Public capital, endogenous growth, and tax concession on savings

Abstract

An endogenous growth model is developed where, on one hand income tax revenue is utilized to finance investment on public capital, and on the other hand tax concession is given on savings. If one of the two instruments—proportional income tax rate and proportional tax-concession rate on savings—is used as the policy variable to maximize the balanced growth rate while the other is treated as a parameter, an exogenous increase in the value of the parameter raises the optimum value of the policy variable and generates a positive effect on endogenous growth rate as well as on the rate of savings in the steady-state equilibrium.

Appendices

Appendix A

The problem is to maximize:

$$u(C) = \frac{{C^{1 - \alpha } - 1}}{1 - \sigma }$$

subject to

$$\dot{K} = \left( {1 + \theta - \tau } \right)sY$$

and

$$C = \left( {1 - \tau } \right)\left( {1 - s} \right)Y.$$

Here, $\mathrm{s}$ is the control variable satisfying $0\le s\le 1$.

The Hamiltonian is given by:

$$He^{\rho t} = U\left( C \right) + q\dot{K}.$$

Here, $q$ is the co-state variable, a function of time.

Maximizing ${He}^{\rho t}$ with respect to $s$ and assuming an interior solution, we have:

$$- \left( {1 - \tau } \right)Y\left[ {\left( {1 - s^{*} } \right)\left( {1 - \tau } \right)Y} \right]^{ - \sigma } + q\left( {1 + \theta - \tau } \right)Y = 0,$$

where ${s}^{*}$ is the optimal savings rate. This equation also implies that:

$$C^{ - \sigma } = \frac{1 + \theta - \tau }{{1 - \tau }}q.$$

(A1)

Here, C and $q$ represent optimum time path of consumption and of the co-state variable.

Given $\theta$ and $\tau$, Eq. (A1) implies that:

$$- \sigma \frac{{\dot{C}}}{C} = \frac{{\dot{q}}}{q}.$$

(A2)

Now, the optimality condition given by:

$$\begin{gathered} \dot{q} = \rho q - \frac{{\partial He^{\rho t} }}{\partial K}, \hfill \\ \Rightarrow \dot{q} = \rho q - \left[ {C^{ - \sigma } \left( {1 - s^{*} } \right)\left( {1 - \tau } \right) + s^{*} \left( {1 + \theta - \tau } \right)q} \right]\frac{\partial Y}{{\partial K}}. \hfill \\ \end{gathered}$$

Using Eq. (A1), we have:

$$\begin{gathered} \dot{q} = \rho q - \left[ {\frac{1 + \theta - \tau }{{1 - \tau }}q(1 - s^{*} )(1 - \tau ) + s^{*} (1 + \theta - \tau )q} \right]\frac{\partial Y}{{\partial K}}, \hfill \\ \Rightarrow \frac{{\dot{q}}}{q} = \rho - \left( {1 + \theta - \tau } \right)\frac{\partial Y}{{\partial K}}. \hfill \\ \end{gathered}$$

(A3)

From Eqs. (A2) and (A3), we have:

$$\frac{{\dot{C}}}{C} = \frac{{\left( {1 + \theta - \tau } \right)\frac{\partial Y}{{\partial K}} - \rho }}{\sigma }.$$

(A4)

This Eq. (A4) is identical to Eq. (6) in the body of the paper.

Appendix B

From Eqs. (3A) and (7), we have:

$$g = \left( {1 + \theta - \tau } \right)s^{*} x^{1 - \alpha } .$$

(B1)

Using Eqs. (6A) and (7), we have:

$$\left( {\rho + \sigma g} \right) = \left( {1 + \theta - \tau } \right)\alpha x^{1 - \alpha } .$$

(B2)

Using Eqs. (B1) and (B2), we have:

$$\begin{gathered} \frac{{s^{*} }}{\alpha } = \frac{g}{{\left( {\rho + \sigma g} \right)}}, \hfill \\ \Rightarrow s^{*} = \frac{\alpha g}{{\rho + \sigma g}}. \hfill \\ \end{gathered}$$

(B3)

This Eq. (B3) is identical to Eq. (9) in the body of the paper.

Appendix C

Using Eqs. (8) and (9), we have:

$$\begin{gathered} g = \left( {\tau - \frac{\theta \alpha g}{{\rho + \sigma g}}} \right)^{1 - \alpha } \left( {1 + \theta - \tau } \right)^{\alpha } \left( {\frac{\alpha g}{{\rho + \sigma g}}} \right)^{\alpha } \hfill \\ \Rightarrow \left( {\rho + \sigma g} \right)g^{1 - \alpha } = \alpha^{\alpha } \left( {1 + \theta - \tau } \right)^{\alpha } \left[ {\rho \tau + \left( {\sigma \tau - \theta \alpha } \right)g} \right]^{1 - \alpha } . \hfill \\ \end{gathered}$$

(C1)

LHS of Eq. (C1) is an increasing function of $g$. Therefore, maximization of $g$ implies the maximization of this LHS. Therefore, maximization of $g$ with respect to $\tau$ implies the maximization of its RHS which depends on $\tau$.

We put:

$$z = \alpha^{\alpha } \left( {1 + \theta - \tau } \right)^{\alpha } \left[ {\rho \tau + \left( {\sigma \tau - \theta \alpha } \right)g} \right]^{1 - \alpha } .$$

Hence:

$$\begin{aligned} \frac{{dz}}{{d\tau }} = & \alpha ^{\alpha } \left( {1 + \theta - \tau } \right)^{{\alpha - 1}} \left[ {\rho \tau + \left( {\sigma \tau - \theta \alpha } \right)g} \right]^{{ - \alpha }} \\ & \times \left\{ {\left( {1 - \alpha } \right)\left( {1 + \theta - \tau } \right)\left( {\rho + \sigma g} \right) + \left( {\sigma \tau - \theta \alpha } \right)\frac{{dg}}{{d\tau }} - \alpha \left( {\rho \tau + \left( {\sigma \tau - \theta \alpha } \right)g} \right)} \right\}. \\ \end{aligned}$$

(C2)

We obtain optimum tax rate maximizing $z$; and maximization of $z$ automatically implies maximization of $g$. When $z$ is maximized, $\frac{dg}{d\tau }=\frac{dz}{d\tau }=0$, and hence, from Eq. (C2), we have:

$$\begin{aligned} 0 = & \alpha ^{\alpha } \left( {1 + \theta - \tau } \right)^{{\alpha - 1}} \left[ {\rho \tau + \left( {\sigma \tau - \theta \alpha } \right)g} \right]^{{ - \alpha }} \\ \quad & \times \left\{ {\left( {1 - \alpha } \right)\left( {1 + \theta - \tau } \right)\left( {\rho + \sigma g} \right)} \right.\left. { - \alpha \left( {\rho \tau + \left( {\sigma \tau - \theta \alpha } \right)g} \right)} \right\} \\ & \Rightarrow \left( {1 - \alpha } \right)\left( {1 + \theta - \tau } \right)\left( {\rho + \sigma g} \right) = \alpha \rho \tau + \alpha \sigma g\tau - \theta \alpha ^{2} g \\ & \Rightarrow \left( {1 - \alpha } \right)\left( {1 + \theta } \right)\left( {\rho + \sigma g} \right) + \theta \alpha ^{2} g = \tau \left( {\rho + \sigma g} \right)\left( {1 - \alpha } \right) + \alpha \tau \left( {\rho + \sigma g} \right) \\ & \Rightarrow \tau = \left( {1 - \alpha } \right)\left( {1 + \theta } \right) + \theta \alpha \frac{{\alpha g}}{{\rho + \sigma g}} = \tau ^{*} . \\ \end{aligned}$$

(C3)

Here, ${\tau }^{*}$ is the balanced growth rate maximizing tax rate. Using Eqs. (9) and (C3), we have:

$\tau^{*} = \left( {1 - \alpha } \right)\left( {1 + \theta } \right) + \theta \alpha s^{*}$ (C4).

This Eq. (C4) is identical to Eq. (10) in the body of the paper.

From Eq. (C2), we have:

$$\frac{{d^{2} z}}{{d\tau^{2} }} = z_{1}^{\prime } \left( \tau \right) \cdot z_{2} \left( \tau \right) + z_{1} \left( \tau \right) \cdot z_{2}^{\prime } \left( \tau \right),$$

where

$$z_{1} \left( \tau \right) = \alpha^{\alpha } \left( {1 + \theta - \tau } \right)^{\alpha - 1} \left\{ {\rho \tau + \left( {\sigma \tau - \theta \alpha } \right)g} \right\}^{ - \alpha } ;$$

and

$$\begin{aligned} z_{2} \left( \tau \right) = & \{ \left( {1 - \alpha } \right)\left( {1 + \theta - \tau } \right)\left( {\rho + \sigma g} \right) \\ & + \left( {\sigma \tau - \theta \alpha } \right)\frac{dg}{{d\tau }} - \alpha (\rho \tau + \left( {\sigma \tau - \theta \alpha } \right)g\} . \\ \end{aligned}$$

When τ = ${\tau }^{*}$, ${ z}_{2}\left(\tau \right)$=0 and $\frac{dg}{d\tau }=0$.

Hence, at τ = ${\tau }^{*}$:

$$z_{2}^{\prime } \left( \tau \right) = - \left( {1 - \alpha } \right)\left( {\rho + \sigma g} \right) - \alpha \left( {\rho + \sigma g} \right) = - \left( {\rho + \sigma g} \right) < 0.$$

Therefore, at τ = ${\tau }^{*}$:

$$\begin{gathered} \frac{{d^{2} z}}{{d\tau^{2} }} = z_{1} (\tau ) \cdot z_{2}^{\prime } (\tau ) \hfill \\ \Rightarrow \frac{{d^{2} z}}{{d\tau^{2} }} = - \left( {\rho + \sigma g} \right) \cdot \alpha^{\alpha } \left( {1 + \theta - \tau } \right)^{\alpha - 1} \cdot \left\{ {\rho \tau + \left( {\sigma \tau - \theta \alpha } \right)g} \right\}^{ - \alpha } < 0. \hfill \\ \end{gathered}$$

Using the second-order derivatives from Eq. (C1) and, then putting $\frac{dg}{d\tau }=0$, we have:

$$\left[ {\sigma g^{1 - \alpha } + \left( {\rho + \sigma g} \right)\left( {1 - \alpha } \right)g^{ - \alpha } } \right]\left( {\frac{{d^{2} g}}{{d\tau^{2} }}} \right) = \frac{{\partial^{2} z}}{{\partial \tau^{2} }} + \frac{\partial z}{{\partial g}}\frac{{d^{2} g}}{{d\tau^{2} }},$$

(C5)

where

$$\frac{\partial z}{{\partial g}} = \frac{{z\left( {1 - \alpha } \right)\left( {\left( {\sigma \tau - \theta \alpha } \right)} \right)}}{{\rho \tau + \left( {\sigma \tau - \theta \alpha } \right)g}}.$$

This Eq. (C5) can further be simplified as:

$$\frac{{d^{2} g}}{{d\tau^{2} }} = \frac{{\frac{{d^{2} z}}{{d\tau^{2} }}}}{{\alpha g^{1 - \alpha } + \frac{{z\left( {1 - \alpha } \right)\rho \tau }}{{\rho \tau + \left( {\sigma \tau - \theta \alpha } \right)g}}}}.$$

Hence, $\frac{{d}^{2}g}{d{\tau }^{2}}<0$ if $\tau >\frac{\alpha \theta g}{\rho +\sigma g}$, and Eq. (C3) shows that ${\tau }^{*}>\frac{\alpha \theta g}{\rho +\sigma g}$ if $\alpha \le \sigma$. Thus, the second-order condition of maximization of $g$ is satisfied.

Appendix D

Using Eqs. (8) and (10), we have:

$$g = \left( {1 - \alpha } \right)\left( {1 + \theta \left( {1 - s^{*} } \right)} \right)\left( {\frac{\alpha }{1 - \alpha }} \right)^{\alpha } s^{*\alpha } .$$

(D1)

Using Eqs. (D1) and (9), we have:

$$\begin{gathered} g = \left( {1 - \alpha } \right)\left( {1 + \theta \left( {1 - \frac{\alpha g}{{\rho + \sigma g}}} \right)} \right)\left( {\frac{\alpha }{1 - \alpha }} \right)^{\alpha } \left( {\frac{\alpha g}{{\rho + \sigma g}}} \right)^{\alpha } , \hfill \\ \Rightarrow g\left( {\frac{\rho + \sigma g}{g}} \right)^{\alpha } = \left( {1 - \alpha } \right)\frac{{\alpha^{\alpha } }}{{\left( {1 - \alpha } \right)^{\alpha } }}\alpha^{\alpha } \left( {1 + \theta \left( {\frac{{\left( {\sigma - \alpha } \right)g + \rho }}{\sigma g + \rho }} \right)} \right), \hfill \\ \Rightarrow g^{1 - \alpha } \left( {\sigma g + \rho } \right)^{\alpha } = \left( {1 - \alpha } \right)^{{\left( {1 - \alpha } \right)}} \alpha^{2\alpha } \left( {1 + \theta \left( {\frac{{\left( {\sigma - \alpha } \right)g + \rho }}{\sigma g + \rho }} \right)} \right), \hfill \\ \Rightarrow N\left( g \right) = M\left( {1 + \theta Q\left( g \right)} \right). \hfill \\ \end{gathered}$$

(D2)

Here:

$$N(g) = g^{(1 - \alpha )} (\sigma g + \rho )^{\alpha } \ge 0,$$

$$M = \left( {1 - \alpha } \right)^{1 - \alpha } \alpha^{2\alpha } > 0,$$

and

$$Q\left( g \right) = \frac{{\left( {\sigma - \alpha } \right)g + \rho }}{\sigma g + \rho }$$

Here:

$$N^{\prime}\left( g \right) = \left( {1 - \alpha } \right)g^{ - \alpha } \left( {\sigma g + \rho } \right)^{\alpha } + \alpha g^{1 - \alpha } \left( {\sigma g + \rho } \right)^{\alpha - 1} \sigma > 0,$$

and

$$\begin{gathered} Q^{\prime}\left( g \right) = \frac{{\left( {\sigma - \alpha } \right)\left( {\sigma g + \rho } \right) - \sigma \left[ {\left( {\sigma - \alpha } \right)g + \rho } \right]}}{{\left( {\sigma g + \rho } \right)^{2} }}, \hfill \\ \Rightarrow Q^{\prime}\left( g \right) = - \frac{\alpha \rho }{{\left( {\sigma g + \rho } \right)^{2} }} < 0. \hfill \\ \end{gathered}$$

This Eq. (D2) is identical to Eq. (11) in the body of the paper.

Appendix E

Here, we maximize $Z$, i.e., the RHS of Eq. (C1) with respect to $\theta$, given $\tau$:

$$\begin{aligned} \frac{dz}{{d\theta }} = & \alpha^{\alpha } \alpha \left( {1 + \theta - \tau } \right)^{\alpha - 1} \left[ {\rho \tau + \left( {\sigma \tau - \theta \alpha } \right)g} \right]^{1 - \alpha } \\ & + \alpha^{\alpha } \left( {1 + \theta - \tau } \right)^{\alpha } \left( {1 - \alpha } \right)\left[ {\rho \tau + \left( {\sigma \tau - \theta \alpha } \right)g} \right]^{ - \alpha } \left[ { - \alpha g + \left( {\sigma \tau - \theta \alpha } \right)\frac{dg}{{d\theta }}} \right]. \\ \end{aligned}$$

(E1)

When $g$ is maximized, $\frac{dg}{d\theta }=\frac{dz}{d\theta }=0$; and hence, from Eq. (E1), we have:

$$\begin{aligned} 0 = \, & \alpha ^{\alpha } \left( {1 + \theta - \tau } \right)^{{\alpha - 1}} \left[ {\rho \tau + \left( {\sigma \tau - \theta \alpha } \right)g} \right]^{{ - \alpha }} \\ \quad & \times \left\{ {\alpha \rho \tau + \left( {\sigma \tau - \theta \alpha } \right)\alpha g - \alpha g\left( {1 - \alpha } \right)\left( {1 + \theta - \tau } \right)} \right\} \\ & \Rightarrow \alpha \rho \tau + \left( {\sigma \tau - \theta \alpha } \right)\alpha g = \alpha g\left( {1 - \alpha } \right)\left( {1 + \theta - \tau } \right) \\ & \Rightarrow \alpha \tau \left( {\rho + \sigma g} \right) - \theta \alpha ^{2} g = \alpha g\left( {1 - \alpha } \right)\left( {1 - \tau } \right) + \alpha g\left( {1 - \alpha } \right)\theta \\ & \Rightarrow \alpha \tau - \frac{{\alpha g}}{{\rho + \sigma g}}\left( {1 - \alpha } \right)\left( {1 - \tau } \right) = \left[ {\frac{{\alpha g}}{{\rho + \sigma g}}\left( {1 - \alpha } \right) + \frac{{\alpha ^{2} g}}{{\rho + \sigma g}}} \right]\theta \\ & \Rightarrow \theta = \frac{{\alpha \tau + s^{*} \left( {1 - \alpha } \right)\left( {\tau - 1} \right)}}{{\left( {1 - \alpha } \right)s^{*} + \alpha s^{*} }} \\ & \Rightarrow \theta = \alpha \frac{\tau }{{s^{*} }} + \left( {1 - \alpha } \right)\left( {\tau - 1} \right) = \theta ^{*} . \\ \end{aligned}$$

(E2)

This Eq. (E2) is identical to Eq. (12) in the body of the paper.

From Eq. (E1), we have:

$$\frac{dz}{d\theta }={z}_{1}\left(\uptheta \right)\cdot{z}_{2}\left(\uptheta \right),$$

where

$$z_{1} \left( \theta \right) = \alpha^{\alpha } \left( {1 + \theta - \tau } \right)^{\alpha - 1} \left\{ {\rho \tau + \left( {\sigma \tau - \theta \alpha } \right)g} \right\}^{ - \alpha } ,$$

and

$$z_{2} \left( \theta \right) = \alpha \rho \tau + \left( {\sigma \tau - \theta \alpha } \right)\alpha g - \alpha g\left( {1 - \alpha } \right)\left( {1 + \theta - \tau } \right).$$

Here:

$$\frac{{d^{2} z}}{{d\theta^{2} }} = z_{1} \left( \theta \right) \cdot z_{2}^{\prime } \left( \theta \right) + z_{1}^{\prime } \left( \theta \right) \cdot z_{2} \left( \theta \right).$$

At $\theta$ =${\theta }^{*}$, ${z}_{2}\left(\uptheta \right)=0$ and $\frac{dg}{d\theta }=0$.

Hence, at $\theta$ =${\theta }^{*}$:

$$\frac{{d^{2} z}}{{d\theta^{2} }} = z_{1} \left( \theta \right) \cdot z_{2}^{\prime } \left( \theta \right).$$

Also, at $\theta$ =${\theta }^{*}$:

$$z_{2}^{\prime } (\theta ) = - \alpha^{2} g - \alpha g(1 - \alpha ) = - \alpha g < 0.$$

Hence:

$$\frac{{d^{2} z}}{{d\theta^{2} }} = - \alpha g\alpha^{\alpha } \left( {1 + \theta - \tau } \right)^{\alpha - 1} \left\{ {\rho \tau + \left( {\sigma \tau - \theta \alpha } \right)g} \right\}^{ - \alpha } < 0.$$

Here also, we have:

$$\frac{{d^{2} g}}{{d\theta^{2} }} = \frac{{\frac{{d^{2} z}}{{d\theta^{2} }}}}{{\alpha g^{1 - \alpha } + \frac{{z\left( {1 - \alpha } \right)\rho \tau }}{{\rho \tau + \left( {\sigma \tau - \theta \alpha } \right)g}}}}.$$

(E3)

From Eq. (E2), we have:

$$\theta^{*} = \frac{{\left( {\rho + \sigma g} \right)\tau }}{g} - \left( {1 - \alpha } \right)\left( {1 - \tau } \right),$$

and for $\theta$ =${\theta }^{*}$:

$$\rho \tau + \left( {\sigma \tau - \theta \alpha } \right)g = \left( {1 - \alpha } \right)\left[ {\left( {\rho + \sigma g} \right)\tau + 1 - \tau } \right] > 0.$$

Hence, $\frac{{d}^{2}g}{d{\theta }^{2}}<0$ at $\theta$ =${\theta }^{*}$.

Appendix F

Using Eqs. (8) and (12), we have:

$$\begin{gathered} g = \left[ {\tau - \tau \left\{ {\alpha + s^{*} \left( {1 - \alpha } \right)} \right\} + \left. {s^{*} \left( {1 - \alpha } \right)} \right]} \right.^{1 - \alpha } \left[ {1 - \tau + \tau \frac{\alpha }{{s^{*} }} + \left( {\tau - 1} \right)} \right.\left. {\left( {1 - \alpha } \right)} \right]^{\alpha } s^{*\alpha } , \hfill \\ \Rightarrow g = \left[ {\tau \left( {1 - \alpha } \right)} \right. + \left. {s^{*} \left( {1 - \alpha } \right)\left( {1 - \tau } \right)} \right]^{1 - \alpha } \left[ {\tau \frac{\alpha }{s} + \left( {1 - \tau } \right)\alpha } \right]s^{*\alpha } , \hfill \\ \Rightarrow g = \left( {1 - \alpha } \right)^{1 - \alpha } \left[ \tau \right. + \left. {s^{*} \left( {1 - \tau } \right)} \right]^{1 - \alpha } \alpha^{\alpha } \left[ {\tau + \left( {1 - \tau } \right)} \right.\left. {s^{*} } \right]^{\alpha } , \hfill \\ \Rightarrow g = \left( {1 - \alpha } \right)^{1 - \alpha } \alpha^{\alpha } \left( {\tau + s^{*} \left( {1 - \tau } \right)} \right), \hfill \\ \Rightarrow g = \left( {1 - \alpha } \right)^{1 - \alpha } \alpha^{\alpha } \left( {s^{*} + \tau \left( {1 - s^{*} } \right)} \right). \hfill \\ \end{gathered}$$

(F1)

Using Eqs. (F1) and (9), we have:

$$\begin{gathered} g = \left( {1 - \alpha } \right)^{1 - \alpha } \alpha^{\alpha } \left( {\frac{\alpha g}{{\sigma g + \rho }} + \tau \frac{{\left( {\sigma - \alpha } \right)g + \rho }}{\sigma g + \rho }} \right), \hfill \\ \Rightarrow g\left( {\sigma g + \rho } \right) = \left( {1 - \alpha } \right)^{1 - \alpha } \alpha^{\alpha } \left[ {\alpha g + \left( {\sigma - \alpha } \right)g\tau + \tau \rho } \right], \hfill \\ \Rightarrow \sigma g + \rho = \left( {1 - \alpha } \right)^{1 - \alpha } \alpha^{\alpha } \left[ {\alpha + \left( {\sigma - \alpha } \right)\tau + \left( {\frac{\rho }{g}} \right)\tau } \right], \hfill \\ \Rightarrow \sigma g + \rho = \widehat{M}\left[ {\alpha + \left( {\sigma - \alpha } \right)\tau + \left( {\frac{\rho }{g}} \right)\tau } \right]. \hfill \\ \end{gathered}$$

(F2)

Here, $\widehat{M}={\left(1-\alpha \right)}^{1-\alpha }{\alpha }^{\alpha }>0$.

From Eq. (F2), we have:

$$\begin{gathered} \sigma dg = \widehat{M}\left( {\sigma - \alpha } \right)d\tau + \rho \frac{gd\tau - \tau dg}{{g^{2} }}\widehat{M}, \hfill \\ \Rightarrow \left( {\sigma + \frac{\rho \tau }{{g^{2} }}\widehat{M}} \right)dg = \widehat{M}\left[ {\left( {\sigma - \alpha } \right) + \frac{\rho }{g}} \right]d\tau , \hfill \\ \Rightarrow \frac{dg}{{d\tau }} = \frac{{\widehat{M}\left[ {\left( {\sigma - \alpha } \right) + \frac{\rho }{g}} \right]}}{{\sigma + \frac{\rho \tau }{{g^{2} }}\widehat{M}}} > 0\quad {\text{if}}\;\sigma \ge \alpha . \hfill \\ \end{gathered}$$

Appendix G

$$\begin{gathered} \frac{{[1 + (\sigma - \alpha ) + \tfrac{\rho }{g}]g^{2} }}{\tau \rho } > \frac{{\widehat{M}[(\sigma - \alpha ) + \tfrac{\rho }{g}]}}{{\sigma + \tfrac{\rho \tau }{{g^{2} }}\widehat{M}}}, \hfill \\ \Rightarrow \frac{{1 + \left( {\sigma - \alpha } \right) + \frac{\rho }{g}}}{\tau \rho } > \frac{{\widehat{M}\left[ {\sigma - \alpha + \frac{\rho }{g}} \right]}}{{\sigma g^{2} + \tau \rho \widehat{M}}}, \hfill \\ \Rightarrow \left( {1 + \left( {\sigma - \alpha } \right) + \frac{\rho }{g}} \right)\sigma g^{2} + \left( {1 + \left( {\sigma - \alpha } \right) + \frac{\rho }{g}} \right)\tau \rho \widehat{M} > \widehat{M}\left( {\sigma - \alpha + \frac{\rho }{g}} \right)\tau \rho , \hfill \\ \Rightarrow \left( {1 + \left( {\sigma - \alpha } \right) + \frac{\rho }{g}} \right)\sigma g^{2} > \tau \rho \left[ {\widehat{M}\left( {\sigma - \alpha + \frac{\rho }{g}} \right) - \widehat{M}\left( {1 + \left( {\sigma - \alpha } \right) + \frac{\rho }{g}} \right)} \right], \hfill \\ \Rightarrow \left( {1 + \left( {\sigma - \alpha } \right) + \frac{\rho }{g}} \right)\sigma g^{2} > - \tau \rho \widehat{M}. \hfill \\ \end{gathered}$$

Here, LHS is positive, and RHS is always negative. Therefore, this inequality is always true.

Appendix H

When $g$ is maximized simultaneously with respect to $\tau$ and $\theta$, the two first-order conditions are given by:

$$\tau^{*} = \left( {1 - \alpha } \right)\left( {1 + \theta^{*} } \right) + \alpha \theta^{*} s^{*}$$

(H1)

and

$$\tau^{*} = \frac{{\left( {1 - \alpha } \right)\left( {1 + \theta^{*} } \right)s^{*} + \alpha \theta^{*} s^{*} }}{{\alpha + s^{*} \left( {1 - \alpha } \right)}}.$$

(H2)

Equation (H1) is same as Eq. (15) and Eq. (H2) is same as Eq. (16). These two are identical when ${s}^{*}=1$. However:

$$s^{*} = \frac{\alpha g}{{\sigma g + \rho }} < 1\quad {\text{if}}\;\sigma \ge \alpha .$$

Solution to these equations does not exist in that case because:

$$\left( {1 - \alpha } \right)\left( {1 + \theta^{*} } \right) + \alpha \theta^{*} s^{*} > \frac{{\left( {1 - \alpha } \right)\left( {1 + \theta^{*} } \right) + \alpha \theta^{*} s^{*} }}{{\alpha + s^{*} \left( {1 - \alpha } \right)}}$$

for any ${\theta }^{*}>0$ and for all ${s}^{*}$ satisfying $0<{s}^{*}<1$, and the proof is shown below.

From Eq. (H1), we have:

$$\begin{gathered} \tau^{*} - \theta^{*} = \left( {1 - \alpha } \right)\left( {1 + \theta^{*} } \right) + \theta^{*} \alpha s^{*} - \theta^{*} , \hfill \\ \Rightarrow \tau^{*} - \theta^{*} = 1 - \alpha - \alpha \theta^{*} + \alpha \theta^{*} s^{*} . \hfill \\ \end{gathered}$$

(H3)

From Eq. (H2), we have:

$$\begin{gathered} \tau^{*} - \theta^{*} = \frac{{\left( {1 - \alpha } \right)\left( {1 + \theta^{*} } \right)s^{*} + \alpha \theta^{*} s^{*} }}{{\alpha + s^{*} \left( {1 - \alpha } \right)}} - \theta^{*} , \hfill \\ \Rightarrow \tau^{*} - \theta^{*} = \frac{{(1 - \alpha )s^{*} + \theta^{*} s^{*} - \theta^{*} \alpha s^{*} + \alpha \theta^{*} s^{*} - \theta^{*} \alpha - \theta^{*} s^{*} (1 - \alpha )}}{{\alpha + s^{*} (1 - \alpha )}}, \hfill \\ \Rightarrow \tau^{*} - \theta^{*} = \frac{{\left( {1 - \alpha } \right)s^{*} - \theta^{*} \alpha + \theta^{*} \alpha s^{*} }}{{\alpha + s^{*} \left( {1 - \alpha } \right)}}. \hfill \\ \end{gathered}$$

(H4)

Here:

$$\begin{gathered} 1 - \alpha - \theta^{*} \alpha + \theta^{*} s^{*} \alpha > \frac{{\left( {1 - \alpha } \right)s^{*} - \theta^{*} \alpha + \theta^{*} \alpha s^{*} }}{{\alpha + s^{*} \left( {1 - \alpha } \right)}}, \hfill \\ \Rightarrow \left( {1 - \alpha } \right)\left[ {\alpha + s^{*} \left( {1 - \alpha } \right)} \right] - \theta^{*} \alpha \left( {1 - s^{*} } \right)\left[ {\alpha + s^{*} \left( {1 - \alpha } \right)} \right] > \left( {1 - \alpha } \right)s^{*} - \theta^{*} \alpha \left( {1 - s^{*} } \right), \hfill \\ \Rightarrow (1 - \alpha )[\alpha + s^{*} (1 - \alpha ) - s^{*} ] + \theta^{*} \alpha (1 - s^{*} )[1 - \alpha - s^{*} (1 - \alpha )] > 0, \hfill \\ \Rightarrow \left( {1 - \alpha } \right)\alpha \left( {1 - s^{*} } \right) + \theta^{*} \alpha \left( {1 - s^{*} } \right)\left( {1 - \alpha } \right)\left( {1 - s^{*} } \right) > 0. \hfill \\ \end{gathered}$$

This is always true for ${\theta }^{*}>0$, because $0<{s}^{*}<1$ and $0<\alpha <1$.

Hence, $\left({\tau }^{*}-{\theta }^{*}\right)$ given by (H3) exceeds $\left({\tau }^{*}-{\theta }^{*}\right)$ given by (H4) for any ${\theta }^{*}>0$.

Appendix I

Using Eqs. (1), (18), and (19) and defining $x=\frac{G}{K}$, we find that in the long-run equilibrium:

$$\frac{{\dot{K}}}{K} = g = \left( {1 - \tau } \right)s^{*} x^{1 - \alpha } + \left( {\frac{\tau \beta }{K}} \right)$$

(I1)

$$\frac{{\dot{G}}}{G} = g = \tau x^{ - \alpha } - \left( {\frac{\tau \beta }{G}} \right)$$

(I2)

and

$$\left( {\rho + \sigma g} \right) = \left( {1 - \tau } \right)\alpha x^{1 - \alpha } .$$

(I3)

Here, $K$ and $G$ grow at the rate, $g>0$.

Hence:

$$K = K(0)e^{g} t,$$

and

$$G = G\left( 0 \right)e^{gt} .$$

Therefore, as $t\to \infty$, $\frac{1}{K}=\frac{1}{G}\to 0$. Since long-run equilibrium is attained at infinite time, from Eqs. (I1) and (I2), we have in the long-run equilibrium:

$$\frac{{\dot{K}}}{K} = g = \left( {1 - \tau } \right)s^{*} x^{1 - \alpha }$$

(I4)

and

$$\frac{{\dot{G}}}{G} = g = \tau x^{ - \alpha } .$$

(I5)

Using these two equations, we have:

$$x = \frac{\tau }{{\left( {1 - \tau } \right)s^{*} }}$$

(I6)

and

$$g = \tau^{1 - \alpha } \left( {1 - \tau } \right)^{\alpha } s^{*\alpha } .$$

(17)

This Eq. (I7) is identical to Eq. (21) in the body of the paper.

Using Eqs. (I4) and (I3), we have:

$$\begin{gathered} \frac{g}{\sigma g + \rho } = \frac{{\left( {1 - \tau } \right)s^{*} x^{1 - \alpha } }}{{\left( {1 - \tau } \right)\alpha x^{1 - \alpha } }} \hfill \\ \Rightarrow s^{*} = \frac{\alpha g}{{\sigma g + \rho }}. \hfill \\ \end{gathered}$$

(I8)

This Eq. (I8) is identical to Eq. (10) in the body of the paper.