Can consumption habit spillovers be a source of equilibrium indeterminacy?

Abstract

This paper investigates whether the external consumption habit can be a source of indeterminacy in a one-sector growth model when the labor supply is elastic. When there is a proper habit effect with a positive intertemporal elasticity of substitution, we find that the model exhibits indeterminacy if the coefficient of the habit formation is sufficiently large that allows for a substantial impact of current consumption on the habit. Indeterminacy arises even though the elasticity of the Frisch labor supply is positive and the elasticity of the labor demand in negative. In a calibrated version, we find that indeterminacy is empirically plausible when the habit effect is negative that features the “catching up with the Joneses” effect. Under given “catching up with the Joneses” effects, the external consumption habit can be a source of indeterminacy even if more than a half of the external consumption habit comes from past average consumption.

Appendices

Appendix 1: The elasticity of the Frisch labor supply

If we differentiate (5a) and keep $\lambda$ fixed, we obtain

$$\begin{aligned} dc=\displaystyle {{u_{13} dl-u_{12} dH} \over {u_{11} }}. \end{aligned}$$

(19)

Under given $\lambda$, differentiating (5b) gives

$$\begin{aligned} \displaystyle {{dw} \over w}=\displaystyle {{u_{31} dc+u_{32} dH-u_{31} dl} \over {u_3 }}. \end{aligned}$$

(20)

Substituting (19) into (20) yields

$$\begin{aligned} \zeta (k,l,H)&\equiv&\displaystyle {{\textstyle {{dw} \over w}} \over {\textstyle {{dl} \over l}}}=\displaystyle {l \over {u_3 }}\left[ {\displaystyle {{u_{13} (u_{31} +u_{32} \textstyle {{dH} \over {dc}})} \over {u_{11} +u_{12} \textstyle {{dH} \over {dc}}}}-u_{33} } \right]\\&= \left[ {\textstyle {{\textstyle {{lu_{13} } \over {u_1 }}(\textstyle {{u_{31} } \over {u_3 }}c+\textstyle {{u_{32} } \over {u_3 }}H\textstyle {c \over H}\textstyle {{dH} \over {dc}})} \over {\textstyle {{u_{11} } \over {u_1 }}c+\textstyle {{u_{12} } \over {u_1 }}H\textstyle {c \over H}\textstyle {{dH} \over {dc}}}}-\displaystyle {{lu_{33} } \over {u_3 }}} \right], \end{aligned}$$

and thus the elasticity of the Frisch labor supply

$$\begin{aligned} \zeta (k,l,H)=\left\{ {\displaystyle {{\varepsilon [\phi -(\varsigma +\chi )\textstyle {C \over H}\textstyle {{dH} \over {dC}}]} \over {-\sigma -\varsigma \textstyle {C \over H}\textstyle {{dH} \over {dC}}}}-\eta } \right\} . \end{aligned}$$

(21)

In order to obtain dH/dC, we rewrite $H_{t}$ as

$$\begin{aligned} H_t =\int _{-\infty }^t {D(t,\tau )C_\tau d\tau ,} \end{aligned}$$

(22)

where $D(t,\tau )=\beta e^{-\beta (\tau -t)}$ which satisfies

$$\begin{aligned} \int _{-\infty }^t {D(t,\tau )d\tau =1.} \end{aligned}$$

(23)

Hence, $D(t,\tau )$ measures the influence of $C_{\tau }$ on $H_{t}$ at time $\tau \in (-\infty ; t]$ and is thus a weight function defined on $\tau \in (-\infty ; t]$ with the total weight equal 1 for all $\beta >0$. $D(t,\tau )$ is a decreasing function of $\beta$ for $\beta \ge \textstyle {1 \over {t-\tau }}$ that approaches 0 as $\beta \rightarrow \infty$. Thus, as $\beta \rightarrow \infty$, $D(t, \tau )$ approaches to the function $\Upsilon _{t }(\tau )$ that satisfies

$$\begin{aligned} \Upsilon _t (\tau )=\left\{ {{\begin{array}{*{20}c} {0,\quad \tau <t,} \\ {\infty ,\quad \tau =t,} \\ \end{array} }} \right. \end{aligned}$$

and $\int _{-\infty }^t {\Upsilon _t (\tau )d\tau =1.}$ This indicates that for $\beta \rightarrow \infty$, $H_{t}=C_{t}$.

To explore further how the change in $C_{t}$ affects $H_{t}$, let us consider the case wherein $C_{\tau }$ is changed to $C_{\tau }+dC_{\tau }$ and $H_{t}$ is changed to $H_{t}+dH_{t}$. Using (22), we obtain

$$\begin{aligned} dH_t =\int _{-\infty }^t {D(t,\tau )dC_\tau } d\tau . \end{aligned}$$

(24)

We assume that for some $t_{1}<t$,

$$\begin{aligned} dC_\tau =\left\{ {{\begin{array}{l@{\quad }l} 0,&\tau <t_1 , \\ q,&t_1 \le \tau \le t.\\ \end{array} }} \right. \end{aligned}$$

Denote $\Delta t=t-t_{1}>0$. Then by (24), $dH_t =(1-e^{-\beta \Delta t})q$ and the effect of the change in $C_{t}$ on $H_{t}$ becomes $x(\beta )\equiv \textstyle {{dH_t } \over {dC_\tau }}=\textstyle {{dH_t } \over q}=(1-e^{-\beta \Delta t}).$ One can see that $0<x(\beta )<1$ and $x(\beta )$ is increasing in $\beta$, with $x(\beta )\rightarrow 0$ as $\beta \rightarrow 0$ and $x (\beta )\rightarrow 1$ as $\beta \rightarrow \infty$.

Therefore, (21) can be rewritten as

$$\begin{aligned} \displaystyle \zeta (\beta )=\left\{ {\textstyle {{\varepsilon [\phi -(\varsigma +\chi )\textstyle {C \over H}x(\beta )]} \over {-\sigma -\varsigma \textstyle {C \over H}x(\beta )}}-\eta } \right\} . \end{aligned}$$

 

Appendix 2: Proof of Theorem 1

In the Appendix, we prove the Theorem 1. Denote $f_{i}$, $i=l$, $k$ and $lj$, $j=C$, $k$, $H$, as partial derivatives with respect to $i$ and $j$. If we take the linear Taylor’s expansion of the dynamic equilibrium system (2), (6c) and (16) in the neighborhood of a steady state, along with the use of (8), we obtain

$$\begin{aligned} \left[ {{\begin{array}{*{20}c} {\dot{C}_t } \\ {\dot{k}_t } \\ {\dot{H}_t } \\ \end{array} }} \right]=\left[ {{\begin{array}{*{20}c} {J_{11} }&\quad {J_{12} }&\quad {J_{13} } \\ {f_l l_c -1}&\quad {f_k -\delta -f_l l_k }&\quad {f_l l_H } \\ \beta&\quad 0&\quad {-\beta } \\ \end{array} }} \right]\left[ {{\begin{array}{*{20}c} {C_t -C^*} \\ {k_t -k^*} \\ {H_t -H^*} \\ \end{array} }} \right], \end{aligned}$$

(25)

where

$$\begin{aligned} J_{11}&= \displaystyle {C \over \Omega }\left\{ [(\alpha +\varepsilon -\eta )\varsigma +\varepsilon \chi ]\displaystyle {\beta \over H}+(\eta -\varepsilon -\alpha )f_{kl} l_c +\displaystyle {{\alpha \varepsilon } \over k}(f_l l_c -1)\right\} ,\\ J_{12}&= \displaystyle {C \over \Omega }\left\{ (\eta -\varepsilon -\alpha )(f_{kl} l_k +f_{kk} )+\displaystyle {{\alpha \varepsilon } \over k}(f_l l_k +f_k -\delta )\right\} ,\\ J_{13}&= \displaystyle {C \over \Omega }\left\{ -[(\alpha +\varepsilon -\eta )\varsigma +\varepsilon \chi ]\displaystyle {\beta \over H}+(\eta -\varepsilon -\alpha )f_{kl} l_H +\displaystyle {{\alpha \varepsilon } \over k}f_l l_H \right\} . \end{aligned}$$

Let $J$ denote the Jacobean matrix in (25) and $\omega$ denote its corresponding eigenvalue. The characteristic polynomial is in (11), with Det($J)$, Tr($J)$ and Ds($J)$ defined in (12a)–(12c).

It is clear from (11) that $G(\omega )=-\infty$ when $\omega =\infty$ and $G(\omega )=\infty$ when $\omega =-\infty$. A sink requires three stable roots. The necessary conditions for the presence of three stable roots are: (i) $G(0)=Det(J)<0$ and (ii) $G^{\prime }(0)=-Ds(J)<0$. Moreover, according to the Routh-Hurwitz theorem, the requirement of no eigenvalues with positive real parts in the above characteristic polynomial suggests no variation in signs in the following series: $\left\{ {-1, Tr( J),_ -Ds( J)+Det( J)/Tr( J), Det( J)} \right\} .$ This indicates the additional requirement of (iii) Tr $(J)<0$ and (iv) $-Ds( J)+Det( J)/Tr( J)<0.$

To investigate these conditions,

• (i) $G( 0)=Det( J)<0$ Since $Det( J)=\beta ( {1-\alpha })( {\rho +\delta })\left[ {\rho +\delta ( {1-\alpha })} \right][(\phi +\sigma +\varepsilon -\eta )-\chi ]$/(-$\Omega \alpha )$ and $\Omega <0$, it is obvious that this requires $\chi >\phi +\sigma -\eta +\varepsilon$.

• (ii) $Tr( J)<0$. As $T>0$, Tr $(J)<0$ requires both

  $$\begin{aligned} \Gamma _{1}<0 \quad \text{ and}\quad \beta >\beta _{a}\equiv T/(-\Gamma _{1})>0. \end{aligned}$$

  (26a)

• (iii) $G^{\prime }(0)= -Ds(J)<0$ As $M<0, Ds(J)\,>\,0$ requires both

  $$\begin{aligned} \Gamma _2 >0 \quad \text{ and}\quad \beta >\beta _b \equiv M/( {-\Gamma _2 })>0. \end{aligned}$$

  (26b)

• (iv) $-Ds(J)+Det(J)/Tr(J)<0$ Under Tr $(J)<0$ in (ii), condition (iv) is equivalent to $-Ds(J)Tr(J)+Det(J)>0$. Using (12a)–(12c), this requires

  $$\begin{aligned} L(\beta )=\beta ^2-\beta \left\{ \frac{M}{-\Gamma _2}+\frac{T}{-\Gamma _1}+ \frac{N}{-\Gamma _1 \Gamma _2}\right\} + \frac{MT}{\Gamma _1 \Gamma _2}>0, \end{aligned}$$

  (27a)

  where $N=\textstyle {{(1-\alpha )(\rho +\delta )} \over {(-\Omega )}}\textstyle {{\rho +\delta (1-\alpha )} \over \alpha }(\phi +\sigma -\eta +\varepsilon -\chi )<0,$ whose the negative sign comes from using (i).

When $L({\beta })=0$ the polynomial has two roots ${\beta }_{1}$ and ${\beta }_{2 }, {\beta }_{1}\ge {\beta }_{2}$, as follows.

$$\begin{aligned} \displaystyle {1 \over 2}\left\{ \displaystyle {\mathrm{N} \over {\Gamma _1 \Gamma _2 }}+\displaystyle {\mathrm{M} \over {-\Gamma _2 }}+\displaystyle {\mathrm{T} \over {-\Gamma _1 }}\pm \left[(\displaystyle {\mathrm{N} \over {\Gamma _1 \Gamma _2 }}+\displaystyle {\mathrm{M} \over {-\Gamma _2 }}+\displaystyle {\mathrm{T} \over {-\Gamma _1 }})^2-4\displaystyle {{\mathrm{M}\mathrm{T}} \over {\Gamma _1 \Gamma _2 }}\right]^{1 / 2}\right\} . \end{aligned}$$

Under (i) Det $(J)<0$, (ii) Tr $(J)<0$ and (iii) Ds $(J)>0$, both ${\beta }_{1}$ and ${\beta }_{2}$ are positive, as verified by

$$\begin{aligned} \beta _{1}{\beta }_{2}={MT}/(\Gamma _{1}\Gamma _{2})>0 \quad \text{ and}\quad {\beta }_{1}+{\beta }_{2}=\{ \Gamma _{1}M+\Gamma _{2}T- N \}/(-\Gamma _{1}\Gamma _{2})>0. \end{aligned}$$

The inequality sign in (26a) is satisfied if any one of the following two cases holds: (a) $\beta >\beta _{1}\ge \beta _{2}$ or (b) $\beta <\beta _{2}\le \beta _{1}$. However, case (b) is impossible as case (b) implies $\beta _{2}<T/(-\Gamma _{1})\equiv \beta _{a}$, which is against the requirement of $\beta >\beta _{a}$ for Tr $(J)<0$ in (ii).

Therefore, (27a) and $-Ds(J)Tr(J)+Det(J)>0$ both can be met only if

$$\begin{aligned} \beta >\beta _1. \end{aligned}$$

(27b)

It is straightforward to show that $\beta _{1}>\beta _{a}$ and $\beta _{1}>\beta _{b}$. Thus, (26a), (26b) and (27b) indicate that the requirement of $\beta >\beta _{1}$.

Therefore, under $\beta >\beta _{1}$, the conditions of a sink are: $\chi >\phi +\sigma -\eta +\varepsilon$, $\Gamma _{1}<0$ and $\Gamma _{2}>0$.