Redistributive politics under ambiguity

Abstract

The conflicting views that agents and voters have about redistributive taxation have been broadly studied. The literature has focused on situations where the counterfactual outcomes that would have occurred had other actions been chosen are observable or point identified. I analyze this problem from an econometric standpoint, in a context of ambiguity. The extent to which individuals are responsible for their own fate is partially identified. Agents have partial knowledge of the relative importance of effort in the generation of income inequality and, therefore, the magnitude of the incentive costs. I present a simple model of redistribution and show that multiple equilibria might arise even in the presence of ambiguity: One where the rate of redistribution is high, agents are pessimistic, and exert low effort (Pessimism/Welfare State), and another where the redistribution tax rate is low, agents are optimistic, and exert high effort (Optimism/Laissez Faire).

Appendices

Appendix

A Data

This data appendix describes the data sources and variables used in Fig. 1, Social Spending and Belief that Luck Determines Income.

1.1 Social spending data

The dataset was obtained from Persson and Tabellini (2005),¹² for the period 1960-1998. The variable used is SSW, which is defined in Persson and Tabellini’s data appendix as:

Social spending. Consolidated central government expenditures on social services and welfare as a percentage of GDP as reported in the IMF GFS Yearbook divided by GDP and multiplied by 100. Source: IMF - GFS Yearbook 2000 and IMF-IFS CD-Rom.

1.2 Survey data

The dataset was obtained from the World Values Survey (WVS),¹³ for the period 1981–1997. The question used corresponds to E040, where responders were asked whether they agree (higher value) or disagree (lower value) with the statement that: “Hard work doesn’t generally bring success.” Specifically, the variable is constructed as follows.

Belief that luck determines income. Average responses by country to the WVS question:

Now I’d like you to tell me your views on various issues. How would you place your views on this scale? 1 means you disagree completely with the statement there; 10 means you agree completely with the statement there; and if your views fall somewhere in between, you can choose any number in between. “In the long run, hard work usually brings a better life” vs. “Hard work doesn’t generally bring success—it’s more a matter of luck and connections:

1: In the long run, hard work usually brings a better life.

2: $\dots$

$\vdots$

9: $\dots$

10: Hard work doesn’t generally bring success—it’s more a matter of luck and connections.

The WVS data in Fig. 1 correspond to the average E040 responses by country divided by 10.

B Proofs

1.1 B.1 Maximin criterion

Proof optimal effort Maximin Criterion

The agent acts as if $\beta =\beta _L$ and solves:

$$\begin{aligned} \max _{\delta \in [0,1]} (1-\tau )\bigg [\alpha +\delta [\beta _L - \alpha ]\bigg ]-C_j(\delta ). \end{aligned}$$

Assumption 1 implies that $\beta _L<\alpha$ and, thus, the solution to the previous problem is trivially solved by setting $\delta _{MMin}^*=0$. $\square$

Proof that effort does not depend on tax Maximin Criterion

The agent’s welfare under Maximin criterion is given by:

$$\begin{aligned} \begin{aligned} U_j^{MMin}(\lambda ,\delta ,\tau )&=(1-\tau )\bigg [\alpha +\delta [\beta _L - \alpha ]\bigg ]-C_j(\delta )+\tau \bar{Y} + M(\lambda ),\\&=\alpha +M(\lambda ), \end{aligned} \end{aligned}$$

where the last equality is obtained by replacing the maximin solution $\delta _{MMin}^*=0$.

Then:

$$\begin{aligned} \begin{aligned} \frac{\partial U_j^{Mmin}(\lambda ,\delta ,\tau )}{\partial \tau } = 0.\\ \end{aligned} \end{aligned}$$

$\square$

1.2 B.2 Maximax Criterion

Proof optimal effort Maximax Criterion

Now the agent sets $\beta =\beta _U$ and solves:

$$\begin{aligned} \max _{\delta \in [0,1]} (1-\tau )\bigg [\alpha +\delta [\beta _U - \alpha ]\bigg ]-C_j(\delta ). \end{aligned}$$

Assumption 1 implies that $\beta _U>\alpha$. Therefore, the maximax solution is $\delta _{MMax}^*=1$. $\square$

Proof that effort does not depend on tax Maximax Criterion

The agent’s welfare under Maximax criterion is given by:

$$\begin{aligned} \begin{aligned} U_j^{MMax}(\lambda ,\delta ,\tau )&=(1-\tau )\bigg [\alpha +\delta [\beta _U - \alpha ]\bigg ]-C_j(\delta )+\tau \bar{Y} + M(\lambda ),\\&=(1-\tau )\bigg [\alpha +[\beta _U - \alpha ]\bigg ]-\frac{1}{2 a_{j}(\lambda )}-\tau \bigg [\alpha +[\beta _U - \alpha ]\bigg ]+M(\lambda ), \end{aligned} \end{aligned}$$

where the last equality is obtained by replacing the maximin solution $\delta _{MMax}^*=1$ and $\bar{Y}=\alpha +\left( \beta _{U}-\alpha \right)$.

Then:

$$\begin{aligned} \frac{\partial U_j^{MMax}(\lambda ,\delta ,\tau )}{\partial \tau } = 0. \end{aligned}$$

$\square$

1.3 B.3 Hurwicz Criterion

Proof optimal effort Hurwicz Criterion

According to this criterion the agent solves the following problem:

$$\begin{aligned} \begin{aligned} U_j^H(\delta _H^*;\lambda _j)&=\max _{\delta \in [0,1]}\lambda _j\bigg \{\inf _{\beta \in [\beta _L,\beta _U]}\bigg (\mathbb {E}[(1-\tau )y(e_j)]-C_j(\delta ) \bigg ) \bigg \}\\&\quad + (1-\lambda _j)\bigg \{\sup _{\beta \in [\beta _L,\beta _U]}\bigg (\mathbb {E}[(1-\tau )y(e_j)]-C_j(\delta ) \bigg ) \bigg \}, \\&=\max _{\delta \in [0,1]} (1-\tau )\bigg \{\alpha +\bigg [\bigg (\lambda _j\beta _L+(1-\lambda _j)\beta _U \bigg )- \alpha \bigg ]\delta _j\bigg \}-C_j(\delta ),\\&=\max _{\delta \in [0,1]} (1-\tau )\big \{\alpha +\big [\bar{\beta _{\lambda }} - \alpha \big ]\delta _j\big \}-C_j(\delta ), \end{aligned} \end{aligned}$$

where $\bar{\beta }_{\lambda }:= \lambda _j\beta _L+(1-\lambda _j)\beta _U$.

The first-order necessary condition for an interior solution is:

$$\begin{aligned} \begin{aligned} (1-\tau )[\bar{\beta }_{\lambda }-\alpha ]&=C_j'(\delta _H),\\&=\frac{\delta _H}{a_j}. \end{aligned} \end{aligned}$$

Solving for $\delta _H$ yields $\delta _H=a_j(1-\tau )[\bar{\beta }_{\lambda } - \alpha ]$. Thus, the solution is:

$$\begin{aligned} \delta _H^* ={\left\{ \begin{array}{ll} 0 &{} \text {if}\, \delta _H\le 0,\\ 1 &{} \text {if}\,\delta _H\ge 1,\\ \delta _H &{} \text {if}\,\delta _H\in (0,1). \end{array}\right. } \end{aligned}$$

The second-order sufficient condition is satisfied because $\frac{\partial ^2U_j^H(\delta )}{\partial \delta ^2}=-\frac{\partial ^2C_j(\delta )}{\partial \delta ^2}:= -C_j''(\delta )= -\frac{1}{a_j(\lambda )}<0$ by Assumption 3. (The second-order sufficient condition only requires $C_j(\delta )$ to be convex.) $\square$

Proof of interior condition for pessimism rate, $\lambda$. For $\overline{\lambda }$, $\nicefrac {\partial \delta _{H}}{\partial \lambda } < 0$, then $\overline{\lambda }=\{\lambda : \delta _{H} = a_{j} (1-\tau ) [\lambda \beta _{L} + (1-\lambda )\beta _{U} - \alpha ] = 0\}.$ Then, $\overline{\lambda }= \frac{\beta _{U} - \alpha }{\beta _{U} - \beta _L} \in (0,1)$. For $\underline{\lambda }$, let $\underline{\lambda }^*=\{\lambda : a_{j} (1-\tau ) [\lambda \beta _{L} + (1-\lambda )\beta _{U} - \alpha ] = 1\}$, yielding $\underline{\lambda }^*= \frac{\beta _{U} - \alpha }{\beta _{U} - \beta _{L}} - \frac{1}{a_{j} (1-\tau )}\lessgtr 0$. Then, $\underline{\lambda } = \max \{0, \lambda : \underline{\lambda }^*<1\}$. $\square$

Proof of Lemma 1

By backwards induction, the optimal level of effort is the value of $\delta$ that maximizes Eq. (5) because each agent chooses effort optimally using the Hurwicz criterion. Knowing the degree of pessimism, $\lambda$, and the equilibrium tax rate, $\tau$, this optimization problem is simplified to Eq.  (2) and the optimal level of effort is given by Eq.  (3). That is, ${\textrm{arg}\,\textrm{max}}_{\delta \in [0,1]}U_j^{CH}(\lambda ,\delta ,\tau )={\textrm{arg}\,\textrm{max}}_{\delta \in [0,1]}U_j^H(\delta ;\lambda _j)$. Then, Eq. (6) follows from the optimal effort under the Hurwicz criterion. $\square$

Proof of Lemma 2

By Lemma 1, the optimal level of effort is given by equation (6). If $(\bar{\beta }_{\lambda }-\alpha )\le 0$, then $\delta _H\le 0$ and $\delta _H^*=0$ (corner solution), so $\frac{\partial \delta _H}{\partial \tau }=0$. If $(\bar{\beta }_{\lambda }-\alpha )>0$, there are two possibilities. If $(\bar{\beta }_{\lambda }-\alpha )\ge \frac{1}{a_j(1-\tau )}$, then $\delta _H\le 1$ and $\delta _H^*=1$ (corner solution), so $\frac{\partial \delta _H}{\partial \tau }=0$. If $0<(\bar{\beta }_{\lambda }-\alpha )< \frac{1}{a_j(1-\tau )}$, we have an interior solution because $\delta _H\in (0,1)$ and $\delta _H^*=\delta _H$. Taking the derivative with respect to the tax rate yields $\frac{\partial \delta _H}{\partial \tau }=-(\bar{\beta }_{\lambda }-\alpha )<0$. $\square$

Proof of Lemma 3

The agent’s welfare is given by Eq. (5):

$$\begin{aligned} U_j^{CH}(\lambda ,\delta ,\tau ) = (1-\tau )(\alpha + (\bar{\beta }_{\lambda } - \alpha )\delta ) - C(\delta ) + \tau \bar{Y} + M(\lambda ). \end{aligned}$$

Maximizing the previous expression with respect to $\tau$:

$$\begin{aligned} \begin{aligned} \frac{\partial U_j^{CH}(\lambda ,\delta ,\tau )}{\partial \tau }&= - (1-\tau )(\bar{\beta }- \alpha )^2 a_{j}(\lambda ) + \bar{a}(\lambda )(1-\tau )^2 G - 2 \tau (1-\tau ) \bar{a}(\lambda ) G, \\ \end{aligned} \end{aligned}$$

where $G:= [(1-\phi ) a_L (\bar{\beta }_{\lambda _L} - \alpha ) + \phi a_H (\bar{\beta }_{\lambda _H} - \alpha ) ] [(1-\phi )(\bar{\beta }_{\lambda _L} - \alpha ) + \phi (\bar{\beta }_{\lambda _H} - \alpha ) ]$.

In an interior solution:

$$\begin{aligned} \begin{aligned} \tau ^*&= \frac{\bar{a}(\lambda ) G - (\bar{\beta } - \alpha )^2 a_{j}(\lambda )}{3\bar{a}(\lambda ) G}. \end{aligned} \end{aligned}$$

The second-order sufficient condition is $\frac{\partial ^2 U_j^{CH}(\lambda ,\delta , \tau ^*)}{\partial \tau ^2}= -5 (\bar{\beta } - \alpha )^2 a_{L}(\lambda ) - 2 \bar{a}(\lambda ) G < 0$ because $G>0$ and $a_{L}(\lambda )> 0$. $\square$

Proof of Lemma 4

Voters’ preferences over $\tau$ are single peaked and, thus, the median voter theorem applies. Because $\phi <\frac{1}{2}$, the equilibrium tax outcome, T, is the optimal desired tax from the disadvantaged group, $\tau _L^*(\lambda )$. Denote $\bar{\beta }_{L} = \lambda _{L} \beta _{L} + (1-\lambda _{L}) \beta _{U}$, where $\lambda _{L}$ is the pessimism rate of disadvantaged agents; similarly, for advantaged agents. The lemma holds trivially if the equilibrium tax rate is 0. If the equilibrium tax rate is interior, then by Lemma 3, $\tau ^{*} = \frac{1}{3} - \frac{(\bar{\beta }_{L} - \alpha )^2 a_{j}(\lambda )}{3\bar{a}(\lambda ) P}$, where $P:=a_{L}(\lambda _{L})(1-\phi )^2 (\bar{\beta }_{\lambda _L}-\alpha )^2 + \phi (1-\phi )(a_{L}(\lambda _{L}) + a_{H}(\lambda _{H}))(\bar{\beta }_{\lambda _H} - \alpha ) (\bar{\beta }_{\lambda _L} - \alpha ) + a_{H}(\lambda _{H})\phi ^2 (\bar{\beta }_{\lambda _H} - \alpha )^2$. Then:

$$\begin{aligned} \begin{aligned} \frac{\partial T}{\partial \lambda }&=\frac{\partial \tau _L^*(\lambda )}{\partial \lambda },\\&=-\left[ \underbrace{\left( 3\bar{a}(\lambda ) P\right) ^{-1}}_{:= A} \underbrace{2 a_{j}(\lambda )(\bar{\beta }_{L} - \alpha )(\beta _{L} -\beta _{U}) + (\bar{\beta }_{L} - \alpha )^2 a'(\lambda )}_{:= B}\right] . \end{aligned} \end{aligned}$$

Note that:

• $A>0$ because $\bar{a}(\lambda ) > 0$ and $P>0$.

• $B<0$ because $\beta _{L}- \beta _{U} <0$, $a'(\lambda )<0$, $2a_{j}(\lambda )>0, \bar{\beta }_{L}-\alpha > 0$.

Then, $\frac{\partial T}{\partial \lambda }>0$.

Next, we note that pessimism is decreasing in agents’ cost parameters. Let Assumptions 1, 2, 3 hold and $\lambda \in (\underline{\lambda },\bar{\lambda })$. Consider the problem of choosing $\lambda$ in period 0:

$$\begin{aligned} \begin{aligned} \max _{\lambda \in (\underline{\lambda }, \overline{\lambda })} (1-\tau )\big [\alpha +\big (\bar{\beta }_{\lambda }- \alpha \big )\delta ]-C_j(\delta )+\tau \bar{Y} + M(\lambda ). \end{aligned} \end{aligned}$$

Or, after replacing $\delta _H^*$ and the equilibrium tax rate:

$$\begin{aligned} \begin{aligned} \max _{\lambda \in (\underline{\lambda }, \overline{\lambda })} F_j(\lambda ,T), \end{aligned} \end{aligned}$$

(B.1)

where $F_j(\lambda ,T):=(1-T)\big [\alpha +\big (\bar{\beta }_{\lambda }- \alpha \big )\delta _H^*\big ]-C_j(\delta _H^*)+T \bar{Y} + M(\lambda )$.

Let $J_j(\Lambda ,a_{j}):=(1-T)\alpha + \frac{1}{2} (a_{j} + a(-\Lambda )) (1-T)^2(\bar{\gamma }_{\Lambda } - \alpha )^2 + T \bar{Y} + M(-\Lambda )$, where $\Lambda := - \lambda$, and $\bar{\gamma }_{\Lambda }:= - \beta _{L} \Lambda + (1+\Lambda )\beta _{U} = \beta _{L} \lambda + (1-\lambda ) \beta _{U} = \bar{\beta }_{\lambda }$.

Then:

$$\begin{aligned} \frac{\partial J_j(\Lambda ,a_{j})}{\partial a_{j}}= & {} \frac{1}{2} (1-T)^2 (\bar{\gamma }_{\Lambda } - \alpha )^{2}.\\ \frac{\partial ^2 J_j(\Lambda ,a_{j})}{\partial a_{j} \partial \Lambda }= & {} \underbrace{ (1-T)(\bar{\gamma }_{\Lambda } - \alpha )^2 \frac{\partial T}{\partial \lambda }}_{:= C} + \underbrace{(1-T) \frac{\delta _H^*}{a_j+a(\lambda )} (\beta _{U} - \beta _{L})}_{:= D}. \end{aligned}$$

Note that:

• $C>0$ because $(1-T) \in (0,1)$ and T is increasing in $\lambda$.

• $D>0$ because $\delta _H^*\in (0,1)$.

Therefore, $J_j(\Lambda ,a_{j})$ has increasing differences in $(\Lambda ,a_{j})$, and pessimism being decreasing in agents’ cost parameters is established using monotone comparative statics.

Finally, note that the preferred tax rate of the disadvantaged agents is strictly increasing with the pessimism rate. It then follows that disadvantaged agents prefer an interior tax rate.

Let Assumptions 1, 2, and 3 hold, and $\lambda \in (\underline{\lambda },\bar{\lambda })$. Agent j preferred tax is $\tau ^{*} = \frac{1}{3} [1 - \frac{(\bar{\beta } - \alpha )^2 a_{j}(\lambda )}{\overline{a(\lambda )} P}]$, where $P:= a_{L}(\lambda _{L})(1-\phi )^2 (\bar{\beta }_{\lambda _L}-\alpha )^2 + \phi (1-\phi )(a_{L}(\lambda _{L}) + a_{H}(\lambda _H))(\bar{\beta }_{\lambda _H} - \alpha ) (\bar{\beta }_{\lambda _L} - \alpha ) + a_{H}(\lambda _H)\phi ^2 (\bar{\beta }_{\lambda _H} - \alpha )^2$. The disadvantaged agents choose an interior tax rate if $(\bar{\beta }_{L} - \alpha )^2 a_{j}(\lambda ) < \overline{a(\lambda )} P$. Note that $\overline{a(\lambda )}$ is a weighted average of $a_{L}(\lambda _{L})$ and $a_{H}(\lambda _{H})$, where $\lambda _{H}>\lambda _{L}$, $\overline{a(\lambda )} > a_{L}(\lambda )$ (decreasing pessimism as a function of cost). Note also that $a_{H} > a_{L}$ and $(\bar{\beta }_{\lambda _{H}} - \alpha ) > (\bar{\beta }_{\lambda _{L} }- \alpha )$. Then:

$$\begin{aligned} \begin{aligned}&a_{L}(\lambda _L)(1-\phi )^2 (\bar{\beta }_{\lambda _L}-\alpha )^2 + \phi (1-\phi )(a_{L}(\lambda _L) \\&\quad +a_{H}(\lambda _H))(\bar{\beta }_{\lambda _H} - \alpha ) (\bar{\beta }_{\lambda _L} - \alpha ) + a_{H}(\lambda _H)\phi ^2 (\bar{\beta }_{\lambda _H} - \alpha )^2 \\&\quad> a_{L}(\lambda _L)(1-\phi )^2 (\bar{\beta }_{\lambda _L} - \alpha )^2 + \phi (1-\phi )(2a_{L}(\lambda _L) ) (\bar{\beta }_{\lambda _L} - \alpha )^2 + a_{L}(\lambda _L)\phi ^2 (\bar{\beta }_{\lambda _L} - \alpha )^2, \\&\quad > (\bar{\beta }_{\lambda _L} - \alpha )^2, \end{aligned} \end{aligned}$$

by Assumption 3. Then, the result follows because $P > (\bar{\beta }-\alpha )^2$. $\square$

Proof of Lemma 5

As before, the problem of choosing $\lambda$ in period 0 is:

$$\begin{aligned} \max _{\lambda \in (\underline{\lambda }, \overline{\lambda })} (1-T)\left[ \alpha +(\bar{\beta }_{\lambda } - \alpha )\delta _H^*\right] - C_{j}(\delta _H^*) + T\bar{Y} + M(\lambda ). \end{aligned}$$

Or:

$$\begin{aligned} \max _{\lambda \in (\underline{\lambda }, \overline{\lambda })} F_j(\lambda , T), \end{aligned}$$

where $F_j(\lambda , T):= (1-T) \alpha + \frac{1}{2} a_{j}(\lambda )(1-T)^2(\bar{\beta }_\lambda - \alpha )^2 + T\bar{Y} + M(\lambda )$.

Then:

$$\begin{aligned} \begin{aligned} \frac{\partial {F_j(\lambda , T)}}{\partial {T}}&= -\alpha - a_{j}(\lambda ) (1-T) (\bar{\beta }_{\lambda } - \alpha )^2 + \bar{Y} + T \frac{\partial {\bar{Y}}}{\partial {T}}.\\ \frac{\partial ^{2}F_j(\lambda , T)}{\partial T\partial \lambda }&= \underbrace{- a_{j}'(\lambda ) (1-T)(\bar{\beta }_{\lambda } - \alpha )^2}_{:= E} + \underbrace{(-2) a_{j}(\lambda )(1-T)(\bar{\beta }_{\lambda } - \alpha ) (\beta _{L}- \beta _{U})}_{:= F} \end{aligned} \end{aligned}$$

• $E>0$ because $a'(\lambda ) < 0$ by Assumption 3 and $1-T>0$ (interior solution).

• $F>0$ because $\bar{\beta }_{\lambda } - \alpha > 0$.

Therefore, $F_j(\lambda ,T)$ has increasing differences in $(\lambda ,T)$ and the result follows from monotone comparative statics. $\square$

Proof of Lemma 6

It is sufficient to show that $\exists \ \bar{T}=\tau _L^*(\bar{\lambda }^*_L) \wedge \ \underline{T}=\tau _L^*(\underline{\lambda }^*_L)$ with $\bar{\lambda }=\bar{\lambda }^*_L>\underline{\lambda }=\underline{\lambda }^*_L$ such that:

$$\begin{aligned}&F_L(\bar{\lambda }^*_L,\bar{T})> F_L(\underline{\lambda }^*_L,\bar{T}), \end{aligned}$$

(B.2a)

$$\begin{aligned}&F_H(\bar{\lambda }^*_H,\bar{T})> F_H(\underline{\lambda }^*_H,\bar{T}),\end{aligned}$$

(B.2b)

$$\begin{aligned}&F_L(\bar{\lambda }^*_L,\underline{T})< F_L(\underline{\lambda }^*_L,\underline{T}),\end{aligned}$$

(B.2c)

$$\begin{aligned}&F_H(\bar{\lambda }^*_H,\underline{T})< F_H(\underline{\lambda }^*_H,\underline{T}). \end{aligned}$$

(B.2d)

Expressions (B.2a) and (B.2c) hold because L-types are optimizing and they are the pivotal group. We know that $F_H(\bar{\lambda },\bar{T})>F_L(\bar{\lambda },\bar{T})$ because H-types have lower costs than L-types and $F_L(\bar{\lambda },\bar{T})> F_L(\underline{\lambda },\bar{T})$ (equation B.2a). Using the H-types desired tax for $\bar{\lambda }^*_H$, $F_H(\bar{\lambda }^*_H,\bar{\tau }^*)> F_H(\underline{\lambda }^*_H,\bar{\tau }^*)$. Then, $F_H(\bar{\lambda }^*_H,\bar{T})>F_H(\underline{\lambda }^*_H,\bar{T})$ because of Lemma 4 and because the L-types are the pivotal group. An analogue argument shows that $F_H(\bar{\lambda }^*_H,\underline{T}>F_H(\underline{\lambda }^*_H,\underline{T})$ holds. Then, the result follows by Lemma 5 because the solution to (B.2), $\lambda _j^*(T)$, is a continuous function of the equilibrium tax rate, T. $\square$

Proof of Proposition 1

We know that $\tau _L^*(\tilde{\lambda })>\tau _L^*(\hat{\lambda })$ because $\underline{\lambda }<\hat{\lambda }<\tilde{\lambda }<\bar{\lambda }$. Take $\tilde{T} \in (\bar{T}=\tau _L^*(\bar{\lambda }),\underline{T}=\tau _L^*(\underline{\lambda }))$. Such a $\tilde{T}$ exists by continuity. Then, the result follows by Lemmas 2, 4 and 6. $\square$