Nash welfarism and the distributive implications of informational constraints

Abstract

We study two informational simplicity conditions for aggregating von Neumann–Morgenstern preferences. When the best relevant alternative for each individual cannot be ascertained with confidence (as when allocating an uncertain endowment of goods), Independence of Harmless Expansions requires that the social ranking of lotteries be unaffected by the addition of any alternative that every agent deems at least as good as the one she originally found worst. This axiom, along with the Weak Pareto Principle and Anonymity, characterizes bottom-calibrated Nash welfarism: utilities are calibrated so that the worst alternative is worth zero and lotteries are ranked according to the product of such bottom-calibrated utilities. When the worst relevant alternatives are difficult to identify, replacing Independence of Harmless Expansions by the dual axiom of Independence of Useless Expansions yields a characterization of top-calibrated Nash welfarism: lotteries are ranked according to the opposite of the product of the absolute values of top-calibrated utilities. The distributive implications of our two informational simplicity axioms are thus drastically different: while bottom-calibrated Nash welfarism recommends randomizing between two alternatives that it deems equally good, top-calibrated Nash welfarism is randomization-averse.

Appendix

This Appendix is devoted to the proof of our Theorem.

Pareto Indifference and either (i) Independence of Harmless Expansions or (ii) Independence of Useless Expansions together imply a strong form of neutrality. Let $\Pi ({\mathbb {X}})$ denote the set of permutations on ${\mathbb {X}}.$ If $(X,R_{N})\in {\mathcal {P}},$$\pi \in \Pi ({\mathbb {X}}),$$a\in \Delta (X)$, and $R_{N}\in {\mathcal {R}}^{*}(X)^{N},$ denote by $a^{\pi }$ the lottery on $\pi (X)$ given by $a^{\pi }(\pi (x))=a(x)$ for all $x\in X$, and denote by $R_{N}^{\pi }$ the preference profile on $\pi (X)$ given by $a^{\pi }R_{i}^{\pi }b^{\pi }\Leftrightarrow aR_{i}b$ for all $i\in N$ and all $a,b\in \Delta (X).$

Neutrality. For all $(X,R_{N})\in {\mathcal {P}}$, $a,b\in \Delta (X)$ and $\pi \in \Pi ({\mathbb {X}}),$$a{\mathbf {R}}(X,R_{N})b \Leftrightarrow a^{\pi }{\mathbf {R}}(\pi (X),$$R_{N}^{\pi })b^{\pi }.$

Lemma

If an aggregation rule satisfies Pareto Indifference and Independence of Harmless Expansions, then it satisfies Neutrality.

This result is a corollary to Lemma 1 in Sprumont (2013); we include a full proof here to make the presentation self-contained.

Proof

Let ${\mathbf {R}}$ satisfy Pareto Indifference and either (i) Independence of Harmless Expansions or (ii) Independence of Useless Expansions. Let $(X,R_{N})\in {\mathcal {P}}$, $a,b\in \Delta (X)$ and $\pi \in \Pi ({\mathbb {X}}).$ We prove that $a{\mathbf {R}}(X,R_{N})b\Rightarrow a^{\pi }{\mathbf {R}}(\pi (X),R_{N}^{\pi })b^{\pi }.$ The converse implication follows immediately since $a=(a^{\pi })^{\pi ^{-1}},$$b=(b^{\pi })^{\pi ^{-1}},$$X=\pi ^{-1}(\pi (X)),$ and $R_{N}=(R_{N}^{\pi })^{\pi ^{-1}}.$ Let us thus assume that

$$\begin{aligned} a{\mathbf {R}}(X,R_{N})b. \end{aligned}$$

(3)

Step 1. $a^{\pi }{\mathbf {R}}(\pi (X),R_{N}^{\pi })b^{\pi }$if $\pi (X)\cap X=\emptyset .$

Let ${\overline{X}}=X\cup \pi (X).$ For each $i\in N$, let ${\overline{R}}_{i}$ be the vN–M preference ordering over $\Delta ({\overline{X}})$ which coincides with $R_{i}$ on $\Delta (X)$ and is such that $x{\overline{I}}_{i}\pi (x)$ for all $x\in X.$ This is well defined because $\pi (X)\cap X=\emptyset .$ Observe that ${\overline{R}}_{i}$ coincides with $R_{i}^{\pi }$ on $\Delta (\pi (X)).$ Moreover, we have (i) $x{\overline{R}}_{i}a$ for all $x\in {\overline{X}}$ and all $a\in \Delta _{w}(X,R_{i})\cup \Delta _{w}(\pi (X),R_{i}^{\pi })$ and (ii) $a{\overline{R}}_{i}x$ for all $x\in {\overline{X}}$ and all $a\in \Delta _{b}(X,R_{i})\cup \Delta _{b}(\pi (X),R_{i}^{\pi }).$ Let ${\overline{R}}_{N}=({\overline{R}}_{1},\ldots ,{\overline{R}}_{n}).$ Applying either Independence of Harmless Expansions or Independence of Useless Expansions to (3),

$$\begin{aligned} a{\mathbf {R}}({\overline{X}},{\overline{R}}_{N})b. \end{aligned}$$

(4)

Since $a^{\pi }{\overline{I}}_{i}a$ and $b^{\pi }{\overline{I}}_{i}b$ for all $i\in N,$ Pareto Indifference implies $a^{\pi }{\mathbf {I}}({\overline{X}}, {\overline{R}}_{N})a$ and $b^{\pi }{\mathbf {I}}({\overline{X}},{\overline{R}}_{N})b.$ Hence, from (4),

$$\begin{aligned} a^{\pi }{\mathbf {R}}({\overline{X}},{\overline{R}}_{N})b^{\pi }. \end{aligned}$$

(5)

Applying Independence of Harmless Expansions or Independence of Useless Expansions to (5) and recalling that ${\overline{R}}_{N}$ coincides with $R_{N}^{\pi }$ on $\Delta (\pi (X)),$ we obtain $a^{\pi }{\mathbf {R}}(\pi (X),R_{N}^{\pi })b^{\pi }.$

Step 2. $a^{\pi }{\mathbf {R}}(\pi (X),R_{N}^{\pi })b^{\pi }.$

Choose $\rho \in \Pi ({\mathbb {X}})$ such that $\rho (X)\cap X=\rho (X)\cap \pi (X)=\emptyset .$ By Step 1, (3) implies

$$\begin{aligned} a^{\rho }\mathbf {R(}\rho (X),R_{N}^{\rho })b^{\rho }. \end{aligned}$$

(6)

Next consider the permutation $\pi \circ \rho ^{-1}\in \Pi ({\mathbb {X}}).$ Since $(\pi \circ \rho ^{-1})(\rho (X))\cap \rho (X)=\emptyset$, Step 1 and (6) imply

$$\begin{aligned} (a^{\rho })^{\pi \circ \rho ^{-1}}\mathbf {R(}(\pi \circ \rho ^{-1})(\rho (X)),(R_{N}^{\rho })^{\pi \circ \rho ^{-1}})(b^{\rho })^{\pi \circ \rho ^{-1}}. \end{aligned}$$

(7)

By definition, $(\pi \circ \rho ^{-1})(\rho (X))=\pi (X).$ Moreover, $(a^{\rho })^{\pi \circ \rho ^{-1}}=a^{\pi }$ since $(a^{\rho })^{\pi \circ \rho ^{-1}}(\pi (x))$$=(a^{\rho })^{\pi \circ \rho ^{-1}}((\pi \circ \rho ^{-1})(\rho (x)))$$=a^{\rho }(\rho (x))$$=a(x)$ for all $x\in X.$ Likewise, $(b^{\rho })^{\pi \circ \rho ^{-1}}=b^{\pi }$ and $(R_{N}^{\rho })^{\pi \circ \rho ^{-1}}=R_{N}^{\pi }.$ Hence, (7) reduces to $a^{\pi }{\mathbf {R}}(\pi (X),R_{N}^{\pi })b^{\pi }.$${{\square }}$

With the above Lemma in hand, we now turn to the proof of our main result.

Proof of the Theorem

Proof of statement (a). The proof of the “if” part is straightforward. To prove the converse statement, fix an aggregation rule ${\mathbf {R}}$ satisfying the Weak Pareto Principle, Anonymity, and Independence of Harmless Expansions. This rule satisfies Pareto Indifference, hence also Neutrality, by the above lemma.

If $(X,R_{N})\in {\mathcal {P}}$ and $u_{i}\in {\mathcal {U}}_{0}(X,R_{i})$ for each $i\in N,$ define $u_{N}:\Delta (X)\rightarrow {\mathbb {R}} _{+}^{N}$ by $u_{N}(a)=(u_{1}(a),\ldots ,u_{n}(a))$ for all $a\in \Delta (X).$ With some abuse of notation, let ${\mathcal {U}}_{0}(X,R_{N})=\prod \limits _{i \in N}{\mathcal {U}}_{0}(X,R_{i}).$ Define the binary relations $\succ ,$$\sim ,$ and $\succsim$ on ${\mathbb {R}} _{+}^{N}$ as follows:

(i) $v\succ w$ if and only if there exist $(X,R_{N})\in {\mathcal {P}} ,$$u_{N}\in {\mathcal {U}}_{0}(X,R_{N})$, and $a,b\in \Delta (X)$ such that $u_{N}(a)=v,$$u_{N}(b)=w,$ and $a{\mathbf {P}}(X,R_{N})b,$

(ii) $v\sim w$ if and only if there exist $(X,R_{N})\in {\mathcal {P}} ,$$u_{N}\in {\mathcal {U}}_{0}(X,R_{N})$, and $a,b\in \Delta (X)$ such that $u_{N}(a)=v,$$u_{N}(b)=w,$ and $a{\mathbf {I}}(X,R_{N})b,$

(iii) $v\succsim w$ if and only if $v\succ w$ or $v\sim w.$

We note that an equivalent definition of the relations $\succ ,\sim ,$ and $\succsim$ is obtained by replacing $\Delta (X)$ with X in statements (i) and (ii). To see this, fix $v,w\in {\mathbb {R}} _{+}^{N}$ and suppose there exist $(X,R_{N})\in {\mathcal {P}},$$u_{N}\in {\mathcal {U}}_{0}(X,R_{N})$, and $a,b\in \Delta (X)$ such that $u_{N}(a)=v,$$u_{N}(b)=w,$ and $a{\mathbf {P}}(X,R_{N})b$ (respectively, $a{\mathbf {I}} (X,R_{N})b$). Choose distinct pure alternatives $x,y\in {\mathbb {X}}\setminus X$ and let $X^{\prime }=X\cup \left\{ x,y\right\} .$ For each $i\in N,$ let $R_{i}^{\prime }$ be the vN–M preference ordering on $\Delta (X^{\prime })$ which coincides with $R_{i}$ on $\Delta (X)$ and is such that $xI_{i}^{\prime }a$ and $yI_{i}^{\prime }b,$ and let $R_{N}^{\prime }=(R_{1}^{\prime },\ldots ,R_{n}^{\prime }).$ Then $(X^{\prime },R_{N}^{\prime })\in {\mathcal {P}},$$x,y\in X^{\prime },$ and $u_{N}(x)=v,$$u_{N}(y)=w.$ Independence of Harmless Expansions and Pareto Indifference imply $x\mathbf {I }(X^{\prime },R_{N}^{\prime })a{\mathbf {P}}(X^{\prime },R_{N}^{\prime })b {\mathbf {I}}(X^{\prime },R_{N}^{\prime })y$ (respectively, $x{\mathbf {I}} (X^{\prime },R_{N}^{\prime })a{\mathbf {I}}(X^{\prime },R_{N}^{\prime })b {\mathbf {I}}(X^{\prime },R_{N}^{\prime })y$).

Step 1. (i) The binary relation $\succsim$ is an ordering and (ii) for all $v,w\in {\mathbb {R}} _{+}^{N},$ one and only one of the following statements holds: (a) $v\succ w,$ (b) $w\succ v,$ (c) $v\sim w.$

To prove reflexivity and completeness of $\succsim$, fix two (possibly equal) vectors $v,w\in {\mathbb {R}} _{+}^{N}.$ Let $x_{0},x_{1},x_{2},x_{3}\in {\mathbb {X}}$ be four distinct alternatives and let $X=\left\{ x_{0},x_{1},x_{2},x_{3}\right\} .$ For each $i\in N,$ choose a number $z_{i}\in {\mathbb {R}} _{+}$ such that $z_{i}\ne v_{i},w_{i},$ and let $u_{i}:\Delta (X)\rightarrow {\mathbb {R}} _{+}$ be the (unique) vN–M function such that $u_{i}(x_{0})=v_{i},$$u_{i}(x_{1})=w_{i},$$u_{i}(x_{2})=0,$ and $u_{i}(x_{3})=z_{i}.$ Let $R_{i}$ be the preference ordering on $\Delta (X)$ represented by $u_{i}:$ by construction, $R_{i}\in {\mathcal {R}}^{*}(X)$. Letting $u_{N}:=(u_{1},\ldots ,u_{n})$ and $R_{N}=(R_{1},\ldots ,R_{N}),$ we have $(X,R_{N})\in {\mathcal {P}}$ and $u_{N}\in {\mathcal {U}}_{0}(X,R_{N}).$ Since ${\mathbf {R}}(X,R_{N})$ is complete and reflexive, $x_{1}{\mathbf {R}} (X,R_{N})x_{2}$ or $x_{2}{\mathbf {R}}(X,R_{N})x_{1},$ implying that $v\succsim w$ or $w\succsim v.$

To prove transitivity of $\succsim ,$ fix $v^{1},v^{2},v^{3}\in {\mathbb {R}} _{+}^{N}$ such that $v^{1}\succsim v^{2}\succsim v^{3}.$ By definition, there exist $(X^{1},R_{N}^{1}),(X^{2},R_{N}^{2})\in {\mathcal {P}},$$u_{N}^{1}\in {\mathcal {U}}_{0}(X^{1},R_{N}^{1}),$$u_{N}^{2}\in {\mathcal {U}} _{0}(X^{2},R_{N}^{2}),$$x^{1},y^{1}\in X^{1},$ and $x^{2},y^{2}\in X^{2}$ such that

$$\begin{aligned} u_{N}^{1}\left( x^{1}\right) =v^{1},\text { }u_{N}^{1}\left( y^{1}\right) =v^{2}=u_{N}^{2}\left( x^{2}\right) ,\text { and }u_{N}^{2}(y^{2})=v^{3}, \end{aligned}$$

(8)

and

$$\begin{aligned} x^{1}{\mathbf {R}}(X^{1},R_{N}^{1})y^{1}\text { and }x^{2}{\mathbf {R}} (X^{2},R_{N}^{2})y^{2}. \end{aligned}$$

(9)

By Neutrality, we may assume that $X^{1}\cap X^{2}=\varnothing .$ Let $X=X^{1}\cup X^{2}.$ For each $i\in N,$ let $u_{i}:\Delta (X)\rightarrow {\mathbb {R}} _{+}$ be the vN–M function such that

$$\begin{aligned} u_{i}(x)=\left\{ \begin{array}{c} u_{i}^{1}(x)\quad \text { if }\;x\in X^{1}, \\ u_{i}^{2}(x)\quad \text { if }\;x\in X^{2}. \end{array} \right. \end{aligned}$$

(10)

Let $R_{i}$ be the vN–M preference ordering on $\Delta (X)$ represented by $u_{i},$ let $u_{N}=(u_{1},\ldots ,u_{n}),$ and let $R_{N}=(R_{1},\ldots ,R_{n}).$

Note that $R_{N}$ coincides with $R_{N}^{1}$ on $\Delta (X^{1})$ and with $R_{N}^{2}$ on $\Delta (X^{2}).$ Moreover, because $u_{N}^{1}\in {\mathcal {U}} _{0}(X^{1},R_{N}^{1})$ and $u_{N}^{2}\in {\mathcal {U}}_{0}(X^{2},R_{N}^{2}),$ ( 10) implies that $xR_{i}a_{i}$ for all $x\in X,$ all $a_{i}\in \Delta _{w}(X^{1},R_{i}^{1})\cup \Delta _{w}(X^{2},R_{i}^{2}),$ and all $i\in N.$ We may, therefore, apply Independence of Harmless Expansions to (9) and conclude

$$\begin{aligned} x^{1}{\mathbf {R}}(X,R_{N})y^{1}\text { and }x^{2}{\mathbf {R}}(X,R_{N})y^{2}. \end{aligned}$$

On the other hand, (8) and (10) imply $y^{1}I_{i}x^{2}$ for all $i\in N$, hence by Pareto Indifference,

$$\begin{aligned} y^{1}{\mathbf {I}}(X,R_{N})x^{2}. \end{aligned}$$

Transitivity of ${\mathbf {R}}(X,R_{N})$ now implies $x^{1}{\mathbf {R}} (X,R_{N})y^{2}.$ Since $(X,R_{N})\in {\mathcal {P}},$$u_{N}\in {\mathcal {U}} _{0}(X,R_{N})$, and $u_{N}(x^{1})=v^{1}$ and $u_{N}(y^{2})=v^{3},$ the definition of $\succsim$ gives us $v^{1}\succsim v^{3}.$ This completes the proof of statement (i).

The proof of statement (ii) is similar to the proof of transitivity of $\succsim$ and, therefore, omitted.

Step 2. $v\succsim w$ $\Leftrightarrow$ $\prod \limits _{i\in N}v_{i}\ge \prod \limits _{i\in N}w_{i}$ for all $v,w\in {\mathbb {R}} _{+}^{N}.$

We use $\ge ,>,\gg$ to write inequalities in ${\mathbb {R}} _{+}^{N}.$ We begin by establishing three properties of $\succsim .$

Step 2.1. $\succsim$ is scale invariant: $v\succsim w\Leftrightarrow \lambda *v\succsim \lambda *w$ for all$\lambda \in {\mathbb {R}} _{++}^{N},$where$\lambda *v=(\lambda _{1}v_{1},\ldots ,\lambda _{n}v_{n}).$

To check this point, fix $v,w\in {\mathbb {R}} _{+}^{N},$$\lambda \in {\mathbb {R}} _{++}^{N},$ and suppose $v\succ w$ (respectively, $v\sim w$). By definition, there exist $(X,R_{N})\in {\mathcal {P}},$$u_{N}\in {\mathcal {U}}_{0}(X,R_{N}),$ and $a,b\in \Delta (X)$ such that $u_{N}(a)=v,$$u_{N}(b)=w,$ and $a\mathbf {P }(X,R_{N})b$ (respectively, $a{\mathbf {I}}(X,R_{N})b$). Observe that $\lambda *u_{N}:=(\lambda _{1}u_{1},\ldots ,\lambda _{n}u_{n})\in {\mathcal {U}} _{0}(X,R_{N})$ and $(\lambda *u_{N})(a)=\lambda *v,$$(\lambda *u_{N})(b)=\lambda *w.$ Since $(X,R_{N})\in {\mathcal {P}}$ and $a{\mathbf {P}} (X,R_{N})b$ (respectively, $a{\mathbf {I}}(X,R_{N})b$), the definition of $\succ$ (respectively, $\sim$) implies $\lambda *v\succ \lambda *w$ (respectively, $\lambda *v\sim \lambda *w$), as desired.

Step 2.2. $\succsim$ is weakly monotonic: $v\gg w\Rightarrow v\succ w.$

This follows immediately from the fact that ${\mathbf {R}}$ satisfies (part (i) of) the Weak Pareto Principle.

Step 2.3. $\succsim$is symmetric: $v\sim \sigma v$ for all $\sigma \in \Pi (N),$ where $\sigma v$ is the vector defined by $(\sigma v)_{\sigma (i)}=v_{i}$ for all $i\in N.$

Check first that for all $v,w\in {\mathbb {R}} _{+}^{N}$ and $\sigma \in \Pi (N),$$v\succ w\Rightarrow \sigma v\succ \sigma w$. Indeed, if $v\succ w,$ then by definition, there exist $(X,R_{N})\in {\mathcal {P}},$$u_{N}\in {\mathcal {U}}_{0}(X,R_{N}),$ and $a,b\in \Delta (X)$ such that $u_{N}(a)=v,$$u_{N}(b)=w,$ and $a{\mathbf {P}} (X,R_{N})b.$ By Anonymity,

$$\begin{aligned} a{\mathbf {P}}(X,\sigma R_{N})b. \end{aligned}$$

(11)

Defining $\sigma u_{N}=((\sigma u)_{1},\ldots ,(\sigma u)_{n})$ by $(\sigma u)_{\sigma (i)}=u_{i}$ for all $i\in N,$ we have $\sigma u_{N}\in {\mathcal {U}} _{0}(X,\sigma R_{N}).$ Moreover, $(\sigma u_{N})(a)=\sigma v$ and $(\sigma u_{N})(b)=\sigma w.$ It then follows from (11) and the definition of $\succ$ that $\sigma v\succ \sigma w,$ as desired.

To prove Step 2.3, fix now $v\in {\mathbb {R}} _{+}^{N}$ and $\sigma \in \Pi (N).$ Because every bijection on N is a composition of permutations exchanging only two individuals, and because $\sim$ is transitive, we may assume without loss of generality that there exist $i,j\in N$ such that $\sigma (i)=j,$$\sigma (j)=i,$ and $\sigma (k)=k$ for all $k\ne i,j.$ Note that, since $\sigma ^{-1}=\sigma ,$ we have $\sigma \sigma v=v.$ Suppose, contrary to the claim, that, say, $v\succ \sigma v.$ By the result established in the previous paragraph, it follows that $\sigma v\succ \sigma \sigma v=v.$ Thus, $v\succ \sigma v$ and $\sigma v\succ v,$ contradicting Step 1, statement (ii).

Step 2.4. $v\sim (\underbrace{1,\ldots ,1}_{n-1},\prod \limits _{i\in N}v_{i})$ for all $v\in {\mathbb {R}} _{+}^{N}.$

For $k=0,\ldots ,n,$ define $V^{k}=\left\{ v\in {\mathbb {R}} _{+}^{N}\mid \left\{ i\in N\mid v_{i}\ne 1\right\} \le k\right\} ,$ the set of vectors in ${\mathbb {R}} _{+}^{N}$ having at most k coordinates different from 1. Note that $V^{0}=\left\{ (1,\ldots ,1)\right\}$ and $V^{n}= {\mathbb {R}} _{+}^{N}.$ Trivially,

$$\begin{aligned} v\sim (\underbrace{1,\ldots ,1}_{n-1},\prod \limits _{i\in N}v_{i})\text { for all }v\in V^{0}. \end{aligned}$$

Next, we proceed by induction: we fix k such that $0\le k<n,$ make the induction hypothesis

$$\begin{aligned} v\sim (\underbrace{1,\ldots ,1}_{n-1},\prod \limits _{i\in N}v_{i})\text { for all }v\in V^{k}. \end{aligned}$$

and prove that $v\sim (\underbrace{1,\ldots ,1}_{n-1},\prod \limits _{i\in N}v_{i})$ for all $v\in V^{k+1}.$

Let $v\in V^{k+1}$ and, to avoid triviality, suppose $v\notin V^{k}:$ exactly $k+1$ coordinates of v differ from 1. Without loss of generality, say $v=(v_{1},\ldots ,v_{k+1},\underbrace{1,\ldots ,1}_{n-k-1}),$ with $v_{i}\ne 1$ for $i=1,\ldots ,k+1.$ By symmetry of $\sim ,$

$$\begin{aligned} (v_{1},\ldots ,v_{k},\underbrace{1,\ldots ,1}_{n-k})\sim (v_{1},\ldots ,v_{k-1},1,v_{k}, \underbrace{1,\ldots ,1}_{n-k-1}). \end{aligned}$$

(12)

Since $\succsim$ is scale invariant, (12) implies

$$\begin{aligned} v\sim (v_{1},\ldots ,v_{k-1},1,v_{k}\cdot v_{k+1},\underbrace{1,\ldots ,1}_{n-k-1}). \end{aligned}$$

By the induction hypothesis,

$$\begin{aligned} (v_{1},\ldots ,v_{k-1},1,v_{k}\cdot v_{k+1},\underbrace{1,\ldots ,1}_{n-k-1})\sim ( \underbrace{1,\ldots ,1}_{n-1},\prod \limits _{i\in N}v_{i}); \end{aligned}$$

hence, by transitivity of $\sim ,$

$$\begin{aligned} v\sim (\underbrace{1,\ldots ,1}_{n-1},\prod \limits _{i\in N}v_{i}). \end{aligned}$$

Step 2.5. $v\succsim w$$\Leftrightarrow$$\prod \limits _{i\in N}v_{i}\ge \prod \limits _{i\in N}w_{i}$for all $v,w\in {\mathbb {R}} _{+}^{N}.$

First, observe that

$$\begin{aligned} v\sim (\prod \limits _{i\in N}v_{i}^{\frac{1}{n}},\ldots ,\prod \limits _{i\in N}v_{i}^{\frac{1}{n}})\text { for all }v\in {\mathbb {R}} _{+}^{N}. \end{aligned}$$

(13)

This is simply because Step 2.4 implies both $v\sim \left( \underbrace{1,\ldots ,1} _{n-1},\prod \limits _{i\in N}v_{i}\right)$ and $\left( \prod \limits _{i\in N}v_{i}^{ \frac{1}{n}},\ldots ,\prod \limits _{i\in N}v_{i}^{\frac{1}{n}}\right)$$\sim \left( \underbrace{1,\ldots ,1}_{n-1},\prod \limits _{i\in N}\prod \limits _{i\in N}v_{i}^{\frac{1}{n}}{}_{i}\right) =\left( \underbrace{1,\ldots ,1}_{n-1},\prod \limits _{i\in N}v_{i}\right)$.

Next, fix $v,w\in {\mathbb {R}} _{+}^{N}.$ It is enough to show that $\left[ \prod \limits _{i\in N}v_{i}=\prod \limits _{i\in N}w_{i}\right] \Rightarrow \left[ v\sim w\right]$ and $\left[ \prod \limits _{i\in N}v_{i}>\prod \limits _{i\in N}w_{i}\right] \Rightarrow \left[ v\succ w\right] .$ If $\prod \limits _{i\in N}v_{i}=\prod \limits _{i\in N}w_{i},$ then (13) implies $v\sim (\prod \limits _{i\in N}v_{i}^{\frac{1}{n}},\ldots ,\prod \limits _{i\in N}v_{i}^{ \frac{1}{n}})=(\prod \limits _{i\in N}w_{i}^{\frac{1}{n}},\ldots ,\prod \limits _{i\in N}w_{i}^{\frac{1}{n}})\sim w.$ If $\prod \limits _{i\in N}v_{i}>\prod \limits _{i\in N}w_{i},$ then (13) implies $v\sim (\prod \limits _{i\in N}v_{i}^{\frac{1}{n}},\ldots ,\prod \limits _{i\in N}v_{i}^{ \frac{1}{n}})\gg (\prod \limits _{i\in N}w_{i}^{\frac{1}{n} },\ldots ,\prod \limits _{i\in N}w_{i}^{\frac{1}{n}})\sim w$. Since $\succsim$ is weakly monotonic, $(\prod \limits _{i\in N}v_{i}^{\frac{1}{n} },\ldots ,\prod \limits _{i\in N}v_{i}^{\frac{1}{n}})\succ (\prod \limits _{i\in N}w_{i}^{\frac{1}{n}},\ldots ,\prod \limits _{i\in N}w_{i}^{\frac{1}{n}})$ and, by transitivity of $\succsim ,$$v\succ w.$

Step 3. For all $(X,R_{N})\in {\mathcal {P}}$and all$a,b\in \Delta (X),$

$$\begin{aligned} a{\mathbf {R}}(X,R_{N})b\Leftrightarrow \prod \limits _{i\in N}u_{i}(a)\ge \prod \limits _{i\in N}u_{i}(b)\text { for all }u_{N}\in {\mathcal {U}} _{0}(X,R_{N}). \end{aligned}$$

Fix $(X,R_{N})\in {\mathcal {P}}$ and $a,b\in \Delta (X).$

Suppose $\prod \limits _{i\in N}u_{i}(a)\ge \prod \limits _{i\in N}u_{i}(b)$ for all $u_{N}\in {\mathcal {U}}_{0}(X,R_{N}).$ Fix any $u_{N}\in {\mathcal {U}} _{0}(X,R_{N}).$ By Step 2, $u_{N}(a)\succsim u_{N}(b).$ If we had $b\mathbf {P }(X,R_{N})a,$ the definition of $\succ$ would entail $u_{N}(b)\succ u_{N}(a);$ hence, by Step 2 $\prod \limits _{i\in N}u_{i}(b)>\prod \limits _{i\in N}u_{i}(a),$ a contradiction. Therefore, $a {\mathbf {R}}(X,R_{N})b.$

Next, suppose $\prod \limits _{i\in N}u_{i}(b)>\prod \limits _{i\in N}u_{i}(a)$ for some $u_{N}\in {\mathcal {U}}_{0}(X,R_{N}).$ Then Step 2 implies $u_{N}(b)\succ u_{N}(a).$ If we had $a{\mathbf {R}}(X,R_{N})b,$ the definition of $\succsim$ would entail $u_{N}(a)\succsim u_{N}(b);$ hence, by Step 2 $\prod \limits _{i\in N}u_{i}(a)\ge \prod \limits _{i\in N}u_{i}(b),$ a contradiction. Therefore, $b{\mathbf {P}}(X,R_{N})a.$

Proof of statement (b). Again, the proof of the “if” part is straightforward. To prove the converse statement, fix an aggregation rule ${\mathbf {R}}$ satisfying the Weak Pareto Principle, Anonymity, and Independence of Useless Expansions. For any $X\in {\mathcal {X}}$, any $R_{N}=(R_{1},\ldots ,R_{n})\in {\mathcal {R}} ^{*}(X)^{N},$ and any $i\in N$, let $R_{i}^{-1}\in {\mathcal {R}}^{*}(X)$ be the opposite of $R_{i}$ (defined by $aR_{i}^{-1}b\Leftrightarrow bR_{i}a$ for all $a,b\in \Delta (X)$) and let $R_{N}^{-1}:=(R_{1}^{-1},\ldots ,R_{n}^{-1}).$

Define the rule $\widetilde{{\mathbf {R}}}$ as follows: for all $(X,R_{N})\in {\mathcal {P}}$ and all $a,b\in \Delta (X),$

$$\begin{aligned} a\widetilde{{\mathbf {R}}}(X,R_{N})b\Leftrightarrow b{\mathbf {R}}(X,R_{N}^{-1})a. \end{aligned}$$

It is straightforward to check that $\widetilde{{\mathbf {R}}}$ satisfies the Weak Pareto Principle, Anonymity, and Independence of Harmless Expansions. To complete the proof, it, therefore, suffices to invoke statement (a). For all $(X,R_{N})\in {\mathcal {P}}$ and $a,b\in \Delta (X),$

$$\begin{aligned}&a{\mathbf {R}}(X,R_{N})b \\&\quad \Leftrightarrow b\widetilde{{\mathbf {R}}}(X,R_{N}^{-1})a \\&\quad \Leftrightarrow \prod \limits _{i\in N}u_{i}(b)\ge \prod \limits _{i\in N}u_{i}(a)\text { for all }u_{N}\in {\mathcal {U}}_{0}(X,R_{N}^{-1}), \\&\quad \Leftrightarrow \prod \limits _{i\in N}(-u_{i}(b))\ge \prod \limits _{i\in N}(-u_{i}(a))\text { for all }u_{N}\in {\mathcal {U}}^{0}(X,R_{N}) \\&\quad \Leftrightarrow -\prod \limits _{i\in N}(-u_{i}(a))\ge -\prod \limits _{i\in N}(-u_{i}(b))\text { for all }u_{N}\in {\mathcal {U}} ^{0}(X,R_{N}) \\&\quad \Leftrightarrow a{\mathbf {R}}^{0}(X,R_{N})b, \end{aligned}$$

where the second equivalence holds by statement (a) (because $\widetilde{ {\mathbf {R}}}$ satisfies the Weak Pareto Principle, Anonymity, and Independence of Harmless Expansions) and the third because, for all $(X,R_{N})\in {\mathcal {P}}$, $u_{N}\in {\mathcal {U}}_{0}(X,R_{N}^{-1})$ if and only if $-u_{N}\in {\mathcal {U}}^{0}(X,R_{N}).$${{\square }}$