Decomposition-integral: unifying Choquet and the concave integrals

Abstract

This paper introduces a novel approach to integrals with respect to capacities. Any random variable is decomposed as a combination of indicators. A prespecified set of collections of events indicates which decompositions are allowed and which are not. Each allowable decomposition has a value determined by the capacity. The decomposition-integral of a random variable is defined as the highest of these values. Thus, different sets of collections induce different decomposition-integrals. It turns out that this decomposition approach unifies well-known integrals, such as Choquet, the concave and Riemann integral. Decomposition-integrals are investigated with respect to a few essential properties that emerge in economic contexts, such as concavity (uncertainty-aversion), monotonicity with respect to stochastic dominance and translation-covariance. The paper characterizes the sets of collections that induce decomposition-integrals, which respect each of these properties.

Appendix

Proof of Proposition 1

We show first that the transition from Eq. (3) to Eq. (4) is true. Let $X$ be a variable and let $Y=\sum _{i=1}^{k}\alpha _{i}\mathbb {I}_{A_{i}}$ have the following properties: (a) it is ${\mathcal{F}^\mathrm{Ch}}$-allowable sub-decomposition of ${X}$ (satisfying $A_1\subseteq \cdots \subseteq A_k$) that achieves the maximum of the RHS of Eq. (3); (b) $\alpha _{i}>0$ for every $i=1,\ldots ,k$. We can assume that there is no other sub-decomposition that satisfies (a) and (b) and (weakly) dominates $Y$. That is, for every variable $Z\not =Y$ that satisfies (a) and (b), there is $\ell \in N$ such that $Y(\ell ) > Z(\ell )$. We show that $Y$ is actually a decomposition of $X$.

Assume, on the contrary, that there is $j\in N$ such that $Y(j)<X(j)$. If $j\in A_{i}$ for every $i=1,\ldots ,k$, then the set $\left\{ A_1,\ldots ,A_k, \{j\}\right\}$ is a chain and $Y'=Y + (X(j)-Y(j))\mathbb {I}_{\{j\}}$ satisfies (a) and (b). Moreover, $Y'\not = Y$ and $Y'(\ell ) \ge Y(\ell )$ for every $\ell \in N$, which contradicts the choice of $Y$. We may therefore assume that there is an index $i$ such that $j\not \in A_{i}$. Let $i_0$ be the smallest index that $j \not \in A_{i_0}$. The set $\left\{ A_1,\ldots , A_{i_0}\cup \{j\}, A_{i_0},A_{i_0+1},\ldots , A_k\right\}$ is a chain. Furthermore, $Y'=\sum _{i\not =i_0}\alpha _{i}\mathbb {I}_{A_{i}}+ \beta \mathbb {I}_{A_{i_0}\cup \{j\}}+(\alpha _{i_0}-\beta )\mathbb {I}_{A_{i_0}}$, where $\beta =\min [X(j)-Y(j), \alpha _{i_0}]$, satisfies (a) and (b). Since $Y'\not = Y$ and $Y'(\ell ) \ge Y(\ell )$ for every $\ell \in N$, we obtain a contradiction to the choice of $Y$. We thus conclude that $Y$ is a decomposition of $X$ (i.e., $X=Y$).

It remains to show that, ignoring indicators whose coefficients are zero, $X$ has only one decomposition. That is, $\{A_1,\ldots ,A_k\}=\{A_1(X),\ldots ,A_n(X)\}$. By definition, $A_i(X)=\{j\in N; \, X(j)\ge X_{\sigma (i)}\}$. Thus, it is enough to show that if $m,\ell \in N$ satisfy $X(m)=X(\ell )$, then $m\in A_i \Leftrightarrow \ell \in A_i$. Let $m,\ell \in N$ satisfy $X(m)=X(\ell )$. If there is $i_0$ such that $\ell \in A_{i_0}$ and $m \not \in A_{i_0}$, then due to property (b) and the fact that $X=Y,\,X(m)=Y(m)<Y(\ell )=X(\ell )$, in contradiction with the choice of $m$ and $\ell$. $\square$

Proof of Lemma 1

Suppose that there exists an optimal sub-decomposition of $X$ w.r.t. $\mathcal{F}$ is obtained by a $D$-sub-decomposition ($D\in \mathcal{F}$), $\sum _{A\in D}\alpha _{A}\mathbb {I}_{A}$. Define the set $D_{X}=\left\{ A\in D\mid \alpha _{A}>0\right\}$. We may choose a sub-decomposition such that $|D_{X}|$ is minimal. If $D_{X}$ is an independent collection, the proof is complete. Otherwise, the variables $\mathbb {I}_{A}, A\in D_{X}$ are linearly dependent. Meaning that there is a linear combination $\sum _{A\in D_{X}}\delta _{A}\mathbb {I}_{A}=0$ where at least one $\delta _{A}\not =0$. Without loss of generality, we may assume that $\sum _{A\in D_{X}}\delta _{A}v(A)\le 0$. Otherwise we could consider $\sum _{A\in D_{X}}(-\delta _{A})\mathbb {I}_{A}$ instead. Since all $\mathbb {I}_{A}$ are non-negative, the fact that at least one $\delta _{A}\not =0$ implies that there is at least one $\delta _{A}>0$. Let $\varepsilon =\min _{A\in D_{X},\delta _{A}>0}\frac{\alpha _{A}}{\delta _{A}}$. Since all the coefficients $\alpha _{A}-\varepsilon \delta _{A}$ are greater than or equal to 0, $\sum _{A\in D}\alpha _{A}\mathbb {I}_{A}-\varepsilon \sum _{A\in D_{X}}\delta _{A}\mathbb {I}_{A}= \sum ){A\in D}(\alpha _{A}-\varepsilon \delta _{A})\mathbb {I}_{A}$ is a sub-decomposition of $X$. As for optimality of this sub-decomposition, $\sum _{A\in D}(\alpha _{A}-\varepsilon \delta _{A})v(A)= \sum _{A\in D}\alpha _{A}v(A)-\varepsilon \sum _{A\in D_{X}}\delta _{A}v(A)\ge \sum _{A\in D}\alpha _{A}v(A)$. Therefore, this is an optimal sub-decomposition. Moreover, for at least one $A\in D_{X}$, the coefficient $\alpha _{A}-\varepsilon \delta _{A}=0$, implying that $\sum _{A\in D}(\alpha _{A}-\varepsilon \cdot \delta _{A})\mathbb {I}_{A}$ is an optimal sub-decomposition of $X$ that involves a smaller number of indicators than does $D_{X}$, contradicting the choice of $D_{X}$. It implies that $D_{X}$ is indeed an independent collection. $\square$

Proof of Lemma 2

Suppose that for every independent collection $C\subseteq \cup \mathcal{F}$, there exists $D\in \mathcal{F}$ such that $C\subseteq D$. Define the following set consisting of one collection: $\mathcal{F}'=\{\cup \mathcal{F}\}$. By assumption, for every $D'\in \mathcal{F}'$ and every independent collection $C\subseteq D'$ (i.e., for every independent $C\subseteq \cup \mathcal{F}$), there is $D\in \mathcal{F}$ such that $C\subseteq D$. Thus, from Proposition 4, $\int _{\mathcal{F}'}\cdot \,\mathrm{d}v\le \int _{\mathcal{F}}\cdot \,\mathrm{d}v$. On the other hand, from the definition of $\mathcal{F}'$, for every $D\in \mathcal{F}$ and every independent collection $C\subseteq D$, there is $D'\in \mathcal{F}'$ such that $C\subseteq D'$. Thus, again, due to Proposition 4, $\int _{\mathcal{F}}\cdot \,\mathrm{d}v\le \int _{\mathcal{F}'}\cdot \,\mathrm{d}v$, which leads us to conclude that $\int _{\mathcal{F}}\cdot \,\mathrm{d}v=\int _{\mathcal{F}'}\cdot \,\mathrm{d}v$.

As for the inverse direction, suppose $\int _{\mathcal{F}'}\cdot \,\mathrm{d}v=\int _{\mathcal{F}}\cdot \,\mathrm{d}v$, and $\mathcal{F}'=\{D'\}$ (i.e., $\mathcal{F}'$ is a singleton). We show that $\cup \mathcal{F}\subseteq D'$. Assume to the contrary that $\cup \mathcal{F}\not \subseteq D'$. Then, there exists $D_{1}\in \mathcal{F}$ such that $D_{1}\nsubseteq D'$. By assumption, $\int _{\mathcal{F}'}\cdot \,\mathrm{d}v\ge \int _{\mathcal{F}}\cdot \,\mathrm{d}v$, and from Proposition 4, we infer that for every independent collection $C\subseteq D_{1}$, there is $D\in \mathcal{F}'$ such that $C\subseteq D$. Any event in $D_{1}$ is an independent collection; thus, $D'$ must include any event in $D_{1}$ and thus must contain $D_{1}$ itself (i.e., $D_{1}\subseteq D'$).

Finally, consider an independent $C\subseteq \cup \mathcal{F}$. By the previous argument $C\subseteq D'$. Since $\int _{\mathcal{F}'}\cdot \,\mathrm{d}v\le \int _{\mathcal{F}}\cdot \,\mathrm{d}v$, we obtain from Proposition 4 that there exists $D\in \mathcal{F}$ such that $C\subseteq D$, which completes the proof. $\square$