The tyranny puzzle in social preferences: an empirical investigation

Abstract

When forming their preferences about the distribution of income, rational people may be caught between two opposite forms of “tyranny.” Giving absolute priority to the worst-off imposes a sort of tyranny on the rest of the population, but giving less than absolute priority imposes a reverse form of tyranny where the worst-off may be sacrificed for the sake of small benefits to many well-off individuals. We formally show that this intriguing dilemma is more severe than previously recognised, and we examine how people negotiate such conflicts with a questionnaire-experimental study. Our study shows that both tyrannies are rejected by a majority of the participants, which makes it problematic for them to define consistent distributive preferences on the distribution.

Appendices

Appendix 1: proof

The following is the proof of the proposition in Sect. 2.2.

Proof

Let $N$ be such that $\left| N\right| =n$. For simplicity of notation, we assume $N=\left\{ 1,...,n\right\}$.

Consider an allocation $x\in \mathbb {R}_{+}^{n}$ such that for all $i,j>1,\, x_{i}=x_{j}$ and for all $j>1,\, x_{j}=r+1>r>q=x_{1}>0$.

Let $y\in \mathbb {R}_{+}^{n}$ be such that:

• for all $i,j>1,\, y_{i}=y_{j}$;

• for all $j>1,\, y_{j}=x_{j}+\alpha ^{\prime }>r>q=x_{1}>y_{1}=x_{1}-\beta ^{\prime }>0$ (case $q>\beta ^{\prime }$);

• for all $j>1,\, y_{j}=x_{j}+\alpha ^{\prime }>r>q=x_{1}>y_{1}=0$ (case $q\le \beta ^{\prime }$).

By Aggregation, $y\ R^{N}\ x$.

Let $m=\left[ \frac{\alpha ^{\prime }+1}{\beta }\right] ^{+}$ (the first integer that is at least as great as $\frac{\alpha ^{\prime }+1}{\beta }$) and $\gamma =\frac{\alpha ^{\prime }+1}{m}$. This guarantees that $\gamma \le \beta$ and for all $j>1,\, r\le y_{j}-m\gamma =r+1+\alpha ^{\prime }-m\gamma \le x_{j}=r+1$. Let $\delta =(x_{1}-y_{1})/\left( m+1\right)$ and $p=\left[ \alpha /\delta \right] ^{+}$.

We now consider the following sequence, where the first allocation is a $p$-replica of $y$. We assume for the moment that the population of this size is in the domain $\mathcal {N}$.

$$\begin{aligned} p*y= & {} \left( \underset{p}{\underbrace{y_{1},\ldots ,y_{1}}},\underset{p}{\underbrace{\left( y_{j},\ldots ,y_{j}\right) _{j>1}}}\right) ,\\ z^{1}= & {} \left( y_{1}+p\delta ,\underset{p-1}{\underbrace{y_{1},\ldots ,y_{1}}},\underset{p}{\underbrace{\left( y_{j}-\gamma ,\ldots ,y_{j}-\gamma \right) _{j>1}}}\right) ,\\ w^{1}= & {} \left( \underset{p}{\underbrace{y_{1}+\delta ,\ldots ,y_{1}+\delta }},\underset{p}{\underbrace{\left( y_{j}-\gamma ,\ldots ,y_{j}-\gamma \right) _{j>1}}}\right) , \end{aligned}$$

and for $t=1,...,m-1$,

$$\begin{aligned} z^{t+1}= & {} \left( y_{1}+t\delta +p\delta ,\underset{p-1}{\underbrace{y_{1}+t\delta ,\ldots ,y_{1}+t\delta }},\right. \\&\times \left. \underset{p}{\underbrace{\left( y_{j}-(t+1)\gamma ,\ldots ,y_{j}-(t+1)\gamma \right) _{j>1}}}\right) ,\\ w^{t+1}= & {} \left( \underset{p}{\underbrace{y_{1}+(t+1)\delta ,\ldots ,y_{1}+(t+1)\delta }},\right. \\&\left. \times \underset{p}{\underbrace{\left( y_{j}-(t+1)\gamma ,\ldots ,y_{j}-(t+1)\gamma \right) _{j>1}}}\right) . \end{aligned}$$

By Non-Aggregation, $z^{1}\ R^{p*N}\ p*y$ and, for $t=1,...,m-1,\, z^{t+1}\ R^{p*N}\ w^{t}$. Observe that for all $j>1$, $y_{j}-m\gamma >r$, so that in this sequence the best-off are always better-off than $r$, as requested for the application of Non-Aggregation. Similarly, $y_{1}+m\delta <q$, meaning that the worst-off is always below $q$.

For every $t=1,...,m$, by applying Pigou–Dalton $p-1$ times (between the first individual and the next $p-1$ individuals), one has $w^{t}\ R^{p*N}\ z^{t}$.

By transitivity, it follows that $w^{m}\ R^{p*N}\ p*y$, where $w^{m}$ is equal to:

$$\begin{aligned} w^{m}=\left( \underset{p}{\underbrace{y_{1}+m\delta ,\ldots ,y_{1}+m\delta }},\underset{p}{\underbrace{\left( y_{j}-m\gamma ,\ldots ,y_{j}-m\gamma \right) _{j>1}}}\right) . \end{aligned}$$

One has $y_{1}+m\delta <x_{1}$ and for all $j>1$, $y_{j}-m\gamma <x_{j}$, so that by Weak Pareto, $p*x\ P^{p*N}\ w^{m}$. Hence, by transitivity, $p*x\ P^{p*N}\ p*y$. By Replication Invariance, $x\ P^{N}\ y$, which contradicts the supposition in the first part of this step of the proof.

The dimension of $y$ is $n$. The value of $p$ is no greater than

$$\begin{aligned} \frac{\alpha }{\delta }+1= & {} \frac{\alpha \left( m+1\right) }{x_{1}-y_{1}}+1 \\\le & {} 1+\frac{\alpha }{x_{1}-y_{1}}\left( \frac{\alpha ^{\prime }+1}{\beta }+2\right) \\= & {} 1+\frac{\alpha }{\beta ^{\prime }}\left( \frac{\alpha ^{\prime }+1}{\beta }+2\right) \qquad (\text {case }q>\beta ^{\prime }) \\= & {} 1+\frac{\alpha }{q}\left( \frac{\alpha ^{\prime }+1}{\beta }+2\right) \qquad (\text {case }q\le \beta ^{\prime }) \end{aligned}$$

which implies that the possible size of a $p$-replica of $y$ is at most $n$ times this quantity. Therefore the above contradiction will occur if ${\mathcal {N}}$ contains all populations of that size or less. $\square$

Appendix 2: questionnaire

The following is the standard version of the questionnaire used in this study. About half of the respondents received an alternate version that presented the second scenario before the first.