Subgame Consistent Cooperative Provision of Public Goods Under Accumulation and Payoff Uncertainties

Abstract

The provision of public goods constitutes a classic case of market failure which calls for cooperative optimization. However, cooperation cannot be sustainable unless there is guarantee that the agreed-upon optimality principle can be maintained throughout the planning duration. This paper derives subgame consistent cooperative solutions for public goods provision by asymmetric agents in a discrete-time dynamic game framework with uncertainties in stock accumulation dynamics and future payoff structures. In particular, subgame consistency ensures that as the game proceeds agents are guided by the same optimality principle and hence they do not possess incentives to deviate from the previously adopted optimal behavior. A “payoff distribution procedure” leading to subgame-consistent solutions is derived and an illustration is presented. This is the first time that subgame consistent cooperative provision of public goods with uncertainties in stock dynamics and future payoffs is analyzed.

Appendix

Proof of Proposition 1

Consider first the last stage, that is stage 3, when $\theta_{3}^{\sigma _{3}}$ occurs. Invoking that

$$\begin{aligned} &V^{(\sigma_3 )i} (3,K) = \bigl[A_3^{ (\sigma_3 ) i} K + C_3^{ (\sigma_3 ) i} \bigr] (1+r)^{-2} \quad\mbox{and}\\ &V^{ (\sigma_4 ) i} (4, K_4 ) = \bigl(q^i K + m^i \bigr) (1+r)^{-3} \end{aligned}$$

from Proposition 1, the condition governing t=3 in equation (30) becomes

$$\begin{aligned} & \bigl[A_3^{ (\sigma_3 ) i} K + C_3^{ (\sigma_3 ) i} \bigr] (1+r)^{-2} \\ &\quad =\max_{I_3^i} \Biggl\{ \bigl[ \alpha_3^{ (\sigma_3 ) i} K - c_3^{ (\sigma_3 ) i} \bigl(I_3^i\bigr)^2 \bigr] (1+r)^{-2} \\ &\quad\quad{} +\sum_{y=1}^3 \gamma_3^y \sum_{\sigma_4 = 1}^4 \lambda_4^{\sigma_4 } \Biggl[ q^i \Biggl( K + \sum _{\substack{j = 1\\ j \ne i}}^4 \phi_3^{ (\sigma_3 ) j^*} (K) + I_3^i - \delta K + \vartheta_3^y \Biggr) +m^i \Biggr] \\ &\quad\quad{}\times (1+r)^{-3} \Biggr\} , \quad \mbox{for } i\in N. \end{aligned}$$

(39)

Performing the indicated maximization in (39) yields the game equilibrium strategies in stage 3 as:

$$ \phi_3^{ (\sigma_3 ) i^*} (K) = \frac{q^i (1+r)^{-1}}{ 2 c_3^{ (\sigma_3 ) i}},\quad \mbox{for } i\in N. $$

(40)

Substituting (40) into (39) yields:

$$\begin{aligned} & \bigl[A_3^{ (\sigma_3 ) i} K + C_3^{ (\sigma_3 ) i} \bigr] \\ &\quad= \alpha_3^{ (\sigma_3 ) i} K - \frac{ (q^i )^2 (1+r)^{-2}}{ 4 c_3^{ (\sigma_3 ) i} } \\ &\quad\quad {} +\sum_{y=1}^3 \gamma_3^y \Biggl[ q^i \Biggl( K + \sum_{j = 1}^n \frac{q^j (1+r)^{-1}}{2 c_3^{ (\sigma_3 ) j}} - \delta K + \vartheta_t^y \Biggr) + m^i \Biggr] \\ &\quad\quad{}\times(1+r)^{-1}, \quad\mbox{for } i \in N. \end{aligned}$$

(41)

Note that both sides of equation (41) are linear expressions of K. For (41) to hold it is required that:

$$ \begin{aligned} & A_3^{ (\sigma_3 ) i} = \alpha_3^{ (\sigma_3 ) i} + q^i (1-\delta) (1+r)^{-1} ,\quad \mbox{and} \\ & C_3^{ (\sigma_3 ) i} = - \frac{ (q^i )^2 (1+r)^{-2}}{ 4 c_3^{ (\sigma_3 ) i}} \\ &\hphantom{C_3^{ (\sigma_3 ) i} =} {} + \Biggl[ q^i \sum_{j = 1}^n \frac{q^j (1+r)^{-1}}{2 c_3^{ (\sigma_3 ) j}} + q^i \varpi_3 + m^i \Biggr] (1+r)^{-1}, \quad\mbox{for } i \in N. \end{aligned} $$

(42)

Now we proceed to stage 2, using $V^{ (\sigma_{3} ) i} (3, K)= [A_{3}^{ (\sigma_{3} ) i} K + C_{3}^{ (\sigma_{3} ) i} ] (1+r)^{-2}$ with $A_{3}^{ (\sigma_{3} ) i}$ and $C_{3}^{ (\sigma_{3} ) i}$ given in (42), the conditions in equation (30) become

$$\begin{aligned} & \bigl[A_2^{ (\sigma_2 ) i} K + C_2^{ (\sigma_2 ) i} \bigr] (1+r)^{-1} \\ &\quad= \max_{I_2^i} \Biggl\{ \bigl[ \alpha_2^{ (\sigma_2 )i} K - c_2^{ (\sigma_2 ) i} \bigl(I_2^i \bigr)^2 \bigr] (1+r)^{-1} \\ &\quad\quad {} + \sum_{y = 1}^3 \gamma_2^y \sum_{\sigma_3 = 1}^4 \lambda_3^{\sigma_3 } \Biggl[ A_3^{ (\sigma_3 ) i} \Biggl( K + \mathop{\sum_{j = 1}}_{ j \ne i}^n \phi_2^{ (\sigma_2 ) j^*} (K) + I_2^i - \delta K + \vartheta_2^y \Biggr) \\ &\quad\quad{} + C_3^{ (\sigma_3 ) i} \Biggr] (1+r)^{-2} \Biggr\} , \quad \mbox{for } i\in N. \end{aligned}$$

(43)

Performing the indicated maximization in (43) yields the game equilibrium strategies in stage 2 as:

$$ \phi_2^{ (\sigma_2 ) i^*} (K) = \sum _{\sigma_3 = 1}^4 \lambda_3^{\sigma_3} \frac{A_3^{ (\sigma_3 ) i} (1+r)^{-1} }{ 2 c_2^{ (\sigma_2 ) i}},\quad \mbox{for } i\in N. $$

(44)

Substituting (44) into (43) yields:

$$\begin{aligned} & \bigl[A_2^{ (\sigma_2 ) i} K + C_2^{ (\sigma_2 ) i} \bigr] \\ &\quad = \alpha_2^{ (\sigma_2 ) i} K - \frac{1}{4 c_2^{ (\sigma _2 ) i} } \Biggl( \sum_{\sigma_3 = 1}^4 \lambda_3^{\sigma_3 } A_3^{ (\sigma_3 ) i} (1+r)^{-1} \Biggr)^2 \\ &\quad\quad {} +\sum_{y = 1}^3 \gamma_2^y \sum_{\sigma_3 = 1}^4 \lambda_3^{\sigma_3} \Biggl[ A_3^{ (\sigma_3 ) i} \Biggl( K + \sum_{j = 1}^n \sum _{\hat{\sigma}_3 = 1}^4 \lambda_3^{\hat{\sigma}_3} \frac{A_3^{ (\hat{\sigma}_3 ) j} (1+r)^{-1}}{ 2 c_2^{ (\sigma_2 ) j}} -\delta K + \vartheta_2^y \Biggr) \\ &\quad\quad {} + C_3^{ (\sigma_3 ) i} \Biggr] (1+r)^{-1} \quad\mbox{for } i\in N. \end{aligned}$$

(45)

Both sides of equation (45) are linear expressions of K. For (45) to hold it is required that:

$$ \begin{aligned} &A_2^{ ( \sigma_2 ) i} = \alpha_2^{ ( \sigma_2 ) i} + \sum_{\sigma_3 = 1}^4 \lambda_3^{\sigma_3} A_3^{ ( \sigma_3 ) i} (1-\delta) (1+r)^{-1},\quad \mbox{and} \\ &C_2^{ (\sigma_2 ) i} = -\frac{1}{4 c_2^{ (\sigma_2 ) i} } \Biggl( \sum _{\sigma_3 = 1}^4 \lambda_3^{\sigma_3 } A_3^{ (\sigma_3 ) i} (1+r)^{-1} \Biggr)^2 \\ & \hphantom{C_2^{ (\sigma_2 ) i} =} {} +\sum_{\sigma_3 = 1}^4 \lambda_3^{\sigma_3} \Biggl[ A_3^{ (\sigma_3 ) i} \Biggl( \sum_{j = 1}^n \sum _{\hat{\sigma}_3 = 1}^4 \lambda_3^{\hat{\sigma}_3 } \frac{A_3^{ (\hat{\sigma}_3 ) j} (1+r)^{-1}}{2 c_2^{ (\sigma_2 ) j}} + \varpi_2 \Biggr) \\ & \hphantom{C_2^{ (\sigma_2 ) i} =}{}+ C_3^{ (\sigma_3 ) i} \Biggr] (1+r)^{-1}, \quad\mbox{for } i\in N. \end{aligned} $$

(46)

Now we proceed to stage 1, using $V^{ (\sigma_{2} ) i} (2, K)= [A_{2}^{ (\sigma_{2} ) i} K + C_{2}^{ (\sigma_{2} ) i} ] (1+r)^{-1}$ with $A_{2}^{ (\sigma_{2} ) i}$ and $C_{2}^{ (\sigma_{2} ) i}$ given in (46), the conditions in equation (30) become

$$\begin{aligned} & \bigl[A_1^{ (\sigma_1 ) i} K + C_1^{ (\sigma_1 ) i} \bigr] \\ &\quad=\max_{I_1^i } \Biggl\{ \bigl[\alpha_1^{ (\sigma_1 ) i} K - c_1^{ (\sigma_1 ) i} \bigl( I_1^i \bigr)^2 \bigr] \\ &\quad\quad{} + \sum_{y = 1}^3 \gamma_1^y \sum_{\sigma_2 = 1}^4 \lambda_2^{\sigma_2} \Biggl[A_2^{ (\sigma_2 ) i} \Biggl( K + \mathop{\sum_{j = 1 }}_{ j \ne i} ^n \phi_1^{ (\sigma_1 ) j^*} (K) + I_1^i - \delta K + \vartheta_1^y \Biggr) \\ &\quad\quad{} + C_2^{ (\sigma_2 ) i} \Biggr] (1+r)^{-1} \Biggr\} ,\quad \mbox{for } i\in N. \end{aligned}$$

(47)

Performing the indicated maximization in (47) yields the game equilibrium strategies in stage 1 as:

$$ \phi_1^{ (\sigma_1 ) i^*} (K) = \sum _{\sigma_2 = 1}^4 \lambda_2^{\sigma_2} \frac{A_2^{ (\sigma_2 ) i} (1+r)^{-1} }{ 2 c_1^{ (\sigma_1 ) i}}, \quad\mbox{for } i\in N. $$

(48)

Substituting (48) into (47) yields:

$$\begin{aligned} &\bigl[A_1^{ (\sigma_1 ) i} K + C_1^{ (\sigma_1 ) i} \bigr] \\ &\quad= \alpha_1^{ (\sigma_1 ) i} K - \frac{1}{4 c_1^{ (\sigma _1 ) i} } \Biggl( \sum_{\sigma_2 = 1}^4 \lambda_2^{\sigma_2} A_2^{ (\sigma_2 ) i} (1+r)^{-1} \Biggr)^2 \\ &\quad\quad {} +\sum_{y = 1}^3 \gamma_1^y \sum_{\sigma_2 = 1}^4 \lambda_2^{\sigma_2} \Biggl[ A_2^{ (\sigma_2 ) i} \Biggl( K+\sum_{j = 1}^n \sum _{\hat{\sigma}_2 = 1}^4 \lambda_2^{\hat{\sigma}_2} \frac{A_2^{ (\hat{\sigma}_2 ) j} (1+r)^{-1} }{2c_1^{ (\sigma_1 ) j}} - \delta K + \vartheta_1^y \Biggr) \\ &\quad\quad {} + C_2^{ (\sigma_2 ) i} \Biggr] (1+r)^{-1}, \quad\mbox{for } i\in N. \end{aligned}$$

(49)

Both sides of equation (49) are linear expressions of K. For (49) to hold it is required that:

$$\begin{aligned} &A_1^{ (\sigma_1 ) i} = \alpha_1^{ (\sigma_1 ) i} + \sum_{\sigma_2 = 1}^4 \lambda_2^{\sigma_2} A_2^{ (\sigma_2 ) i} (1-\delta) (1+r)^{-1}, \quad\mbox{and} \\ &C_1^{ (\sigma_1 ) i} = - \frac{1}{4 c_1^{ (\sigma_1 ) i}} \Biggl( \sum _{\sigma_{1} = 1}^4 \lambda_2^{\sigma_2} A_2^{ (\sigma_2 ) i} (1+r)^{-1} \Biggr)^2 \\ & \hphantom{C_1^{ (\sigma_1 ) i} =}{} +\sum_{\sigma_2 = 1}^4 \lambda_2^{\sigma_2} \Biggl[ A_2^{ (\sigma_2 ) i} \Biggl( \sum_{j = 1}^n \sum _{\hat{\sigma}_2 = 1}^4 \lambda_2^{\hat{\sigma}_2} \frac{A_2^{ (\hat{\sigma}_2 ) j} (1+r)^{-1} }{ 2c_1^{ (\sigma_1 ) j}} + \varpi_1 \Biggr) \\ & \hphantom{C_1^{ (\sigma_1 ) i} =}{} + C_2^{ (\sigma _2 ) i} \Biggr] (1+r)^{-1}, \quad \mbox{for } i\in N. \end{aligned}$$

Hence Proposition 1 follows. □