A Way to Play Claims Problems

Abstract

Commitment among agents is always difficult, especially when a scarce resource is to be shared. On the one hand, there are many possible ways to assign the available amount; on the other hand, each agent is motivated to propose a distribution that maximizes her award. In this paper, we propose a mechanism that combines the diminishing claims (Chun in Math Soc Sci 17(3):245–261, 1989) and the unanimous concessions (Herrero in Advances in economic design. Springer, Berlin, 2003) procedures, thereby obtaining a new justification of rules based on averaging.

Appendix

Proof of Theorem 1

The proof of this result is based on a fact, two lemmas and a remark.

Fact 1

By the definition of the double concessions procedure, it can be easily seen that, for any $P$-admissible rule, the weakly dominant strategies for the smallest and the highest claimant are dual. In other words, if the weakly dominant strategy for the smallest claimant is $f$, then the weakly dominant strategy for the highest agents will be $f^{d}$, and vice versa.

The first lemma shows that, in any step $m\in \mathbb N$, $m>1$, the sum of minimum and maximum amounts recommended by these rules coincides with the sum of the claims.

Lemma 1

For each $(E,c)\in \mathcal B$ and each $i\in N$, such that $\varphi ^{i} \in \Phi (P,f)$,and $m\in \mathbb N ,m>1$,

$$\begin{aligned} \sum \limits _{i\in N}\left[ ce_{i}(E^{m},c^{m})+s_{i}(E^{m},c^{m})\right] =C^{m}. \end{aligned}$$

Proof

Let each $(E,c)\in \mathcal B$ and each $i\in N$, such that $\varphi ^{i} \in \Phi (P,f)$ and $m\in \mathbb N ,m>1$. Then, $s_{i}(E,c)=\min \left\{ f_{i}\left( E,c\right) ,f^{d}_{i}\left( E,c\right) \right\}$, and $ce_{i}(E,c)=\max \left\{ f_{i}\left( E,c\right) ,f^{d}_{i}\left( E,c\right) \right\}$.

By duality, for each agent we are adding $f$ and $f^{d}$ rules. Hence, the next expression comes straightforwardly

$$\begin{aligned} \sum \limits _{i\in N}\left[ \frac{ ce_{i}\left( E^{m},c^{m}\right) +s_{i}\left( E^{m},c^{m}\right) }{2}\right] =E^{m}. \end{aligned}$$

Finally, we know that

$$\begin{aligned} E^{m}&= E^{m-1}-\sum \limits _{i\in N}s_{i}\left( E^{m-1},c^{m-1}\right) \\&= \sum \limits _{i\in N}\left[ \frac{ ce_{i}\left( E^{m-1},c^{m-1}\right) +s_{i}\left( E^{m-1},c^{m-1}\right) }{2}\right] \\&\quad - \sum \limits _{i\in N}s_{i}\left( E^{m-1},c^{m-1}\right) \\&= \sum \nolimits _{i\in N}\left[ \frac{ ce_{i}\left( E^{m-1},c^{m-1}\right) -s_{i}\left( E^{m-1},c^{m-1}\right) }{2}\right] =C^{m}/2, \end{aligned}$$

by the definition of the double concessions procedure.$\square$

The following remark is a direct consequence of Lemma 1 and it states that for each claims problem at any step $m\in \mathbb N ,m>1$, the half of the claims sum at every step of the double concessions procedure coincides with both the endowment and the total loss at every step of the process.

Remark 1

For each $(E,c)\in \mathcal B$ and each $i\in N$, such that $\varphi ^{i} \in \Phi (P,f)$ and $m\in ,m>1$, $E^{m}=L^{m}=C^{m}/2$.

Proof

Let each $(E,c)\in \mathcal B$ and each $i\in N$, such that $\varphi ^{i} \in \Phi (P,f)$ and $m>1\in \mathbb N$. We know that, $L^{m}=C^{m}-E^{m}$. By Lemma 1, $E^{m}=C^{m}/2$. Therefore, $L^{m}=C^{m}-C^{m}/2=C^{m}/2$. $\square$

Finally, the next lemma states that each agent’s claim at each step, different from the initial one, coincides with the sum of both the minimum and the maximum amounts recommended by these two focus.

Lemma 2

For each $(E,c)\in \mathcal B$ and each $i\in N$, such that $\varphi ^{i} \in \Phi (P,f)$ and $m>1\in \mathbb N$,

$$\begin{aligned} c_{i}^{m}=ce_{i}(E^{m},c^{m})+s_{i}(E^{m},c^{m}). \end{aligned}$$

Proof

Let each $(E,c)\in \mathcal B _{P}$ and each $i\in N$, such that $\varphi ^{i} \in \Phi (P,f)$ and $m>1\in \mathbb N$, by Remark 1 we know that for $m>1\in \mathbb N$, $L^{m}=E^{m}$, so, $s_{i}(E^{m},c^{m})=s_{i}(\left( L^{m},c^{m}\right) ^{d})$. By duality $ce_{i}(E^{m},c^{m})=c_{i}^{m}-s_{i}(\left( L^{m},c^{m}\right) )=c_{i}^{m}-s_{i}(E^{m},c^{m})$, then, $c_{i}^{m}=ce_{i}(E^{m},c^{m})+s_{i}(E^{m},c^{m})$. $\square$

Proof of Theorem 1

Let $(E,c)\in \mathcal B$ such that $\varphi ^{i} \in \Phi (P,f)$, for each $i\in N$, and each $m\in \mathbb N$, given that $\psi =(\varphi ^1,\varphi ^2,\dots ,\varphi ^n)$,

$$\begin{aligned} DU_{i}\left[ \psi ,\left( E,c\right) \right] =\underset{m\rightarrow \infty }{\lim }\underset{k=1}{\overset{m}{\sum }}s\left( E^{k},c^{k}\right) =s_{i}(E,c)+\underset{m=2}{\overset{\infty }{\sum }}s_{i}\left( E^{m},c^{m}\right) . \end{aligned}$$

By the definition of the double concessions procedure

$$\begin{aligned} \overset{\infty }{\underset{m=2}{\sum }}c_{i}^{m}&= \overset{\infty }{\underset{m=2}{\sum }}\left[ ce_{i}\left( E^{m-1},c^{m-1}\right) -s_{i}\left( E^{m-1},c^{m-1}\right) \right] \\&= ce_{i}\left( E^{m},c^{m}\right) +\overset{\infty }{\underset{m=2}{\sum }} ce_{i}\left( E^{m},c^{m}\right) -s_{i}\left( E^{m},c^{m}\right) \\&\quad -\overset{\infty }{\underset{m=2}{\sum }}s_{i}\left( E^{m},c^{m}\right) . \end{aligned}$$

By Lemma 2,

$$\begin{aligned} \overset{\infty }{\underset{m=2}{\sum }}c_{i}^{m}=\overset{\infty }{\underset{m=2}{\sum }}\left[ ce_{i}\left( E^{m},c^{m}\right) +s_{i}\left( E^{m},c^{m}\right) \right] . \end{aligned}$$

Hence,

$$\begin{aligned} ce_{i}\left( E,c\right) +\overset{\infty }{\underset{m=2}{\sum }} ce_{i}\left( E^{m},c^{m}\right) -s_{i}\left( E,c\right) -\overset{\infty }{\underset{m=2}{\sum }}s_{i}\left( E^{m},c^{m}\right) \\ =\overset{\infty }{\underset{m=2}{\sum }}\left[ ce_{i}\left( E^{m},c^{m}\right) +s_{i}\left( E^{m},c^{m}\right) \right] . \end{aligned}$$

Thus,

$$\begin{aligned} \overset{\infty }{\underset{m=2}{\sum }}s_{i}\left( E,c\right) =\left( ce_{i}(E^{m},c^{m}\right) -s_{i}\left( E,c\right) /2, \text{ and }\\ s_{i}\left( E,c\right) +\frac{ce_{i}\left( E,c\right) -s_{i}\left( E,c\right) }{2} =\frac{s_{i}\left( E,c\right) +ce_{i}\left( E,c\right) }{2}. \end{aligned}$$

Therefore, by Fact 1

$$\begin{aligned} DU\left[ \psi ,\left( E,c\right) \right]&= \frac{s\left( E,c\right) +ce\left( E,c\right) }{2}\\&= \frac{f\left( E,c\right) +f^{d}\left( E,c\right) }{2} \end{aligned}$$

$\square$