Public Good Provision Games on Networks with Resource Pooling

Abstract

We consider the interaction of strategic agents and their decision-making process toward the provision of a public good. In this interaction, each user exerts a certain level of effort to improve his own utility. At the same time, the agents are interdependent and the utility of each agent depends not only on his own effort but also on the other agents’ effort level. As the agents have a limited budget and can exert limited effort, question arises as to whether there is advantage to agents pooling their resources. In this study, we show that resource pooling may or may not improve the agents’ utility when they are driven by self-interest. We identify some scenarios where resource pooling does lead to social welfare improvement as compared to without resource pooling. We also propose a taxation–subsidy mechanism that can effectively incentivize the agents to exert socially optimal effort under resource pooling.

This work is supported by the NSF under grant CNS-1422211, CNS-1616575 and CNS-1739517.

Appendix

Proof

(Lemma 2) Let $\overline{\pmb {m}}$ be a Nash equilibrium of the game induced by proposed mechanism, and $\overline{{m}}_i = ( \overline{\pmb {x}}^{(i)} , \overline{\pmb {\pi }}^{(i)})$. We need to show that the following term is equal to zero at NE:

$$\begin{aligned} \begin{array}{ll} &{}(\overline{\pmb {x}}^{(i)} - \overline{\pmb {x}}^{(i+1)} )^T diag(\overline{\pmb {\pi }}^{(i)} ) (\overline{\pmb {x}}^{(i)} - \overline{\pmb {x}}^{(i+1)} )\\ &{} - (\overline{\pmb {x}}^{(i+1)} - \overline{\pmb {x}}^{(i+2)} )^T diag(\overline{\pmb {\pi }}^{(i+1)} ) (\overline{\pmb {x}}^{(i+1)} - \overline{\pmb {x}}^{(i+2)})~. \end{array} \end{aligned}$$

(37)

Because $\overline{\pmb {m}}$ is the Nash equilibrium, we have

$$\begin{aligned} r_i(\hat{X}(m_i , \overline{\pmb {m}}_{-i}),\hat{t}_i( m_i , \overline{\pmb {m}}_{-i}) ) \le r_i( \hat{X}(\overline{\pmb {m}}),\hat{t}_i(\overline{\pmb {m}})), \forall m_i\in \mathcal {M}_i~. \end{aligned}$$

(38)

We substitute $m_i = (\overline{\pmb {x}}^{(i)} , \pmb {\pi }^{(i)} )$ in (38). Using (17) and (18), we have,

$$\begin{aligned} \begin{array}{ll} r_i(\hat{X}( ( \overline{\pmb {x}}^{(i)} , \pmb {\pi }^{(i)}), \overline{\pmb {m}}_{-i}),\hat{t}_i( ( \overline{\pmb {x}}^{(i)} , \pmb {\pi }^{(i)}), \overline{\pmb {m}}_{-i}) ) =\\ r_i(\hat{X}(\overline{\pmb {m}}),\hat{t}_i( ( \overline{\pmb {x}}^{(i)} , \pmb {\pi }^{(i)}), \overline{\pmb {m}}_{-i}) ) \le r_i( \hat{X}(\overline{\pmb {m}}),\hat{t}_i(\overline{\pmb {m}})), \forall \pmb {\pi }^{(i)}\in R^{N(N-1)}_+~. \end{array} \end{aligned}$$

(39)

Because $r_i(.,.)$ is a decreasing function in $t_i$, (39) implies that,

$$\begin{aligned} \begin{array}{ll} \hat{t}_i( ( \overline{\pmb {x}}^{(i)} , \pmb {\pi }^{(i)}), \overline{\pmb {m}}_{-i}) \ge \hat{t}_i(\overline{\pmb {m}}), \forall \pmb {\pi }^{(i)}\in R^{N(N-1)}_+~. \end{array} \end{aligned}$$

(40)

In other words,

$$\begin{aligned} \begin{array}{ll} &{}(\overline{\pmb {\pi }}^{(i+1)} - \overline{\pmb {\pi }}^{(i+2)})^T \underline{\pmb {x}} \\ &{}+ (\overline{\pmb {x}}^{(i)} - \overline{\pmb {x}}^{(i+1)} )^T diag(\overline{\pmb {\pi }}^{(i)} ) (\overline{\pmb {x}}^{(i)} - \overline{\pmb {x}}^{(i+1)} )\\ &{} - (\overline{\pmb {x}}^{(i+1)} - \overline{\pmb {x}}^{(i+2)} )^T diag(\overline{\pmb {\pi }}^{(i+1)} ) (\overline{\pmb {x}}^{(i+1)} - \overline{\pmb {x}}^{(i+2)} )\le \\ &{}(\overline{\pmb {\pi }}^{(i+1)} -\overline{ \pmb {\pi }}^{(i+2)})^T \underline{\pmb {x}} \\ &{}+ (\overline{\pmb {x}}^{(i)} - \overline{\pmb {x}}^{(i+1)} )^T diag(\pmb {\pi }^{(i)} ) (\overline{\pmb {x}}^{(i)} - \overline{\pmb {x}}^{(i+1)} )\\ &{} - (\overline{\pmb {x}}^{(i+1)} - \overline{\pmb {x}}^{(i+2)} )^T diag(\overline{\pmb {\pi }}^{(i+1)} ) (\overline{\pmb {x}}^{(i+1)} - \overline{\pmb {x}}^{(i+2)} )~, \\ &{} \forall \pmb {\pi }^{(i)}\in R^{N\cdot (N-1)}_+~. \end{array} \end{aligned}$$

(41)

By simplifying the above equation, we have

$$\begin{aligned} \begin{array}{ll} (\overline{\pmb {x}}^{(i)} - \overline{\pmb {x}}^{(i+1)} )^T diag(\pmb {\pi }^{(i)} - \overline{\pmb {\pi }}^{(i)} ) (\overline{\pmb {x}}^{(i)} - \overline{\pmb {x}}^{(i+1)} ) \ge 0~, \forall \pmb {\pi }^{(i)}\in R^{N\cdot (N-1)}_+ ~. \end{array} \end{aligned}$$

(42)

Because the above equation is valid for all $\pmb {\pi }^{(i)}\in R^{N(N-1)}_+$, it implies

$$\begin{aligned} (\overline{\pmb {x}}^{(i)} - \overline{\pmb {x}}^{(i+1)} )^T diag(\overline{\pmb {\pi }}^{(i)} ) (\overline{\pmb {x}}^{(i)} - \overline{\pmb {x}}^{(i+1)}) = 0, ~ \forall i\in V~. \end{aligned}$$

(43)

Therefore, at the NE we have

$$\begin{aligned} \hat{t}_i (\overline{\pmb {m}}) = (\overline{\pmb {\pi }}^{(i+1)} - \overline{\pmb {\pi }}^{(i+2)})^T \underline{\pmb {x}},~ \forall i\in V~. \end{aligned}$$

(44)

   $\blacksquare$

Proof

(Theorem 3) Let $\overline{\pmb {m}}$ be a NE of the proposed mechanism, and $\overline{{m}}_i = ( \overline{\pmb {x}}^{(i)} , \overline{\pmb {\pi }}^{(i)})$ and $\underline{\pmb {x}} = \hat{\pmb {x}} (\overline{\pmb {m}})$. By definition, we have

$$\begin{aligned} \begin{array}{ll} r_i(\hat{X}( ( \pmb {x}^{(i)} , \pmb {\pi }^{(i)}), \overline{\pmb {m}}_{-i}) ,\hat{t}_i( ( \pmb {x}^{(i)} , \pmb {\pi }^{(i)}), \overline{\pmb {m}}_{-i}) ) \le r_i( \hat{X}(\overline{\pmb {m}}),\hat{t}_i(\overline{\pmb {m}})),\\ \forall m_i = ( \pmb {x}^{(i)} ,\pmb {\pi }^{(i)} )\in \mathcal {M}_i \end{array} \end{aligned}$$

(45)

Let $\tilde{\pmb {x}}^{(i)}$ be a vector such that $\frac{1}{N}(\tilde{\pmb {x}}^{(i)}+\sum _{k\in V-\{i\}} \overline{\pmb {x}}^{(k)}) = 0$. Moreover, we set $\pmb {\pi }^{(i)}= \pmb {0}$. By Lemma 2, we have

$$\begin{aligned} \begin{array}{ll} \hat{\pmb {x}}( (\tilde{\pmb {x}}^{(i)} , \pmb {0}), \overline{\pmb {m}}_{-i}) =\pmb {0}\\ \hat{X}( (\tilde{\pmb {x}}^{(i)} , \pmb {0}), \overline{\pmb {m}}_{-i}) =diag(B_1,B_2,\cdots ,B_N)\\ \hat{t}_i( ( \tilde{\pmb {x}}^{(i)} , \pmb {0}), \overline{\pmb {m}}_{-i}) = \pmb {0}\\ r_i(\hat{X}( (\tilde{\pmb {x}}^{(i)} , \pmb {0}), \overline{\pmb {m}}_{-i}),\hat{t}_i( ( \tilde{\pmb {x}}^{(i)} , \pmb {0}), \overline{\pmb {m}}_{-i}) ) = v_i^o~. \end{array} \end{aligned}$$

(46)

Equations (45) and (46) together imply that $r_i(\hat{X}(\overline{\pmb {m}}) , \hat{t}_i(\overline{\pmb {m}})) \ge v_i^o$.    $\blacksquare$

Proof

(Theorem 5) Let us assume $X^*$ is a socially optimal effort profile. Let $\underline{\pmb {x}} = [x_{12}^*, x_{13}^*, \cdots , x_{1N}^*,x_{21}^*,x_{23}^*\cdots , x^*_{N(N-1)}]$. First we show that there is vector $\overline{\pmb {l}}_i$ such that

$$\begin{aligned} \underline{\pmb {x}} \in \text {arg} \max \limits _{\pmb {x}\in R^{N(N-1)},A(\pmb {x})\in S} -\overline{\pmb {l}}_i^T \pmb {x}+v_i(A(\pmb {x}))~. \end{aligned}$$

(47)

As $A(\underline{\pmb {x}})$ is the socially optimal effort profile, we have

$$\begin{aligned} \begin{array}{ll} \underline{\pmb {x}} = \text {arg max}_{\{\pmb {x}\in R^{N(N-1)},A(\pmb {x})\in S\}} \sum _{i=1}^N v_i(A(\pmb {x})) \rightarrow \text{ KKT } \text{ Conditions: }\\ -\left( \sum _{i\in V} \bigtriangledown _{\pmb {x}} v_i(A(\underline{\pmb {x}}))\right) - \pmb {\lambda } + \pmb {\theta } = 0\\ \pmb {\lambda } ^T \underline{\pmb {x}} = 0,\\ \theta _j (-B_j+ \sum _{k\in V-\{ j \} } \underline{x}_{jk} ) = 0\\ \pmb {\lambda } \ge 0\\ \pmb {\theta } = \left[ \underbrace{\theta _1,\cdots ,\theta _1}_{N-1 ~times},\underbrace{ \theta _2,\cdots ,\theta _2}_{N-1 ~times},\cdots ,\underbrace{ \theta _N,\cdots ,\theta _N}_{N-1 ~ times}\right] ^T \ge 0~. \end{array} \end{aligned}$$

(48)

We can define $\overline{\pmb {l}}_i = \bigtriangledown _{\pmb {x}} v_i(A(\underline{\pmb {x}})) + \pmb {\lambda }/N - \pmb {\theta }/N$. Then we have

$$\begin{aligned} \overline{\pmb {l}}_i - \bigtriangledown _{\pmb {x}} v_i(A(\underline{\pmb {x}})) - \pmb {\lambda }/N + \pmb {\theta }/N = \pmb {0}~, \end{aligned}$$

(49)

which implies that $\underline{\pmb {x}} , \pmb {\lambda }/N ,\pmb {\theta }/N$ satisfies the KKT conditions for the following optimization problem:

$$\begin{aligned} \begin{array}{ll} \text {arg max}_{\{\pmb {x}\in R^{N(N-1)},A(\pmb {x})\in S\}} -\overline{\pmb {l}}_i^T \pmb {x}+v_i(A(\pmb {x}))~. \end{array} \end{aligned}$$

(50)

As the above optimization problem is convex and the KKT conditions are necessary and sufficient for optimality, $\underline{\pmb {x}}$ is the solution to (50).

Now let us assume that we have already found $\overline{\pmb {l}}_i, \forall i\in V$. Consider following system of equations,

(51)

First, we show that the above system of equations has at least one solution.

If we set $\overline{\pmb {x}}^{(i)} = \underline{\pmb {x}}$, then Eqs. (51.a), (51.c) are satisfied. Moreover, the summation of left-hand side and right-hand side of (51.b) is zero which implies that one of the equations of type (51.b) is redundant. Therefore, if we choose an arbitrary value for $\overline{\pmb {\pi }}^{(1)}$, then $\overline{\pmb {\pi }}^{(2)}, \overline{\pmb {\pi }}^{(3)}, \cdots , \overline{\pmb {\pi }}^{(N)}$ can be determined accordingly based on (51.b). Moreover, notice that if we add all $\overline{\pmb {\pi }}^{(i)}$ by a constant vector $\pmb {c}$, then they still satisfy (51.a), (51.b), (51.c). Therefore, we can select an appropriate constant vector $\pmb {c}$ to satisfy (51.d).

Now, we show the solution introduced above is a Nash equilibrium of the proposed mechanism. We chose $\overline{\pmb {l}}_i$ such that it satisfies the following:

$$\begin{aligned} \begin{array}{ll} \overline{\pmb {x}}^{(i)} = \underline{\pmb {x}} \in \text {arg max}_{\pmb {x}\in R^{N\cdot (N-1)}} -\overline{\pmb {l}}_i^T \pmb {x}+v_i(A(\pmb {x}))~. \end{array} \end{aligned}$$

(52)

We use the following change of variable for the above optimization problem: $N\pmb {x} - \sum _{j\in V-\{i\}} \overline{\pmb {x}}^{(j)} = \pmb {x}^{(i)}$. We have

(53)

By (51.c) and the fact that the users’ utility function is decreasing in tax, we have

(54)

The last equation implies that the solution to (51) is the fixed point of the best-response mapping. Therefore, the solution to (51) is a NE of the proposed mechanism.    $\blacksquare$