Let $J$ be an integer and let $I$ be an integer multiple of $J$. Let ${\cal I}= \lbrace 1,2,\ldots, I\rbrace$ and ${\cal J}= \lbrace 1,2,\ldots, J\rbrace$. The set $X_{I,J}$ of all $\frac{I}{J}$-subsets of $\cal I$ contains exactly $N=\binom{I}{\frac{I}{J}}$ elements. Now, consider the set $Y_{I,J}$ of all permutations of $X_{I,J}$, viewed as columns enumerating the elements of $X_{I,J}$. There are $N!$ permutations in $Y_{I,J}$. Next, let $V$ be a $N\times J$ matrix whose entries are sets in $X_{I,J}$, and all of whose columns are in $Y_{I,J}$. Let $\cal V$ denote the set of all such matrices. The cardinality of $\cal V$ is $N!^{J}$.

The goal is to prove that for any $V \in \mathcal{V}$, and for any $m \in \mathbb{R}_+^J$ such that $m$ is injective ($m_a \ne m_b$ if $a \ne b$) , there exists an $s=(s_1,\ldots,s_J)$ and a $p=(p_1,\ldots,p_I)$ such that \begin{align} \forall a \ne b \in \mathcal{J}, & \:\:V_{s_a,a} \cap V_{s_b,b} = \varnothing \\ \forall j \in \mathcal{J}, \forall k \in \{1,\ldots,s_j-1\}, &\sum_{i \in V_{k,j}} p_i > m_j \\ \forall j \in \mathcal{J}, &\sum_{i \in V_{s_j,j}} p_i \leq m_j \end{align}

Motivation: If I can prove the above, I can argue existence of equilibria in a pretty unique context with indivisible objects, fiat money and quotas.

This is something I believe to be true. I even have a rough Matlab algorithm based on Walrasian Tatonnement that finds $s$ and $m$ given $V$, and I've found $s$ and $m$ for hundreds of thousands of randomly-generated $V$ without failing to find one, but it often requires a lot of manual tinkering (messing with increments and such) and the algorithm doesn't have nice properties that lend it to an existence proof.

Anwyay, I'm stumped as to how to prove it. Even if you cannot prove it, any suggestions as to what I might try would be greatly appreciated.

Example and Interpretation: Let $I=6$ and $J=3$. For expositional purposes, let's put $\mathcal{I}$ in letters, i.e. $\mathcal{I}=\{A,B,C,D,E,F\}$. Then $V_{example} =$

\begin{array}{ccc} \{B,C\} & \{A,B\} & \{C,D\} \\ \mathbf{\{B,D\}} & \{B,C\} & \{B,C\} \\ \{A,C\} & \{B,D\} & \{A,C\} \\ \{C,D\} & \{B,E\} & \{C,E\} \\ \{A,B\} & \{B,F\} & \{C,F\} \\ \{C,E\} & \{C,D\} & \{A,B\} \\ \{C,F\} & \mathbf{\{C,E\}} & \{A,D\} \\ \{B,E\} & \{C,F\} & \{E,F\} \\ \{B,F\} & \{A,C\} & \mathbf{\{A,F\}} \\ \{D,E\} & \{A,D\} & \{B,D\} \\ \{D,F\} & \{A,E\} & \{B,E\} \\ \{E,F\} & \{A,F\} & \{B,F\} \\ \{A,D\} & \{D,E\} & \{D,E\} \\ \{A,E\} & \{D,F\} & \{D,F\} \\ \{A,F\} & \{E,F\} & \{A,E\} \\ \end{array} If $m=(3,2.9,2.8)$. Then $s=(2,7,9)$ -- the cells are bolded above -- and $p=(p_A,\ldots,p_F) = (1.375,1.55,1.5,1.45,1.4,1.425)$ meet the above conditions.

You can interpret the conditions as follows. Condition 1 states that no letter may be used in more than one column of $V$. Condition 2 states that all the cells above the bolded ones contain elements whose prices in terms of $p$ sum to greater than $m_j$. Condition 3 states that each bolded cell is such that prices in terms of $p$ sum to less than or equal to $m_j$.

Note that serial dictatorships will often work. By that, I mean that $s_1=1$ and that each subsequent agent $j$ gets the best available bundle that doesn't conflict with any of the players before. I even have conditions under which such allocations will always be an equilibrium, but in some cases it won't. I intentionally put this example in because it is one case in which there exists an $m$ such that there exists no equilibrium where $s_1=1$.

  • 1
    $\begingroup$ So the integer $J$ must always divide the integer $I$ ? $\endgroup$ Commented Nov 10, 2014 at 17:40
  • $\begingroup$ Yes. That's correct. I'm looking for even allocations of objects to people. So each of the $J$ agents will end up with $I/J$ objects. $\endgroup$
    – Shane
    Commented Nov 10, 2014 at 17:46
  • $\begingroup$ You also need to assume $(I,J)\neq (4,2)$, right ? Because in that case $V_{s_1,1}$ and $V_{s_2,2}$ can be disjoint only if $s_1=s_2$. $\endgroup$ Commented Nov 12, 2014 at 8:32
  • $\begingroup$ It's actually fine if $s_1=s_2$, but I don't think I agree with the statement. Consider $ V= [\{A,B\},\{C,D\};\{A,B\},\{C,D\};...]$ Here you could have $p=(m_1/2,m_1/2,m_2/2,m_2/2, ... )$ and $s=(1,2)$. Then $V_{s_1,1}=\{A,B\}$ and $V_{s_2,2}-\{C,D\}$. Does that clarify, or is my question unclear (i.e. require editing)? $\endgroup$
    – Shane
    Commented Nov 12, 2014 at 14:49
  • 1
    $\begingroup$ The cardinality of $\cal V$ is $N!^{J}$ because there are $J$ columns and $N!$ possible values for each column. $\endgroup$ Commented Nov 12, 2014 at 17:30

1 Answer 1


It took me a while, but I've finally shown the conjecture is not true by generating a counter-example:

\begin{array}{|c|c|c|} \hline EF & CF & CF \\ \hline CD & BE & DE \\ \hline AB & DF & AB \\ \hline AC & AC & BE \\ \hline BE & BF & EF \\ \hline BD & BC & BC \\ \hline AE & AF & CE \\ \hline AF & DE & AE \\ \hline DF & CE & DF \\ \hline AD & AE & AC \\ \hline CF & AD & BF \\ \hline BC & BD & CD \\ \hline CE & CD & AF \\ \hline DE & EF & BD \\ \hline BF & AB & AD \\ \hline \end{array} Let $m=(11,10,9)$. The proof that no $p$ and $s$ exist to clear this as follows:

Step 1: If such a $p$ and $s$ exist, $s_1<4$.

Proof: If $s_1 \geq 4$, then

$$p_E+p_F>11,\: p_C+p_D>11,\: p_A+p_B>11 \Rightarrow p_A+p_B+p_C+p_D+p_E+p_F>33$$

But $p_A+p_B+p_C+p_D+p_E+p_F \leq 11+10+9=30$ for $p$ and $s$ to clear. A contradiction.

Step 2: If such a $p$ and $s$ exist, $s_1 \ne 3$.

Proof: If $s_1 = 3$, then

$$p_E+p_F>11,\: p_C+p_D>11,\: \Rightarrow p_C+p_D+p_E+p_F>22$$

But then players 2 and 3 must end up with bundles from $\left\{C,D,E,F\right\}$, so $p_C+p_D+p_E+p_F \leq 10+9=19$. A contradiction.

Step 3: Consider all remaining cases, i.e. those in which $s_1\leq2$.

Technically there are 12 remaining cases to check. Because once we fix $s_1$, there are four remaining letters to be assigned across two people. $ {4 \choose 2} = 6$. Since we allow for two possible $s_1$'s, there are 12 total possibilities. I find contradictions for the most salient ones below -- contradictions for the remaining ones are even easier to come by.

3.a (EF-AC-BD): Then $p_B+p_E>10$ and $p_D+p_F>10$ so $p_B+p_D+p_E+p_F>20$. But $p_E+p_F \leq 11$ so $p_B+p_D>9$. A contradiction.

3.b (EF-BC-AD): $p_A$ and $p_D$ must be lowest two prices. But $p_A+p_C>p_B+p_C$. A contradiction.

3.c (EF-CD-AB): $p_B+p_E>10$ and $p_A+p_F>10$, so $p_A+p_B+p_E+p_F>20$. But we need $p_E+p_F\leq 11$ and $p_A+p_B<9$ so $p_A+p_B+p_E+p_F\leq20$. A contradiction.

3.d (CD-BE-AF): $p_A+p_B>p_A+p_F \Rightarrow p_B>p_F$. But $p_E+p_F>11$. Therefore, $p_B+p_E>11$. But we needed $p_B+p_E\leq 10$. A contradiction.


You'll note here that the problem arises from the fact that some players are seeing some sets of goods as complements where others are seeing the same set of goods as substitutes. In any case, thanks to all those that commented and thought about the question. While it turns out the conjecture was incorrect, I'm hoping that slightly weaker conjectures may hold, and working on those currently.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .