EEF doesn't imply EF for additive valuations

Dependencies:

1. Fair division
2. Envy-freeness
3. Epistemic fairness
4. Additive set function

There exists a fair division instance with 3 agents and additive valuations, and an allocation $A$ for that instance that is epistemic EF and PROP but not EF. This holds for both goods division and chores division, and both divisible and indivisible items.

Proof

Let $M = \bigcup_{i=1}^6 S_i$, where $S_i \cap S_j = \emptyset$ for all $i \neq j$.

Define additive set functions $f_1, f_2, f_3$ as follows:

	$S\_1$	$S\_2$	$S\_3$	$S\_4$	$S\_5$	$S\_6$
$f\_1$	2	2	3	3	1	1
$f\_2$	1	1	2	2	3	3
$f\_3$	3	3	1	1	2	2

Note that $f_1(M) = f_2(M) = f_3(M) = 12$.

Let $A = (S_1 \cup S_2, S_3 \cup S_4, S_5 \cup S_6)$ and $B^{(1)} = (S_1 \cup S_2, S_3 \cup S_5, S_4 \cup S_6)$.

If $v_1 = f_1$, then agent 1 envies agent 2 in $A$, but $B^{(1)}$ is agent 1's epistemic EF certificate for $A$. Similarly, we can show that agents 2 and 3 are also envious in $A$ but they too have epistemic EF certificates for $A$. Hence, $A$ is EEF but not EF. Also, $v_i(A_i) = 4$ for all $i$, so $A$ is PROP.

A similar argument holds when $v_1 = -f_1$.

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. Additive set function