Bankruptcy problems with non-integer claims: definition and characterizations of the ICEA Solution

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.

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

$$\begin{aligned} \Delta ^*=E^*-\sum _{j\in N}[\mathrm{CEL}_j(N,[c],E^*)] \end{aligned}$$

(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

$$\begin{aligned} \mathrm{ICEL}_j(N,c,E^*)=\mathrm{ICEL}_j(N,[c],E^*)=[\mathrm{CEL}_j(N,[c],E^*)]+d^*_j. \end{aligned}$$

(10.1)

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),

$$\begin{aligned}{}[\mathrm{CEL}_j(N,[c],E^*)]=\left\{ \begin{array}{l@{\quad }l} 0 &{} \mathrm{if}\quad j\not \in P\\ {[c_j]-[\beta ]-1} &{} \mathrm{if}\quad j\in P. \end{array}\right. \end{aligned}$$

Since $\beta \not \in \mathbb N$, it follows that

$$\begin{aligned} \mathrm{ICEL}_j(N,c,E^*)=\left\{ \begin{array}{l@{\quad }l} 0 &{} \mathrm{if}\quad [c_j]<\beta \\ \mathrm{either}\ {[c_j]-[\beta ]-1}\ \mathrm{or}\ [c_j]-[\beta ]&{} \mathrm{if}\quad [c_j]>\beta ; \end{array}\right. \end{aligned}$$

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

$$\begin{aligned} \begin{array}{rl} (a) &{} \min \, z\ge \min \, \mathrm{ICEA}(N,c,E),\\ (b) &{} \max \, z\le \max \, \mathrm{ICEA}(N,c,E).\\ \end{array} \end{aligned}$$

(10.2)

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

$$\begin{aligned} \begin{array}{rl} (i) &{} m-1=s_0=\min \, z,\\ (ii) &{} {{{ mg\;}}}(z)=1, \end{array} \end{aligned}$$

(10.3)

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

$$\begin{aligned} E=\sum _{j\in N}\mathrm{ICEA}_j(N,c,E)\ge nm\ge \sum _{k\in N}z_k=E, \end{aligned}$$

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

$$\begin{aligned} s_0+1=m=\max \, z>\max \, \mathrm{ICEA}(N,c,E) \end{aligned}$$

(10.4)

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

$$\begin{aligned} F=\cup \{B_s:~s\le s_0-1\},\qquad G=\{j\in N:~ \mathrm{ICEA}_j(N,c,E)\le [\beta ]-1\}\!: \end{aligned}$$

we claim that $G\subseteq F$. In fact, by Lemma 4.1 again, if $j\in G$, then $j\not \in P$, hence

$$\begin{aligned} z_j\le [c_j]=\mathrm{ICEA}_j(N,c,E)\le [\beta ]-1\le M-1\le s_0-1, \end{aligned}$$

therefore, $j\in F$, whence the inclusion. In particular,

$$\begin{aligned} \begin{array}{rl} (i) &{} \mathrm{for~ all}\ j\in G, \ z_j=[c_j]=\mathrm{ICEA}_j(N,c,E),\\ (ii) &{} \mathrm{for~ all}\ j\in F\backslash G, \ z_j=[c_j]\ge \mathrm{ICEA}_j(N,c,E),\\ (iii) &{} \mathrm{for~ all}\ j\in N\backslash F, \ z_j\ge s_0\ge M\ge \mathrm{ICEA}_j(N,c,E),\\ (iv) &{} \mathrm{there~ exists}\ j\in N\backslash F, \ z_j> s_0\ge M\ge \mathrm{ICEA}_j(N,c,E)\\ \end{array} \end{aligned}$$

(10.5)

(the strict inequality in (10.5) $\;(iv)$ holds because $\max \, z=m=s_0+1$). Thus,

$$\begin{aligned} \begin{array}{rl} E &{}= \displaystyle {\sum _{j\in G}[c_j]+\sum _{j\in F\backslash G}[c_j]+\sum _{j\in N\backslash F}z_j>\sum _{j\in G}[c_j]+\sum _{j\in F\backslash G}[c_j]+\sum _{j\in N\backslash F}M}\\ &{}\ge \displaystyle {\sum _{j\in N}\mathrm{ICEA}_j(N,c,E)=E,}\\ \end{array} \end{aligned}$$

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

$$\begin{aligned} F=\cup \{B_s:~s\le s_0-1\},\quad F^T=\cup \{B_s^T:~s\le s_0^T-1\}\!: \end{aligned}$$

(10.6)

by definition of $s_0$ and of $s_0^T$,

$$\begin{aligned} \begin{array}{rl} (i)&{} j\in F\ \mathrm{if~and~only~if}\ z_j=[c_j]<s_0,\\ (ii)&{} j\in F^T\ \mathrm{if~and~only~if}\ z_j^T=[c_j]<s_0^T.\\ \end{array} \end{aligned}$$

(10.7)

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

$$\begin{aligned} \begin{array}{rl} E^T&{}= \displaystyle {\sum _{j\in T}z_j=\sum _{j\in T\cap F}[c_j]+s_0\, card\,(T\backslash F)+ card\,(T\cap B_{s_0+1})}\\ &{}= \displaystyle {\sum _{j\in T}z_j^T=\sum _{j\in F^T}[c_j]+s_0^T\, card\,(T\backslash F^T)+ card\,(B_{s_0^T+1}^T)\!:}\\ \end{array} \end{aligned}$$

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.

• 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.

• 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

$$\begin{aligned} s_L=\min \,\{s:\ \exists j\in L\cap B_s\, :\ z_j<[c_j]\} \end{aligned}$$

is well defined, it satisfies $s_L\ge s_0$ and there exists $i_L\in L$ such that

$$\begin{aligned} \begin{array}{ll} &{} z_{i_L}=s_L<[c_{i_L}];\ \mathrm{moreover,}\\ \mathrm{if\ }j\in N\ \mathrm{is\ such\ that}\ z_j\le s_L,\ \mathrm{then\ }&{} z_j<[c_{i_L}];\\ \mathrm{if\ }j\in N\ \mathrm{is\ such\ that}\ z_j> s_L,\ \mathrm{then\ }&{} z_j=m=s_L+1< [c_{i_L}]+1\le [c_{i_L}].\\ \end{array} \end{aligned}$$

(10.17)

Since $z_j\le [c_j]$ for every $j\in N$, it follows that

$$\begin{aligned} E = \sum _{j:\ z_j\le s_L}z_j+\sum _{j:\ z_j> s_L}z_j < \sum _{j\in N}\min \,\{[c_j],[c_{i_L}]\}:\end{aligned}$$

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

$$\begin{aligned} s_0\ge m-1. \end{aligned}$$

(10.18)

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)

$$\begin{aligned} 1\ge {{{ rmg\;}}}(z)\ge z_i-z_k=m-s_0\, \end{aligned}$$

whence (10.18) and the proof is complete.