Fair division with uncertain needs

Abstract

Imagine agents having uncertain needs for a resource when the resource has to be divided before uncertainty resolves. In this situation, waste occurs when an agent’s assignment turns out to exceed his realized need. How should the resource be divided in the face of possible waste? This is a question out of the scope of the existing rationing literature. Our main axiom to address the issue is no domination. It requires that no agent receive more of the resource than another while producing a larger expected waste, unless the other agent has been fully compensated. Together with conditional strict endowment monotonicity, consistency, and strong upper composition, a class of rules which we call expected-waste constrained uniform gains rules is characterized. Any such rule is associated with a function that aggregates the two types of cost generated by an agent at an allocation: the amount of the resource assigned to him and the expected waste he generates. The rule selects the allocation that equalizes as much as possible the cost generated by each agent. The subclasses of rules associated with homothetic and linear cost functions are also characterized. Lastly, to appreciate the role of no domination, we establish all the characterizations with a decomposition of no domination into two axioms: risk aversion and no reversal. They respectively capture the ideas that a rule should not be unresponsive to uncertainty, and that neither should it be overly responsive to it.

Appendix

Given $I,I'\in \mathcal N$, $F\in \mathcal F^I$ and $F'\in \mathcal F^{I'}$ with $I\cap I'=\emptyset$, let $(F,F')$ denote the claim profile in $\mathcal F^{I\cup I'}$ such that for each $i\in I$, agent i’s claim is $F_i$, and for each $j\in I'$, agent j’s claim is $F'_j$.

$\displaystyle \mathbf{Anonymity: }$  For each $I\in \mathcal N$, each $(F,T)\in \mathcal C^I$ and each $\pi :I\rightarrow {\mathbb N}$ which is injective, if $(F',T)\in \mathcal C^{\pi (I)}$ is such that for each $i\in I$, $F'_{\pi (i)}=F_i$, then for each $i\in I$, $r_i(F,T)=r_{\pi (i)}(F',T)$.

Anonymity requires the name of agents to have no impact on allocation. It is implied by symmetry and consistency together. For a proof, see Lemma 3 in Chambers and Thomson (2002). We state some useful intermediate results below. Their proofs are in an online appendix.

Lemma 2

If r is endowment monotonic, then (1) for each $I\in \mathcal N$, each $F\in \mathcal F^I$, each $i\in I$, and each $t_i\in [0,\overline{\text {supp }}\ F_i]$, there is a smallest $T\in [0,\sum \overline{\text {supp }}\ F_j]$ such that $r_i(F,T)=t_i$; (2) for each $I\in \mathcal N$, each $(F,T)\in \mathcal C^I$, and each $i\in I$, if $T>0$, $r_i(F,T)=\sup \{r_i(F,T'):T'\in [0,T)\}$.

Lemma 3

Let r be a rule satisfying symmetry and either (i) strong upper composition, or (ii) consistency, lower composition, and claims truncation invariance. For each $I\in \mathcal N$, each $(F,T)\in \mathcal C^I$, and each pair $\{i,j\}\subseteq I$, if there is $c\in [0,\min \{\overline{\text {supp }}\ F_i,\overline{\text {supp }}\ F_j\}]$ such that $F_i$ and $F_j$ agree on $(-\infty ,c)$ and $r_i(F,T)<c$, then $r_i(F,T)=r_j(F,T)$.

Lemma 4

If a rule satisfies positivity and lower composition, then it is conditionally strictly endowment monotonic.

Lemma 5

Let a rule satisfy consistency and strong upper composition. If it additionally satisfies either no domination and conditional strict endowment monotonicity, or risk aversion and no reversal, then it is symmetric.

Lemma 6

Let a rule satisfy consistency, lower composition, and claims truncation invariance. If it additionally satisfies either no domination and positivity, or risk aversion, then it is symmetric.

Proof of Theorem 1

The “if” direction is readily verified, so the proof is omitted. To show the “only if” direction, let r be a rule satisfying no domination, conditional strict endowment monotonicity, consistency, and strong upper composition. By strong upper composition, r is endowment monotonic. By Lemma 5, r is symmetric. By symmetry and consistency, r is anonymous.

$\displaystyle \mathbf{Step~1 }$  For each $I\in \mathcal N$, each $(F,T)\in \mathcal C^I$, each $J\subseteq I$, and each $G\in \mathcal F^J$, if for each $T'\in [0,T)$ and each $i\in J$, $r_i(F,T')<r_i(F,T)$, and for each $i\in J$, $\overline{\text {supp}}\ G_i\ge r_i(F,T)$ and $w(F_i,r_i(F,T))=w(G_i,r_i(F,T))$, then $r((G,F_{I{\setminus } J}),T)=r(F,T)$.

Let $I\in \mathcal N$, $(F,T)\in \mathcal C^I$, $J\subseteq I$, and $G\in \mathcal F^J$ satisfy the required conditions. Let $t:=r(F,T)$. Let $\pi :J\rightarrow {\mathbb N}{\setminus } I$ be an injective mapping, and $F_{\pi (J)}$ a claim profile such that for each $i\in J$, $F_{\pi (i)}=G_i$. Consider $((F,F_{\pi (J)}),T+\sum \limits _{i\in J}t_i)\in \mathcal C^{I\cup \pi (J)}$ and let $t':=r((F,F_{\pi (J)}),T+\sum \nolimits _{i\in J}t_i)$. By consistency and anonymity, it suffices to show that for each $i\in J$, $t'_{\pi (i)}=t_i$. Let $j\in J$. Suppose that $t'_{\pi (j)}> t_j$. Then $t'_j\ge t_j$. To see this, assume that $t_j'<t_j$. Then $t'_j<t'_{\pi (j)}$ and $w(F_j,t'_j)\le w(F_j,t_j)=w(G_j,t_j)\le w(F_{\pi (j)},t'_{\pi (j)})$. If $w(F_j,t_j')<w(F_{\pi (j)},t'_{\pi (j)})$, then by no domination, $t_j'=\overline{\text {supp}}\ F_j$. However, $t_j'<t_j\le \overline{\text {supp}}\ F_j$, which is a contradiction. If $w(F_j,t_j')=w(F_{\pi (j)},t'_{\pi (j)})$, then $w(F_j,t_j')=w(F_j,t_j)=0$ and thus $w(G_j,t_j)=0$. Hence, $F_j$ and $F_{\pi (j)}$ agree on $(-\infty ,t_j)$. Since $t_j'<t_j$, then by Lemma 3, $t_j'=t_{\pi (j)}'$, which is a contradiction. Thus, $t_j'\ge t_j$. Then by consistency, $r_j(F,\sum \nolimits _{i\in I}t_i')=t_j'\ge t_j$. Since for each $T'\in [0,T)$, $r_j(F,T')<t_j$, then $\sum \nolimits _{i\in I}t_i'\ge T$. Hence, $\sum \nolimits _{i\in J}t_{\pi (i)}'\le \sum \nolimits _{i\in J}t_i$. Since $t'_{\pi (j)}>t_j$, then there is $k\in J$ such that $t'_{\pi (k)}<t_k$. By endowment monotonicity and consistency, $t_k\le r_k(F,\sum \nolimits _{i\in I}t_i')=t_k'$. Thus, $t'_{\pi (k)}<t_k$ and $w(F_{\pi (k)},t'_{\pi (k)})\le w(G_k,t_k)=w(F_k,t_k)\le w(F_k,t_k')$. If $w(F_{\pi (k)},t'_{\pi (k)})<w(F_k,t_k')$, then by no domination, $t_{\pi (k)}'=\overline{\text {supp}}\ F_{\pi (k)}$. However, $t_{\pi (k)}'<t_k\le \overline{\text {supp}}\ G_k=\overline{\text {supp}}\ F_{\pi (k)}$, which is a contradiction. If $w(F_{\pi (k)},t'_{\pi (k)})=w(F_k,t_k')$, then $w(F_{\pi (k)},t_{\pi (k)}')=w(G_k,t_k)=0$ and $w(F_k,t_k)=0$. Hence, $F_k$ and $F_{\pi (k)}$ agree on $(-\infty ,t_k)$. Since $t_{\pi (k)}'<t_k$, then $t'_{\pi (k)}=t'_k$, which is again a contradiction. Hence, $t'_{\pi (j)}>t_j$ is not possible. Similarly, $t'_{\pi (j)}<t_j$ is not possible, either.

$\displaystyle \mathbf{Step~2 }$  Define a binary relation $\simeq$ of “as costly as” between vectors in D as follows. For each pair $(x_1,x_2),(x_1',x_2')\in D$, $(x_1,x_2)\simeq (x_1',x_2')$ if there are $I\in \mathcal N$, $(F,T)\in \mathcal C^I$ and $\{i,j\}\subseteq I$ with $w(F_i,x_1)=x_2$ and $w(F_j,x'_1)=x'_2$, and satisfying

• (a) $r_i(F,T)=x_1$, $r_j(F,T)=x_1'$;

• (b) for each $T'\in [0,T)$, $r_i(F,T')<x_1$, $r_j(F,T')<x_1'$.

Intuitively, $(x_1,x_2)$ seems revealed to be as costly as $(x_1',x_2')$ if in a two-agent problem, one agent is assigned $x_1$, generating expected waste $x_2$, and the other agent is assigned $x_1'$, generating expected waste $x_2'$. This is actually what condition (a) says. But this condition is not sufficient. If the endowment decreases and one agent’s assignment, say $x_1$, remains unchanged, then his expected waste, $x_2$, is also unchanged, whereas the other agent’s assignment must decrease and his expected waste not increase. In this case, if we imposed only condition (a) when defining the relation $\simeq$, $(x_1,x_2)$ would then be as costly as two vectors, one of which dominates the other, violating the monotonicity property that we intend to have. To avoid that, condition (b) further requires that as the endowment decreases, neither agent’s assignment remain unchanged, which is equivalent, in light of endowment monotonicity, to saying that both of their assignments decrease.

We claim that $\simeq$ is an equivalence relation. Moreover, for each pair $(x_1,x_2)$ and $(x_1',x_2')$ in D, if $x_1<x_1'$ and either $x_2<x_2'$ or $x_2=x_2'=0$, then $(x_1,x_2)\not \simeq (x_1',x_2')$.

By symmetry, $\simeq$ is reflexive. By definition, $\simeq$ is symmetric. To show $\simeq$ is transitive, let $(x_1,x_2),(x'_1,x'_2),(x_1'',x_2'')\in D$ be such that $(x_1,x_2)\simeq (x'_1,x'_2)$ and $(x_1',x_2')\simeq (x_1'',x_2'')$. By endowment monotonicity, consistency, and anonymity, there are $(F,T)\in \mathcal C^{\{1,2\}}$ and $(F',T')\in \mathcal C^{\{3,4\}}$ such that $w(F_1,x_1)=x_2$, $w(F_2,x'_1)=w(F_3,x_1')=x_2'$, $w(F_4,x_1'')=x_2''$, and condition (a) and (b) hold for both (F, T) and $(F',T')$.

Let $((F,F'),T+T')\in \mathcal C^{\{1,2,3,4\}}$ and $t:=r((F,F'),T+T')$. We claim that $t_1=x_1$, $t_2=t_3=x_1'$, and $t_4=x_1''$. Suppose that $t_1<x_1$. Then by endowment monotonicity, consistency, and condition (b), $t_2<x'_1\le t_3$. Thus, $t_2<\overline{\text {supp}}\ F_2$ and $w(F_2,t_2)\le w(F_2,x_1')=w(F_3,x'_1)\le w(F_3,t_3)$. If $w(F_2,t_2)<w(F_3,t_3)$, then no domination is violated. If $w(F_2,t_2)=w(F_3,t_3)$, then $w(F_2,t_2)=w(F_2,x_1')=0$, and thus $w(F_3,x_1')=0$. Hence, $F_2$ and $F_3$ agree on $(-\infty ,x_1')$. Since $t_2<x_1'$, then by Lemma 3, $t_2=t_3$, which contradicts $t_2<t_3$. Suppose that $t_1>x_1$. Then by endowment monotonicity, consistency, and condition (b), $t_3<x'_1\le t_2$, which will result in similar contradictions. Hence, $t_1=x_1$. Analogously, $t_4=x_1''$. Moreover, since $t_2+t_3=x_1+2x_1'+x_1''-t_1-t_4=2x_1'$, then by the previous arguments, $t_2=t_3=x_1'$.

Next, let $T''\in [0,T+T')$ and $t':=r((F,F'),T'')$. Since $\sum \nolimits _{i=1}^4t_i'=T''<T+T'$, then either $t'_1+t_2'<T$ or $t'_3+t'_4<T'$. In the former case, by consistency and condition (b), for $i\in \{1,2\}$, $t_i'<t_i$. Since $t_2'<t_2=x_1'$, then by the arguments in the last paragraph, $t_3'<x_1'=t_3$. By endowment monotonicity, consistency, and condition (b), $t_4'<t_4$. Similarly, in the latter case, we can show that $t_1'<t_1$ and $t_4'<t_4$.

Lastly, abusing notation, let $(x_1,x_2),(x_1',x_2')\in D$ be such that $x_1<x_1'$ and $(x_1,x_2)\simeq (x_1',x_2')$. By the definition of $\simeq$, $x_1>0$. Let $(F,T)\in \mathcal C^{\{1,2\}}$ be as above. Let $G_1\in \mathcal {F}$ assign probability $\frac{x_2}{x_1}$ to 0 and $1-\frac{x_2}{x_1}$ to $x_1'$. Note that $\overline{\text {supp}}\ G_1>x_1$ and $w(G_1,x_1)=x_2$. By Step 1, $r((G_1,F_2),T)=r(F,T)$. Then $r_1((G_1,F_2),T)=x_1<x_1'=r_2((G_1,F_2),T)$. Suppose that $x_2<x_2'$. Then $w(G_1,x_1)=x_2<x_2'=w(F_2,x_1')$. Since $r_1((G_1,F_2),T)<r_2((G_1,F_2),T)$ and $w(G_1,x_1)<w(F_2,x_1')$, by no domination, $r_1((G_1,F_2),T)=\overline{\text {supp}}\ G_1$. Thus, $\overline{\text {supp}}\ G_1=x_1$, which contradicts $\overline{\text {supp}}\ G_1>x_1$. Suppose that $x_2=x_2'=0$. Then $w(G_1,x_1)=x_2=x_2'=w(F_2,x_1')=0$. Since $x_1<x_1'$, then $G_1$ and $F_2$ agree on $(-\infty ,x_1)$. Let $\epsilon >0$ be such that $x_1<x_1'-\epsilon$. By conditional strict endowment monotonicity, $r_1((G_1,F_2),T-\epsilon )< x_1$. Then $r_2((G_1,F_2),T-\epsilon )=T-\epsilon -r_1((G_1,F_2),T-\epsilon )=x_1+x_1'-\epsilon -r_1((G_1,F_2),T-\epsilon )>x_1'-\epsilon$. Thus, $r_1((G_1,F_2),T-\epsilon )<r_2((G_1,F_2),T-\epsilon )$. However, by Lemma 3, $r_1((G_1,F_2),T-\epsilon )=r_2((G_1,F_2),T-\epsilon )$, which is a contradiction.

$\displaystyle \mathbf{Step~3 }$  Let $f:{\mathbb R}_+\rightarrow {\mathbb R}_+$ be such that for each $c\in {\mathbb R}_+$, $f(c)=\frac{c}{2}$. For each $(x_1,x_2)\in D$, there is a unique $c\in {\mathbb R}_+$ such that $(x_1,x_2)\simeq (c,f(c))$.

Let $(x_1,x_2)\in D$. Let $F\in \mathcal F^{\{1,2\}}$ be such that $\overline{\text {supp}}\ F_1>x_1$, $w(F_1,x_1)=x_2$, and $F_2$ assigns probability 0.5 respectively to 0 and some $\bar{c}\in {\mathbb R}_+$ satisfying $(\bar{c},f(\bar{c}))>(x_1,x_2)$. Note that for each $t_2\in [0,\overline{\text {supp}}\ F_2]$, $w(F_2,t_2)=f(t_2)$. By Lemma 2, there is $T\in [0,\overline{\text {supp}}\ F_1+\overline{\text {supp}}\ F_2]$ such that $r_1(F,T)=x_1$. Let $c:=r_2(F,T)$. By no domination, $c<\bar{c}$. By conditional strict endowment monotonicity, for each $T'<T$, $r(F,T)<r(F,T')$. Thus, $(x_1,x_2)\simeq (c,f(c))$. By Step 2, such a c is unique.

$\displaystyle \mathbf{Step~4 }$  Define $U:D\rightarrow {\mathbb R}_+$ by setting for each $(x_1,x_2)\in D$, $U(x_1,x_2)=c$ if $(x_1,x_2)\simeq (c,f(c))$ for some $c\in {\mathbb R}_+$. The function U belongs to $\mathcal U$.

Clearly, $U(0,0)=0$. To show the monotonicity of U, let $(x_1,x_2),(x_1',x_2')\in D$ be such that $x_1<x_1'$ and either $x_2<x_2'$ or $x_2=x_2'=0$. Let $c:=U(x_1,x_2)$ and $c':=U(x_1',x_2')$. Suppose to the contrary that $c\ge c'$. Let $F\in \mathcal F^{\{1,2\}}$ be as in the proof of Step 3, so $c<\overline{\text {supp}}\ F_2$ and by the definition of U, $r_1(F,x_1+c)=x_1$ and $r_2(F,x_1+c)=c$. By Lemma 2, there is $T\in [0,\overline{\text {supp}}\ F_1+\overline{\text {supp}}\ F_2]$ such that $r_2(F,T)=c'$. Let $x''_1:=r_1(F,T)$. By endowment monotonicity and conditional endowment monotonicity, $x''_1\le x_1$ and $(c',w(F_2,c'))\simeq (x''_1,w(F_1,x''_1))$. Since $w(F_2,c')=f(c')$, then $(c',f(c'))\simeq (x_1'',w(F_1,x_1''))$. Since $\simeq$ is transitive, $(x_1',x_2')\simeq (x_1'',w(F_1,x_1''))$. But $x_1''\le x_1<x_1'$, and either $w(F_1,x_1'')\le x_2<x_2'$ or $w(F_1,x_1'')=x_2=x_2'=0$, which contradicts Step 2.

Lastly, we show that U is continuous. To see its upper semi-continuity, let $\epsilon >0$ and $(x_1,x_2)\in D$. We claim that there is an open set O in D such that $(x_1,x_2)\in O$ and $O\subseteq U^{-1}((-\infty ,U(x_1,x_2)+\epsilon ))$. Abusing notation, let $c:=U(x_1,x_2)$. Consider $F\in \mathcal F^{\{1,2\}}$ as in the proof of Step 3 and such that $x_1\ge \underline{\text {supp}}\ F_1$, so $w(F_1,\cdot )$ is increasing on $[x_1,\overline{\text {supp}}\ F_1]$. Let $\delta \in (0,\min \{\epsilon ,\overline{\text {supp}}\ F_1-x_1,\overline{\text {supp}}\ F_2-c\})$, $x_1':=r_1(F,x_1+c+\delta )$ and $c':=r_2(F,x_1+c+\delta )$. By endowment monotonicity and conditional strict endowment monotonicity, $x_1'\in (x_1,\overline{\text {supp}}\ F_1)$, $c'\in (c,\min \{c+\epsilon ,\overline{\text {supp}}\ F_2\})$, and $(x_1',w(F_1,x_1'))\simeq (c',w(F_2,c'))$. Since $w(F_2,c')=f(c')$, then $(x_1',w(F_1,x_1'))\simeq (c',f(c'))$. Thus, $U(x_1',w(F_1,x_1'))=c'\in (c,c+\epsilon )$. Since $(x_1,x_2)<(x_1',w(F_1,x_1'))$, then there is an open neighborhood O of $(x_1,x_2)$ such that for each $(x_1'',x_2'')\in O$, $(x_1'',x_2'')<(x_1',w(F_1,x_1'))$. By the monotonicity of U, $O\subseteq U^{-1}((-\infty ,U(x_1,x_2)+\epsilon ))$. Lower semicontinuity follows from a similar argument.

$\displaystyle \mathbf{Step~5 }$  The rule r is the expected-waste constrained uniform gains rule associated with U.

Let $I\in \mathcal N$, $(F,T)\in \mathcal C^I$, and $c^*\in {\mathbb R}_+$ be such that $c^*$ solves $\sum U^{-1}_{F_i}(c)=T$. Suppose to the contrary that for some $\{j,k\}\subseteq I$, $r_j(F,T)>U^{-1}_{F_j}(c^*)$ and $r_k(F,T)<U^{-1}_{F_k}(c^*)$. Thus, $T>0$, $c^*>0$, $0<U^{-1}_{F_j}(c^*)<r_j(F,T)\le \overline{\text {supp}}\ F_j$ and $r_k(F,T)<\overline{\text {supp}}\ F_k$. By endowment monotonicity and Lemma 2, there is $T'\in (0,T)$ such that $r_j(F,T')=U^{-1}_{F_j}(c^*)$. By endowment monotonicity and conditional strict endowment monotonicity, $r_k(F,T')<r_k(F,T)$ and $(r_j(F,T'),w(F_j,r_j(F,T')))\simeq (r_k(F,T'),w(F_k,r_k(F,T')))$. By Step 2 and the definition of U, $U(r_j(F,T'),w(F_j,r_j(F,T')))=U(r_k(F,T'),w(F_k,r_k(F,T')))$. However, $U(r_j(F,T'),w(F_j,r_j(F,T')))=c^*>U(r_k(F,T),w(F_k,r_k(F,T)))>U(r_k(F,T'),w(F_k,r_k(F,T')))$, which is a contradiction.

Proof of Proposition 1

The “if” direction is readily verified, so the proof is omitted. To show the “only if” direction, let $U\in \mathcal U$ and r be the expected-waste constrained uniform gains rule associated with U such that r satisfies scale invariance. Let $a>0$, and $(x_1,x_2),(x_1',x_2')\in D$ be such that $(x_1,x_2)\ne (x_1',x_2')$ and $U(x_1,x_2)=U(x_1',x_2')$. We claim that $U(ax_1,ax_2)=U(ax_1',ax_2')$. By the monotonicity of U, $x_1>0$ and $x_1'>0$. Let $(F,T)\in \mathcal C^{\{1,2\}}$ be such that $F_1$ assigns probability $\frac{x_2}{x_1}$ to 0 and $1-\frac{x_2}{x_1}$ to $x_1+x_1'+1$, $F_2$ assigns probability $\frac{x_2'}{x_1'}$ to 0 and $1-\frac{x_2'}{x_1'}$ to $x_1+x_1'+1$, and $T=x_1+x_1'$. Thus, $r_1(F,T)=x_1$ and $r_2(F,T)=x_1'$. Let $F'\in \mathcal F^{\{1,2\}}$ be such that for each $i\in \{1,2\}$ and each $y_i\in {\mathbb R}$, $F'_i(ay_i)=F_i(y_i)$. By scale invariance, $r_1(F',aT)=ax_1$ and $r_2(F',aT)=ax_1'$. Note that for each $i\in \{1,2\}$, $ax_i<\overline{\text {supp}}\ F'_i$. Hence, $U(ax_1,ax_2)=U(ax_1,w(F_1',ax_1))=U(ax_1',w(F_2',ax_1'))=U(ax_1',ax'_2)$.

Proof of Theorem 2

The “if” direction is readily verified, so the proof is omitted. To show the “only if” direction, let r be a rule satisfying no domination, positivity, consistency, lower composition, and either strong upper composition or claims truncation invariance. By lower composition, r is endowment monotonic. Since r satisfies positivity and lower composition, by Lemma 4, it is conditionally strictly endowment monotonic. By Lemmas 5 and 6, r is symmetric. Recall that by symmetry and consistency, r is anonymous. Since r satisfies the axioms required in Lemma 3, the conclusion of Lemma 3 holds. Note that our proof of the “only if” direction of Theorem 1 only relies on no domination, symmetry, endowment monotonicity, conditional strict endowment monotonicity, consistency, and Lemma 3. Thus, by applying the same proof, we know that there is $U\in \mathcal U$ such that r is the expected-waste constrained uniform gains rule associated with U.

Let ${\ \succsim \ }$ be the weak order on D represented by U, i.e, for each pair $(x_1,x_2),(x_1',x_2')\in D$, $(x_1,x_2){\ \succsim \ }(x_1',x_2')$ if and only if $U(x_1,x_2)\ge U(x_1',x_2')$. We claim that ${\ \succsim \ }$ is homothetic. Let $a>0$, and $(x_1,x_2),(x_1',x_2')\in D$ be such that $(x_1,x_2)\ne (x_1',x_2')$ and $U(x_1,x_2)=U(x_1',x_2')$. We need to show that $U(ax_1,ax_2)=U(ax_1',ax_2')$. By the monotonicity of U, $x_1>0$ and $x_1'>0$. Let $(F,T), (F',aT)\in \mathcal C^{\{1,2\}}$ be as in the proof of Proposition 1. Thus, $r_1(F,T)=x_1$ and $r_2(F,T)=x_1'$. Suppose that $a=n$ where $n\in {\mathbb N}$. For each $m\in {\mathbb N}$ with $m\le n$, let $F^m_1$ assign probability $\frac{x_2}{x_1}$ to 0 and $1-\frac{x_2}{x_1}$ to $n(x_1+x_1'+1)-(m-1)x_1$, and $F_2$ assign probability $\frac{x_2'}{x_1'}$ to 0 and $1-\frac{x_2'}{x_1'}$ to $n(x_1+x_1'+1)-(m-1)x_1'$. By lower composition, $r(F',nT)=\sum \nolimits _{m=1}^n r(F^m,T)=nr(F,T)$. Thus, $U(ax_1,ax_2)=U(ax_1',ax_2')$. A similar argument applies when $a=\frac{1}{n}$, $n\in {\mathbb N}$, and hence when $a=\frac{m}{n}$, $m,n\in {\mathbb N}$. The case of a general a follows from the continuity of U.

Moreover, ${\ \succsim \ }$ is quasi-linear in the first coordinate. Let $a>0$, $(x_1,x_2),(x_1',x_2')\in D$ be such that $(x_1,x_2)\ne (x_1',x_2')$ and $U(x_1,x_2)=U(x_1',x_2')$. Thus, $x_1>0$ and $x_1'>0$. Let $(F,T)\in \mathcal C^{\{1,2\}}$ be as in the proof of Proposition 1. Thus, $r_1(F,T)=x_1$ and $r_2(F,T)=x_1'$. Abusing notation, let $F'\in \mathcal F^{\{1,2\}}$ be such that for each $i\in \{1,2\}$ and each $y_i\in {\mathbb R}$, $F_i'(y_i+a)=F_i(y_i)$. By lower composition, for each $i\in \{1,2\}$, $r_i(F',T+2a)=r_i(F,T)+a$. Since $w(F_1',x_1+a)=x_2$ and $w(F_2',x_1'+a)=x_2'$, then $U(x_1+a,x_2)=U(x_1'+a,x'_2)$. By the monotonicity and continuity properties of U, the same result holds if $a<0$ and $(x_1+a,x_2),(x_1'+a,x'_2)\in D$.

Let $U':D\rightarrow {\mathbb R}_+$ be defined by setting for each $(x_1,x_2)\in D$, $U'(x_1,x_2)=c$ if and only if $U(x_1,x_2)=U(c,0)$. By the monotonicity, continuity, and homotheticity properties of U, $U'$ is well-defined and ${\ \succsim \ }$ is represented by $U'$. Since only the ordinal properties of U matter when defining r, then r is the expected-waste constrained uniform gains rule with respect to $U'$. We claim that $U'$ is linear. Define $u:{\mathbb R}_+\rightarrow {\mathbb R}$ by setting for each $y\in {\mathbb R}_+$, $u(y)=U'(2y,y)-2y$. Let $(x_1,x_2)\in D$. By the definition of $U'$, $U(x_1,x_2)=U(U'(x_1,x_2),0)$ and $U(2x_2,x_2)=U(U'(2x_2,x_2),0)$. By quasi-linearity of ${\ \succsim \ }$ and the monotonicity of U, $2x_2-x_1=U'(2x_2,x_2)-U'(x_1,x_2)$. Thus, $U'(x_1,x_2)=x_1+u(x_2)$. Since ${\ \succsim \ }$ is homothetic, when $x_2>0$, $u(x_2)=U'(2x_2,x_2)-2x_2=x_2[U'(2,1)-2]=u(1)x_2$. When $x_2=0$, $u(x_2)=U'(0,0)-0=0=u(1)x_2$. Hence, $U'(x_1,x_2)=x_1+u(1)x_2$. Lastly, note that by the monotonicity of U, $u(1)\ge 0$.

Proof of Lemma 1

Let r be a rule satisfying consistency and the axioms either in (i) or (ii) in Lemma 1. Then r is endowment monotonic. Moreover, if r satisfies positivity and lower composition, then by Lemma 4, it is conditionally strictly endowment monotonic.

Suppose that r satisfies no domination. By Lemmas 5 and 6, r is symmetric. Recall that by symmetry and consistency, r is anonymous. To see that it is risk averse, let $I\in \mathcal N$, $(F,T)\in \mathcal C^I$, and $\{i,j\}\subseteq I$ be such that $\overline{\text {supp}}\ F_i=\overline{\text {supp}}\ F_j$ and $F_i$ is riskier than $F_j$. Suppose to the contrary that $r_i(F,T)>r_j(F,T)$. Then $T>0$, $r_j(F,T)<\overline{\text {supp}}\ F_j$ and

$$\begin{aligned} w(F_i,r_i(F,T))&\ge w(F_i,r_j(F,T))=\int _{-\infty }^{r_j(F,T)}F_i(x_i)dx_i\\&\ge \int _{-\infty }^{r_j(F,T)}F_j(x_j)dx_j=w(F_j,r_j(F,T)). \end{aligned}$$

If $w(F_i,r_i(F,T))>w(F_j,r_j(F,T))$, then no domination is violated. If $w(F_i,r_i(F,T))=w(F_j,r_j(F,T))$, then $w(F_i,r_i(F,T))=w(F_i,r_j(F,T))=0$ and thus $w(F_j,r_j(F,T))=0$. Hence, $F_i$ and $F_j$ agree on $(-\infty ,r_j(F,T))$. Since $T>0$ and $\overline{\text {supp}}\ F_j>0$, by conditional strict endowment monotonicity, $r_j(F,T)>0$. By endowment monotonicity and Lemma 2, there is $T'\in [0,T)$ such that $r_i(F,T')=\frac{1}{2}[r_i(F,T)+r_j(F,T)]$. By endowment monotonicity and conditional strict endowment monotonicity, $r_j(F,T')<r_j(F,T)$. Thus, $r_j(F,T')<r_i(F,T')$. Since $F_i$ and $F_j$ agree on $(-\infty ,r_j(F,T))$ and $r_j(F,T')<r_j(F,T)$, then by Lemma 3, $r_j(F,T')=r_i(F,T')$, which contradicts $r_j(F,T')<r_i(F,T')$. Hence, $r_i(F,T)\le r_j(F,T)$ as desired.

To see that r satisfies no reversal, let $I\in \mathcal N$, $(F,T)\in \mathcal C^I$, and $\{i,j\}\subseteq I$ be such that $\overline{\text {supp}}\ F_i=\overline{\text {supp}}\ F_j$ and $F_i$ is riskier than $F_j$. Suppose to the contrary that $w(F_i,r_i(F,T))<w(F_j,r_j(F,T))$. If $r_i(F,T)\ge r_j(F,T)$, then $w(F_i,r_i(F,T))\ge w(F_i,r_j(F,T))$. Since $F_i$ is riskier than $F_j$, then $w(F_i,r_j(F,T))\ge w(F_j,r_j(F,T))$. Thus, $w(F_i,r_i(F,T))\ge w(F_j,r_j(F,T))$, which contradicts $w(F_i,r_i(F,T))<w(F_j,r_j(F,T))$. Hence, $r_i(F,T)<r_j(F,T)$ and $r_i(F,T)<\overline{\text {supp}}\ F_j=\overline{\text {supp}}\ F_i$. This contradicts no domination.

Conversely, suppose that r satisfies risk aversion and no reversal. By Lemmas 5 and 6, r is symmetric. Recall that by symmetry and consistency, r is anonymous. Suppose to the contrary that it violates no domination. By consistency and anonymity, there is $(F,T)\in \mathcal C^{\{1,2\}}$ such that $r_1(F,T)>r_2(F,T)$, $w(F_1,r_1(F,T))>w(F_2,r_2(F,T))$, and $r_2(F,T)<\overline{\text {supp}}\ F_2$. Then $T>0$ and $w(F_1,r_1(F,T))>0$. Let $t:=r(F,T)$. By endowment monotonicity and Lemma 2, there is $\bar{T}\in (0,T]$ such that $r_1(F,\bar{T})=t_1$ and for each $T'\in [0,\bar{T})$, $r_1(F,T')<t_1$. By endowment monotonicity, there is $T'\in (0,\bar{T})$ such that $r_1(F,T')>r_2(F,T')$, $w(F_1,r_1(F,T'))>w(F_2,r_2(F,T'))$, $r_1(F,T')<\overline{\text {supp}}\ F_1$ and $r_2(F,T')<\overline{\text {supp}}\ F_2$. Hence, it is without loss of generality to assume that $t_1<\overline{\text {supp}}\ F_1$. We shall derive a contradiction in six steps.

$\displaystyle \mathbf{Step~1 }$  For each $F'\in \mathcal F^{\{1,2\}}$ such that for each $i\in \{1,2\}$, $F_i'$ and $F_i$ agree on $(-\infty ,t_i)$, $r(F',T)=t$.

Let $G\in \mathcal F^{\{3,4\}}$ be such that $G_3=F'_1$ and $G_4=F'_2$. Let $t':= r((F,G),2T)$. We claim that $t'_3=t_1$ and $t'_4=t_2$. To see this, suppose first that $t'_3<t_1$. Then by Lemma 3, $t'_1=t'_3<t_1$. By endowment monotonicity, conditional strict endowment monotonicity, and consistency, $t'_2<t_2$. Thus, by Lemma 3, $t'_4=t'_2<t_2$. Hence, $\sum \nolimits _{i=1}^4 t'_i<2T$, which is a contradiction. Suppose now that $t'_3>t_1$. Then by Lemma 3, $t'_1\ge t_1$. By endowment monotonicity, conditional strict endowment monotonicity, and consistency, $t'_2\ge t_2$. Thus, by Lemma 3, $t'_4\ge t_2$. Hence, $\sum \limits _{i=1}^4 t'_i>2T$, which is a contradiction. Thus, $t'_3=t_1$. Similarly, $t'_4=t_2$. By consistency and anonymity, $r(F',T)=t$.

$\displaystyle \mathbf{Step~2 }$  For each $I\in \mathcal N$, each $(F',T)\in \mathcal F^{I}$, and each pair $\{i,j\}\subseteq I$ such that $\overline{\text {supp}}\ F_i'=\overline{\text {supp}}\ F_j'$ and $F_i'$ is riskier than $F_j'$, if $w(F'_i,r_i(F',T))=w(F'_j,r_i(F',T))$, then $r_j(F',T)=r_i(F',T)$.

By risk aversion, $r_j(F',T)\ge r_i(F',T)$. Suppose to the contrary that $r_j(F',T)>r_i(F',T)$. Then $T>0$. If $w(F'_j,r_j(F',T))>0$, then $w(F'_j,r_i(F',T))<w(F'_j,r_j(F',T))$, and since $w(F'_i,r_i(F',T))=w(F'_j,r_i(F',T))$, then $w(F'_i,r_i(F',T))<w(F'_j,r_j(F',T))$, violating no reversal. Assume that $w(F'_j,r_j(F',T))=0$. Then $w(F'_j,r_i(F',T))=0$. Since $w(F'_i,r_i(F',T))=w(F'_j,r_i(F',T))$, then $w(F'_i,r_i(F',T))=0$. Hence, $F'_i$ and $F'_j$ agree on $(-\infty ,r_i(F',T))$. By endowment monotonicity and Lemma 2, there is $T'\in [0,T)$ such that $r_j(F',T')\in (r_i(F',T),r_j(F',T))$. Since $r_i(F',T)<r_j(F',T)\le \overline{\text {supp}}\ F_j'=\overline{\text {supp}}\ F_i'$, by endowment monotonicity and conditional strict endowment monotonicity, $r_i(F',T')<r_i(F',T)$. By Lemma 3, $r_j(F',T')=r_i(F',T')$. But $r_i(F',T')<r_i(F',T)<r_j(F',T')$, which is a contradiction.

$\displaystyle \mathbf{Step~3 }$  Let $p_1\in (F_1(t_1),1)$ and $p_2\in (F_2(t_2),1)$ be such that when $\frac{w(F_1,t_1)-w(F_2,t_2)}{t_1-t_2}<1$, $p_1>p_2>\frac{w(F_1,t_1)-w(F_2,t_2)}{t_1-t_2}$, and when $\frac{w(F_1,t_1)-w(F_2,t_2)}{t_1-t_2}\ge 1$, $p_2>p_1$. Then for each $i\in \{1,2\}$, $t_i-\frac{w(F_i,t_i)}{p_i}\ge 0$, and $\frac{1}{p_1-p_2}[t_1p_1-t_2p_2-w(F_1,t_1)+w(F_2,t_2)]>t_1$.

For each $i\in \{1,2\}$, $w(F_i,t_i)\le t_i F_i(t_i)\le t_ip_i$, so $t_i-\frac{w(F_i,t_i)}{p_i}\ge 0$. By our conditions on $p_1$ and $p_2$, $\frac{1}{p_1-p_2}[t_1p_1-t_2p_2-w(F_1,t_1)+w(F_2,t_2)]-t_1 =\frac{t_1-t_2}{p_1-p_2}(p_2-\frac{w(F_1,t_1)-w(F_2,t_2)}{t_1-t_2})>0.$ Hence, $\frac{1}{p_1-p_2}[t_1p_1-t_2p_2-w(F_1,t_1)+w(F_2,t_2)]>t_1$.

$\displaystyle \mathbf{Step~4 }$  Let $F'\in \mathcal F^{\{1,2\}}$ be such that $F'_1$ assigns probability $p_1$ to $t_1-\frac{w(F_1,t_1)}{p_1}$ and probability $1-p_1$ to $\frac{1}{p_1-p_2}[t_1p_1-t_2p_2-w(F_1,t_1)+w(F_2,t_2)]$, and $F'_2$ assigns probability $p_2$ to $t_2-\frac{w(F_2,t_2)}{p_2}$ and probability $1-p_2$ to $\frac{1}{p_1-p_2}[t_1p_1-t_2p_2-w(F_1,t_1)+w(F_2,t_2)]$. When $\frac{w(F_1,t_1)-w(F_2,t_2)}{t_1-t_2}<1$, $F'_2$ is riskier than $F'_1$, and when $\frac{w(F_1,t_1)-w(F_2,t_2)}{t_1-t_2}\ge 1$, $F'_1$ is riskier than $F'_2$.

Since $t_1>t_2$, by Step 3, $\overline{\text {supp}}\ F_1'=\overline{\text {supp}}\ F_2'=\frac{1}{p_1-p_2}[t_1p_1-t_2p_2-w(F_1,t_1)+w(F_2,t_2)]>\max \{t_i-\frac{w(F_i,t_i)}{p_i}:i\in \{1,2\}\}$. For each $i\in \{1,2\}$ and each $c\in {\mathbb R}$,

$$\begin{aligned}&\int _{-\infty }^cF_i'(x_i)dx_i\\&\quad =\left\{ \begin{array}{ll} 0 &{} \quad \text{ if } c\in \left( -\infty ,t_i-\frac{w(F_i,t_i)}{p_i}\right) , \\ p_i\left[ c-\left( t_i-\frac{w(F_i,t_i)}{p_i}\right) \right] &{} \quad \text{ if } \,c\in \left[ t_i-\frac{w(F_i,t_i)}{p_i},\overline{\text {supp}}\ F_i'\right) ,\\ p_i\left[ \overline{\text {supp}}\ F_i'-\left( t_i-\frac{w(F_i,t_i)}{p_i}\right) \right] +c-\overline{\text {supp}}\ F_i' &{}\quad \text{ if } \,c\in [\overline{\text {supp}}\ F_i',\infty ). \end{array} \right. \end{aligned}$$

Moreover,

$$\begin{aligned}&p_1\left[ \overline{\text {supp}}\ F_1'-\left( t_1-\frac{w(F_1,t_1)}{p_1}\right) \right] \\&\quad =\frac{p_1}{p_1-p_2}\left[ \phantom {\left. -t_1(p_1-p_2)+\left( 1-\frac{p_2}{p_1}\right) w(F_1,t_1)\right] }t_1p_1-t_2p_2-w(F_1,t_1)+w(F_2,t_2)\right. \\&\qquad \left. -t_1(p_1-p_2)+\left( 1-\frac{p_2}{p_1}\right) w(F_1,t_1)\right] \\&\quad =\frac{p_1p_2}{p_1-p_2}\left[ t_1-\frac{w(F_1,t_1)}{p_1}-t_2+\frac{w(F_2,t_2)}{p_2}\right] , \end{aligned}$$

and similarly,

$$\begin{aligned} p_2\left[ \overline{\text {supp}}\ F_2'-\left( t_2-\frac{w(F_2,t_2)}{p_2}\right) \right] =\frac{p_1p_2}{p_1-p_2}\left[ t_1-\frac{w(F_1,t_1)}{p_1}-t_2+\frac{w(F_2,t_2)}{p_2}\right] . \end{aligned}$$

Thus, $p_1[\overline{\text {supp}}\ F_1'-(t_1-\frac{w(F_1,t_1)}{p_1})]=p_2[\overline{\text {supp}}\ F_2'-(t_2-\frac{w(F_2,t_2)}{p_2})]$. Since $\overline{\text {supp}}\ F_1'=\overline{\text {supp}}\ F_2'$, then for each $c\in [\overline{\text {supp}}\ F_1',\infty )$, $\int _{-\infty }^c F_1'(x_1)dx_1=\int _{-\infty }^c F_2'(x_2)dx_2$, and $F_1'$ and $F_2'$ have the same mean.

Suppose that $\frac{w(F_1,t_1)-w(F_2,t_2)}{t_1-t_2}<1$. Then by the conditions on $p_1$ and $p_2$,

$$\begin{aligned} t_1-\frac{w(F_1,t_1)}{p_1}-\left( t_2-\frac{w(F_2,t_2)}{p_2}\right)>t_1-t_2-\left( \frac{w(F_1,t_1)}{p_2}-\frac{w(F_2,t_2)}{p_2}\right) >0. \end{aligned}$$

Thus, $t_1-\frac{w(F_1,t_1)}{p_1}>t_2-\frac{w(F_2,t_2)}{p_2}$, and for each $c\in (-\infty ,t_1-\frac{w(F_1,t_1)}{p_1})$, $\int _{-\infty }^cF_1'(x_1)dx_1\le \int _{-\infty }^cF_2'(x_2)dx_2$. Since $\overline{\text {supp}}\ F_1'=\overline{\text {supp}}\ F_2'$ and $p_1>p_2$, then for each $c\in [t_1-\frac{w(F_1,t_1)}{p_1},\overline{\text {supp}}\ F_1')$,

$$\begin{aligned}&p_1\left[ c-\left( t_1-\frac{w(F_1,t_1)}{p_1}\right) \right] -p_2 \left[ c-\left( t_2-\frac{w(F_2,t_2)}{p_2}\right) \right] \\&\quad =(p_1-p_2)(c-\overline{\text {supp}}\ F_1') <0, \end{aligned}$$

and thus $\int _{-\infty }^c F'_1(x_1)dx_1<\int _{-\infty }^c F'_2(x_2)dx_2$. Hence, if $\frac{w(F_1,t_1)-w(F_2,t_2)}{t_1-t_2}<1$, then $F_2'$ is riskier than $F_1'$.

Suppose that $\frac{w(F_1,t_1)-w(F_2,t_2)}{t_1-t_2}\ge 1$. Then by the conditions on $p_1$ and $p_2$,

$$\begin{aligned} t_1-\frac{w(F_1,t_1)}{p_1}- \left( t_2-\frac{w(F_2,t_2)}{p_2}\right)&<t_1-t_2- \left( \frac{w(F_1,t_1)}{p_2}-\frac{w(F_2,t_2)}{p_2}\right) \\&<t_1-t_2-[w(F_1,t_1)-w(F_2,t_2)]\le 0. \end{aligned}$$

Thus, $t_1-\frac{w(F_1,t_1)}{p_1}<t_2-\frac{w(F_2,t_2)}{p_2}$, and for each $c\in (-\infty ,t_2-\frac{w(F_2,t_2)}{p_2})$, $\int _{-\infty }^cF_2'(x_2)dx_2\le \int _{-\infty }^cF_1'(x_1)dx_1$. Since $\overline{\text {supp}}\ F_1'=\overline{\text {supp}}\ F_2'$ and $p_1<p_2$, then for each $c\in [t_2-\frac{w(F_2,t_2)}{p_2},\overline{\text {supp}}\ F_2')$,

$$\begin{aligned}&p_1\left[ c-\left( t_1-\frac{w(F_1,t_1)}{p_1}\right) \right] -p_2 \left[ c-\left( t_2-\frac{w(F_2,t_2)}{p_2}\right) \right] \\&\quad =(p_1-p_2)(c-\overline{\text {supp}}\ F_2') >0, \end{aligned}$$

and thus $\int _{-\infty }^c F'_2(x_2)dx_2<\int _{-\infty }^c F'_1(x_1)dx_1$. Hence, if $\frac{w(F_1,t_1)-w(F_2,t_2)}{t_1-t_2}\ge 1$, $F_1'$ is riskier than $F_2'$.

$\displaystyle \mathbf{Step~5 }$  Let $I:=\{1,2,3,4\}$ and $F'\in \mathcal F^{I}$ be such that for each $i\in \{1,2\}$, $F_i'$ is defined as in Step 4, and $F_{i+2}'$ agrees with $F_i$ on $(-\infty ,t_i)$ and agrees with $F_i'$ on $[t_i,\infty )$.¹⁴ Then for each $i\in \{1,2\}$, $r_i(F',2T)=r_{i+2}(F',2T)=t_i$.

Let $t':=r(F',2T)$. By Step 3, for each $i\in \{1,2\}$ and each $c\in {\mathbb R}$,

$$\begin{aligned} \int _{-\infty }^cF_i'(x_i)dx_i=\left\{ \begin{array}{ll} 0 &{} \quad \text{ if } \,c\in \left( -\infty ,t_i-\frac{w(F_i,t_i)}{p_i}\right) , \\ w(F_i,t_i)-(t_i-c)p_i &{} \quad \text{ if } \,c\in \left[ t_i-\frac{w(F_i,t_i)}{p_i},t_i\right) ,\\ w(F_i,t_i)+\int _{t_i}^cF_i'(x_i)dx_i &{} \quad \text{ if } \,c\in [t_i,\infty ). \end{array} \right. \end{aligned}$$

Besides, for each $i\in \{1,2\}$, by the definition of $F_{i+2}'$,

$$\begin{aligned} \int _{-\infty }^cF_{i+2}'(x_{i+2})dx_{i+2}=\left\{ \begin{array}{ll} w(F_i,t_i)-\int _{c}^{t_i} F_i(x_i)dx_i &{} \quad \text{ if } \,c\in (-\infty ,t_i), \\ w(F_i,t_i)+\int _{t_i}^cF_i'(x_i)dx_i &{} \quad \text{ if } \,c\in [t_i,\infty ). \end{array} \right. \end{aligned}$$

By Step 3, for each $i\in \{1,2\}$, $\overline{\text {supp}}\ F_i'=\overline{\text {supp}}\ F_{i+2}'>t_i$, so $\int _{-\infty }^{\overline{\text {supp}}\ F_i'}F_i'(x_i)dx_i=\int _{-\infty }^{\overline{\text {supp}}\ F_{i+2}'}F_{i+2}'(x_{i+2})dx_{i+2}$, and thus $F_i'$ and $F_{i+2}'$ have the same mean. For each $i\in \{1,2\}$, since $p_i>F_i(t_i)$, then for each $c\in {\mathbb R}$,

$$\begin{aligned} \int _{-\infty }^cF_i'(x_i)dx_i\le \int _{-\infty }^cF_{i+2}'(x_{i+2})dx_{i+2}, \end{aligned}$$

(2)

and (2) holds with strict inequality when $c\in [t_i-\frac{w(F_i,t_i)}{p_i},t_i)$. Since $w(F_1,t_1)>0$, $[t_1-\frac{w(F_1,t_1)}{p_1},t_1)\ne \emptyset$. Hence, $F_3'$ is riskier than $F_1'$. Similarly, if $w(F_2,t_2)>0$, then $F_4'$ is riskier than $F_2'$. If $w(F_2,t_2)=0$, then $\underline{\text {supp}}\ F_2\ge t_2=\underline{\text {supp}}\ F_2'$, and thus $F_4'=F_2'$.

By anonymity and Step 1, for each $i\in \{1,2\}$, $r_{i+2}((F_3',F_4'),T)=t_i$. Suppose that $t_3'<t_1$. Then by endowment monotonicity and consistency, $t_4'\le t_2$. Since $\sum \nolimits _{i=1}^4t_i'=2T=2(t_1+t_2)$, then either $t_1'>t_1$ or $t_2'>t_2$. Let $j\in \{1,2\}$ be such that $t_j'>t_j$. Then by endowment monotonicity and Lemma 2, there is $T'\in [2T,2\overline{\text {supp}}\ F_1'+2\overline{\text {supp}}\ F_2']$ such that $r_{j+2}(F',T')=t_j$. By endowment monotonicity, $r_j(F',T')>t_j$. Thus, $r_j(F',T')>r_{j+2}(F',T')$. Moreover, $w(F_{j+2}',r_{j+2}(F',T'))=w(F_{j+2}',t_j)=w(F_j,t_j)=w(F_j',t_j)=w(F_j',r_{j+2}(F',T'))$. By the conclusion in the previous paragraph, either $F_{j+2}'$ is riskier than $F_{j}'$ or $F_{j+2}'=F_j'$. If $F_{j+2}'$ is riskier than $F_{j}'$, since $\overline{\text {supp}}\ F_j'=\overline{\text {supp}}\ F_{j+2}'$ and $w(F_{j+2}',r_{j+2}(F',T'))=w(F_j',r_{j+2}(F',T'))$, then by Step 2, $r_j(F',T')=r_{j+2}(F',T')$. This contradicts $r_j(F',T')>r_{j+2}(F',T')$. If $F_{j+2}'=F_j'$, by symmetry, $r_j(F',T')=r_{j+2}(F',T')$, which is again a contradiction. Hence, $t_3'\ge t_1$. Similarly, $t_4'\ge t_2$. Suppose that $t_1'<t_1$. Then $t_3'\ge t_1>t_1'$. Since $\overline{\text {supp}}\ F_1'=\overline{\text {supp}}\ F_3'$ and $F_3'$ is riskier than $F_1'$, then by risk aversion, $t_3'\le t_1'$. This contradicts $t_3'>t_1'$. Hence, $t_1'\ge t_1$. Similarly, if $t_2'<t_2$ and $F_4'$ is riskier than $F_2'$, there is a contradiction. If $t_2'<t_2$ and $F_4'=F_2'$, then $t_4'\ge t_2>t_2'$ and by symmetry, $t_4'=t_2'$, which is also not possible. Hence, $t_2'\ge t_2$. Since for each $i\in \{1,2\}$, $t_i'\ge t_i$ and $t_{i+2}'\ge t_i$, then for each $i\in \{1,2\}$, $t_i'=t_{i+2}'=t_i$.

$\displaystyle \mathbf{Step~6 }$  When $\frac{w(F_1,t_1)-w(F_2,t_2)}{t_1-t_2}<1$, no reversal is violated. When $\frac{w(F_1,t_1)-w(F_2,t_2)}{t_1-t_2}\ge 1$, risk aversion is violated.

Let $I\in \mathcal N$ and $F'\in \mathcal F^I$ be defined as in Step 5. By Step 5, $r_1(F',2T)=t_1$ and $r_2(F',2T)=t_2$. Then $r_1(F',2T)>r_2(F',2T)$ and $w(F_1',r_1(F',2T))=w(F_1,t_1)>w(F_2,t_2)=w(F_2',r_2(F',2T))$. When $\frac{w(F_1,t_1)-w(F_2,t_2)}{t_1-t_2}<1$, by Step 3 and Step 4, $\overline{\text {supp}}\ F_1'=\overline{\text {supp}}\ F_2'$ and $F_2'$ is riskier than $F_1'$. Thus, no reversal is violated. When $\frac{w(F_1,t_1)-w(F_2,t_2)}{t_1-t_2}\ge 1$, by Steps  3 and 4, $\overline{\text {supp}}\ F_1'=\overline{\text {supp}}\ F_2'$ and $F_1'$ is riskier than $F_2'$. Thus, risk aversion is violated.

Proof of Proposition 2

Let r be a rule satisfying symmetry and strong upper composition. Then r is endowment monotonic. Let $I\in \mathcal N$ and $(F,T)\in \bar{\mathcal C}^I$. Let $t:=r(F,T)$ and $c^*\in {\mathbb R}_+$ be such that $\sum \min \{\overline{\text {supp}}\ F_i,c^*\}=T$. We claim that for each $i\in I$, $t_i=\min \{\overline{\text {supp}}\ F_i,c^*\}$. Suppose to the contrary that there are $j,k\in I$ such that $t_j<\min \{\overline{\text {supp}}\ F_j,c^*\}$ and $t_k>\min \{\overline{\text {supp}}\ F_k,c^*\}$. Then $c^*\le \overline{\text {supp}}\ F_k$ and $t_j<t_k$. Thus, $\min \{\overline{\text {supp}}\ F_j,c^*\}\le \overline{\text {supp}}\ F_k$, and $F_j$ and $F_k$ agree on $(-\infty ,\min \{\overline{\text {supp}}\ F_j,c^*\})$. Since $t_j<\min \{\overline{\text {supp}}\ F_j,c^*\}$, then by Lemma 3, $t_j=t_k$, which contradicts $t_j<t_k$.

Let r be a rule satisfying symmetry, lower composition, and claims truncation invariance. By a similar argument as in the first paragraph of Case 2 in the proof of Lemma 3, it can be shown that for each $I\in \mathcal N$ and each $(F,T)\in \bar{\mathcal C}^I$, if $c\in [0,\min \{\overline{\text {supp}}\ F_i:i\in I\}]$ is such that all the claims agree on $(-\infty ,c)$ and $T\le |I|c$, then for each $i\in I$, $r_i(F,T)=\frac{T}{|I|}$. Lastly, by lower composition, for each $i\in I$, $r_i(F,T)=\min \{\overline{\text {supp}}\ F_i,c^*\}$ where $c^*\in {\mathbb R}_+$ is such that $\sum \min \{\overline{\text {supp}}\ F_i,c^*\}=T$.