As seen in Example 3.3, the Talmud rule \(\tau \) allows for multiple \(\tau \)-transfer schemes for a non-hierarchical mutual liability problem. For an arbitrary bankruptcy rule \(\varphi \), however, there is always a unique \(\varphi \)-transfer allocation.
Theorem 5.1
Let \(C\in {\mathcal {L}}^N\), let \(\varphi \) be a bankruptcy rule and let \(P,{\tilde{P}}\in {\mathcal {P}}^{\varphi }(C)\). Then,
$$\begin{aligned} \alpha ^{P}=\alpha ^{{\tilde{P}}}. \end{aligned}$$
Proof
On the contrary suppose that \(\alpha ^{P}\ne \alpha ^{{\tilde{P}}}\). For notational convenience, set \(\alpha ^{P}=\alpha \), \(\alpha ^{{\tilde{P}}}={\tilde{\alpha }}\), \(E_i = \sum _{j\in N} p_{ji}\) and \({{\tilde{E}}}_i = \sum _{j\in N} {{\tilde{p}}}_{ji}\).
By estate monotonicity, for all \(i\in N\): \(E_i\le {{\tilde{E}}}_i\) implies that for all \(j\in N\) we have \(p_{ij}\le {{\tilde{p}}}_{ij}\). As a consequence, for all \(i\in N\) we have
$$\begin{aligned} p_{ij}<{{\tilde{p}}}_{ij} \text { for some } j\in N \Longrightarrow p_{ij}\le {{\tilde{p}}}_{ij} \text { for all } j\in N. \end{aligned}$$
(12)
Let \(N=\{1,\ldots ,n\}\). Without loss of generality we assume that, \(\alpha _1<{\tilde{\alpha }}_1\). Then \({\tilde{\alpha }}_1>0\), so Lemma 3.5 implies that for all \(j\in N\),
$$\begin{aligned} {\tilde{p}}_{1j}=c_{1j}. \end{aligned}$$
(13)
We will show by induction that for all \(i\ge 2\) we have:
$$\begin{aligned} \alpha _i=0 \text { and } p_{ij}\le {{\tilde{p}}}_{ij} \text { for all } j\in N. \end{aligned}$$
(14)
For this, let \(j\in \{2, \ldots ,n\}\) and suppose that (14) is valid for all \(i\in \{2, \ldots ,j-1\}\). Note that this is a void assumption in case \(j=2\). The agents in \(\{1, \ldots ,j-1\}\) together possess less after performing transfer scheme P than after performing \({{\tilde{P}}}\), i.e. \(\alpha _1\) vs. at least \({{\tilde{\alpha }}} _1\). Hence, there must be agents \(k\in \{1, \ldots ,j-1\}\) and \(\ell \in \{j, \ldots ,n\}\) such that the net payment from \(\ell \) to k is greater when \({{\tilde{P}}}\) is applied than when P is applied, i.e., such that \( p_{\ell k}-p_{k\ell }< {\tilde{p}}_{\ell k}- {\tilde{p}}_{k\ell }. \) Without loss of generality we assume that \(\ell =j\). If \(k=1\), we have \(p_{kj}\le {{\tilde{p}}}_{kj}\) because of (13). If \(k>1\), we have \(p_{kj}\le {{\tilde{p}}}_{kj}\) by the induction hypothesis for \(i=k\). In both cases we find that \( p_{jk} < {\tilde{p}}_{jk} \) and hence \(\alpha _j =0\) (Lemma 3.5) and \(p_{jm}\le {{\tilde{p}}}_{jm}\text { for all }m\in N\) [by (12)]. This proves the induction step and completes the verification of (14).
Equation (14), however, leads to the contradiction
$$\begin{aligned} \sum _{j\in N}c_{jj} = \sum _{j\in N}\alpha _{j} = \alpha _1<{{\tilde{\alpha }}}_1 \le \sum _{j\in N}{{\tilde{\alpha }}}_{j}= \sum _{j\in N}c_{jj}, \end{aligned}$$
which finishes the proof. \(\square \)
Theorem 5.1 allows for the following definition.
Definition F
Let \(\varphi \) be a bankruptcy rule. The corresponding \(\varphi \)-based mutual liability rule \(\rho ^{\varphi }:{\mathcal {L}}^{N}\rightarrow {\mathbb {R}}^N\) is for all \(C\in {\mathcal {L}}^{N}\) defined by
$$\begin{aligned} \rho ^{\varphi }(C)=\alpha ^{P}, \end{aligned}$$
where P is a \(\varphi \)-transfer scheme for C.
The final part of this section will provide an axiomatic characterization of \(\rho ^{\tau }\) as a \(\varphi \)-based mutual liability rule on the class \({\mathcal {L}}\) of all mutual liability problems with an arbitrary but finite set of players by extending the concede and divide-principle and consistency for bankruptcy rules to the setting of mutual liability.
In bankruptcy problems the principle of concede and divide is defined for problems with two claimants. However, in a mutual liability problem with two agents, every agent faces only one (possible) claimant. For such mutual liability problems the allocation prescribed by any \(\varphi \)-based mutual liability rule is unique, as the following lemma states (the proof of this lemma is straightforward and therefore omitted).
Lemma 5.2
Let \({\mathcal {C}}\in {\mathcal {L}}^N\) with \(N=\{1,2\}\). Let \(\varphi ^I\) and \(\varphi ^{II}\) be bankruptcy rules. Then,
$$\begin{aligned} \rho ^{\varphi ^I}(C)=\rho ^{\varphi ^{II}}(C). \end{aligned}$$
Three entities are involved in a classical bankruptcy situation with two agents, i.e., the bank and the two agents. Since in mutual liability situations agents have multiple roles, it is natural to define a c&d type of axiom in the three agents setting.
Definition G
A mutual liability rule f satisfies the concede & divide-principle (\( c \& d \)) if for each N with \(|N|=3\) and for each \(C\in {\mathcal {L}}^N\), there exists an underlying transfer scheme \(P\in {\mathcal {P}}(C)\) such that \(f(C)=\alpha ^P\) and for each player \(i\in N\), his ‘estate’ \(e^i=c_{ii}+\sum _{\ell \ne i}p_{\ell i}\) is allocated among the remaining two players, j, k, respecting the bankruptcy concede and divide-principle, i.e.
$$\begin{aligned} p_{ij}={\left\{ \begin{array}{ll} c_{ij} &{}\text { if } e^i\ge c_{ij}+c_{ik},\\ (e^i-c_{ik})^+ +\frac{e^i-(e^i-c_{ik})^+-(e^i-c_{ij})^+}{2} &{}\text { otherwise. } \end{array}\right. } \end{aligned}$$
(15)
Example 5.3
Reconsider the mutual liability problem \(C\in {\mathcal {L}}^N\) of Example 2.1 with \(N=\{1,2,3\}\) and C given by
$$\begin{aligned} C=\begin{bmatrix} 3&1&4 \\ 2&2&6 \\ 1&0&1 \end{bmatrix}. \end{aligned}$$
Take \(P\in {\mathcal {P}}^{\tau }(C)\) given by
$$\begin{aligned} P=\begin{bmatrix} 3&1&4\\ 1&2&2\\ 1&0&1 \end{bmatrix} \end{aligned}$$
with \(\rho ^{\tau }(C)=\alpha ^{P}=(0,0,6)\). We check that the entries in P satisfy (15). Here, \(e^1=p_{11}+p_{21}+p_{31}=5\), \(e^2=3\) and \(e^3=7\). Both player 1’s and player 3’s estate are sufficient to satisfy their claimants, hence \(p_{12}=c_{12}=1\), \(p_{13}=4\) and \(p_{31}=1\). Player 2’s estate is not sufficient, therefore
$$\begin{aligned} p_{21}=(e^2-c_{23})^+ +\frac{e^2-(e^2-c_{21})^+ -(e^2-c_{23})^+}{2}=0+\frac{3-1-0}{2}=1 \end{aligned}$$
and \(p_{23}=2\). \(\square \)
Next, we define the property of consistency for a mutual liability rule. This property is defined on the class \({\mathcal {L}}\) of mutual liability problems with arbitrary but finite N. The consistency property requires that a reallocation of the total amount which has been allocated to a coalition T, on the basis of that rule and an underlying transfer scheme, does not change the initial individual allocations within this coalition.
Definition H
A mutual liability rule f for \({\mathcal {L}}\) is called consistent if for all N and for all \(C\in {\mathcal {L}}^N\) there exists a \(P\in {\mathcal {P}}(C)\) such that \(f(C)=\alpha ^P\) and such that for all \(T\in 2^N\backslash \{\emptyset \}\) with \(C^{T,P}\in {\mathcal {L}}^T\),
$$\begin{aligned} f(C^{T,P})=f(C)_{|T}, \end{aligned}$$
(16)
where \(C^{T,P}\in {\mathbb {R}}^{T\times T}\) is defined, for all \(i,j\in T\), by
$$\begin{aligned} c^{T,P}_{ij}={\left\{ \begin{array}{ll} c_{ij} &{}\text { if } i\ne j,\\ c_{ii}+\sum _{k\in N\backslash T}(p_{ki}-p_{ik}) &{}\text { if } i=j. \end{array}\right. } \end{aligned}$$
(17)
Note that there is only the consistency requirement (16) for T if \(C^{T,P}\in {\mathcal {L}}^T\). As is seen in the following example, it can indeed happen that \(C^{T,P}\notin {\mathcal {L}}^T\).
Example 5.4
Let \(N=\{1,2,3,4\}\). Reconsider the hierarchical mutual liability problem \(C\in {\mathcal {L}}^N\) of Example 2.2, given by
$$\begin{aligned} C=\begin{bmatrix} 4&2&4&4\\0&3&0&1\\ 0&0&2&3\\ 0&0&0&2\end{bmatrix}. \end{aligned}$$
As recursively determined in Example 4.2, the unique \(\tau \)-transfer scheme P for C is given by
$$\begin{aligned} P=\begin{bmatrix} 4&1&1.5&1.5 \\ 0&3&0&1 \\ 0&0&2&3\\ 0&0&0&2 \end{bmatrix} \end{aligned}$$
and \(\rho ^{\tau }=(0,3,0.5,7.5)\).
With \(T=\{1,2,4\}\) we have
$$\begin{aligned} C^{T,P}=\begin{bmatrix} 2.5&2&4\\0&3&1\\ 0&0&5 \end{bmatrix}, \end{aligned}$$
which is a mutual liability problem and the unique \(\tau \)-transfer scheme \(P^T\) for \(C^{T,P}\) is given by
$$\begin{aligned} P^T=\begin{bmatrix} 2.5&1&1.5 \\ 0&3&1 \\ 0&0&5 \end{bmatrix}, \end{aligned}$$
while \(\rho ^{\tau }(C^{T,P})=(0,3,7.5)\). We see that the consistency requirement for this T is satisfied. However, with \(T=\{1,2,3\}\), we obtain
$$\begin{aligned} C^{T,P}=\begin{bmatrix} 2.5&2&4\\0&2&0\\ 0&0&-1 \end{bmatrix}. \end{aligned}$$
Since \(C^{T,P}\) contains negative entries, it is not a mutual liability problem and therefore does not impose a consistency requirement. \(\square \)
The \(\tau \)-based mutual liability rule \(\rho ^\tau \) satisfies consistency and \( c \& d \).
Theorem 5.5
\(\rho ^{\tau }\) is consistent and satisfies \( c \& d \) on \({\mathcal {L}}\).
Proof
We start with proving \( c \& d \). Let \(C\in {\mathcal {L}}^N\) with \(|N|=3\). Let \(i\in N\) and set \(N\backslash \{i\}=\{j,k\}\). Consider an arbitrary \(P\in {\mathcal {P}}^{\tau }(C)\). Obviously \(\rho ^{\tau }(C)=\alpha ^P\) by Theorem 5.1. Moreover,
$$\begin{aligned} p_{ij}&=\tau _j(p_{ii}+p_{ji}+p_{ki},d^i(C))\\&=\tau _j(e^i,(c_{ij},c_{ik})). \end{aligned}$$
Since the bankruptcy rule \(\tau \) satisfies the \( c \& d \) principle for bankruptcy problems, we find that
$$\begin{aligned} p_{ij}={\left\{ \begin{array}{ll} c_{ij} &{}\text { if } e^i\ge c_{ij}+c_{ik},\\ (e^i-c_{ik})^+ +\frac{e^i-(e^i-c_{ik})^+-(e^i-c_{ij})^+}{2} &{}\text { otherwise. } \end{array}\right. } \end{aligned}$$
Next, we show consistency. For this, let \(C\in {\mathcal {L}}^N\), consider an arbitrary \(P\in {\mathcal {P}}^{\tau }(C)\) and let \(T\in 2^N\backslash \{\emptyset \}\) be such that \(C^{T,P}\in {\mathcal {L}}^T\). It suffices to show that \(\rho ^{\tau }(C)_{|T}=\rho ^{\tau }(C^{T,P})\).
Define \(P^T=(p_{ij}^T)\in {\mathbb {R}}^{T\times T}\) by
$$\begin{aligned} p_{ij}^T={\left\{ \begin{array}{ll} p_{ij} &{}\text { if } i\ne j\\ p_{ii} +\sum _{k\in N\backslash T}(p_{ki}-p_{ik}) &{}\text { if } i=j. \end{array}\right. } \end{aligned}$$
(18)
We first show that \(P^T\in {\mathcal {P}}^{\tau }(C^{T,P})\), which implies that \(\alpha ^{P^T}=\rho ^{\tau }(C^{T,P})\).
For this, note that \(c^{T,P}_{ii}=p_{ii}^T\) for all \(i\in T\). It remains to prove that for all \(i\in T\) and \(j\in T\backslash \{i\}\),
$$\begin{aligned} p_{ij}^T=\tau _j\left( p_{ii}^T+\sum _{k\in T\backslash \{i\}}p_{ki}^T,d^i(C^{T,P})\right) . \end{aligned}$$
This is true because for each \(i\in T\) and \(j\in T\backslash \{i\}\)
$$\begin{aligned} p^T_{ij}=p_{ij}&=\tau _j \left( p_{ii}+\sum _{k\in N\backslash \{i\}}p_{ki}, d^i(C)\right) \\&=\tau _j \left( p_{ii}+\sum _{k\in N\backslash \{i\}}p_{ki}-\sum _{k\in N\backslash T}\tau _k\bigg (p_{ii}+\sum _{k\in N\backslash \{i\}}p_{ki},d^i(C)\bigg ), d^i(C)_{|T}\right) \\&=\tau _j \left( p_{ii}+\sum _{k\in N\backslash \{i\}}p_{ki}-\sum _{k\in N\backslash T}p_{ik}, d^i(C)_{|T}\right) \\&=\tau _j\left( p_{ii} +\sum _{k\in N\backslash T}(p_{ki}-p_{ik})+\sum _{k\in T\backslash \{i\}} p_{ki}, d^i(C)_{|T}\right) \\&=\tau _j\left( p^T_{ii}+\sum _{k\in T\backslash \{i\}} p^{T}_{ki}, d^i(C^{T,P})\right) , \end{aligned}$$
where the third equality follows from consistency of the Talmud rule, the fourth equality follows from the fact that \(P\in {\mathcal {P}}^{\tau }(C)\), while the last equality follows from (18).
The proof is finished if we show that \(\alpha ^{P^T}=\rho ^{\tau }(C)_{|T}\). For this, note that with \(i\in T\)
$$\begin{aligned} \alpha ^{P^T}_i&=p_{ii}^T+\sum _{j\in T\backslash \{i\}}(p^T_{ji}-p_{ij}^T)\\&=p_{ii}+\sum _{j\in N\backslash T}(p_{ji}-p_{ij})+\sum _{j\in T\backslash \{i\}}(p_{ji}-p_{ij})\\ \phantom {\alpha ^{P^T}_i}&=p_{ii}+\sum _{j\in N\backslash \{i\}}(p_{ji}-p_{ij})\\&=\alpha ^{P}_i=\rho _i^{\tau }(C). \end{aligned}$$
\(\square \)
We conclude this section with a characterization of the \(\tau \)-based mutual liability rule.
Theorem 5.6
Let \(\varphi \) be a bankruptcy rule. Then, \(\rho ^{\varphi }(C)=\rho ^{\tau }(C)\) for all \(C\in {\mathcal {L}}\) if and only if \(\rho ^{\varphi }\) satisfies consistency and \( c \& d \).
Proof
For the ‘only if’-part, we refer to Theorem 5.5. To prove the ‘if’-part, let \(\varphi \) be a bankruptcy rule such that \(\rho ^{\varphi }\) satisfies consistency and \( c \& d \). As we have seen before, the class \({\mathcal {B}}\) of bankruptcy problems is a subclass of \({\mathcal {L}}\) by identifying each \((E,d)\in {\mathcal {B}}^N\) with \(N=\{1,\ldots ,n\}\), with \(C(E,d)\in {\mathcal {L}}^{N\cup \{0\},\Delta }\) given by
Let P be the unique \(\varphi \)-transfer scheme for C(E, d). Then,
$$\begin{aligned} P= \begin{bmatrix}E&p_{01}&\cdots&p_{0n} \\0&0&\cdots&0\\\vdots&\ddots&\vdots \\0&\cdots&0\end{bmatrix}, \end{aligned}$$
with
$$\begin{aligned} \alpha ^{P}_i={\left\{ \begin{array}{ll}p_{0i} &{} \text { if } i \in N,\\ E-\sum _{j\in N}p_{0j} &{} \text { if } i=0. \end{array}\right. } \end{aligned}$$
Moreover, for all \(i\in N\)
$$\begin{aligned} \rho ^{\varphi }_i(C(E,d))=\alpha ^{P}_i=p_{0i}=\varphi _i(c_{00},d^0(C))=\varphi _i(E,d). \end{aligned}$$
(19)
Thus \(\varphi (E,d)=\rho ^{\varphi }(C(E,d))_{|N}\).
If we can show that
-
(I)
\( c \& d \) of \(\rho ^{\varphi }\) on \({\mathcal {L}}\) implies \( c \& d \) of \(\varphi \) on \({\mathcal {B}}\),
-
(II)
consistency of \(\rho ^{\varphi }\) on \({\mathcal {L}}\) implies consistency of \(\varphi \) on \({\mathcal {B}}\),
then, \(\varphi =\tau \) (cf. Aumann and Maschler 1985) and consequently \(\rho ^{\varphi }=\rho ^{\tau }\).
For this, we first show that P is the unique transfer scheme for C(E, d) that leads to the transfer allocation \(\alpha ^P\) and for this reason \( c \& d \) and consistency of \(\rho ^{\varphi }\) can only have implications on P.
Let \({\tilde{P}}=({\tilde{p}}_{ij})\in {\mathcal {P}}(C(E,d))\) be an arbitrary transfer scheme for C(E, d) with \({\tilde{P}}\ne P\). Then,
$$\begin{aligned} {\tilde{P}}= \begin{bmatrix}E&{\tilde{p}}_{01}&\cdots&{\tilde{p}}_{0n} \\0&0&\cdots&0\\\vdots&\ddots&\vdots \\0&\cdots&0\end{bmatrix} \end{aligned}$$
and there must be a player \(i\in N\) with \({\tilde{p}}_{0i}\ne p_{0i}\). Hence, \(\alpha ^{{\tilde{P}}}\ne \alpha ^{P}\).
With respect to (I), let \(N=\{1,2\}\) and \((E,d)\in {\mathcal {B}}^N\). Let \(i\in N\) and \(\{j\}=N\backslash \{i\}\). We need to show that
$$\begin{aligned} \varphi _i(E,d)={\left\{ \begin{array}{ll} d_i &{} \text { if } E\ge d_{1}+d_{2},\\ (E-d_{j})^+ +\frac{E-(E-d_{i})^+-(E-d_{j})^+}{2} &{}\text { otherwise.} \end{array}\right. } \end{aligned}$$
\( C \& d \) on \({\mathcal {L}}\) and (19) imply that, with \(c_{0i}=C_{0i}(E,d)\) and \(c_{0j}=C_{0j}(E,d)\),
$$\begin{aligned} \varphi _i(E,d){=}\rho ^{\varphi }_i(C(E,d))= & {} {\left\{ \begin{array}{ll} c_{0i} &{} \text { if } e^0{\ge } c_{01}+c_{02},\\ (e^0-c_{0j})^+ +\frac{e^0-(e^0-c_{0i})^+-(e^0-c_{0j})^+}{2} &{}\text { otherwise, } \end{array}\right. }\\= & {} {\left\{ \begin{array}{ll} d_i &{} \text { if } E\ge d_{1}+d_{2},\\ (E-d_{j})^+ +\frac{E-(E-d_{i})^+-(E-d_{j})^+}{2} &{}\text { otherwise.} \end{array}\right. } \end{aligned}$$
With respect to (II), let \((E,d)\in {\mathcal {B}}^N\) and \(T\in 2^N\backslash \{\emptyset \}\). We have to prove that
$$\begin{aligned} \varphi (E,d)_{|_T}=\varphi \left( \sum _{j\in T}\varphi _j(E,d),d_{|_T}\right) . \end{aligned}$$
Let \(T=\{k_1,\ldots ,k_t\}\). Then, using (17) and (19),
Clearly, \(C^{T\cup \{0\},P}(E,d)\in {\mathcal {L}}^{T\cup \{0\},\Delta }\) and
$$\begin{aligned} C^{T\cup \{0\},P}(E,d) =C\left( E-\sum _{j\in N\backslash T}\varphi _j(E,d),d_{|T}\right) . \end{aligned}$$
Using consistency, we find for all \(i\in T\) that
$$\begin{aligned} \rho ^{\varphi }_i(C(E,d))_{|T\cup \{0\}}=\rho ^{\varphi }_i\left( C\left( E-\sum _{j\in N\backslash T}\varphi _j(E,d),d_{|T}\right) \right) . \end{aligned}$$
By Eq. (19), for all \(i\in T\)
$$\begin{aligned} {\left\{ \begin{array}{ll} \rho ^{\varphi }_i(C(E,d))&{}=\varphi _i(E,d)\\ \rho ^{\varphi }_i(C(E-\sum _{j\in N\backslash T}\varphi _j(E,d),d_{|T}))&{}=\varphi _i(E-\sum _{j\in N\backslash T}\varphi _j(E,d),d_{|T}) \end{array}\right. } \end{aligned}$$
and therefore,
$$\begin{aligned} \varphi ((E,d))_{|T}=\varphi \left( E-\sum _{j\in N\backslash T}\varphi _j(E,d),d_{|T}\right) =\varphi \left( \sum _{j\in T}\varphi _j(E,d),d_{|T}\right) , \end{aligned}$$
where the last equality follows from (1). \(\square \)