Abstract
In Fragnelli et al. (TOP 22:892–933, 2014), 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 question of existence and uniqueness of a solution and proposed the “box method” that is strongly oriented towards the constrained equal losses solution; uniqueness may be guaranteed by introducing a ranking on the claimants. Here, we introduce a solution oriented towards the constrained equal awards solution and give three characterizations and a simple method for determining the solution.
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The authors are grateful to two anonymous reviewers for their remarks that allowed to improve the paper.
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Appendix: proofs
Appendix: proofs
This section is devoted to the proofs of the technical statements presented in the previous sections.
Proof of Lemma 4.1. If \(\beta \in \mathbb N\), then \(\mathrm{CEA}(N,[c],E)\) has integer components, therefore, by Proposition 3.2 \(\mathrm{ICEA}_j(N,c,E)=\mathrm{CEA}_j(N,[c],E)=\min \;\{[c_j],\beta \}\) for all \(j\in N\) and this case is complete. Suppose \(\beta \not \in \mathbb N\): then the quantity
(namely, the remainder that has to be assigned to the agents after truncation of \(\mathrm{CEL}(N,[c],E^*)\)) is strictly positive. It was shown in Fragnelli et al. (2014) that for every \(j\in N\) there exists an integer number \(d^*_j\le 1\) such that
In particular, if \(d^*_j=1\) then \(j\in P\); moreover, there exist \(i,j\in P\) such that \(d^*_i=0 \) and \(d^*_j=1\) (because \(0<\Delta ^*<card\,P\)): in this case \(j\displaystyle {\mathop \succ \nolimits _\mathrm{dec}}i\). Recalling (2.3),
Since \(\beta \not \in \mathbb N\), it follows that
moreover, there exist \(i,j\in P\) such that \(\mathrm{ICEL}_i(N,c,E^*)=[c_i]-[\beta ]-1\) and \(\mathrm{ICEL}_j(N,c,E^*)=[c_j]-[\beta ]\); finally, whenever \(i,\;j\) are such that this occurs, then \(j\displaystyle {\mathop \succ \nolimits _\mathrm{dec}}i\). To complete the proof, it is enough to read these properties in terms of \(ICEA\) and to recall Proposition 3.2 and (4.1).
Proof of Lemma 6.4. Let \(\phi \) be an integer division rule satisfying the \(TEST\) algorithm and let \((N,c,E)\) be any bankruptcy problem satisfying Assumption 5.1, then set \(z=\phi (N,c,E)\). If \(E=0\) or \(E=[[C]]\) then the integer solution to \((N,c,E)\) is unique and nothing has to be proved, therefore, we assume (6.1). Since the minimum value of the quantity \({{{ mg\;}}}\) defined in (4.3) is \({{{ mg\;}}}(\mathrm{ICEA}(N,c,E))\), due to Proposition 4.2, it is sufficient to prove that \({{{ mg\;}}}(z)\le {{{ mg\;}}}(\mathrm{ICEA}(N,c,E))\). This is trivial if \(s_0=m\), for in this case \(z_j=[c_j]\) for every \(s<m\) and every \(j\in B_s\,\): this means that the maximum gap of the awards is the lowest possible.
The argument is more complicated if \(s_0=m-1\). In this case, we claim that
Inequality (10.2)\(\;(a)\) is immediate if \(\min \, z=[c_k]\) for some \(k\in N\) (in this case, \(\min \, \mathrm{ICEA}(N,c,E)\) cannot be strictly larger than \([c_k]\)). Thus, let us suppose that \(\min \, z<[c_k]\) for every \(k\in N\), whence
because \(m=\max \, z\). By contradiction, if the inequality \((a)\) in (10.2) does not hold, then (10.3)\(\;(i)\) implies that, for every \(j\in N\), \(\mathrm{ICEA}_j(N,c,E)\ge \min \, z+1=s_0+1=m=\max \, z\ge z_k\), for every \(k\in N\). By (2.2), it follows that
therefore, the two inequalities must be equalities, hence \(z_k=m\), for every \(k\in N\), which contradicts (10.3)\(\;(ii)\). The conclusion is the validity of (10.2)\(\;(a)\) in all cases.
Let us prove (10.2)\(\;(b)\). By contradiction, let us suppose that
and denote the last quantity by \(M\): by Lemma 4.1, either \(M=[\beta ]\) or \(M=[\beta ]+1\), moreover \(s_0\ge M\), by (10.4). Let
we claim that \(G\subseteq F\). In fact, by Lemma 4.1 again, if \(j\in G\), then \(j\not \in P\), hence
therefore, \(j\in F\), whence the inclusion. In particular,
(the strict inequality in (10.5) \(\;(iv)\) holds because \(\max \, z=m=s_0+1\)). Thus,
a contradiction, therefore (10.2)\(\;(b)\) holds and the proof of (10.2) is complete.
Clearly, the inequality \({{{ mg\;}}}(z)\le {{{ mg\;}}}(\mathrm{ICEA}(N,c,E))\) is a direct consequence of (10.2), hence the proof is complete.
Proof of Lemma 6.5. Let \(\phi \), \((N,c,E)\), \(z\), \(T\), \((T,c^T,E^T)\), \(z^T\), \(B_s^T\), \(m^T\), \(s_0^T\) be as in the statement and let us introduce the following sets
by definition of \(s_0\) and of \(s_0^T\),
Let us discuss the consequences of a comparison between \(s_0\) and \(s_0^T\).
-
(a)
Suppose \(s_0\le s_0^T\). We claim that
$$\begin{aligned} T\cap F\subseteq F^T. \end{aligned}$$(10.8)Indeed, if \(j\) is any element of \(T\cap F\), then by (10.7)\(\;(i)\) \(z_j=[c_j]<s_0\le s_0^T\), whence \(z_j^T=[c_j]\) and \(j\in F^T\) by (10.7)\(\;(ii)\), whence (10.8). The following are consequences of (10.8):
$$\begin{aligned} \begin{array}{rl} (i) &{} \mathrm{for~every}\ j\in T\cap F,\ z_j^T=z_j=[c_j],\\ (ii)&{} \mathrm{for~every}\ j\in F^T \backslash T\cap F,\ z_j^T=[c_j]\ge z_j,\\ (iii)&{} \mathrm{for~every}\ j\in T \backslash F^T ,\ z_j^T\ge s_0^T.\\ \end{array} \end{aligned}$$(10.9)In particular, if \(s_0< s_0^T\) then in (10.9)\(\;(iii)\) \(z_j^T\ge s_0+1\ge z_j\), therefore, (10.9) implies
$$\begin{aligned} z_j^T\ge z_j,\quad \mathrm{for\ every}\quad j\in T. \end{aligned}$$(10.10)This implies
$$\begin{aligned} E^T=\sum _{j\in T}z_j^T\ge \sum _{j\in T}z_j=E^T \end{aligned}$$(10.11)and (6.5).
-
(b)
Suppose \(s_0^T\le s_0\). In analogy with case (a), we may obtain
$$\begin{aligned} T\cap F= F^T\, \end{aligned}$$(10.12)and
$$\begin{aligned} \begin{array}{rl} (i) &{} \mathrm{for~every}\ j\in F^T,\ z_j=z_j^T=[c_j],\\ (ii)&{} \mathrm{for~every}\ j\in T\cap F \backslash F^T,\ z_j=[c_j]\ge z_j^T,\\ (iii)&{} \mathrm{for~every}\ j\in T \backslash F ,\ z_j\ge s_0.\\ \end{array} \end{aligned}$$(10.13)In particular, if \(s_0^T< s_0\) then in (10.13)\(\;(iii)\) \(z_j\ge s_0^T+1\ge z_j^T\), therefore, (10.13) implies
$$\begin{aligned} z_j\ge z_j^T,\quad \mathrm{for\ every}\quad j\in T, \end{aligned}$$(10.14)whence (6.5).
It follows from (a) and (b) that (6.5) holds if \(s_0^T\ne s_0\).
Next, let us suppose \(s_0^T= s_0\). If \(T\cap B_{s_0}=\emptyset \) then for every \(j\in T\backslash F\) \(z_j=s_0+1\), hence \(z_j\ge z_j^T\) by (10.13)\(\;(iii)\), whence (10.14) and (6.5). If \(T\cap B_{s_0+1}=\emptyset \) then for every \(j\in T \backslash F^T\) \(z_j\le s_0\), hence \(z_j^T\ge z_j\) by (10.9)\(\;(iii)\), whence (10.10) and (6.5). It only remains to prove (3).
Suppose \({s_0}={s_0^T}\), \(T\cap B_{s_0}\ne \emptyset \) and \(T\cap B_{s_0+1}\ne \emptyset \). By (10.7), \(z\) (respectively, \(z^T\)) takes both values \(s_0\) and \(s_0+1\) (respectively, \(s_0^T\) and \(s_0^T+1\)) in such a way that
by (10.8) and (10.12) it follows that (6.6) holds.
Proof of Lemma 7.3.
-
1.
It is a mere reformulation of CFC (see Definition 7.1);
-
2.
It follows immediately by definition of \(L\) and implies \(3.\) straightforwardly.
-
4.
By definition of \(s_0\), there exists \(k\in B_{s_0}\) such that \(z_k<[c_k]\), whence \(k\in N\backslash L\) by \((1)\).
-
5.
Let \(s^*\in \mathbb N\) be such that \(s^*<s_0\) and \(B_{s^*}\ne \emptyset \), then consider \(i^*\in B_{s^*}\). By definition of \(s_0\), we have that \(z_{i^*}=s^*=[c_{i^*}]\) and
$$\begin{aligned} \begin{array}{rl} \mathrm{if\ }j\in N\ \mathrm{is\ such\ that}\ z_j\le s^*, &{} \mathrm{then\ }z_j=[c_j]\le [c_{i^*}];\\ \mathrm{if\ }j\in N\ \mathrm{is\ such\ that}\ z_j> s^*,&{} \mathrm{then\ }[c_j]\ge z_j>[c_{i^*}].\\ \end{array} \end{aligned}$$It follows that
$$\begin{aligned} \begin{array}{rl} E&{}= \displaystyle {\sum _{j:\ z_j\le s^*}z_j+\sum _{j:\ z_j> s^*}z_j\ge \sum _{j:\ z_j\le s^*}[c_j]+\sum _{j:\ z_j> s^*}[c_{i^*}]} = \displaystyle {\sum _{j\in N}\min \,\{[c_j],[c_{i^*}]\}},\\ \end{array} \end{aligned}$$whence \(i^*\in L\) and inclusion \((a)\). Inclusion \((b)\) holds by difference.
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6.
Let us assume that \(\phi \) satisfies MRMG and, by contradiction, let \(j\in L\) be such that \(j\not \in \cup \{B_{s},~s\le s_0\}\). It follows that \(z_j=[c_j]>s_0\). If \(j_1\in L\) is such that \([c_{j_1}]\ge [c_i]\) for all \(i\in L\), then by \((1)\) \(z_{j_1}=[c_{j_1}]\ge [c_j]>s_0\). Let us denote
$$\begin{aligned} s_1=z_{j_1}=[c_{j_1}],\quad \mathrm{so\ that}\quad s_1> s_0\, \end{aligned}$$(10.15)and let us distinguish two cases.
-
(a)
For every \(i\in N\), \(z_i\le s_1\). Let \(k\) be an element of \((N\backslash L)\cap B_{s_0}\) (it exists, by 4): then by definition of \(j_1\)
$$\begin{aligned} \begin{array}{rl} E&{}= \displaystyle {\sum _{i\in L}z_i+\sum _{i\in N\backslash L}z_i\le \sum _{i\in L}[c_i]+\sum _{i\in N\backslash L,\;i\ne k}s_1+s_0}\\ &{}\displaystyle {\mathop =^{3}} \displaystyle {\sum _{i\in L}\min \, \{[c_{j_1}],[c_i]\}+\sum _{i\in N\backslash L,\;i\ne k}\min \, \{[c_{j_1}],[c_i]\}}\\ &{}\quad +\,s_0\displaystyle {\mathop <^{(10.15)}} \displaystyle {\sum _{i\in N}\min \, \{[c_{j_1}],[c_i]\}\le E} \end{array} \end{aligned}$$(last inequality holds because \(j_1\in L\)). This is a contradiction, therefore, 6 holds in this case: notice that the additional assumption on \(\phi \) has not been used here.
-
(b)
There exists \(i\in N\), \(z_i> s_1\). By \((1)\), by definition of \(j_1\) and by (10.15), it cannot be \(i\in L\), hence \(i\in N\backslash L\); if \(k\) is an element of \((N\backslash L)\cap B_{s_0}\) (which exists, by \(4.\)), then
$$\begin{aligned} {{{ rmg\;}}}(z)\ge z_i-z_k\ge s_1+1-s_0\displaystyle {\mathop \ge ^{(10.15)}}2, \end{aligned}$$violating Definition 7.2 (which we assumed to apply), hence a contradiction again and \(6.\) holds also in this case.
-
(a)
-
7.
Assuming (5.1), let us suppose \(\phi =\mathrm{ICEA}\) and argue by contradiction as in the proof of \(6.\) By Proposition 6.3, TEST holds, hence \(s_0\le m\) and case (\(b\)) in the proof of \(6.\) cannot occur, therefore, the contradiction is reached independently of the assumption made in \(6.\) Finally, if the parameter \(\beta \) characterizing \(\mathrm{CEL}(N,[c],E^*)\) does not belong to \(\mathbb N\), then by Lemma 4.1 \(m=[\beta ]+1\); moreover, as shown in the proof of Proposition 6.3, the second alternative in (6.3) does not occur, hence \(s_0=\beta \). Now, let \(i\in (N\backslash L)\cap B_{s_0}\) be such that, by contradiction, \(z_i=[c_i]\). Then,
$$\begin{aligned} \begin{array}{rl} \mathrm{for\ every}\ j\in L,&{} [c_i]>[c_j]\ \mathrm{by}\ 3;\\ \mathrm{for\ every}\ j\in B_{s_0},&{} [c_j]\ge z_j=s_0=[c_{i}];\\ \mathrm{for\ every}\ j\in B_{s_0+1},&{} [c_j]\ge z_j=s_0+1=[c_{i}]+1.\\ \end{array} \end{aligned}$$(10.16)Therefore,
$$\begin{aligned} \begin{array}{rl} E&{}= \displaystyle {\sum _{j\in L}z_j+\sum _{j\in N\backslash L}z_j\mathop =^{(1)} \sum _{j\in L}[c_j]+\sum _{j\in (N\backslash L)\cap B_{s_0}}s_0+\sum _{j\in (N\backslash L)\cap B_{s_0+1}}(s_0+1)}\\ &{}\displaystyle {\mathop \ge ^{(10.16)}} \displaystyle {\sum _{j\in L}\min \, \{[c_i],[c_{j}]\}+\sum _{j\in N\backslash L}\min \, \{[c_i],[c_{j}]\}}= \displaystyle {\sum _{j\in N}\min \, \{[c_i],[c_{j}]\},} \end{array} \end{aligned}$$whence \(i\in L\), a contradiction. This completes the proof of 7., hence of the lemma.
Proof of Lemma 7.5. To show that CFC holds, let \((N,c,E)\) be a bankruptcy problem verifying Assumption 5.1, then set \(z=\phi (N,c,E)\). By contradiction, suppose that there exists some \(i\in L\) such that \(z_i<[c_i]\). Then, the quantity
is well defined, it satisfies \(s_L\ge s_0\) and there exists \(i_L\in L\) such that
Since \(z_j\le [c_j]\) for every \(j\in N\), it follows that
the inequality is strict because \(z_j<\min \,\{[c_j],[c_{i_L}]\}\) at least for \(j=i_L\). Thus, \(i_L\not \in L\) by Definition 7.1, a contradiction: this proves CFC.
To show that \(\phi \) satisfies MRMG, let \((N,c,E)\) and \(z\) be as before, then suppose (6.1) (otherwise nothing has to be proved). Consider any \(j\in N\backslash L\): by Lemma 7.3 \(\;5\), \(z_j\ge s_0\), hence either \(z_j=s_0\) or \(z_j=s_0+1\) (by \(TEST(B)\), see Proposition 6.3), whence \({{{ rmg\;}}}(z)\le 1\).
Proof of Lemma 7.6. Let \(\phi \) be an integer division rule satisfying the assumptions and let \((N,c,E)\) be any bankruptcy problem verifying Assumption 5.1 and (6.1) (otherwise nothing has to be proved), then set \(z=\phi (N,c,E)\). In particular, the quantities \(m\) and \(s_0\) of \(\mathrm{TEST}(A)\) may be defined: our aim is to show that the alternative \(\mathrm{TEST}(B)\) holds true, that is
To this end, consider any \(i\in B_m\) and any \(k\in (N\backslash L)\cap B_{s_0}\) (which exists, by Lemma 7.3 \(\;4.\)): then by (7.2)
whence (10.18) and the proof is complete.
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Fragnelli, V., Gagliardo, S. & Gastaldi, F. Bankruptcy problems with non-integer claims: definition and characterizations of the ICEA Solution. TOP 24, 88–130 (2016). https://doi.org/10.1007/s11750-015-0376-x
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DOI: https://doi.org/10.1007/s11750-015-0376-x