MEFS but not EEF

Dependencies:

1. Fair division
2. Epistemic fairness
3. Minimum fair share
4. Envy-freeness

$\newcommand{\defeq}{:=}$ Consider a fair division instance with 6 goods and 3 agents having equal entitlements and additive valuations. Valuations are given by this table:

$j$	1	2	3	4	5	6
$v$1( $j$ )	20	20	20	10	10	10
$v$2( $j$ ), $v$3( $j$ )	20	10	10	1	1	1

Then the allocation $A \defeq (\{4, 5, 6\}, \{1\}, \{2, 3\})$ is MEFS, but no allocation is epistemic EF.

Proof

Agents 2 and 3 are envy-free in $A$. Agent 1 has $B \defeq (\{1, 4\}, \{2, 5\}, \{3, 6\})$ as her MEFS-certificate for $A$. Hence, $A$ is MEFS.

Suppose an epistemic EF allocation $X$ exists. Let $Y^{(i)}$ be each agent $i$'s epistemic-EF-certificate. For agent 2 to be envy-free in $Y^{(2)}$, we require $Y^{(2)}_2 \supseteq \{1\}$ or $Y^{(2)}_2 \supseteq \{2, 3\}$. Similarly, $Y^{(3)}_3 \supseteq \{1\}$ or $Y^{(3)}_3 \supseteq \{2, 3\}$. Since $Y^{(i)}_i = X_i$ for all $i$, we get $X_2 \cup X_3 \supseteq \{1, 2, 3\}$. Hence, $X_1 \subseteq \{4, 5, 6\}$. But then no epistemic-EF-certificate exists for agent 1 for $X$, contradicting our assumption that $X$ is epistemic EF. Hence, no epistemic EF allocation exists.

Dependency for: None

Info:

• Depth: 5
• Number of transitive dependencies: 7

Transitive dependencies:

1. /sets-and-relations/countable-set
2. σ-algebra
3. Set function
4. Fair division
5. Envy-freeness
6. Epistemic fairness
7. Minimum fair share