The link between the Shapley value and the beta factor

Abstract

In this article, we provide a link between the Shapley value in cooperative game theory and the capital asset pricing model (CAPM) in finance. In particular, the Shapley value of a suitably defined cooperative game is closely related to the beta factor in the CAPM. The beta factor for any given security may be interpreted as the asset’s fairly allocated share of the market risk or as the asset’s average marginal contribution to the market risk, respectively. Other fairness properties and axioms of the Shapley value may be reinterpreted in this context to attain a deeper understanding of the beta factor and the connotation of systematic risk. Our game theoretic approach further allows for a generalisation of the CAPM with respect to arbitrary risk measures other than variance. Last but not least, we discuss the volatility of an asset’s theoretical fair assessment of risk and of its systematic risk, respectively. This result lends itself to face the challenge of an empirical investigation on real stock markets.

Appendix

Proof of Lemma 1

Note that the set of symmetric $(n\times n)$-matrices is a linear vector space. A basis is given by the set of symmetric basis matrices $E_{ij}$ having coefficients

$$\begin{aligned} e_{kl} =\left\{ {\begin{array}{ll} 1 &{} \hbox {if }\{k,l\}=\{i,j\} \\ 0 &{} \hbox {otherwise} \\ \end{array}} \right. . \end{aligned}$$

These matrices are obviously linear independent. Note that the number of such linear independent matrices is $\left( {\begin{array}{l} n \\ 2 \\ \end{array}} \right) +n=\frac{n(n+1)}{2}$ . It is equal to the dimension of the vector space of symmetric $(n\times n)$ matrices. By means of an illustration, for $n=3$ said basis is given by

$$\begin{aligned}&\left\{ E_{11} =\left( {{\begin{array}{c@{\quad }c@{\quad }c} 1&{} 0&{} 0 \\ 0&{} 0&{} 0 \\ 0&{} 0&{} 0 \\ \end{array} }} \right) ;E_{12} =\left( {{\begin{array}{c@{\quad }c@{\quad }c} 0&{} 1&{} 0 \\ 1&{} 0&{} 0 \\ 0&{} 0&{} 0 \\ \end{array} }} \right) ;E_{13} =\left( {{\begin{array}{c@{\quad }c@{\quad }c} 0&{} 0&{} 1 \\ 0&{} 0&{} 0 \\ 1&{} 0&{} 0 \\ \end{array} }} \right) ; \right. \\&\quad \left. E_{22} =\left( {{\begin{array}{c@{\quad }c@{\quad }c} 0&{} 0&{} 0 \\ 0&{} 1&{} 0 \\ 0&{} 0&{} 0 \\ \end{array} }} \right) ;E_{23} =\left( {{\begin{array}{c@{\quad }c@{\quad }c} 0&{} 0&{} 0 \\ 0&{} 0&{} 1 \\ 0&{} 1&{} 0 \\ \end{array} }} \right) ;E_{33} =\left( {{\begin{array}{c@{\quad }c@{\quad }c} 0&{} 0&{} 0 \\ 0&{} 0&{} 0 \\ 0&{} 0&{} 1 \\ \end{array} }} \right) \right\} . \end{aligned}$$

For a fixed market portfolio M, the set of capital market games $\Gamma =\left( {M,v_A } \right)$ is a linear vector space, too, if we apply point-by-point addition and scalar multiplication of characteristic functions. Note that the basis games $(M,v_{E_{ij}})$ are exactly the unanimity games with at most 2 relevant players. By means of an illustration for $n=3$, we see that $v_{E_{12} } \left( {\{1,3\}} \right) =v_{E_{12} } \left( {\{2,3\}} \right) =v_{E_{12} } \left( {\{1\}} \right) =v_{E_{12} } \left( {\{2\}} \right) =v_{E_{12} } \left( {\{3\}} \right) =v_{E_{12} } \left( \varnothing \right) =0$ and $v_{E_{12} } \left( {\{1,2,3\}} \right) =v_{E_{12} } \left( {\{1,2\}} \right) =2$. We then easily deduce that $\Phi _1 (M,v_{E_{12} } )=\Phi _2 (M,v_{E_{12} } )=1$ and $\Phi _3 (M,v_{E_{12}})=0$.

In general, we observe that for any basis game, it holds true that

$$\begin{aligned} \Phi _i (M,v_{E_{ij} } )= & {} 1=\sum _{k\in M} {e_{ik} } \\ \Phi _j (M,v_{E_{ij} } )= & {} 1=\sum _{k\in M} {e_{kj} } \\ \Phi _k (M,v_{E_{ij} } )= & {} 0\hbox { for }k\ne i,j. \end{aligned}$$

Now, for any given symmetric matrix $A=(a_{ij} )$, the linear combination is known. It follows that the characteristic function is given by . By linearity of the Shapley value, it is true that

$$\begin{aligned} \Phi _k (M,v_A )= & {} \sum _{i\in M} {\mathop {\mathop {\sum \limits _{j\in M}}\limits _{j\ge i}}} {a_{ij} \Phi _k (M,v_{E_{ij} } )} \\= & {} \sum _{i\in M} {\mathop {\mathop {\sum \limits _{j\in M}}\limits _{j\ge i}}} {a_{ij} (\delta _{ik} +\delta _{kj} -\delta _{ij} )} \\= & {} {\mathop {\mathop {\sum \limits _{i\in M}}\limits _{i=k}}} {\mathop {\mathop {\sum \limits _{j\in M}}\limits _{j\ge i}}} {a_{kj} } +\sum _{i\in M} {\mathop {\mathop {\mathop {\sum \limits _{j\in M}}\limits _{j\ge i}}\limits _{j=k}}} {a_{ik} } -\sum _{i\in M} {\mathop {\mathop {\mathop {\sum \limits _{j\in M}}\limits _{j\ge i}}\limits _{j=i}}}{a_{ii}} \\= & {} {\mathop {\mathop {\sum \limits _{j\in M}}\limits _{j\ge k}}} {a_{kj} } + {\mathop {\mathop {\sum \limits _{i\in M}}\limits _{i\le k}}} {a_{ik}} -\sum \limits _{i\in M} {a_{ii} } \\= & {} {\mathop {\mathop {\sum \limits _{j\in M}}\limits _{j\ge k}}} {a_{kj} } + {\mathop {\mathop {\sum \limits _{j\in M}}\limits _{j\le k}}} {a_{kj} } -\sum _{i\in M} {a_{ii} } \\= & {} \sum _{j\in M} {a_{kj} } \\ \end{aligned}$$

where we have used the Kronecker delta: $\delta _{ik} =1$ for $i=k$ and $\delta _{ik} =0$ otherwise. $\square$

Proof of Lemma 2

The starting point for the computation of the volatility is its definition as per Eq. (9):

$$\begin{aligned} \sigma _k^2 (\Gamma )= {\mathop {\mathop {\sum \limits _{S\subseteq M}}\limits _{k\in S}}} {p_S } \cdot \left( {v_A (S)-v_A \left( {S\backslash \{k\}} \right) } \right) ^{2}-\left( {\Phi _k (\Gamma _A )} \right) ^{2} \end{aligned}$$

First we note that

$$\begin{aligned} \left( {\Phi _k (\Gamma _A )} \right) ^{2}= & {} \left( {\sum _{j\in M} {a_{kj} } } \right) ^{2}=\left( {a_{kk} +\sum _{j\in M\backslash \{k\}} {a_{kj} } } \right) ^{2} \\= & {} a_{kk}^2 +2a_{kk} \cdot \sum _{j\in M\backslash \{k\}} {a_{kj} } +\left( {\sum _{j\in M\backslash \{k\}} {a_{kj} } } \right) ^{2}. \end{aligned}$$

In addition, it holds true for any marginal difference with $S\subseteq M,i\in S$ that

$$\begin{aligned} v_A (S)-v_A \left( {S\backslash \{k\}} \right)= & {} \sum _{i,j\in S} {a_{ij} } -\sum _{i,j\in S\backslash \{k\}} {a_{ij} } \\= & {} a_{kk} +\sum _{j\in S\backslash \{k\}} {a_{kj} } +\sum _{i\in S\backslash \{k\}} {a_{ik} } +\sum _{i,j\in S\backslash \{k\}} {a_{ij} } -\sum _{i,j\in S\backslash \{k\}} {a_{ij} } \\= & {} a_{kk} +2\sum _{i\in S\backslash \{k\}} {a_{ik} } . \end{aligned}$$

It follows that

$$\begin{aligned} \left( {v_A (S)-v_A \left( {S\backslash \{k\}} \right) } \right) ^{2}= & {} \left( {a_{kk} +2\sum _{i\in S\backslash \{k\}} {a_{ik} }} \right) ^{2} \\= & {} a_{kk}^2 +4a_{kk} \cdot \sum _{i\in S\backslash \{k\}} {a_{ik} } +4\left( {\sum _{i\in S\backslash \{k\}} {a_{ik} } } \right) ^{2} . \end{aligned}$$

Using the abbreviations $|S|=s$ and $|M|=n$, we note that

$$\begin{aligned} {\mathop {\mathop {\sum \limits _{S\subseteq M}}\limits _{i\in S}}} {p_S } a_{kk}^2= & {} a_{kk}^2 {\mathop {\mathop {\sum \limits _{S\subseteq M}}\limits _{i\in S}}} {\frac{\left( {s-1} \right) ! \, \cdot \, \left( {n-s} \right) !}{n!}} \\= & {} a_{kk}^2 \sum _{s=1}^n {\left( {\begin{array}{l} n-1 \\ s-1 \\ \end{array}} \right) \frac{\left( {s-1} \right) ! \, \cdot \, \left( {n-s} \right) !}{n!}} \\= & {} a_{kk}^2 \sum _{s=1}^n {\frac{(n-1)!}{\left( {n-1-(s-1)} \right) ! \, \cdot \, \left( {s-1} \right) !}\cdot \frac{\left( {s-1} \right) ! \, \cdot \, \left( {n-s} \right) !}{n!}} \\= & {} a_{kk}^2 \sum _{s=1}^n {\frac{1}{n}=a_{kk}^2 } . \end{aligned}$$

Likewise, it holds true that

$$\begin{aligned} {\mathop {\mathop {\sum \limits _{S\subseteq M}}\limits _{i\in S}}} {p_S 4a_{kk} \sum _{i\in S\backslash \{k\}} {a_{ik} } }= & {} 4a_{kk} {\mathop {\mathop {\sum \limits _{S\subseteq M}}\limits _{i\in S}}} {\frac{\left( {s-1} \right) ! \, \cdot \, \left( {n-s} \right) !}{n!}} \, \sum _{i\in S\backslash \{k\}} {a_{ik} } \\= & {} 4a_{kk} \sum _{s=2}^n {\frac{(s-1)! \, \cdot \, (n-s)!}{n!}} \sum \limits _{\begin{array}{c} i=1 \\ i\ne k \end{array}}^n {\left( {\begin{array}{l} n-2 \\ s-2 \\ \end{array}} \right) a_{ik} } \\= & {} 4a_{kk} \sum _{s=2}^n {\frac{(s-1)}{n(n-1)}} \sum \limits _{\begin{array}{c} i=1\\ i\ne k \end{array}}^{n} {a_{ik} } \\= & {} 4a_{kk} \frac{1}{n(n-1)}\cdot \frac{(n-1)n}{2} \sum \limits _{\begin{array}{c} i=1\\ i\ne k \end{array}}^{n} {a_{ik} } \\= & {} 2a_{kk} \sum _{i\in M\backslash \{k\}} {a_{ik} } . \end{aligned}$$

In addition, we see that

$$\begin{aligned}&{\mathop {\mathop {\sum \limits _{S\subseteq M}}\limits _{k\in S}}} {p_S } 4\left( {\sum _{i\in S\backslash \{k\}} {a_{ik} } } \right) ^{2} \\&\quad =4 {\mathop {\mathop {\sum \limits _{S\subseteq M}}\limits _{k\in S}}} {\frac{\left( {|S|-1} \right) ! \, \cdot \, \left( {|M|-|S|} \right) !}{|M|!}} \left( {\sum _{i\in S\backslash \{k\}} {a_{ik} } } \right) ^{2} \\&\quad =4 {\mathop {\mathop {\sum \limits _{S\subseteq M}}\limits _{k\in S}}} {\frac{\left( {|S|-1} \right) ! \, \cdot \, \left( {|M|-|S|} \right) !}{|M|!}} \left( {\sum _{i\in S\backslash \{k\}} {a_{ik}^2 } + {\mathop {\mathop {\sum \limits _{i,j\in S\backslash \{k\}}}\limits _{i\ne j}}} {a_{ik} a_{jk} } } \right) \\&\quad =4\sum _{s=2}^n {\frac{(s-1)! \, \cdot \, (n-s)!}{n!} \sum \limits _{\begin{array}{c} i=1\\ i\ne k \end{array}}^{n} {\left( {\begin{array}{l} n-2 \\ s-2 \\ \end{array}} \right) a_{ik}^2 } } \\&\qquad +8\sum _{s=3}^n {\frac{(s-1)! \, \cdot \, (n-s)!}{n!} \sum \limits _{\begin{array}{c} i=1\\ i\ne k \end{array}}^{n} \sum \limits _{\begin{array}{c} j=1\\ j\ne i,k \end{array}}^{n} {\frac{1}{2}\left( {\begin{array}{l} n-3 \\ s-3 \\ \end{array}} \right) a_{ik} a_{jk} } } \\&\quad =4\sum _{s=2}^n {\frac{(s-1)}{n(n-1)} \sum \limits _{\begin{array}{c} i=1\\ i\ne k \end{array}}^{n} {a_{ik}^2 }} +4\sum _{s=3}^n{\frac{(s-1)(s-2)}{n(n-1)(n-2)} \sum \limits _{\begin{array}{c} i=1\\ i\ne k \end{array}}^{n} \sum \limits _{\begin{array}{c} j=1\\ j\ne i,k \end{array}}^{n} {a_{ik} a_{jk}}} \\&\quad =2 \sum \limits _{\begin{array}{c} i=1\\ i\ne k \end{array}}^{n} {a_{ik}^2 } +4\sum _{s=3}^n {\frac{s^{2}-3s+2}{n(n-1)(n-2)} \sum \limits _{\begin{array}{c} i=1 \\ i\ne k \end{array}}^{n} \sum \limits _{\begin{array}{c} j=1\\ j\ne i,k \end{array}}^{n} {a_{ik} a_{jk} } } \\&\quad =2 \sum \limits _{\begin{array}{c} i=1\\ i\ne k \end{array}}^{n} {a_{ik}^2 } +\frac{4}{n(n-1)(n-2)} \\&\qquad \quad \cdot \left( {\frac{n(n+1)(2n+1)}{6}-5-3\frac{n(n+1)}{2}+9+2(n-2)} \right) \sum \limits _{\begin{array}{c} i=1\\ i\ne k \end{array}}^{n} \sum \limits _{\begin{array}{c} j=1\\ j\ne i,k \end{array}}^{n} {a_{ik} a_{jk} } \\&\quad =2 \sum \limits _{\begin{array}{c} i=1\\ i\ne k \end{array}}^{n} {a_{ik}^2 } +\frac{4}{n(n-1)(n-2)}\cdot \frac{n(n+1)(2n-8)+12n}{6} \sum \limits _{\begin{array}{c} i=1\\ i\ne k \end{array}}^{n} \sum \limits _{\begin{array}{c} j=1\\ j\ne i,k \end{array}}^{n} {a_{ik} a_{jk} } \\&\quad =2 {\mathop {\mathop {\sum \limits _{i\in M}}\limits _{i\ne k}}} {a_{ik}^2 } +\frac{(2n+2)(2n-8)+24}{3(n-1)(n-2)} {\mathop {\mathop {\sum \limits _{i\in M}}\limits _{i\ne k}}} {\mathop {\mathop {\sum \limits _{j\in M}}\limits _{j\ne i,k}}} {a_{ik} a_{jk}}. \end{aligned}$$

Putting it all together, we find that

$$\begin{aligned} \sigma _i^2 (\Gamma )= & {} a_{kk}^2 +2a_{kk} \cdot \sum _{i\in M\backslash \{k\}} {a_{ik} } +2\sum _{i\in M\backslash \{k\}} {a_{ik}^2 } \\&+\frac{(2n+2)(2n-8)+24}{3(n-1)(n-2)}\sum _{i\in M\backslash \{k\}} {\sum _{j\in M\backslash \{i,k\}} {a_{ik} a_{jk}}}\\&-a_{kk}^2 -2a_{kk} \cdot \sum _{j\in M\backslash \{k\}} {a_{kj} } -\left( {\sum _{j\in M\backslash \{k\}} {a_{kj} } } \right) ^{2} \\= & {} 2\sum _{i\in M\backslash \{k\}} {a_{ik}^2 } +\frac{4n^{2}-12n+8}{3n^{2}-9n+6}\sum _{i\in M\backslash \{k\}} {\sum _{j\in M\backslash \{i,k\}} {a_{ik} a_{jk} } } \\&-\left( {\sum _{i\in M\backslash \{k\}} {a_{ik}^2 } +\sum _{i\in M\backslash \{k\}} {\sum _{j\in M\backslash \{i,k\}} {a_{ik} a_{jk} } } } \right) \\= & {} \sum _{i\in M\backslash \{k\}} {a_{ik}^2 } +\frac{n^{2}-3n+2}{3n^{2}-9n+6}\sum _{i\in M\backslash \{k\}} {\sum _{j\in M\backslash \{i,k\}} {a_{ik} a_{jk} } } \\= & {} \sum _{i\in M\backslash \{k\}} {a_{ik}^2 } +\frac{1}{3}\sum _{i\in M\backslash \{k\}} {\sum _{j\in M\backslash \{i,k\}} {a_{ik} a_{jk}}}. \end{aligned}$$

$\square$