The Owen–Shapley Spatial Power Index in Three-Dimensional Space

Abstract

Inspired by Owen’s (Nav Res Logist Quart 18:345–354, 1971) previous work on the subject, Shapley (A comparison of power indices and a non-symmetric generalization. Rand Corporation, Santa Monica, 1977) introduced the Owen–Shapley spatial power index, which takes the ideological location of individuals into account, represented by vectors in the Euclidean space ${\mathbb {R}}^{m}$, to measure their power. In this work we study the Owen–Shapley spatial power index in three-dimensional space. Peters and Zarzuelo (Int J Game Theory 46:525–545, 2017) carried out a study of this index for individuals located in two-dimensional space, but pointed out the limitation of the two-dimensional feature. In this work focusing on three-dimensional space, we provide an explicit formula for spatial unanimity games, which makes it possible to calculate the Owen–Shapley spatial power index of any spatial game. We also give a characterization of the Owen–Shapley spatial power index employing two invariant positional axioms among others. Finally, we calculate this power index for the Basque Parliament, both in the two-dimensional and three-dimensional cases. We compare these positional indices against each other and against those that result when classical non-positional indices are considered, such as the Shapley–Shubik power index (Am Polit Sci Rev 48(3):787–792, 1954) and the Banzhaf-normalized index (Rutgers Law Rev 19:317–343, 1965).

Appendix

Proof of Lemma 5

By 3DPI, without a loss of generality, we can suppose that $\measuredangle aed=\measuredangle aec=\pi /2$ (fixing e,  and moving c and d towards or outwards a). By DP and 3DPI, we have to prove

$$\begin{aligned}&\varphi _{0}\left( \left\{ 0,1,2,3\right\} ,u_{\left\{ 0,1,2,3\right\} },p_{\left\{ 0,1,2,3\right\} }\right) \\&\quad =\varphi _{0}\left( \left\{ 0,1,2,3\right\} ,u_{\left\{ 0,1,2,3\right\} },q_{\left\{ 0,1,2,3\right\} }\right) +\varphi _{0}\left( \left\{ 0,1,2,3\right\} ,u_{\left\{ 0,1,2,3\right\} },s_{\left\{ 0,1,2,3\right\} }\right) -\frac{1}{4}. \end{aligned}$$

Let $d^{\prime }=d+(b-a)$, $e^{\prime }=e+(b-a)$ and $c^{\prime }=c+(b-a)$ (Fig. 13). Consider ${\widetilde{N}}=\left\{ 0,1,2,3,4,5\right\}$ and ${\widetilde{p}},{\widetilde{q}},{\widetilde{s}}\in C ^{{\widetilde{N}}}$ such that ${\widetilde{p}}_{\left\{ 0,1,2,3\right\} }=p_{\left\{ 0,1,2,3\right\} }$ , ${\widetilde{q}}_{\left\{ 0,1,2,3\right\} }=q_{\left\{ 0,1,2,3\right\} }$, ${\widetilde{s}}_{\left\{ 0,1,2,3\right\} }=s$, $\left( {\widetilde{p}}_{4}, {\widetilde{p}}_{5}\right) =\left( c^{\prime },d^{\prime }\right)$, $\left( {\widetilde{q}}_{4},{\widetilde{q}}_{5}\right) =\left( e^{\prime },d^{\prime }\right)$ and $\left( {\widetilde{s}}_{4},{\widetilde{s}}_{5}\right) =\left( c^{\prime },e^{\prime }\right) .$

Fig. 13

In this figure c′, d′ and e′ are added

By DP and 3DPI, the equality to prove is

$$\begin{aligned} \varphi _{0}\left( {\widetilde{N}},u_{{\widetilde{N}}},{\widetilde{p}}\right) =\varphi _{0}\left( {\widetilde{N}},u_{{\widetilde{N}}},{\widetilde{q}}\right) +\varphi _{0}\left( {\widetilde{N}},u_{{\widetilde{N}}},{\widetilde{s}}\right) - \frac{1}{4}. \end{aligned}$$

(6)

Since $\varphi$ is a power index,

$$\begin{aligned} \sum \limits _{j=0}^{5}\varphi _{j}\left( {\widetilde{N}},u_{{\widetilde{N}} },{\widetilde{p}}\right) =1. \end{aligned}$$

(7)

By AN and RI, taking Q the plane through $a+0.5\left( b-a\right)$ with normal vector $b-a$, equality (7) turns into

$$\begin{aligned} 2\varphi _{0}\left( {\widetilde{N}},u_{{\widetilde{N}}},{\widetilde{p}}\right) +2\varphi _{1}\left( {\widetilde{N}},u_{{\widetilde{N}}},{\widetilde{p}}\right) +2\varphi _{2}\left( {\widetilde{N}},u_{{\widetilde{N}}},{\widetilde{p}}\right) =1. \end{aligned}$$

(8)

(i) We obtain another expression for $\varphi _{2}\left( {\widetilde{N}},u_{ {\widetilde{N}}},{\widetilde{p}}\right)$. DP and 3DPI imply

$$\begin{aligned} \varphi _{2}\left( {\widetilde{N}},u_{{\widetilde{N}}},{\widetilde{p}}\right) =\varphi _{2}\left( {\widetilde{N}},u_{{\widetilde{N}}},{\widetilde{q}}\right) \text {,} \end{aligned}$$

(9)

and since $\varphi$ is a power index,

$$\begin{aligned} \sum \limits _{j=0}^{5}\varphi _{j}\left( {\widetilde{N}},u_{{\widetilde{N}} },{\widetilde{q}}\right) =1. \end{aligned}$$

Again, AN and RI, with Q the plane through $a+0.5\left( b-a\right)$ with normal vector $\left( b-a\right)$, transform this equality into

$$\begin{aligned} 2\varphi _{0}\left( {\widetilde{N}},u_{{\widetilde{N}}},{\widetilde{q}}\right) +2\varphi _{1}\left( {\widetilde{N}},u_{{\widetilde{N}}},{\widetilde{q}}\right) +2\varphi _{2}\left( {\widetilde{N}},u_{{\widetilde{N}}},{\widetilde{q}}\right) =1. \end{aligned}$$

Taking into account that $\left( \measuredangle {\widetilde{q}}_{0}\widetilde{q }_{1}{\widetilde{q}}_{2},\measuredangle {\widetilde{q}}_{0}{\widetilde{q}}_{1} {\widetilde{q}}_{4},\measuredangle {\widetilde{q}}_{2}{\widetilde{q}}_{1} {\widetilde{q}}_{4}\right) =\left( \pi /2,\pi /2,\pi /2\right)$, Lemma 4 implies

$$\begin{aligned} \varphi _{1}\left( {\widetilde{N}},u_{{\widetilde{N}}},{\widetilde{q}}\right) = \frac{1}{8}, \end{aligned}$$

and hence

$$\begin{aligned} 2\varphi _{2}\left( {\widetilde{N}},u_{{\widetilde{N}}},{\widetilde{q}}\right) = \frac{3}{4}-2\varphi _{0}\left( {\widetilde{N}},u_{{\widetilde{N}}},{\widetilde{q}} \right) . \end{aligned}$$

(10)

Therefore, this equality and (9) imply

$$\begin{aligned} 2\varphi _{2}\left( {\widetilde{N}},u_{{\widetilde{N}}},{\widetilde{p}}\right) = \frac{3}{4}-2\varphi _{0}\left( {\widetilde{N}},u_{{\widetilde{N}}},{\widetilde{q}} \right) . \end{aligned}$$

(11)

(ii) We obtain another expression for $\varphi _{1}\left( {\widetilde{N}},u_{ {\widetilde{N}}},{\widetilde{p}}\right)$. Similarly, by DP and 3DPI it holds

$$\begin{aligned} \varphi _{1}\left( {\widetilde{N}},u_{{\widetilde{N}}},{\widetilde{p}}\right) =\varphi _{1}\left( {\widetilde{N}},u_{{\widetilde{N}}},{\widetilde{s}}\right) , \end{aligned}$$

and reasoning as in (i) we obtain

$$\begin{aligned} 2\varphi _{1}\left( {\widetilde{N}},u_{{\widetilde{N}}},{\widetilde{s}}\right) = \frac{3}{4}-2\varphi _{0}\left( {\widetilde{N}},u_{{\widetilde{N}}},{\widetilde{s}} \right) . \end{aligned}$$

And hence,

$$\begin{aligned} 2\varphi _{1}\left( {\widetilde{N}},u_{{\widetilde{N}}},{\widetilde{p}}\right) = \frac{3}{4}-2\varphi _{0}\left( {\widetilde{N}},u_{{\widetilde{N}}},{\widetilde{s}} \right) . \end{aligned}$$

(12)

Substituting (11) and ( 12) in (8), equality (6) is obtained.

Finally, if $\measuredangle cae=\measuredangle dae$, the equality to prove is

$$\begin{aligned} \varphi _{0}\left( {\widetilde{N}},u_{{\widetilde{N}}},{\widetilde{p}}\right) =2\varphi _{0}\left( {\widetilde{N}},u_{{\widetilde{N}}},{\widetilde{q}}\right) - \frac{1}{4}. \end{aligned}$$

By AN and RI, with Q the plane through e with normal vector $c-e$, we have

$$\begin{aligned} \varphi _{2}\left( {\widetilde{N}},u_{{\widetilde{N}}},{\widetilde{q}}\right) =\varphi _{1}\left( {\widetilde{N}},u_{{\widetilde{N}}},{\widetilde{s}}\right) , \end{aligned}$$

which, together with (10) and ( 12), implies

$$\begin{aligned} \varphi _{0}\left( {\widetilde{N}},u_{{\widetilde{N}}},{\widetilde{q}}\right) =\varphi _{0}\left( {\widetilde{N}},u_{{\widetilde{N}}},{\widetilde{s}}\right) . \end{aligned}$$

Substituting this equality in (6), we get the result we require. $\square$

Proof of Lemma 7

By 3DPI, without loss of generality, we can suppose that $\measuredangle aed=\measuredangle aec=\pi /2$. Let $a^{\prime }=e+(e-a)$, $b^{\prime }=b+2(e-a)$, $c^{\prime }=c+(b-a)$, $d^{\prime }=d+(b-a)$ and $e^{\prime }=e+(b-a)$ (Fig. 14), and consider ${\widetilde{N}} =\left\{ 0,1,2,3,4,5,6,7\right\}$, ${\widetilde{p}}\in C ^{{\widetilde{N}} }$ such that ${\widetilde{p}}_{\left\{ 0,1,2,3\right\} }=p_{\left\{ 0,1,2,3\right\} }$, $\left( {\widetilde{p}}_{4},{\widetilde{p}}_{5},{\widetilde{p}} _{6},{\widetilde{p}}_{7}\right) =\left( a^{\prime },b^{\prime },c^{\prime },d^{\prime }\right)$, and ${\widetilde{q}}\in C ^{{\widetilde{N}} \backslash \{1,6\}}$ such that ${\widetilde{p}}_{{\widetilde{N}}\backslash \{1,6\}}={\widetilde{q}}_{{\widetilde{N}}\backslash \{1,6\}}$.

Fig. 14

In this figure a′, b′, c′, d′ and e′ are added

By DP and 3DPI,

$$\begin{aligned} \varphi _{2}\left( {\widetilde{N}},u_{{\widetilde{N}}},{\widetilde{p}}\right) =\varphi _{2}\left( {\widetilde{N}}\backslash \{1,6\},u_{{\widetilde{N}} \backslash \{1,6\}},{\widetilde{q}}\right) . \end{aligned}$$

(13)

RI and AN imply

$$\begin{aligned} \varphi _{0}\left( {\widetilde{N}},u_{{\widetilde{N}}},{\widetilde{p}}\right) =\varphi _{3}\left( {\widetilde{N}},u_{{\widetilde{N}}},{\widetilde{p}}\right) =\varphi _{4}\left( {\widetilde{N}},u_{{\widetilde{N}}},{\widetilde{p}}\right) =\varphi _{5}\left( {\widetilde{N}},u_{{\widetilde{N}}},{\widetilde{p}}\right) \end{aligned}$$

and

$$\begin{aligned} \varphi _{2}\left( {\widetilde{N}},u_{{\widetilde{N}}},{\widetilde{p}}\right) =\varphi _{7}\left( {\widetilde{N}},u_{{\widetilde{N}}},{\widetilde{p}}\right) ,\varphi _{1}\left( {\widetilde{N}},u_{{\widetilde{N}}},{\widetilde{p}}\right) =\varphi _{6}\left( {\widetilde{N}},u_{{\widetilde{N}}},{\widetilde{p}}\right) , \end{aligned}$$

and therefore, taking the definition of a power index into account,

$$\begin{aligned} 2\varphi _{2}\left( {\widetilde{N}},u_{{\widetilde{N}}},{\widetilde{p}}\right) =1-4\varphi _{0}\left( {\widetilde{N}},u_{{\widetilde{N}}},{\widetilde{p}}\right) -2\varphi _{1}\left( {\widetilde{N}},u_{{\widetilde{N}}},{\widetilde{p}}\right) . \end{aligned}$$

(14)

RI and AN imply $\varphi _{2}\left( {\widetilde{N}}\backslash \{1,6\},u_{ {\widetilde{N}}\backslash \{1,6\}},{\widetilde{q}}\right) =\varphi _{7}\left( {\widetilde{N}}\backslash \{1,6\},u_{{\widetilde{N}}\backslash \{1,6\}}, {\widetilde{q}}\right)$ and

$$\begin{aligned} \varphi _{0}\left( {\widetilde{N}}\backslash \{1,6\},u_{{\widetilde{N}} \backslash \{1,6\}},{\widetilde{q}}\right)= & {} \varphi _{3}\left( {\widetilde{N}} \backslash \{1,6\},u_{{\widetilde{N}}\backslash \{1,6\}},{\widetilde{q}}\right) \\= & {} \varphi _{4}\left( {\widetilde{N}}\backslash \{1,6\},u_{{\widetilde{N}} \backslash \{1,6\}},{\widetilde{q}}\right) =\varphi _{5}\left( {\widetilde{N}} \backslash \{1,6\},u_{{\widetilde{N}}\backslash \{1,6\}},{\widetilde{q}}\right) , \end{aligned}$$

and hence, taking the definition of a power index into account,

$$\begin{aligned} 2\varphi _{2}\left( {\widetilde{N}}\backslash \{1,6\},u_{{\widetilde{N}} \backslash \{1,6\}},{\widetilde{q}}\right) =1-4\varphi _{0}\left( {\widetilde{N}} \backslash \{1,6\},u_{{\widetilde{N}}\backslash \{1,6\}},{\widetilde{q}}\right) . \end{aligned}$$

(15)

Substituting ( 14) and ( 15) in ( 13) , we get:

$$\begin{aligned} -2\varphi _{0}\left( {\widetilde{N}},u_{{\widetilde{N}}},{\widetilde{p}}\right) -\varphi _{1}\left( {\widetilde{N}},u_{{\widetilde{N}}},{\widetilde{p}}\right) =-2\varphi _{0}\left( {\widetilde{N}}\backslash \{1,6\},u_{{\widetilde{N}} \backslash \{1,6\}},{\widetilde{q}}\right) . \end{aligned}$$

Since $\varphi _{1}\left( {\widetilde{N}},u_{{\widetilde{N}}},{\widetilde{p}} \right) \ge 0$, then $\varphi _{0}\left( {\widetilde{N}},u_{{\widetilde{N}}}, {\widetilde{p}}\right) \le \varphi _{0}\left( {\widetilde{N}}\backslash \{1,6\},u_{{\widetilde{N}}\backslash \{1,6\}},{\widetilde{q}}\right)$. By DP and 3DPI, it holds $\varphi _{0}\left( {\widetilde{N}},u_{{\widetilde{N}}}, {\widetilde{p}}\right) =\varphi _{0}\left( N,u_{S},p\right)$ and $\varphi _{0}\left( {\widetilde{N}}\backslash \{1,6\},u_{{\widetilde{N}}\backslash \{1,6\}},{\widetilde{q}}\right) =\varphi _{0}\left( N,u_{S},q\right)$, and hence $\varphi _{0}\left( N,u_{S},p\right) \le \varphi _{0}\left( N,u_{S},q\right)$. $\square$