Remarks on the integer Talmud solution for integer bankruptcy problems

Abstract

In Fragnelli et al. (TOP 22:892–933, 2014; TOP 24:88–130, 2016), we considered a bankruptcy problem with the additional constraint that the estate has to be assigned in integer unities, allowing for non-integer claims; we dealt with the extension to our setting of the constrained equal losses solution and of the constrained equal awards solution. Here, we analyze the possibilities of extending the Talmud solution to the integer situation, starting from the existing approaches for the non-integer case; some of these approaches are compatible with the non-integer claims, but in order to comply with as much as possible of the approaches it is necessary to switch to integer claims.

Appendix

In this section, we provide motivations and counterexamples for the properties considered in Sect. 8.

• Efficiency (EFF) By construction, all solutions are efficient.

• Weak equal treatment (WET) $\hbox {ITAL}^a, \hbox {ITAL}^b$ and $\hbox {ITAL}^c$ satisfy WET because ICEA and ICEL satisfy it. $\hbox {DFL}(a), \hbox {DFL}(b)$ and $\hbox {DFL}(c)$ satisfy WET because the vessels of the equal claimants have identical structure so their level may differ at most by a unity. Integer nucleolus: if two agents i and j have equal claims they are symmetric in the game; if a solution assigns $x_i = k, x_j = k + 2$ then also $x_i = k + 2, x_j = k$ produces the same excesses, but $x_i = x_j = k + 1$ produces weakly better excesses.

• Two-player consistency (2-CONS) $\hbox {ITAL}^a, \hbox {ITAL}^b$ and $\hbox {ITAL}^c$ satisfy 2-CONS by definition of $\hbox {ICG}, \hbox {ICG}^*, \hbox {ICG}^c$, with the hypothesis that the priority of the players is coherent with the ordering according the values of the claims. $\hbox {DFL}(a), \hbox {DFL}(b)$ and $\hbox {DFL}(c)$ satisfy 2-CONS because restricting to any subset of two players, the geometry of their vessels remains unchanged and consequently the possible distribution of the unities of the liquid. Integer nucleolus does not satisfy 2-CONS. Consider $I\nu (\{1,2,3\},(1,3,5),5) = \{(0,2,3), (1,1,3)\}$ the restriction to $\{1,3\}$ of the first solution produces $I\nu (\{1,3\},(1,5),3) = \{(0,3), (1,2)\}$ and the restriction to $\{1,3\}$ of the second solution produces $I\nu (\{1,3\},(1,5),4) = \{(0,4), (1,3)\}$.

• Independence of claims truncation (ICT) Fix the priority ordering $1 \succ 2 \succ 3$. $\hbox {ITAL}^a$ and $\hbox {ITAL}^c$ do not satisfy ICT; in fact, $\hbox {ITAL}^a(\{1,2,3\},(1,3,10),7) = (0,1,6)$, while $\hbox {ITAL}^a(\{1,2,3\},(1,3,7),7) = (0,2,5)$ and $\hbox {ITAL}^c(\{1,2,3\},(1,3,8),5) = (0,1,4)$, while $\hbox {ITAL}^c(\{1,2,3\},(1,3,5),5) = (1,2,2)$. $\hbox {DFL}(a)$ and $\hbox {DFL}(c)$ do not satisfy ICT because the solution of the sets in the two cases may not coincide; in fact $\hbox {DFL}(a)(\{1,2,3\},(1,3,8),5) = \{(0,1,4)\}$, while $\hbox {DFL}(a)(\{1,2,3\},(1,3,5),5) = \{(0,2,3), (0,1,4)\}$ and $\hbox {DFL}(c)(\{1,2,3\},(1,3,8),5) = \{(0,1,4)\}$, while $\hbox {DFL}(c)(\{1,2,3\},(1,3,5),5) = \{(0,2,3), (1,1,3), (1,2,2)\}$. Integer nucleolus satisfies ICT because the two games coincide.

• Composition from minimal rights (CMR) Fix the priority ordering $1 \succ 2 \succ 3$. $\hbox {ITAL}^b$ and $\hbox {ITAL}^c$ do not satisfy CMR; in fact, $\hbox {ITAL}^b(\{1,2,3\},(1,5,9),10) = (1,3,6)$, while $(0,0,4) + \hbox {ITAL}^b((1,5,5),6) = (1,2,7)$ when $2 \succ 3$ and $\hbox {ITAL}^c(\{1,2,3\},(1,2,5),5) = (1,1,3)$, while $(0,0,2) + \hbox {ITAL}^c((1,2,3),3) = (0,1,4)$. $\hbox {DFL}(b)$ and $\hbox {DFL}(c)$ do not satisfy CMR because the solution of the sets in the two cases may not coincide; in fact $\hbox {DFL}(b)(\{1,2,3\},(1,5,9),10) = \{(1,2,7)\}$, while $(0,0,4) + \hbox {DFL}(b)(\{1,2,3\},(1,5,5),6) = \{(1,2,7),(1,3,6)\}$ and $\hbox {DFL}(c)(\{1,2,3\},(1,2,5),5) = \{(1,1,3)\}$, while $(0,0,2) + \hbox {DFL}(c)(\{1,2,3\},(1,2,3),3) = \{(0,1,4), (1,1,3)\}$.

• Self-duality (SD) Fix the priority ordering $3 \succ 2 \succ 1$. $\hbox {ITAL}^a,\hbox {ITAL}^b$ and $\hbox {ITAL}^c$ do not satisfy SD; in fact, $\hbox {ITAL}^a(\{1,2,3\},(1,3,5),5) = (0,1,4)$, while $(\hbox {ITAL}^a)^*(\{1,2,3\},(1,3,5),5) = (1,3,5) - \hbox {ITAL}^a(\{1,2,3\},(1,3,5),4) = (1,2,2)$, $\hbox {ITAL}^b(\{1,2,3\},(1,3,5),5) = (1,2,2)$, while $(\hbox {ITAL}^b)^*(\{1,2,3\},(1,3,5),5) = (1,3,5) - \hbox {ITAL}^b(\{1,2,3\},(1,3,5),4) = (0,1,4)$ and $\hbox {ITAL}^c(\{1,2,3\},(1,3,5),5) = (1,2,2)$, while $(\hbox {ITAL}^c)^*(\{1,2,3\},(1,3,5),5) = (1,3,5) - \hbox {ITAL}^b((1,3,5),4) = (0,2,3)$. In order to discuss SD for the solution sets $\hbox {DFL}(a,b,c)$, we ought to define what “duality” means in these cases. Not pretending exhaustiveness, we confine our attention to the case at hand, observing that:

  • 1. $\hbox {DFL}(a)(N,c,E)$ and $\hbox {DFL}(b)(N,c,C-E)$ have the same cardinality and the same occurs when interchanging a and b;

  • 2. $\hbox {DFL}(c)(N,c,E)$ and $\hbox {DFL}(c)(N,c,C-E)$ have the same cardinality;

  • 3. for every $x\in \hbox {DFL}(a)(N,c,E)$ there exists a unique $y\in \hbox {DFL}(b)(N,c,C-E)$ such that $x+y=c$ (the analogous happens interchanging a and b);

  • 4. for every $x\in \hbox {DFL}(c)(N,c,E)$ there exists a unique $y\in \hbox {DFL}(c)(N,c,C-E)$ such that $x+y=c$.

  These properties are a consequence of the geometry of the three vessel systems considered in Approach 2 of Sect. 2: in case c, the symmetry of the top and bottom vessels for each agent causes that the parts filled by the estate E are equal to the parts that remain empty when the estate is $C-E$; in cases a and b, the properties follow by the mirror symmetry top/bottom of one system with respect to the other. In analogy with the definition of duality for division rules, we may say that $\hbox {DFL}(a)$ and $\hbox {DFL}(b)$ are dual of each other, while $\hbox {DFL}(c)$ satisfies $\hbox {SD}$. In particular, it is enough to take an example where $\hbox {DFL}(a)\ne \hbox {DFL}(b)$, e.g., $\hbox {DFL}(a)(\{1,2\},(1,2),1)= \{(0,1)\}$ while $\hbox {DFL}(b)(\{1,2\},(1,2),1)= \{(0,1),(1,0)\}$, to show that neither $\hbox {DFL}(a)$ nor $\hbox {DFL}(b)$ satisfy SD.