On recursive solutions to simple allocation problems

Abstract

We propose and axiomatically analyze a class of rational solutions to simple allocation problems where a policy-maker allocates an endowment $E$ among $n$ agents described by a characteristic vector c. We propose a class of recursive rules which mimic a decision process where the policy-maker initially starts with a reference allocation of $E$ in mind and then uses the data of the problem to recursively adjust his previous allocation decisions. We show that recursive rules uniquely satisfy rationality, c-continuity, and other-c monotonicity. We also show that a well-known member of this class, the Equal Gains rule, uniquely satisfies rationality, c-continuity, and equal treatment of equals.

Appendix

The proofs presented below make explicit use of contraction independence which, by Remark 1, is equivalent to rationality.

The following lemmata are useful in the proof of Theorem 1. The first lemma shows under contraction independence and c-continuity that, if $c_{i}$ decreases below agent $i$’s current share, agent $i$’s updated share should be his new characteristic value.

Lemma 1

Assume that $F$ is contraction independent and c-continuous. Let $(c,E), (c^{\prime },E) \in \mathcal{C }$, and $i\in N$ be such that $c_{-i}=c_{-i}^{\prime }, \,c_{i}>c_{i}^{\prime },$ and $F_{i}(c,E) >c_{i}^{\prime }.$ Then $F_{i}(c^{\prime },E) =c_{i}^{\prime }.$

Proof

Suppose $F_{i}( c^{\prime },E) <c_{i}^{\prime }.$ By c-continuity, there is $c_{i}^{\prime \prime }\in \mathbb{R }_{+}$ such that $c_{i}^{\prime }<c_{i}^{\prime \prime }<c_{i}$ and $F_{i}(c_{i}^{\prime \prime },c_{-i},E) =F_{i}(c_{i}^{\prime \prime },c_{-i}^{\prime },E) =c_{i}^{\prime }.$ Then, $F(c_{i}^{\prime \prime },c_{-i}^{\prime },E) \leqq c^{\prime }\leqq (c_{i}^{\prime \prime },c_{-i}^{\prime }),$ by contraction independence, implies $F(c^{\prime },E) =F(c_{i}^{\prime \prime },c_{-i}^{\prime },E) ,$ a contradiction. $\square$

The following lemma states that if $c_{i}$ decreases, the share of agent $i$ cannot increase and the shares of other agents cannot decrease in response.

Lemma 2

Assume that $F$ is contraction independent, c-continuous, and other-c monotonic. Let $(c,E), (c^{\prime },E) \in \mathcal{C }$, and $i\in N$ be such that $c_{-i}=c_{-i}^{\prime }$ and $c_{i}>c_{i}^{\prime }.$ Then $F_{i}(c,E) \geqq F_{i}(c^{\prime },E)$ and for each $j\in N\setminus \{i\}, \,F_{j}(c^{\prime },E) \geqq F_{j}(c,E).$

Proof

If $F_{i}(c,E) \leqq c_{i}^{\prime },$ by contraction independence, we have $F(c^{\prime },E) =F(c,E).$ Thus, the result trivially holds. Alternatively assume $F_{i}(c,E) >c_{i}^{\prime }.$ Then by Lemma 1, $F_{i}(c^{\prime },E) =c_{i}^{\prime }<F_{i}(c,E)$. Thus, $\sum _{N\setminus \{i\}}F_{j}(c,E)<\sum _{N\setminus \{i\} }F_{j}(c^{\prime },E)$. Therefore, there is $k\in N\setminus \{i\}$ such that $F_{k}(c,E) <F_{k}(c^{\prime },E).$ By other-c monotonicity, this implies for each $j\in N\setminus \{i\},\,F_{j}(c,E) \leqq F_{j}(c^{\prime },E).$ $\square$

We next present the proofs of Theorems 1 and 2.

Proof of Theorem 1

$\left( \varvec{\Leftarrow }\right)$ Let $F^{g,r}$ be a recursive rule. Since $g^{n}$ is continuous in $c, \,F^{g,r}$ is c-continuous. Let $(c,E), (c^{\prime },E) \in \mathcal{C }$ and $i\in N$ be such that $c_{-i}=c_{-i}^{\prime }.$

Claim 1

If $F_{i}^{g,r}(c,E) \leqq c_{i}^{\prime }\leqq c_{i}$ or $F_{i}^{g,r}(c,E) <c_{i}\leqq c_{i}^{\prime },$ we have $F^{g,r}(c,E) =F^{g,r}(c^{\prime },E) .$ If $c_{i}=c_{i}^{\prime },$ the claim trivially holds. If $F_{i}^{g,r}(c,E) <c_{i}^{\prime }<c_{i}$ or $F_{i}^{g,r}(c,E) <c_{i}<c_{i}^{\prime },$ the claim follows from Property $3$ of $g$. Finally if $F_{i}^{g,r}(c,E) =c_{i}^{\prime }<c_{i},$ the claim follows from the previous case and c-continuity of $F^{g,r}.$

Claim 2

$F^{g,r}$ is rational. Applying Claim 1 to each $i\in N$ shows that it is contraction independent. Then, by Remark 1, $F$ is rational.

Claim 3

$F^{g,r}$ is other-c monotonic. Assume $c_{i}\not =c_{i}^{\prime }.$ Then by Property 4, either [for each $j\in N\setminus \{i\},\,F_{j}^{g,r}(c,E) \geqq F_{j}^{g,r}(c^{\prime },E)$] or [for each $j\in N\setminus \{i\},\,F_{j}^{g,r}(c,E)\leqq F_{j}^{g,r}(c^{\prime },E)$].

$\left( \varvec{\Rightarrow }\right)$ Let $F$ satisfy the given properties.

Step 1: defining $g$ and $r$. For each $E\in \mathbb{R }_{+}$, let $r(E) =F(E_{N},E).$ For each $(c,E)\in \mathcal{C }$ and $x\in \mathbb{R }_{+}^{N}$ such that $\sum _{N}x_{i}=E,$ let $g(x,c,E) =F(( c_{M(x,c) },E_{N\setminus M(x,c) }),E)$ where $M(x,c)=\left\{ i\in N\mid c_{i}\leqq x_{i}\right\}$.

For the following steps, we will introduce some notation. Let $x^{0}=r(E)$ and for $t\in \{1,\ldots ,n\},$ let $x^{t}=g(x^{t-1},c,E) =g^{t}(r(E),c,E).$ Let $M^{-1}=\emptyset$ and for each $t\in \{0,\ldots ,n\}, \$let $M^{t}=M(x^{t},c).$

Step 2: if $t\in \left\{ 0,\ldots ,n\right\}$ and $i\in M^{t-1},$ then  $x_{i}^{t}=c_{i}$ and so $i\in M^{t}.$ The proof is by induction. For $t=0, \,M^{-1}=\emptyset$ implies the desired conclusion. Now assume $M^{-1}\subseteq \cdots \subseteq M^{t-1}.$ Let $i\in M^{t-1}.$ Then $x_{i}^{t-1}\geqq c_{i}.$ Let $K=M^{t-2}\cup \left\{ i\right\}$ and note that $K\subseteq M^{t-1}.$ If $x_{i}^{t-1}=c_{i},$ by contraction independence, $x^{t-1}=F\left( \left( c_{K},E_{N\setminus K}\right) , E\right) .$ Thus, $F_{i}\left( \left( c_{K},E_{N\setminus K}\right) , E\right) =c_{i}.$ Alternatively, if $x_{i}^{t-1}>c_{i},$ by Lemma 1, $F_{i}\left( \left( c_{K},E_{N\setminus K}\right) ,E\right) =c_{i}.$ If $K=M^{t-1},$ by construction of $g$ in Step 1, we have $i\in M^{t}$ and $x_{i}^{t}=c_{i}.$ Otherwise, $F_{i}\left( \left( c_{K},E_{N\setminus K}\right) , E\right) =c_{i}$ and Lemma 2 imply $x_{i}^{t}\geqq c_{i}.$ Thus $i\in M^{t}.$ Since $i\in M^{t-1}$, by definition, $x_{i}^{t}\leqq c_{i}.$ Thus overall, $x_{i}^{t}=c_{i}.$

Step 3: $x^{n}\leqq c.$ First assume $M^{t-1}=M^{t}$ for some $t\in \left\{ 0,\ldots ,n-1\right\} .$ Then by definition, $x^{t}=x^{t+1}$ and thus, $M^{t}=M^{t+1}.$ Iterating, $x^{t}=x^{n}.$ Also, by Step $2$, for each $i\in M^{t}, \,x_{i}^{t}=c_{i}.$ Thus, $x^{t}=x^{n}\leqq c.$ Alternatively, assume $M^{t-1}\not =M^{t}$ for each $t\in \left\{ 0,\ldots ,n-1\right\} .$ By Step 2, $M^{n-1}=N.$ Thus $x^{n}=F(c,E) \leqq c.$

Step 4: $F=F^{g,r}.$ Let $(c,E) \in \mathcal C .$ By Remark 1, assume $c\leqq E_{N}$.¹⁴ Note that $F^{g,r}(c,E) =x^{n}=F\left( \left( c_{M^{n-1}},E_{N\setminus M^{n-1}}\right) , E\right) .$ By Step 3, $x^{n}\leqq c\leqq \left( c_{M^{n-1}},E_{N\setminus M^{n-1}}\right)$. Then, by contraction independence, $x^{n}=F(c,E).$

Step 5: $g$ is a recursive adjustment function. Since $F$ is c-continuous, $g^{n}$ is continuous in $c.$ Also, Step $2$ above proves Property $1$. Now let $i\in N$ and $t\in \left\{ 1,\ldots ,n\right\} .$

For Property 2, assume $x_{i}^{t-1}<c_{i}$. Then $i\not \in M^{t-1}$ implies $i\not \in M^{t-2}.$ If $M^{t-2}=M^{t-1},$ by definition, $x^{t}=x^{t-1}$ and thus, $x_{i}^{t}\geqq x_{i}^{t-1}.$ Otherwise, by Lemma 2, $x_{i}^{t}\geqq x_{i}^{t-1}.$

For Property 3, assume $x_{i}^{t-1}<\widetilde{c}_{i}<c_{i}.$ Let $\widetilde{c}=\left( \widetilde{c}_{i},c_{-i}\right) .$ Then $i\not \in M(x^{t-1},c) =M(x^{t-1},\widetilde{c}).$ Thus, by definition of $g$, we have $g(x^{t-1},c,E) =g(x^{t-1}, \widetilde{c},E)$.

For Property 4, assume $c_{i}<\widetilde{c}_{i}.$ Let $\widetilde{c} =(\widetilde{c}_{i},c_{-i}), \,x=g^{n}(r(E), c,E)$ and $\widetilde{x}=g^{n}(r(E), \widetilde{c}, E).$ Then, $x=F(c,E)$ and $\widetilde{x}=F(\widetilde{c},E).$ By Lemma 2, $x_{i}\leqq \widetilde{x}_{i}$ and for each $j\in N\setminus \{i\}, \,x_{j}\geqq \widetilde{x}_{j}.$ $\square$

Proof of Theorem 2

It is straightforward to show that Equal Gains satisfies the given properties. Conversely, let $F$ be a rule that satisfies rationality (thus by Remark 1, contraction independence), c-continuity, and equal treatment of equals. We next show $F=\text{ EG}.$ Let $(c,E) \in \mathcal{C }$. By contraction independence, assume $c\leqq E_{N}$ (see Footnote 14). Without loss of generality, assume $c_{1}\leqq \cdots \leqq c_{n}$. Let $c^{0}=E_{N}, \, c^{n}=c,$ and for each $k\in \{1,\ldots ,n-1\},$ let $c^{k}=(c_{\{ 1,\ldots ,k\}},E_{\{k+1,\ldots ,n\}}).$

We inductively show that for each $k\in \{0,\ldots ,n\},$ we have $F(c^{k},E) =\text{ EG}(c^{k},E).$ For $k=n,$ this will imply the desired conclusion. Initially, let $k=0.$ By equal treatment of equals, $F(c^{0},E) =\text{ EG}(c^{0},E).$Now let $k\in \{1,\ldots ,n\}$ and assume that the statement holds for each $l<k.$

Case 1

There is $l<k$ such that $F( c^{k},E)=F(c^{l},E).$ Then, by our assumption, $F(c^{l},E) =\text{ EG}(c^{l},E).$ Thus, $\text{ EG}(c^{l},E) \leqq c^{k}\leqq c^{l}.$ This, by contraction independence, implies $\text{ EG} (c^{l},E) =\text{ EG} (c^{k},E).$ Combining the equalities, we then have $F(c^{k},E) =\text{ EG}(c^{k},E).$

Case 2

For each $l<k, \,F(c^{k},E) \not =F(c^{l},E).$ Thus, $F(c^{k},E)$ is first obtained at $c^{k}$.

We first show that $F_{k}(c^{k},E) =c_{k}.$ For this, note that $F(c^{k},E)\not =F(c^{k-1},E)$ implies, by contraction independence, $F_{k}(c^{k-1},E) >c_{k}.$ Thus, by Lemma 1, $F_{k}(c^{k},E) =c_{k}.$

We next show that for each $l<k, \,F_{l}(c^{k},E) =c_{l}.$ Let $\overline{c}=c^{k}+(c_{k}-c_{l}) e_{l}$ and note that $\overline{c}\geqq c^{k}.$ First assume $F(\overline{c},E) \not =F(c^{k},E).$ Then $c_{l}<c_{k}.$ Thus, by contraction independence, $F_{l}(\overline{c},E) >c_{l}.$ This, by Lemma 1, implies $F_{l}(c^{k},E) =c_{l}$. Next, assume $F(\overline{c},E) =F(c^{k},E).$ By equal treatment of equals, $F_{l}(\overline{c},E) =F_{k}(\overline{c},E).$ Then, $F_{l}(c^{k},E) =F_{k}(c^{k},E).$ This implies $F_{l}(c^{k},E) =c_{k}\geqq c_{l}$. Thus, $F_{l}(c^{k},E) =c_{l}$.

Overall, for each $l\in \left\{ 1,\ldots ,k\right\} , \,F_{l}(c^{k},E) =c_{l}.$ This, by equal treatment of equals, implies for each $i\in \left\{ k+1,\ldots ,n\right\} ,\,F_{i}(c^{k},E) = \frac{E-\sum _{j=1}^{k}c_{j}}{n-k}.$ Applying the same arguments to Equal Gains shows that it picks the same allocation. Thus, $F(c^{k},E) =\text{ EG}(c^{k},E).$ $\square$