Private provision of public good and endogenous income inequality

Abstract

We model a simple economy with privately provided public good and with two groups of individuals—rich and poor. Only rich people contribute to the public good and they own capital input that is used to produce varieties of private goods under monopolistically competitive market structure. Income inequality is endogenously determined in the unique Nash equilibrium of the economy. We show that government intervention in the form of higher tax rate and/or larger transfer of the tax-revenue from rich to poor would be able to reduce the degree of income inequality. However, the impact of such a policy on the aggregate provision level of the public good is, in general, ambiguous. Income inequality in the market economy is too high and there will be under provision of the public good compared to the planner’s economy even if the planner is only interested in maximizing the welfare of the rich people. We derived the optimal capital and wage income tax rate that implement the first best solution.

Appendices

Appendices

Appendix I

Here we are to show that $\frac{\partial {(Gini)}}{\partial {\delta }}>0$. For $t\ne 0$, the expression of Gini obtained from Eq. (23) is given by

$$\begin{aligned} Gini = \frac{1-t+\delta t - \frac{H}{H+L} }{1+\frac{H+L}{rK-G}}. \end{aligned}$$

(I.1)

By assumption, $Gini\ge 0$. Using Eqs. (19) and (20), we have

$$\begin{aligned} rK-G= & {} \frac{ (1-\theta )(H+L)\delta t + \{ (1-\theta )(H+L)(1-t) H - H \} }{ (1+\theta )\delta t + (1-t)(1+H\theta ) } \nonumber \\ \Rightarrow \; \frac{rK-G}{H+L}= & {} \frac{ (1-\theta )\delta t + \{ (1-\theta )(1-t) H - \frac{H}{H+L} \} }{ (1+\theta )\delta t + (1-t)(1+H\theta ) } \nonumber \\ \Rightarrow \; \frac{H+L}{rK-G}= & {} \frac{ (1+\theta )\delta t + (1-t)(1+H\theta ) }{ (1-\theta )\delta t + \{ (1-\theta )(1-t) H - \frac{H}{H+L} \} } \nonumber \\ \Rightarrow \; 1+ \frac{H+L}{rK-G}= & {} 1+ \frac{ (1+\theta )\delta t + (1-t)(1+H\theta ) }{ (1-\theta )\delta t + \{ (1-\theta )(1-t) H - \frac{H}{H+L} \} } \nonumber \\ \Rightarrow \; 1+ \frac{H+L}{rK-G}= & {} \frac{ 2\delta t + (1+H) (1-t) -\frac{H}{H+L} }{ (1-\theta )\delta t + (1-\theta )(1-t) H - \frac{H}{H+L} }. \end{aligned}$$

(I.2)

Using the above expression and the expression of Gini, we have

$$\begin{aligned} Gini = \frac{\left( 1-t+\delta t - \frac{H}{H+L} \right) \left[ (1-\theta )\delta t + (1-\theta )(1-t) H - \frac{H}{H+L} \right] }{ 2\delta t + (1+H) (1-t) -\frac{H}{H+L}} \end{aligned}$$

(I.3)

We introduce the following notations:

$$\begin{aligned} T_1= & {} 1-t+\delta t - \frac{H}{H+L};\\ T_2= & {} (1-\theta )\delta t + (1-\theta )(1-t) H - \frac{H}{H+L}; \end{aligned}$$

and

$$\begin{aligned} T_3=2\delta t + (1+H) (1-t) -\frac{H}{H+L}. \end{aligned}$$

Then,

$$\begin{aligned} Gini=\frac{T_1T_2}{T_3}. \end{aligned}$$

So, we have the following,

$$\begin{aligned} \frac{T_3}{T_1}= & {} \frac{2\delta t + (1+H) (1-t) -\frac{H}{H+L} }{ 1-t+\delta t - \frac{H}{H+L} } \nonumber \\{} & {} \quad \Rightarrow \quad \frac{1}{2}\frac{T_3}{T_1} = \frac{2\delta t + (1+H) (1-t) -\frac{H}{H+L} }{ 2\delta t +2(1-t) - \frac{2H}{H+L}} \nonumber \\{} & {} \quad \Rightarrow \quad \frac{1}{2}\frac{T_3}{T_1} -1 = \frac{ (H-1)(1-t) + \frac{H}{H+L}}{2\delta t +2(1-t) - \frac{2H}{H+L} } >0. \end{aligned}$$

(I.4)

The expression $\left( \frac{1}{2}\frac{T_3}{T_1} -1\right)$ is clearly a decreasing function of $\delta$. Hence, $\frac{T_3}{T_1}$ falls as $\delta$ increases. Also $T_2$ is an increasing function of $\delta$. Therefore, Gini is increased when $\delta$ is increased.

Appendix II

Here, we are to show that

$$\begin{aligned} \frac{\partial {G}}{\partial {\delta }}{<}{(}{=}{)}{>}0, \quad \text{ iff, }\quad \frac{(H+L)(H-1)}{H} {<}{(}{=}{)}{>} \frac{1+\theta }{(1-\theta )(1-t)}. \end{aligned}$$

We write the expression of G obtained from Eq. (19) as follows:

$$\begin{aligned} G= & {} \frac{[(1-t) (H+L) (1-\theta ) + H\theta ] + \delta t (H+L)(1-\theta ) }{ (1-t)(1+H\theta ) +\delta t (1+\theta ) }\nonumber \\{} & {} \quad {}\Rightarrow \quad G = \frac{a_1+\delta b_1}{c_1+\delta d_1}; \end{aligned}$$

(II.1)

where

$$\begin{aligned} a_1= & {} (1-t) (H+L) (1-\theta ) + H\theta ;\\ b_1= & {} t (H+L)(1-\theta );\\ c_1= & {} (1-t)(1+H\theta ); \end{aligned}$$

and

$$\begin{aligned} d_1= t(1+\theta ). \end{aligned}$$

From Eq. (II.1), we obtain

$$\begin{aligned} \frac{\partial {G}}{\partial {\delta }} = \frac{\partial {}}{\partial {\delta }} \left( \frac{a_1+\delta b_1}{c_1+\delta d_1}\right) = \frac{b_1c_1-a_1d_1}{(c_1+\delta d_1 )^2}. \end{aligned}$$

(II.2)

Therefore,

$$\begin{aligned}{} & {} \frac{\partial {}}{\partial {\delta }} \left( \frac{a_1+\delta b_1}{c_1+\delta d_1}\right)< (=)> 0, \nonumber \\{} & {} \quad \text{ iff, } \quad b_1c_1<(=)>a_1d_1, \nonumber \\{} & {} \quad \Rightarrow \text{ iff, }\quad t (H+L)(1-\theta )(1-t)(1+H\theta )<(=)> [ (1-t) (H+L) (1-\theta ) + H\theta ]t(1+\theta ),\nonumber \\{} & {} \quad \Rightarrow \text{ iff, }\quad (H+L)(1-\theta )(1-t)[H\theta -\theta ]<(=)> H\theta (1+\theta ), \nonumber \\{} & {} \quad \Rightarrow \text{ iff, }\quad (H+L)(1-\theta )(1-t)(H-1)<(=)> H (1+\theta ), \nonumber \\{} & {} \quad \Rightarrow \text{ iff, }\quad \frac{(H+L)(H-1)}{H} <(=)> \frac{1+\theta }{(1-\theta )(1-t)}. \end{aligned}$$

(II.3)

This proves the result that

$$\begin{aligned} \frac{\partial {G}}{\partial {\delta }}<(=)>0, \quad \text{ iff, }\quad \frac{(H+L)(H-1)}{H} <(=)> \frac{1+\theta }{(1-\theta )(1-t)}. \end{aligned}$$

$\square$

Appendix III

Here, we are to show that $\frac{\partial {G}}{\partial {t}}>0$. Using Eq. (19), we write G as follows:

$$\begin{aligned} G= & {} \frac{H\theta + (H+L)(1-\theta ) - t(1-\delta )(H+L)(1-\theta )}{ 1+H\theta - t [1+H\theta -\delta -\delta \theta ] } \nonumber \\{} & {} \quad \Rightarrow G = \frac{a_2-t b_2}{c_2- t d_2}; \end{aligned}$$

(III.1)

where

$$\begin{aligned} a_2\equiv & {} H\theta + (H+L)(1-\theta );\\ b_2\equiv & {} (1-\delta )(H+L)(1-\theta );\\ c_2\equiv & {} 1+H\theta ; \end{aligned}$$

and

$$\begin{aligned} d_2\equiv 1+H\theta -\delta -\delta \theta . \end{aligned}$$

From Eq. (III.1), we get:

$$\begin{aligned} \frac{\partial {G}}{\partial {t}} = \frac{\partial {}}{\partial {t}} \left( \frac{a_2-t b_2}{c_2- t d_2} \right) = \frac{a_2d_2 - b_2c_2}{(c_2- t d_2 )^2}. \end{aligned}$$

(III.2)

Therefore,

$$\begin{aligned}{} & {} \frac{\partial {}}{\partial {t}} \left( \frac{a_2-t b_2}{c_2- t d_2} \right)> 0, \nonumber \\{} & {} \quad \text{ if, } \quad a_2d_2>b_2c_2,\nonumber \\{} & {} \quad \Rightarrow \text{ if, } \quad [H\theta + (H+L)(1-\theta )][1+H\theta -\delta -\delta \theta ]> (1-\delta )(H+L)(1-\theta ) (1+H\theta ), \nonumber \\{} & {} \quad \Rightarrow \text{ if, }\quad H\theta [1+H\theta -\delta -\delta \theta ]> (H+L)(1-\theta ) [(1-\delta )(1+H\theta ) - 1-H\theta +\delta +\delta \theta ], \nonumber \\{} & {} \quad \Rightarrow \text{ if, } \quad H\theta [1+H\theta -\delta -\delta \theta ]> (H+L)(1-\theta ) [ \delta \theta (1-H) ], \nonumber \\{} & {} \quad \Rightarrow \text{ if, } \quad H\theta [1-\delta +\theta (H-\delta )] > - (H+L)(1-\theta ) [ \delta \theta (H-1) ]. \end{aligned}$$

(III.3)

The left hand side of inequality (III.3) is always positive while its right hand side is always negative for $H\ge 1$. Hence, inequality (III.3) always holds true for all $H\ge 1$. This proves the result $\frac{\partial {G}}{\partial {t}}>0$. $\square$

Appendix IV

We are to show that $\frac{\partial {(Gini)}}{\partial {t}}<0$. We write Gini from Eq. (23) as follows:

$$\begin{aligned} Gini = \frac{1-t(1-\delta ) - \frac{H}{H+L} }{1+\frac{H+L}{rK-G}}. \end{aligned}$$

We take the expression of $\left( 1+ \frac{H+L}{rK-G}\right)$ as derived in Eq. (I.2) in appendix III, and express it as follows:

$$\begin{aligned} 1+ \frac{H+L}{rK-G}= & {} \frac{ 2\delta t + (1+H) (1-t) -\frac{H}{H+L} }{ (1-\theta )\delta t + (1-\theta )(1-t) H - \frac{H}{H+L} }, \nonumber \\{} & {} \quad \Rightarrow \quad 1+ \frac{H+L}{rK-G} = \frac{ \left[ 1+H-\frac{H}{H+L}\right] - t (1+H-2\delta )}{ \left[ (1-\theta )H-\frac{H}{H+L} \right] - t(1-\theta )(H-\delta ) } \nonumber \\{} & {} \quad \Rightarrow \quad 1+ \frac{H+L}{rK-G} = \frac{ a_3 - t b_3}{ c_3 - td_3 }. \end{aligned}$$

(IV.1)

where,

$$\begin{aligned} a_3\equiv & {} 1+H-\frac{H}{H+L};\\ b_3\equiv & {} 1+H-2\delta ;\\ c_3\equiv & {} (1-\theta )H-\frac{H}{H+L}; \end{aligned}$$

and

$$\begin{aligned} d_3\equiv (1-\theta )(H-\delta ). \end{aligned}$$

Differentiating the R.H.S. expression in Eq. (IV.1) with respect to t, we get;

$$\begin{aligned} \frac{\partial {}}{\partial {t}} \left( \frac{ a_3 - t b_3}{ c_3 - td_3 }\right) = \frac{a_3d_3-b_3c_3}{(c_3 - td_3 )^2}. \end{aligned}$$

(IV.2)

Therefore,

$$\begin{aligned}{} & {} \frac{\partial {}}{\partial {t}} \left( \frac{ a_3 - t b_3}{ c_3 - td_3 }\right)> 0, \nonumber \\{} & {} \quad \text{ iff, } \quad a_3d_3>b_3c_3,\nonumber \\{} & {} \quad \Rightarrow \quad \text{ iff, }\quad \left( 1+H-\frac{H}{H+L}\right) (1-\theta )(H-\delta )> (1+H-2\delta ) \left( (1-\theta )H-\frac{H}{H+L} \right) ,\nonumber \\{} & {} \quad \Rightarrow \quad \text{ iff, } \quad \frac{H}{H+L } \left[ 1-\delta +\theta (H -\delta ) \right] > - (1-\theta )\delta (H-1). \end{aligned}$$

(IV.3)

However, this inequality is always true since its left hand side is always positive and the right hand side is always negative for $H\ge 1$. Hence, $\frac{\partial {}}{\partial {t}} \left( \frac{ a_3 - t b_3}{ c_3 - td_3 }\right) >0$; and this implies that, $\frac{\partial {}}{\partial {t}} \left( 1+ \frac{H+L}{rK-G} \right) >0$. This further implies that $\frac{\partial {(Gini)}}{\partial {t}}<0$ because the numerator of the Gini expression is always a negative function of t. $\square$

Appendix V

Here we are to show that,

$$\begin{aligned} 0\le \frac{(H+L-\theta L)(H-1)}{(H+L-\theta L)(H-\delta ) - (1-\delta )\{L-\theta (H+L)\}}\le 1, \quad \text{ for } \text{ all }\;\; \delta \in [0,1]. \end{aligned}$$

First we show that it does not exceed unity.

$$\begin{aligned}{} & {} \frac{(H+L-\theta L)(H-1)}{(H+L-\theta L)(H-\delta ) - (1-\delta )\{L-\theta (H+L)\}} \le 1 \\{} & {} \quad \Rightarrow \quad (H+L-\theta L)(H-1) \le (H+L-\theta L)(H-\delta ) - (1-\delta )\{L-\theta (H+L)\}\\{} & {} \quad \Rightarrow \quad -(H+L-\theta L)(1-\delta ) \le -(1-\delta ) \{L-\theta (H+L)\}\\{} & {} \quad \Rightarrow \quad -(1-\delta ) \le \theta (1-\delta ). \end{aligned}$$

This last expression is always true for all $\delta \in [0,1]$. Next, we are to show that

$$\begin{aligned} \frac{(H+L-\theta L)(H-1)}{(H+L-\theta L)(H-\delta ) - (1-\delta )\{L-\theta (H+L)\}}\ge 0. \end{aligned}$$

Since the numerator of this expression is always non-negative for $H\ge 1$, we are to show that the denominator is positive. This implies that

$$\begin{aligned}{} & {} (H+L-\theta L)(H-\delta ) - (1-\delta )\{L-\theta (H+L)\}>0\\{} & {} \quad \Rightarrow (H+L) [H-\delta + (1-\delta )\theta ]>L [ 1-\delta + \theta (H-\delta ) ] \\{} & {} \quad \Rightarrow H [H-\delta + (1-\delta )\theta ]> L [ 1-\delta + \theta (H-\delta ) -H+\delta -\theta +\theta \delta ] \\{} & {} \quad \Rightarrow H [H-\delta + (1-\delta )\theta ]> L [ 1-\theta +\theta H -H ] \\{} & {} \quad \Rightarrow H [H-\delta + (1-\delta )\theta ]> L (1-\theta )(1-H) \\{} & {} \quad \Rightarrow H [H-\delta + (1-\delta )\theta ] > - L (1-\theta )(H-1). \end{aligned}$$

However, this is always true for $H\ge 1$ and $\delta \in [0,1]$. Hence, the proof follows. $\square$

Appendix VI

Here we show that $\frac{\partial {t^{opt}}}{\partial {\delta }}>0$.

Using Eqs. (19) and (28), we find that

$$\begin{aligned}{} & {} G^*=G^m\\{} & {} \quad \Rightarrow \frac{(1-t+\delta t)(H+L)(1-\theta )+H\theta }{(1-t)(1+H\theta )+\delta t (1+\theta )} = \frac{H+L(1-\theta )}{1+\theta }. \end{aligned}$$

With some rearrangement, we solve for $t=t^{opt}$ as follows:

$$\begin{aligned} t^{opt}=\frac{[H+L(1-\theta )]\theta (H-1)}{(H+L)\theta (H+\theta ) - L\theta (H\theta +1) - \delta (1+\theta )H\theta }. \end{aligned}$$

(VI.1)

When $\delta =1$, the denominator of the above expression becomes

$$\begin{aligned}{} & {} (H+L)\theta (H+\theta ) - \theta \left[ L(H\theta +1) - (1+\theta )H\right] \\{} & {} \quad = \theta (H-1)(H+L(1-\theta ))\ge 0. \end{aligned}$$

Therefore, we get $t^{opt}=1$ when evaluated at $\delta =1$. From the expression of $t^{opt}$ in (VI.1), we can see that $\delta$ only affect the denominator and an increase in $\delta$ lowers the denominator monotonically. Therefore, we can clearly state that

$$\begin{aligned} \frac{\partial {t^{opt}}}{\partial {\delta }}>0. \end{aligned}$$

$\square$