Tax progressivity and social welfare with a continuum of inequality views

Abstract

We develop a framework for analysing the loss of social welfare due to taxation, where tax progressivity and regressivity, measured by the Kakwani index, are given interpretation in terms of welfare gain and loss, respectively. Our framework generalises the framework of Kakwani and Son (J Econ Inequal 19:185–212, 2021), by taking account of the intermediate inequality views, a continuum of combinations of the relative view (equal relative change of all incomes does not change inequality) and the absolute view (equal absolute change of all incomes does not change inequality). The welfare loss of taxation is decomposed into three components, one being a generalised Kakwani index, which accommodates the intermediate inequality views. We show that for a progressive (regressive) tax, moving closer to the relative view reduces (increases) the importance of progressivity (regressivity) for the welfare impact of the tax. The perception of the importance of progressivity and regressivity is thus affected by the inequality view taken. The empirical application considers the welfare loss of taxation in Croatia.

Appendix

1.1 A.1 Proof

Here we prove that $D_{{t_{\alpha } }} = \alpha G_{x}$. By analogy with (7b) in Sect. 2.1, $D_{{t_{\alpha } }}$ is given by

$$\begin{array}{*{20}c} {D_{{t_{\alpha } }} = 1 - \frac{{\Psi_{{t_{\alpha } }} }}{{\mu_{{t_{\alpha } }} }},} \\ \end{array}$$

(22)

where

$$\begin{aligned} \Psi_{{t_{\alpha } }} & = \mathop \smallint \limits_{0}^{\infty } t_{\alpha } \left( x \right)\omega \left( {F\left( x \right),\rho } \right)f\left( x \right){\text{d}}x = \mathop \smallint \limits_{0}^{\infty } \left[ {\alpha \tau x + \left( {1 - \alpha } \right)\tau \mu_{x} } \right]\omega \left( {F\left( x \right),\rho } \right)f\left( x \right){\text{d}}x \\ & = \alpha \tau \underbrace {{\mathop \smallint \limits_{0}^{\infty } x\omega \left( {F\left( x \right),\rho } \right)f\left( x \right){\text{d}}x}}_{{ = W_{x} }} + \left( {1 - \alpha } \right)\tau \mu_{x} \underbrace {{\mathop \smallint \limits_{0}^{\infty } \omega \left( {F\left( x \right),\rho } \right)f\left( x \right){\text{d}}x}}_{ = 1} \\ & \begin{array}{*{20}c} { = \alpha \tau W_{x} + \left( {1 - \alpha } \right)\tau \mu_{x} .} \\ \end{array} \\ \end{aligned}$$

(23)

Plugging (23) into (22) and using the fact that $\mu_{t\alpha } = \mu_{t}$, we get

$$\begin{aligned} D_{{t_{\alpha } }} & = 1 - \frac{{\alpha \tau W_{x} + \left( {1 - \alpha } \right)\tau \mu_{x} }}{{\mu_{{t_{\alpha } }} }} = 1 - \frac{{\alpha \tau W_{x} + \left( {1 - \alpha } \right)\tau \mu_{x} }}{{\mu_{t} }} = 1 - \frac{{\alpha \tau W_{x} + \tau \mu_{x} - \alpha \tau \mu_{x} }}{{\tau \mu_{x} }} \\ & \begin{array}{*{20}c} { = \frac{{ - \alpha \tau W_{x} + \alpha \tau \mu_{x} }}{{\tau \mu_{x} }} = \alpha \left( {1 - \frac{{W_{x} }}{{\mu_{x} }}} \right) = \alpha G_{x} ,} \\ \end{array} \\ \end{aligned}$$

(24)

where the last equality is due to (6a) in Sect. 2.1. ■

1.2 A.2 Proof

Here we prove that

$$\begin{array}{*{20}c} {\frac{1}{{\mu_{t} }}\left( {\Omega_{\alpha } - W_{x} } \right) = - \left( {1 - \alpha G_{x} } \right),} \\ \end{array}$$

(25)

$$\begin{array}{*{20}c} {\frac{1}{{\mu_{t} }}\left( {\Psi_{y} - \Omega_{\alpha } } \right) = D_{t} - \alpha G_{x} ,} \\ \end{array}$$

(26)

$$\begin{array}{*{20}c} {\frac{1}{{\mu_{t} }}\left( {W_{y} - \Psi_{y} } \right) = \frac{{\mu_{y} }}{{\mu_{t} }}\left( {D_{y} - G_{y} } \right).} \\ \end{array}$$

(27)

We first show that (25) holds:

$$\begin{aligned} \frac{1}{{\mu_{t} }}\left( {\Omega_{\alpha } - W_{x} } \right) = & \frac{1}{{\mu_{t} }}\left( {\mathop \smallint \limits_{0}^{\infty } y_{\alpha } \left( x \right)\omega \left( {F\left( x \right),\rho } \right)f\left( x \right)dx - \mathop \smallint \limits_{0}^{\infty } x\omega \left( {F\left( x \right),\rho } \right)f\left( x \right)dx} \right) \\ = & \frac{1}{{\mu_{t} }}\left( {\mathop \smallint \limits_{0}^{\infty } \left[ {x - t_{\alpha } \left( x \right)} \right]\omega \left( {F\left( x \right),\rho } \right)f\left( x \right)dx - \mathop \smallint \limits_{0}^{\infty } x\omega \left( {F\left( x \right),\rho } \right)f\left( x \right)dx} \right) \\ = & - \frac{1}{{\mu_{t} }}\mathop \smallint \limits_{0}^{\infty } t_{\alpha } \left( x \right)\omega \left( {F\left( x \right),\rho } \right)f\left( x \right)dx = - \frac{1}{{\mu_{t} }}\Psi_{t\alpha } = - \frac{1}{{\mu_{t} }}\mu_{{t_{\alpha } }} \left( {1 - D_{{t_{\alpha } }} } \right) = - \left( {1 - \alpha G_{x} } \right), \\ \end{aligned}$$

(28)

where the last two equalities are due to Eqs. (22) and (24), respectively. Next, we show that (26) holds:

$$\begin{aligned} \frac{1}{{\mu_{t} }}(\Psi_{y} - \Omega_{\alpha } ) = & \frac{1}{{\mu_{t} }}\left( {\mathop \smallint \limits_{0}^{\infty } y\left( x \right)\omega \left( {F\left( x \right),\rho } \right)f\left( x \right){\text{d}}x - \mathop \smallint \limits_{0}^{\infty } y_{\alpha } \left( x \right)\omega \left( {F\left( x \right),\rho } \right)f\left( x \right){\text{d}}x} \right) \\ = & \frac{1}{{\mu_{t} }}\mathop \smallint \limits_{0}^{\infty } \left[ {y\left( x \right) - \left( {x - t_{\alpha } \left( x \right)} \right)} \right]\omega \left( {F\left( x \right),\rho } \right)f\left( x \right){\text{d}}x \\ = & \frac{1}{{\mu_{t} }}\left( { - \mathop \smallint \limits_{0}^{\infty } t\left( x \right)\omega \left( {F\left( x \right),\rho } \right)f\left( x \right){\text{d}}x + \mathop \smallint \limits_{0}^{\infty } t_{\alpha } \left( x \right)\omega \left( {F\left( x \right),\rho } \right)f\left( x \right){\text{d}}x} \right) \\ = & \frac{1}{{\mu_{t} }}\mathop \smallint \limits_{0}^{\infty } \left[ { - t\left( x \right) + t_{\alpha } \left( x \right)} \right]\omega \left( {F\left( x \right),\rho } \right)f\left( x \right){\text{d}}x = \frac{1}{{\mu_{t} }}\left( { - \Psi_{t} + \Psi_{{t_{\alpha } }} } \right) \\ = & \frac{1}{{\mu_{t} }}\left[ { - \mu_{t} \left( {1 - D_{t} } \right) + \mu_{{t_{\alpha } }} \left( {1 - D_{{t_{\alpha } }} } \right)} \right] = D_{t} - D_{{t_{\alpha } }} = D_{t} - \alpha G_{x} \\ \end{aligned}$$

(29)

where the last three equalities are due to Eqs. (7b), (22), (24), and the fact that $\mu_{t\alpha } = \mu_{t}$. Finally, we show that (27) holds:

$$\begin{array}{*{20}c} {\frac{1}{{\mu_{t} }}\left( {W_{y} - \Psi_{y} } \right) = \frac{1}{{\mu_{t} }}\left[ {\mu_{y} \left( {1 - G_{y} } \right) - \mu_{y} \left( {1 - D_{y} } \right)} \right] = \frac{{\mu_{y} }}{{\mu_{t} }}\left( {D_{y} - G_{y} } \right),} \\ \end{array}$$

(30)

where the first equality is due to Eqs. (6b) and (7a). ■

1.3 A.3 Proof

Here we prove that an $\alpha$-neutral tax cannot cause reranking. Without loss of generality, consider any two individuals, called $A$ and $B$, with the pre-tax incomes $x_{A}$ and $x_{B}$ such that $x_{A} \ge x_{B}$. Suppose an $\alpha$-neutral tax is levied, with the mean tax liability $\mu_{t}$ and the tax ratio $\tau \equiv \mu_{t} /\mu_{x}$. Their respective post-tax incomes are:

$$\begin{array}{*{20}c} {y_{i} = x_{i} - \alpha \tau x_{i} - \left( {1 - \alpha } \right)\mu_{t} , i = A,B.} \\ \end{array}$$

(31)

Now suppose that the tax reranks $A$ and $B$ so that $y_{A} < y_{B}$. We have:

$$\begin{array}{*{20}c} {x_{A} - \alpha \tau x_{A} - \left( {1 - \alpha } \right)\mu_{t} < x_{B} - \alpha \tau x_{B} - \left( {1 - \alpha } \right)\mu_{t} } \\ \end{array}$$

(32)

$$x_{A} \left( {1 - \alpha \tau } \right) < x_{B} \left( {1 - \alpha \tau } \right)$$

$$x_{A} < x_{B}$$

which contradicts $x_{A} \ge x_{B}$. ■

1.4 A.4 Proof

Here we prove that ${\Delta }W \le 0$. Suppose the opposite holds:

$$\begin{gathered} \begin{array}{*{20}c} {\Delta W = N\left( \alpha \right) + P\left( \alpha \right) + H > 0} \\ \end{array} \hfill \\ - \left( {1 - \alpha G_{x} } \right) + \left( {D_{t} - \alpha G_{x} } \right) + \frac{{\mu_{y} }}{{\mu_{t} }}\left( {D_{y} - G_{y} } \right) > 0 \hfill \\ \end{gathered}$$

(33)

$$\begin{array}{*{20}c} {D_{t} - 1 + \frac{{\mu_{y} }}{{\mu_{t} }}\left( {D_{y} - G_{y} } \right) > 0.} \\ \end{array}$$

(34)

By definition of the concentration coefficient, $D_{t} \in \left[ { - 1,1} \right]$; thus, $D_{t} - 1 \le 0$. K&S (Appendix, Theorem 2) prove that $\Psi_{y} \ge W_{y}$, which implies $D_{y} \le G_{y}$; thus, $\frac{{\mu_{y} }}{{\mu_{t} }}\left( {D_{y} - G_{y} } \right) \le 0$. Therefore, (33) cannot hold. ■

1.5 A.5 Proof

Here we prove that ${\Delta }W = 0$ only if $D_{t} = 1$ and $H = 0$. Suppose that ${\Delta }W = 0$. If so, then the left-hand side of (34) is equal to zero:

$$\begin{array}{*{20}c} {D_{t} - 1 + \frac{{\mu_{y} }}{{\mu_{t} }}\left( {D_{y} - G_{y} } \right) = 0,} \\ \end{array}$$

(35)

which is true only if $D_{t} = 1$ and $H: = \frac{{\mu_{y} }}{{\mu_{t} }}\left( {D_{y} - G_{y} } \right) = 0$. ■

1.6 A.6 Proof

Here we prove that $\partial \pi \left( \alpha \right)/\partial \alpha < 0$, except if $D_{t} = 1$.

$$\begin{aligned} \frac{\partial \pi \left( \alpha \right)}{{\partial \alpha }} = & \frac{\partial }{\partial \alpha }\left( {\frac{P\left( \alpha \right)}{{\left| {N\left( \alpha \right)} \right|}}} \right) = \frac{{\frac{\partial P\left( \alpha \right)}{{\partial \alpha }}\left| {N\left( \alpha \right)} \right| - P\left( \alpha \right)\frac{{\partial \left| {N\left( \alpha \right)} \right|}}{\partial \alpha }}}{{\left| {N\left( \alpha \right)} \right|^{2} }} = \frac{{ - G_{x} \left| {N\left( \alpha \right)} \right| - P\left( \alpha \right)\frac{N\left( \alpha \right)}{{\left| {N\left( \alpha \right)} \right|}}\frac{\partial N\left( \alpha \right)}{{\partial \alpha }}}}{{\left| {N\left( \alpha \right)} \right|^{2} }} \\ = & \frac{{ - G_{x} \left| {N\left( \alpha \right)} \right| - P\left( \alpha \right)\frac{N\left( \alpha \right)}{{\left| {N\left( \alpha \right)} \right|}}G_{x} }}{{\left| {N\left( \alpha \right)} \right|^{2} }} = \frac{{ - G_{x} \left| {N\left( \alpha \right)} \right| - P\left( \alpha \right)\left( { - 1} \right)G_{x} }}{{\left| {N\left( \alpha \right)} \right|^{2} }} = \frac{{G_{x} \left( {P\left( \alpha \right) - \left| {N\left( \alpha \right)} \right|} \right)}}{{\left| {N\left( \alpha \right)} \right|^{2} }} \\ = & \frac{{G_{x} \left( {D_{t} - \alpha G_{x} - \left| { - \left( {1 - \alpha G_{x} } \right)} \right|} \right)}}{{\left| {N\left( \alpha \right)} \right|^{2} }} = \frac{{G_{x} \left( {D_{t} - \alpha G_{x} - \left( {1 - \alpha G_{x} } \right)} \right)}}{{\left| {N\left( \alpha \right)} \right|^{2} }} = \frac{{G_{x} \left( {D_{t} - 1} \right)}}{{\left| {N\left( \alpha \right)} \right|^{2} }}.\# \left( {36} \right) \\ \end{aligned}$$

(36)

Assuming $G_{x} > 0$, the derivative cannot be positive, as that would require $D_{t} > 1$, which cannot be true because by definition of the concentration coefficient, $D_{t} \in \left[ { - 1,1} \right]$. The fraction can be either negative, when $D_{t} < 1$, or zero, when $D_{t} = 1$. ■

1.7 A.7 Proof

Here we prove that $\partial \eta \left( \alpha \right)/\partial \alpha < 0$, except if $H = 0$.

$$\begin{array}{*{20}c} {\frac{\partial \eta \left( \alpha \right)}{{\partial \alpha }} = \frac{\partial }{\partial \alpha }\left( {\frac{H}{{\left| {N\left( \alpha \right)} \right|}}} \right) = \frac{{\frac{\partial H}{{\partial \alpha }}\left| {N\left( \alpha \right)} \right| - H\frac{{\partial \left| {N\left( \alpha \right)} \right|}}{\partial \alpha }}}{{\left| {N\left( \alpha \right)} \right|^{2} }} = \frac{{ - H\frac{{\partial \left( {1 - \alpha G_{x} } \right)}}{\partial \alpha }}}{{\left| {N\left( \alpha \right)} \right|^{2} }} = \frac{{ - H\left( { - 1} \right)G_{x} }}{{\left| {N\left( \alpha \right)} \right|^{2} }} = \frac{{HG_{x} }}{{\left| {N\left( \alpha \right)} \right|^{2} }}.} \\ \end{array}$$

(37)

Assuming, $G_{x} > 0$, the derivative is negative if $H < 0$, and zero if $H = 0$. ■

1.8 A.8 Proof

Here we prove that $\tau /\tau^{*} \left( \alpha \right) = 1/\left( { - \delta \left( \alpha \right)} \right)$. Multiplying Eq. (18a) by $\mu_{t}$, we obtain

$$\begin{array}{*{20}c} \begin{aligned} \mu_{t} \Delta W = & \mu_{t} N\left( \alpha \right) + \mu_{t} P\left( \alpha \right) + \mu_{t} H \\ W_{y} - W_{x} = & - \mu_{t} \left( {1 - \alpha G_{x} } \right) + \mu_{t} \left( {D_{t} - \alpha G_{x} } \right) + \mu_{y} \left( {D_{y} - G_{y} } \right). \\ \end{aligned} \\ \end{array}$$

(38)

The welfare loss $W_{y} - W_{x}$ can as well be brought about by an $\alpha$-neutral tax. By definition of $\alpha$-neutral tax, this tax must be such that the average tax liability, $\mu_{t}^{*} \left( \alpha \right)$, is implicitly given by

$$\begin{array}{*{20}c} {W_{y} - W_{x} = - \mu_{t}^{*} \left( \alpha \right)\left( {1 - \alpha G_{x} } \right).} \\ \end{array}$$

(39)

Thus, combining (38) and (39),

$$\begin{array}{*{20}c} { - \mu_{t}^{*} \left( \alpha \right)\left( {1 - \alpha G_{x} } \right) = - \mu_{t} \left( {1 - \alpha G_{x} } \right) + \mu_{t} \left( {D_{t} - \alpha G_{x} } \right) + \mu_{y} \left( {D_{y} - G_{y} } \right).} \\ \end{array}$$

(40)

Dividing (40) by $- \mu_{x} \left( {1 - \alpha G_{x} } \right)$, we get:

$$\begin{aligned} \frac{{\mu_{t}^{*} \left( \alpha \right)}}{{\mu_{x} }} = & \frac{{\mu_{t} }}{{\mu_{x} }} + \frac{{\mu_{t} }}{{\mu_{x} }}\frac{{D_{t} - \alpha G_{x} }}{{ - \left( {1 - \alpha G_{x} } \right)}} + \frac{{\mu_{y} }}{{\mu_{x} }}\frac{{D_{y} - G_{y} }}{{ - \left( {1 - \alpha G_{x} } \right)}} = \tau - \tau \frac{{D_{t} - \alpha G_{x} }}{{1 - \alpha G_{x} }} - \tau \frac{{\mu_{y} }}{{\mu_{t} }}\frac{{D_{y} - G_{y} }}{{1 - \alpha G_{x} }} \\ = & \tau - \tau \frac{P\left( \alpha \right)}{{\left| {N\left( \alpha \right)} \right|}} - \tau \frac{H}{{\left| {N\left( \alpha \right)} \right|}} = \tau \left( {1 - \pi \left( \alpha \right) - \eta \left( \alpha \right)} \right) = - \left( { - 1 + \pi \left( \alpha \right) + \eta \left( \alpha \right)} \right)\tau = - \delta \left( \alpha \right)\tau . \\ \end{aligned}$$

(41)

Denoting $\mu_{t}^{*} \left( \alpha \right)/\mu_{x} : = \tau^{*} \left( \alpha \right)$, a straightforward reorganisation of (41) gives (21). ■

1.9 A.9 Additional tables

See Tables 2 and 3.

Table 2 Data plotted on panel A of Fig. 1

Table 3 Data plotted on Figs. 5 and 6