Warm-glow giving in networks with multiple public goods

Abstract

This paper explores a voluntary contribution game in the presence of warm-glow effects. There are many public goods and each public good benefits a different group of players. The structure of the game induces a bipartite network structure, where players are listed on one side and the public good groups they form are listed on the other side. The main result of the paper shows the existence and uniqueness of a Nash equilibrium. The unique Nash equilibrium is also shown to be asymptotically stable. Then the paper provides some comparative statics analysis regarding pure redistribution, taxation and subsidies. It appears that small redistributions of wealth may sometimes be neutral, but generally, the effects of redistributive policies depend on how public good groups are related in the contribution network structure.

Appendix

Given a contribution structure g, let ${\mathbf {x}}_g$ stand for the column vector of contributions: ${\mathbf {x}}_g$ is the link by link profile of contributions and has size r(g). The links in ${\mathbf {x}}_g$ are sorted in lexicographic order: the contribution $x_{ij}$ is listed above the contribution $x_{kl}$ when $i<k$ or when $i = k$ and $j < l$. For the contribution structures $g_1$ and $g_2$ given in Fig. 2,

$$\begin{aligned} {\mathbf {x}}_{g_1}= \left( \begin{array}{l} x_{11} \\ x_{12} \\ x_{21} \\ x_{22} \end{array} \right) \quad \text {and} \quad {\mathbf {x}}_{g_2}= \left( \begin{array}{l} x_{11} \\ x_{21} \\ x_{22} \\ x_{32} \end{array} \right) . \end{aligned}$$

The Nash equilibrium of the multiple public goods game is noted ${\mathbf {x}}_g^*$.

Proof of Theorem 1

Because of the budget constraints, the allowed contributions are limited by the requirement that ${\mathbf {x}}_g$ be selected from a convex and compact set S such that

$$\begin{aligned} S = \prod \limits _{ij \in L}{[0,w_i]} \subset {\mathbb {R}}_+^{r(g)}. \end{aligned}$$

Then, the existence of a Nash equilibrium follows from fixed point arguments (such as Kakutani fixed point theorem) as in Theorem 1 of Rosen (1965).

To prove the uniqueness of the Nash equilibrium, Theorems 2 and 6 of Rosen (1965) are applied, which entail that the Nash equilibrium of the multiple public goods game is unique whenever the $r(g) \times r(g)$ Jacobian matrix of marginal utilities ${\mathbf {J}}({\mathbf {x}}_g)$ is a symmetric negative definite matrix for all ${\mathbf {x}}_g \in S$. Observe that, for all $ij \in L$,

$$\begin{aligned} \frac{\partial ^2 U_i}{\partial x_{kl} \partial x_{\textit{ij}}} \left( {\mathbf {x}}_g \right) = \left\{ \begin{array}{ll} b''_j \left( G_j \right) + \delta ''_{ij} \left( x_{ij} \right) + c''_i \left( w_i - X_i \right) ,&{}\text { for } kl \in L \text { s.t. } kl = \textit{ij};\\ c''_i \left( w_i - X_i \right) , &{}\text { for } kl \in L \text { s.t. } k = i \hbox { and } l \ne j;\\ b''_j \left( G_j \right) , &{} \text { for } kl \in L \text { s.t. } k \ne i\hbox { and }l = j;\\ 0, &{}\text { for } kl \in L \text { s.t. } k \ne i\hbox { and }l \ne j, \end{array} \right. \end{aligned}$$

so ${\mathbf {J}}( {\mathbf {x}}_g)$ is a symmetric matrix which can be decomposed as

$$\begin{aligned} {\mathbf {J}} \left( {\mathbf {x}}_g \right) = {\mathbf {B}}\left( {\mathbf {x}}_g \right) + \mathbf {\Delta } \left( {\mathbf {x}}_g \right) + {\mathbf {C}} \left( {\mathbf {x}}_g \right) , \end{aligned}$$

where ${\mathbf {B}}({\mathbf {x}}_g)$ is the Jacobian matrix of marginal collective benefits, $\mathbf {\Delta }({\mathbf {x}}_g)$ is the Jacobian matrix of marginal warm-glow, and ${\mathbf {C}}({\mathbf {x}}_g)$ is the Jacobian matrix of marginal private consumption. Both ${\mathbf {B}}({\mathbf {x}}_g)$, $\mathbf {\Delta }({\mathbf {x}}_g)$ and ${\mathbf {C}}({\mathbf {x}}_g)$ are symmetric matrices. Moreover, $\mathbf {\Delta }({\mathbf {x}}_g)$ is a diagonal matrix with all diagonal elements negative since under Assumption 1, $\delta ''_{ij}(.) < 0$ for all $ij \in L$. Then, $\mathbf {\Delta }({\mathbf {x}}_g)$ is negative definite for all ${\mathbf {x}}_g \in S$. In the following lemmas, it is shown that both ${\mathbf {B}}({\mathbf {x}}_g)$ and ${\mathbf {C}}({\mathbf {x}}_g)$ are negative semidefinite for all ${\mathbf {x}}_g \in S$, so ${\mathbf {J}}({\mathbf {x}}_g)$ is a sum of a symmetric negative definite matrix and two symmetric negative semidefinite matrices. Hence, ${\mathbf {J}}({\mathbf {x}}_g)$ is symmetric negative definite for all ${\mathbf {x}}_g \in S$, and uniqueness is established. $\square$

Lemma 1

${\mathbf {B}}({\mathbf {x}}_g)$ is negative semidefinite for all ${\mathbf {x}}_g \in S$.

Proof

To show that ${\mathbf {B}}({\mathbf {x}}_g)$ is negative semidefinite for all ${\mathbf {x}}_g \in S$, it is proved that there exists a matrix ${\mathbf {R}}_g$, with possibly dependent columns, such that $-{\mathbf {B}}({\mathbf {x}}_g) = {{\mathbf {R}}_g}^\mathsf{T} {\mathbf {R}}_g$ (see Strang 1988, p. 333). Observe that, for all $ij \in L$,

$$\begin{aligned} - \frac{\partial ^2 b_j}{\partial x_{kl} \partial x_{ij}} \left( {\mathbf {x}}_g \right) = \left\{ \begin{array}{ll} - b''_j \left( G_j \right) , &{} \text { for } kl \in L \text { s.t. } l = j;\\ 0, &{}\text { for } kl \in L \text { s.t. } l \ne j, \end{array} \right. \end{aligned}$$

so $- {\mathbf {B}}({\mathbf {x}}_g)$ is a symmetric matrix. For $s \in \{1, \ldots ,m\}$, let ${\mathbf {v}}^s \in {\mathbb {R}}_+^{r(g)}$ be such that

$$\begin{aligned} v_{ij}^s = \left\{ \begin{array}{ll} \sqrt{-b''_j ( G_j )}, &{} \text { for } ij \in L \text { s.t. } j = s;\\ 0, &{} \text { for } ij \in L \text { s.t. } j \ne s. \end{array} \right. \end{aligned}$$

Define ${\mathbf {R}}_g$ as a partitioned matrix such that

$$\begin{aligned} {{\mathbf {R}}_g}^\mathsf{T} = \left( \begin{array}{ccc} {\mathbf {v}}^1&\ldots&{\mathbf {v}}^m \end{array} \right) _{r(g) \times m}. \end{aligned}$$

It is straightforward to check that $-{\mathbf {B}}({\mathbf {x}}_g) = {{\mathbf {R}}_g}^\mathsf{T} {\mathbf {R}}_g$, so ${\mathbf {B}}({\mathbf {x}}_g)$ is negative semidefinite for all ${\mathbf {x}}_g \in S$. $\square$

Lemma 2

${\mathbf {C}}({\mathbf {x}}_g)$ is negative semidefinite for all ${\mathbf {x}}_g \in S$.

Proof

Let’s prove that there exists a matrix ${\mathbf {R}}_g$ such that $- {\mathbf {C}}({\mathbf {x}}_g) = {{\mathbf {R}}_g}^\mathsf{T} {\mathbf {R}}_g$. Observe that, for all $ij \in L$,

$$\begin{aligned} - \frac{\partial ^2 c_i}{\partial x_{kl} \partial x_{ij}} \left( {\mathbf {x}}_g \right) = \left\{ \begin{array}{ll} - c''_i \left( w_i - X_i \right) , &{} \text { for } kl \in L \text { s.t. } k = i;\\ 0, &{}\text { for } kl \in L \text { s.t. } k \ne i, \end{array} \right. \end{aligned}$$

so $- {\mathbf {C}}({\mathbf {x}}_g)$ is a symmetric matrix. For $t \in \{1 , \ldots , n\}$, let ${\mathbf {w}}^t \in {\mathbb {R}}_+^{r(g)}$ be such that

$$\begin{aligned} w_{ij}^t = \left\{ \begin{array}{ll} \sqrt{- c''_i ( w_i - X_i )}, &{} \text { for } ij \in L \text { s.t. } i = t; \\ 0, &{} \text { for } ij \in L \text { s.t. } i \ne t. \end{array} \right. \end{aligned}$$

Define ${\mathbf {R}}_g$ as a partitioned matrix such that

$$\begin{aligned} {{\mathbf {R}}_g}^\mathsf{T} = \left( \begin{array}{ccc} {\mathbf {w}}^1&\ldots&{\mathbf {w}}^n \end{array} \right) _{r(g) \times n}. \end{aligned}$$

It is straightforward to check that $- {\mathbf {C}}({\mathbf {x}}_g) = {{\mathbf {R}}_g}^\mathsf{T} {\mathbf {R}}_g$, so ${\mathbf {C}}({\mathbf {x}}_g)$ is negative semidefinite for all ${\mathbf {x}}_g \in S$. $\square$

Proof of Theorem 2

Since asymptotic stability is a local property, it can be assumed that clearly inactive links remain inactive following a small change of contributions at the other links (see, e.g.,Bramoullé et al. 2014; Allouch 2015). Hence, under Assumption 2, the dynamic system reduces to

$$\begin{aligned} {\dot{x}_{\textit{ij}}} = \phi _{ij} \left( G_{-i,j} , w_i - X_{i,-j} \right) - x_{\textit{ij}}, \quad \text {for all }ij \in B. \end{aligned}$$

Let ${\mathbf {Z}} ( {\mathbf {x}}_B )$ be the $r(B) \times r(B)$ Jacobian matrix of the function $z_{ij}({\mathbf {x}}_B ) = \phi _{ij} ( G_{-i,j} , w_i - X_{i,-j} ) - x_{ij}$ for all $ij \in B$. To prove the asymptotic stability of the Nash equilibrium, Lyapunov’s indirect method is applied. This entails that the Nash equilibrium of the multiple public goods game is asymptotically stable whenever the real part of each eigenvalue of ${\mathbf {Z}} ( {\mathbf {x}}^*_B )$ is negative.²²

For all $ij \in B$, observe that

$$\begin{aligned} \frac{\partial z_{ij}}{\partial x_{kl}} \left( {\mathbf {x}}_B \right) = \left\{ \begin{array}{ll} - 1 , &{} \text { for } kl \in B \text { s.t. } kl = \textit{ij};\\ \frac{- c''_i \left( w_i - X_i \right) }{b''_j \left( G_j \right) + \delta ''_{\textit{ij}} \left( x_{ij} \right) + c''_i \left( w_i - X_i \right) }, &{} \text { for } kl \in B \text { s.t. } k = i\hbox { and }l \ne j;\\ \frac{- b''_j \left( G_j \right) }{b''_j \left( G_j \right) + \delta ''_{ij} \left( x_{ij} \right) + c''_i \left( w_i - X_i \right) }, &{} \text { for } kl \in B \text { s.t. } k \ne i\hbox { and } l = j;\\ 0, &{} \text { for } kl \in B \text { s.t. } k \ne i\hbox { and }l \ne j, \end{array} \right. \end{aligned}$$

so ${\mathbf {Z}} ( {\mathbf {x}}_B )$ is an asymmetric matrix which can be decomposed as

$$\begin{aligned} {\mathbf {Z}} \left( {\mathbf {x}}_B \right) = {\mathbf {Y}} \left( {\mathbf {x}}_B \right) {\mathbf {J}} ( {\mathbf {x}}_B ) , \end{aligned}$$

where ${\mathbf {J}} ( {\mathbf {x}}_B )$ is the Jacobian matrix of marginal utilities and ${\mathbf {Y}} ( {\mathbf {x}}_B )$ is a diagonal matrix with all diagonal elements positive, i.e.,

$$\begin{aligned} \left[ {\mathbf {Y}} \left( {\mathbf {x}}_B \right) \right] _{ij,ij} = - \frac{1}{b''_j \left( G_j \right) + \delta ''_{ij} \left( x_{ij} \right) + c''_i \left( w_i - X_i \right) } > 0, \quad \text {for all } ij \in B. \end{aligned}$$

Then, ${\mathbf {Y}} ( {\mathbf {x}}_B )$ is a symmetric positive definite matrix for all ${\mathbf {x}}_B \in S_B = \prod \nolimits _{ij \in B}{[0,w_i]}$. It has been shown in the proof of Theorem 1 that under Assumption 1, ${\mathbf {J}} ( {\mathbf {x}}_g )$ is a symmetric negative definite matrix for all ${\mathbf {x}}_g \in S$. Given that any principal submatrix of a symmetric negative definite matrix is symmetric negative definite, ${\mathbf {J}} ( {\mathbf {x}}_B )$ is a symmetric negative definite matrix for all ${\mathbf {x}}_B \in S_B$. It follows that $- {\mathbf {Z}}({\mathbf {x}}_B)$ is the product of two symmetric positive definite matrices, ${\mathbf {Y}}({\mathbf {x}}_B)$ and $-{\mathbf {J}}({\mathbf {x}}_B)$. By Theorem 2 in Ballantine (1968), all the eigenvalues of $- {\mathbf {Z}}({\mathbf {x}}_B)$ are real and positive for all ${\mathbf {x}}_B \in S_B$. Thus, all the eigenvalues of ${\mathbf {Z}}({\mathbf {x}}^*_B)$ are real and negative, and asymptotic stability of the Nash equilibrium is established. $\square$

Proof of Proposition 1

Totally differentiating the best-response functions at each link $ij \in B$ yields

$$\begin{aligned} dx_{ij} = \frac{\partial \phi _{ij}}{\partial G_{-i,j}}dG_{-i,j} + \frac{\partial \phi _{ij}}{\partial (w_i - X_{i,-j})} \left( dw_i - dX_{i,-j} \right) . \end{aligned}$$

It follows that

$$\begin{aligned} dx_{ij} = - \frac{b''_j}{b''_j + \delta ''_{ij} + c''_i} dG_{-i,j} + \frac{c''_i}{b''_j + \delta ''_{ij} + c''_i} \left( dw_i - dX_{i,-j} \right) , \end{aligned}$$

or equivalently, since $dG_{-i,j} = dG_j - dx_{ij}$,

$$\begin{aligned} dx_{ij} = - \frac{b''_j}{\delta ''_{ij} + c''_i} dG_j + \alpha _{ij} \left( dw_i - dX_{i,-j} \right) . \end{aligned}$$

Since $C=D=\emptyset$, all links are clearly active, i.e., $L=B$. Hence, summing across all $a_i \in N_g(p_j)$ and solving for $d G_j$ yields

$$\begin{aligned} d G_j = k_j \sum \limits _{a_i \in N_g(p_j)}{\left\{ \alpha _{ij} \left( dw_i - d X_{i, -j} \right) \right\} , \quad \text { for all }\ p_j \in P, } \end{aligned}$$

(1)

where

$$\begin{aligned} k_j = \left( 1 + \sum \limits _{a_i \in N_g (p_j)}{\frac{b''_j}{\delta ''_{ij} + c''_i}} \right) ^{-1} \in \left( 0 , 1 \right] . \end{aligned}$$

Since $\alpha _{ij} = \alpha _j$ for all $ij \in L$, Eq. (1) becomes

$$\begin{aligned} dG_j = k_j \alpha _j \sum \limits _{a_i \in N_g(p_j)}{\{ dw_i - dX_{i,-j} \} }, \quad \text {for all }p_j \in P. \end{aligned}$$

Moreover, since g is a complete bipartite graph, it holds that $N_g(a_i) = P$ for all $a_i \in A$, and equivalently $N_g(p_j) = A$ for all $p_j \in P$. Hence,

$$\begin{aligned} \sum \limits _{a_i \in N_g(p_j)} dw_i = \sum \limits _{a_i \in A} dw_i = 0 \end{aligned}$$

and

$$\begin{aligned} \sum \limits _{a_i \in N_g(p_j)}{dX_{i,-j}} = \sum \limits _{a_i \in A}{dX_{i,-j}} = \sum \limits _{p_l \in P \backslash \{p_j \}}{dG_l}. \end{aligned}$$

It follows that, for all $p_j \in P$,

$$\begin{aligned} dG_j = - k_j \alpha _j \sum \limits _{p_l \in P \backslash \{p_j \}}{dG_l}. \end{aligned}$$

From this last equation, it appears that

$$\begin{aligned} \sum \limits _{p_l \in P}{dG_l} = \left( 1 - \frac{1}{k_1 \alpha _1} \right) dG_1 = \ldots = \left( 1 - \frac{1}{k_m \alpha _m} \right) dG_m, \end{aligned}$$

so it holds that

$$\begin{aligned} \text {sign}\left( dG_1\right) = \ldots = \text {sign}\left( dG_m\right) . \end{aligned}$$

Then, for all $p_j \in P$,

$$\begin{aligned} \text {sign}\left( dG_j\right)= & {} \text {sign}\left( \sum \limits _{p_l \in P \backslash \{p_j \}}{dG_l}\right) \\= & {} \text {sign}\left( k_j \alpha _j \sum \limits _{p_l \in P \backslash \{p_j \}}{dG_l}\right) \\= & {} \text {sign}\left( -dG_j\right) \end{aligned}$$

if and only if $dG_j = 0$. $\square$

Proof of Proposition 2

When the contribution structure is complete, a best-response function of the form given is sufficient since identical values of the altruism coefficient among all agents with respect to each public good is sufficient. The remainder of the proof is therefore devoted to the necessary condition.

Since $C=D=\emptyset$, $x_{ij} = \phi _{ij}(G_{-i,j} , w_i - X_{i,-j})$ holds for all $ij \in L$. Moreover, since $dG_j = 0$ for all $p_j \in P$, the total differential of the best-response functions given in the proof of Proposition 1 yields

$$\begin{aligned} dx_{ij} = \alpha _j \left( dw_i - dX_{i,-j} \right) , \quad \text { for all }\ ij \in L, \end{aligned}$$

where $\alpha _j = \alpha _j({\mathbf {x}}_g^*)$. This implies that $\phi _{ij}(G_{-i,j} , w_i - X_{i,-j})$ is linear in $w_i - X_{i,-j}$. Then, it holds that

$$\begin{aligned} x_{ij} = \phi _{ij} \left( G_{-i,j} , w_i - X_{i,-j} \right) = \phi ^*_{ij} \left( G_{-i,j} \right) + \alpha _j \left( w_i - X_{i,-j} \right) , \quad \text { for all }\ ij \in L, \end{aligned}$$

where $\phi ^*_{ij}$ is decreasing since $\partial \phi _{ij} / \partial G_{-i,j} = -b''_j / (b''_j + \delta ''_{ij} + c''_i) \le 0$. $\square$

Proof of Proposition 3

Totally differentiating the best-response functions at each link $ij \in B$ while keeping $ds_{ij} = dw_i = 0$ yields

$$\begin{aligned} d {{\tilde{x}}_{ij}}= & {} \frac{\partial \phi _{ij}}{\partial G_{-i,j}}dG_{-i,j} + \frac{\partial \phi _{ij}}{\partial (\frac{\tau _{ij}}{1-s_{ij}})} \times \frac{1}{1-s_{ij}} d\tau _{ij} - \frac{\partial \phi _{ij}}{\partial (w_i - X_{i,-j})} dX_{i,-j}, \end{aligned}$$

or equivalently,

$$\begin{aligned} d {{\tilde{x}}_{ij}}= & {} -\frac{b''_j}{b''_j + \frac{\delta ''_{ij}}{(1-s_{ij})^2} + c''_i}dG_{-i,j} + \frac{\frac{\delta ''_{ij}}{(1-s_{ij})^2}}{b''_j + \frac{\delta ''_{ij}}{(1-s_{ij})^2} + c''_i}d\tau _{ij} - \frac{c''_i}{b''_j + \frac{\delta ''_{ij}}{(1-s_{ij})^2} + c''_i} dX_{i,-j}. \end{aligned}$$

Since $C=D=\emptyset$, all links are clearly active, i.e., $L=B$. Hence, rearranging as in the proof of Proposition 1 yields

$$\begin{aligned} d {{\tilde{G}}_j} = {{\tilde{k}}_j} \sum \limits _{a_i \in N_g(p_j)}{\left\{ \left( 1 - {{{\tilde{\alpha }}}_{ij}} \right) d\tau _{ij} - {{{\tilde{\alpha }}}_{ij}} d{{\tilde{X}}_{i,-j}} \right\} }, \quad \text { for all }\ p_j \in P, \end{aligned}$$

(2)

where

$$\begin{aligned} {{\tilde{k}}_j} = \left( 1 + \sum \limits _{a_i \in N_g (p_j)}{\frac{b''_j}{\frac{\delta ''_{ij}}{(1-s_{ij})^2} + c''_i}} \right) ^{-1} \in \left( 0 , 1 \right] . \end{aligned}$$

Let $\tau _j = \sum \nolimits _{a_i \in N_g(p_j)}{\tau _{ij}}$ denote the total lump sum taxes with respect to public good $p_j$. Since the contribution structure is complete and ${{{\tilde{\alpha }}}_{ij}} = {{{\tilde{\alpha }}}_j}$ for all $ij \in L$, Equation (2) can be rearranged as

$$\begin{aligned} d {{\tilde{G}}_j} = {{\tilde{k}}_j} \left( 1 - {{{\tilde{\alpha }}}_j} \right) d\tau _j - {{\tilde{k}}_j} {{{\tilde{\alpha }}}_j} \sum \limits _{p_l \in P \backslash \{p_j\}}{d{{\tilde{G}}_l}}, \quad \text { for all }\ p_j \in P. \end{aligned}$$

Hence, assuming that $d\tau _1 \ne 0$ and $d\tau _l = 0$ for all $p_l \in P \backslash \{p_1\}$ yields

$$\begin{aligned} d {{\tilde{G}}_1} = {{\tilde{k}}_1} \left( 1 - {{{\tilde{\alpha }}}_1} \right) d\tau _1 - {{\tilde{k}}_1} {{{\tilde{\alpha }}}_1} \sum \limits _{p_l \in P \backslash \{p_1\}}{d{{\tilde{G}}_l}} \end{aligned}$$

and

$$\begin{aligned} d {{\tilde{G}}_l} = - {{\tilde{k}}_l} {{{\tilde{\alpha }}}_l} \sum \limits _{p_j \in P \backslash \{p_l\}}{d{{\tilde{G}}_j}}, \quad \text { for all }\ p_l \in P \backslash \{p_1\}. \end{aligned}$$

From this last equation, it appears that

$$\begin{aligned} \sum \limits _{p_j \in P}{d{{\tilde{G}}_j}} = \left( 1 - \frac{1}{{{\tilde{k}}_2}{{{\tilde{\alpha }}}_2}} \right) d{{\tilde{G}}_2} = ... = \left( 1 - \frac{1}{{{\tilde{k}}_m}{{{\tilde{\alpha }}}_m}} \right) d{{\tilde{G}}_m}. \end{aligned}$$

(3)

Hence, it holds that

$$\begin{aligned} d{{\tilde{G}}_l} = \beta _l d{{\tilde{G}}_1}, \quad \text {for all} p_l \in P \backslash \{p_1\}, \end{aligned}$$

(4)

where

$$\begin{aligned} \beta _l = \left( -\frac{1}{{{\tilde{k}}_l}{{{\tilde{\alpha }}}_l}} - \sum \limits _{p_j \in P \backslash \{p_1,p_l\}}{\left\{ \frac{1 - \frac{1}{{{\tilde{k}}_l}{{{\tilde{\alpha }}}_l}}}{1 - \frac{1}{{{\tilde{k}}_j}{{{\tilde{\alpha }}}_j}}}\right\} } \right) ^{-1} \in (-1,0). \end{aligned}$$

Now, let $d\tau _1 > 0$ and suppose that $d{{\tilde{G}}_1} \le 0$. Then, from Equation (4), $d{{\tilde{G}}_l} \ge 0$ for all $p_l \in P \backslash \{p_1\}$, and therefore, from Equation (3), $\sum \nolimits _{p_j \in P}{d{{\tilde{G}}_j}} \le 0$. Hence,

$$\begin{aligned} -d{{\tilde{G}}_1} \ge \sum \limits _{p_l \in P \backslash \{p_1\}}{d{{\tilde{G}}_l}} \ge 0. \end{aligned}$$

It follows that

$$\begin{aligned} \begin{array}{rcl} d{{\tilde{G}}_1} &{} = &{} {{\tilde{k}}_1}\left( 1 - {{{\tilde{\alpha }}}_1} \right) d\tau _1 - {{\tilde{k}}_1}{{{\tilde{\alpha }}}_1}\sum \limits _{p_l \in P \backslash \{p_1\}}{d{{\tilde{G}}_l}}\\ &{}\ge &{} {{\tilde{k}}_1}\left( 1 - {{{\tilde{\alpha }}}_1} \right) d\tau _1 - {{\tilde{k}}_1}{{{\tilde{\alpha }}}_1} \left( -d{{\tilde{G}}_1} \right) \\ &{} = &{} {{\tilde{k}}_1}\left( 1 - {{{\tilde{\alpha }}}_1} \right) d\tau _1 + {{\tilde{k}}_1}{{{\tilde{\alpha }}}_1} d{{\tilde{G}}_1}. \end{array} \end{aligned}$$

Then, it appears that

$$\begin{aligned} d{{\tilde{G}}_1} \left( 1 - {{\tilde{k}}_1}{{{\tilde{\alpha }}}_1} \right) \ge {{\tilde{k}}_1} \left( 1 - {{{\tilde{\alpha }}}_1} \right) d\tau _1 \iff d{{\tilde{G}}_1} \ge \frac{{{\tilde{k}}_1} ( 1 - {{{\tilde{\alpha }}}_1} )}{ 1 - {{\tilde{k}}_1}{{{\tilde{\alpha }}}_1} }d\tau _1 > 0, \end{aligned}$$

a contradiction. The same contradiction can easily be obtained under the assumption that $d{{\tilde{G}}_1} \ge 0$ when $d\tau _1 < 0$. Hence, sign$(d\tau _1)$ = sign$(d{{\tilde{G}}_1})$ = sign$(-d{{\tilde{G}}_l})$ for all $p_l \in P \backslash \{p_1\}$ = sign$(\sum \nolimits _{p_j \in P}{d{{\tilde{G}}_j}})$. $\square$

Proof of Proposition 4

Totally differentiating the best-response functions at each link $ij \in B$ while keeping $d\tau _{ij} = dw_i = 0$ yields

$$\begin{aligned} d {{\tilde{x}}_{ij}}= & {} \frac{\partial \phi _{ij}}{\partial G_{-i,j}}dG_{-i,j} + \frac{\partial \phi _{ij}}{\partial s_{ij}}ds_{ij} + \frac{\partial \phi _{ij}}{\partial (\frac{\tau _{ij}}{1-s_{ij}})} \times \frac{\tau _{ij}}{(1-s_{ij})^2} ds_{ij} - \frac{\partial \phi _{ij}}{\partial (w_i - X_{i,-j})} dX_{i,-j}, \end{aligned}$$

or equivalently,

$$\begin{aligned} d {{\tilde{x}}_{ij}}= & {} -\frac{b''_j}{b''_j + \frac{\delta ''_{ij}}{(1-s_{ij})^2} + c''_i}dG_{-i,j} - \frac{\frac{\delta '_{ij}}{(1-s_{ij})^2} - \frac{\delta ''_{ij} \tau _{ij} }{(1-s_{ij})^3}}{b''_j + \frac{\delta ''_{ij}}{(1-s_{ij})^2} + c''_i} ds_{ij} - \frac{c''_i}{b''_j + \frac{\delta ''_{ij}}{(1-s_{ij})^2} + c''_i} dX_{i,-j}. \end{aligned}$$

Since $C=D=\emptyset$, all links are clearly active, i.e., $L=B$. Hence, rearranging as in the proof of Proposition 1 yields

$$\begin{aligned} d {{\tilde{G}}_j}= & {} {{\tilde{k}}_j} \sum \limits _{a_i \in N_g(p_j)}{\left\{ \left( {{{\tilde{\alpha }}}_{ij}} \kappa _{ij} + \left( 1 - {{{\tilde{\alpha }}}_{ij}} \right) \frac{\tau _{ij}}{1-s_{ij}} \right) ds_{ij} - {{{\tilde{\alpha }}}_{ij}} d{{\tilde{X}}_{i,-j}} \right\} }, \nonumber \\&\text { for all }\ p_j \in P, \end{aligned}$$

(5)

where

$$\begin{aligned} \kappa _{ij} = \frac{\frac{\partial \phi _{ij}}{\partial s_{ij}}}{\frac{\partial \phi _{ij}}{\partial (w_i - {{\tilde{X}}_{i,-j}})}} = \frac{- \frac{\delta '_{ij}}{(1-s_{ij})^2}}{c''_i} > 0, \end{aligned}$$

and ${{\tilde{k}}_j} \in ( 0 , 1 ]$ as in the proof of Proposition 3. Since the contribution structure is complete and ${{{\tilde{\alpha }}}_{ij}} = {{{\tilde{\alpha }}}_j}$ for all $ij \in L$, Equation (5) can be rearranged as

$$\begin{aligned} d {{\tilde{G}}_j}= & {} {{\tilde{k}}_j} \sum \limits _{a_i \in A}{\left\{ \left( {{{\tilde{\alpha }}}_j} \kappa _{ij} + \left( 1 - {{{\tilde{\alpha }}}_j} \right) \frac{\tau _{ij}}{1-s_{ij}} \right) ds_{ij}\right\} } - {{\tilde{k}}_j} {{{\tilde{\alpha }}}_j} \sum \limits _{p_l \in P \backslash \{p_j\}}{d{{\tilde{G}}_l}}, \\&\text {for all }p_j \in P. \end{aligned}$$

Hence, assuming that $ds_{i1} \ne 0$ for at least one agent $a_i \in N_g(p_1)$ and $ds_{il} = 0$ for all $a_i \in N_g(p_l)$ for all $p_l \in P \backslash \{p_1\}$ yields

$$\begin{aligned} d {{\tilde{G}}_1} = {{\tilde{k}}_1} \sum \limits _{a_i \in A}{\left\{ \left( {{{\tilde{\alpha }}}_1} \kappa _{i1} + \left( 1 - {{{\tilde{\alpha }}}_1} \right) \frac{\tau _{i1}}{1-s_{i1}} \right) ds_{i1}\right\} } - {{\tilde{k}}_1} {{{\tilde{\alpha }}}_1} \sum \limits _{p_l \in P \backslash \{p_1\}}{d{{\tilde{G}}_l}} \end{aligned}$$

and

$$\begin{aligned} d {{\tilde{G}}_l} = - {{\tilde{k}}_l} {{{\tilde{\alpha }}}_l} \sum \limits _{p_j \in P \backslash \{p_l\}}{d{{\tilde{G}}_j}}, \quad \text { for all }\ p_l \in P \backslash \{p_1\}. \end{aligned}$$

From this last equation, observe that Equations (3) and (4) hold, and since

$$\begin{aligned} {{{\tilde{\alpha }}}_1} \kappa _{i1} + \left( 1 - {{{\tilde{\alpha }}}_1} \right) \frac{\tau _{i1}}{1-s_{i1}} > 0, \quad \text { for all }\ a_i \in A, \end{aligned}$$

the same contradiction as in the proof of Proposition 3 can easily be obtained. $\square$