Consumption tax, seigniorage tax and tax switch in a cash-in-advance economy of endogenous growth

Abstract

This paper studies the effects of alternative tax policies (a consumption tax, seigniorage tax, and tax switch) on economic growth under different methods of government budget adjustment in a monetary endogenous growth model with a labor–leisure choice and a cash-in-advance constraint which is only imposed on consumption. It is found that the validity of both the Mundell–Tobin effect and the consumption tax neutrality crucially depends on the adjustment methods used to maintain the balanced government budget. In addition, we find that a switch from a consumption tax to a seigniorage tax unambiguously enhances economic growth. This result stands in sharp contrast to that of Ho et al. (J Money Credit Bank 39:105–131, 2007), in that it does confirm the validity of the qualitative equivalence between the money-in-the-utility-function and cash-in-advance approaches in terms of the effect of tax shifting away from a consumption tax towards a seigniorage tax.

Appendices

Appendix A (The proof of Theorem 1)

By substituting (15c) and (16) into (15d) and (15e), the dynamic system can be represented as follows:

$$\begin{aligned} \left[ {\begin{array}{l} \dot{x}_t \\ \dot{z}_t \\ \end{array}} \right] =\left[ \begin{array}{ll} {\Sigma _1 l_x \hat{x}} &{} {\Sigma _3 \hat{x}} \\ {\Sigma _2 l_x \hat{z}} &{} {\Sigma _4 \hat{z}} \\ \end{array} \right] \left[ {\begin{array}{l} dx_t \\ dz_t \\ \end{array}} \right] +\left[ \begin{array}{l} \Gamma _1 \hat{x} \\ \Gamma _2 \hat{z} \\ \end{array} \right] d\mu +\left[ \begin{array}{l} \Pi _1 \hat{x} \\ \Pi _2 \hat{z} \\ \end{array} \right] d\tau _c , \end{aligned}$$

where $\Sigma _1 =(\sigma -\alpha )\beta A(1-\hat{l})^{\beta -1}>0,\, \Sigma _2 =(1-\alpha )\beta A(1-\hat{l})^{\beta -1}-\Lambda \frac{1-\beta \hat{l}}{\hat{l}(1-\hat{l})}, \Sigma _3 =\Sigma _1 l_z +\sigma [\frac{\theta }{1+\tau _c }+(1-\theta )(1+\mu )], \Sigma _4 =\Sigma _2 l_z +[\frac{\theta }{1+\tau _c }+(1-\theta )(1+\mu )+\frac{\Lambda }{\hat{z}}]$, $\Gamma _1 =(1-\theta )\sigma \hat{z}$, $\Gamma _2 =1+(1-\theta )\hat{z}$, $\Pi _1 =\Sigma _1 l_{\tau _c } -\theta \frac{\sigma }{(1+\tau _c )^2}\hat{z}$, $\Pi _2 =\Sigma _2 l_{\tau _c } -\theta \frac{\hat{z}}{(1+\tau _c )^2}$, and $\Lambda =\frac{\delta }{1-\delta }\frac{1}{\hat{z}}\beta A\hat{l}(1-\hat{l})^{\beta -1}>0$. Thus, we can compute the Jacobian matrix evaluated at the steady state and, accordingly, have:

$$\begin{aligned} \begin{array}{l} Det(J)\equiv \Delta =\hat{x}\hat{z}l_x \left\{ {(\Sigma _1 \frac{\Lambda }{\hat{z}}+(\Sigma _1 -\sigma \Sigma _2 )\left[ {\frac{\theta }{1+\tau _c }+(1-\theta )(1+\mu )} \right] } \right\} >0, \\ Tr(J)=\hat{x}l_x \left[ {\Sigma _1 +(1-\sigma )\Sigma _2 } \right] +\hat{z}\left[ {\frac{\theta }{1+\tau _c }+(1-\theta )(1+\mu )+\frac{\Lambda }{\hat{z}}} \right] >0, \\ \end{array} \end{aligned}$$

where

$$\begin{aligned} \begin{array}{l} \Sigma _1 -\sigma \Sigma _2 =\alpha (\sigma -1)\beta A(1-\hat{l})^{\beta -1}+\sigma \Lambda \frac{1-\beta \hat{l}}{\hat{l}(1-\hat{l})}>0, \\ \Sigma _1 +(1-\sigma )\Sigma _2 =[(1-\alpha )+\alpha (\sigma -1)]\beta A(1-\hat{l})^{\beta -1}+(\sigma -1)\Lambda \frac{1-\beta \hat{l}}{\hat{l}(1-\hat{l})}>0. \\ \end{array} \end{aligned}$$

This implies that regardless of Case $\theta =1$ or Case $\theta =0$, the dynamic system has two roots with real positive parts. Since the dynamic system has two jump variables $x_t$ and $z_t$, there is a unique path leading to the determinate steady-state equilibrium.

By focusing on Case $\theta =1$, the steady-state equilibrium is characterized by $\dot{x}_t =0$ and $\dot{z}_t =0$ in (15d) and (15e), i.e.:

$$\begin{aligned}&\displaystyle \frac{\sigma }{(1+\tau _c )}\hat{z}+(\alpha -\sigma )A(1-\hat{l})^\beta -\rho =0,\end{aligned}$$

(19)

$$\begin{aligned}&\displaystyle \mu +1-(1-\alpha )A(1-\hat{l})^\beta -\frac{\delta }{1-\delta }\frac{1}{\hat{z}}\beta \hat{l}A(1-\hat{l})^{\beta -1}+\frac{\hat{z}}{1+\tau _c }=0, \end{aligned}$$

(20)

where $\hat{l}$ and $\hat{z}$ are the steady-state values of $l_t$ and $z_t$, respectively. For ease of exposition, combining (19) with (20) yields:

$$\begin{aligned} \sigma [1+\mu -\frac{\delta }{1-\delta }\frac{1}{\hat{z}}\beta A\hat{l}(1-\hat{l})^{\beta -1}]+\rho +\alpha (\sigma -1)A(1-\hat{l})^\beta =0. \end{aligned}$$

(21)

In what follows, we will use (19) and (21) to solve $\hat{l}$ and $\hat{z}$. In the ($z$, $l)$ space, the XX locus characterizes all the combinations which satisfy (19) and the XZ locus describes all the combinations which satisfy (20). From (19), the XX locus is downward sloping, i.e., $\frac{dz_t }{dl_t }\vert _{XX} =\frac{-\Sigma _1 (1+\tau _c )}{\sigma }<0$, and it intersects the $z_t$-coordinate at $\frac{(1+\tau _c )}{\sigma }\left[ {(\sigma -\alpha )A+\rho } \right] >0$ when $l_t =0$, as shown in Fig. 1 below. Moreover, given the restriction of non-negative working hours ($0\le l_t \le 1)$, the ratio of real money balances to capital converges to $z_t =\frac{(1+\tau _c )}{\sigma }\rho >0$, if the working hours approach $l_t =1$.

Fig. 1

The existence and uniqueness of the steady-state equilibrium: case $\theta =0$

In addition, (21) tells us that the XZ locus is upward sloping ($\frac{dz_t }{dl_t }\vert _{XZ} =\frac{\hat{z}(\Sigma _1 -\sigma \Sigma _2 )}{\sigma \Lambda }>0)$ and satisfies $\lim _{l_t \rightarrow 0} z_t =0$ and $\lim _{l_t \rightarrow 1} z_t =\infty$. Thus, as is evident in Fig. 1, the steady-state equilibrium exists and is unique.

We then turn our focus to Case $\theta =0$. With the government’s budget constraint (15c) $( {( {\frac{\tau _c }{1+\tau _c }+\mu })z_t =\frac{G_t }{k_t }})$, we have the following relationship of the steady-state equilibrium from (15d) and (15e) with $\dot{x}_t =0$ and $\dot{z}_t =0$:

$$\begin{aligned}&\displaystyle \sigma (1+\mu )\hat{z}+(\alpha -\sigma )A(1-\hat{l})^\beta -\rho =0,\end{aligned}$$

(22)

$$\begin{aligned}&\displaystyle \mu +1-(1-\alpha )A(1-\hat{l})^\beta -\frac{\delta }{1-\delta }\frac{1}{\hat{z}}\beta A\hat{l}(1-\hat{l})^{\beta -1}+(1+\mu )\hat{z}=0. \end{aligned}$$

(23)

By putting the above equations together, we further have:

$$\begin{aligned} \frac{1+\hat{z}}{\sigma \hat{z}}[(\sigma -\alpha )A(1-\hat{l})^\beta +\rho ]-(1-\alpha )A(1-\hat{l})^\beta -\frac{\delta }{1-\delta }\frac{1}{\hat{z}}\beta A\hat{l}(1-\hat{l})^{\beta -1}=0.\nonumber \\ \end{aligned}$$

(24)

Similar to Case $\theta =1$, we use (22) and (24) to characterize the steady-state equilibrium. It follows from (22) that the XX locus is downward sloping, i.e., $\frac{dz_t }{dl_t }\vert _{XX} =\frac{-\Sigma _1 }{\sigma (1+\mu )}<0$, and it intersects the $z_t$-coordinate at $\frac{1}{\sigma (1+\mu )}[(\sigma -\alpha )A+\rho ]>0$ when $l_t =0$. As shown in Fig. 2, the ratio of real money balances to capital converges to $z_t =\frac{\rho }{\sigma (1+\mu )}>0$, if the working hours approach $l_t =1$. On the other hand, it follows from (24) that the XZ locus is upward sloping, i.e., $\frac{dz_t }{dl_t }\vert _{XZ} =\frac{\Sigma _1 +\hat{z}(\Sigma _1 -\sigma \Sigma _2 )}{\alpha (\sigma -1)A(1-\hat{l})^\beta +\rho }>0$. Moreover, the XZ locus intersects the $z_t$-coordinate at a negative value (i.e., $-\frac{(\sigma -\alpha )A+\rho }{\alpha (\sigma -1)A+\rho }<0$ when $l_t =0)$ and the ratio of real money balances to capital approaches $z_t =\infty$ as $l_t =1$.

Fig. 2

The existence and uniqueness of the steady-state equilibrium: case $\theta =0$

As shown in Fig. 2, in Case $\theta =0$the steady-state equilibrium also exists and is unique. $\square$

Appendix B (The proof of Propositions 1 and 2)

Given (16), by using (15c)–(15e) with $\dot{x}_t =0$ and $\dot{z}_t =0$, the steady-state effects of $\mu$ on the leisure time $\hat{l}$ under Cases $\theta =1$ and $\theta =0$, respectively, are given by:

$$\begin{aligned} \frac{d\hat{l}}{d\mu }\vert _{\theta =1} =\frac{\hat{x}\hat{z}}{\Delta }\frac{\sigma }{1+\tau _c }l_x >0\,\,\text{ and }\,\,\frac{d\hat{l}}{d\mu }\vert _{\theta =0} =-\frac{\hat{x}\hat{z}}{\Delta }[\alpha (\sigma -1)A(1-\hat{l})^\beta +\rho ]l_x <0. \end{aligned}$$

In addition, it follows from (10) that the balanced-growth rate is given by:

$$\begin{aligned} \hat{\varphi }=\frac{1}{\sigma }\left[ {\alpha A(1-\hat{l})^\beta -\rho } \right] . \end{aligned}$$

(25)

Accordingly, the steady-state effects of $\mu$on the growth rate under Cases $\theta =1$ and $\theta =0$ are, respectively:

$$\begin{aligned} \frac{d\hat{\phi }}{d\mu }\vert _{\theta =1}&= -\frac{1}{\sigma }\alpha \beta A(1-\hat{l})^{\beta -1}\cdot \frac{d\hat{l}}{d\mu }\vert _{\theta =1} <0\,\,\text{ and }\,\frac{d\hat{\phi }}{d\mu }\vert _{\theta =0}\\&= -\frac{1}{\sigma }\alpha \beta A(1-\hat{l})^{\beta -1}\cdot \frac{d\hat{l}}{d\mu }\vert _{\theta =0} >0. \end{aligned}$$

By applying a similar procedure, we can easily derive the steady-state effects of $\tau _c$ on $\hat{l}$and $\hat{\phi }$ under both Cases $\theta =1$ and $\theta =0$:

$$\begin{aligned}&\frac{d\hat{l}}{d\tau _c }\vert _{\theta =1} =\frac{\hat{x}\hat{z}}{\Delta }\frac{\sigma \Lambda }{(1+\tau _c )^2}l_x >0,\,\,\frac{d\hat{l}}{d\tau _c }\vert _{\theta =0} =0,\\&\frac{d\hat{\phi }}{d\tau _c }\vert _{\theta =1} =-\frac{1}{\sigma }\alpha \beta A(1-\hat{l})^{\beta -1}\frac{d\hat{l}}{d\tau _c }\vert _{\theta =1} <0,\,\,\text{ and }\,\,\frac{d\hat{\phi }}{d\tau _c }\vert _{\theta =0} =0. \end{aligned}$$

$\square$

Appendix C (The proof of Proposition 3)

Expanding (18a) and (18b) in the neighborhood of the steady-state equilibrium yields:

$$\begin{aligned} \left[ {\begin{array}{l} {\dot{x}_t } / {x_t } \\ {\dot{z}_t }/{z_t } \\ \end{array}} \right] =\left[ \begin{array}{ll} {\Phi _1 l_x } &{} {\Phi _3 } \\ {\Phi _2 l_x } &{} {\Phi _4 } \\ \end{array} \right] \left[ \begin{array}{l} dx_t \\ dz_t \\ \end{array} \right] +\left[ \begin{array}{l} \mathrm{H}_1 \\ \mathrm{H}_2 \\ \end{array} \right] d\tau _c , \end{aligned}$$

where $\Phi _1 =[\sigma (1-\bar{g})-\alpha ]\beta A(1-\hat{l})^{\beta -1}$, $\Phi _2 =[(1-\bar{g})-\alpha -\frac{\bar{g}}{\hat{z}}]\beta A(1-\hat{l})^{\beta -1}-\frac{\Lambda (1-\beta \hat{l})}{\hat{l}(1-\hat{l})}$, $\Phi _3 =\Phi _1 l_z +\frac{\sigma }{1+\tau _c }$, $\Phi _4 =\Phi _2 l_z +\frac{1}{1+\tau _c }+\frac{\Lambda }{\hat{z}}-\frac{\bar{g}}{\hat{z}^2}A(1-\hat{l})^\beta$, $\mathrm{H}_1 =\Phi _1 l_{\tau _c } -\frac{\sigma \hat{z}}{(1+\tau _c )^2}$, and $\mathrm{H}_2 =\Phi _2 l_{\tau _c } -\frac{1+\hat{z}}{(1+\tau _c )^2}$.We impose a sufficient condition of $1-\alpha =\beta >\bar{g}$ so that the steady-state equilibrium is determinate. Given the fact that the labor income share $1-\alpha =\beta$ is around 0.6 and the government spending to GDP ratio is around 0.2, this sufficient condition is empirically plausible. Accordingly, this above equation allows us to solve the steady state $\hat{z}$ and $\hat{x}$. Thus, substituting $\hat{z}$ and $\hat{x}$ into (16) yields the steady-state effect of a tax switch on $\hat{l}$:

$$\begin{aligned} \frac{d\hat{l}}{d\tau _c }\vert _{switch} =\frac{\hat{z}[\alpha (\sigma -1)A(1-\hat{l})^\beta +\rho ]/(1+\tau _c )^2}{\frac{\hat{z}}{(1+\tau _c )}(\Phi _1 -\sigma \Phi _1 )+\Phi _1 [\Lambda -\frac{\bar{g}}{\hat{z}}A(1-\hat{l})^\beta ]}>0, \end{aligned}$$

where

$$\begin{aligned}&\Phi _1 -\sigma \Phi _2 =[\alpha (\sigma -1)+\sigma \frac{\bar{g}}{\hat{z}}]\beta A(1-\hat{l})^{\beta -1}+\sigma \frac{\Lambda (1-\beta \hat{l})}{\hat{l}(1-\hat{l})}>0 \\&\Lambda -\frac{\bar{g}}{\hat{z}}A(1-\hat{l})^\beta =\frac{1}{1+\tau _c }+\frac{\rho +\alpha (\sigma -1)A(1-\hat{l})^\beta }{\sigma }>0. \\ \end{aligned}$$

Accordingly, we can further derive the growth effect of a tax switch from (25):

$$\begin{aligned} \frac{d\hat{\phi }}{d\tau _c }\vert _{switch} =-\frac{1}{\sigma }\alpha \beta A(1-\hat{l})^{\beta -1}\frac{d\hat{l}}{d\tau _c}\vert _{switch} <0. \end{aligned}$$

$\square$