Envy-free and budget-balanced assignment of identical objects

Abstract

Strategy-proof, budget-balanced, and envy-free rank mechanisms assign q identical objects to n agents. The efficiency loss is the largest ratio of surplus loss to efficient surplus, over all profiles of non-zero valuations. The smallest efficiency loss $\frac{n-q}{n^{2}-n}$ is uniquely achieved by the following simple allocation rule: assign one object to each of the $q-1$ agents with the highest valuations, a large probability to the agent with the qth highest valuation, and the remaining probability to the agent with the $(q+1)$th highest valuation.

Appendix

1.1 Lemma 2 and its proof

Lemma 2

Suppose a symmetric function $h:\mathbb {R}_{+}^{n-1}\rightarrow \mathbb {R}$ satisfies the following functional equation :  for any $x=(x_{1},x_{2},\ldots ,x_{n})\in {\mathcal {V}},$

$$\begin{aligned} \sum _{i}h(x_{-i})=\sum _{i=2}^{n}\lambda _{i}x_{i}, \end{aligned}$$

where $\lambda _{i}\ge 0$ for all i. Then there exist $(\delta _{1},\ldots ,\delta _{n-1})\in \mathbb {R}^{n-1}$ such that for any $z\in \mathbb {R}_{+}^{n-1},$ $h(z)=\delta _{1}z_{*1}+\delta _{2}z_{*2}+\cdots +\delta _{n-1}z_{*(n-1)}.$

Proof

Note that the lemma states only a necessary condition of the solutions. We show it by induction. Since

$$\begin{aligned} \lambda _{2}x_{2}+\lambda _{3}x_{3}+\cdots +\lambda _{n}x_{n}= & {} h(x_{2},x_{3},\ldots ,x_{n})+h(x_{1},x_{3},\ldots ,x_{n})+\cdots \\&+\,h(x_{1},x_{2},\ldots ,x_{n-1}), \end{aligned}$$

taking $x_{1}=\cdots =x_{n}=0$, we get $h(0,0,\ldots ,0)=0$. Taking $x_{1}\ge x_{2}=\cdots =x_{n}=0$, we get $h(x_{1},0,\ldots ,0)=0$. Now suppose $h(x_{1},x_{2},\ldots ,x_{k},0,\ldots ,0)=\delta _{1}^{\prime }x_{1}+\delta _{2}^{\prime }x_{2}+\cdots +\delta _{k}^{\prime }x_{k}$ for any $x_{1}\ge x_{2}\ge \cdots \ge x_{k}\ge 0$. We show that $h(x_{1},x_{2},\ldots ,x_{k},x_{k+1}0,\ldots ,0)=\delta _{1}^{\prime \prime }x_{1}+\delta _{2}^{\prime \prime }x_{2}+\cdots +\delta _{k}^{\prime \prime }x_{k}+\delta _{k+1}^{\prime \prime }x_{k+1}$ for any $x_{1}\ge x_{2}\ge \cdots \ge x_{k}\ge x_{k+1}\ge 0$.

Note that

$$\begin{aligned}&\lambda _{2}x_{2}+\lambda _{3}x_{3}+\cdots +\lambda _{k}x_{k}+\lambda _{k+1}x_{k+1} \\&\quad =h(x_{2},x_{3},\ldots ,x_{k+1},0,\ldots ,0)\\&\qquad +\,h(x_{1},x_{3},\ldots ,x_{k+1},0,\ldots ,0)\\&\qquad +\cdots \\&\qquad +\,h(x_{1},x_{2},\ldots ,x_{k},0,\ldots ,0)\\&\qquad +\,(n-k-1)h(x_{1},x_{2},\ldots ,x_{k},x_{k+1},0,\ldots ,0)\\&\quad =\sum _{l=1}^{k+1}[(l-1)\delta _{l-1}^{\prime }+(k-l+1)\delta _{l}^{\prime }]x_{l}\\&\qquad +\,(n-k-1)h(x_{1},x_{2},\ldots ,x_{k},x_{k+1},0,\ldots ,0), \end{aligned}$$

where $\delta _{0}^{\prime }=0$.

Therefore $h(x_{1},x_{2},\ldots ,x_{k},x_{k+1},0,\ldots ,0)=\delta _{1}^{\prime \prime }x_{1}+\delta _{2}^{\prime \prime }x_{2}+\cdots \delta _{k}^{\prime \prime }x_{k}+\delta _{k+1}^{\prime \prime }x_{k+1}$, where

$$\begin{aligned} \delta _{l}^{\prime \prime }=\frac{\lambda _{l}-[(l-1)\delta _{l-1}^{\prime }+(k-l+1)\delta _{l}^{\prime }]}{n-k+1}, \end{aligned}$$

and $\lambda _{1}=0$. $\square$

1.2 Lemma 3 and its proof

Lemma 3

The system of linear equations (4) has a unique solution $\{\delta _{k}\}_{k=1}^{n-1}$ if and only if Eq. (6) holds.

Proof

Note that the system of linear equations (4) have $n-1$ unknowns and n equations. We show the “only if” part. We write out the equations:

Note that according to the first three equations, $\delta _{1}=0$, $\delta _{2}=\frac{a_{1}-a_{2}}{n-2}$, and

$$\begin{aligned} \delta _{3}=-\frac{2}{(n-2)(n-3)}a_{1} +\left[ \frac{2}{(n-2)(n-3)}+\frac{2}{n-3}\right] a_{2}-\frac{2}{n-3}a_{3}. \end{aligned}$$

We show by induction that for each k, $2\le k\le n-1$,

$$\begin{aligned} \delta _{k}= & {} (-1)^{k}\cdot \prod _{i=n-k}^{n-2}\lambda _{i}\cdot a_{1}+(-1)^{k-1}\cdot \left( \prod _{i=n-k}^{n-2}\lambda _{i}+ \prod _{i=n-k}^{n-3}\lambda _{i}\right) \cdot a_{2}\nonumber \\&+\cdots +(-1)^{2}\cdot \left( \prod _{i=n-k}^{n-k+1} \lambda _{i}+\lambda _{n-k}\right) \cdot a_{k-1}+(-1)\cdot \lambda _{n-k}\cdot a_{k}, \end{aligned}$$

(9)

where $\lambda _{n-1}=0$, and for each i, $1\le i\le n-2$, $\lambda _{i}=\frac{n-i-1}{i}$.

Let $k<n-1$. Suppose for each l, $2\le l\le k$, Eq. (9) holds. Then according to the $(k+1)$th equation of the linear system (4),

$$\begin{aligned} \delta _{k+1}= & {} \frac{k}{n-k-1}(a_{k}-a_{k+1})-\frac{k}{n-k-1}\delta _{k}\\= & {} {\uplambda }_{n-k-1}(a_{k}-a_{k+1})-\lambda _{n-k-1}\delta _{k}. \end{aligned}$$

Now it is easy to calculate $\delta _{k+1}$. And Eq. (9) holds for $k+1$, as desired.

Hence according to the first $n-1$ equations of the system,

$$\begin{aligned} \delta _{n-1}= & {} (-1)^{n-1}\cdot \prod _{i=1}^{n-2}\lambda _{i}\cdot a_{1}+(-1)^{n-2}\cdot \left( \prod _{i=1}^{n-2}\lambda _{i}+\prod _{i=1}^{n-3}\lambda _{i}\right) \cdot a_{2}\nonumber \\&+\cdots +(-1)^{2}\cdot \left( \prod _{i=1}^{2}\lambda _{i}+\lambda _{1}\right) \cdot a_{n-2}+(-1)\cdot \lambda _{1}\cdot a_{n-1} . \end{aligned}$$

(10)

Note that $\prod _{i=1}^{m}\lambda _{i}=\mathcal {C}_{m}^{n-2}=\mathcal {C}_{n-2-m}^{n-2}=\prod _{i=1}^{n-2-m}\lambda _{i}$.

According to the last equation of the system, we have

$$\begin{aligned} \delta _{n-1}=a_{n-1}-a_{n}. \end{aligned}$$

(11)

Combining (10) and (11), we have (6).

Now the “if” part is easy to see. $\square$