Sentiment-driven limit cycles and chaos

Abstract

A recent strand of macroeconomic literature has placed sentiment fluctuations at the forefront of the academic debate about the foundations of business cycles. Waves of optimism and pessimism influence the decisions of investors and consumers, and they might therefore be interpreted as a driving force for the performance of the economy in the short term. In this context, two questions regarding the formation and evolution of psychological moods in an economic setting gain relevance: First, how can we model the process of transmission of sentiments across economic agents? Second, is this process capable of generating endogenous and persistent fluctuations? This paper answers these two questions by proposing a simple and intuitive continuous-time dynamic sentiment spreading model based on the rumor propagation literature. As agents contact with one another, endogenous fluctuations are likely to emerge, with trajectories of sentiment shares potentially exhibiting periodic cycles and chaotic behavior.

Appendix

Proof of Proposition 1

Applying the equilibrium condition$\overset {\text {{\large .}}}{y}=\overset {\text {{\large .}}}{v}=\overset {\text {{\large .}}}{z}=\overset {\text {{\large .}}}{w}=0$to system (2) gives

$$\left\{\begin{array}{l} \left[ 1-(y^{\ast }+z^{\ast }+v^{\ast }+w^{\ast })\right] y^{\ast }=y^{\ast }(y^{\ast }+z^{\ast }) \\ \left[ 1-(y^{\ast }+z^{\ast }+v^{\ast }+w^{\ast })\right] v^{\ast }=v^{\ast }(v^{\ast }+w^{\ast }) \\ y^{\ast }(y^{\ast}+z^{\ast })=z^{\ast }\left[ 1-(y^{\ast }+z^{\ast }+v^{\ast }+w^{\ast })\right]\\ v^{\ast }(v^{\ast }+w^{\ast})=w^{\ast }\left[ 1-(y^{\ast }+z^{\ast }+v^{\ast }+w^{\ast })\right] \end{array}\right. $$

Thesolutions of the above system are the four equilibrium points and the equilibrium line in the proposition. □

Proof of Proposition 2

The Jacobian matrix for system (2) has the general form

$$\begin{array}{@{}rcl@{}} J &=&\zeta \left[ \begin{array}{cc} 1-4y^{\ast }-2z^{\ast }-v^{\ast }-w^{\ast } & -y^{\ast } \\ -v^{\ast } & 1-y^{\ast }-z^{\ast }-4v^{\ast }-2w^{\ast } \\ 2y^{\ast }+2z^{\ast } & z^{\ast } \\ w^{\ast } & 2v^{\ast }+2w^{\ast } \end{array}\right. \\ &&\left. \begin{array}{cc} -2y^{\ast } & -y^{\ast } \\ -v^{\ast } & -2v^{\ast } \\ -(1-2y^{\ast }-2z^{\ast }-v^{\ast }-w^{\ast }) & z^{\ast } \\ w^{\ast } & -(1-y^{\ast }-z^{\ast }-2v^{\ast }-2w^{\ast}) \end{array}\right] \end{array} $$

For each of the equilibria,

$$e_{1}:J=\frac{\zeta }{6}\left[ \begin{array}{cccc} -2 & -1 & -2 & -1 \\ -1 & -2 & -1 & -2 \\ 4 & 1 & 0 & 1 \\ 1 & 4 & 1 & 0 \end{array}\right] $$

$$e_{2}:J=\frac{\zeta }{4}\left[ \begin{array}{cccc} -2 & -1 & -2 & -1 \\ 0 & 2 & 0 & 0 \\ 4 & 1 & 0 & 1 \\ 0 & 0 & 0 & -2 \end{array}\right] $$

$$e_{3}:J=\frac{\zeta }{4}\left[ \begin{array}{cccc} 2 & 0 & 0 & 0 \\ -1 & -2 & -1 & -2 \\ 0 & 0 & -2 & 0 \\ 1 & 4 & 1 & 0 \end{array}\right] $$

$$e_{4}:J=\zeta \left[ \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{array}\right] $$

$$e_{5}:J=\zeta \left[ \begin{array}{cccc} -z^{\ast } & 0 & 0 & 0 \\ 0 & -(1-z^{\ast }) & 0 & 0 \\ 2z^{\ast } & z^{\ast } & z^{\ast } & z^{\ast } \\ 1-z^{\ast } & 2(1-z^{\ast }) & 1-z^{\ast } & 1-z^{\ast } \end{array}\right] $$

Fromthe Jacobian matrices, the respective eigenvalues are

$$\begin{array}{@{}rcl@{}} \begin{array}{c} e_{1}:\lambda_{1,2}=\left( -1\pm i\sqrt{3}\right) \frac{\zeta }{6};\lambda_{3,4}=\left( -1\pm i\sqrt{11}\right) \frac{\zeta }{6}\\ e_{2},e_{3}:\lambda_{1,2}=\left( -1\pm i\sqrt{7}\right) \frac{\zeta }{4} ;\lambda_{3}=-\frac{\zeta }{2};\lambda_{4}=\frac{\zeta }{2}\\ e_{4}:\lambda_{1,2}=-\zeta ;\lambda_{3,4}=\zeta\\ e_{5}:\lambda_{1}=-z^{\ast }\zeta ;\lambda_{2}=-(1-z^{\ast })\zeta ;\lambda_{3}=0;\lambda_{4}=\zeta \end{array} \end{array} $$

The signs of the eigenvalues show that the equilibrium points e ₂, e ₃, e ₄and e ₅are all unstable since at least one of the eigenvalues has a non-negative sign. Point e ₁is locallystable because the respective eigenvalues are two pairs of complex conjugates with negative realparts. □

Proof of Proposition 3

Equilibrium condition$\overset {\text {{\large .}}}{y}=\overset {\text {{\large .}}}{v}=\overset {\text {{\large .}}}{z} =\overset {\text {{\large .}}}{w}=0$is now applied to system (5) with the result,

$$\left\{\begin{array}{l} x^{\ast }y^{\ast }=\frac{\zeta }{2}\left\{ 1-\tanh \left[ \zeta x^{\ast }(y^{\ast }-z^{\ast })\right] \right\} y^{\ast }(y^{\ast }+z^{\ast }) \\ x^{\ast }v^{\ast }=\frac{\zeta }{2}\left\{ 1-\tanh \left[ \zeta x^{\ast}(v^{\ast }-w^{\ast })\right] \right\} v^{\ast }(v^{\ast }+w^{\ast }) \\ \frac{\zeta }{2}\left\{ 1-\tanh \left[ \zeta x^{\ast }(y^{\ast }-z^{\ast}) \right] \right\} y^{\ast }(y^{\ast }+z^{\ast })=z^{\ast }x^{\ast } \\ \frac{\zeta }{2}\left\{ 1-\tanh \left[ \zeta x^{\ast }(v^{\ast }-w^{\ast }) \right] \right\} v^{\ast }(v^{\ast }+w^{\ast })=w^{\ast }x^{\ast } \end{array}\right. $$

Given that tanh(0) = 0,the solution of the above system gives the set of equilibrium points claimed in the proposition. □

Proof of Proposition 4

The Jacobian matrix of system (5) is a modified version of the one presented in the proofof Proposition 2 given by

$$\begin{array}{@{}rcl@{}} \widehat{J} &=&\zeta \left[ \begin{array}{c} 1-3y^{\ast }-\frac{3}{2}z^{\ast }-v^{\ast }-w^{\ast }+\frac{\chi (y,y)}{2} y^{\ast }(y^{\ast }+z^{\ast }) \\ -v^{\ast }+\frac{\chi (v,y)}{2}v^{\ast }(v^{\ast }+w^{\ast })\\ y^{\ast }+\frac{3}{2}z^{\ast }-\frac{\chi (y,y)}{2}y^{\ast }(y^{\ast }+z^{\ast }) \\ w^{\ast }-\frac{\chi (v,y)}{2}v^{\ast}(v^{\ast }+w^{\ast }) \end{array}\right. \\ && \\ && \begin{array}{c} -y^{\ast }+\frac{\chi (y,v)}{2}y^{\ast }(y^{\ast }+z^{\ast }) \\ 1-y^{\ast }-z^{\ast }-3v^{\ast }-\frac{3}{2}w^{\ast }+\frac{\chi (v,v)}{2} v^{\ast }(v^{\ast }+w^{\ast }) \\ z^{\ast}-\frac{\chi (y,v)}{2}y^{\ast }(y^{\ast }+z^{\ast }) \\ v^{\ast }+\frac{3}{2}w^{\ast }-\frac{\chi (v,v)}{2}v^{\ast }(v^{\ast }+w^{\ast }) \end{array}\\ && \\ && \begin{array}{c} -\frac{3}{2}y^{\ast }+\frac{\chi (y,z)}{2}y^{\ast }(y^{\ast }+z^{\ast }) \\ -v^{\ast }+\frac{\chi (v,z)}{2}v^{\ast }(v^{\ast }+w^{\ast }) \\ -\left( 1-\frac{3}{2}y^{\ast }-2z^{\ast }-v^{\ast }-w^{\ast }\right) -\frac{ \chi (y,z)}{2}y^{\ast }(y^{\ast }+z^{\ast }) \\ w^{\ast }-\frac{\chi (v,z)}{2}v^{\ast }(v^{\ast }+w^{\ast }) \end{array}\\ && \\ &&\left. \begin{array}{c} -y^{\ast }+\frac{\chi (y,w)}{2}y^{\ast }(y^{\ast }+z^{\ast }) \\ -\frac{3}{2}v^{\ast }+\frac{\chi (v,w)}{2}v^{\ast }(v^{\ast }+w^{\ast }) \\ z^{\ast }-\frac{\chi (y,w)}{2}y^{\ast }(y^{\ast }+z^{\ast }) \\ -\left( 1-y^{\ast }-z^{\ast }-\frac{3}{2}v^{\ast }-2w^{\ast }\right) -\frac{ \chi (v,w)}{2}v^{\ast }(v^{\ast }+w^{\ast }) \end{array} \right] \end{array} $$

with

$$\begin{array}{@{}rcl@{}} \chi (y,y) &=&\left. \frac{\partial \tanh \left[ \zeta x(y-z)\right] }{ \partial y}\right\vert_{(y^{\ast },z^{\ast },v^{\ast },w^{\ast })}=\left. \frac{\frac{\partial \left[ \zeta x(y-z)\right] }{\partial y}}{\cosh^{2} \left[ \zeta x(y-z)\right] }\right\vert_{(y^{\ast },z^{\ast },v^{\ast },w^{\ast })} \\ &=&\zeta (1-2y^{\ast }-v^{\ast }-w^{\ast }) \end{array} $$

$$\begin{array}{@{}rcl@{}} \chi (y,v) &=&\left. \frac{\partial \tanh \left[ \zeta x(y-z)\right] }{ \partial v}\right\vert_{(y^{\ast },z^{\ast },v^{\ast },w^{\ast })}=\left. \frac{\frac{\partial \left[ \zeta x(y-z)\right] }{\partial v}}{\cosh^{2} \left[ \zeta x(y-z)\right] }\right\vert_{(y^{\ast },z^{\ast },v^{\ast },w^{\ast })} \\ &=&-\zeta \left( y^{\ast }-z^{\ast }\right)\\ \chi (y,z) &=&\left. \frac{\partial \tanh \left[ \zeta x(y-z)\right] }{ \partial z}\right\vert_{(y^{\ast },z^{\ast },v^{\ast },w^{\ast })}=\left. \frac{\frac{\partial \left[ \zeta x(y-z)\right] }{\partial z}}{\cosh^{2} \left[ \zeta x(y-z)\right] }\right\vert_{(y^{\ast },z^{\ast },v^{\ast },w^{\ast })} \\ &=&-\zeta \left( 1-2z^{\ast }-v^{\ast }-w^{\ast }\right)\\ \chi (y,w) &=&\left. \frac{\partial \tanh \left[ \zeta x(y-z)\right] }{ \partial w}\right\vert_{(y^{\ast },z^{\ast },v^{\ast },w^{\ast })}=\left. \frac{\frac{\partial \left[ \zeta x(y-z)\right] }{\partial w}}{\cosh^{2} \left[ \zeta x(y-z)\right] }\right\vert_{(y^{\ast },z^{\ast },v^{\ast },w^{\ast })} \\ &=&-\zeta \left( y^{\ast }-z^{\ast }\right)\\ \chi (v,y) &=&\left. \frac{\partial \tanh \left[ \zeta x(v-w)\right] }{ \partial y}\right\vert_{(y^{\ast },z^{\ast },v^{\ast },w^{\ast })}=\left. \frac{\frac{\partial \left[ \zeta x(v-w)\right] }{\partial y}}{\cosh^{2} \left[ \zeta x(v-w)\right] }\right\vert_{(y^{\ast },z^{\ast },v^{\ast },w^{\ast })} \\ &=&-\zeta \left( v^{\ast }-w^{\ast }\right)\\ \chi (v,v) &=&\left. \frac{\partial \tanh \left[ \zeta x(v-w)\right] }{ \partial v}\right\vert_{(y^{\ast },z^{\ast },v^{\ast },w^{\ast })}=\left. \frac{\frac{\partial \left[ \zeta x(v-w)\right] }{\partial v}}{\cosh^{2} \left[ \zeta x(v-w)\right] }\right\vert_{(y^{\ast },z^{\ast },v^{\ast },w^{\ast })} \\ &=&\zeta \left( 1-y^{\ast }-z^{\ast }-2v^{\ast }\right)\\ \chi (v,z) &=&\left. \frac{\partial \tanh \left[ \zeta x(v-w)\right] }{ \partial z}\right\vert_{(y^{\ast },z^{\ast },v^{\ast },w^{\ast })}=\left. \frac{\frac{\partial \left[ \zeta x(v-w)\right] }{\partial z}}{\cosh^{2} \left[ \zeta x(v-w)\right] }\right\vert_{(y^{\ast },z^{\ast },v^{\ast },w^{\ast })} \\ &=&-\zeta \left( v^{\ast }-w^{\ast }\right)\\ \chi (v,w) &=&\left. \frac{\partial \tanh \left[ \zeta x(v-w)\right] }{ \partial w}\right\vert_{(y^{\ast },z^{\ast },v^{\ast },w^{\ast })}=\left. \frac{\frac{\partial \left[ \zeta x(v-w)\right] }{\partial w}}{\cosh^{2} \left[ \zeta x(v-w)\right] }\right\vert_{(y^{\ast },z^{\ast },v^{\ast },w^{\ast })} \\ &=&-\zeta \left( 1-y^{\ast }-z^{\ast }-2w^{\ast }\right) \end{array} $$

For $e_{1}^{\prime }$, observethat $\chi (y,y)=\chi (v,v)=\frac {\zeta }{5}$, χ(y, v) = χ(y, w) = χ(v, y) = χ(v, z) = 0, and$\chi (y,z)=\chi (v,w)=-\frac {\zeta }{5}$. Giventhe values of the derivatives of the hyperbolic tangent functions evaluated in the equilibrium, it isstraightforward to compute the respective Jacobian matrix,

$$e_{1}^{\prime }:\widehat{J}=\frac{\zeta }{10}\left[ \begin{array}{cccc} -3+\frac{2}{25}\zeta & -2 & -3-\frac{2}{25}\zeta & -2 \\ -2 & -3+\frac{2}{25}\zeta & -2 & -3-\frac{2}{25}\zeta \\ 5-\frac{2}{25}\zeta & 2 & 1+\frac{2}{25}\zeta & 2 \\ 2 & 5-\frac{2}{25}\zeta & 2 & 1+\frac{2}{25}\zeta \end{array}\right] $$

The eigenvalues of $\widehat {J}$for $e_{1}^{\prime }$are

$$\begin{array}{@{}rcl@{}} \lambda_{1,2} &=&\left[ -\left( 1-\frac{2}{25}\zeta \right) \pm \sqrt{ \left( 1-\frac{2}{25}\zeta \right)^{2}-4}\right] \frac{\zeta }{10}; \\ \lambda_{3,4} &=&\left[ -\left( 1-\frac{2}{25}\zeta \right) \pm \sqrt{ \left( 1-\frac{2}{25}\zeta \right)^{2}-20}\right] \frac{\zeta }{10} \end{array} $$

Relative to the above eigenvalues, four different cases are identifiable (excludingthe border cases that imply the existence of bifurcation points): (i) if$\zeta >\frac {25}{2}\left (1+\sqrt {20}\right )$then the four eigenvalues havepositive real values; (ii) if $\frac {75}{2}<\zeta <\frac {25 }{2}\left (1+\sqrt {20}\right )$then two of the eigenvalues are positive real roots, while the other two area pair of complex conjugate eigenvalues with a positive real part; (iii) if$\frac {25}{2}<\zeta <\frac {75}{2}$, thenthe eigenvalues are two pairs of complex conjugates with positive real parts; (iv) if$\zeta <\frac {25}{2}$thenthe eigenvalues are two pairs of complex conjugates with negative real parts. Only in the lastcase will stability hold, and thus the condition for stability is the one claimed in theproposition.

Next, we analyze the stability of the other equilibrium points and confirm that they are alsounstable. Observe that:

• For$e_{2}^{\prime }$and$e_{3}^{\prime }:\chi (y,y)=\chi (v,v)= \frac {\zeta }{3}$, χ(y, v) = χ(y, w) = χ(v, y) = χ(v, z) = 0,$\chi (y,z)=\chi (v,w)=-\frac {\zeta }{3};$

• For e ₄ : χ(y, y) = χ(v, v) = ζ, χ(y, v) = χ(y, w) = χ(v, y) = χ(v, z) = 0, χ(y, z) = χ(v, w) = −ζ;

• For e ₅ : χ(y, y) = χ(y, z) = z ^∗ ζ, χ(y, v) = χ(y, w) = χ(v, y) = χ(v, z) = 0, χ(v, v) = χ(v, w) = (1 − z ^∗)ζ.

With these derivatives, matrix $\widehat {J}$ in each case will be

$$e_{2}^{\prime }:\widehat{J}=\frac{\zeta }{6}\left[ \begin{array}{cccc} -3+\frac{2}{9}\zeta & -2 & -3-\frac{2}{9}\zeta & -2 \\ 0 & 2 & 0 & 0 \\ 5-\frac{2}{9}\zeta & 2 & 1+\frac{2}{9}\zeta & 2 \\ 0 & 0 & 0 & -2 \end{array}\right] $$

$$e_{3}^{\prime }:\widehat{J}=\frac{\zeta }{6}\left[ \begin{array}{cccc} 2 & 0 & 0 & 0 \\ -2 & -3+\frac{2}{9}\zeta & -2 & -3-\frac{2}{9}\zeta \\ 0 & 0 & -2 & 0 \\ 2 & 5-\frac{2}{9}\zeta & 2 & 1+\frac{2}{9}\zeta \end{array}\right] $$

$$e_{4}:\widehat{J}=\zeta \left[ \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{array}\right] $$

$$e_{5}:\widehat{J}=\zeta \left[ \begin{array}{cccc} -\frac{1}{2}z^{\ast } & 0 & 0 & 0 \\ 0 & -\frac{1}{2}(1-z^{\ast }) & 0 & 0 \\ \frac{3}{2}z^{\ast } & z^{\ast } & z^{\ast } & z^{\ast } \\ 1-z^{\ast } & \frac{3}{2}(1-z^{\ast }) & 1-z^{\ast } & 1-z^{\ast } \end{array}\right] $$

As in the original model, the eigenvalues of the pair of equilibrium points$e_{2}^{\prime }$and$e_{3}^{\prime }$areidentical and are given in this case by

$$e_{2}^{\prime },e_{3}^{\prime }:\lambda_{1,2}=\left[ -\left( 1-\frac{2}{9} \zeta \right) \pm \sqrt{\left( 1-\frac{2}{9}\zeta \right)^{2}-12}\right] \frac{\zeta }{6};\lambda_{3}=-\frac{\zeta }{3};\lambda_{4}=\frac{\zeta }{3} $$

At least one ofthe eigenvalues of $\widehat {J}$for $e_{2}^{\prime }$and$e_{3}^{\prime }$has a positive sign,regardless of the value of ζ,and therefore the instability of the respective equilibria is confirmed. For e ₄, the Jacobianmatrices J and $\widehat {J}$are identical. Thus the corresponding eigenvalues are also the same, namely λ _1,2 = −ζ; λ _3,4 = ζ,and local stability is absent in this case as well. Finally, note that the eigenvalues for e ₅are$\lambda _{1}=-\frac {1}{2} z^{\ast }\zeta ;\lambda _{2}=-\frac {1}{2}(1-z^{\ast })\zeta ;\lambda _{3}=0;\lambda _{4}=\zeta$; forthe equilibrium line, one of the eigenvalues is positive and one other is zero, which again impliesinstability. □