On stability of economic networks

Abstract

In the spirit of Von Neumann and Morgenstern (Theory of games and economic behavior, Princeton University Press, Princeton, 1944), we introduce a notion of network stability. We study the structure of stable economic networks and their associated stable payoff allocations by analyzing the conditions under which complete networks and star networks (both with desirable property of inclusiveness) are stable. We also address conditions for existence and uniqueness of stable set of networks.

Appendix: Proofs

Proof of Claim 1

Let V be any stable set of payoff allocation and network pairs. Assume in negation that there exists a pair of core payoff allocation and network $(x, g) \in {\mathbb {C}} \setminus V$. Then, by external stability of V, there must exist a payoff allocation and network pair $(y, {\hat{g}}) \in V$ such that $(y, {\hat{g}}) \in \Delta (x, g)$. This, in turn, implies that there exist a subset of players $S \subset N$, ${\hat{g}}\in G$, and $y\in A({\hat{g}})$ such that $g \longmapsto _{S} {\hat{g}}$, ${\hat{g}}(S)\in C({\hat{g}})$, $\sum _{i\in S}y^i \le \nu ({\hat{g}}(S))$, $y^i\ge x^i \forall i\in S$, and $y^i> x^i$ for some $i\in S$. Then, we conclude that $\sum _{i\in S}x^i <\sum _{i\in S}y^i \le \nu ({\hat{g}}(S))$. Therefore, there exist $\hat{g}\in G$, $S\subset N$, $g\longmapsto _{S}\hat{g}$ and $\hat{g}(S)\in C(\hat{g})$ such that $\sum _{i\in S}x^{i}< \nu (\hat{g}(S))$. This contradicts our negation assumption that $(x, g) \in {\mathbb {C}}$. $\square$

Proof of Theorem 1

Assume in negation that network g is not complete for some $(x, g)\in {\mathbb {C}}$. By convexity of $\nu$ we have $\nu (g_c)>\nu (g)\ge \sum _{i\in N}x^i$ since $g\subset g_c$. Finally, it is direct that $g\longmapsto _N g_c$, $g_c(N)\in C(g_c)$, ${\tilde{x}}_i=x_i+[\nu (g_c)-\nu (g)]/|N|>x_i, \forall i\in N,$ and $\sum _{i\in N}{\tilde{x}}_i\le \nu (g_c)$. All these imply that $(x, g)\in \Omega \setminus {\mathbb {C}}$, which is a contradiction. $\square$

Proof of Lemma 1

We prove this by induction. Let $N\setminus S=\{ k_1, k_2, k_3, \ldots ,k_J\}$. Construct network $g_{k1}\equiv g(S)\cup \{ik_1\}$, where $ik_1\in g\setminus g(S)$, for some $i\in N$. Note that as there is no isolated player under g, such a link exists. By convexity, $\nu (g_{k1})-\nu (g(S)) \ge \nu (g(S)) -\nu (\emptyset )$, indicating that $\nu (g_{k1})/|S\cup \{k_1\}| \ge 2\nu (g(S))/|S\cup \{k1\}|\ge \nu (g(S))/|S|$, where the last inequality is due to the fact that $2/|S\cup \{k1\}|=2/(|S|+1)\ge 1/|S|\Leftrightarrow 2|S|\ge |S|+1$ since $|S|\ge 1$. Thus, $\nu (g_{k1})/|S\cup k_1| \ge \nu (g(S))/|S|$. Similarly, construct $g_{k2}\equiv g_{k1}\cup \{ik_2\}$, where $ik_2\in g\setminus g(S)$ for some $i\in N$. Again by convexity, we have $\nu (g_{k2})-\nu (g_{k1}) \ge \nu (g_{k1})- \nu (g(S))$. This implies that $\nu (g_{k2}) \ge 2\nu (g_{k1})- \nu (g(S)) \ge \nu (g_{k1})+\nu (g(S))\ge 3\nu (g(S))$, where the last two inequalities are due to the fact that $\nu (g_{k1})\ge 2\nu (g(S))$. It then follows that $\nu (g_{k2})/|S\cup \{k_1,k2\}| \ge 3\nu (g(S)))/|S\cup \{k_1,k_2\}|$. On the other hand, $3\nu (g(S)))/|S\cup \{k_1,k_2\}|\ge \nu (g(S)))/|S|$ because $3/|S\cup \{k_1,k_2\}|=3/(|S|+2)\ge 1/|S|\Leftrightarrow 3|S|\ge |S|+2$ since $|S|\ge 1$. All these imply that $\nu (g_{k2})/(|S\cup \{k_1,k_2\}|)\ge \nu (g(S))/|S|$. This sufficiently concludes our base case. To show our inductive step, assume that $\nu (g_{kJ-1})/|N\setminus \{k_J\}| \ge \nu (g(S))/|S|$ where $g_{kJ-1}$ is defined as $g_{k1}, g_{k2}$, etc. as above. Similar to the argument we just made for $g_{k1}$ and $g_{k2}$, it can be shown that $\nu (g_{kJ})/|N| \ge \nu (g(S))/|S|$ where $g_{kJ}\equiv g_{kJ-1}\cup \{ik_J\}$ and $ik_J\in g\setminus g(S)$. Therefore, since N is a finite set and by induction, for $\forall l\in N$ and $g_{lm} \subset g$ (as constructed above), $lm \in g\setminus g(S)$, and $m\in N\setminus S$, we have $\nu (g_{lm})/|S\cup \{m\}| \ge \nu (g(S))/|S|$. This concludes the lemma. $\square$

Proof of Theorem 2

In light of Claim 1 it is enough to show that $(x_e, g_c )\in {\mathbb {C}}$. To prove this, assume in negation that $(x_e, g_c )\in \Omega \setminus {\mathbb {C}}$. That is, there exist $S\subset N$ and ${\hat{g}}\in G$ such that $g_c \longmapsto _{S} {\hat{g}}$, ${\hat{g}}(S)\in C({\hat{g}})$, $y\in A({\hat{g}})$, $\sum _{i\in S}y^i\le \nu ({\hat{g}}(S))$ and $y^i>x_e^i, \forall i\in S$. As $g_c$ is a complete network, it follows that $ij\in g_c\setminus {\hat{g}}$ for either some $i,j \in S$ or some $i\in S, j\in N \setminus S$. That is, members of S can induce ${\hat{g}}$ by only severing links among themselves or links with other players. It also follows from our negation assumption that for all $i\in S$, $y^i>x_e^i=\nu (g_c)/|N|$. Let $y^k=\min \{y^i:i\in S\}$. Such a minimum exists since N is finite. By the definition of domination, we conclude that $|S| y^k \le \nu ({\hat{g}}(S))$ since $\sum _{i\in S}y^i\le \nu ({\hat{g}}(S))$ and $y^k\le y^i, \forall i\in S\setminus \{k\}$. As ${\hat{g}} \subset g_c$, it follows from convexity of $\nu$ that $\nu (g_c(S)) \ge \nu ({\hat{g}}(S))$. Then, it follows that $\nu (g_c(S))/|S| \ge \nu ({\hat{g}}(S))/|S| \ge y^k>x_e^k=\nu (g_c)/|N|$. That is, $\nu (g_c(S))/|S| >\nu (g_c)/|N|$. However, this contradicts Lemma 1 since $\nu$ is convex and there is no isolated player under a complete network. $\square$

Proof of Theorem 3

Assume the negation. That is, assume an anonymous convex value function $\nu$ and let V be a stable set of payoff allocation and network pairs where $(x, g_s) \in V$ for some $x\in A(g_s)$. Without loss of generality let player 1 be the central player. Construct the following sequence of payoff allocation and network pairs.

• $g_1=g_s$, $y_1=x$

• $g_2=g_1\bigcup \{23\}$, $y_2\in A(g_2)$ such that $y_2^j=y_1^j+\frac{\nu (g_2)-\nu (g_1)}{2} \quad \forall j\in \{2,3\}$ and $y_2^j=y_1^j \quad \forall j\in N\setminus \{2,3\}$

• $g_3=g_2\bigcup \{24\}$, $y_3\in A(g_2)$ such that $y_3^j=y_2^j+\frac{\nu (g_3)-\nu (g_2)}{2} \quad \forall j\in \{2,4\}$ and $y_3^j=y_2^j \quad \forall j\in N\setminus \{2,4\}$

• ...

• $g_{n-1}=g_{n-2}\bigcup \{2n\}$, $y_{n-1}\in A(g_{n-1})$ such that $y_{n-1}^j=y_{n-2}^j+\frac{\nu (g_{n-1})-\nu (g_{n-2})}{2} \quad \forall j\in \{2,n\}$ and $y_{n-1}^j=y_{n-2}^j \quad \forall j\in N\setminus \{2,n\}$

• ...

• $g_k=g_{k-1}\bigcup \{i(i+1)\}$, $y_k\in A(g_k)$ such that $y_k^j=y_{k-1}^j+\frac{\nu (g^k)-\nu (g^{k-1})}{2} \quad \forall j\in \{i, i+1\}$ and $y^k_j=y^{k-1}_j \quad \forall j\in N\setminus \{i, i+1\}$

• ...

• $g_M=g_{M-1}\bigcup \{(n-1)n\}$, $y_M\in A(g_M)$ such that $y_M^j=y_{M-1}^j+\frac{\nu (g_M)-\nu (g_{M-1})}{2} \quad \forall j\in \{(n-1), n\}$ and $y_M^j=y_{M-1}^j \quad \forall j\in N\setminus \{(n-1), n\}$,

where $M=[n!/(n-2)!]/2 -(n-2)$ is finite since n is finite. Note also that $g_M=g_c$. Due to convexity, we then have $\sum _{i\in \{2,3\}} y_2^i \ge \nu (\{23\})$ because $\nu (g_2)-\nu (g_1)\ge \nu (g_1)\ge \nu (g_1(\{1i\}) \ge \nu (\{1i\})= \nu (\{23\})$. These inequalities follow from convexity and the last equality is due to anonymity. Similarly, we conclude that $\sum _{i\in \{2,4\}} y_3^i \ge \nu (\{24\})$ because $\nu (g_3)-\nu (g_2)\ge \nu (g_2)\ge \nu (g_1)\ge \nu (g_1(\{1i\}))\ge \nu (\{1i\}) = \nu (\{24\}) \quad \forall i\in N\setminus \{1\}$. Again, the last equality is due to anonymity. Moreover, $\sum _{i\in \{2,3,4\}} y_3^i \ge \nu (\{23,24\})$ because $\nu (g_3)-\nu (g_1)\ge \nu (g_1)\ge \nu (g_1(\{1i,1j\})) \ge \nu (\{23,24\}) \quad \forall i,j\in N\setminus \{1\}$. In the same fashion, we conclude that $\sum _{i\in \{2,n \}} y_n^i \ge \nu (\{2n\})$ because $\nu (g_n)-\nu (g_{n-1})\ge \nu (g_{n-1})\ge \nu (g_1)\ge \nu (g_1(\{1i\})) \ge \nu (\{1i\}))=\nu (\{2n\}) \quad \forall i\in N\setminus \{1\}$. In addition, $\sum _{i\in N\setminus \{1\}} y_n^i \ge \nu (\{2i|i\in N \setminus \{1\} \})$, $\sum _{i\in N\setminus \{1,n\}} y_n^i \ge \nu (\{2i|i\in N\setminus \{1,n\} \})$, ..., $\sum _{i\in \{2,3,4\}} y_n^i \ge \nu (\{23, 24\})$. By continuing with this line of argument we have $\sum _{i\in \{n,n-1 \}} y_M^i \ge \nu (\{n,n-1\})$ because $\nu (g_M)-\nu (g_{M-1})\ge \nu (g_{M-1})\ge \nu (g_1)\ge \nu (g_1(\{1i\})) \ge \nu (\{1i\})=\nu (\{n(n-1)\}) \quad \forall i\in N\setminus \{1\}$ and $\sum _{i\in N\setminus \{1\}} y_M^i \ge \nu (\{ni:i\in N\setminus \{1, n \})$, $\sum _{i\in N\setminus \{1,n\}} y_M^i \ge \nu (\{(n-1)i:i\in N\setminus \{1, n-1\} \})$, ..., $\sum _{i\in \{2,3,4\}} y_M^i \ge \nu (\{23, 24\})$. That is, by construction, we have $\sum _{i\in S}y_M^i \ge \nu (g(S))$, $\forall g\in G$, $\forall S\subset N$. This implies by external stability of V that $(y_M, g_M) \in V$. It also follows from our construction that $y_M^i>y_1^i, \forall i\in N\setminus {1}$ and $y_M^1=y_1^1$. Moreover, by the definition of a complete network we have $g_c=g_c(N)\in C(g_c)$, i.e., set N can induce $g_c$ from $g_s$. Thus, $(y_M,g_M) \in \Delta (x,g_s)\cap V$, which contradicts the internal stability of V. $\square$

Proof of Theorem 4

First, Let $x\in {\mathbb {R}}^n_+$ be a core payoff vector for game $(N, \omega ^\nu )$. In negation, assume that there does not exist $g\in G$ such that $(x, g) \in {\mathbb {C}}$. Our negation assumption implies that $\exists S\subset N, {\hat{g}}\in G$ such that $g\longmapsto _S {\hat{g}}, {\hat{g}}(S)\in C({\hat{g}}), \sum _{i\in S} x^i< \nu ({\hat{g}}(S))$. However, since $\omega ^\nu (S)=\max _{g\in G^S} \nu (g)$, it must be the case that $\omega ^\nu (S)\ge \nu ({\hat{g}}(S))$, implying that $\sum _{i\in S} x^i<\omega ^\nu (S)$. In turn, this implies for $y^i=x^i + [\omega ^\nu (S)-\sum _{i\in S} x^i]/|S|>x^i, \forall i\in S$. Note also that by construction $\sum _{i\in \in S} y^i=\omega ^\nu (S)$. Hence, there exists $S\subset N$, a feasible payoff $y\in {\mathbb {R}}^n_+$ such that $y^i>x^i, \forall i\in S$ and $\sum _{i\in \in S} y^i\le \omega ^\nu (S)$. This contradicts x being in the core of $(N, \omega ^\nu )$.

Next, assume that $(x, g) \in {\mathbb {C}}$ and let x not be a core payoff for game $(N, \omega ^\nu )$. By our negation assumption there exist $S\subset N$ and a feasible payoff y such that $y^i>x^i, \forall i\in S$ and $\sum _{i\in S}y^i\le \omega ^\nu (S)$. By definition of $\omega ^\nu$, there must exist a network ${\hat{g}}\in G$ and $S\subset N$ such that ${\hat{g}}(S)\in \arg \max _{g\in G^S} \nu (g)$. Thus, ${\hat{g}}(S)\in C({\hat{g}})$ and $\sum _{i\in S} x_i<\nu ({\hat{g}}(S)) =\omega ^\nu (S)$, contradicting $(x, g) \in {\mathbb {C}}$. Note that it is trivial to show that $g\longmapsto _{S}\hat{g}$. $\square$

Proof of Theorem 5

First, we show that ${\mathbb {C}}_\omega$ is a unique vN &M solution of game $(N, \omega ^\nu )$ if $\nu$ is convex. Since it is well known that the set of core payoffs of a cooperative game is its unique vN &M solution if its characteristic function is convex, it is sufficient to show that if the value function $\nu$ is convex then $\omega ^\nu$ is convex.⁷ To prove this, let $\nu$ be convex and fix any arbitrary $S\subset T\subset N$. By convexity of $\nu$, for any $g\in G$ and $i\in S\subset T$ we have $\nu (g(T))\ge 2\nu (g(T\setminus \{i\}))$ and $\nu (g(T))\ge 2\nu (g(S))$. These inequalities imply that $\nu (g(T))\ge \nu (g(T\setminus \{i\}))+\nu (g(S))$. Since $\nu (g(S\setminus \{i\})\ge 0$, we conclude that $\nu (g(T))\ge \nu (g(T\setminus \{i\}))+\nu (g(S))-\nu (g(S\setminus \{i\})\iff \nu (g(T))-\nu (g(T\setminus \{i\}))\ge \nu (g(S))-\nu (g(S\setminus \{i\})$. Since this inequality is true for any arbitrary $g\in G$ and $i\in S\subset T\subset N$, it implies that $\omega ^\nu (T)-\omega ^\nu (T\setminus \{i\}) \ge \omega ^\nu (S)-\omega ^\nu (S\setminus \{i\})$. That is, $\omega ^\nu$ is convex.

Next, we shall show that ${\mathbb {C}}$ is stable. Since $(x,g)\in {\mathbb {C}}$ if $\Delta (x,g)=\emptyset$, internal stability of ${\mathbb {C}}$ is direct. Therefore, it is enough to show that ${\mathbb {C}}$ is externally stable. Let $(x,g)\in \Omega \setminus {\mathbb {C}}$. It follows from Theorem 4 that $x\not \in {\mathbb {C}}_\omega$. By external stability of ${\mathbb {C}}_\omega$ there must exist $S\subset N, y\in {\mathbb {C}}_\omega$ such that $\sum _{i\in S}y_i\le \omega ^\nu (S)$ and $y_i>x_i, \forall i\in S$. Then, Claim 1 and Theorem 4 imply that $(y,g_c) \in {\mathbb {C}}$. If $S=N$, then $(y,g_c)\in {\mathbb {C}}\cap \Delta (x,g)$, i.e., ${\mathbb {C}}$ is externally stable. To show that $S=N$, assume the negation: $S\not =N$. Construct payoff $z\in {\mathbb {R}}^n_+$ such that $z^i=y^i+[\omega ^\nu (N)-\omega ^\nu (S)]/|N|, \forall i\in N$. Since $\max _{g\in G^N} \nu (g) >\max _{g\in G^S} \nu (g)$ due to the convexity of $\nu$, by construction we have $z^i>y^i,\forall i\in N$ implying that $y\not \in {\mathbb {C}}_\omega$, which is a contradiction.

Finally, we need to show that ${\mathbb {C}}$ is unique and non-empty. In light of Theorem 2, note that $(x_e,g_c) \in {\mathbb {C}}$. Thus, we conclude that ${\mathbb {C}}\not =\emptyset$. To prove the uniqueness, assume the negation. In particular, assume that there exists another stable set of payoff allocation and network pairs $V\not ={\mathbb {C}}$. It follows from Claim 1 that ${\mathbb {C}}\subset V$. Consider a payoff allocation and network pair $(x, g)\in V\setminus {\mathbb {C}}$. By external stability of ${\mathbb {C}}$, there must exist a payoff allocation and network pair $(y,{\hat{g}})$ such that $(y, {\hat{g}})\in {\mathbb {C}}\cap \Delta (x,g)$. But, since ${\mathbb {C}}\subset V$, then $(y, {\hat{g}})\in V$. That is, we have $(x,g)\in V$ and $(y, g)\in V$, where $(y,g)\in \Delta (x,g)$, contradicting internal stability of V. $\square$