_{2}-1)} f^{\prime}_{2}(0)\\ \frac{\partial F_{2}}{\partial y_{2}}\left( y^{*}_{1},y^{*}_{2},E^{*}\right)= &1-b_{2}f^{\prime}_{2}(0)\\ \frac{\partial F_{2}}{\partial E}\left( y^{*}_{1},y^{*}_{2},E^{*}\right)= & \frac{a_{1}}{(b_{1}-1)}f^{\prime}_{2}(0)\\ \frac{\partial F_{3}}{\partial y_{1}}\left( y^{*}_{1},y^{*}_{2},E^{*}\right)= & -f^{\prime}_{3}(0)\frac{a_{2}(b_{1}-1)^{2}}{{a_{1}^{2}}(b_{2}-1)}\gamma{\sum}_{j=1}^{n}\lambda_{j}(0) \frac{\partial k_{j}}{\partial z_{1}} \left( \frac{a_{2}(b_{1}-1)}{a_{1}(b_{2}-1)}, \frac{a_{2}(b_{1}-1)}{a_{1}(b_{2}-1)}\right)\\ \frac{\partial F_{3}}{\partial y_{2}}\left( y^{*}_{1},y^{*}_{2},E^{*}\right)= & f^{\prime}_{3}(0)\frac{(b_{1}-1)}{a_{1}}\gamma{\sum}_{j=1}^{n}\lambda_{j}(0) \frac{\partial k_{j}}{\partial z_{1}} \left( \frac{a_{2}(b_{1}-1)}{a_{1}(b_{2}-1)}, \frac{a_{2}(b_{1}-1)}{a_{1}(b_{2}-1)}\right)\\ \frac{\partial F_{3}}{\partial E}\left( y^{*}_{1},y^{*}_{2},E^{*}\right)= &1+ f^{\prime}_{3}(0) \gamma{\sum}_{j=1}^{n}\lambda_{j}(0) \frac{\partial k_{j}}{\partial z_{2}} \left( \frac{a_{2}(b_{1}-1)}{a_{1}(b_{2}-1)}, \frac{a_{2}(b_{1}-1)}{a_{1}(b_{2}-1)}\right) \end{array} $$

Note that

$$R=f^{\prime}_{3}(0)\gamma{\sum}_{j=1}^{n}\lambda_{j}(0) \frac{\partial k_{j}}{\partial z_{1}} \left( \frac{a_{2}(b_{1}-1)}{a_{1}(b_{2}-1)}, \frac{a_{2}(b_{1}-1)}{a_{1}(b_{2}-1)}\right), $$

where R is defined in Eq. 12, and, for every j ∈{1,…,n} and \(s\in \mathbb {R}\),

$$\frac{\partial k_{j}}{\partial z_{1}} (s,s)=-\frac{\partial k_{j}}{\partial z_{2}} (s,s). $$

Then we have

$$\frac{\partial F_{3}}{\partial y_{1}}\left( y^{*}_{1},y^{*}_{2},E^{*}\right)= - \frac{a_{2}(b_{1}-1)^{2}}{{a_{1}^{2}}(b_{2}-1)}R $$
$$\frac{\partial F_{3}}{\partial y_{2}}\left( y^{*}_{1},y^{*}_{2},E^{*}\right)= \frac{(b_{1}-1)}{a_{1}}R $$
$$\frac{\partial F_{3}}{\partial E}\left( y^{*}_{1},y^{*}_{2},E^{*}\right)=R $$

In order to study the stability of Eq. 17, we study the stability of the polynomial

$$\begin{array}{@{}rcl@{}} &&P(s)=\det \left( sI-D(F_{1},F_{2},F_{3})\left( y^{*}_{1},y^{*}_{2},E^{*}\right)\right)\\&&\qquad= \det\left[\begin{array}{ccc} s-1 +b_{1}f^{\prime}_{1}(0) & -\frac{a_{1}(b_{2}-1)}{a_{2}(b_{1}-1)} f^{\prime}_{1}(0) & \frac{{a^{2}_{1}}(b_{2}-1)}{a_{2}(b_{1}-1)^{2}}f^{\prime}_{1}(0)\\ -\frac{a_{2}(b_{1}-1)}{a_{1}(b_{2}-1)} f^{\prime}_{2}(0) & s-1+b_{2}f^{\prime}_{2}(0) & -\frac{a_{1}}{(b_{1}-1)}f^{\prime}_{2}(0)\\ \frac{a_{2}(b_{1}-1)^{2}}{{a_{1}^{2}}(b_{2}-1)}R & - \frac{(b_{1}-1)}{a_{1}}R & s-1+R \end{array}\right] \end{array} $$

that is, we study whether all of its roots lie inside the unit circle or there is at least one root lying outside the unit circle. A computation shows that

$$P(s)=(s-1)^{3}+\rho_{1}(s-1)^{2}+\rho_{2}(s-1)+\rho_{3} $$

where

$$\begin{array}{l} \rho_{1}=R+b_{2}f_{2}^{\prime}(0)+b_{1}f_{1}^{\prime}(0),\\ \rho_{2}=Rb_{2}f_{2}^{\prime}(0)+Rb_{1}f_{1}^{\prime}(0)+ b_{1}b_{2}f_{1}^{\prime}(0)f_{2}^{\prime}(0)-f_{1}^{\prime}(0)R-f_{1}^{\prime}(0)f_{2}^{\prime}(0)-f_{2}^{\prime}(0)R,\\ \rho_{3}=Rf_{1}^{\prime}(0)f_{2}^{\prime}(0) (b_{1}b_{2}-b_{1}-b_{2}+1). \end{array} $$

A simple condition which is sufficient for instability is P(1) < 0, that is, ρ 3 < 0. Since each factor in \(f_{1}^{\prime }(0)f^{\prime }_{2}(0)(b_{1}b_{2}-b_{1}-b_{2}+1)\) is positive, we have that P(1) < 0 if and only if R < 0. That proves Statement A1 of Proposition 1.

Another simple condition for instability is P(−1) > 0, that is, − 8 + 4ρ 1 − 2ρ 2 + ρ 3 > 0. Note that,

$$-8+4\rho_{1}-2\rho_{2}+\rho_{3}>0 $$

if and only if

$$\begin{array}{@{}rcl@{}} &&R\left[4\,-\,2b_{2}f^{\prime}_{2}(0)\,-\,2b_{1}f^{\prime}_{1}(0)\!+2f^{\prime}_{2}(0)+2f^{\prime}_{2}(0)+\! f_{1}^{\prime}(0)f_{2}^{\prime}(0) (b_{1}b_{2}\,-\,b_{1}\,-\,b_{2}+\!1)\right]\\&&\qquad>8-4b_{2}f^{\prime}_{2}(0)-4b_{1}f^{\prime}_{1}(0)+ 2f_{1}^{\prime}(0)f_{2}^{\prime}(0)b_{1}b_{2}-2f_{1}^{\prime}(0)f_{2}^{\prime}(0) \end{array} $$
(18)

Since

$$\begin{array}{@{}rcl@{}} &&4-2b_{2}f^{\prime}_{2}(0)-2b_{1}f^{\prime}_{1}(0)+2f^{\prime}_{2}(0)+2f^{\prime}_{2}(0)+ f_{1}^{\prime}(0)f_{2}^{\prime}(0) (b_{1}b_{2}-b_{1}-b_{2}+1)\\&&\qquad= \left( 2-f^{\prime}_{1}(0)(b_{1}-1)\right)\left( 2-f^{\prime}_{2}(0)(b_{2}-1)\right), \end{array} $$

and \(f^{\prime }_{1}(0),f^{\prime }_{2}(0)>0\), we have that Eq. 18 is implied by

$$R\!\left( 2\,-\,f^{\prime}_{1}(0)(b_{1}\,-\,1)\right)\left( 2\,-\,f^{\prime}_{2}(0)(b_{2}\,-\,1)\right)\!\!>\!8-4b_{2}f^{\prime}_{2}(0)-4b_{1}f^{\prime}_{1}(0)+ 2f_{1}^{\prime}(0)f_{2}^{\prime}(0)b_{1}b_{2} $$

that is

$$ R\left( 2-f^{\prime}_{1}(0)(b_{1}-1)\right)\left( 2-f^{\prime}_{2}(0)(b_{2}-1)\right)>2\left( 2-f^{\prime}_{1}(0)b_{1}\right)\left( 2-f^{\prime}_{2}(0)b_{2}\right). $$
(19)

It can be verified that Eq. 19 holds true if one of the following set of conditions is satisfied:

  • \(\frac {2}{b_{1}}<f_{1}^{\prime }(0)<\frac {2}{b_{1}-1}\), \(f_{2}^{\prime }(0)<\frac {2}{b_{2}}\), and R > 0;

  • \(f_{1}^{\prime }(0)<\frac {2}{b_{1}}\), \(\frac {2}{b_{2}}<f_{2}^{\prime }(0)<\frac {2}{b_{2}-1}\), and R > 0;

  • \(f_{1}^{\prime }(0)>\frac {4}{b_{1}-1}\), \(f_{2}^{\prime }(0)>\frac {4}{b_{2}-1}\), \(R>2\frac {(b_{1}+1)(b_{2}+1)}{(b_{1}-1)(b_{2}-1)}\).

Those conditions are exactly the ones in the Statements A2, A3 and A4 of Proposition 1.

In order to prove the last part of the proposition, assume m p c 1 = m p c 2, m p i 1 = m p i 2, \(g^{\prime }_{1}(0)=g^{\prime }_{2}(0)\), and R > 0. That implies b 1 = b 2 and \(f_{1}^{\prime }(0)=f_{2}^{\prime }(0)\) and

$$\begin{array}{@{}rcl@{}} P(s)&=&(s-1)\left[ (s-1)^{2}+2bf(s-1) +f^{2}(b^{2}-1) \right]\\ &&+R\left[(s-1)^{2}+2f(b-1)(s-1)+f^{2}(b-1)^{2} \right] \end{array} $$

where we set b = b 1 and \(f=f^{\prime }_{1}(0)\). Let P(s) = (s − 1)K(s) + R J(s), where

$$K(s)= (s-1)^{2}+2bf(s-1) +f^{2}(b^{2}-1), $$

and

$$J(s)=(s-1)^{2}+2f(b-1)(s-1)+f^{2}(b-1)^{2}. $$

The equation (s − 1)K(s) = 0 has three distinct solutions given by

$$s_{1}=1-f(b+1), \quad s_{2}=1-f(b-1),\quad 1, $$

with s 1 < s 2 < 1, and since J(s) = [(s − 1) + f(b − 1)]2 = (ss 2)2, we also have that P(s 2) = 0, P (s 2) < 0, P(s 1) > 0 and, for every s ∈ [1, + ), P(s) > 0. As a consequence P has three distinct real roots \(s^{*}_{1}<s^{*}_{2}<s^{*}_{3}\) such that

$$s^{*}_{1}<s_{1},\quad s^{*}_{2}=s_{2},\quad s^{*}_{3}\in(s_{2},1). $$

As a consequence, if \(f\ge \frac {2}{b+1}\), then \(s^{*}_{1}<s_{1}\le -1\) and Eq. 17 is unstable: that proves Statement B1 of Proposition 1.

Assuming instead \(f<\frac {2}{b+1}\), we get s 1 > −1 and then, \(s^{*}_{1}<-1\) if and only if P(−1) > 0, while \(s^{*}_{1}\in (-1,1)\) if and only if P(−1) < 0.

A computation shows that

$$P(-1)=-8+8fb-2f^{2}b^{2}+2f^{2}+R(2-f(b-1))^{2}, $$

and then P(−1) > 0 if and only if

$$R(2-f(b-1))^{2}>8-8fb+2f^{2}b^{2}-2f^{2}=2(2-f(b+1))(2-f(b-1)) $$

if and only if

$$\frac{R}{2}>\frac{2-f(b+1)}{2-f(b-1)}. $$

Analogously, we have P(−1) < 0 if and only if

$$\frac{R}{2}<\frac{2-f(b+1)}{2-f(b-1)}. $$

In the first case we get instability while in the second case we get asymptotic stability: by a substitution we obtain conditions of Statements B2 and B3 of Proposition 1.

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Gori, M., Ricchiuti, G. A dynamic exchange rate model with heterogeneous agents. J Evol Econ 28, 399–415 (2018). https://doi.org/10.1007/s00191-017-0513-9

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