The Shapley value in the Knaster gain game

Abstract

In Briata et al. (AUCO Czech Econ Rev 6:199–208, 2012), the authors introduce a cooperative game with transferable utility for allocating the gain of a collusion among completely risk-averse agents involved in the fair division procedure introduced by Knaster (Ann Soc Pol Math 19:228–230, 1946). In this paper we analyze the Shapley value (Shapley, in: Kuhn, Tucker (eds) Contributions to the theory of games II (Annals of Mathematics Studies 28), Princeton University Press, Princeton, 1953) of the game and propose its use as a measure of the players’ attitude towards collusion. Furthermore, we relate the sign of the Shapley value with the ranking order of the players’ evaluation, and show that some players in a given ranking will always deter collusion. Finally, we characterize the coalitions that maximize the gain from collusion, and suggest an ad-hoc coalition formation mechanism.

A Appendix: Proof of Theorem 3.1

Finding an explicit expression for the Shapley value is hampered by the fact that the marginal contribution of an agent to a coalition has a complicated espression. In fact, if $i \in N$ and $S \subseteq N \setminus \{i\}$, the following holds

$$\begin{aligned} v_g(S \cup \{i\})-v_g(S) = \frac{n-s-1}{n^2} \; \max \left\{ s(v_i - b^S),b^S - v_i \right\} - \frac{1}{n^2} \sum _{k \in S}(b^S-v_k). \end{aligned}$$

(24)

We will consider, instead, a general formula about the difference of Shapley values for adjacent players

$$\begin{aligned} \phi _j(v_g) - \phi _{j+1}(v_g) = \sum _{S \subseteq N \setminus \{ j,j+1 \}} \frac{s!(n-s-2)!}{(n-1)!} \left[ v_g(S \cup \{j\}) - v_g(S \cup \{j+1\}) \right] . \end{aligned}$$

(25)

The following lemma shows that the difference between the value of the coalition joined by two successive players results in a formula much simpler than (24). For any $j \in N \setminus \{n\}$, let $J= \{1,\ldots ,j\}$ and $J^c= \{j+1,\ldots ,n\}$.

Lemma A.1

For any $j \in N \setminus \{n\}$ and $S \subseteq N \setminus \{j,j+1\}$,

$$\begin{aligned} v_g(S \cup \{j\}) - v_g(S \cup \{j+1\})= {\left\{ \begin{array}{ll} \frac{(n-s-1)s}{n^2}(v_j-v_{j+1}) &{} \hbox {if}\quad S \cap J = \varnothing \\ -\frac{n-s-1}{n^2}(v_j-v_{j+1}) &{} \hbox {if}\quad S \cap J \ne \varnothing \end{array}\right. }. \end{aligned}$$

(26)

Proof

If $S \cap J = \varnothing$, then $S \subseteq J^c \setminus \{j+1\}$, and

$$\begin{aligned} v_g(S \cup \{j\}) - v_g(S \cup \{j+1\})= & {} \frac{n-s-1}{n^2} \left( \sum _{i \in S \cup \{j\}} \left( b_{S \cup \{j\}} - v_i \right) - \sum _{i \in S \cup \{j+1\}} \left( b_{S \cup \{j+1\}} - v_i \right) \right) \\= & {} \frac{n-s-1}{n^2} \left( \sum _{i \in S } \left( v_j - v_i \right) - \sum _{i \in S } \left( v_{j+1} - v_i \right) \right) \\= & {} \frac{n-s-1}{n^2} s (v_j - v_{j+1}). \end{aligned}$$

Otherwise, $S \cap J \ne \varnothing$, and

$$\begin{aligned} v_g(S \cup \{j\}) - v_g(S \cup \{j+1\})= & {} \frac{n-s-1}{n^2} \left( \sum _{i \in S \cup \{j\}} \left( b_S - v_i \right) - \sum _{i \in S \cup \{j+1\}} \left( b_S - v_i \right) \right) \\= & {} - \frac{n-s-1}{n^2} (v_j - v_{j+1}). \end{aligned}$$

$\square$

Proof of Theorem 3.1

Each coalition worth in the gain game can be written as

$$\begin{aligned} v_g(S) = \frac{n-s}{n^2}\sum _{i \in S} ( b_S - v_i ) = \sum _{j=1}^{n-1} c_{S,j} \left( v_j - v_{j+1} \right) , \end{aligned}$$

(27)

where

$$\begin{aligned} c_{S,j}= \frac{n-s}{n^2} \sum _{i \in S} I_{S,i}(j), \quad I_{S,i}(j)= {\left\{ \begin{array}{ll} 1 &{} \hbox {if}\quad \min \nolimits _{h \in S} h \le j < i\\ 0 &{} \hbox {otherwise} \end{array}\right. }, \end{aligned}$$

and the coefficients $c_{S,j}$ do not depend on the actual values of the $v_i$’s (within a fixed ranking). The Shapley value for a player is a linear combination of differences of coalitions’ worths, and, therefore, a linear combination of the differences in valuations between successive agents, explaining (2).

The gain game is such that $v_g(N)=0$ and, therefore,

$$\begin{aligned} 0 = v_g(N) = \sum _{i =1}^n \phi _i(v_g) = \sum _{i = 1}^n \sum _{j =1}^{n-1} \psi _{ij}(v_{j} - v_{j+1}) = \sum _{j=1}^{n-1} ( v_{j} - v_{j+1}) \sum _{i =1}^n \psi _{ij}. \end{aligned}$$

Since this holds for any choice of the $v_i$, $i \in N$, it must be that

$$\begin{aligned} \sum _{i=1}^n \psi _{ij}=0 \quad \hbox {for any }j \in N \setminus \{n\}. \end{aligned}$$

(28)

In order to have a simple expression for the coefficients $\psi _{ij}$, we introduce a particular set of evaluation for the agents, and then we consider the related gain game. For any $j \in N \setminus \{n\}$, let the evaluations be as follows:

$$\begin{aligned} v_h= {\left\{ \begin{array}{ll} 1 &{} \hbox {if}\quad h \in J\\ 0 &{} \hbox {if}\quad h \in J^c \end{array}\right. }. \end{aligned}$$

(29)

Let $v_{g,j}$ be the corresponding gain game. Clearly, from (2) we have:

$$\begin{aligned} \phi _i(v_{g,j}) = \psi _{ij} \quad \hbox {for any }i,j \in N. \end{aligned}$$

Moreover, by the symmetry¹ of the players in J and in $J^c$, and, recalling that the Shapley value assigns equal amounts to symmetric players, we have

$$\begin{aligned} \psi _{1j}= & {} \psi _{2j} = \cdots = \psi _{jj} =a_j\nonumber \\ \psi _{j+1,j}= & {} \psi _{j+2,j} = \cdots = \psi _{nj} =b_j. \end{aligned}$$

(30)

To compute every $\psi _{ij}$ we only need to determine the differences between $\psi _{jj}$ and $\psi _{j+1,j}$, which we can write as:

$$\begin{aligned} \psi _{jj}- \psi _{j+1,j} =a_j - b_j= \phi _j(v_{g,j}) - \phi _{j+1}(v_{g,j}). \end{aligned}$$

(31)

We now apply (25), together with Lemma (A.1). Noting that $v_j - v_{j+1}=1$ in the game $v_{gj}$, the Lemma distinguishes between two cases:

Case 1::

$S \cap J = \varnothing$ and, therefore, $S \subseteq J^c \setminus \{j+1\}$. The part of formula (25) pertaining these coalitions is present only when $j \in N \setminus \{n-1,n\}$ and it is given by

$$\begin{aligned}&\sum _{S \subseteq J^c \setminus \{j+1\}} \frac{s!(n-s-2)!}{(n-1)!} \left( v_g(S \cup \{j\}) - v_g(S \cup \{j+1\})\right) \\&\quad =\sum _{S \subseteq J^c \setminus \{j+1\}} \frac{s!(n-s-2)!}{(n-1)!} \frac{(n-s-1)s}{n^2} = \frac{1}{n^2} \sum _{S \subseteq J^c \setminus \{j+1\}} \frac{s!(n-s-2)!(n-s-1)}{(n-1)!} s\\&\quad =\frac{1}{n^2} \sum _{S \subseteq J^c \setminus \{j+1\}} \frac{1}{\left( {\begin{array}{c}n-1\\ s\end{array}}\right) } s = \frac{1}{n^2} \sum _{s=1}^{n-j-1} \frac{\left( {\begin{array}{c}n-j-1\\ s\end{array}}\right) }{\left( {\begin{array}{c}n-1\\ s\end{array}}\right) } s. \end{aligned}$$

Case 2::

$S \cap J \ne \varnothing$. The part of formula (25) pertaining this case is given by

$$\begin{aligned}&\quad \sum _{S \cap J \ne \varnothing , S \subseteq N \setminus \{j,j+1\}} -\ \frac{s! (n-s-2)!}{(n-1)!} \frac{n-s-1}{n^2}\\&\quad \quad = \ - \frac{1}{n^2} \sum _{S \cap J \ne \varnothing , S \subseteq N \setminus \{j,j+1\}} \frac{s! (n-s-1)!}{(n-1)!} = - \frac{1}{n^2} \sum _{S \cap J \ne \varnothing , S \subseteq N \setminus \{j,j+1\}} \frac{1}{\left( {\begin{array}{c}n-1\\ s\end{array}}\right) }. \end{aligned}$$

Now, we can choose a set of s units, with at least 1 unit from the first $j-1$ in a number of ways given by

$$\begin{aligned} \sum _{t = \max \{1, s+j+1-n\}}^{\min \{j-1,s\}} \left( {\begin{array}{c}j-1\\ t\end{array}}\right) \left( {\begin{array}{c}n-j-1\\ s-t\end{array}}\right) ={\left\{ \begin{array}{ll} \left( {\begin{array}{c}n-2\\ s\end{array}}\right) - \left( {\begin{array}{c}n-j-1\\ s\end{array}}\right) &{} \hbox {if}\quad s \le n - j -1\\ \left( {\begin{array}{c}n-2\\ s\end{array}}\right)&\hbox {if}\quad s > n - j -1 \end{array}\right. }. \end{aligned}$$

Therefore, the part of (25) pertaining this case becomes

Joining the results for Cases 1 and 2 when $j < n-1$, we have

$$\begin{aligned} \psi _{jj} - \psi _{j+1,j} = \frac{1}{n^2}\,\left[ \sum _{s=1}^{n-j-1} \frac{\left( {\begin{array}{c}n-j-1\\ s\end{array}}\right) }{\left( {\begin{array}{c}n-1\\ s\end{array}}\right) } s - \sum _{s=1}^{n-2} \frac{\left( {\begin{array}{c}n-2\\ s\end{array}}\right) }{\left( {\begin{array}{c}n-1\\ s\end{array}}\right) } + \sum _{s=1}^{n-j-1} \frac{\left( {\begin{array}{c}n-j-1\\ s\end{array}}\right) }{\left( {\begin{array}{c}n-1\\ s\end{array}}\right) }\right] . \end{aligned}$$

(32)

Applying Lemma 3.3 to the r.h.s. of (32), it is easy to check that

$$\begin{aligned} \psi _{jj} - \psi _{j+1,j} = \frac{1}{n^2}\,\left[ \frac{n\,(n-j-1)}{(j+1)\,(j+2)} - \frac{n-2}{2} + \frac{n-j-1}{j+1}\right] = \frac{2n - 3j - j^2}{2n(j+1)(j+2)}.\nonumber \\ \end{aligned}$$

(33)

To prove (3), we recall (28), (30) and solve the following system of linear equations in the variables $a_j$ and $b_j$

$$\begin{aligned} \left\{ \begin{array}{l} j a_j + (n-j) b_j =0\\ a_j - b_j = \frac{2n - 3j - j^2}{2n(j+1)(j+2)}. \end{array}\right. \end{aligned}$$

$\square$