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