No-envy and egalitarian-equivalence under multi-object-demand for heterogeneous objects

Abstract

We study the problem of allocating heterogeneous indivisible tasks in a multi-object-demand model (i.e., each agent can be assigned multiple objects) where monetary transfers are allowed. Agents’ costs for performing tasks are their private information and depend on what other tasks they are obtained with. First, we show that when costs are unrestricted or superadditive, then there is no envy-free and egalitarian-equivalent mechanism that assigns the tasks efficiently. Then, we characterize the class of envy-free and egalitarian-equivalent Groves mechanisms when costs are subadditive. Finally, within this class, under a bounded-deficit condition, we identify the Pareto-dominant subclass. We show that the mechanisms in this subclass are not Pareto-dominated by any other Groves mechanism satisfying the same bounded-deficit condition.

Appendix

Proof of Lemma 2

(a) Follows from (1) and the definition of egalitarian-equivalence.

(b) Let $\varphi ^{\tau } \equiv (A^{\tau } ,t^{\tau })$ be an assignment-efficient mechanism. First, suppose that $\varphi ^{\tau }$ is egalitarian-equivalent and budget-balanced. Then, there is $T\in \mathbb {R}$ and for each $N\in \mathcal {N}$ and each $c\in \mathcal {C}^{N},$ there are $R(c)\in 2^{ \mathbb {A}}$ and $r(c)\in \mathbb {R}$ such that (5) holds and $\sum \nolimits _{i\in N} t_{i}^{\tau } (c)= \sum \nolimits _{i\in N} c_{i}(A_{i}^{\tau } (c))-\underset{i\in N}{\sum } c_{i}(R(c))+nr(c)=T.$ That is, $r(c)=\frac{1}{n}\left[ T-W(c)+\sum \nolimits _{i\in N} c_{i}(R(c))\right]$. The converse direction follows similarly.

(c) Let $\varphi \equiv (A,t)$ be egalitarian-equivalent and envy-free. By egalitarian-equivalence, for each $N\in \mathcal {N}$ and each $c\in \mathcal {C}^{N},$ there are $R(c)\in 2^{ \mathbb {A}}$ and $r(c)\in \mathbb {R}$ such that (5) holds. By no-envy, (1) and (5 ), for each $N\in \mathcal {N}$ , each $c\in \mathcal {C} ^{N},$ and each pair $\{i,j\}\subseteq N,$ $-c_{i}(R(c))+r(c)\ge -c_{i}(A_{j}(c))+c_{j}(A_{j}(c))-c_{j}(R(c))+r(c).$ That is, (7) holds. Conversely, let $\varphi \equiv (A,t)$ be such that for each $N\in \mathcal {N}$ and each $c\in \mathcal {C}^{N},$ there are $R(c)\in 2^{ \mathbb {A}}$ and $r(c)\in \mathbb {R}$ such that for each pair $\{i,j\}\subseteq N,$ (5) and (7) holds. By part (a), $\varphi \equiv (A,t)$ is egalitarian-equivalent. By (5), for each $N\in \mathcal {N},$ each $c\in \mathcal {C}^{N},$ and each $\{i,j\}\subseteq N,$ $c_{i}(R(c))-c_{j}(R(c))=c_{i}(A_{i}(c))-c_{j}(A_{j}(c))-t_{i}(c)+t_{j}(c).$ This equality and (7) together imply that for each $N\in \mathcal {N},$ each $c\in \mathcal {C}^{N},$ and each $\{i,j\}\subseteq N,$ $-c_{i}(A_{i}(c))+t_{i}(c)\ge -c_{i}(A_{j}(c))+t_{j}(c).$ By (1 ), $\varphi$ is envy-free.

(d) Follows from (1) and (5). $\square$

Proof of Proposition 1

(a) Let $\mathcal {C\in }\{\mathcal {C}_{un},\mathcal {C}_{ sup }\}.$ Assume, by contradiction, that on $\bigcup \nolimits _{N\in \mathcal {N}} \mathcal {C}^{N}$, $\varphi ^{\tau } \equiv (A^{\tau } ,t^{\tau } )$ is assignment-efficient, envy-free, and egalitarian-equivalent. Let $N=\{1,2,3\},$ $\mathbb {A} =\{\alpha ,\beta ,\theta \},$ and $c\in \mathcal {C}^{N}$ be such that

• (i) $c_{1}(\{\alpha \})=c_{1}(\{\beta \})=c_{3}(\{\alpha \})=c_{3}(\{\theta \})=2,$

• (ii) $c_{1}(\{\theta \})=c_{2}(\{\beta \})=c_{2}(\{\theta \})=c_{3}(\{\beta \})=3,$

• (iii) $c_{2}(\{\alpha \})=4,$

• (iv) for each pair $\{i,j\}\subset N$ and each $A\subseteq \mathbb {A}$ with $|A|\ge 2,$ $c_{i}(A)\ge 8,$ and $c_{i}(A)\ne c_{j}(A),$

• (v) for each $i\in N,$ $c_{i}( \mathbb {A} )\ge \max \nolimits _{A\subset \mathbb {A}} \{c_{i}(A)+c_{i}( \mathbb {A} \backslash A)\}$.

Let $A^{\prime } =(A_{1}^{\prime } ,A_{2}^{\prime },A_{3}^{\prime } )=(\{\alpha \},\{\beta \},\{\theta \})$ and $A^{\prime \prime } =(\{\beta \},\{\theta \},\{\alpha \}).$ Note that $W(c)=7=\sum \nolimits _{i\in N} c_{i}(A_{i}^{\prime })=\sum \nolimits _{i\in N} c_{i}(A_{i}^{\prime \prime })$.

By assignment-efficiency, $A^{\tau } (c)\in \{A^{\prime } ,A^{\prime \prime } \}.$ Then, $u(\varphi _{2}^{\tau }(c);c_{2})=-3+t_{2}^{\tau } (c)$ and for $i\in \{1,3\},$ $u(\varphi _{i}^{\tau } (c);c_{i})=-2+t_{i}^{\tau } (c)$.

If $A^{\tau } (c)=A^{\prime } ,$ then by no-envy, $u(\varphi _{1}^{\tau } (c);c_{1})\ge -c_{1}(A_{2}^{\tau }(c))+t_{2}^{\tau } (c),$ $u(\varphi _{2}^{\tau } (c);c_{2})\ge -c_{2}(A_{3}^{\tau } (c))+t_{3}^{\tau } (c),$ and $u(\varphi _{3}^{\tau } (c);c_{3})\ge -c_{3}(A_{1}^{\tau } (c))+t_{1}^{\tau }(c).$ Since $c_{1}(A_{1}^{\prime } )=c_{1}(A_{2}^{\prime } ),$ $c_{2}(A_{2}^{\prime } )=c_{2}(A_{3}^{\prime } ),$ and $c_{3}(A_{3}^{\prime } )=c_{3}(A_{1}^{\prime } ),$ we have $t_{1}^{\tau } (c)\ge t_{2}^{\tau } (c)\ge t_{3}^{\tau } (c)\ge t_{1}^{\tau } (c)$.

If $A^{\tau } (c)=A^{\prime \prime } ,$ then by no-envy, $u(\varphi _{1}^{\tau } (c);c_{1})\ge -c_{1}(A_{3}^{\tau } (c))+t_{3}^{\tau } (c),$ $u(\varphi _{2}^{\tau }(c);c_{2})\ge -c_{2}(A_{1}^{\tau } (c))+t_{1}^{\tau } (c),$ and $u(\varphi _{3}^{\tau } (c);c_{3})\ge -c_{3}(A_{2}^{\tau }(c))+t_{2}^{\tau } (c).$ Since $c_{1}(A_{1}^{\prime \prime })=c_{1}(A_{3}^{\prime \prime } ),$ $c_{2}(A_{2}^{\prime \prime })=c_{2}(A_{1}^{\prime \prime } ),$ and $c_{3}(A_{3}^{\prime \prime })=c_{3}(A_{2}^{\prime \prime } ),$ we have $t_{1}^{\tau } (c)\ge t_{3}^{\tau } (c)\ge t_{2}^{\tau } (c)\ge t_{1}^{\tau } (c)$.

In both cases, there is $k^{\tau } \in \mathbb {R}$ such that for each $i\in N,$ $t_{i}^{\tau } (c)=k^{\tau }$. By Lemma 2a, there are $R(c)\in 2^{\mathbb {A}}$ and $r(c)\in \mathbb {R}$ such that

$$\begin{aligned} u(\varphi _{1}^{\tau } (c);c_{1})= & {} -2+k^{\tau } =-c_{1}(R(c))+r(c). \end{aligned}$$

(14)

$$\begin{aligned} u(\varphi _{2}^{\tau } (c);c_{2})= & {} -3+k^{\tau } =-c_{2}(R(c))+r(c). \end{aligned}$$

(15)

$$\begin{aligned} u(\varphi _{3}^{\tau } (c);c_{3})= & {} -2+k^{\tau } =-c_{3}(R(c))+r(c). \end{aligned}$$

(16)

By (14) and (16), $c_{1}(R(c))=c_{3}(R(c))$. Then, $R(c)\in \{\alpha ,\emptyset \}$. Suppose $R(c)=\{\alpha \}$. By (14), $r(c)=k^{\tau }$. Since $c_{2}(\{\alpha \})=4,$ by (15), $-3+k^{\tau } =-4+r(c),$ a contradiction. Suppose $R(c)=\emptyset$. By (14), $r(c)=k^{\tau } -2,$ which contradicts (15). This completes the proof for part (a). $\Diamond$

(b) Let $\mathcal {C\in }\{\mathcal {C}_{un},\mathcal {C} _{ad},\mathcal {C}_{sub},\mathcal {C}_{sup }\}$ and consider mechanisms $\varphi ^{\tau }$ defined on $\bigcup \nolimits _{N\in \mathcal {N}:|N|=2} \mathcal {C}^{N}.$

Let $\varphi ^{\tau } \equiv (A^{\tau } ,t^{\tau } )$ be an assignment-efficient and egalitarian-equivalent mechanism such that for each $N\in \mathcal {N}$ with $|N|=2$ and each $c\in \mathcal {C}^{N},$ $R(c)=A_{i}^{\tau } (c)$ for some $i\in N.$ Assume, by contradiction, that $\varphi ^{\tau }$ is not envy-free on $\bigcup \nolimits _{N\in \mathcal {N}:|N|=2} \mathcal {C}^{N}$. Then, there is $N=\{i,j\}\in \mathcal {N}$ and $c\in \mathcal {C}^{N}$ such that $u(A_{i}^{\tau } (c),t_{i}^{\tau }(c);c_{i})<-c_{i}(A_{j}^{\tau } (c))+t_{j}^{\tau } (c).$ That is, by (7),

$$\begin{aligned} c_{i}(A_{j}^{\tau } (c))+c_{j}(R(c))<c_{i}(R(c))+c_{j}(A_{j}^{\tau }(c)). \end{aligned}$$

(17)

If $R(c)=A_{j}^{\tau } (c),$ then by (17), $c_{i}(A_{j}^{\tau } (c))+c_{j}(A_{j}^{\tau } (c))<c_{i}(A_{j}^{\tau }(c))+c_{j}(A_{j}^{\tau } (c)),$ a contradiction. If $R(c)=A_{i}^{\tau } (c),$ then by (17) , $c_{i}(A_{j}^{\tau } (c))+c_{j}(A_{i}^{\tau } (c))<c_{i}(A_{i}^{\tau }(c))+c_{j}(A_{j}^{\tau } (c)),$ which contradicts that $(A_{i}^{\tau }(c),A_{j}^{\tau } (c))$ is an efficient assignment for c . Hence, $\varphi ^{\tau }$ must be envy-free.

(c) Let $\mathcal {C}\in \{\mathcal {C}_{ad},\mathcal {C}_{sub}\}.$ On $\bigcup \nolimits _{N\in \mathcal {N}} \mathcal {C}^{N},$ let $\varphi ^{\tau } \equiv (A^{\tau },t^{\tau } )$ be assignment-efficient and welfare-egalitarian. By Lemma 2d, $\varphi ^{\tau }$ is egalitarian-equivalent such that for each $N\in \mathcal {N}$ and each $c\in \mathcal {C}^{N},$ $R(c)=\emptyset$. Assume, by contradiction, that $\varphi ^{\tau }$ is not envy-free on $\bigcup \nolimits _{N\in \mathcal {N}} \mathcal {C}^{N}$ . Then, there exist $N\in \mathcal {N}$, $c\in \mathcal {C}^{N},$ and a pair of agents $\{i,j\}\subseteq N$ such that $u(A_{i}^{\tau } (c),t_{i}^{\tau } (c);c_{i})<-c_{i}(A_{j}^{\tau }(c))+t_{j}^{\tau } (c).$ That is, by (7) in Lemma 2c,

$$\begin{aligned} c_{i}(A_{j}^{\tau } (c))<c_{j}(A_{j}^{\tau } (c)). \end{aligned}$$

(18)

This inequality implies that $A_{j}^{\tau } (c)\ne \emptyset .$ Hence, if $| \mathbb {A} |=1$, then j is the agent who is assigned the only task which, by (18), contradicts that $A^{\tau }(c)$ is an efficient assignment for c. If $| \mathbb {A} |\ge 2,$ then by (18),

$$\begin{aligned} c_{i}(A_{i}^{\tau } (c))+c_{i}(A_{j}^{\tau } (c))<c_{i}(A_{i}^{\tau }(c))+c_{j}(A_{j}^{\tau } (c)). \end{aligned}$$

(19)

Note that on the subadditive domain, $c_{i}(A_{i}^{\tau }(c)\cup A_{j}^{\tau } (c))\le c_{i}(A_{i}^{\tau }(c))+c_{i}(A_{j}^{\tau } (c)),$ which holds as an equality on the additive domain. Hence, by (19), it is less costly to assign both $A_{i}^{\tau } (c)$ and $A_{j}^{\tau } (c)$ to agent i rather than assigning these sets to i and j,  respectively. This contradicts that $A^{\tau } (c)$ is an efficient assignment for c. This completes the proof for part (c). $\Diamond$

(d) Let $N=\{1,2,3\},$ $\mathbb {A} =\{\alpha ,\beta ,\theta \},$ and $c^{\prime } \in \mathcal {C}_{ad}^{N}$ be such that

• (i) $c_{1}^{\prime } (\{\alpha \})=c_{1}^{\prime } (\{\beta \})=10,$ $c_{1}^{\prime } (\{\theta \})=15,$

• (ii) $c_{2}^{\prime } (\{\alpha \})=11,$ $c_{2}^{\prime } (\{\beta \})=12,$ $c_{2}^{\prime } (\{\theta \})=14,$

• (iii) $c_{3}^{\prime } (\{\alpha \})=12,$ $c_{3}^{\prime } (\{\beta \})=11,$ $c_{3}^{\prime } (\{\theta \})=12.$

Let $\mathcal {C\in }\{\mathcal {C}_{ad},\mathcal {C}_{sub}\}$ and $\varphi ^{\tau }\equiv (A^{\tau } ,t^{\tau } )$ be an assignment-efficient mechanism such that for each $N\in \mathcal {N}$, each $c\in \mathcal {C}^{N}$ with $c\ne c^{\prime } ,$ and each $i\in N,$ $t_{i}^{\tau } (c)=\frac{30-W(c)}{n}+c_{i}(A_{i}^{\tau } (c));$ and for each $i\in \{1,2,3\},$ $t_{i}^{\tau } (c^{\prime } )=c_{i}^{\prime }(A_{i}^{\tau } (c^{\prime } ))-c_{i}^{\prime } (\{\alpha ,\theta \})+20.$ We will show that $\varphi ^{\tau }$ is envy-free and egalitarian-equivalent, but not welfare-egalitarian on $\bigcup \nolimits _{ N\in \mathcal {N}}\mathcal {C}^{N}$.

By Lemma 2a, for each $N\in \mathcal {N}$ and each $c\in \mathcal {C}^{N}$ with $c\ne c^{\prime } ,$ $\varphi ^{\tau }$ chooses an egalitarian-equivalent allocation with reference set of tasks $R(c)=\emptyset$ and reference transfer $r(c)=\frac{30-W(c)}{n}.$ By Proposition 1c, $\varphi ^{\tau } (c)$ is an envy-free allocation.

Now, consider $c^{\prime } .$ By Lemma 2a, $\varphi ^{\tau }$ chooses an egalitarian-equivalent allocation for $c^{\prime }$ where the reference set is $R(c^{\prime } )=\{\alpha ,\theta \}$ and the reference transfer is $r(c^{\prime } )=20.$

The efficient assignment at $c^{\prime }$ is $A^{\tau }(c^{\prime } )=(\{\alpha ,\beta \},\emptyset ,\{\theta \})$ and $W(c^{\prime } )=32.$ Agents’ transfers at $c^{\prime }$ are $t_{1}^{\tau } (c^{\prime } )=c_{1}^{\prime } (A_{1}^{\tau } (c^{\prime }))-c_{1}^{\prime } (\{\alpha ,\theta \})+20=15,$ $t_{2}^{\tau }(c^{\prime } )=-5,$ and $t_{3}^{\tau } (c^{\prime } )=8$.

Note that $u(\varphi _{1}^{\tau } (c^{\prime });c_{1})=-c_{1}^{\prime } (\{\alpha ,\beta \})+t_{1}^{\tau }(c^{\prime } )=-5;$ $u(\varphi _{2}^{\tau } (c^{\prime });c_{2})=-c_{2}^{\prime } (\emptyset )+t_{2}^{\tau } (c^{\prime })=-5;$ and $u(\varphi _{3}^{\tau } (c^{\prime });c_{3})=-c_{3}^{\prime } (\{\theta \})+t_{3}^{\tau } (c^{\prime })=-4.$ Since $u(\varphi _{2}^{\tau } (c^{\prime } );c_{2})\ne$ $u(\varphi _{3}^{\tau } (c^{\prime } );c_{3})$, $\varphi ^{\tau }$ is not welfare-egalitarian. Note that

• $u(\varphi _{1}^{\tau } (c^{\prime } );c_{1})\ge -c_{1}(\emptyset )+t_{2}^{\tau } (c^{\prime } )=-5$  and  $u(\varphi _{1}^{\tau } (c^{\prime } );c_{1})\ge -c_{1}(\{\theta \})+t_{3}^{\tau } (c^{\prime } )=-15+8;$

• $u(\varphi _{2}^{\tau } (c^{\prime } );c_{2})\ge -c_{2}(\{\alpha ,\beta \})+t_{1}^{\tau } (c^{\prime } )=-23+15$   and   $u(\varphi _{2}^{\tau } (c^{\prime } );c_{2})\ge -c_{2}(\{\theta \})+t_{3}^{\tau } (c^{\prime } )=-14+8;$

• $u(\varphi _{3}^{\tau } (c^{\prime } );c_{3})\ge -c_{3}(\{\alpha ,\beta \})+t_{1}^{\tau } (c^{\prime } )=-23+15$   and   $u(\varphi _{3}^{\tau } (c^{\prime } );c_{3})\ge -c_{3}(\emptyset )+t_{2}^{\tau } (c^{\prime } )=-5.$

Hence, $\varphi ^{\tau } (c^{\prime } )$ is an envy-free allocation. Altogether, on the additive or the subadditive domain, $\varphi ^{\tau }$ is assignment-efficient, egalitarian-equivalent, and envy-free, but not welfare-egalitarian. $\Diamond$

(e) Let $\mathcal {C\in }\{\mathcal {C}_{ad},\mathcal {C} _{sub}\}$. On $\bigcup \nolimits _{N\in \mathcal {N}} \mathcal {C}^{N},$ let $\varphi ^{\tau } \equiv (A^{\tau },t^{\tau } )$ be assignment-efficient and egalitarian-equivalent. Assume that for each $N\in \mathcal {N}$ and each $c\in \mathcal {C}^{N}$, $R(c)\ne \emptyset .$ We will show that $\varphi ^{\tau }$ is not envy-free on $\bigcup \nolimits _{N\in \mathcal {N}} \mathcal {C}^{N}$.

Let $N=\{1,2,3\}, \mathbb {A} =\{\alpha ,\beta ,\theta \},$ and $c^{\prime } \in \mathcal {C}_{ad}^{N}$ be as in Proposition 1d. Let $\widehat{c}_{3}\in \mathcal {C}_{ad}$ be such that $\widehat{c}_{3}(\{\alpha \})=14,$ $\widehat{c}_{3}(\{\beta \})=15,$ $\widehat{c}_{3}(\{\theta \})=12.$ Let $\widehat{c}\in \mathcal {C} _{ad}^{N}$ be such that $\widehat{c}=(\widehat{c}_{3},c_{-3}^{\prime } ).$ Note that $A_{1}^{\tau } (\widehat{c})=\{\alpha ,\beta \},$ $A_{2}^{\tau } ( \widehat{c})=\emptyset ,$ $A_{3}^{\tau } (\widehat{c})=\{\theta \},$ and $W( \widehat{c})=32.$ Assume, by contradiction, that $\varphi ^{\tau }$ is envy-free on $\bigcup \nolimits _{N\in \mathcal {N}} \mathcal {C} ^{N}.$ Then, by (7), for $j=2$ and for each $i\in \{1,3\},$ $\widehat{c}_{i}(A_{j}^{\tau }(\widehat{c}))+\widehat{c}_{j}(R(\widehat{c} ))\ge \widehat{c}_{j}(A_{j}^{\tau } (\widehat{c}))+\widehat{c}_{i}(R( \widehat{c})).$ This inequality and $A_{2}^{\tau }(\widehat{c})=\emptyset$ together imply that for each $i\in \{1,3\},$

$$\begin{aligned} \widehat{c}_{2}(R(\widehat{c}))\ge \widehat{c}_{i}(R(\widehat{c})). \end{aligned}$$

(20)

Note that $\widehat{c}_{2}(\{\theta \})<\widehat{c}_{1}(\{\theta \})$ and for each $A\in (2^{ \mathbb {A}} \backslash \{\emptyset ,\{\theta \}\}),$ $\widehat{c}_{2}(A)<\widehat{c} _{3}(A)$. These inequalities and the fact that $R(\widehat{c})\ne \emptyset$ together contradict (20). $\square$

Proof of Theorem 1

If Part Pick an assignment-efficient and strategy-proof mechanism. By Lemma 1, it is a Groves mechanism $G^{h,\tau }$ for some $h\in \mathcal {H}$ and $\tau \in \mathcal {T} .$ Let $G^{h,\tau } \in \widehat{\mathcal {E}}$. Then, there is $\gamma : \mathcal {N}\mathbb {\rightarrow } \mathbb {R}$ such that $G^{h,\tau } =\widehat{E}^{\gamma ,\tau }$ where (11), (12), and (13) hold.

We will show that $G^{h,\tau }$ is (a) envy-free on the subadditive domain, and (b) egalitarian-equivalent on every domain.

(a) Assume, by contradiction, that $G^{h,\tau }$ is not envy-free on the subadditive domain. Then, there are $N\in \mathcal {N},$ $c\in \mathcal {C}_{sub}^{N},$ and $\{i,j\}\subseteq N$ such that $u(G_{i}^{h,\tau }(c);c_{i})<u(G_{j}^{h,\tau } (c);c_{i}).$ This inequality, (2), and (3) together imply

$$\begin{aligned} -W(c)+h_{i}(c_{-i})< & {} -c_{i}(A_{j}^{\tau } (c))+t_{j}^{h,\tau } (c), \nonumber \\= & {} -c_{i}(A_{j}^{\tau } (c))-W(c)+c_{j}(A_{j}^{\tau }(c))+h_{j}(c_{-j}). \end{aligned}$$

(21)

First, consider equations (11) and (12). By (21), $c_{i}(A_{j}^{\tau } (c))<c_{j}(A_{j}^{\tau }(c)).$ This inequality implies that $A_{j}^{\tau } (c)\ne \emptyset$ and since c is subadditive, we have

$$\begin{aligned} c_{i}(A_{i}^{\tau } (c)\cup A_{j}^{\tau } (c))\le c_{i}(A_{i}^{\tau }(c))+c_{i}(A_{j}^{\tau } (c))<c_{i}(A_{i}^{\tau }(c))+c_{j}(A_{j}^{\tau } (c)). \end{aligned}$$

Then, it would be less costly than W(c),  if i was assigned $(A_{i}^{\tau } (c)\cup A_{j}^{\tau } (c))$ and j was assigned no task,  which contradicts that $A^{\tau } (c)$ is an efficient assignment.

Now, consider equation (13). By (21),

$$\begin{aligned} c_{j}( \mathbb {A} )<-c_{i}(A_{j}^{\tau } (c))+c_{j}(A_{j}^{\tau }(c))+c_{i}( \mathbb {A} ). \end{aligned}$$

(22)

By (22), $A_{j}^{\tau } (c)\ne \mathbb {A} .$ Since c is subadditive and $\mathbb {A} =A_{i}^{\tau } (c)\cup A_{j}^{\tau } (c),$ then $c_{i}( \mathbb {A} )\le c_{i}(A_{i}^{\tau }(c))+c_{i}(A_{j}^{\tau } (c)).$ This inequality and ( 22) together imply $c_{j}( \mathbb {A} )<c_{i}(A_{i}^{\tau }(c))+c_{j}(A_{j}^{\tau } (c)).$ Then, it would be less costly than W(c),  if j was assigned all the tasks,  which contradicts that $A^{\tau } (c)$ is an efficient assignment.

(b) Now, we show that $G^{h,\tau }$ is egalitarian-equivalent. Let $N\in \mathcal {N}$ and $c\in \mathcal {C}^{N}$.

First, consider (11) and (12). By (3), for each $i\in N,$ $u(G_{i}^{h,\tau }(c);c_{i})=-W(c)+\gamma (N).$ Let $R(c)=\emptyset$ and $r(c)=-W(c)+\gamma (N).$ Then, for each $i\in N,$ $u(G_{i}^{h,\tau }(c);c_{i})=-c_{i}(R(c))+r(c).$ Hence, $G^{h,\tau }$ is egalitarian-equivalent.

Next, consider (13). By (3), for each $i\in N,$ $u(G_{i}^{h,\tau } (c);c_{i})=-W(c)+\gamma (N)+c_{j}( \mathbb {A} )$ for $j\in N\backslash \{i\}.$ Let $R(c)= \mathbb {A}$ and $r(c)=-W(c)+\sum \nolimits _{i\in N} c_{i}( \mathbb {A} )+\gamma (N)$. Then, for each $i\in N,$ $u(G_{i}^{h,\tau }(c);c_{i})=-c_{i}(R(c))+r(c).$ Hence, $G^{h,\tau }$ is egalitarian-equivalent.

Only-if Part Pick an assignment-efficient and strategy-proof mechanism. By Lemma 1, it is a Groves mechanism $G^{h,\tau }$ for some $h\in \mathcal {H}$ and $\tau \in \mathcal {T} .$ Let $G^{h,\tau }$ be egalitarian-equivalent and envy-free on the subadditive domain.

By Theorem 1 in Yengin (2012a), if a Groves mechanism is egalitarian-equivalent, then for economies with different populations, different reference set of tasks can be chosen; but for economies with the same population N,  the same reference set of tasks $\overline{R}(N)$ must work. Moreover, by Yengin (2012a), a Groves mechanism is egalitarian-equivalent if and only if for each $N\in \mathcal {N},$ there are a real number $\gamma (N)\in \mathbb {R}$ and a reference set $\overline{R}(N)\in 2^{ \mathbb {A}}$ such that for each $i\in N,$

$$\begin{aligned} h_{i}(c_{-i})=\gamma (N)+\underset{j\in N\backslash \{i\}}{\sum }c_{j}( \overline{R}(N)) \quad \text { for each } \; c\in \mathcal {C}^{N}. \end{aligned}$$

(23)

By Theorem 1 in Pápai (2003) (adapted to our setting), if $G^{h,\tau }$ is envy-free on the subadditive domain, then there is a list of functions indexed by populations, $\{\sigma _{N}\}_{_{N\in \mathcal {N}}}$ with $\sigma _{N}: \mathbb {R} _{+}\rightarrow \mathbb {R}$ such that for each $N\in \mathcal {N},$ each $i\in N,$ and each $c\in \mathcal {C}_{sub}^{N},$

$$\begin{aligned} h_{i}(c_{-i})=\sigma _{N}(W(c_{-i})). \end{aligned}$$

(24)

By (23) and (24), for each $N\in \mathcal {N}$ and each pair $\{i,j\}\subseteq N,$

$$\begin{aligned} \sigma _{N}(W(c_{-i}))-\sigma _{N}(W(c_{-j}))=c_{j}(\overline{R}(N))-c_{i}( \overline{R}(N)) \quad \text {for each } \; c\in \mathcal {C}_{sub}^{N}. \end{aligned}$$

(25)

Using (23), (24), and, (25), we will prove that $G^{h,\tau } =\widehat{E}^{\gamma ,\tau }$ by showing, on the subadditive domain, the equivalence of equation (23) to (11) when $|N|>2$; and (23) to (12) or (13) when $|N|=2$. To achieve this, we need to prove the following two cases:

Case 1 For each $N\in \mathcal {N}$ with $|N|>2,$ there is $\gamma (N)\in \mathbb {R}$ such that (23) holds for $\overline{R}(N)=\emptyset$ for each$\ c\in \mathcal {C}_{sub}^{N}.$ That is, equality (23) is equivalent to (11).

Proof of Case 1

Let $N\in \mathcal {N}$ with $|N|>2.$ By egalitarian-equivalence, there are $\gamma (N)\in \mathbb {R}$ and $\overline{R}(N)\in 2^{ \mathbb {A}}$ such that (23) holds for each $c\in \mathcal {C}_{sub}^{N}$ .

Claim For each $c\in \mathcal {C}_{sub}^{N}$ and each pair $\{i,j\}\subset N$, $c_{i}(\overline{R}(N))=c_{j}(\overline{R}(N))$.

Note that the Claim holds for each $c\in \mathcal {C}_{sub}^{N};$ and by (23), $\overline{R}(N)$ is same for each $c\in \mathcal {C} _{sub}^{N}.$ This is possible if and only if $\overline{R}(N)=\emptyset .$ This would prove that Case 1 holds.

Now, we will show that the Claim is true:

Assume, by contradiction to the claim, that there is $c\in \mathcal {C}_{sub}^{N}$ such that for some pair $\{i,j\}\subset N$,

$$\begin{aligned} c_{i}(\overline{R}(N))\ne c_{j}(\overline{R}(N)). \end{aligned}$$

(26)

Without loss of generality, let $W(c_{-j})\le W(c_{-i}).$ Let $\Delta \equiv c_{j}(\overline{R}(N))-c_{i}(\overline{R}(N))$. Let $\widehat{c}\in \mathcal {C}_{ad}^{N}$ be as follows:

(i) for each $A\in 2^{ \mathbb {A}} ,$ $\widehat{c}_{i}(A)=\frac{|A|W(c_{-j})}{| \mathbb {A} |},$

(ii) there is  $\varepsilon >\max \{-\Delta ,0\}$ such that for each $A\in 2^{ \mathbb {A}} ,$ $\widehat{c}_{j}(A)=|A|\left( \frac{W(c_{-i})}{| \mathbb {A} |}+\Delta +\varepsilon \right) ,$

(iii) for each $k\in N\backslash \{i,j\}$ and each $A\in 2^{ \mathbb {A}} ,$ $\widehat{c}_{k}(A)=\frac{|A|W(c_{-i})}{| \mathbb {A} |}.$

Note that $\Delta +\varepsilon >\max \{\Delta ,0\}.$ Hence, for each $A\in (2^{ \mathbb {A}} \backslash \emptyset ),$ $\widehat{c}_{j}(A)>\widehat{c}_{k}(A).$

By (i) and (ii), for each $A\in (2^{ \mathbb {A} }\backslash \emptyset ),$

$$\begin{aligned} \widehat{c}_{j}(A)-\widehat{c}_{i}(A)= & {} |A|\left( \frac{W(c_{-i})-W(c_{-j})}{| \mathbb {A}|}+\Delta +\varepsilon \right) , \nonumber \\> & {} \Delta . \end{aligned}$$

(27)

By (ii) and (iii), (I) $W(\widehat{c}_{-i})=\widehat{c}_{k}( \mathbb {A} )=W(c_{-i})$ for some $k\in N\backslash \{i,j\}$.

Since $W(c_{-j})\le W(c_{-i}),$ by (i) and (iii), (II) $W( \widehat{c}_{-j})=\widehat{c}_{i}( \mathbb {A} )=W(c_{-j})$.

By (I), (II), and (25), $c_{j}(\overline{R}(N))-c_{i}( \overline{R}(N))=\Delta =\widehat{c}_{j}(\overline{R}(N))-\widehat{c}_{i}( \overline{R}(N)).$ This equality and (27) together imply $\overline{R} (N)=\emptyset .$ This implies $c_{j}(\overline{R}(N))=c_{i}(\overline{R} (N)),$ which contradicts (26). Hence, Claim must be true.

Case 2 For each $N\in \mathcal {N}$ with $|N|=2,$ there is $\gamma (N)\in \mathbb {R}$ such that (23) holds either for $\overline{R}(N)=\emptyset$ for each $c\in \mathcal {C}_{sub}^{N};$ or for $\overline{R}(N)= \mathbb {A}$ for each $c\in \mathcal {C}_{sub}^{N}.$ That is, equality (23) is equivalent to (12) or (13).

Proof of Case 2

Let $N\in \mathcal {N}$ with $|N|=2.$ Without loss of generality, let $N=\{i,j\}.$ By egalitarian-equivalence, there are $\gamma (N)\in \mathbb {R}$ and $\overline{R}(N)\in 2^{ \mathbb {A}}$ such that (23) holds for each $c\in \mathcal {C}_{sub}^{N}.$ We will show that $\overline{R}(N)\in \{\emptyset , \mathbb {A}\}$.

Let $c\in \mathcal {C}_{sub}^{N}$ and $\widehat{c}=(c_{i},\widehat{c} _{j})\in \mathcal {C}_{sub}^{N}$ be such that (I) $\widehat{c}_{j}( \mathbb {A} )=c_{j}( \mathbb {A} )$ and (II) for each $A\in (2^{ \mathbb {A}} \backslash \{\emptyset , \mathbb {A} \}),$ $\widehat{c}_{j}(A)\ne$ $c_{j}(A)$. By (I), $W(\widehat{c} _{-i})=c_{j}( \mathbb {A} )=W(c_{-i}).$ This equality and (24) together imply $h_{i}( \widehat{c}_{-i})=h_{i}(c_{-i}).$ This equality and (23) together imply $\widehat{c}_{j}(\overline{R}(N))=c_{j}(\overline{R}(N)).$ This equality and (II) together imply that $\overline{R}(N)\in \{\emptyset , \mathbb {A} \}$. This proves Case 2. $\square$

Proof of Proposition 2

Let $T\in \mathbb {R}$.

(a) Mechanism $E^{\gamma ,\tau }$ satisfies T- bounded-deficit if and only if for each $N\in \mathcal {N}$ and each $c\in \mathcal {C}^{N},$ $\sum \nolimits _{i\in N} t_{i}^{\gamma ,\tau } (c)=-(n-1)W(c)+n\gamma (N)\le T$. That is, for each $N\in \mathcal {N},$ $\gamma (N)\le \min \nolimits _{c\in \mathcal {C}^{N}} \{\frac{T+(n-1)W(c)}{n} \}=T/n$. Note that the right-hand-side-of the inequality achieves its minimal value at an economy $c\in \mathcal {C}^{N}$ where $W(c)=0$ (for instance, if $c_{i}( \mathbb {A} )=0$ for some $i\in N)$.

(b) Let $\widehat{E}^{\gamma ,\tau } \in \widehat{\mathcal {E} }^{T-BD}$. Assume, by contradiction, that $\widehat{E}^{\gamma ,\tau } \in \widehat{\mathcal {E}}^{T-BD}\backslash \mathcal {E}^{T-BD}$. Then, by (10), there is $N=\{i,j\}\in \mathcal {N}$ such that $\widehat{t}_{i}^{ \gamma ,\tau } (c)=c_{j}( \mathbb {A} )-c_{j}(A_{j}^{\tau } (c))+\gamma (N)$ and $\widehat{t}_{j}^{\gamma ,\tau } (c)=c_{i}( \mathbb {A} )-c_{i}(A_{i}^{\tau } (c))+\gamma (N)$ for each $(c_{i},c_{j})\in \mathcal {C} ^{N}$. Let $c^{\prime }=(c_{i}^{\prime } ,c_{j}^{\prime } )\in \mathcal {C}^{N}$ be such that $c_{i}^{\prime } ( \mathbb {A} )>T-2\gamma (N)$ and $c_{j}^{\prime } ( \mathbb {A} )=0.$ Since $W(c^{\prime } )=c_{j}^{\prime } ( \mathbb {A} )=0,$ $\sum \nolimits _{l\in N} \widehat{t}_{l}^{\gamma ,\tau }(c^{\prime } )=c_{i}^{\prime } ( \mathbb {A} )+2\gamma (N)>T,$ which contradicts T-bounded-deficit. $\square$

Proof of Theorem 2

Let $T\in \mathbb {R}$.

(a) Let $\widehat{E}^{\gamma ,\tau } \in \widehat{\mathcal {E} }^{T-BD}\backslash \mathcal {E}^{\gamma ^{T}}.$ Then, by Lemma 2 and Proposition 2b, for each $N\in \mathcal {N},$ $\gamma (N)<T/n.$ By (3), for each $N\in \mathcal {N},$ each $c\in \mathcal {C}^{N},$ and each $i\in N$, $u(\widehat{E}_{i}^{\gamma ,\tau } (c);c_{i})=-W(c)+\gamma (N)<-W(c)+T/n.$ For each $E^{\gamma ^{T},\tau } \in \mathcal {E}^{\gamma ^{T}},$ since for each $N\in \mathcal {N},$ each $c\in \mathcal {C}^{N},$ and each $i\in N$, $u(E_{i}^{\gamma ^{T},\tau } (c);c_{i})=-W(c)+T/n,$ $E^{\gamma ^{T},\tau }$ Pareto-dominates $\widehat{E} ^{\gamma ,\tau }$.

(b) By Proposition 2 in Yengin (2012b), an egalitarian-equivalent Groves mechanism $G^{h,\tau }$ satisfies T-bounded-deficit if and only if $G^{h,\tau }\in \mathcal {E}^{T-BD}$. Let $G^{h,\tau } \in \mathcal {E}^{T-BD}\backslash \mathcal {E}^{\gamma ^{T}}.$ By Proposition 2b, $G^{h,\tau } \in \widehat{\mathcal {E}} ^{T-BD}\backslash \mathcal {E}^{\gamma ^{T}}.$ By part (a), for each $E^{\gamma ^{T},\tau } \in \mathcal {E}^{\gamma ^{T}},$ $E^{\gamma ^{T},\tau }$ Pareto-dominates $G^{h,\tau }$.

Let $G^{h,\tau }$ generate the minimal surplus among all Groves mechanisms that satisfy egalitarian-equivalence and T-bounded-deficit. For each $N\in \mathcal {N}$ and each $c\in \mathcal {C}^{N}$ , the budget surplus generated by $G^{h,\tau }$ is $-\sum \nolimits _{i\in N} t_{i}^{h,\tau }(c)=(n-1)W(c)-\sum \nolimits _{i\in N} h_{i}(c_{-i})$. Hence, the surplus is minimal if and only if $\sum \nolimits _{i\in N} h_{i}(c_{-i})$ is maximal. Since $G^{h,\tau } \in \mathcal {E}^{T-BD},$ $\sum \nolimits _{i\in N} h_{i}(c_{-i})=n\gamma (N)\le T$ is maximal if and only if for each $N\in \mathcal {N},$ $\gamma (N)=\frac{T}{|N|}$. Hence, $G^{h,\tau } \in \mathcal {E}^{\gamma ^{T}}$.

(c) By Corollary 1 in Yengin (2012a), on the subadditive domain, a Groves mechanism $G^{h,\tau }$ satisfies no-envy and solidarity if and only if $G^{h,\tau } \in \mathcal {E}.$ Hence, on this domain, a Groves mechanism $G^{h,\tau }$ satisfies no-envy, solidarity, and T-bounded-deficit if and only if $G^{h,\tau } \in \mathcal {E}^{T-BD}.$ The rest of the proof is as in part (b).

(d) By Proposition 2b in Yengin (2012a), a Groves mechanism $G^{h,\tau }$ satisfies order preservation and solidarity if and only if $G^{h,\tau } \in \mathcal {E}.$ The rest of the proof is as in part (c).

(e) Follows from Proposition 4b in Yengin (2012a).

(f) Let $E^{\gamma ^{T},\tau } \in \mathcal {E}^{\gamma ^{T}}.$ Assume, by contradiction, that there is a Groves mechanism $G^{h,\tau }$ which satisfies T-bounded-deficit and Pareto-dominates $E^{\gamma ^{T},\tau } .$ By (3), for each $N\in \mathcal {N},$ each $c\in \mathcal {C}^{N},$ and each $i\in N$, $u(G_{i}^{h,\tau }(c);c_{i})=-W(c)+h_{i}(c_{-i})$ and $u(E_{i}^{\gamma ^{T},\tau }(c);c_{i})=-W(c)+\frac{T}{n}.$ Since $G^{h,\tau }$ Pareto-dominates $E^{\gamma ^{T},\tau } ,$ there is $N\in \mathcal {N},$ $c\in \mathcal {C}^{N},$ and $j\in N$ such that $h_{j}(c_{-j})>\frac{T}{n}$. Let $c_{j}^{0}\in \mathcal {C}$ be such that $c_{j}^{0}( \mathbb {A} )=0$ and $c^{\prime }=(c_{j}^{0},c_{-j})\in \mathcal {C}^{N}.$ Since $c_{-j}^{\prime }=c_{-j},$ $h_{j}(c_{-j}^{\prime } )=h_{j}(c_{-j})>\frac{T}{n} .$ Since $W(c^{\prime } )=c_{j}^{0}( \mathbb {A} )=0,$ then $\sum \nolimits _{i\in N} t_{i}^{h,\tau } (c^{\prime } )=-(n-1)W(c^{\prime })+\sum \nolimits _{i\in N}h_{i}(c_{-i}^{\prime } )= \sum \nolimits _{i\in N\backslash \{j\}} h_{i}(c_{-i}^{\prime } )+h_{j}(c_{-j}).$ Since $G^{h,\tau }$ Pareto-dominates $E^{\gamma ^{T},\tau } ,$ for each $i\in N\backslash \{j\}$, $h_{i}(c_{-i}^{\prime } )\ge \frac{T}{ n}.$ Then, $\sum \nolimits _{i\in N} t_{i}^{h,\tau } (c^{\prime } )>T,$ which contradicts T-bounded-deficit. This completes the proof. $\square$