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Procedural fairness and redistributive proportional tax

Abstract

We study the implications of procedural fairness on the share of income that should be redistributed. We formulate procedural fairness as a particular non-cooperative bargaining game and examine the stationary subgame perfect equilibria of the game. The equilibrium outcome is called tax equilibrium and is shown to be unique. The procedurally fair tax rate is defined as the tax rate that results in the limit of tax equilibria when the probability that negotiations break down converges to zero. The procedurally fair tax rate is shown to be unique. We also provide a characterization of the procedurally fair tax rate that involves the probability mass of below average income citizens and a particular measure of the citizens’ boldness. This characterization is then used to show that in a number of interesting cases, the procedurally fair tax rate equals the probability mass of below average income citizens.

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Notes

  1. A procedurally fair procedure should offer equal opportunities to all involved parties. This implies in particular that, in the absence of agreement, all citizens consume the same amount. We choose this amount to be zero to ensure that all citizens prefer any positive tax rate to the disagreement point.

  2. Suitable measurability conditions are that \( p : N \rightarrow [0,1] \) be \({{\mathcal {A}}}\)-measurable, and \( \cap _{t \in N} A_{t} \) be Borel-measurable.

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Correspondence to P. Jean-Jacques Herings.

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The authors would like to thank the Netherlands Organization for Scientific Research (NWO) for financial support.

Appendices

Appendix A The proof of theorem 5.2

Theorem 5.2 follows immediately from Lemmas 8.1 and 8.2 below.

Lemma 8.1

The function \(d_{1}\) is a strictly increasing function on \((0,1)\) assuming only finite values.

Proof

Consider some \(t \in N^{+}\) and some \( \beta \in (0,1). \) Using \(v_{t}(1) \ge 0\) and the concavity of \(v_{t}\), we find that

$$\begin{aligned} -(1 - \beta ) \partial _{-} v_{t}(\beta ) \le v_{t}(\beta ). \end{aligned}$$

Rearranging yields the inequality

$$\begin{aligned} d_{1t}(\beta ) \le \frac{1}{1 - \beta }. \end{aligned}$$

This establishes the finiteness of \(d_{1}\) on \( (0,1). \) We show that \(d_{1}\) is strictly increasing. Let \(\beta \) and \(\beta ^{\prime }\) be elements of \( (0,1) \) such that \(\beta < \beta ^{\prime }\). For \(t \in N^{+}\) we have

$$\begin{aligned} -\partial _{-} v_{t}(\beta ) (\beta ^{\prime } - \beta ) \le v_{t}(\beta ) - v_{t}(\beta ^{\prime }) \end{aligned}$$

We therefore obtain

$$\begin{aligned} \begin{array}{rcl} -\partial _{-} v_{t}(\beta )(v_{t}(\beta ^{\prime }) - \partial _{-} v_{t}(\beta ^{\prime }) (\beta ^{\prime } - \beta )) &{} \le &{} -\partial _{-} v_{t}(\beta ) v_{t}(\beta ^{\prime }) - \partial _{-} v_{t}(\beta ^{\prime })(v_{t}(\beta ) - v_{t}(\beta ^{\prime })) \\ &{} = &{} (\partial _{-} v_{t}(\beta ^{\prime }) -\partial _{-} v_{t}(\beta )) v_{t}(\beta ^{\prime }) - \partial _{-} v_{t}(\beta ^{\prime }) v_{t}(\beta ) \\ &{} \le &{} -\partial _{-} v_{t}(\beta ^{\prime }) v_{t}(\beta ), \end{array} \end{aligned}$$

where the last inequality uses \(v_{t}(\beta ^{\prime })\ge 0\) and concavity. Dividing both sides of the inequality by \(v_{t}(\beta ) v_{t}(\beta ^{\prime })\) gives

$$\begin{aligned} d_{1t}(\beta ) (1+(\beta ^{\prime }-\beta )d_{1t}(\beta ^{\prime })) \le d_{1t}(\beta ^{\prime }). \end{aligned}$$

Dividing by \(d_{1t}(\beta ^{\prime })\) yields the inequality

$$\begin{aligned} d_{1t}(\beta ) \le \frac{d_{1t}(\beta ^{\prime })}{1+(\beta ^{\prime } - \beta ) d_{1t}(\beta ^{\prime })}. \end{aligned}$$

Taking the supremum with respect to \(t \in N^{+}\) yields

$$\begin{aligned} d_{1}(\beta ) = \sup _{t \in N^{+}} d_{1t}(\beta ) \le \sup _{t \in N^{+}} \frac{d_{1t}(\beta ^{\prime })}{1+(\beta ^{\prime } - \beta ) d_{1t}(\beta ^{\prime })} \le \frac{d_{1}(\beta ^{\prime })}{1+(\beta ^{\prime } - \beta ) d_{1}(\beta ^{\prime })} < d_{1}(\beta ^{\prime }). \end{aligned}$$

This completes the proof. \(\square \)

The proof that the function \(d_{2}\) is strictly decreasing follows by a similar argument and is therefore omitted.

Lemma 8.2

The function \(d_{2}\) is a strictly decreasing function on \( (0,1) \) assuming only finite values.

Appendix B The proof of theorem 6.3

Consider the bargaining equilibrium \( (\beta ^{-}, \beta ^{+}) \) and consider the bargaining power of the poor and the rich, respectively, at the upper bound \( \beta ^{+} \) of the social acceptance set. Lemma 8.3 states that the size of the social acceptance set \( \beta ^{+} - \beta ^{-} \) multiplied by the bargaining power of the poor \( m^{-} d_{2}(\beta ^{+}) \) is bounded from above by \( (1-\delta )/\delta , \) and is greater than or equal to this number when multiplied by the bargaining power \( m^{+} d_{1}(\beta ^{-}) \) of the rich.

Lemma 8.3

Consider the tax equilibrium \((\beta ^{-},\beta ^{+})\). If \(\beta ^{+} < 1\), then

  1. 1.

    \(\delta m^{-} d_{2}(\beta ^{+})(\beta ^{+}-\beta ^{-}) \le (1-\delta )\).

  2. 2.

    \(\delta m^{+} d_{1}(\beta ^{+})(\beta ^{+}-\beta ^{-}) \ge (1-\delta )\).

Proof

For \( t \in N^{-}, \) it holds that the proposal \( \beta ^{-} \) is accepted, so \( \delta (m^{-} v_{t}(\beta ^{+}) + m^{+} v_{t}(\beta ^{-})) \le v_{t}(\beta ^{-}). \) Rewriting this inequality results in

$$\begin{aligned} \delta m^{-} (v_{t}(\beta ^{+}) - v_{t}(\beta ^{-})) \le (1-\delta ) v_{t}(\beta ^{-}) \le (1-\delta ) v_{t}(\beta ^{+}). \end{aligned}$$

Since \( v_{t} \) is concave, we find that

$$\begin{aligned} \delta m^{-} \partial _{+} v_{t}(\beta ^{+})(\beta ^{+} - \beta ^{-}) \le (1-\delta ) v_{t}(\beta ^{+}), \end{aligned}$$

and therefore \( \delta m^{-} d_{2t}(\beta ^{+})(\beta ^{+}-\beta ^{-}) \le 1-\delta . \) Since this inequality holds for all \( t \in N^{-}, \) we obtain

$$\begin{aligned} \delta m^{-} d_{2}(\beta ^{+}) (\beta ^{+}-\beta ^{-}) = \sup _{t \in N^{-}} \delta m^{-} d_{2t}(\beta ^{+})(\beta ^{+}-\beta ^{-}) \le 1-\delta . \end{aligned}$$

This proves Lemma 8.3.1.

Consider some \( t \in N^{+}. \) By concavity of \( v_{t} \) we have that \( -\partial _{-} v_{t}(\beta ^{+}) (\beta ^{+} - \beta ^{-}) \ge v_{t}(\beta ^{-}) - v_{t}(\beta ^{+}). \) It then follows that

$$\begin{aligned} d_{1t}(\beta ^{+})(\beta ^{+}-\beta ^{-}) \ge \frac{v_{t}(\beta ^{-})}{v_{t}(\beta ^{+})} - 1. \end{aligned}$$

We take the supremum over all \( t \in N^{+} \) and find that

$$\begin{aligned} d_{1}(\beta ^{+})(\beta ^{+}-\beta ^{-}) \ge \sup _{t \in N^{+}} \frac{v_{t}(\beta ^{-})}{v_{t}(\beta ^{+})} - 1. \end{aligned}$$

We complete the proof of Lemma 8.3.2 by showing that

$$\begin{aligned} \delta m^{+} \left( \sup _{t \in N^{+}} \frac{v_{t}(\beta ^{-})}{v_{t}(\beta ^{+})}-1\right) = 1-\delta , \end{aligned}$$

or equivalently

$$\begin{aligned} \sup _{t \in N^{+}} \frac{\delta (m^{-} v_{t}(\beta ^{+}) + m^{+} v_{t}(\beta ^{-}))}{v_{t}(\beta ^{+})} = 1. \end{aligned}$$

Since all \( t \in N^{+} \) accept the proposal \( \beta ^{+}, \) we have

$$\begin{aligned} \frac{\delta (m^{-} v_{t}(\beta ^{+}) + m^{+} v_{t}(\beta ^{-}))}{v_{t}(\beta ^{+})} \le 1, \quad t \in N^{+}. \end{aligned}$$

Suppose there is an \(\varepsilon > 0\) with the property

$$\begin{aligned} \frac{\delta (m^{-} v_{t}(\beta ^{+}) + m^{+} v_{t}(\beta ^{-}))}{v_{t}(\beta ^{+})} \le 1-\varepsilon \end{aligned}$$

for all \(t \in N^{+}\). Then

$$\begin{aligned} \begin{array}{rcl} v_{t}((1-\varepsilon )\beta ^{+} + \varepsilon ) &{} \ge &{} (1-\varepsilon ) v_{t}(\beta ^{+}) + \varepsilon v_{t}(1) \ge (1- \varepsilon ) v_{t}(\beta ^{+})\\ &{} \ge &{} \delta (m^{-} v_{t}(\beta ^{+}) + m^{+} v_{t}(\beta ^{-})). \end{array} \end{aligned}$$

All citizens \(t \in N^{+}\) (as well as citizens in \(N^{-} \cup N^{0} \)) therefore accept the proposal \((1 - \varepsilon ) \beta ^{+} + \varepsilon > \beta ^{+}\) thereby contradicting that \(\beta ^{+}\) is the upper bound of the social acceptance set. \(\square \)

We are now in a position to prove the first half of Theorem 6.3.

Lemma 8.4

The tax equilibrium \((\beta ^{-},\beta ^{+})\) satisfies \(\beta ^{*} \le \beta ^{+}\), where \(\beta ^{*}\) is the generalized zero point of \(z\).

Proof

The result is obviously true when \(\beta ^{+} = 1\). In particular, if \(\delta = 0\) then \(\beta ^{+} = 1\) and \(\beta ^{-} = 0\) by Equation (4.3). Suppose now \(\delta > 0\) and \( \beta ^{+} < 1\). Subtracting the second inequality of Lemma 8.3 from the first one gives \(\delta z(\beta ^{+})(\beta ^{+} - \beta ^{-}) \le 0\). Since by (4.3) \( 0 < \beta ^{+} - \beta ^{-}, \) it follows that \(z(\beta ^{+}) \le 0\). The result follows since \(z\) is a decreasing function. \(\square \)

The proof of the second half of Theorem 6.3 follows from an analogous argument and is therefore omitted.

Appendix C The derivation of Equations (7.1)–(7.2)

We provide a derivation of Equations (7.1)–(7.2).

For each \(t \in N\) and each \( c_{t} > 0,\) we have \(0 = u_{t}(0) \le u_{t}(c_{t}) - c_{t}\partial _{+}u_{t}(c_{t})\), where the inequality holds by the concavity of \(u_{t}\). Rearranging this expression, we obtain that

$$\begin{aligned} b_{t}(c_{t}) \le \frac{1}{c_{t}}, \quad c_{t} > 0. \end{aligned}$$
(8.1)

For each \( t \in N^{+}\cap S, \) we have

$$\begin{aligned} d_{1t}(\beta ) = \frac{w_{t} - \bar{w}}{w_{t} + \beta (\bar{w} - w_{t})}, \quad \beta \in (0,1) \end{aligned}$$

This expression is non–decreasing in \(w_{t}\) on \([\bar{w},+\infty )\), and it converges to \( 1/(1 - \beta ) \) as \(w_{t}\) approaches infinity. As the set \(\{w_{t} \mid t \in S\}\) is unbounded from above, it holds that \(d_{1}(\beta ) \ge \tfrac{1}{1 - \beta }\). On the other hand, for each \(t \in N^{+}\) it holds that

$$\begin{aligned} d_{1t}(\beta ) = (w_{t} - \bar{w})b_{t}(w_{t} + \beta (\bar{w} - w_{t})) \le \frac{w_{t} - \bar{w}}{w_{t} + \beta (\bar{w} - w_{t})} \le \frac{1}{1 - \beta }, \quad \beta \in (0,1), \end{aligned}$$

where the first inequality is implied by (8.1). It follows that \(d_{1}(\beta ) \le 1/(1 - \beta ), \) which proves Eq. (7.1).

Similarly for each citizen \(t \in N^{-} \cap S,\) we have

$$\begin{aligned} d_{2t}(\beta ) = \frac{\bar{w} - w_{t}}{w_{t} + \beta (\bar{w} - w_{t})}, \quad \beta \in (0,1). \end{aligned}$$

Since the infimum of \(\{w_{t} \mid t \in S\}\) is zero, it holds that \( d_{2}(\beta ) \ge 1/\beta . \) On the other hand, for each \(t \in N^{-}\) it holds that

$$\begin{aligned} d_{2t}(\beta ) = (\bar{w} - w_{t})b_{t}(w_{t} + \beta (\bar{w} - w_{t})) \le \frac{\bar{w} - w_{t}}{w_{t} + \beta (\bar{w} - w_{t})} \le \frac{1}{\beta }, \quad \beta \in (0,1), \end{aligned}$$

where the first inequality follows by (8.1). It follows that \(d_{2}(\beta ) \le 1/\beta , \) which proves Eq. (7.2).

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Herings, P.JJ., Predtetchinski, A. Procedural fairness and redistributive proportional tax. Econ Theory 59, 333–354 (2015). https://doi.org/10.1007/s00199-014-0852-9

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Keywords

  • Procedural fairness
  • Tax rate
  • Bargaining

JEL Classification

  • C78
  • D63
  • H24