Shapley values, different versions of the symmetry axiom

Question

I have come across different versions of the axioms, from which the Shapley values (https://en.wikipedia.org/wiki/Shapley_value) are derived. I want to prove that these are equivalent, but I miss an argument for the symmetry axiom.

The first version of the axioms is from the Shapley paper (1953) (http://library.fa.ru/files/Roth2.pdf page 33):

Axiom 1 (symmetry): For each $\pi \in \Pi (U)$, $\phi_{\pi i} [\pi v] = \phi_i[v]$

Axiom 2 (efficiency): For each carrier N of $v$, $\Sigma_{i \in N} \phi_i[v] = v(N)$

Axiom 3 (law of aggregation): For any two games $v$ and $w$, $\phi[v + w] = \phi[v] + \phi[w]$

• Here v is the game, i.e. a function from the subsets of U to the real numbers,
• $U$ is the set of players
• $\Pi(U)$ is the set of permutations on $U$,
• $\pi v$ is defined by $\pi v(\pi S) = v(S)$ for all $S \subseteq U$,
• $\phi_i$ is the attribution to player i,
• and a carrier $N$ is a subset of $U$ such that $v(N \cap S) = v(S)$ for all $S \subseteq U$.

Another version uses a modified set of axioms. Let's refer to these as properties to distinguish them:

Property 1 (symmetry): If $v(S \cup i) = v(S \cup j)$ for all $S \subseteq U\backslash \{i,j\}$, then $\phi_i[v]=\phi_j[v]$

Property 2 (efficiency): $\Sigma_{i \in U} \phi_i[v] = v(U)$

Property 3 (dummy): If $v(S \cup j) = v(S)$ for all $S \subseteq \ {j}$, then $\phi_i[v] = 0$

Property 4 (law of aggregation): For any two games $v$ and $w$, $\phi[v + w] = \phi[v] + \phi[w]$

I have shown that the three axioms are equivalent to the four properties, except I need to show that Axiom 1 (symmetry) is implied by the properties (I guess, just by property 1).

Can someone see how this follows?

Answer 1

I stumbled on your question while looking for its answer.

One can go about it by repeating Shapley's proof with P1-P4 instead of the original axioms, i.e. proving that all value functions which satisfy P1-P4 coincide with Shapley's values for games with finite carriers.

Since Shapley's values satisfy Axiom 1, it follows that P1-P4 imply the original axioms for games $v$ with finite carriers.

Personally I don't find this satisfactory - maybe there is a proof that doesn't require going through all the steps of Shapley's proof - but I couldn't find it. I provide the details of how the proof is modified below.

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Lemma 1 follows from the 'dummy' property. For any $i \notin N$, where $N$ is a carrier:

$$ v(S \cup \left\{i\right\}) = v\left((S \cup \left\{i\right\}) \cap N\right) = v(S \cap N) = v(S) $$

Lemma 2 holds unchanged since it does not involve the notion of value function.

For Lemma 3 ($\phi_i[c\, v_R ] = \frac{c}{|R|} \mathbb{1}(i \in R)$):

• for $i \notin R$, $\phi_i[c\, v_R ] = 0\;$ follows directly from the 'dummy' property
• if $i,\,j \in R$, the 'symmetry' property implies that $\phi_i[c\, v_R ] = \phi_j[c\, v_R ]$, so all elements of $R$ have equal values and they should sum up to $c$ according to the 'efficiency property'

The rest of the proof makes use of the 'law of aggregation' which is in common between Axioms and Properties.