Optimal income taxation with Kalai wage bargaining and endogenous participation

Abstract

This paper characterizes the optimal redistributive tax schedule in a search–matching framework where (voluntary) nonparticipation and (involuntary) unemployment are endogenous and wages are determined by proportional bargaining à la Kalai. The optimal employment tax rate is given by an inverse elasticity rule. This rule depends on the global response of the employment rate, which depends not only on the participation (labor supply) responses, but also on the vacancy posting (labor demand) responses and on the product of these two responses. For plausible values of the parameters, our matching environment induces much lower employment tax rates than the usual competitive model with endogenous participation only. However, optimal employment tax rates are larger (in absolute value) when a given level of the global elasticity of employment is more due to search frictions and less due to participation responses.

Appendices

Appendix 1: Link between the elasticity of the labor demand and the elasticity of the matching function

Let $\mu _{a}\left( .\right)$ denote the elasticity of the matching function $M_{a}\left( .,.\right)$ with respect to the mass of job-seekers $U_{a}$. Because the matching function is increasing in both arguments and exhibits constant returns to scale, $\mu _{a}$ depends only on the level of tightness and one has $\mu _{a}\left( \theta \right) \in \left( 0,1\right)$ for all $\theta$. From $m_{a}\left( \theta \right) =M_{a}\left( 1,1/\theta \right)$, the elasticity of the probability of filling a vacancy to the tightness level (i.e. $\left( \theta _{a}/m_{a}\right) \left( \partial m_{a}\left( \theta \right) /\partial \theta _{a}\right)$) equals $-\mu _{a}\left( \theta \right)$. Hence the elasticity of the reciprocal $m_{a}^{-1}\left( .\right)$ equals $-1/\mu _{a}\left( m_{a}^{-1}\left( .\right) \right)$. The log-differentiation of the $L_{a}$ function (4) with respect to the firm’s surplus $a-w_{a}$ gives:

$$\begin{aligned} \frac{dL_{a}}{L_{a}}=\left( -1+\frac{1}{\mu _{a}\left( \theta _{a}\right) }\right) \cdot \frac{d\left( a-w_{a}\right) }{a-w_{a}} \end{aligned}$$

which leads to the second equality in (5). The inequality holds because $\mu _{a}\left( \theta \right) \in \left( 0,1\right)$.

Appendix 2: Proof of Proposition 1

The Lagrangian of the optimal tax problem is

$$\begin{aligned} \int \limits _{a_{0}}^{a_{1}}\mathcal L \left( \tau _{a},b,\lambda \right) \cdot dF\left( a\right) -\lambda b-\lambda R \end{aligned}$$

where

$$\begin{aligned}&\mathcal L \left( \tau _{a},b,\lambda \right) \overset{\text{ def }}{\equiv } \int \limits _{0}^{\gamma _{a}\cdot \left( a-\tau _{a}\right) \cdot L_{a}\left[ \left( 1-\gamma _{a}\right) \left( a-\tau _{a}\right) \right] }\Phi \left( \gamma _{a}\cdot \left( a-\tau _{a}\right) \cdot L_{a}\left[ \left( 1-\gamma _{a}\right) \left( a-\tau _{a}\right) \right] \right. \\&\quad \left. +b-\chi \right) \cdot dH\left( \chi \left| a\right. \right) \\&\quad +\Phi \left( b\right) \cdot \left( 1-H\left( \gamma _{a}\cdot \left( a-\tau _{a}\right) \cdot L_{a}\left[ \left( 1-\gamma _{a}\right) \left( a-\tau _{a}\right) \right] \left| a\right. \right) \right) \\&\quad +\lambda \cdot \tau _{a}\cdot L_{a}\left[ \left( 1-\gamma _{a}\right) \left( a-\tau _{a}\right) \right] \cdot H\left( \gamma _{a}\cdot \left( a-\tau _{a}\right) \cdot L_{a}\left[ \left( 1-\gamma _{a}\right) \left( a-\tau _{a}\right) \right] \left| a\right. \right) \end{aligned}$$

The first-order condition with respect to $b$ is:

$$\begin{aligned} \int \limits _{a_{0}}^{a_{1}}\left\{ \int \limits _{0}^{\gamma _{a}\cdot \left( a-\tau _{a}\right) \cdot L_{a}\left[ \left( 1-\gamma _{a}\right) \left( a-\tau _{a}\right) \right] }\Phi ^{\prime }\left( \Sigma _{a}+b-\chi \right) \cdot dH\left( \chi \left| a\right. \right) +\Phi ^{\prime }\left( b\right) \cdot \left( 1-h_{a}\right) \right\} dF\left( a\right) =\lambda \end{aligned}$$

Using (17) and (18) gives (19a). The first-order condition with respect to $\tau _{a}$ writes $0=\frac{\partial \mathcal L }{\partial \tau _{a}}\left( \tau _{a},b,\lambda \right)$. Using (3) and (5), this leads to:

$$\begin{aligned} 0&= -\gamma _{a}\cdot \left( 1+\eta _{a}^{D}\right) \cdot \ell _{a}\cdot \left( \int \limits _{0}^{\Sigma _{a}}\Phi ^{\prime }\left( \Sigma _{a}+b-\chi \right) \cdot dH\left( \chi \left| a\right. \right) \right) \\&\quad +\lambda \left\{ 1-\frac{\tau _{a}}{a-\tau _{a}}\eta _{a}^{D}-\frac{\tau _{a}}{a-\tau _{a}}\left( 1+\eta _{a}^{D}\right) \cdot \eta _{a}^{P}\right\} \ell _{a}\cdot h_{a} \end{aligned}$$

Dividing both sides by $\lambda h_{a}\ell _{a}=\lambda e_{a}$, using (17) and $w_{a}-\tau _{a}=\gamma _{a}\left( a-\tau _{a}\right)$ [from (7)] gives (19b).