Decomposition of interaction indices: alternative interpretations of cardinal–probabilistic interaction indices

Abstract

In cooperative game theory, the concept of interaction index is an extension of the concept of one-point solution that takes into account interactions among players. In this paper, we focus on cardinal–probabilistic interaction indices that generalize the class of semivalues. We provide two types of decompositions. With the first one, a cardinal–probabilistic interaction index for a given coalition equals the difference between its external interaction index and a weighted sum of the individual impact of the remaining players on the interaction index of the considered coalition. The second decomposition, based on the notion of the "decomposer", splits an interaction index into a direct part, the decomposer, which measures the interaction in the coalition considered, and an indirect part, which indicates how all remaining players individually affect the interaction of the coalition considered. We propose alternative characterizations of the cardinal–probabilistic interaction indices.

Appendix: Proofs and complementary results

For any coalition $S\in 2^N\backslash \emptyset$, the $S-$derivative of a TU game can be represented by the dividends of coalitions they belong to, as shown in the following result.

Proposition 9

Given two games $v,w \in TU(N)$ and $S\in 2^N\backslash \emptyset$, the following two statements imply each other.

• (i) For all $T\subseteq N\backslash S$, we have

  $$\begin{aligned} \Delta _{S}v(T)=\Delta _{S}w(T). \end{aligned}$$

• (ii) For all $T\subseteq N\backslash S$, we have

  $$\begin{aligned} \lambda _v(S\cup T)=\lambda _w(S\cup T). \end{aligned}$$

Proof

In view of (Casajus, 2020, Appendix A) and using the induction method on the cardinally of S, we have the result. $\square$

The next proposition gives a necessary and sufficient condition under which an AID interaction index is a cardinal–probabilistic interaction index.

Proposition 10

Let $\Psi$ be an AID interaction index with the associated family of real numbers $\{f^{s}_{n}(t)\}_{t=0,..., n-s-1}$. For any non-empty coalition $S\subseteq N$, we have

$$\begin{aligned} \Psi (v,S)=\displaystyle \sum _{T\subseteq N\backslash S}P^s_n(t)\Delta _{S} v(T), \end{aligned}$$

where

$$\begin{aligned} P^s_n(t)= & {} \big [(n-s-t)f^{s}_{n}(t)-tf^{s}_{n}(t-1)\big ]q^s_n(t)+(n-s)f^{s}_{n}(0)r^s_n(t)\\{} & {} \quad +\big [1-(n-s)f^{s}_{n}(n-s-1)\big ]k^s_n(t), \end{aligned}$$

and

$q^s_n(t)=\left\{ \begin{array}{ll} 1~~~if~~ 0< t<n-s\\ 0~~~\textrm{otherwise} \end{array} \right.$; $k^s_n(t)=\left\{ \begin{array}{ll} 1 ~~~if~~ t=n-s\\ 0 ~~~\textrm{otherwise} \end{array} \right.$; $r^s_n(t)=\left\{ \begin{array}{ll} 1 ~~~~if~~ t=0\\ 0 ~~~~\textrm{otherwise} \end{array} \right.$

$\Psi$ is a cardinal–probabilistic interaction index if and only if $f^{s}_{n}(n-s-1)\le \frac{2}{n-s}$, $f^{s}_{n}(0)\ge 0$ and $f^{s}_{n}(t)\ge \frac{t}{n-t-s}f^{s}_{n}(t-1)$ with $0<t<n-s-1$.

Proof of Proposition 10

This result follows from the proof of Theorem 1 below. $\square$

Proof of Theorem 1

Grabisch and Roubens (1999a) have been proved that the following statements are equivalent.

• (i) $\Psi$ satisfies the Linearity, Dummy and Symmetry axioms.

• (ii) For any game $v\in TU(N)$, $n\ge 1$, and $S\subseteq N$, $s\ge 1$, there exists a collection of real constants $\{P^s_n(t)\}_{t=0,...,n-s}$, such that

  $$\begin{aligned} \Psi (v,S)=\displaystyle \sum _{T\subseteq N\backslash S}P^s_n(t)\Delta _{S}v(T), \end{aligned}$$

  and for any $S\nsubseteq N$, we have $\Psi (v,S)=0.$

It suffices to show that (ii) is equivalent to $\Psi$ is AID.

$\Psi {\textbf { is AID}}\Longrightarrow {\textbf {(ii)}}$ For any game $v\in TU(N)$, $n\ge 1$, if $\Psi$ is AID, then, for any $S\subseteq N$, $s\ge 1$, there exists a collection of real constants $\{f^{s}_{n}(t)\}_{t=0,...,n-s-1}$ such that,

$$\begin{aligned} \Psi (v, S)=\Psi ^{ext}(v,S)-\displaystyle \sum _{j\in N\backslash S}~\displaystyle \sum _{T\subseteq N\backslash {(S\cup j)}} f^{s}_{n}(t)\times \Delta _{S\cup j} v(T). \end{aligned}$$

We know that,

$$\begin{aligned} \Delta _{S}v(T)=\displaystyle \sum _{L\subseteq S}(-1)^{s-l}v(L\cup T). \end{aligned}$$

For any player $i\in S$, $\Delta _{S}v(T)$ can be expressed recursively as follows:

$$\begin{aligned} \Delta _{S}v(T)=\Delta _{S\backslash i}v(T\cup i)-\Delta _{S\backslash i}v(T). \\ \Psi (v, S)= & {} \Psi ^{ext}(v,S)-\displaystyle \sum _{j\in N\backslash S}~\displaystyle \sum _{T\subseteq N\backslash {S\cup j}} f^{s}_{n}(t)\times \Delta _{S\cup j} v(T)\\= & {} \Delta _{S} v(N\backslash S)-\displaystyle \sum _{j\in N\backslash S}~\displaystyle \sum _{T\subseteq N\backslash {S\cup j}} f^{s}_{n}(t)\times \left[ \Delta _{S\backslash i} v(T\cup ij) - \Delta _{S\backslash i} v(T\cup j) \right. \\{} & {} -\left. \Delta _{S\backslash i} v(T\cup i)+ \Delta _{S\backslash i} v(T) \right] \\= & {} \Delta _{S} v(N\backslash S)-a+c+b-d, \end{aligned}$$

where

$$\begin{aligned} a= & {} \displaystyle \sum _{j\in N\backslash S}~\displaystyle \sum _{T\subseteq N\backslash {(S\cup j)}}f^{s}_{n}(t)\times \Delta _{S\backslash i} v(T\cup ij) =\displaystyle \sum _{j\in N\backslash S}\displaystyle \sum _{\begin{array}{c} T\subseteq N\backslash S \\ {j\in T} \end{array} }f^{s}_{n}(t-1)\times \Delta _{S\backslash i} v(T\cup i) \\= & {} \displaystyle \sum _{\begin{array}{c} T\subseteq N\backslash S\\ { T\ne \emptyset } \end{array}}\displaystyle \sum _{j\in T}f^{s}_{n}(t-1)\times \Delta _{S\backslash i} v(T\cup i) =\displaystyle \sum _{\begin{array}{c} T\subseteq N\backslash S \\ { T\ne \emptyset } \end{array}}tf^{s}_{n}(t-1)\times \Delta _{S\backslash i} v(T\cup i)\\ b= & {} \displaystyle \sum _{j\in N\backslash S}~\displaystyle \sum _{T\subseteq N\backslash {S\cup j}}f^{s}_{n}(t)\times \Delta _{S\backslash i} v(T\cup i) =\displaystyle \sum _{j\in N\backslash S}\displaystyle \sum _{\begin{array}{c} T\subseteq N\backslash S \\ {j \notin T} \end{array} }f^{s}_{n}(t)\times \Delta _{S\backslash i} v(T\cup i) \\= & {} \displaystyle \sum _{T\subset N\backslash S}\displaystyle \sum _{j\notin T\cup S}f^{s}_{n}(t)\times \Delta _{S\backslash i} v(T\cup i) =\displaystyle \sum _{T\subset N\backslash S}(n-s-t)f^{s}_{n}(t)\times \Delta _{S\backslash i} v(T\cup i) \\ c= & {} \displaystyle \sum _{j\in N\backslash S}~\displaystyle \sum _{T\subseteq N\backslash {(S\cup j)}}f^{s}_{n}(t)\times \Delta _{S\backslash i} v(T\cup j) =\displaystyle \sum _{j\in N\backslash S}\displaystyle \sum _{\begin{array}{c} T\subseteq N\backslash S \\ {j\in T} \end{array} }f^{s}_{n}(t-1)\times \Delta _{S\backslash i} v(T)\\= & {} \displaystyle \sum _{\begin{array}{c} T\subseteq N\backslash S \\ { T\ne \emptyset } \end{array}}\displaystyle \sum _{j\in T}f^{s}_{n}(t-1)\times \Delta _{S\backslash i} v(T) =\displaystyle \sum _{\begin{array}{c} T\subseteq N\backslash S \\ { T\ne \emptyset } \end{array}}tf^{s}_{n}(t-1)\times \Delta _{S\backslash i} v(T)\\ d= & {} \displaystyle \sum _{j\in N\backslash S}~\displaystyle \sum _{T\subseteq N\backslash {S\cup j}}f^{s}_{n}(t)\times \Delta _{S\backslash i} v(T) =\displaystyle \sum _{j\in N\backslash S}\displaystyle \sum _{\begin{array}{c} T\subseteq N\backslash S \\ {j \notin T} \end{array} }f^{s}_{n}(t)\times \Delta _{S\backslash i} v(T)\\= & {} \displaystyle \sum _{T\subset N\backslash S}\displaystyle \sum _{j\notin T\cup S}f^{s}_{n}(t)\times \Delta _{S\backslash i} v(T) =\displaystyle \sum _{T\subset N\backslash S}(n-s-t)f^{s}_{n}(t)\times \Delta _{S\backslash i} v(T). \end{aligned}$$

Using the expressions of a, b, c and d, $\Psi (v, S)$ can be written as:

$$\begin{aligned} \Psi (v, S)= & {} \Delta _{S} v(N\backslash S)-\displaystyle \sum _{\begin{array}{c} T\subseteq N\backslash S \\ { T\ne \emptyset } \end{array}}tf^{s}_{n}(t-1)\times \Delta _{S\backslash i} v(T\cup i) +\displaystyle \sum _{\begin{array}{c} T\subseteq N\backslash S \\ { T\ne \emptyset } \end{array}}tf^{s}_{n}(t-1)\times \Delta _{S\backslash i} v(T) \\{} & {} +\displaystyle \sum _{T\subset N\backslash S}(n-s-t)f^{s}_{n}(t)\times \Delta _{S\backslash i} v(T\cup i)-\displaystyle \sum _{T\subset N\backslash S}(n-s-t)f^{s}_{n}(t)\times \Delta _{S\backslash i} v(T) \\= & {} \Delta _{S} v(N\backslash S)-\displaystyle \sum _{\begin{array}{c} T\subseteq N\backslash S \\ { T\ne \emptyset } \end{array}}tf^{s}_{n}(t-1)\big [\Delta _{S\backslash i} v(T\cup i)-\Delta _{S\backslash i} v(T)\big ]\\{} & {} +\displaystyle \sum _{T\subset N\backslash S}(n-s-t)f^{s}_{n}(t)\big [\Delta _{S\backslash i} v(T\cup i)-\Delta _{S\backslash i} v(T)\big ]\\= & {} \displaystyle \sum _{\begin{array}{c} T\subset N\backslash S \\ { T\ne \emptyset } \end{array}}\big [(n-s-t)f^{s}_{n}(t)-tf^{s}_{n}(t-1)\big ] \Delta _{S} v(T)+(n-s)f^{s}_{n}(0)\times \Delta _{S} v(\emptyset )\\{} & {} +\big [1-(n-s)\times f^{s}_{n}(n-s-1)\big ]\Delta _{S} V(N\backslash S) \\= & {} \displaystyle \sum _{T\subseteq N\backslash S }\big [(n-s-t)f^{s}_{n}(t)-tf^{s}_{n}(t-1)\big ]q^s_n(t) \Delta _{S} v(T)+\displaystyle \sum _{T\subseteq N\backslash S}(n-s)f^{s}_{n}(0)r^s_n(t)\Delta _{S} v(T)\\{} & {} +\displaystyle \sum _{T\subseteq N\backslash S}\big [1-(n-s)\times f^{s}_{n}(n-s-1)\big ]k^s_n(t)\Delta _{S} v(T)\\= & {} \displaystyle \sum _{T\subseteq N\backslash S} P^s_n(t)\Delta _{S} v(T) \end{aligned}$$

where $q^s_n(t)=\left\{ \begin{array}{ll} 1~~~if~~ 0< t<n-s\\ 0~~~\textrm{otherwise} \end{array} \right.$; $k^s_n(t)=\left\{ \begin{array}{ll} 1 ~~~if~~ t=n-s\\ 0 ~~~\textrm{otherwise} \end{array} \right.$; $r^s_n(t)=\left\{ \begin{array}{ll} 1 ~~~~if~~ t=0\\ 0 ~~~~\textrm{otherwise}\end{array} \right.$ and

$$\begin{aligned} P^s_n(t)=\big [(n-s-t)f^{s}_{n}(t)-tf^{s}_{n}(t-1)\big ]q^s_n(t)+(n-s)f^{s}_{n}(0)r^s_n(t)+\big [1-(n-s)f^{s}_{n}(n-s-1)\big ]k^s_n(t). \end{aligned}$$

Finally, for all $S\nsubseteq N$, for all $T\subseteq N\backslash S$, we have $\Delta _{S} v(T)=0$ (Fujimoto et al. 2006, page 76). We can deduce that $\Psi (v,S)=0$.

$\Psi {\textbf {is AID}}\Longleftarrow {\textbf {(ii)}}$ We suppose that, for any game $v\in TU(N)$, $n\ge 1$, and $S\subseteq N$, $s\ge 1$, there exists a collection of real constants $\{P^s_n(t)\}_{t=0,...,n-s}$, such that,

$$\begin{aligned} \Psi (v,S)=\displaystyle \sum _{T\subseteq N\backslash S}P^s_n(t)\Delta _{S} v(T). \end{aligned}$$

We set,

$$\begin{aligned} f^{s}_{n}(t)=\left\{ \begin{array}{llll} \frac{ P^s_n(0)}{n-s} ~~~~~~~~~~~~if~~ t=0\\ \\ \frac{1-P^s_n(n-s)}{n-s}~~~~~~~~if~~ t=n-s-1\\ \\ \frac{t \times f^{s}_{n}(t-1)+P^s_n(t)}{n-t-s} ~~~~otherwise \end{array} \right. . \end{aligned}$$

$P^s_n(t)$ can be expressed as follows:

$$\begin{aligned} P^s_n(t)=\big [(n-s-t)f^{s}_{n}(t)-tf^{s}_{n}(t-1)\big ]q^s_n(t)+(n-s)f^{s}_{n}(0)r^s_n(t)+\big [1-(n-s)f^{s}_{n}(n-s-1)\big ]k^s_n(t). \end{aligned}$$

Finally, using the reverse of the first part of this proof, we can deduce that $\Psi$ is an AID interaction index.

$f^{s}_{n}(t)$ is simply given as follows:

$$\begin{aligned} f^{s}_{n}(t)=\left\{ \begin{array}{lll} \frac{ P^s_n(0)}{n-s} ~~~~~~~~~~~~if~~ t=0\\ \\ \frac{t \times f^{s}_{n}(t-1)+P^s_n(t)}{n-t-s} ~~~~otherwise \end{array} \right. \end{aligned}$$

since by linearity, everything is determined based on dirac games ($e_K (S) = 1$ if $S = K$ and 0 otherwise).

$$\begin{aligned} \Psi (e_N,S)=P^s_n(n-s)=1-(n-s)f^{s}_{n}(n-s-1) \end{aligned}$$

and then,

$$\begin{aligned} f^{s}_{n}(n-s-1)=\frac{1-P^s_n(n-s)}{n-s}. \end{aligned}$$

We conclude that, an interaction index is AID if and only if it satisfies the three axioms: Linearity, Dummy and Symmetry. $\square$

Proof of Proposition 2

• (i) The Shapley interaction index is AID and the associated coefficient is $f^{s}_{n}(t)= \frac{(t+1)!(n-t-s-1)!}{(n-s+1)!}$, with $0\le t\le n-s-1$. Moreover,

  $$\begin{aligned} \displaystyle \sum _{t=0}^{n-s-1}\left( ^{n-s-1}_{~~t}\right) 2f^{s}_{n}(t)= & {} 2\displaystyle \sum _{t=0}^{n-s-1}\left( ^{n-s-1}_{~~t}\right) \frac{(t+1)!(n-t-s-1)!}{(n-s+1)!}\\= & {} \frac{2}{(n-s+1)(n-s)}\displaystyle \sum _{t=0}^{n-s-1}(t+1)\\= & {} \frac{2}{(n-s+1)(n-s)}\bigg [ n-s+\frac{(n-s)(n-s-1)}{2}\bigg ] =1. \end{aligned}$$

• (ii) The Banzhaf interaction index is AID and the associated coefficient is

  $$\begin{aligned} f^{s}_{n}(t)=\left\{ \begin{array}{ll} \frac{1}{(n-s)2^{n-s}} ~~~~~~~~~~~~if~~ t=0\\ \\ \frac{t\times f^{s}_{n}(t-1)\times 2^{n-s}+1}{(n-t-s)2^{n-s}} ~~~~\textrm{otherwise} \end{array} \right. . \end{aligned}$$

  Moreover,

  $$\begin{aligned}{} & {} \displaystyle \sum _{t=0}^{n-s-1}\left( ^{n-s-1}_{~~t}\right) 2f^{s}_{n}(t)\\{} & {} \quad =\frac{1}{(n-s)2^{n-s-1}}+\displaystyle \sum _{t=1}^{n-s-1}\left( ^{n-s-1}_{~~t}\right) \frac{t\times f^{s}_{n}(t-1)\times 2^{n-s}+1}{(n-t-s)2^{n-s-1}}\\{} & {} \quad = \frac{1}{(n-s)2^{n-s-1}}+\displaystyle \sum _{t=1}^{n-s-1}\frac{\left( ^{n-s-1}_{~~t}\right) }{(n-t-s)2^{n-s-1}}+2\displaystyle \sum _{t=1}^{n-s-1}\left( ^{n-s-1}_{t-1}\right) f^{s}_{n}(t-1)\\{} & {} \quad = \frac{1}{(n-s)2^{n-s-1}}+\displaystyle \sum _{t=1}^{n-s-1}\frac{\left( ^{n-s-1}_{~~t}\right) }{(n-t-s)2^{n-s-1}}+2\displaystyle \sum _{t=0}^{n-s-2}\left( ^{n-s-1}_{~~t}\right) f^{s}_{n}(t)\\{} & {} \quad =\frac{1}{(n-s)2^{n-s-1}}+\displaystyle \sum _{t=1}^{n-s-1}\frac{\left( ^{n-s-1}_{~~t}\right) }{(n-t-s)2^{n-s-1}}+\frac{1}{(n-s)2^{n-s-1}}+\displaystyle \sum _{t=1}^{n-s-2}\frac{\left( ^{n-s-1}_{~~t}\right) }{(n-t-s)2^{n-s-1}}\\{} & {} \qquad +2\displaystyle \sum _{t=0}^{n-s-3}\left( ^{n-s-1}_{~~t}\right) f^{s}_{n}(t)\\{} & {} \quad =\frac{1}{(n-s)2^{n-s-1}}+\displaystyle \sum _{t=1}^{n-s-1}\frac{\left( ^{n-s-1}_{~~t}\right) }{(n-t-s)2^{n-s-1}}+\frac{1}{(n-s)2^{n-s-1}}+\displaystyle \sum _{t=1}^{n-s-2}\frac{\left( ^{n-s-1}_{~~t}\right) }{(n-t-s)2^{n-s-1}}\\{} & {} \qquad +\cdots +\frac{1}{(n-s)2^{n-s-1}}+\displaystyle \sum _{t=1}^{1}\frac{\left( ^{n-s-1}_{~~t}\right) }{(n-t-s)2^{n-s-1}}+\frac{1}{(n-s)2^{n-s-1}}\\{} & {} \quad =\left( ^{n-s-1}_{~~0}\right) \frac{1}{2^{n-s-1}}+\left( ^{n-s-1}_{~~1}\right) \frac{1}{2^{n-s-1}}+\left( ^{n-s-1}_{~~2}\right) \frac{1}{2^{n-s-1}}+\cdots +\left( ^{n-s-1}_{n-s-1}\right) \frac{1}{2^{n-s-1}}\\{} & {} \quad =\frac{1}{2^{n-s-1}}\displaystyle \sum _{t=0}^{n-s-1}\left( ^{n-s-1}_{~~t}\right) \\{} & {} \quad =1. \end{aligned}$$

$\square$

Proof of Proposition 4

Following Proposition 3, if a one-point solution $\Psi$ is CH-decomposable, then there exists a unique CH-decomposer $\varphi$ that is itself CH-decomposable. For any game $v\in TU(N)$, i and j two players, Theorem 4 (page 40) of Casajus and Huettner (2018) shows that $\left[ \varphi (v, i)-\varphi (v^{N\backslash j}, i)\right] =\left[ \varphi (v, j)-\varphi (v^{N\backslash i}, j)\right] .$ Thus, $\Psi (v,i)=\varphi (v,i) +\displaystyle \sum _{j\in N\backslash i}\left[ \varphi (v, i)-\varphi (v^{N\backslash j}, i)\right]$. $\Psi$ is I-decomposable and $\varphi$ is an I-decomposer.

It is easier to show by induction on n that if a one-point solution is I-decomposable, then its I-decomposer is unique (see the proof of Proposition 5 below). $\square$

Proof of Proposition 5

Let $\varphi$ and $\varphi ^{'}$ be two C-decomposers of $\Psi$. Given a game $v \in TU (N)$ and for all $S\in 2^N\backslash \emptyset$ a coalition, we will show by induction on the cardinality of n that $\varphi (v,S)=\varphi ^{'}(v,S).$

Induction basis: given a game $v \in TU (N)$ and for all $S\in 2^N\backslash \emptyset$, if $n=1$ then, $\Psi (v,S)=\varphi (v,S)=\varphi ^{'}(v,S)$.

Induction hypothesis: given a game $v \in TU (N)$ and for all $S\in 2^N\backslash \emptyset$ with $2\le n\le k$ $(k\in {\mathbb {N}})$, we suppose that $\varphi (v,S)=\varphi ^{'}(v,S)$

Induction step: given a game $v \in TU (N)$ with $n=k+1$ and for all $S\in 2^N\backslash \emptyset$,

$$\begin{aligned} \Psi (v,S)= & {} \varphi (v,S) +\displaystyle \sum _{j\in N\backslash S}\left[ \varphi (v, S)-\varphi (v^{N\backslash j}, S)\right] \\= & {} \varphi ^{'}(v,S) +\displaystyle \sum _{j\in N\backslash S}\left[ \varphi ^{'}(v^N, S)-\varphi ^{'}(v^{N\backslash j}, S)\right] .\\ (n-s+1)\varphi (v,S) -\displaystyle \sum _{j\in N\backslash S}\varphi (v^{N\backslash j}, S)= & {} (n-s+1)\varphi ^{'}(v,S) -\displaystyle \sum _{j\in N\backslash S}\varphi ^{'}(v^{N\backslash j}, S) \\ (n-s+1)\varphi (v,S)= & {} (n-s+1)\varphi ^{'}(v,S) \\ \varphi (v,S)= & {} \varphi ^{'}(v,S). \end{aligned}$$

$\square$

Lemma 1

An interaction index $\Psi$ satisfies the Recursive axiom if and only if $\Psi$ is defined as follows.

For every game $v \in TU (N)$ with $n\ge 2$, $S\subseteq N$ a coalition of N with $2 \le s\le n$ and any player $i\in S,$ we have

$$\begin{aligned} \Psi (v,S)=\displaystyle \sum _{\beta =0}^{s-1}(-1)^{\beta }\displaystyle \sum _{\begin{array}{c} T\subseteq S\backslash i \\ {t=\beta } \end{array}}~\Psi \big (v^{N\backslash (S\backslash i)}_{\cup ((S\backslash i)\backslash T)}, i\big ). \end{aligned}$$

Proof of Lemma 1

$\Longleftarrow )$ We will show that the interaction index $\Psi$ satisfies the Recursive axiom.

For all game $v \in TU (N)$ with $n\ge 2$, $S\subseteq N$ a coalition of N with $2 \le s\le n$ and any player $i\in S.$ W.l.o.g. we suppose that $S:=\{i_1,i_2,...,i_s\}$ and $i=i_1$. We have,

$$\begin{aligned} \Psi (v^{N\backslash i_s}_{\cup i_s},S\backslash i_s)=\displaystyle \sum _{\beta =0}^{s-2}(-1)^{\beta }\displaystyle \sum _{\begin{array}{c} T\subseteq (S\backslash i_1)\backslash i_s \\ {t=\beta } \end{array}}~\Psi \big (v^{N\backslash (S\backslash i_1)}_{\cup (S\backslash i_1)\backslash T}, i_1\big ) \end{aligned}$$

and

$$\begin{aligned} \Psi (v^{N\backslash i_s},S\backslash i_s)=\displaystyle \sum _{\beta =0}^{s-2}(-1)^{\beta }\displaystyle \sum _{\begin{array}{c} T\subseteq (S\backslash i_1)\backslash i_s \\ {t=\beta } \end{array}}~\Psi \big (v^{N\backslash (S\backslash i_1)}_{\cup ((S\backslash i_1)\backslash (T\cup i_s))}, i_1\big ). \end{aligned}$$

Furthermore,

$$\begin{aligned}\psi(v,S)= & {} \displaystyle \sum _{\beta =0}^{s-1}(-1)^{\beta }\displaystyle \sum _{\begin{array}{c} T\subseteq (S\backslash i_1)\\ {t=\beta } \end{array}}~\Psi \big (v^{N\backslash (S\backslash i_1)}_{\cup ((S\backslash i_1)\backslash T)}, i_1\big )\\= & {} \displaystyle \sum _{\beta =0}^{s-2}(-1)^{\beta }\displaystyle \sum _{\begin{array}{c} T\subset S\backslash i_1 \\ {t=\beta ;~i_s\notin T } \end{array}}~\Psi \big (v^{N\backslash (S\backslash i_1)}_{\cup ((S\backslash i_1)\backslash T)}, i_1\big ) +\displaystyle \sum _{\beta =1}^{s-1}(-1)^{\beta }\displaystyle \sum _{\begin{array}{c} T\subseteq S\backslash i_1 \\ {t=\beta };~ i_s\in T \end{array}}~\Psi \big (v^{N\backslash (S\backslash i_1)}_{\cup ((S\backslash i_1)\backslash T)}, i_1\big ) \\= & {} \displaystyle \sum _{\beta =0}^{s-2}(-1)^{\beta }\displaystyle \sum _{\begin{array}{c} T\subseteq (S\backslash i_1)\backslash i_s \\ {t=\beta } \end{array}}~\Psi \big (v^{N\backslash (S\backslash i_1)}_{\cup ((S\backslash i_1)\backslash T)}, i_1\big )+\displaystyle \sum _{\beta =1}^{s-1}(-1)^{\beta }\displaystyle \sum _{\begin{array}{c} T\subseteq (S\backslash i_1)\backslash i_s \\ {t=\beta -1} \end{array}}~\Psi \big (v^{N\backslash (S\backslash i_1)}_{\cup ((S\backslash i_1)\backslash (T\cup i_s))}, i_1\big ) \\= & {} \displaystyle \sum _{\beta =0}^{s-2}(-1)^{\beta }\displaystyle \sum _{\begin{array}{c} T\subseteq (S\backslash i_1)\backslash i_s\\ {t=\beta } \end{array}}~\Psi \big (v^{N\backslash (S\backslash i_1)}_{\cup ((S\backslash i_1)\backslash T)}, i_1\big )+\displaystyle \sum _{\beta =0}^{s-2}(-1)^{\beta +1}\displaystyle \sum _{\begin{array}{c} T\subseteq (S\backslash i_1)\backslash i_s \\ {t=\beta } \end{array}}~\Psi \big (v^{N\backslash (S\backslash i_1)}_{\cup ((S\backslash i_1)\backslash (T\cup i_s))}, i_1\big )\\= & {} \displaystyle \sum _{\beta =0}^{s-2}(-1)^{\beta }\displaystyle \sum _{\begin{array}{c} T\subseteq (S\backslash i_1)\backslash i_s \\ {t=\beta } \end{array}}~\Psi \big (v^{N\backslash (S\backslash i_1)}_{\cup ((S\backslash i_1)\backslash T)}, i_1\big )-\displaystyle \sum _{\beta =0}^{s-2}(-1)^{\beta }\displaystyle \sum _{\begin{array}{c} T\subseteq (S\backslash i_1)\backslash i_s \\ {t=\beta } \end{array}}~\Psi \big (v^{N\backslash (S\backslash i_1)}_{\cup ((S\backslash i_1)\backslash (T\cup i_s))}, i_1\big )\\= & {} \Psi (v^{N\backslash i_s}_{\cup i_s},S\backslash i_s)-\Psi (v^{N\backslash i_s},S\backslash i_s). \end{aligned}$$

$\Psi$ satisfies the Recursive axiom.

$\Longrightarrow )$ We suppose that an interaction index $\Psi$ satisfies the Recursive axiom. Let us show by induction on the cardinality of the coalition that $\Psi$ is defined as follows:

For all game $v \in TU (N)$ with $n\ge 2$ and $S\subseteq N$ a coalition of N $(2 \le s\le n)$. W.l.o.g., we suppose that$\{i_1,i_2,...,i_s\}$.

$$\begin{aligned} \Psi (v,S)=\displaystyle \sum _{\beta =0}^{s-1}(-1)^{\beta }\displaystyle \sum _{\begin{array}{c} T\subseteq (S\backslash i_1) \\ {t=\beta } \end{array}}~\Psi \big (v^{N\backslash (S\backslash i_1)}_{\cup ((S\backslash i_1)\backslash T)}, i_1\big ). \end{aligned}$$

Induction basis: for all game $v \in TU (N)$ and S any coalition of N containing two players. Let $i_1$ and $i_2$ be these two players. Since $\Psi$ satisfies the Recursive axiom, we have

$$\begin{aligned} \Psi (v,S)= & {} \Psi (v^{N\backslash i_2}_{\cup i_2}, i_1)-\Psi (v^{N\backslash i_2}, i_1) =\displaystyle \sum _{\beta =0}^{2-1}(-1)^{\beta }\displaystyle \sum _{\begin{array}{c} T\subseteq i_2 \\ {t=\beta } \end{array}}~\Psi \big (v^{N\backslash i_2}_{\cup (i_2\backslash T)}, i_1\big ). \end{aligned}$$

Induction hypothesis: for all game $v \in TU (N)$ with $n\ge 2$, let $k\in {\mathbb {N}}$ be any integer such that $k\le n$ and let $S=\{i_1,i_2,...,i_s\}$ be any coalition of N with $2<s\le k$. We suppose that

$$\begin{aligned} \Psi (v,S)=\displaystyle \sum _{\beta =0}^{s-1}(-1)^{\beta }\displaystyle \sum _{\begin{array}{c} T\subseteq (S\backslash i_1) \\ {t=\beta } \end{array}}~\Psi \big (v^{N\backslash (S\backslash i_1)}_{\cup ((S\backslash i_1)\backslash T)}, i_1\big ). \end{aligned}$$

Induction step: for all game $v \in TU (N)$ with $n\ge 2$, $n>k$; $S:=\{i_1,i_2,...,i_s\}$ any coalition of N with $s=k+1$. Let us show that,

$$\begin{aligned} \Psi (v,S)=\displaystyle \sum _{\beta =0}^{s-1}(-1)^{\beta }\displaystyle \sum _{\begin{array}{c} T\subseteq (S\backslash i_1) \\ {t=\beta } \end{array}}~\Psi \big (v^{N\backslash (S\backslash i_1)}_{\cup ((S\backslash i_1)\backslash T)}, i_1\big ). \end{aligned}$$

$$\begin{aligned} \Psi (v,S)= & {} \Psi (v^{N\backslash i_s}_{\cup i_s},S\backslash i_s)-\Psi (v^{N\backslash i_s},S\backslash i_s)\\= & {} \displaystyle \sum _{\beta =0}^{s-2}(-1)^{\beta }\displaystyle \sum _{\begin{array}{c} T\subseteq (S\backslash i_1)\backslash i_s \\ {t=\beta } \end{array}}~\Psi \big (v^{N\backslash (S\backslash i_1)}_{\cup ((S\backslash i_1)\backslash T)}, i_1\big )-\displaystyle \sum _{\beta =0}^{s-2}(-1)^{\beta }\displaystyle \sum _{\begin{array}{c} T\subseteq (S\backslash i_1)\backslash i_s \\ {t=\beta } \end{array}}~\Psi \big (v^{N\backslash (S\backslash i_1)}_{\cup ((S\backslash i_1)\backslash (T\cup i_s))}, i_1\big )\\= & {} \displaystyle \sum _{\beta =0}^{s-1}(-1)^{\beta }\displaystyle \sum _{\begin{array}{c} T\subseteq S\backslash i_1 \\ {t=\beta } \end{array}}~\Psi \big (v^{N\backslash (S\backslash i_1)}_{\cup ((S\backslash i_1)\backslash T)}, i_1\big ). \end{aligned}$$

$\square$

Proof of Theorem 2

Let $\Psi$ be a CH-decomposable interaction index. From Proposition 4, the I-decomposer is unique and coincides with the unique CH-decomposable CH-decomposer. Using Proposition 3, the I-decomposer is the interaction index $\varphi$ given by: for all game $v\in TU (N)$ and for all $j\in N$,

$$\begin{aligned} \varphi (v, j)=\displaystyle \sum _{\begin{array}{c} T\subseteq N \\ {j\in T} \end{array}}\frac{\lambda _{v^{\Psi }}(T) }{t^{2}}. \end{aligned}$$

(A1)

Let us show by induction that the unique C-decomposer of the interaction index $\Psi$ is the interaction index $\Psi ^{d}$ defined as follows:

For all game $v \in TU (N)$, for all coalition $S\in 2^N\backslash \emptyset$ and for any player $i\in S$,

$$\begin{aligned} \Psi ^{d}(v,S)=\left\{ \begin{array}{lll}\displaystyle \sum _{\beta =0}^{s-1}(-1)^{\beta }\displaystyle \sum _{\begin{array}{c} T\subseteq S\backslash i \\ {t=\beta } \end{array}}~\varphi \big (v^{N\backslash (S\backslash i)}_{\cup ((S\backslash i)\backslash T)}, i\big )~~~~~~~~if~~ s\ge 2\\ \\ \varphi (v,i)~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~if~~ s=1 \end{array} \right. . \end{aligned}$$

From Lemma 1, $\Psi ^{d}$ satisfies the Recursive axiom when the coalition contains more than one player.

Induction basis: for all game $v \in TU (N)$, $S\in 2^N\backslash \emptyset$, if $s=1$ ( that is $S=i$) then $\Psi$ is I-decomposable from the hypothesis and the I-decomposer is the interaction index $\varphi$.

$$\begin{aligned} \Psi ^{d}(v, i)=\varphi (v, i). \end{aligned}$$

Induction hypothesis: for all game $v \in TU (N)$, let $k\in {\mathbb {N}}$ be any integer such that $k\le n$ and let S be any coalition of N with $2\le s\le k$. We suppose that:

$$\begin{aligned} \Psi (v,S)=\Psi ^{d}(v,S)+\displaystyle \sum _{j\in N\backslash S}\left[ \Psi ^{d}(v,S)-\Psi ^{d}(v^{N\backslash j},S)\right] . \end{aligned}$$

Induction step: for all game $v \in TU (N)$, $n> k$ and $S\in 2^N$ such that $s=k+1$. Let us show that:

$$\begin{aligned} \Psi (v,S)=\Psi ^{d}(v,S)+\displaystyle \sum _{j\in N\backslash S}\left[ \Psi ^{d}(v,S)-\Psi ^{d}(v^{N\backslash j},S)\right] . \end{aligned}$$

Let $i\in S$. Since $\Psi$ satisfies the Recursive axiom, we have

$$\begin{aligned} \Psi (v,S)= & {} \Psi (v^{N\backslash i}_{\cup i},S\backslash i)-\Psi (v^{N\backslash i},S\backslash i)\\= & {} \Psi ^{d}(v^{N\backslash i}_{\cup i},S\backslash i)+\displaystyle \sum _{j\in N\backslash S}\left[ \Psi ^{d}(v^{N\backslash i}_{\cup i},S\backslash i)-\Psi ^{d}(v^{N\backslash ij}_{\cup i},S\backslash i)\right] \\{} & {} -\Psi ^{d}(v^{N\backslash i},S\backslash i)-\displaystyle \sum _{j\in N\backslash S}\left[ \Psi ^{d}(v^{N\backslash i},S\backslash i)-\Psi _{d}(v^{N\backslash ij},S\backslash i)\right] \\= & {} \Psi ^{d}(v^{N\backslash i}_{\cup i},S\backslash i)-\Psi ^{d}(v^{N\backslash i},S\backslash i)+\displaystyle \sum _{j\in N\backslash S}\left[ \left( \Psi ^{d}(v^{N\backslash i}_{\cup i},S\backslash i)\right. \right. \\{} & {} \left. \left. -\Psi ^{d}(v^{N\backslash i},S\backslash i)\right) -\left( \Psi ^{d}(v^{N\backslash ij}_{\cup i},S\backslash i)-\Psi ^{d}(v^{N\backslash ij},S\backslash i)\right) \right] \\= & {} \Psi ^{d}(v,S)+\displaystyle \sum _{j\in N\backslash S}\left[ \Psi ^{d}(v,S)-\Psi ^{d}(v^{N\backslash j},S)\right] . \end{aligned}$$

Since $\Psi ^{d}$ satisfies the Recursive axiom. $\Psi ^{d}$ is a C-decomposer of the interaction index $\Psi$. $\Psi$ is C-decomposable and from Proposition 5, the C-decomposer is unique. $\square$

Proof of Proposition 7

Let $v \in TU(N)$ ($n\ge 2$) be a game on N and $S\in 2^N$ with $s\ge 2$. Let $j\in S$,

$$\begin{aligned} \begin{array}{lll} \Psi ^{Ch}(v,S)&{}=&{}\displaystyle \sum _{T\subseteq N\backslash S}\frac{s(s+t-1)!(n-t-s)!}{n!}\Delta _{S}v(T)\\ &{} =&{}\displaystyle \sum _{T\subseteq N\backslash S}\frac{s(s+t-1)!(n-t-s)!}{n!}\Delta _{S\backslash j}v(T\cup j)- \displaystyle \sum _{T\subseteq N\backslash S}\frac{s(s+t-1)!(n-t-s)!}{n!}\Delta _{S\backslash j}v(T)\\ &{}=&{}\displaystyle \sum _{\begin{array}{c} T\subseteq N\backslash (S\backslash j) \\ {j\in T} \end{array}}\frac{s(s+t-2)!(n-t-s+1)!}{n!}\Delta _{S\backslash j}v(T)- \displaystyle \sum _{T\subseteq N\backslash S}\frac{s(s+t-1)!(n-t-s)!}{n!}\Delta _{S\backslash j}v(T)\\ &{}=&{}\displaystyle \sum _{T\subseteq N\backslash (S\backslash j)}\frac{s(s+t-2)!(n-t-s+1)!}{n!}\Delta _{S\backslash j}v(T)\\ &{}-&{} \displaystyle \sum _{\begin{array}{c} T\subseteq N\backslash (S\backslash j) \\ {j\notin S} \end{array}}\frac{s(s+t-2)!(n-t-s+1)!}{n!}\Delta _{S\backslash j}v(T) - \displaystyle \sum _{T\subseteq N\backslash S}\frac{s(s+t-1)!(n-t-s)!}{n!}\Delta _{S\backslash j}v(T)\\ &{}=&{}\displaystyle \sum _{T\subseteq N\backslash (S\backslash j)}\frac{s(s+t-2)!(n-t-s+1)!}{n!}\Delta _{S\backslash j}v(T)\\ &{}&{}-\displaystyle \sum _{T\subseteq N\backslash S}\frac{s(s+t-2)!(n-t-s)!(n-t-s+1+s+t-1)}{n!}\Delta _{S\backslash j}v(T)\\ &{}=&{} s\displaystyle \sum _{T\subseteq N\backslash (S\backslash j)}\frac{(s+t-2)!(n-t-s+1)!}{n!}\Delta _{S\backslash j}v(T)- s\displaystyle \sum _{T\subseteq N\backslash S}\frac{(s+t-2)!(n-t-s)!}{(n-1)!}\Delta _{S\backslash j}v(T)\\ &{}=&{}\displaystyle \frac{s}{s-1}\left[ \Psi ^{Ch}(v,S\backslash j)-\Psi ^{Ch}(v^{N\backslash j},S\backslash j)\right] . \end{array} \end{aligned}$$

For any $i\in S$, we repeatedly apply this process and we obtain,

$$\begin{aligned} \Psi ^{Ch}(v,S)=\left( \displaystyle \prod ^{s-2}_{l=0}\frac{s-l}{s-l-1}\right) \displaystyle \sum _{\beta =0}^{s-1}(-1)^{\beta }\displaystyle \sum _{\begin{array}{c} T\subseteq S\backslash i \\ {t=s-\beta -1} \end{array}}~\Psi ^{Ch}(v^{N\backslash ((S\backslash i)\backslash T)}, i). \end{aligned}$$

Finally, by induction on the cardinality of S, we obtain

$$\begin{aligned} \Psi ^{Ch}(v,S)= s\Psi ^{dc}(v,S)+\displaystyle \sum _{j\in N\backslash S}\left[ \Psi ^{dc}(v,S)-\Psi ^{dc}(v^{N\backslash j},S)\right] , \end{aligned}$$

where

$$\begin{aligned} \Psi ^{dc}(v,S)=\left\{ \begin{array}{lll}\left( \displaystyle \prod ^{s-2}_{l=0}\frac{s-l}{s-l-1}\right) \displaystyle \sum _{\beta =0}^{s-1}(-1)^{\beta }\displaystyle \sum _{\begin{array}{c} T\subseteq S\backslash i \\ {t=s-\beta -1} \end{array}}~\varphi (v^{N\backslash ((S\backslash i)\backslash T)}, i)~~~~~~~~if~~ s\ge 2\\ \\ \varphi (v,i)~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~if~~ s=1 \end{array} \right. . \end{aligned}$$

$\varphi$ is the unique I-decomposer of the Shapley solution and is given by Eq. (A1). The chaining interaction index is not C-decomposable. $\square$

Proof of Proposition 8

Let $\Psi ^c$ be a cardinal–probabilistic interaction index applied to a game $v\in TU (N)$. From Proposition 1, for any non-empty coalition $S\subseteq N$,

$$\begin{aligned} \Psi ^c(v,S)=\Psi ^{ext}(v,S)-\displaystyle \sum _{j\in N\backslash S}~\displaystyle \sum _{T\subseteq N\backslash {S\cup j}} f^{s}_{n}(t)\times \Delta _{S\cup j} v(T) \end{aligned}$$

where $f^{s}_{n}(t)$ is given by

$$\begin{aligned} f^{s}_{n}(t)=\left\{ \begin{array}{lll} \frac{ P^s_n(0)}{n-s} ~~~~~~~~~~~~if~~ t=0\\ \frac{t \times f^{s}_{n}(t-1)+P^s_n(t)}{n-t-s} ~~~~otherwise\\ \end{array} \right. \\ \end{aligned}$$

and the set $\{P^s_n(t)\}_{t=0,...,n-s}$ are nonnegative constants associated to the cardinal–probabilistic interaction index. According to Fujimoto et al. (2006), the coefficients $P^s_n(t)$ obey the recurrence relation:

$$\begin{aligned} P^s_{n}(t)+P^s_{n}(t+1)=P^s_{n-1}(t). \end{aligned}$$

$\Psi ^c(V,S)+ \displaystyle \sum _{j\in N\backslash S}\left[ \Psi ^c(v,S)-\Psi ^c(v^{N\backslash j},S)\right]$

$$\begin{aligned}= & {} \Psi ^{ext}(v,S)-\displaystyle \sum _{j\in N\backslash S}~\displaystyle \sum _{T\subseteq N\backslash {S\cup j}} f^{s}_{n}(t)\times \Delta _{S\cup j} v(T)+\displaystyle \sum _{j\in N\backslash S}~\displaystyle \sum _{T\subseteq N\backslash {S\cup j}} P^s_{n}(t+1)\Delta _{S\cup j} v(T)\\= & {} \Psi ^{ext}(v,S)-\displaystyle \sum _{j\in N\backslash S}~\displaystyle \sum _{T\subseteq N\backslash {S\cup j}} \big [f^{s}_{n}(t)-P^s_{n}(t+1)\big ]\Delta _{S\cup j} v(T)\\= & {} \Psi ^{A}(v,S). \end{aligned}$$

From Proposition 5, the C-decomposer is unique. $\square$