Multidimensional Pigou–Dalton transfers and social evaluation functions

Abstract

We axiomatize, in the multidimensional case, a social evaluation function that can accommodate a natural Pigou–Dalton principle and correlation increasing majorization. This is performed by building upon a simple class of inframodular functions proposed by Müller and Scarsini under risk.

Appendices

Appendix

We split up the Appendix in two parts: Appendix A includes the proof of Theorem 1 and Appendix B gives the proofs of Lemmas 1 and 2.

1.1 Appendix A

Theorem 1

Proof

We give only the sufficient part, since the necessary proof is immediate.

From A.1, A.2, A.3, A.4 (weak order, continuity, monotonicity and independence) and $n\ge 3$, Theorem 3 in Debreu (1960) implies that there exist n increasing¹⁴ and continuous functions $u_{j}:\mathbb {R}^{m}\longrightarrow \mathbb {R}$ such that

$$\begin{aligned} A\succsim B\Longleftrightarrow \sum \limits _{j=1}^{n}u_{j}(A_{j})\ge \sum \limits _{j=1}^{n}u_{j}(B_{j}) \end{aligned}$$

where $u_{j}$ are unique up to affine transformation $\alpha u_{j}+\beta _{j}$ with $\alpha >0$ and $\beta _{j}\in \mathbb {R}$. Thus, we can assume that for all j, $u_{j}(0)=0$.

From A.5 (Anonymity), let us see that we can assume that there exists $u:\mathbb {R}^{m}\longrightarrow \mathbb {R}$, increasing and continuous such that

$$\begin{aligned} A\succsim B\Longleftrightarrow \sum \limits _{j=1}^{n}u(A_{j})\ge \sum \limits _{j=1}^{n}u(B_{j}) \end{aligned}$$

(A.1)

Then, fix $u_{j}$ such that $u_{j}(0)=0$, for all j. By symmetry, we only need to prove that $u_{1}=u_{2}.$ Take any $A_{1}\in \mathbb {R}^{m}$ and consider $\left( A_{1},0,A_{3},\ldots ,A_{n}\right)$ and $\left( 0,A_{1},A_{3},\ldots ,A_{n}\right)$. Through A.5: $u_{1}(A_{1})+u_{2}(0)+\sum _{j=3}^{n}u_{j}(A_{j})= u_{1}(0)+u_{2}(A_{1})+\sum _{j=3}^{n}u_{j}(A_{j})$, this entails straightforwardly $u_{1}(A_{1})=u_{2}(A_{1})$; thus $u_{1}=u_{2}=\cdots =u_{n}=u.$ Therefore, there exists $u:\mathbb {R}^{m}\longrightarrow \mathbb {R}$ increasing continuous (satisfying $u_{j}(0)=0$) such that (A.1) holds. Clearly, u is defined up to a positive affine transformation $\square$

Appendix B

Marinacci and Montrucchio (2005) provided a thorough analysis of ‘Ultramodular Functions’, thus (by reversing the inequality in the definition) of what Müller and Scarsini (2012) called ‘Inframodular Functions’ defined in Sect. 2.

We intend now to prove that inframodular functions agree with our Pigou–Dalton regressive transfers (see Sect. 1).

Lemma 1

if $u:\mathbb {R}^{n}\rightarrow \mathbb {R}$ is inframodular then u satisfies the property (1) quoted in Sect. 2.

Proof

Let $x,y\in \mathbb {R}^{n}$, $x\le y$ and $\varepsilon \ge 0$. Set $x^{\prime }=x-\varepsilon$ and $y^{\prime }=y$, $\varepsilon ^{\prime }=\varepsilon$, so $x^{\prime }\le y^{\prime }$ and $\varepsilon ^{\prime }\ge 0$. Then from the Definition 1 in Section 2 of u inframodular,

$$\begin{aligned} u(x^{\prime }+\varepsilon )-u(x^{\prime })\ge u(y^{\prime }+\varepsilon )-u(y^{\prime }), \end{aligned}$$

i.e.,

$$\begin{aligned} u(x)-u(x-\varepsilon )\ge u(y+\varepsilon )-u(y) \end{aligned}$$

$\square$

Lemma 2

(Proof of Theorem 4) First, it is known that if u is inframodular, then u is submodular, i.e., for all $a,b\in \mathbb {R}^{m}, u(a)+u(b)\ge u(a\wedge b)+u(a\vee b)$ (see, e.g., Marinacci and Montrucchio 2005).

Let us show it again for sake of completeness.

Let $x=a\wedge b$, so $a=a\wedge b+\varepsilon$ with $\varepsilon \ge 0.$

Let $y=b$, one has $x\le y$ and $\varepsilon \ge 0$ then u inframodular implies $u(x+\varepsilon )-u(x)\ge u(y+\varepsilon )-u(y)$, i.e., $u(a)-u(a\wedge b)\ge u(b+a-a\wedge b)-u(b)$. But $b+a-a\wedge b=a\vee b$, hence, the result

$$\begin{aligned} u(a)+u(b)\ge u(a\wedge b)+u(a\vee b). \end{aligned}$$

Thus, one has $u(A_{1})+u(A_{2})\ge u(A_{1}\vee A_{2})+u(A_{1}\wedge A_{2})$.

Since by hypothesis neither $A_{1}\le A_{2}$ nor $A_{2}\le A_{1}$, u strict inframodular implies $u(A_{1})+u(A_{2})>u(A_{1}\vee A_{2})+u(A_{1}\wedge A_{2})$.

Actually, since not $A_{2}\le A_{1}$, we get $A_{1}\wedge A_{2}<A_{2}\$ then letting $x=A_{1}\wedge A_{2}$, $y=A_{2}$, $\varepsilon =A_{1}- A_{1}\wedge A_{2}$, we get

$$\begin{aligned}&u(x+\varepsilon )+u(y)> u(y+\varepsilon )+u(x),\hbox { i.e., } u(A_{1}) \\&\quad + u(A_{2})>u(A_{1}\vee A_{2})+u(A_{1}\wedge A_{2}). \end{aligned}$$

$\square$