Core Allocations in Co-investment Problems

Abstract

In a theoretical co-investment problem, a set of agents face a surplus-sharing situation with a single input and a single output exhibiting increasing average returns. All agents contribute their respective inputs and expect part of the collective output. Focusing on the core of the problem, we analyze whether a core allocation of the output is acceptable or compatible with a variation of input contributions, where larger payoffs are expected by those agents whose contribution has increased. We state a necessary and sufficient condition for a core allocation to be acceptable. We also introduce and study the acceptable core, that is, those core allocations acceptable for any possible increase of inputs. Finally, we axiomatically characterize when a set-solution that contains acceptable core allocations shrinks into the proportional allocation.

Appendix

Given a co-investment problem $(N,f,\omega )$, $\omega '\in {\mathbb {R}}^N$, $\omega '\ge \omega$, $\omega '\ne \omega$ and vector z defined in (7) we have

Claim

For any $R\in {\mathscr {D}}_{S^*}$ and for any pair of coalitions $S_1$ and $S_2$ such that $\varnothing \ne S_1\subseteq S_2\subseteq N\setminus R$ we have

$$\begin{aligned} \frac{f(\omega '_{S_1\cup Q})-z_Q}{\omega '_{S_1}}\le \frac{f(\omega '_{S_2\cup Q})-z_Q}{\omega '_{S_2}}, \text{ for } \text{ any } Q\subseteq R. \end{aligned}$$

(14)

Proof

The proof will be argued by induction on the cardinality of the size of coalition R. If $|R|=1$, i.e. $R=\{j_1\}\in {\mathscr {D}}_{S^*}$ there are just two cases. If $Q=\varnothing$ then (14) is

$$\begin{aligned} \frac{f(\omega '_{S_1})}{\omega '_{S_1}}\le \frac{f(\omega '_{S_2})}{\omega '_{S_2}} \end{aligned}$$

and it follows just by (1). If $Q=\{j_1\}$ then (14) is

$$\begin{aligned} \frac{f(\omega '_{S_1\cup \{j_1\}})-z_{j_1}}{\omega '_{S_1}}\le \frac{f(\omega '_{S_2\cup \{j_1\}})-z_{j_1}}{\omega '_{S_2}}. \end{aligned}$$

To check this last inequality, notice that

$$\begin{aligned} \begin{array}{l}\displaystyle \frac{f(\omega '_{S_1\cup \{j_1\}})-z_{j_1}}{\omega '_{S_1}} = \frac{\displaystyle \frac{f(\omega '_{S_1\cup \{j_1\}})}{\omega '_{S_1\cup \{j_1\}}}\cdot \omega '_{S_1}+\frac{f(\omega '_{S_1\cup \{j_1\}})}{\omega '_{S_1\cup \{j_1\}}}\cdot \omega '_{j_1} \displaystyle -z_{j_1}}{\displaystyle \omega '_{S_1}}\\ \quad = \displaystyle \frac{f(\omega '_{S_1\cup \{j_1\}})}{\omega '_{S_1\cup \{j_1\}}}+\displaystyle \frac{\displaystyle \frac{f(\omega '_{S_1\cup \{j_1\}})}{\omega '_{S_1\cup \{j_1\}}}\cdot \omega '_{j_1} \displaystyle -z_{j_1}}{\displaystyle \omega '_{S_1}} \le \displaystyle \frac{f(\omega '_{S_2\cup \{j_1\}})}{\omega '_{S_2\cup \{j_1\}}}+\displaystyle \frac{\displaystyle \frac{f(\omega '_{S_2\cup \{j_1\}})}{\omega '_{S_2\cup \{j_1\}}}\cdot \omega '_{j_1} \displaystyle -z_{j_1}}{\displaystyle \omega '_{S_1}} \\ \quad \le \displaystyle \frac{f(\omega '_{S_2\cup \{j_1\}})}{\omega '_{S_2\cup \{j_1\}}}+\displaystyle \frac{\displaystyle \frac{f(\omega '_{S_2\cup \{j_1\}})}{\omega '_{S_2\cup \{j_1\}}}\cdot \omega '_{j_1} \displaystyle -z_{j_1}}{\displaystyle \omega '_{S_2}} =\displaystyle \frac{f(\omega '_{S_2\cup \{j_1\}})-z_{j_1}}{\omega '_{S_2}}, \end{array} \end{aligned}$$

where the first inequality follows by (1) and the second one since, by definition of the set ${\mathscr {D}}_{S^*}$, we have $z_{j_1}\ge \displaystyle \frac{f(\omega '_N)}{\omega '_N}\cdot \omega '_{j_1}\ge \frac{f(\omega '_{S_2\cup \{j_1\}})}{\omega '_{S_2\cup \{j_1\}}}\cdot \omega '_{j_1}$, and thus $\displaystyle \frac{f(\omega '_{S_2\cup \{j_1\}})}{\omega '_{S_2\cup \{j_1\}}}\cdot \omega '_{j_1} \displaystyle -z_{j_1}\le 0$.

Suppose now that for any ${\widetilde{R}}\in {\mathscr {D}}_{S^*}$ such that $1\le |{\widetilde{R}}|\le k$ and all $Q \subseteq {\widetilde{R}}$ we have

$$\begin{aligned} \frac{f(\omega '_{{\widetilde{S}}_1\cup Q})-z_Q}{\omega '_{{\widetilde{S}}_1}}\le \frac{f(\omega '_{{\widetilde{S}}_2\cup Q})-z_Q}{\omega '_{{\widetilde{S}}_2}}, \end{aligned}$$

(15)

for all $\varnothing \ne {\widetilde{S}}_1\subseteq {\widetilde{S}}_2\subseteq N\setminus {\widetilde{R}}$. Let $R\in {\mathscr {D}}_{S^*}$, with $|R|=k+1$, that is $R=\{j_1,j_2,\ldots ,j_k, j_{k+1}\}$ where $\theta =(j_1,j_2,\ldots ,j_k, j_{k+1})$ is the ordering in which the inequalities required to be in the set ${\mathscr {D}}_{S^*}$ are satisfied. Let us remark that the coalition $R\setminus \{j_{k+1}\}$ is also in ${\mathscr {D}}_{S^*}$.

We have to prove (14) for all coalitions $\varnothing \ne S_1\subseteq S_2\subseteq N\setminus R$ and coalition $Q\subseteq R=\{j_1,j_2,\ldots ,j_k,j_{k+1}\}$. To this aim consider two cases:

Case 1 Coalition $Q\subseteq R$ satisfies $j_{k+1}\not \in Q$. In this case (14) holds simply by applying (15) to ${\widetilde{R}}=R\setminus \{j_{k+1}\}$ and taking ${\widetilde{S}}_1=S_1$ and ${\widetilde{S}}_2=S_2$.

Case 2 Coalition $Q\subseteq R$ satisfies $j_{k+1}\in Q$. Taking into account $\varnothing \ne S_1\subseteq S_2\subseteq N\setminus R$ we have,

$$\begin{aligned} \begin{array}{l}\displaystyle \frac{f(\omega '_{S_1\cup Q})-z_Q}{\omega '_{S_1}}\\ \quad = \frac{\displaystyle \frac{f(\omega '_{S_1\cup Q})-z_{Q\setminus \{j_{k+1}\}}}{\omega '_{S_1\cup \{j_{k+1}\}}}\cdot \omega '_{S_1}+\frac{f(\omega '_{S_1\cup Q})-z_{Q\setminus \{j_{k+1}\}}}{\omega '_{S_1\cup \{j_{k+1}\}}}\cdot \omega '_{j_{k+1}} \displaystyle -z_{j_{k+1}}}{\displaystyle \omega '_{S_1}}\\ \quad = \displaystyle \frac{f(\omega '_{S_1\cup Q})-z_{Q\setminus \{j_{k+1}\}}}{\omega '_{S_1\cup \{j_{k+1}\}}}+\displaystyle \frac{\displaystyle \frac{f(\omega '_{S_1\cup Q})-z_{Q\setminus \{j_{k+1}\}}}{\omega '_{S_1\cup \{j_{k+1}\}}}\cdot \omega '_{j_{k+1}} \displaystyle -z_{j_{k+1}}}{\displaystyle \omega '_{S_1}} \\ \quad = \displaystyle \frac{f(\omega '_{(S_1\cup \{j_{k+1}\})\cup (Q\setminus \{j_{k+1}\})})-z_{Q\setminus \{j_{k+1}\}}}{\omega '_{S_1\cup \{j_{k+1}\}}}+\\ \quad \displaystyle \frac{\displaystyle \frac{f(\omega '_{(S_1\cup \{j_{k+1}\})\cup (Q\setminus \{j_{k+1}\})})-z_{Q\setminus \{j_{k+1}\}}}{\omega '_{S_1\cup \{j_{k+1}\}}}\cdot \omega '_{j_{k+1}} \displaystyle -z_{j_{k+1}}}{\displaystyle \omega '_{S_1}} \\ \quad \le \displaystyle \frac{f(\omega '_{(S_2\cup \{j_{k+1}\})\cup (Q\setminus \{j_{k+1}\})})-z_{Q\setminus \{j_{k+1}\}}}{\omega '_{S_2\cup \{j_{k+1}\}}}+\\ \quad \displaystyle \frac{\displaystyle \frac{f(\omega '_{(S_2\cup \{j_{k+1}\})\cup (Q\setminus \{j_{k+1}\})})-z_{Q\setminus \{j_{k+1}\}}}{\omega '_{S_2\cup \{j_{k+1}\}}}\cdot \omega '_{j_{k+1}} \displaystyle -z_{j_{k+1}}}{\displaystyle \omega '_{S_1}} \\ \quad \le \displaystyle \frac{f(\omega '_{S_2\cup Q})-z_{Q\setminus \{j_{k+1}\}}}{\omega '_{S_2\cup \{j_{k+1}\}}}+\displaystyle \frac{\displaystyle \frac{f(\omega '_{S_2 \cup Q})-z_{Q\setminus \{j_{k+1}\}}}{\omega '_{S_2\cup \{j_{k+1}\}}}\cdot \omega '_{j_{k+1}} \displaystyle -z_{j_{k+1}}}{\displaystyle \omega '_{S_2}} = \displaystyle \frac{f(\omega '_{S_2\cup Q})-z_Q}{\omega '_{S_2}}, \end{array} \end{aligned}$$

where the first inequality comes from (15) applied to ${\widetilde{R}}=\{j_1,\ldots , j_k\}$, ${\widetilde{S}}_1=S_1\cup \{j_{k+1}\}$ and ${\widetilde{S}}_2=S_2\cup \{j_{k+1}\}$ and the second inequality follows since it can be proved that

$$\begin{aligned} \frac{f(\omega '_{S_2\cup Q })-z_{Q\setminus \{j_{k+1}\}}}{\omega '_{S_2\cup \{j_{k+1}\}}}\cdot \omega '_{j_{k+1}}-z_{j_{k+1}}\le 0. \end{aligned}$$

Indeed, being $R=\{j_1,\ldots , j_k,j_{k+1}\}\in {\mathscr {D}}_{S^*}$, this inequality follows since

$$\begin{aligned} \begin{array}{ll} z_{j_{k+1}}&{}\ge \displaystyle \max _{{\widetilde{Q}}\subseteq \{j_1,\ldots ,j_k\}}\left\{ \displaystyle \frac{f(\omega '_{(N\setminus \{j_1,\ldots ,j_k\})\cup {\widetilde{Q}}})-z_{{\widetilde{Q}}}}{\omega '_{N\setminus \{j_1,\ldots ,j_k\}}}\right\} \cdot \omega '_{j_{k+1}}\\ &{}\ge \displaystyle \frac{f(\omega '_{(N\setminus \{j_1,\ldots ,j_k\})\cup (Q\setminus \{j_{k+1}\})})-z_{Q\setminus \{j_{k+1}\}}}{\omega '_{N\setminus \{j_1,\ldots ,j_k\}}}\cdot \omega '_{j_{k+1}}\\ &{}\ge \displaystyle \frac{f(\omega '_{(S_2\cup \{j_{k+1}\})\cup (Q\setminus \{j_{k+1}\}) })-z_{Q\setminus \{j_{k+1}\}}}{\omega '_{S_2\cup \{j_{k+1}\}}}\cdot \omega '_{j_{k+1}}, \\ &{}\ge \displaystyle \frac{f(\omega '_{S_2\cup Q })-z_{Q\setminus \{j_{k+1}\}}}{\omega '_{S_2\cup \{j_{k+1}\}}}\cdot \omega '_{j_{k+1}}, \\ \end{array} \end{aligned}$$

where the third inequality follows from (15) by taking ${\widetilde{R}}=R\setminus \{j_{k+1}\}=\{j_1,\ldots , j_k\}$, ${\widetilde{S}}_1=S_2\cup \{j_{k+1}\}$ and ${\widetilde{S}}_2=N\setminus \{j_1,\ldots ,j_k\}$. Therefore, the claim is proved.$\square$