Participation and demand levels for a joint project

Abstract

We examine a voluntary participation game in public good provision in which each agent has a demand level for the public good. The agent’s demand level is the minimum level of the public good from which she can receive a positive benefit. In this game, there exists a subgame perfect Nash equilibrium at which the (Pareto) efficient allocation is achieved. The voluntary participation game may also have a subgame perfect Nash equilibrium with underprovision of the public good. However, some subgame perfect Nash equilibrium with the efficient allocation satisfies strong perfection, introduced by Rubinstein (Int J Game Theory 9:1–12, 1980), and strong perfection is satisfied only by the subgame perfect Nash equilibrium with the efficient allocation. Furthermore, all payoffs at strong perfect equilibria belong to the core of the enterprise game. By these results, we conclude that in our case, the voluntary participation problem is not as serious as the earlier studies report. We also discuss the extensibility of these results.

Appendices

Appendix 1: Proofs of propositions

1.1 Proof of Proposition 1

• (i) Suppose that $x_i(y)>0$ for some $i\in P$ such that $Y_i>{\psi }^P({\varvec{x_P}})$ and some $y \in \bar{{\mathcal {Y}}}$. Then, $i$ obtains the payoff $-\sum _{z=1}^{t}x_i(\bar{y}^z)<0$. If she reduces her contribution, then she is made better off.

• (ii) Suppose that $x_i(\bar{y}^l)>0$ for some $i \in P \cap {\mathcal {N}}(0,{\psi }^P({\varvec{x_P}})]$ and some $l \in \{1,\ldots ,t\}$ such that $\bar{y}^l\in \bar{{\mathcal {Y}}}$ and $\bar{y}^l>Y_i$. Note that since $Y_i \ge \bar{y}^1$, $l\ge 2$. Let ${\varvec{x_i^{\prime }}}=(x^{\prime }_i(y))_{y \in \bar{{\mathcal {Y}}}}$ be such that $x_i^{\prime }(\bar{y}^l)=0$ and $x_i^{\prime }(y)=x_i(y)$ for each $y \in \bar{{\mathcal {Y}}}{\setminus } \{\bar{y}^l\}$. We first consider the case of ${\psi }^P({\varvec{x_P}})<\bar{y}^l$. In this case, by the definition of ${\psi }^P$, ${\psi }^P({\varvec{x_P}})={\psi }^P({\varvec{x_i^{\prime }}},{\varvec{x_{P{\setminus } \{i\}}}})$. Thus, $\pi ^P_i({\varvec{x_P}})=\theta _i-\sum _{z=1}^t x_i(\bar{y}^z)<\theta _i-\sum _{z=1}^t x^{\prime }_i(\bar{y}^z)=\pi ^P_i({\varvec{x_i^{\prime }}},{\varvec{x_{P{\setminus } \{i\}}}})$, which implies that ${\varvec{x_P}}$ is not a Nash equilibrium. We also consider the case of ${\psi }^P({\varvec{x_P}}) \ge \bar{y}^l$. If $i$ switches from ${\varvec{x_i}}$ to ${\varvec{x_i^{\prime }}}$, $\bar{y}^l$ may not be provided. By the construction of ${\varvec{x_i^{\prime }}}$ and the definition of ${\psi }^P$, ${\psi }^P({\varvec{x_i^{\prime }}},{\varvec{x_{P{\setminus } \{i\}}}}) \ge \bar{y}^{l-1}$. Since $\bar{y}^l> Y_i$ and $\bar{{\mathcal {Y}}}$ is discrete, then $\bar{y}^{l-1} \ge Y_i$. Then, ${\psi }^P({\varvec{x_i^{\prime }}},{\varvec{x_{P{\setminus } \{i\}}}})\ge Y_i$. Thus, if $i$ switches from ${\varvec{x_i}}$ to ${\varvec{x_i^{\prime }}}$, then her demand level is fulfilled and her contribution declines.

• (iii) Suppose to the contrary that for some $k \in \{1,\ldots ,t\}$ such that $\bar{y}^k \in \bar{{\mathcal {Y}}}$ and $\bar{y}^k \le {\psi }^P({\varvec{x_P}})$, $\sum _{j \in P \cap {\mathcal {N}}(\bar{y}^{k-1},{\psi }^P({\varvec{x_P}})]}x_j(\bar{y}^k)>c(\bar{y}^k)-c(\bar{y}^{k-1})$ or $\theta _i-\sum _{z=k}^t x_i(\bar{y}^z)<0$ for some $i \in P \cap {\mathcal {N}}(\bar{y}^{k-1},{\psi }^P({\varvec{x_P}})]$. If the former holds, there is $j \in P \cap {\mathcal {N}}(\bar{y}^{k-1},{\psi }^P({\varvec{x_P}})]$ such that $x_j(\bar{y}^k)>0$. Agent $j$ can reduce the marginal contribution from $x_j(\bar{y}^k)$ slightly such that ${\psi }^P({\varvec{x_P}})$ is still provided and she is made better off. If the latter holds, $\sum _{z=k}^t x_i(\bar{y}^z)>0$ because $\theta _i>0$. If $i$ sets her total contribution to zero, her payoff is at least 0. Thus, she is made better off. In any case, ${\varvec{x}}_P$ is not a Nash equilibrium.$\square$

1.2 Proof of Proposition 2

We first show (2.2).

(Necessity) Suppose to the contrary that a strong Nash equilibrium ${\varvec{x_P}} \in \mathbb {R}_+^{t|P|}$ does not satisfy GE for $P$. Since ${\varvec{x_P}}$ is a Nash equilibrium and it satisfies the budget balance condition by (2), then ${\psi }^P({\varvec{x_P}}) \notin \arg \max _{y \in \{0\} \cup \bar{{\mathcal {Y}}}}\sum _{i \in P}B_i(y)-c(y)$. Let $\bar{y} \equiv \max \ \arg \max _{y \in \{0\} \cup \bar{{\mathcal {Y}}}}\sum _{i \in P}B_i(y)-c(y)$. Then $\sum _{j \in P \cap {\mathcal {N}}({\psi }^P({\varvec{x_P}}),\bar{y}]}\theta _i>c(\bar{y})-c({\psi }^P({\varvec{x_P}}))$. By (i) of Proposition 1, $P \cap {\mathcal {N}}({\psi }^P({\varvec{x_P}}),\bar{y}]$ can coordinate their contributions in such a way that all of its members are made better off, which is a contradiction. Since ${\varvec{x_P}}$ is a Nash equilibrium, then by (i), (ii), and (3) in (iii) of Proposition 1, IR holds.

(Sufficiency) Suppose to the contrary that there is a coalition $D \subseteq P$ and its deviation ${\varvec{x^{\prime }_D}} \in {\mathbb {R}}_+^{t|D|}$ such that $\pi ^P_i({\varvec{x^{\prime }_D}}, {\varvec{x_{P{\setminus } D}}}) \ge \pi _i^P({\varvec{x_P}})$ for each $i \in P$ with strict inequality for at least one member of $D$. Denote ${\varvec{x_P^{\prime }}}=({\varvec{x^{\prime }_D}}, {\varvec{x_{P{\setminus } D}}})$ and $k \in \{0,1,\ldots ,t\}$ such that $\bar{y}^k \in \{0\} \cup \bar{{\mathcal {Y}}}$ and $\bar{y}^k={\psi }^P({\varvec{x_P^{\prime }}})$. If $\bar{y}^k> {\psi }^P({\varvec{x_P}})$, then $\sum _{j \in P}B_j({\psi }^P\left( {\varvec{x_P}}\right) )-c({\psi }^P({\varvec{x_P}})) \ge \sum _{j \in P}B_j(\bar{y}^k)-c(\bar{y}^k)$; hence, $\sum _{j \in P \cap {\mathcal {N}}\left( {\psi }^P\left( {\varvec{x_P}}\right) ,\bar{y}^k\right] }\theta _j \le c(\bar{y}^k)-c({\psi }^P({\varvec{x_P}}))$. Since by Proposition 1, $x_i(\bar{y})=0$ for each $i \in P$ and each $y \in \bar{{\mathcal {Y}}}$ such that $y>{\psi }^P({\varvec{x_P}})$, some members of $D$ must additionally contribute $c(\bar{y}^k)-c({\psi }^P({\varvec{x_P}}))$ for $D$’s deviation to be improving. On the one hand, for the deviation by $D$ to be improving, its total payment must be less than $\sum _{i \in D}\sum _{y \in \bar{{\mathcal {Y}}}:y \le {\psi }^P({\varvec{x_P}})} x_i(y)+ \sum _{i \in D \cap {\mathcal {N}}\left( {\psi }^P({\varvec{x_P}}),\bar{y}^k\right] }\theta _i$. On the other hand, in order to produce $\bar{y}^k$ by the deviation of $D$, $D$ must pay at least $c(\bar{y}^k)-\sum _{i \in P{\setminus } D}\sum _{y \in \bar{{\mathcal {Y}}}:y \le {\psi }^P({\varvec{x_P}})}x_i(y)$. Since by GE for $P$, $\sum _{i \in P} \sum _{y \in \bar{{\mathcal {Y}}}: y\le {\psi }({\varvec{x_P}})}x_i(y)=c\left( {\psi }^P\left( {\varvec{x_P}}\right) \right)$, then $\sum _{i \in D}\sum _{y\in \bar{{\mathcal {Y}}}:\,y \le {\psi }^P({\varvec{x_P}})} x_i(y)+ \sum _{i \in D \cap {\mathcal {N}}\left( {\psi }^P({\varvec{x_P}}),\bar{y}^k\right] }\theta _i \le c({\psi }^P({\varvec{x_P}}))-\sum _{i \in P{\setminus } D}\sum _{y \in \bar{{\mathcal {Y}}}:\,y \le {\psi }^P({\varvec{x_P}})}x_i(y)+c(\bar{y}^k)-c({\psi }^P({\varvec{x_P}})) = \ c(\bar{y}^k)-\sum _{i \in P{\setminus } D} \sum _{y \in \bar{{\mathcal {Y}}}:\, y \le {\psi }^P({\varvec{x_P}})}x_i(y)$. By this condition, if the total contribution of $i \in D$ decreases by the deviation, the total contribution of another $j \in D$ increases, which induces that $j$ is made worse off.

If $\bar{y}^k = {\psi }^P({\varvec{x_P}})$, $D$ cannot have a profitable deviation, obviously.

If $\bar{y}^k < {\psi }^P({\varvec{x_P}})$, then $c(\bar{y}^{k+1})-c(\bar{y}^k)> \sum _{j \in P} x^{\prime }_j(\bar{y}^{k+1})$. Recall that by Proposition 1, $x_i(\bar{y}^{k+1})>0$ only if $i \in P \cap {\mathcal {N}}\left( \bar{y}^k,{\psi }^P({\varvec{x_P}})\right]$. Hence, $D \cap {\mathcal {N}}\left( \bar{y}^k,{\psi }^P({\varvec{x_P}})\right] \ne \emptyset$ and every $i \in D \cap {\mathcal {N}}\left( \bar{y}^k,{\psi }^P({\varvec{x_P}})\right]$ obtains $\theta _i-\sum _{y \in \bar{{\mathcal {Y}}}} x_i(y)$ before the deviation and $-\sum _{y \in \bar{{\mathcal {Y}}}} x_i^{\prime }(y)$ after the deviation. By IR, $\theta _i-\sum _{y \in \bar{{\mathcal {Y}}}} x_i(y) \ge -\sum _{y \in \bar{{\mathcal {Y}}}} x^{\prime }_i(y)$. Thus, if the deviation by $D$ is profitable, then $\theta _i - \sum _{y \in \bar{{\mathcal {Y}}}} x_i(y)=\sum _{y \in \bar{{\mathcal {Y}}}} x_i^{\prime }(y)=0$ for each $i \in D \cap {\mathcal {N}}\left( \bar{y}^k, {\psi }^P({\varvec{x_P}}) \right]$. Clearly, each $i \in D$ such that $Y_i>{\psi }^P({\varvec{x_P}})$ is not made better off. Therefore, for each $i \in D$, $\pi ^P_i({\varvec{x^{\prime }_D}}, {\varvec{x_{P{\setminus } D}}}) > \pi _i^P({\varvec{x_P}})$ only if $i \in {\mathcal {N}}\left( 0,\bar{y}^k\right]$. Note that for each $i \in D \cap {\mathcal {N}}\left( 0,\bar{y}^k \right]$, $\pi ^P_i({\varvec{x^{\prime }_D}}, {\varvec{x_{P{\setminus } D}}})=\theta _i-\sum _{y \in \bar{{\mathcal {Y}}}}x_i^{\prime }(y)$ and $\pi _i^P({\varvec{x_P}})= \theta _i-\sum _{y \in \bar{{\mathcal {Y}}}}x_i(y)$. If some agent $i\in D\cap {\mathcal {N}}\left( 0,\bar{y}^k\right]$ reduces $i$’s payment, then another agent $j \in D\cap {\mathcal {N}}(0,\bar{y}^k]$ increases $j$’s payment; otherwise, by GE for $P$, $\bar{y}^k$ cannot be provided. Therefore, it is impossible for $D$ to deviate profitably.

We finally show (2.1). A weakly strong Nash equilibrium is immune to all coalition deviations that make all coalition members strictly better off as well as unilateral deviations; hence, our strong Nash equilibrium is a weakly strong Nash equilibrium by definition. Branzei et al. (2005) show the existence of the core of $(N,W)$ and that for each core element, there exists a weakly strong Nash equilibrium whose equilibrium payoff is the core element. Thus, the weakly strong Nash equilibrium exists. We can show that weakly strong and strong Nash equilibria are equivalent in the contribution game as follows. By the sufficiency of (2.2), all Nash equilibria satisfying GE and IR for $P$ are weakly strong Nash equilibria. In a similar way to the necessity in the above proof, we can show that all weakly strong Nash equilibria satisfy GE and IR for $P$. $\square$

1.3 Proof of Proposition 3

Let ${\varvec{x_P}}\in \mathbb {R}_+^{t|P|}$ be a strong Nash equilibrium in the corresponding contribution game. If there exists $i \in P \cap {\mathcal {N}}(0,{\psi }^P({\varvec{x_P}})]$ such that $\sum _{y\in \bar{{\mathcal {Y}}}} x_i(y)=0$, then there is $j \in P \cap {\mathcal {N}}(0,{\psi }^P({\varvec{x_P}})]{\setminus } \{i\}$ such that $x_j(\bar{y})>0$ for some $\bar{y} \le Y_i$. If we reduce the contribution of $j$ slightly and add it to $i$’s contribution, then $i$’s contribution is positive. $\square$

1.4 Proof of Proposition 4

(Necessity) Suppose to the contrary that $x_i^P>0$ and $y^{P{\setminus } \{i \}} \ge Y_i$ for some $i \in P$. By IR, $y^P \ge Y_i$. If $i$ participates, the payoff to $i$ is $\theta _i-x_i^P$. However, if $i$ deviates and chooses not to participate, $i$’s payoff is $\theta _i$. Thus, $i$ is made better off, a contradiction. If $y^{P\cup \{j\}}\ge Y_j>y^{P}$ and $\theta _j - x_j^{P \cup \{i\}}>0$ for some $j \notin P$, then $j$ is better off participating in $P$. Hence, for each $i \notin P$, if $y^{P\cup \{i\}}\ge Y_i>y^{P}$, then $\theta _i - x_i^{P \cup \{i\}}\le 0$. By IR for $P\cup \{i\}$, $\theta _i - x_i^{P \cup \{i\}}= 0$.

(Sufficiency) By IR, $V_i(y^{P},x_i^{P}) \ge 0$ for each $i \in P$. By IS, $V_i(y^{P{\setminus } \{i\}},x_i^{P{\setminus } \{i\}})=0$ for each $i \in P$ such that $x_i^P >0$. For each $k \in P$ such that $x_k^P =0$, $V_k(y^{P},x_i^{P})=0$ if $y^P < Y_k$ and $V_k(y^{P},x_i^{P})=\theta _k$ if $y^P \ge Y_k$. If $y^P < Y_k$, then $y^{P{\setminus } \{k\}} = y^{P}<Y_k$ and $V_k(y^{P{\setminus } \{k\}},x_k^{P{\setminus } \{k\}})=0$. If $y^P \ge Y_k$, then $V_k(y^{P{\setminus } \{k\}},x_k^{P{\setminus } \{k\}})\le \theta _k$. Thus, no agent in $P$ is made better off by switching to not participate. Let $j \notin P$ be such that $y^{P \cup \{j\}} \ge Y_j$. Then, $V_j(y^{P \cup \{j\}},x_j^{P \cup \{j\}})=\theta _j-x_j^{P \cup \{j\}}$. If $y^{P \cup \{j\}}\ge Y_j >y^P$, then by ES, $V_j(y^{P \cup \{j\}},x_j^{P \cup \{j\}})=V_j(y^{P},x_j^{P})=0$. If $y^{P \cup \{j\}}\ge y^P \ge Y_j$, then $V_j(y^{P \cup \{j\}},x_j^{P \cup \{j\}})\le V_j(y^{P},x_j^{P})=\theta _j$. Clearly, no $i \notin P$ such that $y^{P \cup \{i\}} < Y_i$ switches to participate. $\square$

1.5 Proof of Proposition 5

(5.1) Suppose to the contrary that $\sum _{j \in {\mathcal {N}}(y^{l-1},y^l]} \theta _j < c(y^{l})-c(y^{l-1})$ for some $l \in \{1,\ldots ,m\}$ such that $y^l \in {\mathcal {Y}}$. Let $P \subseteq N$ be such that $y^P=y^l$. By GE, $\sum _{j \in P \cap {\mathcal {N}}(y^{l-1},y^l]} \theta _j \ge c(y^{l})-c(y^{l-1})$. However, $\sum _{j \in P \cap {\mathcal {N}}(y^{l-1},y^l]} \theta _j \le \sum _{j \in {\mathcal {N}}(y^{l-1},y^l]} \theta _j < c(y^{l})-c(y^{l-1})$, which is a contradiction.

(5.2) Suppose to the contrary that there exists $y^{\prime } \in \arg \max _{y \in \{0\} \cup {\mathcal {Y}}} \sum _{j \in P} B_j(y)-c(y) {\setminus } \arg \max _{y \in \{0\} \cup \bar{{\mathcal {Y}}}} \sum _{j \in P} B_j(y)-c(y)$ for some $P \subseteq N$. Then, there exists $y^{\prime \prime } \in \{0\} \cup \bar{{\mathcal {Y}}} {\setminus } (\{0\} \cup {\mathcal {Y}})$ such that $\sum _{j \in P} B_j(y^{\prime \prime })-c(y^{\prime \prime })>\sum _{j \in P} B_j(y^{\prime })-c(y^{\prime })$.

We have $\sum _{j \in P} B_j(y^{\prime })-c(y^{\prime }) = \sum _{j \in P} B_j(y^P)-c(y^P)$ since $y^P \!\in \! \arg \max _{y \in \{0\} \cup \bar{{\mathcal {Y}}}} \sum _{j \in P}B_j(y)-c(y)$ by GE for $P$ and $y^P \!\in \! \{0\} \cup {\mathcal {Y}}$ and $y^{\prime } \!\in \! \arg \max _{y \in \{0\} \cup {\mathcal {Y}}} \sum _{j \in P}B_j(y)-c(y)$. In conclusion, $\sum _{j \in P} B_j(y^{\prime \prime })-c(y^{\prime \prime })>\sum _{j \in P} B_j(y^{\prime })-c(y^{\prime }) = \sum _{j \in P} B_j(y^P)-c(y^P),$ which implies that $y^P$ does not maximize $\sum _{j \in P}B_j(y)-c(y)$, a contradiction. $\square$

Appendix 2: Proofs of theorems

1.1 Proof of Theorem 1

Obviously, $(\gamma _i^*(P))_{i \in P}$ is a Nash equilibrium of the contribution game for each $P\subseteq ~N$.

We first prove that $P^*$ provides $y^*$ units of the public good given $(\gamma _i^*)_{i \in N}$: $y^{P^*}=y^*$. By the construction of $P^*$,

$$\begin{aligned} \sum _{j \in P^m\cup \cdots \cup P^k} \theta _j \ge c(y^m)-c(y^{k-1}) \quad \text {for each}\; k \in \{1,\ldots ,m\}\; \text {such that}\; y^k \in {\mathcal {Y}}\end{aligned}$$

where $y^0\equiv 0$. Hence, $y^{m} \in \arg \max _{y \in \{0\} \cup {\mathcal {Y}}} \sum _{j \in P^*}B_j(y)-c(y)$. By (5.2) of Proposition 5, $y^{m} \in \arg \max _{y \in \{0\} \cup \bar{{\mathcal {Y}}}} \sum _{j \in P^*}B_j(y)-c(y)$. By the definition of $(\gamma _i^*)_{i \in N}$, $y^{P^*}=y^m=y^*$.

We next prove that $P^*$ satisfies IS: $y^{P^* {\setminus } \{j\}}<Y_j$ for each $j\in P^*$. Suppose to the contrary that there exists $i\in P^*$ such that $y^{P^* {\setminus } \{i\}} \ge Y_i$. Let $l \in \{1,\ldots ,m\}$ be such that $i \in P^l$. Let $k \in \{1,\ldots ,m\}$ be such that $y^k \in {\mathcal {Y}}$ and $y^k = y^{P^* {\setminus } \{i\}}$. We first show $m>k \ge l >l-1$. Since $P^l \subseteq {\mathcal {N}}(y^{l-1},y^l]$, $y^{l-1}<Y_i \le y^l$. Since $y^{l-1}<Y_i$ and $Y_i \le y^k$, then $l-1<k$. Since ${\mathcal {Y}}$ is discrete and $y^l$ and $y^{l-1}$ lie next to each other, it is impossible that $l> k > l-1$. Thus, we have $k \ge l > l-1$. By the construction of $P^*$, $m>k$.⁸

Since $(\gamma ^*_j(P{\setminus } \{i\}))_{j \in P{\setminus } \{i\}}$ satisfies GE for $P{\setminus } \{i\}$, $y^k \in \arg \max _{y \in \{0\} \cup \bar{{\mathcal {Y}}}}\sum _{j \in P^* {\setminus } \{i\}}B_j(y)-c(y)$. Then, $\sum _{j \in (P^* {\setminus } \{i\})\cap {\mathcal {N}}(y^{l-1},y^k]}\theta _j \ge c(y^k)-c(y^{l-1})$. Since $P^*\cap {\mathcal {N}}(y^{l-1},y^k]=P^l\cup \cdots \cup P^k$,

$$\begin{aligned} \sum _{j \in (P^* {\setminus } \{i\})\cap {\mathcal {N}}(y^{l-1},y^k]}\theta _j&= \sum _{j \in P^* \cap {\mathcal {N}}(y^{l-1},y^k]}\theta _j - \theta _i = \sum _{j \in P^l\cup \cdots \cup P^k}\theta _j - \theta _i. \end{aligned}$$

Hence, $\sum _{j \in P^k \cup \cdots \cup P^l} \theta _j - \theta _i \ge c(y^k)-c(y^{l-1}).$

By the construction of $P^l$, $\sum _{j \in P^m \cup \cdots \cup P^l} \theta _j - \theta _i < c(y^m)-c(y^{l-1}),$ which is rewritten as

$$\begin{aligned} \sum _{j \in P^k \cup \cdots \cup P^l} \theta _j - \theta _i&< c(y^m)-c(y^{l-1})- \left[ \,\sum _{j \in P^{m} \cup \cdots \cup P^{k+1}} \theta _j \right] \\&= c(y^m)-c(y^k)+c(y^k)-c(y^{l-1})- \left[ \,\sum _{j \in P^{m} \cup \cdots \cup P^{k+1}} \theta _j \right] . \end{aligned}$$

Since $\sum _{j \in P^{m} \cup \cdots \cup P^{k+1}} \theta _j\ge c(y^m)-c(y^{k})$, $\sum _{j \in P^k \cup \cdots \cup P^l} \theta _j - \theta _i < c(y^k)-c(y^{l-1}),$ which is a contradiction. Hence, $P^*$ satisfies IS.

We finally confirm that $P^*$ vacuously satisfies ES since for each $Q$, if $P^* \subseteq Q$, $y^* \in \arg \max _{y \in \{0\} \cup \bar{{\mathcal {Y}}}} \sum _{i \in Q}B_i(y)-c(y)$ and $y^Q=y^*$ under $(\gamma ^*_i)_{i \in N}$. By Proposition 4, $P^*$ is supported at an equilibrium. $\square$

1.2 Proof of Theorem 2

We first construct a strategy profile and then show that it is feasible (Lemma 1) and strongly perfect (Lemma 2). Let $y^* \in \arg \max _{y \in \{0\} \cup \bar{{\mathcal {Y}}}} \sum _{j \in N}B_j(y)-c(y)$. We can take a subgame perfect Nash equilibrium $s^{*}=(s^{*1},(\gamma ^{*}_i)_{i \in N}) \in {\mathcal {S}}$ on the path of which $P^*$ provides $y^*$ as the method before Theorem 1. For each $i \in P^*$, denote ${\varvec{g_i}}=(g_i^z)_{z=1}^t \equiv \left( \gamma _i^{*z}\left( P^*\right) \right) _{z=1}^{t}$. Based on $s^*$, we construct another strategy profile $s=(s^1,(\gamma _i)_{i \in N}) \in {\mathcal {S}}$ in which every set of participants $P \subseteq N$ produces the same level of the public good as that at $s^*$, but the cost-shares may be different. Define $P(s) = P^*$. Define $(\gamma _i(P))_{i \in P}$ for each $P \subseteq N$ as follows:

• (a) If $P\cap P^*= \emptyset$, then $\gamma _{i}(P) \equiv \gamma ^*_i(P)$ for each $i \in P$.

• (b) If $P^* \subseteq P$, then for each $i \in P$, $\gamma _i(P) \equiv {\varvec{g_i}}$ if $i \in P^*$ and $\gamma _i(P) \equiv (0,\ldots ,0) \in {\mathbb {R}}_+^t$ otherwise.

• (c) Suppose that $P \subseteq N$ satisfies $P \cap P^* \ne \emptyset$ but not $P^* \subseteq P$. Let $y^P={\psi }^{P}\left( \left( \gamma ^*_i(P) \right) _{i \in P} \right)$. Then, $(\gamma _{i}(P))_{i\in P}$ is defined as follows:

  • (c.1) For each $i \in P$ such that $Y_i> y^{P}$ and each $k \in \{1,\ldots ,t\}$, $\gamma _{i}^k(P) \equiv 0$. For each $i \in P \cap {\mathcal {N}}\left( 0,y^{P}\right]$ and each $k \in \{1,\ldots ,t\}$ such that $\bar{y}^k>Y_i$, $\gamma _{i}^k(P) \equiv 0$.

  • (c.2) For each $k \in \{1,\ldots ,t\}$ such that $\bar{y}^k \in \bar{{\mathcal {Y}}}$ and $\bar{y}^k \le y^{P}$ and each $i \in P \cap {\mathcal {N}}\left( \bar{y}^{k-1}, y^{P}\right]$, $\theta _i-\sum _{z=k}^t \gamma _{i}^z(P) \ge 0$ and $\sum _{i \in P\cap {\mathcal {N}}(\bar{y}^{k-1}, y^{P}]}\gamma _{i}^k(P)=c(\bar{y}^{k})-c(\bar{y}^{k-1})$.

  • (c.3) For each $k \in \{1,\ldots ,t\}$ such that $\bar{y}^k \in \bar{{\mathcal {Y}}}$ and $\bar{y}^k \le y^{P}$ and each $i \in P\cap {\mathcal {N}}\left( \bar{y}^{k-1},y^P\right] \cap P^*$, $\sum _{z=k}^t \gamma _i^z(P) \ge \sum _{z=k}^t g_i^z$.

Properties (c.1) and (c.2) are the same as (i)–(iii) in Proposition 1. In this section, in addition to these conditions, we require (c.3). This property means that if a set of participants switches from $P^*$ to $P$ such that $P \cap P^* \ne \emptyset$ but not $P^* \subseteq P$, then, the common members between these sets whose demand level is met contribute at least the level before the switch. This property plays an important role in proving that $s$ is a strong perfect equilibrium. We first show that $s$ is feasible.

Lemma 1

There exists $(\gamma _i)_{i \in N}$ that satisfies (a), (b), and (c).

Proof

It is clear that $(\gamma _i)_{i \in N}$ can be constructed in a way that satisfies (a) and (b). Thus, we focus on each $P \subseteq N$ such that $P\cap P^* \ne \emptyset$ but not $P^* \subseteq P$.

Let $l \in \{1,\ldots ,t\}$ be such that $\bar{y}^l={\psi }^P\left( \left( \gamma ^*_i(P)\right) _{i \in P} \right)$. We can set that $\gamma ^k_{i}(P)=0$ for each $k$ such that $l<k\le t$ and each $i \in P$, obviously. We next consider the case of $k=l$. Define $\gamma _i^l(P)=0$ for each $i \in P \cap {\mathcal {N}}(0,\bar{y}^{l-1}]$. If $\sum _{i \in P \cap {\mathcal {N}}(\bar{y}^{l-1},\bar{y}^l]{\setminus } P^*}\theta _i\ge c(\bar{y}^l)-c(\bar{y}^{l-1})-\sum _{i \in P \cap {\mathcal {N}}(\bar{y}^{l-1},\bar{y}^l]\cap P^*} g^l_i$, then $\gamma _i^l(P)=g_i^l$ for each $i \in P \cap {\mathcal {N}}(\bar{y}^{l-1},\bar{y}^l]\cap P^*$ and define $(\gamma ^l_i(P))_{i \in P \cap {\mathcal {N}}(\bar{y}^{l-1},\bar{y}^l]{\setminus } P^*}$ such that $\sum _{i \in P \cap {\mathcal {N}}(\bar{y}^{l-1},\bar{y}^l]{\setminus } P^*}\gamma _i^l(P)=c(\bar{y}^l)-c(\bar{y}^{l-1})-\sum _{i \in P \cap {\mathcal {N}}(\bar{y}^{l-1},\bar{y}^l]\cap P^*} g^l_i$ and $\theta _i-\gamma _i^l(P)\ge 0$ for each $i \in P \cap {\mathcal {N}}(\bar{y}^{l-1},\bar{y}^l]{\setminus } P^*$. Otherwise, since $\bar{y}^l$ maximizes $\sum _{i \in N}B_i(y)-c(y)$ within $\{0\} \cup \bar{{\mathcal {Y}}}$, $\sum _{i \in P \cap {\mathcal {N}}(\bar{y}^{l-1},\bar{y}^l]}\theta _i \ge c(\bar{y}^l)-c(\bar{y}^{l-1})$. By this condition,

$$\begin{aligned} \sum _{i \in P \cap {\mathcal {N}}(\bar{y}^{l-1},\bar{y}^l]\cap P^*}\left( \theta _i - g^l_i\right)&\ge \ c(\bar{y}^l)-c(\bar{y}^{l-1})-\sum _{i \in P \cap {\mathcal {N}}(\bar{y}^{l-1},\bar{y}^l]\cap P^*} g^l_i\\&- \sum _{i \in P \cap {\mathcal {N}}(\bar{y}^{l-1},\bar{y}^l]{\setminus } P^*}\theta _i > 0. \end{aligned}$$

For each $i \in P\cap {\mathcal {N}}(\bar{y}^{l-1},\bar{y}^l] \cap P^*$, define $\varepsilon _i^l \ge 0$ as

$$\begin{aligned} \sum _{i \in P \cap {\mathcal {N}}(\bar{y}^{l-1},\bar{y}^l]\cap P^*} \left( \theta _i-g^l_i \right)&\ge \sum _{i \in P \cap {\mathcal {N}}(\bar{y}^{l-1},\bar{y}^l]\cap P^*}\varepsilon ^l_i\\&= c(\bar{y}^l)-c(\bar{y}^{l-1})-\sum _{i \in P \cap {\mathcal {N}}(\bar{y}^{l-1},\bar{y}^l]\cap P^*} g^l_i\\&- \sum _{i \in P \cap {\mathcal {N}}(\bar{y}^{l-1},\bar{y}^l]{\setminus } P^*}\theta _i \end{aligned}$$

and $\theta _i-g^l_i\ge \varepsilon _i^l$ for each $i \in P \cap {\mathcal {N}}(\bar{y}^{l-1},\bar{y}^l]\cap P^*$. Set $\gamma _i^l(P)=g_i^l+\varepsilon _i^l$ for each $i \in P \cap {\mathcal {N}}(\bar{y}^{l-1},\bar{y}^l]\cap P^*$ and $\gamma _i^l(P)=\theta _i$ for each $i \in P \cap {\mathcal {N}}(\bar{y}^{l-1},\bar{y}^l]{\setminus } P^*$. In conclusion, in any case, $\theta _i-\gamma _i^l(P)\ge 0$ for each $i \in P \cap {\mathcal {N}}(\bar{y}^{l-1},\bar{y}^l]$, $\sum _{i \in P \cap {\mathcal {N}}(\bar{y}^{l-1},\bar{y}^l]}\gamma _i^l(P)=c(\bar{y}^l)-c(\bar{y}^{l-1})$, and $\gamma _i^l(P) \ge g^l_i$ for each $i \in P \cap {\mathcal {N}}(\bar{y}^{l-1},\bar{y}^l] \cap P^*$.

Let $k \in \{1,\ldots ,l-1\}$. Suppose that $\gamma _i^z(P)$ is defined in a way that satisfies (c.1)–(c.3) for each $i \in P$ and each $z \in \{k+1,\ldots ,l \}$. We now construct $\gamma _i^k(P)$ for each $i \in P$. Define $\gamma _i^k(P)=0$ for each $i \in P \cap {\mathcal {N}}(0,\bar{y}^{k-1}]$.

Claim 1 provides a sufficient condition for $\gamma _i^k(P)$ for each $i \in P$ to satisfy (c.3) for $k$.

Claim 1

For each $i \in P \cap {\mathcal {N}}(\bar{y}^{k-1},\bar{y}^l] \cap P^*$, if $\gamma _i^k(P) \ge \max \{0,g_i^k-\sum _{z = k+1}^l (\gamma _i^z(P)-g^z_i) \}$, then $\sum _{z=k}^l \gamma _i^z(P) \ge \sum _{z=k}^l g_i^z$.

Proof of Claim 1

If $g_i^k-\sum _{z = k+1}^l (\gamma _i^z(P)-g^z_i)\ge 0$, it is trivial. If $g_i^k-\sum _{z = k+1}^l (\gamma _i^z(P)-g^z_i) < 0$, then $\sum _{z=k+1}^l \gamma _i^z(P) > \sum _{z=k}^l g^z_i$. Since $\gamma _i^k(P)\ge 0$, $\sum _{z=k}^l \gamma _i^z(P) \ge \sum _{z=k+1}^l \gamma _i^z(P) > \sum _{z=k}^l g^z_i$.$\square$

Denote $\Delta \equiv c(\bar{y}^k)-c(\bar{y}^{k-1})-\sum _{i \in P \cap {\mathcal {N}}(\bar{y}^{k-1},\bar{y}^l]\cap P^*} \max \{0,g_i^k-\sum _{z = k+1}^l (\gamma _i^k(P)-g^z_i) \}$. We show $\Delta >0$. Denote $\mathcal {X}\equiv \{i \in P\cap {\mathcal {N}}(\bar{y}^{k-1},\bar{y}^l ] \cap P^*|\, g_i^k-\sum _{z = k+1}^l (\gamma _i^z(P)-g^z_i)>0\}$. Then,

$$\begin{aligned} \Delta&= \underbrace{c(\bar{y}^k)-c(\bar{y}^{k-1})-\sum _{i \in P \cap {\mathcal {N}}(\bar{y}^{k-1},\bar{y}^l]\cap P^* \cap \mathcal {X}} g_i^k}_{(\alpha )}\\&\quad +\underbrace{\sum _{i \in P \cap {\mathcal {N}}(\bar{y}^{k-1},\bar{y}^l]\cap P^* \cap \mathcal {X}}\sum _{z = k+1}^l (\gamma _i^z(P)-g^z_i)}_{(\beta )}. \nonumber \end{aligned}$$

By the definition of ${\varvec{g_i}}$ and the condition that $P^* \subseteq P$ does not hold, we have

$$\begin{aligned} c(\bar{y}^k)-c(\bar{y}^{k-1})&= \sum _{i \in {\mathcal {N}}(\bar{y}^{k-1},\bar{y}^l] \cap P^*}g_i^k > \sum _{i \in P \cap {\mathcal {N}}(\bar{y}^{k-1},\bar{y}^l]\cap P^*}g_i^k \\&\ge \sum _{i \in P \cap {\mathcal {N}}(\bar{y}^{k-1},\bar{y}^l]\cap P^* \cap \mathcal {X}} g_i^k. \end{aligned}$$

Thus, $(\alpha )>0$. By the induction hypothesis, $(\beta )\ge 0$. Hence, $\Delta >0$.

Case 1 $\sum _{i \in P \cap {\mathcal {N}}(\bar{y}^{k-1},\bar{y}^l]{\setminus } P^*}(\theta _i-\sum _{z=k+1}^l \gamma _i^z(P)) \ge \Delta$.

Define $\gamma _i^k(P)$ for each $i \!\in \! P \cap {\mathcal {N}}(\bar{y}^{k-1},\bar{y}^l]{\setminus } P^*$ such that $\sum _{i \in P \cap {\mathcal {N}}(\bar{y}^{k-1},\bar{y}^l]{{\setminus }} P^*} \gamma _i^k(P) = \Delta$ and $\theta _i-\sum _{z=k+1}^l \gamma _i^z(P)\ge \gamma _i^k(P)$ for each $i \in P \cap {\mathcal {N}}(\bar{y}^{k-1},\bar{y}^l]{\setminus } P^*$. Set $\gamma _i^k(P)=\max \{0,g_i^k-\sum _{z = k+1}^l (\gamma _i^z(P)-g^z_i) \}$ for each $i \in P \cap {\mathcal {N}}(\bar{y}^{k-1},\bar{y}^l]\cap P^*$. For each $i \in P \cap {\mathcal {N}}(\bar{y}^{k-1},\bar{y}^l]\cap P^*$, if $g_i^k-\sum _{z = k+1}^l (\gamma _i^z(P)-g^z_i)>0$, then $\theta _i-\sum _{z=k}^l \gamma _i^z(P)=\theta _i-\sum _{z=k}^l g_i^z \ge 0$. Otherwise, $\theta _i-\sum _{z=k}^l \gamma _i^z(P)=\theta _i-\sum _{z=k+1}^l \gamma _i^z(P) \ge 0$ by the induction hypothesis.

Case 2 $\sum _{i \in P \cap {\mathcal {N}}(\bar{y}^{k-1},\bar{y}^l]{\setminus } P^*}(\theta _i-\sum _{z=k+1}^l \gamma _i^z(P)) < \Delta$.

By GE, $\sum _{i \in P\cap {\mathcal {N}}(\bar{y}^{k-1},\bar{y}^l]}\theta _i \ge c(\bar{y}^l)-c(\bar{y}^{k-1})$. By this condition,

$$\begin{aligned} \sum _{i \in P\cap {\mathcal {N}}(\bar{y}^{k-1},\bar{y}^l]}\theta _i -(c(\bar{y}^l)-c(\bar{y}^k)) \ge c(\bar{y}^k)-c(\bar{y}^{k-1}). \end{aligned}$$

By the induction hypothesis,

$$\begin{aligned} \sum _{i \in P\cap {\mathcal {N}}(\bar{y}^{k-1},\bar{y}^l]} \left[ \theta _i -\sum _{z=k+1}^l \gamma _i^z(P)\right] \ge c(\bar{y}^k)-c(\bar{y}^{k-1}), \end{aligned}$$

which is equal to

$$\begin{aligned}&\sum _{i \in P\cap {\mathcal {N}}(\bar{y}^{k-1},\bar{y}^l]\cap P^*} \left[ \theta _i -\sum _{z=k+1}^l \gamma _i^z(P)\right] \ge c(\bar{y}^k)-c(\bar{y}^{k-1})\\&\quad -\sum _{i \in P\cap {\mathcal {N}}(\bar{y}^{k-1},\bar{y}^l]{\setminus } P^*} \left[ \theta _i -\sum _{z=k+1}^l \gamma _i^z(P)\right] . \end{aligned}$$

We finally have

$$\begin{aligned}&\sum _{i \in P\cap {\mathcal {N}}(\bar{y}^{k-1},\bar{y}^l]\cap P^*} \left[ \theta _i -\sum _{z=k+1}^l \gamma _i^z(P)- \max \left\{ 0,g_i^k-\sum _{z = k+1}^l (\gamma _i^k(P)-g^z_i) \right\} \right] \\&\quad \ge \Delta -\sum _{i \in P\cap {\mathcal {N}}(\bar{y}^{k-1},\bar{y}^l]{\setminus } P^*} \left[ \theta _i -\sum _{z=k+1}^l \gamma _i^z(P)\right] . \end{aligned}$$

For each $i \in P\cap {\mathcal {N}}(\bar{y}^{k-1},\bar{y}^l]\cap P^*$, define $\varepsilon _i^k\ge 0$ as

$$\begin{aligned}&\sum _{i \in P\cap {\mathcal {N}}(\bar{y}^{k-1},\bar{y}^l]\cap P^*} \left[ \theta _i -\sum _{z=k+1}^l \gamma _i^z(P)- \max \left\{ 0,g_i^k-\sum _{z = k+1}^l (\gamma _i^k(P)-g^z_i) \right\} \right] \\&\quad \ge \ \sum _{i \in P\cap {\mathcal {N}}(\bar{y}^{k-1},\bar{y}^l]\cap P^*} \varepsilon _i^k \\&\quad = \Delta -\sum _{i \in P\cap {\mathcal {N}}(\bar{y}^{k-1},\bar{y}^l]{\setminus } P^*} \left[ \theta _i -\sum _{z=k+1}^l \gamma _i^z(P)\right] \end{aligned}$$

and $\theta _i-\sum _{z=k+1}^l \gamma _i^z(P) - \max \left\{ 0,g_i^k -\sum _{z = k+1}^l (\gamma _i^k(P)-g^z_i) \right\} \ge \varepsilon _i^k$ for each $i \in P\cap {\mathcal {N}}(\bar{y}^{k-1},\bar{y}^l]\cap P^*$. Let

$$\begin{aligned} \gamma _i^k(P) \!=\! {\left\{ \begin{array}{ll} \theta _i\!-\!\sum _{z=k+1}^l \gamma _i^z(P)&{} \text {for each}\; i \!\in \! P\cap {\mathcal {N}}(\bar{y}^{k-1},\bar{y}^l]{\setminus } P^*\\ \max \left\{ 0,g_i^k -\sum _{z = k+1}^l (\gamma _i^k(P)-g^z_i) \right\} \!+\!\varepsilon _i^k &{} \text {for each}\; i \!\in \! P\cap {\mathcal {N}}(\bar{y}^{k-1},\bar{y}^l]\cap P^* \end{array}\right. }. \end{aligned}$$

In any case, the vector constructed satisfies (c.1)–(c.3). $\square$

Note that $(\gamma _{i})_{i \in N}$ satisfies that for each $P \subseteq N$,

$$\begin{aligned}&{\psi }^P\left( \left( \gamma _i(P) \right) _{i \in P} \right) ={\psi }^P\left( \left( \gamma _i^*(P) \right) _{i \in P} \right) \; \text {and} \nonumber \\&{\psi }^P\left( \left( \gamma _i(P) \right) _{i \in P} \right) =y^*\;\quad \text {if and only if}\;\quad y^* \in \arg \max _{\{0\}\cup \bar{{\mathcal {Y}}}}\sum _{i \in P}B_i(y)-c(y). \nonumber \end{aligned}$$

Note also that $(\gamma _i(P))_{i \in P}$ satisfies GE and IR for each $P \subseteq N$, and hence, by Proposition 2, $(\gamma _i)_{i \in N}$ assigns a strong Nash equilibrium for each set of participants. Clearly, $s$ is one of the efficient subgame perfect Nash equilibria. We now show that $s$ is a strong perfect equilibrium.

Lemma 2

Strategy $s$ is a strong perfect equilibrium.

Proof

Note that $N=\left( P(s) \cap {\mathcal {N}}\left( 0,y^*\right] \right) \cup \left( {\mathcal {N}}\left( 0,y^*\right] {\setminus } P(s) \right) \cup {\mathcal {N}}\left( y^*,\bar{y}^t\right]$. At $s$, for each $i \in N$, $U_i(s)=\theta _i-\sum _{z=1}^t \gamma _i^z(P(s)) \ge 0$ if $i \in P(s) \cap {\mathcal {N}}(0,y^*]$, $U_i(s)=\theta _i$ if $i \in {\mathcal {N}}\left( 0,y^*\right] {\setminus } P(s)$, and $U_i(s)=0$ if $i \in {\mathcal {N}}(y^*,\bar{y}^t]$.

Suppose to the contrary that $s$ is not a strong perfect equilibrium. Then, there are $D \subseteq N$ and $s^{\prime }_D =(s^{\prime 1}_D,(\gamma _i^{\prime })_{i \in D})\in {\mathcal {S}}_D$ such that $U_i(s_D^{\prime },s_{-D}) \ge U_i(s)$ for each $i \in D$ with strict inequality for some $i \in D$. Let us define $s^{\prime } \equiv (s^{\prime }_D,s_{-D})$.

Claim 2

Let $i \in D \cap {\mathcal {N}}(0,y^*] {\setminus } P(s)$. Then, $U_i(s^{\prime })=\theta _i$; thus, $i$ is not made better off by the deviation. If $s_i^{\prime }=1$, then $\gamma _i^{\prime z}\left( P(s^{\prime }) \right) =0$ for each $z \in \{1,\ldots ,t\}$.

Proof of Claim 2

At $s$, each $i \in {\mathcal {N}}(0,\bar{y}^{*}] {\setminus } P(s)$ receives the payoff $\theta _i$, which is the greatest payoff that $i$ can obtain. Thus, even if $i$ joins in the deviation of $D$, then she neither contributes nor is made better off.$\square$

Claim 3

For each $i \in D\cap {\mathcal {N}}(y^{*},\bar{y}^t]$, if $i$ is made better off by the deviation of $D$, then there exists another agent $j \in D$ that is made worse off by the deviation.

Proof of Claim 3

Suppose that $i \in D \cap {\mathcal {N}}(y^{*},\bar{y}^t]$ is made better off by the deviation. Before the deviation, $i$’s payoff is zero. Hence, if $i$ is made better off, then $P(s^{\prime })$ provides at least $Y_{i}$ units of the public good; otherwise, her payoff is not positive. We assume that $P(s^{\prime })$ provides $Y_i$ units of the public good at $s^{\prime }$.⁹ If $P(s^{\prime })$ provides $Y_i$ units of the public good, it contributes $c(Y_i)$ in total. Suppose that at $s$, $P(s^{\prime })$ provides $\bar{y}^l$ for some $l \in \{0,1,\ldots ,t\}$ such that $\bar{y}^l \in \arg \max _{y \in \{0\} \cup \bar{{\mathcal {Y}}}} \sum _{i \in P(s^{\prime })}B_i(y)-c(y)$. Since $y^*$ is the maximal efficient demand level supported at $(\gamma _{j})_{j\in N}$, $\bar{y}^l \le y^*$. Thus, $\bar{y}^l \le y^* < Y_i$. Since $s_{-D}=s^{\prime }_{-D}$, $D\cap P(s^{\prime })$ must contribute $c(\bar{y}^l)-\sum _{j \in P(s^{\prime })\cap {\mathcal {N}}(0,\bar{y}^l]{\setminus } D}\sum _{z=1}^l \gamma _j^z(P(s^{\prime }))$ to the provision of $\bar{y}^l$ units of the public good and $c(Y_i)-c(\bar{y}^l)$ to increase a public good from $\bar{y}^l$ to $Y_i$ units.¹⁰ Hence,

$$\begin{aligned} \sum _{j\in P(s^{\prime })\cap D}\sum ^t_{z=1} \gamma _j^{\prime z}(P(s^{\prime })) \ge c(\bar{y}^l)\!-\!\sum _{j \in P(s^{\prime })\cap {\mathcal {N}}(0,\bar{y}^l]{\setminus } D}\sum _{z=1}^l \gamma _j^z(P(s^{\prime }))\!+c(Y_i)-c(\bar{y}^l).\nonumber \\ \end{aligned}$$

(4)

By the construction of $(\gamma _j)_{j \in P(s^{\prime })}$,

$$\begin{aligned} \sum _{j \in P(s^{\prime })\cap {\mathcal {N}}(0,\bar{y}^l]\cap D} \sum _{z=1}^t \gamma _j^z(P(s^{\prime }))&= \sum _{j \in P(s^{\prime })\cap {\mathcal {N}}(0,\bar{y}^l]\cap D} \sum _{z=1}^l \gamma _j^z(P(s^{\prime }))\nonumber \\&= c(\bar{y}^l)-\sum _{j \in P(s^{\prime })\cap {\mathcal {N}}(0,\bar{y}^l]{\setminus } D}\sum _{z=1}^l \gamma _j^z(P(s^{\prime })).\quad \quad \end{aligned}$$

(5)

By condition (c.3) of $(\gamma _j)_{j \in P(s^{\prime })}$,

$$\begin{aligned}&\sum _{j \in P(s^{\prime })\cap {\mathcal {N}}\left( 0,\bar{y}^l\right] \cap D} \sum _{z=1}^t \gamma _j^z(P(s^{\prime })) \nonumber \\&\quad = \sum _{j \in P(s^{\prime })\cap {\mathcal {N}}\left( 0,\bar{y}^l\right] \cap D \cap P(s)} \sum _{z=1}^t \gamma _j^z(P(s^{\prime }))+\sum _{j \in P(s^{\prime })\cap {\mathcal {N}}\left( 0,\bar{y}^l\right] \cap D {\setminus } P(s)} \sum _{z=1}^t \gamma _j^z(P(s^{\prime }))\nonumber \\&\quad \ge \sum _{j \in P(s^{\prime })\cap {\mathcal {N}}\left( 0,\bar{y}^l\right] \cap D \cap P(s)} \sum _{z=1}^t \gamma _j^z(P(s))=\sum _{j \in P(s^{\prime })\cap {\mathcal {N}}\left( 0,\bar{y}^l\right] \cap D} \sum _{z=1}^t \gamma _j^z(P(s)). \end{aligned}$$

(6)

Note that the final equality in (6) follows from $\sum _{z=1}^t \gamma _j^z(P(s))=0$ for each $j \notin P(s)$ and $\sum _{j \in P(s^{\prime })\cap {\mathcal {N}}\left( 0,\bar{y}^l\right] \cap D \cap P(s)} \sum _{z=1}^t \gamma _j^z(P(s)) + \sum _{j \in P(s^{\prime })\cap {\mathcal {N}}\left( 0,\bar{y}^l\right] \cap D {\setminus } P(s)}\sum _{z=1}^t \gamma _j^z(P(s)) =\sum _{j \in P(s^{\prime })\cap {\mathcal {N}}\left( 0,\bar{y}^l\right] \cap D} \sum _{z=1}^t \gamma _j^z(P(s))$.

For the deviation to be profitable,

$$\begin{aligned} \sum _{j \in P(s^{\prime })\cap {\mathcal {N}}\left( 0,\bar{y}^l\right] \cap D} \sum _{z=1}^t \gamma _j^z(P(s)) \ge \sum _{j \in P(s^{\prime })\cap {\mathcal {N}}(0,\bar{y}^l]\cap D} \sum _{z=1}^t \gamma _j^{\prime z}(P(s)). \end{aligned}$$

(7)

Since $\bar{y}^l \in \arg \max _{y \in \{0\} \cup \bar{{\mathcal {Y}}}} \sum _{i \in P(s^{\prime })}B_i(y)-c(y)$,

$$\begin{aligned} \sum _{j \in P(s^{\prime })\cap {\mathcal {N}}(\bar{y}^l,Y_i]\cap D}\theta _j \le \sum _{j \in P(s^{\prime })\cap {\mathcal {N}}(\bar{y}^l,Y_i]}\theta _j \le c(Y_i)-c(\bar{y}^l). \end{aligned}$$

(8)

By (4)–(8),

$$\begin{aligned} \sum _{j\in P(s^{\prime })\cap D}\sum _{z=1}^t \gamma _j^{\prime z}(P(s^{\prime }))&\ge \sum _{j \in P(s^{\prime })\cap {\mathcal {N}}(0,\bar{y}^l]\cap D} \sum _{z=1}^t \gamma _j^z(P(s^{\prime })) \nonumber \\&+ c(Y_i)-c(\bar{y}^l) \quad \text {(by (4) and (5))} \nonumber \\&\ge \sum _{j \in P(s^{\prime })\cap {\mathcal {N}}(0,\bar{y}^l]\cap D} \sum _{z=1}^t \gamma _j^z(P(s))\nonumber \\&+ \sum _{j \in P(s^{\prime })\cap {\mathcal {N}}(\bar{y}^l,Y_i]\cap D}\theta _j \quad \text {(by (6) and (8))} \nonumber \\&\ge \sum _{j \in P(s^{\prime })\cap {\mathcal {N}}(0,\bar{y}^l]\cap D} \sum _{z=1}^t \gamma _j^{\prime z}(P(s^{\prime }))\nonumber \\&+ \sum _{j \in P(s^{\prime })\cap {\mathcal {N}}(\bar{y}^l,Y_i]\cap D}\theta _j. \quad \text {(by (7))} \end{aligned}$$

(9)

Since

$$\begin{aligned} \sum _{j \in P(s^{\prime })\cap {\mathcal {N}}(0,\bar{y}^l]\cap D} \sum _{z=1}^t \gamma _j^{\prime z}(P(s^{\prime }))&= \sum _{j\in P(s^{\prime })\cap D}\sum _{z=1}^t \gamma _j^{\prime z}(P(s^{\prime }))\\&- \sum _{j\in P(s^{\prime })\cap {\mathcal {N}}(\bar{y}^l,Y_i]\cap D}\sum _{z=1}^t \gamma _j^{\prime z}(P(s^{\prime })), \end{aligned}$$

(9) implies that

$$\begin{aligned} \sum _{j\in P(s^{\prime })\cap {\mathcal {N}}(\bar{y}^l,Y_i]\cap D}\sum _{z=1}^t \gamma _j^{\prime z}(P(s^{\prime })) \ge \sum _{j \in P(s^{\prime })\cap {\mathcal {N}}(\bar{y}^l,Y_i]\cap D}\theta _j. \end{aligned}$$

(10)

By (10), if $\theta _i-\sum _{z=1}^t \gamma _i^{\prime z}(P(s^{\prime })) >0$ for $i \in P(s^{\prime }) \cap {\mathcal {N}}(y^*,Y_i] \cap D$, then there is another $j \in P(s^{\prime })\cap {\mathcal {N}}(\bar{y}^l,Y_i]\cap D$ such that $\theta _j-\sum _{z=1}^t \gamma _j^{\prime z}(P(s^{\prime }))<0$. Since $U_j(s) = 0$ for $j$, $j$ is made worse off by the deviation. $\square$

By Claim 3, the payoff to each $j \in {\mathcal {N}}(y^*,\bar{y}^t] \cap D$ is the same at $s$ and $s^{\prime }$; otherwise, the deviation by $D$ is not profitable. Since the payoff to $j$ at $s$ is zero, $j$’s payoff is also zero at $s'$, which means that the total contribution from $j$ is zero at $s^{\prime }$.

Claim 4

If an agent in $D \cap {\mathcal {N}}(0,y^*]\cap P(s)$ is made better off, then there is another agent in $D$ that is made worse off.

Proof of Claim 4

Suppose that $i \in D \cap {\mathcal {N}}(0,y^*]\cap P(s)$ is made better off. Agent $i$ receives the payoff $\theta _i-\sum _{z=1}^t \gamma _i^z(P) \ge 0$ at $s$. Thus, if $i$ is made better off, her demand level is fulfilled at $s^{\prime }$: otherwise, her payoff at $s^{\prime }$ is at most zero. Since by Claims 2 and 3, no agents outside $P(s) \cap {\mathcal {N}}(0,y^*]$ contribute a positive fee at $s^{\prime }$ and their behaviors have no bearing on whether $Y_i$ is fulfilled, it is sufficient to focus on the strategies of agents in $P(s) \cap {\mathcal {N}}(0,y^*]$. By the change of strategies of agents in $P(s) \cap {\mathcal {N}}(0,y^*]$, $P(s^{\prime }) \subseteq P(s)$.

First, consider the case of $s_i^{\prime }=1$: agent $i$ continues to participate after the deviation. For $i$ to be made better off, $i$’s total contribution at $s^{\prime }$ is less than that at $s$. However, by the construction of $(\gamma _i)_{i \in N}$, if $i \in D$ reduces her contribution, then $i$’s demand level is not fulfilled. Then, agent $k \in D{\setminus } \{i\}$ must increase her contribution to fulfill $i$’s demand level, which implies that $k$ is made worse off.

Second, consider the case of $s_i^{\prime }=0$: agent $i$ switches and decides not to participate. In this case, $P(s^{\prime }) \subsetneq P(s)$. Let $y^{P(s){\setminus } \{i\}} \in \arg \max _{y \in \{0\} \cup \bar{{\mathcal {Y}}}} \sum _{j \in P(s){\setminus } \{i\}}B_j(y) -c(y)$ and $q\in \{0,1,\ldots , t\}$ such that $\bar{y}^q \in \arg \max _{y \in \{0\} \cup \bar{{\mathcal {Y}}}} \sum _{j \in P(s^{\prime })}B_j(y) -c(y)$. Since $P(s)$ satisfies IS, $\bar{y}^q \le y^{P(s){\setminus } \{i\}} < Y_i$. Since $P(s^{\prime })$ produces $\bar{y}^q$ units of the public good at $s$ and $s_{N{\setminus } D}=s^{\prime }_{N {\setminus } D}$, $D \cap P(s^{\prime })$ must pay $\sum _{j \in P(s^{\prime })\cap D \cap {\mathcal {N}}(0,\bar{y}^q]} \sum _{z=1}^{q}\gamma _{j}^z(P(s^{\prime }))$ to produce $\bar{y}^q$ units of the public good at $s^{\prime }$. It also pays $c(Y_i)-c(\bar{y}^q)$ to increase the public good from $\bar{y}^q$ to $Y_i$. If $P(s^{\prime })$ produces $Y_i$ units of the public good at $s^{\prime }$, then $D \cap P(s^{\prime })$ pays at least $c(Y_i)-c(\bar{y}^q)+\sum _{j \in P(s^{\prime })\cap D \cap {\mathcal {N}}(0,\bar{y}^q]} \sum _{z=1}^{q}\gamma _{j}^z(P(s^{\prime }))$; hence,

$$\begin{aligned} \sum _{j \in P(s^{\prime })\cap D \cap {\mathcal {N}}(0,\bar{y}^q]} \sum _{z=1}^{t}\gamma _{j}^{\prime z}(P(s^{\prime })) \!\ge \! c(Y_i)-c(\bar{y}^q)\!+\!\sum _{j \in P(s^{\prime })\cap D \cap {\mathcal {N}}(0,\bar{y}^q]} \sum _{z=1}^{q}\gamma _{j}^z(P(s^{\prime })). \end{aligned}$$

By condition (c.3) of $(\gamma _{i})_{i \in N}$,

$$\begin{aligned} c(Y_i)-c(\bar{y}^q)+\sum _{j \in P(s^{\prime })\cap D \cap {\mathcal {N}}(0,\bar{y}^q]} \sum _{z=1}^{q}\gamma _{j}^z(P(s^{\prime }))> \sum _{j \in P(s^{\prime })\cap D \cap {\mathcal {N}}(0,\bar{y}^q]} \sum _{z=1}^{q}\gamma _{j}^z(P(s)) \end{aligned}$$

Thus,

$$\begin{aligned} \sum _{j \in P(s^{\prime })\cap D \cap {\mathcal {N}}(0,\bar{y}^q]} \sum _{z=1}^{t}\gamma _{j}^{\prime z}(P(s^{\prime }))> \sum _{j \in P(s^{\prime })\cap D \cap {\mathcal {N}}(0,\bar{y}^q]} \sum _{z=1}^{q}\gamma _{j}^z(P(s)). \end{aligned}$$

Note that $\sum _{j \in P(s^{\prime })\cap D \cap {\mathcal {N}}(0,\bar{y}^q]} \sum _{z=1}^{q}\gamma _{j}^z(P(s))=\sum _{j \in P(s^{\prime })\cap D \cap {\mathcal {N}}(0,\bar{y}^q]} \sum _{z=1}^{t}\gamma _{j}^z(P(s))$. Hence, there is $j \in P(s^{\prime }) \cap D \cap {\mathcal {N}}(0,\bar{y}^q]$ such that $\sum _{z=1}^{t}\gamma _{j}^{\prime z}(P(s^{\prime })) > \sum _{z=1}^{t}\gamma _{j}^z(P(s))$, which implies that $j$ is made worse off by this deviation. $\square$

By Claims 2, 3, and 4, it is impossible that each member of $D$ is not made worse off and at least one member is made better off. $\square$ $\square$

1.3 Proof of Theorem 3

Let $s=(s^1_i,\gamma _i)_{i \in N}$ be a strong perfect equilibrium. Denote $y^*\equiv {\psi }^{P(s)}\left( \gamma _i\left( P(s) \right) _{i \in P(s)} \right)$. By Theorem 2, $y^*$ is an efficient level of the public good. Let $u=(u_i)_{i \in N}$ be the payoffs at $s$; then, $u_i=B_i\left( y^* \right) -\sum _{z=1}^t \gamma ^z_{i}\left( P(s) \right)$ if $i \in P(s)$ and $u_i=B_i\left( y^* \right)$ otherwise. We need to show that $\sum _{i \in T}u_i \ge W(T)$ for each $T \subseteq N$ with equality if $T=N$. For each $T \subseteq N$, let $y^T\in \arg \max _{y \in \{0\} \cup \bar{{\mathcal {Y}}}}\sum _{i \in T} B_i(y)-c(y)$. By the direct application of the result of Branzei et al. (2005), $(u_i)_{i \in P(s)}$ belongs to the core of the game $(P(s),W)$ in which $P(s)$ is the set of agents and $W(T)$ is similarly defined for each $T \subseteq P(s)$. Thus, $\sum _{i \in T}u_i \ge W(T)$ for each $T \subseteq P(s)$.

Let $Q \subseteq N$ be such that $P(s) \cap Q = \emptyset$. Then, $u_i=\theta _i$ if $i \in Q$ such that $y^* \ge Y_i$ and $u_i=0$ otherwise. We have $W(Q)=\sum _{i \in Q \cap {\mathcal {N}}\left( 0,y^{Q}\right] }\theta _i - c(y^{Q})$. If $y^* < y^{Q}$, then $\sum _{i \in Q}u_i-W(Q) = -\sum _{i \in Q \cap {\mathcal {N}}\left( y^*,y^{Q} \right] }\theta _i+c(y^{Q})$. By the efficiency of $y^*$, $\sum _{i \in {\mathcal {N}}\left( y^*, y^Q \right] }\theta _i \le c(y^Q)-c(y^*)$. In addition, $\sum _{i \in Q \cap {\mathcal {N}}\left( y^*, y^Q \right] }\theta _i\le \sum _{i \in {\mathcal {N}}\left( y^*, y^Q \right] }\theta _i \le c(y^Q)-c(y^*)<c(y^Q)$. Thus, if $y^* < y^{Q}$, then $\sum _{i \in Q}u_i > W(Q)$. If $y^*=y^{Q}$, then, trivially, $\sum _{i \in Q}u_i > W(Q)$. If $y^* > y^{Q}$, then $\sum _{i \in Q}u_i-W(Q)= \sum _{i \in Q \cap {\mathcal {N}}\left( y^{Q},y^*\right] }\theta _i+c(y^{Q})>0$.

Let $R \subseteq N$ be such that $R \cap P(s) \ne \emptyset$ but not $R \subseteq P(s)$. We have $\sum _{i \in R}u_i = \sum _{i \in R \cap {\mathcal {N}}(0,y^*] \cap P(s)}\left( \theta _i - \sum _{z=1}^t \gamma _{i}^z\left( P(s)\right) \right) +\sum _{i \in R \cap {\mathcal {N}}\left( 0,y^*\right] {\setminus } P(s)} \theta _i$ and $W(R)=\sum _{i \in R \cap {\mathcal {N}}\left( 0,y^{R}\right] }\theta _i - c(y^{R})$. First, suppose that $y^R>y^*$. By the budget balance condition, $\sum _{i \in R \cap {\mathcal {N}}\left( 0,y^{*} \right] \cap P(s)} \sum _{z=1}^t \gamma _{i}^z\left( P(s)\right) \le c(y^*)$. By the efficiency of $y^*$, $\sum _{i \in R \cap {\mathcal {N}}\left( y^*, y^R \right] }\theta _i \le \sum _{i \in {\mathcal {N}}\left( y^*, y^R \right] }\theta _i \le c(y^R)-c(y^*)$. Thus,

$$\begin{aligned} \sum _{i \in R}u_i-W(R)&= -\sum _{i \in R \cap {\mathcal {N}}\left( y^*,y^{R} \right] }\theta _i - \sum _{i \in R \cap {\mathcal {N}}\left( 0,y^{*} \right] \cap P(s)} \sum _{z=1}^t\gamma _{i}^z\left( P(s)\right) +c(y^{R}) \\&\ge c(y^R)-c(y^*)- \sum _{i \in R \cap {\mathcal {N}}\left( y^*,y^{R} \right] }\theta _i \ge 0. \end{aligned}$$

Second, if $y^*=y^{R}$, then $\sum _{i \in R}u_i \ge W(R)$, trivially. Third, suppose that $y^R<y^*$. By IR for $P(s)$, $\sum _{i \in R \cap {\mathcal {N}}\left( y^R, y^* \right] \cap P(s)}\left( \theta _i-\sum _{z =1}^t \gamma _i^z\left( P(s)\right) \right) \ge 0$. Since $\left( \gamma _i(P(s)) \right) _{i \in P(s)}$ is a Nash equilibrium of the contribution game when $P(s)$ is a set of participants, by Proposition 1, $c(y^R) = \sum _{j \in {\mathcal {N}}\left( 0,y^*\right] \cap P(s)} \sum _{z=1}^t \gamma _i^z\left( P(s)\right) = \sum _{i \in {\mathcal {N}}\left( 0,y^R\right] \cap P(s)}\sum _{z=1 }^t \gamma _i^z\left( P(s)\right) +\sum _{i \in {\mathcal {N}}\left( y^R,y^*\right] \cap P(s)}\sum _{z=1 }^t \gamma _i^z\left( P(s)\right) \ge \sum _{i \in {\mathcal {N}}\left( 0,y^R\right] \cap P(s)} \sum _{z=1 }^t \gamma _i^z\left( P(s)\right) \ge \sum _{i \in R \cap {\mathcal {N}}\left( 0,y^R\right] \cap P(s)}\sum _{z=1 }^t \gamma _i^z\left( P(s)\right)$. Thus,

$$\begin{aligned} \sum _{i \in R}u_i- W(R)&= \sum _{i \in R \cap {\mathcal {N}}\left( y^R,y^* \right] \cap P(s)}\theta _i-\sum _{i \in R \cap {\mathcal {N}}\left( 0,y^*\right] \cap P(s)}\sum _{z =1}^t\gamma _i^z\left( P(s)\right) +c(y^R)\\&+ \sum _{i \in R \cap {\mathcal {N}}(y^R,y^*]{\setminus } P(s)} \theta _i \\&= \sum _{i \in R \cap {\mathcal {N}}\left( y^R,y^* \right] \cap P(s)}\left( \theta _i - \sum _{z=1}^t \gamma _i^z\left( P(s)\right) \right) + c(y^R)\\&- \sum _{i \in R \cap {\mathcal {N}}\left( 0,y^R \right] \cap P(s)}\sum _{z=1}^t \gamma _i^z\left( P(s)\right) + \sum _{i \in R \cap {\mathcal {N}}(y^R,y^*] {\setminus } P(s)} \theta _i >0. \end{aligned}$$

$\square$