Bargaining, conditional consistency, and weighted lexicographic Kalai-Smorodinsky Solutions

Abstract

We reconsider the class of weighted Kalai-Smorodinsky solutions of Dubra (Econ Lett 73:131–136, 2001), and using methods of Imai (Econometrica 51:389–401, 1983), extend their characterization to the domain of multilateral bargaining problems. Aside from standard axioms in the literature, this result involves a new property that weakens the axiom Bilateral Consistency (Lensberg, J Econ Theory 45:330–341, 1988), by making the notion of consistency dependent on how ideal values in a reduced problem change relative to the original problem.

Appendix

This Appendix elaborates on several results of Imai (1983). In particular, Appendix 1 corresponds with Imai’s Lemmas 6 and 7, and part of his Proposition 1 (pp. 396–397). Appendix 2 is a visual illustration of Imai’s auxiliary problems. Appendix 3 is a modification of Imai’s Lemma 5, and part of his Proposition 2 (pp. 396, 398). Appendix 4 repeats the final part of that same Proposition 2.

1.1 Appendix 1: Solutions $\varphi \in \mathcal {L}$ satisfy IM

Consider some solution $L^{w}\in \mathcal {L}$ with $w\in \mathbb {R}^{\mathbb {N}}_{++}$. Let $N\in \mathcal {N}$, and let S and T be problems in $\Sigma ^{N}$ with $T\subseteq S$ and $T_{-i}=S_{-i}$ for some $i\in N$. Since $L^{w}$ satisfies SI, we may assume without loss of generality that $u(S)=w^{-1}_{N}$, where $w_{N}$ is the restriction of w to the agents in N. Then

$$\begin{aligned} L^{w}(S)=\xi (S). \end{aligned}$$

(5)

Define $u:=u(S)u(T)^{-1}$, and observe that

$$\begin{aligned} L^{w}(T)=u^{-1}\xi (uT). \end{aligned}$$

(6)

By Equations (5) and (6) it is sufficient to show that $\xi _{i}(S)\geqq \frac{1}{u_{i}}\xi _{i}(uT)$. This is done in two steps. First it is established that $\xi _{i}(S)\geqq \xi _{i}(T)$, and subsequently, that $\xi _{i}(T) \geqq \frac{1}{u_{i}}\xi _{i}(uT)$.

Lemma 6.1

Let $y\in T$ with $i\in Q(T,y)$. Then $Q(T,y)=Q(S,y)$.

Proof

Suppose there is some $i'\in Q(S,y)\setminus Q(T,y)$. Define $x:=x(S,y)$ and $x':=x(T,y)$. Since $S_{-i}=T_{-i}$ and $x\in S$, there is a $z\in T$ such that $z_{j}=x_{j}$ for all $j\in N\setminus i$. By convexity of T it follows that $\lambda x' +(1-\lambda ) z \in T$ for all $\lambda \in [0,1]$. We have $x,x'\geqq y$, $z_{j}=x_{j}$ for all $j\in N\setminus i$, and $x_{i}'>y_{i}$ (since $i\in Q(T,y)$). Hence, there exists a $\lambda \in (0,1)$ such that $z^{*}:=\lambda x' +(1-\lambda ) z\geqq y$ and $z^{*}\in T$. Since $i'\in Q(S,y)\setminus Q(T,y)$, $x_{i'}>y_{i'}=x_{i'}'$. Then $z_{i'}^{*}=\lambda x_{i'}'+(1-\lambda )x_{i'}>y_{i'}$. This implies $i'\in Q(T,y)$, a contradiction. $\square$

In essence, as long as the ith coordinate can be further increased in problem T, it can be further increased in S as well. It follows that $\xi _{i}(S)\geqq \xi _{i}(T)$.

To show that $\xi _{i}(T)\geqq \frac{1}{u_{i}}\xi _{i}(uT)$, define the following:

$$\begin{aligned} \begin{array}{cc} x:=\xi (T)&{}y:=u\xi (T)\\ &{}\\ x':=u^{-1}\xi (uT)&{}\quad y':=\xi (uT) \end{array} \end{aligned}$$

The aim is to show that $x_{i}\geqq x_{i}'$.

Observation 6.2

$x\succ _{N} x'$ and $y'\succ _{N} y$ (or $x=x'$ and $y=y'$).

Proof

Observe that $\xi (T)\in T$, implying $u\xi (T)\in uT$. Since $\xi (uT)\succ _{N} z$ for all $z\in uT$ with $z\ne \xi (uT)$, $\xi (uT)\succ _{N} u\xi (T)$ (or $\xi (uT) = u\xi (T)$). In other words, $y'\succ _{N}y$ (or $y'=y$). Similarly, observe that $\xi (uT)\in uT$, and thus $u^{-1}\xi (uT)\in T$. Since $\xi (T)\succ _{N} z$ for all $z\in T$ with $z\ne \xi (T)$, this implies $\xi (T)\succ _{N} u^{-1}\xi (uT)$ (or $\xi (T)= u^{-1}\xi (uT)$). In other words, $x\succ _{N}x'$ (or $x=x'$). Since $x=x'$ if and only if $y=y'$, the observation follows. $\square$

Lemma 6.3

$x_{i}\geqq x_{i}'$.

Proof

Clearly, if $x=x'$ and $y=y'$, then the inequality holds trivially, and we are done. Hence, assume $x\succ _{N} x'$ and $y'\succ _{N} y$. Since $y'\succ _{N} y$, there is an index $m\leqq |N|$ such that the first $m-1$ elements of $\mu (y)$ and $\mu (y')$ are equal, and $\mu _{m}(y')>\mu _{m}(y)$. Assume first that $x_{i}'\geqq \mu _{m}(y')$. Since $x'$ and $y'$ only differ in the i-th coordinate, $x_{i}'\geqq \mu _{m}(y')$ implies that the first m elements of $\mu (x')$ and $\mu (y')$ coincide. Then the first $m-1$ elements of $\mu (x')$ coincide with the first $m-1$ elements of $\mu (y)$. Observe that x and y only differ in the i-th coordinate and $y_{i}>x_{i}$, so there are two possibilities:

• 1. there is some $k\leqq m-1$ such that the first $k-1$ elements of $\mu (y)$ and $\mu (x)$ coincide, and $\mu _{k}(y)>\mu _{k}(x)$. Then the first $k-1$ elements of $\mu (x')$ and $\mu (x)$ coincide and $\mu _{k}(x')>\mu _{k}(x)$. Then $x'\succ _{N} x$, a contradiction.

• 2. the first m elements of $\mu (y)$ and $\mu (x)$ coincide. Since the first $m-1$ elements of $\mu (x')$ and $\mu (y)$ coincide, this implies that the first $m-1$ elements of $\mu (x)$ and $\mu (x')$ coincide. However, the m-th element of $\mu (x')$ coincides with the m-th element of $\mu (y')$. Since $\mu _{m}(y')>\mu _{m}(y)$ this implies $\mu _{m}(x')>\mu _{m}(y)$. Since $\mu _{m}(x)=\mu _{m}(y)$ this implies $\mu _{m}(x')>\mu _{m}(x)$, and thus $x'\succ _{N} x$. This is a contradiction.

It follows from the above that $x_{i}'<\mu _{m}(y')$. Hence, $x_{i}'=\mu _{m'}(x')$ for some $m'\in \{1,\ldots ,m-1\}$, and take the lowest $m'$ in case of ties. Similarly, $x_{i}=\mu _{k}(x)$ for some $k\in \{1,\ldots ,n\}$, and take the highest possible k in case of ties. There are two possibilities: $k\geqq m'$ or $k<m'$.

• 1. If $k<m'$, then $\mu _{k}(x')=\mu _{k}(y')$ (they only differ in the i-th coordinate; since $\mu _{m'}(x')=x_{i}$, $k<m'$ implies that $\mu _{k'}(x')=\mu _{k'}(y')$ for all $k'=1,\ldots ,k$). Since the first $m-1$ elements of $\mu (y)$ and $\mu (y')$ coincide, and $k<m'\leqq m-1$, the first k elements of $\mu (y)$ and $\mu (y')$ coincide. Since x and y only differ in the i-th coordinate and $x_{i}=\mu _{k}(x)$, the first $k-1$ elements of $\mu (x)$ and $\mu (y)$ coincide. This in turn implies that the first $k-1$ elements of $\mu (x)$ and $\mu (x')$ coincide. Since $y_{i}>x_{i}$ and by the choice of k (it was chosen such that $\mu _{k+1}(x)>\mu _{k}(x)$), $\mu _{k}(y)>\mu _{k}(x)$. Since the first k elements of $\mu (y)$ coincide with the first k elements of $\mu (x')$, this implies $\mu _{k}(x')>\mu _{k}(x)$. Hence, $x'\succ _{N} x$, a contradiction.

• 2. Let $k\geqq m'$. Observe that the first $m' -1$ elements of $\mu (x')$ and $\mu (y')$ coincide (this is so because they only differ in the i-th coordinate and $\mu _{m'}(x')=x_{i}'$). From before, we know that the first $m-1$ elements of $\mu (y)$ and $\mu (y')$ coincide, which implies that their first $m'$ elements coincide as well. Hence, the first $m'-1$ elements of $\mu (x')$ coincide with the first $m'-1$ elements of $\mu (y)$. Since x and y only differ in the i-th coordinate and $x_{i}=\mu _{k}(x)$, the first $m'$ elements of $\mu (x)$ coincide with the first $m'$ elements of $\mu (y)$. Hence, the first $m'-1$ elements of $\mu (x)$ and $\mu (x')$ coincide. If $\mu _{m'}(x')>\mu _{m'}(x)$, then $x'\succ _{N} x$, and we obtain a contradiction. Hence, $\mu _{m'}(x')\leqq \mu _{m'}(x)$. But then $x_{i}'=\mu _{m'}(x')\leqq \mu _{m'}(x)\leqq \mu _{k}(x)=x_{i}$, as desired.$\square$

1.2 Appendix 2: The Auxiliary Problems of Proposition 3.4

This section presents a worked out example of the auxiliary problems used in the proof of Proposition 3.4, for some weights vector w. Suppose that the problem and the solution outcome of a problem S are as in Fig. 12. Then $L^{w}(S)$ is reached in two iterations, i.e., $L^{w}(S)=x^{2}$.

Fig. 12

The problem S and the solution outcome $L^{w}(S)=x^{2}$

Recall that $\underline{S}^{0}=H(e(N),1)$. This problem is depicted in Fig. 13a. From $\underline{S}^{0}$ one can construct the problem $\overline{S}^{1}$: it is given by $H(e(N),\alpha ^{1})\cap (e(N)-\mathbb {R}^{N}_{+})$. Thus, the half-space that determines $\underline{S}^{0}$ slides upwards, and is intersected by a set that limits utilities to 1 (Fig. 13b).

Fig. 13

The problems $\underline{S}^{0}$ and $\overline{S}^{1}$

The problem $S^{1}$ is the intersection of S with the half-space $H(e(N),\alpha ^{1})$ (Fig. 14a). Similarly, $S'^{1}$ is the intersection of the problem S and the problem $\overline{S}^{1}$ (Fig. 14a). Note that these two problems coincide. For the first iteration this is always the case.

Fig. 14

The problems $S^{1}$ and $S'^{1}$

Fig. 15

The problems $\underline{S}^{1}$ and $\overline{S}^{2}$

To determine the auxiliary problems for the second iteration, one must determine $\underline{S}^{1}$. It is equal to $\overline{S}^{1}$, intersected by $H(p^{2},p^{2}\cdot x^{2})$, a half-space that leaves the utilities of agent 3 free, but restricts those of agents 1 and 2 in a $\{1,2\}$-symmetric fashion (Fig. 15a). The problem $\overline{S}^{2}$ is given by $H(e(N),\alpha ^{2})$, intersected by that same half-space $H(p^{2},p^{2}\cdot x^{2})$, and the set $(e(N)-\mathbb {R}^{N}_{+})$ that limits the utilities of all agents to 1 (Fig. 15b).

The problem $S'^{2}$ is the intersection of $\overline{S}^{2}$, and the original problem S (Fig. 16a). The problem $S^{2}$ is given by the intersection of the original problem S and the half-space $H(e(N),\alpha ^{2})$ (Fig. 16b).

Fig. 16

The problems $S'^{2}$ and $S^{2}$

1.3 Appendix 3: Proof of Claim 1

In order to show that $u(\overline{S}^{j})=u(\underline{S}^{j})=u(S^{j})=u(S'^{j})=e(N)$ for all $j=1,\ldots , k$, it is sufficient to show that $e^{i}$ is in $\overline{S}^{j}, \underline{S}^{j}, S^{j}$ and $S'^{j}$ for each j and i. This follows from four observations.

• (a) $e^{i}\in H(p^{j},\ p^{j}\cdot x^{j})$ for each i and j;

• (b) $e^{i}\in H(e(N),\alpha ^{j})$ for each i and j;

• (c) $e^{i}\in S$ for all i;

• (d) $e^{i}\in (e(N)-\mathbb {R}^{N}_{+})$ for all i.

Note that (a), (b) and (d) together imply $e^{i}\in \overline{S}^{j}$ for each i and j. Then by (c), $e^{i}\in S'^{j}$ for each i and j; for $j<k$, it is implied by (a) that each $e^{i}$ is in $\underline{S}^{j}$. Finally, (b) and (c) together imply $e^{i}\in S^{j}$ for each i and j.

Observation (d) is trivial. Observation (c) follows from comprehensiveness of S and the assumption that $u(S)=e(N)$. We now show (a) and (b). Denote $N\setminus Q^{2}$ by Q, $Q^{2}$ by $Q'$, and for $i\in Q$, denote $Q'\cup \{i\}$ by $Q_{i}$. Let $\bar{w}:=(\bar{w}_{i})_{i\in N}$ where $\bar{w}_{i}:=w_{i}/\sum _{i'\in Q}w_{i'}$ for all $i\in N$. Note that $\sum _{i\in Q}\bar{w}_{i}=1$.

By the supporting hyperplane theorem and the definition of $x^{1}$, there is a $p\in \mathbb {R}^{N}_{+}$ with $p_{i}=0$ for all $i\in Q'$, such that $p\cdot z\leqq p\cdot x^{1}$ for all $z\in S$. By observation (c), $p\cdot e^{i}\leqq p\cdot x^{1}$ for all $i\in Q$. Since $p_{i}=0$ for all $i\in Q'$, this implies $p\cdot e(Q_{i})=p\cdot e^{i}\leqq p\cdot x^{1}$ for all $i\in Q$. It follows that $p\cdot \sum _{i\in Q}\bar{w}_{i}e(Q_{i})\leqq p\cdot x^{1}$. Note that $\sum _{i\in Q}\bar{w}_{i}e(Q_{i})=\bar{w}+(e(Q')-\bar{w}e(Q'))$. Since $p_{i}=0$ for all $i\in Q'$, we obtain $p\cdot \bar{w}\leqq p\cdot x^{1}$. Note that $x^{1}=\alpha ^{1}w_{N}=[\alpha ^{1}\sum _{i\in Q}w_{i}]\bar{w}$. Then $p\cdot \bar{w}\leqq p\cdot x^{1}$ is equivalent to $\alpha ^{1}\sum _{i\in Q}w_{i} \geqq 1$. Hence, $x^{1}=\alpha ^{1}w_{N}\geqq \frac{w_{N}}{\sum _{i\in Q}w_{i}}=\bar{w}$.

Since $p^{1}\cdot e^{i}=0$ for all $i\in N$, each $e^{i}$ is trivially in $H(p^{1},\ p^{1}\cdot x^{1})$. Consider some $j\in \{2,\ldots ,k\}$. Note that $p_{i}^{j}=0$ for all $i\in Q^{j}$. Hence,

$$\begin{aligned} p^{j}\cdot \bar{w}=\sum _{i\in N\setminus Q^{j}}\bar{w}_{i}= \frac{\sum _{i\in N\setminus Q^{j}}w_{i}}{\sum _{i\in N\setminus Q^{2}}w_{i}} \geqq \frac{\sum _{i\in N\setminus Q^{2}}w_{i}}{\sum _{i\in N\setminus Q^{2}}w_{i}}=1. \end{aligned}$$

The inequality follows from the observation that $N\setminus Q^{2}$ is a subset of $N\setminus Q^{j}$. By the above, $x^{j}\geqq x^{1}\geqq \bar{w}$. Hence, $p^{j}\cdot x^{j}\geqq p^{j}\cdot x^{1}\geqq p^{j}\cdot \bar{w}\geqq 1$. Since $p^{j}\cdot e^{i}\leqq 1$ for all $i\in N$, we obtain $p^{j}\cdot x^{j}\geqq p^{j}\cdot e^{i}$ for all $i\in N$. This establishes observation (a).

Since $x^{1}\geqq \bar{w}$ and $\sum _{i\in N}\bar{w}_{i} \geqq \sum _{i\in Q}\bar{w}_{i}=1$, $e(N)\cdot e^{i}=1\leqq e(N)\cdot \bar{w}\leqq e(N) \cdot x^{1}= \alpha ^{1}$ for all $i\in N$. Hence, $e^{i}\in H(e(N),\alpha ^{1})$ for all $i\in N$. Since $H(e(N),\alpha ^{1})\subset H(e(N),\alpha ^{j})$ for each $1<j\leqq k$, observation (b) follows. $\square$

1.4 Appendix 4: Proof of Proposition 3.4 (continued)

The proof of Proposition 3.4 is concluded by the following three claims.

Claim 7

$\varphi (\overline{S}^{1})=\varphi (\underline{S}^{1})=\varphi (S^{1})=\varphi (S'^{1})=x^{1}$.

Proof

Since $\varphi (\underline{S}^{0})=x^{0}$, $\varphi (\overline{S}^{1})=x^{1}$ by Claim 6. Since $x^{1}\in \overline{S}^{1}$ and $x^{1}\in S$, $x^{1}\in S'^{1}$. Then $x^{1}=\varphi (\overline{S}^{1})\in S'^{1}\subseteq \overline{S}^{1}$, and by Claim 1, $u(S'^{1})=u(\overline{S}^{1})$. Thus by IIIA, $\varphi (S'^{1})=\varphi (\overline{S}^{1})=x^{1}$. For all $i\in N\setminus Q^{2}$, $x^{1}_{i}=x^{2}_{i}$, implying $p^{2}\cdot x^{1}=p^{2}\cdot x^{2}$. Hence, $x^{1}\in H(p^{2}, p^{2}\cdot x^{2})$. Then $x^{1}=\varphi (\overline{S}^{1})\in \underline{S}^{1}\subseteq \overline{S}^{1}$; by Claim 1, $u(\underline{S}^{1})=u(\overline{S}^{1})$. Then by IIIA, $\varphi (\underline{S}^{1})=\varphi (\overline{S}^{1})=x^{1}$. To see that $\varphi (S^{1})=x^{1}$, observe first that $S'^{1}=S^{1}\cap (e(N)-\mathbb {R}^{N}_{+})$. Since $\varphi (S^{1})\leqq u(S^{1})=e(N)$, $\varphi (S^{1})\in (e(N)-\mathbb {R}^{N}_{+})$. Hence, $\varphi (S^{1})\in S'^{1}\subseteq S^{1}$, and by Claim 1, $u(S'^{1})=u(S^{1})$. Thus, $\varphi (S^{1})=\varphi (S'^{1})=x^{1}$ by IIIA. $\square$

Claim 8

$\varphi (\overline{S}^{j})=\varphi (\underline{S}^{j})=\varphi (S^{j})=\varphi (S'^{j})=x^{j}$ for each $j=1,\ldots ,k$ (or $j=1,\ldots ,k-1$ for $\varphi (\underline{S}^{j})$).

Proof

Consider $j\in \{2,\ldots ,k\}$, and assume $\varphi (\overline{S}^{j-1})=\varphi (\underline{S}^{j-1})=\varphi (S^{j-1})=\varphi (S'^{j-1})=x^{j-1}$. By Claim 6, $\varphi (\overline{S}^{j})=x^{j}$. Then $\varphi (S'^{j})=x^{j}$ follows as in Claim 7. Furthermore, if $j<k$, then also $\varphi (\underline{S}^{j})=x^{j}$ follows as in Claim 7. What is left to show is that $\varphi (S^{j})=x^{j}$. To this end, it is first argued that $\varphi (S^{j})$ is an element of $\overline{S}^{j}$.

• 1. Since $S^{j}\subseteq H(e(N),\alpha ^{j})$, $\varphi (S^{j})\in H(e(N),\alpha ^{j})$.

• 2. Since $\varphi (S^{j})\leqq u(S^{j})=e(N)$, $\varphi (S^{j})\in (e(N)-\mathbb {R}^{N}_{+})$.

• 3. By part (ii) of Lemma 3.5, $S_{-i}^{j-1}=S_{-i}^{j}$ for all $i\in N$. Furthermore, $S^{j-1}\subseteq S^{j}$. Thus by an |N|-fold application of IM, $\varphi (S^{j})\geqq \varphi (S^{j-1})=x^{j-1}$. As in Claim 3, this implies $\varphi _{i}(S^{j})=x_{i}^{j}$ for all $i\in N\setminus Q^{j}$. From this it follows that for all $j'\in \{1,\ldots ,j\}$, $\varphi (S^{j})\in H(p^{j'},p^{j'}\cdot x^{j'})$. Thus, $\varphi (S^{j})\in \bigcap _{j'=1}^{j}H(p^{j'},p^{j'}\cdot x^{j'})$.

It follows that $\varphi (S^{j})\in \overline{S}^{j}$. Since $S^{j}\subseteq S$, we further have $\varphi (S^{j})\in S$. Thus, $\varphi (S^{j})\in S'^{j}\subseteq S^{j}$, and by Claim 1, $u(S^{j})=u(S'^{j})$. Then by IIIA, $\varphi (S^{j})=\varphi (S'^{j})=x^{j}$. $\square$

Claim 9

$\varphi (S)=x^{k}$.

Proof

Since $S^{k}=S\cap H(e(N),\alpha ^{k})$, $\varphi (S)\geqq \varphi (S^{k})$ by an |N|-fold application of IM and part (ii) of Lemma 3.5. Then by Claim 8, $\varphi (S)\geqq x^{k}$. The claim follows from the observation that $x^{k}\in P(S)$. $\square$