Information sharing networks in linear quadratic games

Abstract

We study the bilateral exchange of information in the context of linear quadratic games. An information structure is here represented by a non directed network, whose nodes are agents and whose links represent sharing agreements. We first study the equilibrium use of information given the network, finding that the extent to which a piece of information is observed by others affects the equilibrium use of it, in line with previous results in the literature. We then study the incentives to share information ex-ante, highlighting the role of the elasticity of payoffs to the equilibrium volatility of one’s own strategy and of opponents’ strategies. For the case of uncorrelated signals we fully characterise pairwise stable networks for the general linear quadratic game. For the case of correlated signals, we study pairwise stable networks for three specific linear quadratic games—Cournot Oligopoly, Keynes’ Beauty Contest and a Public Good Game—in which strategies are substitute, complement and orthogonal, respectively. We show that signals’ correlation favours the transmission of information, but may also prevent all information from being transmitted.

Appendices

Appendix 1

Proof of Lemma 1

Assume $p_n=\gamma _A = \gamma _{A \theta }=0$. When passing from network $g$ to $g'=g+ij$ condition (16) reduces to:

$$\begin{aligned} -\gamma _{a}\cdot \left[ var\left( a_{i}^{g^{\prime }}\right) -var\left( a_{i}^{g}\right) \right] >0. \end{aligned}$$

(19)

Using now the expressions for variances:

$$\begin{aligned} var\left( a_{i}^{g^{\prime }}\right) -var\left( a_{i}^{g}\right) = p_{s} \left[ \sum _{h\in N_{i}^{g^{\prime }}}(\beta _{ih}^{g'})^{2} -\sum _{h\in N_{i}^{g}}(\beta _{ih}^{g})^{2}\right] \end{aligned}$$

and the fact that from (18) all equilibrium coefficients applied to signal $y_k$ in $g$ and in $g'$ only depend on the degree of $k$ in $g$ and $g'$, respectively, we conclude that condition (16) is satisfied if and only if the following condition holds:

$$\begin{aligned}&\sum _{h\in N_{i}^{g'}}(\beta _{ih}^{g'})^{2} -\sum _{h\in N_{i}^{g}}(\beta _{ih}^{g})^{2}= \nonumber \\&\quad -\gamma _a\left[ \frac{1}{(2\gamma _a+\gamma _{aA}n_i^g)^2} +\frac{1}{(2\gamma _a+\gamma _{aA}n_j^g)^2} -\frac{1}{(2\gamma _a+\gamma _{aA}(n_i^g-1))^2}\right] \nonumber \\&\quad \times \left[ \dfrac{t \gamma _{a\theta }}{p_{s}}\right] ^2 >0. \end{aligned}$$

(20)

Using now the definition of $\mu$, we then write:

$$\begin{aligned} u_i(g\!+\!ij)\!>\!u_i(g) \iff \frac{1}{(2\!+\!\mu (n_i^g-1))^2} \!-\! \frac{1}{(2+\mu n_i^g)^2} \!<\! \frac{1}{(2+\mu n_j^g)^2}. \end{aligned}$$

(21)

It is clear that, for any given $n_i^g$, if (21) is satisfied for a given $n_j^g=m$ then it remains true for all $n_j^g<m$. The function $f_{\mu }$, acting as a threshold for the formation of a new link $ij\notin g$, is then implicitly defined by the condition:

$$\begin{aligned} \frac{1}{(2+\mu (n_i^g-1))^2} - \frac{1}{(2+\mu n_i^g)^2} = \frac{1}{(2+\mu f_{\mu }(n_i^g))^2}. \end{aligned}$$

(22)

Using similar steps we can define the function $F_{\mu }$, acting as a threshold for the severance of a link $ij\in g$. It is defined as the implicit solution of the following condition:

$$\begin{aligned} \frac{1}{(2+\mu (n_i^g-2))^2} - \frac{1}{(2+\mu (n_i^g-1))^2} = \frac{1}{(2+\mu (F_{\mu }(n_i^g)-1))^2}. \end{aligned}$$

(23)

We now turn to the proof of points (i)–(iv).

• (i) The expression $\frac{1}{(2+\mu n_i^g )^2}$ is decreasing and convex in $n_i^g$. It follows that the LHS of (22) is decreasing in $n_i^g$, which in turns implies that the values of $n_j^g$ that satisfy (22) (that is, the value of $f_{\mu }(n_i^g)$) is increasing. Using condition (23), similar steps show that $F_{\mu }(n_i^g)$ is increasing in $n_i^g$.

• (ii) We show that $f_{\mu }(m)>m$ for all $2\le m$. Let $n_i^g=f_{\mu }(n_i^g)=m$. From (22) $f_{\mu }(m)>m$ if and only if the following condition holds:

  $$\begin{aligned} \frac{1}{(2+\mu (n_i^g-1))^2} - \frac{1}{(2+\mu n_i^g)^2} < \frac{1}{(2+\mu f_{\mu }(n_i^g))^2}. \end{aligned}$$

  (24)

  or, equivalently,

  $$\begin{aligned} 2(2+\mu (m-1))^2 - (2+\mu m)^2>0 \end{aligned}$$

  (25)

  It is directly verifiable that LHS of (25) is convex, the smaller root is negative for $0<\mu \le 1$ and the larger root

  $$\begin{aligned} \frac{2(\mu -1)+\mu \sqrt{2}}{\mu } \end{aligned}$$

  is smaller than $2$ for $0<\mu \le 1$. Therefore condition (24) is satisfied for all $m\ge 2$. When $m=1$, condition (24) is satisfied if and only if $\mu <2(\sqrt{2}-1)<1$.

• (iii) We now show that $f_{\mu }(m-1)=F_{\mu }(m)-1$. The value $f_{\mu }(m-1)$ solves the condition:

  $$\begin{aligned} \frac{1}{(2+\mu (m-2))^2} - \frac{1}{(2+\mu (m-1))^2} = \frac{1}{(2+\mu (f_{\mu }(m-1)))^2}. \end{aligned}$$

  (26)

  The value $F_{\mu }(m)$ solves the condition:

  $$\begin{aligned} \frac{1}{(2+\mu (m-2))^2} - \frac{1}{(2+\mu (m-1))^2} = \frac{1}{(2+\mu (F_{\mu }(m)-1))^2}. \end{aligned}$$

  (27)

  Comparison of conditions (26) and (27) directly implies the result.

• (iv) Now we prove that $f_{\mu }(m)>F_{\mu }(m)$ for $m\ge 1$. First we claim that $f_{\mu }(m)-f_{\mu }(m-1)>1$. From (22) we obtain:

  $$\begin{aligned} f_{\mu }(m)\!=\!\dfrac{2\mu ^2(\mu -4-2n\mu )\!+\!\sqrt{\mu ^3(2+(n-1)\mu )^2 (2\!+\!n\mu )^2 (4+(2n\!-\!1)\mu )}}{\mu ^3 (4+(2n-1)\mu )}\nonumber \\ \end{aligned}$$

  (28)

  Form (28) it is directly verifiable that $f_{\mu }(m)-f_{\mu }(m-1)>1$ for $m\ge 1$ and $0<\mu \le 1$ (the complete proof is available upon request). This result, together with point (iii), imply point (iv).

Proof of Proposition 3

Point (1): If $\mu <0$ (the case of strategic complements), then by direct inspection of condition (21) we conclude that the link $ij$ will form (and will not be severed) for all levels of $n_i^g$ and of $n_j^g$. This implies that the complete network is the unique pairwise stable network.

Point (2). We first show that only regular networks can be pairwise stable when $\mu >0$. Assume $g$ is pairwise stable, and assume that $h$ is a non regular component of $g$. Consider now the node $i$ with maximal degree in $h$, and let $j$ be such that $n_{i}^{g}>n_{j}^{g}$ and $ij\in h$ (such link must exist for at least one node with maximal degree). This means that there exists some node $k\ne j$ such that $ik\in h$ and $jk\notin h$. By Lemma 1, pairwise stability of $g$ imposes the following requirements on the degrees of nodes $i$, $j$ and $k$:

$$\begin{aligned} n_{k}^{g}&\le F_{\mu }(n_{i}^{g})\,\text { and }\quad n_{i}^{g}\le F_{\mu }(n_{k}^{g}); \end{aligned}$$

(29)

$$\begin{aligned} n_{j}^{g}&\le F_{\mu }(n_{i}^{g})\,\text { and }\quad n_{i}^{g}\le F_{\mu }(n_{j}^{g}). \end{aligned}$$

(30)

Note now that since $n_{i}^{g}>n_{j}^{g}$ and since nodes’ degrees are integers, then $n_{i}^{g}-1\ge n_{j}^{g}$. This, together with (29), implies $n_{j}^{g}\le F_{\mu }(n_{k}^{g})-1$ and, together with point (iii) in Lemma 1, that $n_{j}^{g}\le f(n_{k}^{g}-1)$. Finally, from point (i) in Lemma 1, we conclude that $n_{j}^{g}<f_{\mu }(n_{k}^{g}).$ This means that player $k$ has an incentive to form the link $jk$; stability of $g$ requires then that player $j$ has no incentive to form link $jk$, that is $n_{k}^{g}>f_{\mu }(n_{j}^{g})$. This, together with $n_{k}^{g}\le n_{i}^{g}$, implies that $n_{i}^{g}>f_{\mu }(n_{j}^{g})$; using now(30) we obtain that $f_{\mu }(n_{j}^{g})<F_{\mu }(n_{j}^{g})$, contradicting point (iv) of Lemma 1.

Having shown that only regular components can belong to a stable network, the requirement that such components are fully connected comes from point (ii) in Lemma 1, where it shown that two agents with equal degree of at least 2 always have an incentive to form a link.

The required ordering in the sizes $n(h_1),n(h_2),...,n(h_m)$ comes from the observation that stability requires that no link is formed, and that a link joining two components does not form if and only if $n(h_i)>f_{\mu }(n(h_{i-1}))$ for all $1\le k <m$.

Lastly, when $0 < \mu < \frac{2}{1+\sqrt{2}}$ two isolated nodes would form a link (see point (ii) in Lemma 1), implying that two isolated nodes are incompatible with stability of the network. $\square$

Proof of Proposition 4

When $p_n=0$ and $\gamma _{A\theta }=0$, using expressions for variances, condition (16) simplifies as follows:

$$\begin{aligned}{}[u_i^e(g')-u_i^e(g)]&= \gamma _{A} p_{s} \left[ \sum _{k\in N}(B^{g'}_{ik})^{2}-\sum _{k \in N}(B^g_{ik})^{2}\right] \nonumber \\&- \gamma _{a} p_{s}\left[ \sum _{h\in N_{i}^{g'}}(\beta _{ih}^{g'})^{2} -\sum _{h\in N_{i}^{g}}(\beta _{ih}^{g})^{2}\right] >0 \end{aligned}$$

(31)

Consider now the networks $g$ and $g'=g+ij$. Using the characterisation of equilibrium coefficients in (18), we note that $\beta ^{g'}_{ik}=\beta ^{g}_{ik}$ for all $k\ne i,j$ and then $B^{g'}_{ik}=B^{g}_{ik}$ for all $k\ne i,j$, since $n_k^{g'}=n_k^g$ for all $k\ne i,j$. This implies that:

$$\begin{aligned} \sum _{k\in N}(B^{g'}_{ik})^{2}-\sum _{k \in N}(B^g_{ik})^{2}=(B^{g'}_{ii})^{2}-(B^{g}_{ii})^{2}+(B^{g'}_{ij})^{2}-(B^{g}_{ij})^{2} \end{aligned}$$

(32)

Also, using the fact that $n_i^{g'}-1=n_i^g$ and $n_j^{g'}-1=n_j^g$ from (18) we obtain:

$$\begin{aligned} B_{ii}^{g'}=-\frac{n^g_i \gamma _{a \theta }\frac{t}{p_s}}{2 \gamma _a + \gamma _{aA}n^g_i} \end{aligned}$$

(33)

$$\begin{aligned} B_{ii}^{g}=-\frac{(n^g_i-1)\gamma _{a \theta }\frac{t}{p_s}}{2 \gamma _a + \gamma _{aA}(n^g_i-1)} \end{aligned}$$

(34)

$$\begin{aligned} B_{ij}^{g'}=-\frac{n^g_j \gamma _{a \theta }\frac{t}{p_s}}{2 \gamma _a + \gamma _{aA}n^g_j} \end{aligned}$$

(35)

$$\begin{aligned} B_{ij}^{g}=-\frac{n^g_j \gamma _{a \theta }\frac{t}{p_s}}{2 \gamma _a + \gamma _{aA}(n^g_j-1)} \end{aligned}$$

(36)

Condition (31) can now be written as follows:

$$\begin{aligned}&[u_i^e(g+ij)-u_i^e(g)]= \nonumber \\&\quad \quad =\gamma _{a \theta }^2 \frac{t^2}{p_s} \gamma _A\left[ \frac{(n_i^g)^2}{(2\gamma _a+\gamma _{aA}n_i^g)^2}-\frac{(n_i^g-1)^2}{(2\gamma _a+\gamma _{aA}(n_i^g-1))^2} +\frac{(n_j^g)^2}{(2\gamma _a+\gamma _{aA}n_j^g)^2} \right. \nonumber \\&\qquad \left. - \frac{(n_j^g)^2}{(2\gamma _a+\gamma _{aA}(n_j^g-1))^2}\right] - \gamma _{a \theta }^2 \frac{t^2}{p_s}\gamma _{a}\nonumber \\&\qquad \times \left[ \frac{1}{(2\gamma _a+\gamma _{aA}n_i^g)^2} +\frac{1}{(2\gamma _a+\gamma _{aA}n_j^g)^2} -\frac{1}{(2\gamma _a+\gamma _{aA}(n_i^g-1))^2}\right] >0. \end{aligned}$$

(37)

From direct inspection of (37) we see that both terms in squared brackets are positive, which, together with the assumptions that $\gamma _a<0$ and $\gamma _A>0$, implies that (37) is always satisfied and $i$ always has an incentive to form a link with $j$. So we can conclude that all links form. $\square$

Proof of Proposition 5

Agent $i$ has an incentive to link to $j$ iff expression (37) is strictly positive. In the BC we have the following restrictions: $\gamma _{a}=-1$; $\gamma _{A}=-\frac{(1-v)}{(n-1)^2}$; $\gamma _{aA}=\frac{2(1-v)}{n-1}$. Note first that for $n_i^g=n_j^g=1$ (37) is strictly positive iff:

$$\begin{aligned} \frac{(n^2 -2 n + v) (n^2 -2 + 4 v - 2 n v - v^2)}{(4 (n-1)^2 (n + v-2)^2)}>0. \end{aligned}$$

(38)

We then compute the derivative of expression (37) with respect to $n_i^g$ for $n_j^g=1$ and show that is always positive. This implies that agent $i$ has an incentive to link to the singleton agent $j$, independently of $i$’s degree. Then we compute the derivative of expression (37) with respect to $n_j^g$ for any arbitrary value of $n_i^g$ and we show that it is positive for any values of $n_i^g$ and $n_j^g$. This implies that (37) remains strictly positive for all $n_i^g<n$ and $n_j^g<n$, which implies the result. Exact computations of the derivatives involve long expressions and are available upon request. $\square$

Proof of Proposition 6

When $p_n=\gamma _{aA}=0$ and $g'=g+ij$, using (18) and (33)–(36) we can rewrite condition (16) as follows:

$$\begin{aligned}&\gamma _{a \theta }^2 \frac{t^2}{p_s} \gamma _A\left[ \frac{(n_i^g)^2}{2\gamma _a}-\frac{(n_i^g-1)^2}{2\gamma _a}\right] + \gamma _{a \theta } \frac{t^2}{p_s}\gamma _{A \theta } \left[ -\frac{n_i^g}{2\gamma _a}+\frac{n_i^g-1}{2\gamma _a}\right] \nonumber \\&\qquad -\gamma _{a \theta }^2 \frac{t^2}{p_s}\gamma _{a} \left[ \frac{1}{2\gamma _a}\right] >0. \end{aligned}$$

(39)

This is satisfied if and only if:

$$\begin{aligned} \frac{\gamma _{a\theta }\gamma _A}{4\gamma _a^2}(2n_i^g-1)-\frac{\gamma _{A\theta }}{2\gamma _a}-\frac{\gamma _{a\theta }}{4\gamma _a}>0 \end{aligned}$$

(40)

where we have used the assumption that $\gamma _{a\theta }>0$. Condition (40) can be written as:

$$\begin{aligned} \frac{\gamma _A}{\gamma _a}(2n_i^g-1)<\frac{2\gamma _{A\theta }+\gamma _{a\theta }}{\gamma _{a\theta }}. \end{aligned}$$

(41)

Case $\gamma _A=0$. When $2\gamma _{A\theta }+\gamma _{a\theta } >0$, (41) is always satisfied and the complete network is the unique pairwise stable network. When $2\gamma _{A\theta }+\gamma _{a\theta }<0$, (41) is never satisfied and the empty network is the unique pairwise stable network.

Case $\gamma _A>0$. When $2\gamma _{A\theta }+\gamma _{a\theta }>0$, (41) is always satisfied. When $2\gamma _{A\theta }+\gamma _{a\theta }<0$, (41) is satisfied for $n_i^g>\dfrac{1}{2}(\frac{2\gamma _{A\theta }+\gamma _{a\theta }}{\gamma _{a\theta }}\frac{\gamma _a}{\gamma _A}+1)\equiv m >0$. In this case the empty network is trivially stable, as is the complete network. The only other stable architecture is such that all nodes with positive degree have degree larger than $m$, and all such nodes are linked to each other.

Case $\gamma _A<0$. Condition (41) is never satisfied when $2\gamma _{A\theta }+\gamma _{a\theta }<0$ (from which the empty network is the unique pairwise stable network), while if $2\gamma _{A\theta }+\gamma _{a\theta }>0$ the relevant condition for agent $i$ to form a link is the following:

$$\begin{aligned} n_i^g<\dfrac{1}{2}(\frac{2\gamma _{A\theta }+\gamma _{a\theta }}{\gamma _{a\theta }}\frac{\gamma _a}{\gamma _A}+1)\equiv m >0 \end{aligned}$$

Here, only agents with a low enough degree would form a link. If the threshold degree $m$ is less than zero, then the empty network is the unique pairwise stable network; if not, no pairwise stable network exists. To see this, note that two nodes who are linked in a stable network must have degree less than $m$. But in this case they wish to form a link to the agents to which they are not linked. If these agents have degree less than $m$ they also want to link, then we contradict the stability of the network. If they do not wish to link, then they must have a degree which is larger than $m$, in which case they wish to sever a link. Finally, the empty network is not stable since two nodes of degree zero wish to form a link. $\square$

Proof of Proposition 7

Proof of point 1: The proof is organised in several steps, and goes by studying the difference in expected profits of two firms, 1 and 2, in the complete networks $g^{c}$ and in the network $g^{-12}\equiv \left\{ g^{c}-12\right\}$. We first compute equilibrium strategies in $g^{c}.$ The updating coefficient in $g^{c}$ is for every $i$:

$$\begin{aligned} k_{1}^{ig^{c}}=\frac{t}{p_{s}+\left( n-1\right) p_{n}}. \end{aligned}$$

(42)

We obtain the following common equilibrium coefficient:

$$\begin{aligned} \beta ^{g^{c}}=-\frac{t\gamma _{a\theta }}{\left( p_{s}+\left( n-1\right) p_{n}\right) \left( 2 \gamma _a+\gamma _{aA} \left( n-1\right) \right) }. \end{aligned}$$

For $g^{-12}\equiv \left\{ g^{c}-12\right\}$, the updating coefficients are:

$$\begin{aligned} k_{1}^{ig^{-12}}&=\frac{t}{p_{s}+\left( n-2\right) p_{n}},\quad k_{2}^{i}=\frac{p_{n}}{p_{s}+\left( n-2\right) p_{n}},\quad i=1,2 \nonumber \\ k_{1}^{ig^{-12}}&=\frac{t}{p_{s}+\left( n-1\right) p_{n}},\quad \forall i\ge 3 \end{aligned}$$

(43)

We obtain the following equilibrium coefficients for firms 1 and 2:

$$\begin{aligned} \beta _{11}^{g^{-12}}&=\beta _{22}^{g^{-12}}=-\frac{t \gamma _{a\theta }}{2 \gamma _a \left( p_{s}+\left( n-2\right) p_{n}\right) +\gamma _{aA} \left( \left( n-2\right) p_{s}+\left( 5+n\left( n-4\right) \right) p_{n}\right) } \\ \beta _{1j}^{g^{-12}}&=\beta _{2j}^{g^{-12}}\\&=\!-\!\frac{t\left( 2\gamma _a \!+\!\left( n-2\right) \gamma _{aA} \right) \gamma _{a\theta }}{\left( 2 \gamma _a\!+\!\left( n\!-\!1\right) \gamma _{aA} \right) \left( 2 \left( p_{s}\!+\!\left( n\!-\!2\right) p_{n}\right) \gamma _a \!+\!\gamma _{aA} \left( \left( n\!-\!2\right) p_{s}\!+\!\left( 5\!+\!n\left( n\!-\!4\right) \right) p_{n}\right) \right) },\ \forall j\!\ge \! 3 \end{aligned}$$

From (16), we can express the difference $u^e(g^{c})-u^e(g^{-12})$ in the expected profits of firm 1 (and, by symmetry, of firm 2) in $g^{c}$ and in $g^{-12}$ as proportional to:

$$\begin{aligned}&n\cdot \left( \beta _{ij}^{g^{c}}\right) ^{2}\left( p_{s}+\left( n-1\right) p_{n}\right) -\left( \left( \beta _{11}^{g^{-12}}\right) ^{2}+\left( n-2\right) \left( \beta _{1j}^{g^{-12}}\right) ^{2}\right) p_{s} - \nonumber \\&\quad -\left( n-2\right) \beta _{1j}^{g^{-12}}\left( 2\beta _{11}^{g^{-12}}+\left( n-3\right) \beta _{1j}^{g^{-12}}\right) p_{n}. \end{aligned}$$

(44)

Plugging in the values of the $\beta$ coefficients, we obtain the following expression:

$$\begin{aligned} \frac{\left( p_{s}\!-\!p_{n}\right) t^{2}\left( 4\left( p_{s}\!+\!\left( n\!-\!2\right) p_{n}\right) \left( 1\!+\!\left( n\!-\!3\right) \mu \right) \!+\!\left( \left( 7\!+\!\left( n\!-\!6\right) n\right) p_{s}\!+\!\left( n\left( 19\!+\!\left( n\!-\!8\right) n\right) \!-\!16\right) p_{n}\right) \mu ^{2}\right) \gamma _{a\theta }^2 }{\left( p_{s}\!+\!\left( n\!-\!1\right) p_{n}\right) \left( 2 \!+\!\left( n\!-\!1\right) \mu \right) ^{2}\left( 2p_{s} \!+\!2\left( n\!-\!2\right) p_{n} \!+\!\left( n\!-\!2\right) p_{s}\mu \!+\!\left( 5\!+\!\left( n\!-\!4\right) n\right) p_{n}\mu \right) ^{2}\gamma _{a}^2}\nonumber \\ \end{aligned}$$

(45)

The denominator of the above equation is always strictly positive for all admissible values of the parameters; moreover the sign is the same as the sign of the following expression:

$$\begin{aligned}&4\left( p_{s}+\left( n-2\right) p_{n}\right) \left( 1+\left( n-3\right) \mu \right) + \nonumber \\&\quad + \left( \left( 7+\left( n-6\right) n\right) p_{s}+\left( n\left( 19+\left( n-8\right) n\right) -16\right) p_{n}\right) \mu ^{2} \end{aligned}$$

(46)

We divide it in two terms. The first, $4\left( p_{s}+\left( n-2\right) p_{n}\right) \left( 1+\left( n-3\right) \mu \right)$, is always positive: indeed by assumption $\mu >0$ and the proof follows directly; it can be directly verified that the second term is positive for $n\ge 5$. Therefore (46) could be negative only for $n=3$ and $n=4$. But for $n=3$ (46) becomes $4\left( p_{s}+p_{n}\right) -2\left( p_{s}+2p_{n}\right) \mu ^{2}$ and for $n=4$ (46) becomes $4\left( p_{s}+2p_{n}\right) \left( 1+\mu \right) -\left( p_{s}+4p_{n}\right) \mu ^{2}$ and, by the assumption that $0<\mu <1$, both terms are strictly positive.

Proof of Point 2

We study the difference in expected profits of two agents, 1 and 2, in the empty network $g^{\varnothing }$ and in the network $g^{12}\equiv \left\{ 12\right\}$. The updating coefficients for $g^{\varnothing }$ are:

$$\begin{aligned} k_{1}^{ig^{\varnothing }}=\frac{t}{p_{s}}, \qquad k_{2}^{ig^{\varnothing }}=\frac{p_{n}}{p_{s}}, \qquad \forall i \end{aligned}$$

(47)

from which we obtain the common coefficient of agents’ equilibrium strategies:

$$\begin{aligned} \beta _{ii}^{g^{\varnothing }}=-\frac{\gamma _{a\theta }t}{ 2 \gamma _{a} p_{s}+\gamma _{aA}\left( n-1\right) p_{n}}, \quad \forall i \end{aligned}$$

The updating coefficients for $g^{12}\equiv \left\{ 12\right\}$ are:

$$\begin{aligned}&k_{1}^{ig^{12}}=\frac{t}{p_{s}+p_{n}}\quad \hbox {for} \quad i=1,2 \\&k_{1}^{ig^{12}}=\frac{t}{p_{s}}\quad \hbox {for all} \quad i\ge 3 \\&k_{2}^{ig^{12}}=\frac{p_{n}}{p_{s}+p_{n}}\quad \hbox {for} \quad i=1,2 \\&k_{2}^{ig^{12}}=\frac{p_{n}}{p_{s}}\quad \hbox {for} \quad i\ge 3 \end{aligned}$$

We obtain the following equilibrium coefficients for agents 1 and 2:

$$\begin{aligned} \beta _{11}^{g^{12}}&=\beta _{12}^{g^{12}}=\beta _{21}^{g^{12}}=\beta _{22}^{g^{12}}\\&=\!-\!\frac{t\left( 2 p_{s} \gamma _{a}\!-\!\gamma _{aA} p_{n}\right) \gamma _{a\theta }}{4p_{s} \left( p_{s}\!+\!p_{n}\right) \gamma _{a}^{2}\!+\!2\left( p_{s}\!+\!p_{n}\right) \left( p_{s}\!+\!(n\!-\!3)p_{n}\right) \gamma _{a} \gamma _{aA}\!+\!p_{n}(\left( n\!-\!3\right) p_{s}\!-\! \left( 3n\!-\!5\right) p_{n})\gamma _{aA}^{2}}. \end{aligned}$$

From (16), we can express the difference in profits of agent 1 (and, by symmetry, of agent 2) in $g^{\varnothing }$ and in $g^{12}$ as:

$$\begin{aligned} \left( \beta _{ii}^{g^{\varnothing }}\right) ^{2}p_{s}-2\left( \beta _{11}^{g^{12}}\right) ^{2}\left( p_{s}+p_{n}\right) . \end{aligned}$$

(48)

Plugging in (48) the values of the $\beta$ coefficients, recalling the definition of $\mu$ and letting $p\equiv p_{s}+p_{n}$ we obtain the following expression:

$$\begin{aligned}&\frac{t^{2}\gamma _{a\theta }^2}{\gamma _{a}^2}\left[ \dfrac{p_{s}}{ \left( 2p_{s}+\left( n-1\right) p_{n}\mu \right) ^{2}}-\right. \nonumber \\&\quad \left. \dfrac{2p\left( p_{n}\mu -2p_{s}\right) ^{2}}{\left( 4p_{s}p+2p\left( p_{s}+\left( n-3\right) p_{n}\right) \mu -p_{n}\left( \left( 3n-5\right) p_{n}-\left( n-3\right) p_{s}\right) \mu ^{2}\right) ^{2}}\right] \end{aligned}$$

It can be shown that the denominator of the above expression is strictly positive. Its sign of is therefore the sign of the numerator of the above expression, which can be written in the following form:

$$\begin{aligned} a\cdot n^{2}+b\cdot n+c \end{aligned}$$

(49)

where

$$\begin{aligned} a&=\left( p_{s}-p_{n}\right) p_{n}^{2}\mu ^{2}\left( 4p_{s}p\left( \mu -1\right) + \left( p_{s}^{2}-5p_{s}p_{n}+2p_{n}^{2}\right) \mu ^{2}\right) \\ b&=2\left( p_{s}-p_{n}\right) p_{n}\mu \cdot \cdot \left( -8p_{s}^{2}p+4p_{s}p\left( 2p_{s}+3p_{n}\right) \mu \right. \\&\quad \left. +2p_{s}\left( p_{s}-8p_{n}\right) p\mu ^{2}-p_{n} \left( 3p_{s}^{2}-11p_{s}p_{s}+2p_{n}^{2}\right) \mu ^{3}\right) \\ c&=\left( p_{s}-p_{n}\right) \left[ 2p_{n}^{4}\mu ^{4}+p_{s}p_{n}^{3}\mu ^{2}\left( \left( 44-21\mu \right) \mu -36\right) +4p_{s}^{4}\left( \mu \left( 4+\mu \right) -4 \right) \right. \\&\quad \left. -4p_{s}^{3}p_{n}\left( \mu -1\right) \left( 3\mu \left( 4+\mu \right) -4 \right) +p_{s}^{2}p_{n}^{2}\mu \left( 48+\mu \left( \mu \left( 32+9\mu \right) -76\right) \right) \right] \end{aligned}$$

The proof continues now by studying the sign of (49).

We first note that the roots $(n_{-},n_{+})$ of (49) are real (since $b^{2}-4ac\ge 0$), distinct and finite as long as $a\ne 0$. With this in mind, we now look for conditions under which expression (49) is concave. Such conditions will tell us whether the sign of (49) becomes negative for $n$ large enough. $\square$

Lemma 2

If$\mu <\frac{2}{1+\sqrt{2}}$ then (49) is concave. If $\mu >\frac{2}{1+\sqrt{2}}$ then there exists $p_{n}^{*}$ such that for all $p_{n}>p_{n}^{*}$ (49) is concave, otherwise it is convex.

Proof of Lemma 2

Note that concavity of (49) depends on the sign of term $a$ in (49). This term is negative for $\mu <0$. Moreover, the sign of $a$ is the sign of the following term:

$$\begin{aligned} \left( 4p_{s}p\left( \mu -1\right) +\left( p_{s}^{2}-5p_{s}p_{n}+2p_{n}^{2}\right) \mu ^{2}\right) . \end{aligned}$$

(50)

Let us evaluate the roots of (50) as a function of $p_{n}$. We find:

$$\begin{aligned} \frac{4p_{s}\left( 1-\mu \right) +5p_{s}\mu ^{2}\pm p_{s}\left( \mu -2\right) \sqrt{4\left( 1-\mu \right) +17\mu ^{2}}}{4\mu ^{2}} \end{aligned}$$

(51)

Since the largest root yields a value which exceeds $p_{s}$, we only consider the smaller root denoted by $p_{n}^{*}$. Note here that the second derivative of (50) with respect to $p_{n}$ is positive (so that $a$ is a convex function of $p_{n}$). This directly implies that $a$ is negative for all $p_{n}>p_{n}^{*}$. We then turn to the analysis of the root $p_{n}^{*}$ in relation to the parameter $\mu$. We show that if $\mu < \frac{2}{1+\sqrt{2}}$ then $p_{n}^{*}<0$, implying that $a<0$ for all parameters’ values; moreover, when $\mu >\frac{2}{1+\sqrt{2}}$, we show that $p_{n}^{*}>0$ and that $p_{n}^{*}$ is increasing in $\mu$. In this latter case, $a<0$ for all values $p_{n}^{*}<p_{n}<p_{s}$.

Consider again the smaller root in (51):

$$\begin{aligned} p_{n}^{*}=\frac{4p_{s}\left( 1-\mu \right) +5p_{s}\mu ^{2}+p_{s}\left( \mu -2\right) \sqrt{4\left( 1-\mu \right) +17\mu ^{2}}}{4\mu ^{2}}. \end{aligned}$$

(52)

Expression (52) is null for the following values of $\mu$:

$$\begin{aligned} \mu _{-}=2\left( -1-\sqrt{2}\right) ;\, \, \mu _{+}=\frac{2}{1+\sqrt{2}}. \end{aligned}$$

(53)

Moreover, the expression (52) is strictly increasing in $\mu$ for all values of $\mu$ in the range $\left( 0,1\right]$. This implies that $p_{n}^{*}<0$ for all $0<\mu <\mu _{+}$, and that $p_{s}>p_{n}^{*}>0$ for all $\mu _{+}<\mu \le 1$. $\square$

Having established conditions under which (49) is concave in $n$, we study its sign by establishing a few facts about the behaviour of (49) at the point $n=2$.

Lemma 3

At $n=2$: i) expression (49) is negative for $\mu <\frac{2}{3}$, is positive for $\mu >\frac{2}{1+\sqrt{2}}$ and in the intermediate range is positive if and only if $p_{n}>\dfrac{p_{s}\left( 4-4\mu -\mu ^{2}\right) }{2\mu ^{2}}\equiv p_{n}^{**}$. ii) Moreover, there exists $\hat{p}_{n}>0$ such that (49) is increasing in $n$ if $p_{n}<\hat{p}_{n}$ and $\mu >\frac{2}{1+\sqrt{2}}$, otherwise (49) is decreasing in $n$.

Proof of Lemma 3

Point (i) follows from direct computation, and is consistent with Proposition 4.4 in Raith (1996) for the specific case of Cournot oligopoly, setting $n=2$. Point (ii) is proved as follows. The first derivative of (49) at $n=2$ is given by:

$$\begin{aligned} 2\left( p_{s}-p_{n}\right) p_{n}p\mu \left( 2p_{s}-p_{n}\mu \right) \left( p_{s}\left( \mu \left( 4+\mu \right) -4\right) -2p_{n}\mu ^{2}\right) . \end{aligned}$$

(54)

The sign of (54) is the same as the sign of the following expression:

$$\begin{aligned} \mu \left( p_{s}\left( \mu \left( 4+\mu \right) -4\right) -2p_{n}\mu ^{2}\right) . \end{aligned}$$

(55)

The expression in brackets in (55) is positive for $p_{n}<\dfrac{p_{s}\left( \mu \left( 4+\mu \right) -4\right) }{2\mu ^{2}} \equiv \hat{p}_{n}$. It is directly verifiable that $\hat{p}_{n}$ is negative for $\mu <\frac{2}{1+\sqrt{2}}$ and positive for $\mu >\frac{2}{1+\sqrt{2}}$. $\square$

We are now ready to prove Proposition 7.

Point (i) $(\mu <\frac{2}{3})$. we know from Lemma 3 that at $n=2$ (49) is negative and decreasing in $n$, and from Lemma 2 we know that (49) is concave in $n$. This two facts tell us the all points $n\ge 2$ are in the right (and decreasing) branch of the parabola (49). We conclude that (49) is negative for all $n\ge 2$.

Point (ii) ($\frac{2}{3}<\mu <\frac{2}{1+\sqrt{2}}$). From Lemma 2 and Lemma 3 we know that (49) is concave and decreasing in $n$ at $n=2$. These two facts imply that all points $n\ge 2$ are in the right (and decreasing) branch of the parabola (49). In this range of values for $\mu$, however, (49) can be either positive or negative at $n=2$, depending on the value of $p_{n}$ (see Lemma 3 point (i)). Suppose first that (49) is negative at $n=2$; in this case, the two real roots of (49) are strictly smaller than 2, and (49) remains negative for all $n\ge 2$. Suppose then that (49) is positive at $n=2$; in this case, the larger real root $n_{+}$ must be larger than 2, so that (49) is negative for all $n>n_{+}$.

Point (iii) ($\mu >\frac{2}{1+\sqrt{2}}$). In this range, (49) is concave in $n$ if and only if if $p_{n}>p_{n}^{*}>0$, otherwise it is convex (Lemma 2). Moreover, we know from Lemma 3 that at $n=2$ (49) is positive. Consider first the case $p_{n}>p_{n}^{*}$ ((49) concave). Here, the larger real root $n_{+}$ must be larger than 2, so that for all $n>n_{+}$ (49) is negative. Consider then the case $p_{n}<p_{n}^{*}$ ((49) convex). Here, at $n=2$ (49) is increasing in $n$ if $p_{n}<\hat{p}_{n}$. Since $\hat{p} _{n}=$ $p_{n}^{*}$ for $\mu =\frac{2}{1+\sqrt{2}}$ and for $\mu >\frac{2}{1+ \sqrt{2}}$ the difference $\hat{p}_{n}-$ $p_{n}^{*}$ is increasing in $\mu$ ², it follows that $p_{n}^{*}<\hat{p}_{n}$ for all $\mu >\frac{2}{1+\sqrt{2}}$ and that (49) is increasing in $n$ at $n=2$. Since in this case (49) is convex, we conclude that the two real roots $(n_{-},n_{+})$ are smaller than 2, and that (49) is positive for $n\ge 2$. $\square$

Proof of Proposition 8

We replace the parameters in (1) with those specific for the BC: $\gamma _{a}=-1$; $\gamma _{\theta }=-v$; $\gamma _{a\theta }=2v$; $\gamma _{A}=-\frac{(1-v)}{(n-1)^2}$; $\gamma _{aA}=\frac{2(1-v)}{n-1}$.

Point 1. We study again the difference in expected profits of two firms, 1 and 2, in the complete networks $g^{c}$ and in the network $g^{-12}\equiv \left\{ g^{c}-12\right\}$. The updating coefficient in $g^{c}$ is:

$$\begin{aligned} k_{1}^{ig^{c}}=\frac{t}{p_{s}+\left( n-1\right) p_{n}}, \end{aligned}$$

(56)

from which we obtain the following common equilibrium coefficient:

$$\begin{aligned} \beta ^{g^{c}}=-\frac{t}{\left( p_{s}+\left( n-1\right) p_{n}\right) }. \end{aligned}$$

For $g^{-12}\equiv \left\{ g^{c}-12\right\}$, the updating coefficients are:

$$\begin{aligned} k_{1}^{ig^{-12}}&=\frac{t}{p_{s}+\left( n-2\right) p_{n}}, \quad k_{2}^{i}=\frac{p_{n}}{p_{s}+\left( n-2\right) p_{n}}, \quad i=1,2 \nonumber \\ k_{1}^{ig^{-12}}&=\frac{t}{p_{s}+\left( n-1\right) p_{n}}, \quad \forall \quad i\ge 3 \end{aligned}$$

(57)

We obtain the following equilibrium coefficients for firms 1 and 2 and for firms $j,k>2$:

$$\begin{aligned} \beta _{11}^{g^{-12}}&=\beta _{22}^{g^{-12}}=-\frac{t (n-1)v}{p_s + (n-2) p_s v + p_n (n-3 + (5 + (n-4) n) v) } \\ \beta _{1j}^{g^{-12}}&=\beta _{2j}^{g^{-12}}\\&=-\frac{t+(n-2)tv}{p_s + (n-2) p_s v + p_n (n-3+ (5 + (n-4) n) v) },\\ \beta _{j1}^{g^{-12}}&=\beta _{j2}^{g^{-12}}\\&=-\frac{(n-1) ((n-2) p_n + p_s) t v}{((n-1) p_n + p_s) (p_s + (n-2) p_s v + p_n (n-3 + (5 + (n-4) n) v))}\\ \beta _{jj}^{g^{-12}}&=\beta _{jk}^{g^{-12}}\\&=-\frac{t (p_s + (n-2) p_s v + (n-1) p_n (1 + (n-3) v))}{((n-1) p_n + p_s) (p_s + (n-2) p_s v + p_n (n-3 + (5 + (n-4) n) v))} \end{aligned}$$

We will now write down the change in expected payoff of agent $1$ moving from $g^c$ to $g^{-12}$ following condition (17–19) and (21). The terms used to compute the other players’ aggregate volatility are given by:

$$\begin{aligned} B_{1k}^{g^c}&=(n - 1) \beta _{11}^{g^c}; \end{aligned}$$

(58)

$$\begin{aligned} B_{11}^{g^{-12}}&= (n - 2) \beta _{31}^{g^{-12}}; \end{aligned}$$

(59)

$$\begin{aligned} B_{12}^{g^{-12}}&= \beta ^{g^{-12}}_{11} + (n - 2) \beta ^{g^{-12}}_{31}; \end{aligned}$$

(60)

$$\begin{aligned} B_{1k}^{g^{-12}}&=\beta ^{g^{-12}}_{13} + (n - 2) \beta ^{g^{-12}}_{33}. \end{aligned}$$

(61)

From (21) we express the change in payoff moving from $g^{-12}$ to $g^c$ as follows:

$$\begin{aligned}&-\gamma _a [(p_s n \!+\!n(n\!-\!1)p_n) (\beta _{11}^{g^{c}})^2\!-\! p_s (\beta _{11}^{g^{-12}})^2\!-\!(n\!-\!2)(\beta _{1k}^{g^{-12}})^2 p_s\!-\!(2(n-2)\beta _{11}^{g^{-12}}\\&\quad \quad \beta _{1k}^{g^{-12}}+(n-2)(n-3)(\beta _{1k}^{g^{-12}})^2)p_n] +\gamma _A [n p_s (B_{21}^{g^c})^2+n(n-1)(B_{ik}^{g^c})^2 p_n\\&\quad \quad -p_s((B_{12}^{g^{-12}})^2+ (B_{11}^{g^{-12}})^2+(n-2)(B_{13}^{g^{-12}})^2)-2p_n ((B_{11}^{g^{-12}}(B_{12}^{g^{-12}}\\&\quad \quad +(n-2)B_{13}^{g^{-12}})+B_{12}^{g^{-12}}(n-2)B_{13}^{g^{-12}}+(B_{13}^{g^{-12}})^2(n-2)(n-3)/2)] \end{aligned}$$

Now we show that this expression is never negative for all values of $v$, $p_n$ and $p_s$ in the ranges $0 < v \le 1$ and $p_n < p_s$. Replacing the coefficients we get the following expression:

$$\begin{aligned} \dfrac{\left( p_s\!-\!p_n\right) t^2 v\left( p_n \left( 2 n\!+\!12 v\!-\!6\!+\!n\cdot v \left( n \left( n\!-\!2 \right) \!-\!4\right) \!+\!v^2 \left( n\left( 7\!-\!2 n \right) \!-\!8 \right) \right) \!+\!p_s \left( 2\!+\!v\left( n^2-4\!+\!3 v\!-\!2 n v\right) \right) \right) }{\left( \left( n\!-\!1 \right) p_n\!+\!p_s \right) \left( p_s\!+\!\left( n\!-\!2\right) p_s v\!+\!p_n\left( n\!-\!3\!+\!\left( 5\!+\!\left( n\!-\!4\right) n\right) v\right) \right) ^2}\nonumber \\ \end{aligned}$$

(62)

It can be shown that the denominator of (62) is strictly positive. Then the sign of (62) is therefore the sign of its numerator, which can be written in the following form:

$$\begin{aligned} a\cdot v^{2}+b\cdot v+c \end{aligned}$$

(63)

where

$$\begin{aligned}&a=p_n\left( n\left( 7-2n\right) -8\right) +p_s\left( 3-2n\right) \\&b=12p_n+n\cdot p_n\left( n\left( n-2\right) -4\right) +p_s\left( n^2-4\right) \\&c =2p_s+p_n\left( 2n-6\right) \end{aligned}$$

The proof continues now by studying the sign of (63).

We first note that the roots $(n_{-},n_{+})$ of (63) are real (since $b^{2}-4ac\ge 0$), distinct and finite (since $a\ne 0$). Moreover by a direct inspection of (63) we see that it is concave and that its smaller root is negative and the larger one is greater than 1. Then (62) is positive for all parameter values, implying that the complete network is always stable.

Point 2: We study the difference in expected profits of two agents, 1 and 2, in the empty network $g^{\varnothing }$ and in the network $g^{12}\equiv \left\{ 12\right\}$. The updating coefficients for $g^{\varnothing }$ are:

$$\begin{aligned} k_{1}^{ig^{\varnothing }}=\frac{t}{p_{s}}, \quad k_{2}^{ig^{\varnothing }}=\frac{p_{n}}{p_{s}}, \quad \forall i \end{aligned}$$

(64)

from which we obtain the common coefficient of agents’ equilibrium strategies:

$$\begin{aligned} \beta ^{g^{\varnothing }}=-\frac{t v}{ ps + pn (v-1)}. \end{aligned}$$

and

$$\begin{aligned}&B_{ii}^{g^{\varnothing }} = 0 \end{aligned}$$

(65)

$$\begin{aligned}&B_{ij}^{g^{\varnothing }} = \beta ^{g^{\varnothing }}\quad \mathrm{for}\,\, i \ne j \end{aligned}$$

(66)

The updating coefficients for $g^{12}\equiv \left\{ 12\right\}$ are:

$$\begin{aligned}&k_{1}^{ig^{12}}=\frac{t}{p_{s}+p_{n}}\quad \hbox {for } i=1,2 \\&k_{1}^{ig^{12}}=\frac{t}{p_{s}}\quad \hbox {for all } i\ge 3 \\&k_{2}^{ig^{12}}=\frac{p_{n}}{p_{s}+p_{n}}\quad \hbox {for } i=1,2 \\&k_{2}^{ig^{12}}=\frac{p_{n}}{p_{s}}\quad \hbox {for } i\ge 3 \\ \end{aligned}$$

We obtain the following equilibrium coefficients:

$$\begin{aligned}&\beta _{11}^{g^{12}} =\beta _{12}^{g^{12}}=\beta _{21}^{g^{12}}=\beta _{22}^{g^{12}} \\&=\!-\!\dfrac{(n\!-\!1) t v (p_n(1\!-\!v) \!+\! (n\!-\!1) p_s )}{(n\!-\!1) p_s^2 (n \!+\! v\!-\!2) \!+\! p_n^2 (v\!-\!1) ( n(n\!-\!1 \!-\! 3 v) \!+\! 5 v\!-\!2)\!+\! p_n p_s (n\!-\!2 \!+\! v) (2 \!+\! (n\!-\!3) v)}; \\&\beta _{kk}^{g^{12}}\!=\!\dfrac{(n\!-\!1) t v (p_n (2 \!+\! n \!-\! 3 v) \!+\! p_s (n\!-\!2 \!+\! v))}{(n\!-\!1) p_s^2 (n\!-\!2 \!+\! v) \!+\! p_n^2 (v\!-\!1) (n\!-\!2 (n\!-\!1 \!-\! 3 v) \!+\! 5 v) \!+\! p_n p_s (n\!-\!2 \!+\! v) (2 \!+\! (n\!-\!3) v)}, \end{aligned}$$

and:

$$\begin{aligned} B_{11}^{g^{12}}&= \beta _{21}^{g^{12}} \end{aligned}$$

(67)

$$\begin{aligned} B_{1k}^{g^{12}}&= \beta _{kk}^{g^{12}}. \end{aligned}$$

(68)

Using (16) we can express the difference in profits of agent 1 (and, by symmetry, of agent 2) in $g^{\varnothing }$ and in $g^{12}$ as:

$$\begin{aligned}&-\gamma _a[ (\beta _{ii}^{g^{\varnothing }})^{2}p_{s}-2(\beta _{11}^{g^{12}})^{2}(p_{s}+p_{n})] +\gamma _A [(n-1)B_{ij}^{g^{\varnothing }2} p_s\nonumber \\&\quad \quad +(n-1)(n-2)B_{ij}^{g^{\varnothing }2} p_n - ( 2 (B_{11}^{g^{12}} )^2+ (n-2 ) (B_{1k}^{g^{12}} )^2 ) p_s \nonumber \\&\quad \quad - ( 2 (B_{11}^{g^{12}} )^2+( 4n-8)B_{11}^{g^{12}}B_{1k}^{g^{12}}+( n-2) ( n-3)(B_{1k}^{g^{12}} )^2 ) p_n ] \end{aligned}$$

(69)

The proof then goes through the following steps (complete proofs from authors upon request): we plug in the coefficients’ expressions, so we get an expression in $n$, $p_s$, $p_n$ and $v$. Then we find that: i) (69) is strictly positive for $p_n=0$ for all parameters’ values; ii) (69) is equal zero for $p_n=p_s$ for all parameters’ values; iii) the derivative of (69) respect to $p_n$ computed in $p_n =p_s$ is strictly negative for all parameters’ values; iv) (69) never is negative for $p_n\in (0,p_s)$ and for all other parameters’ value. All these evidences are enough to say that (69) is positive for all parameters’ values and, consequently, the empty network is not pairwise stable. $\square$

Proof of Proposition 9

We replace the parameters in (1) with those specific for the Public Good Game: $\gamma _{a}=-b$; $\gamma _{A\theta }=1$; $\gamma _{a\theta }=1$. Expression (6) becomes:

$$\begin{aligned} \beta _{ih}^{g}=\frac{t}{2b(p_{s}+\left( n_{i}^{g}-1\right) p_{n})}, \quad \forall h\in N_{i}^g ; \end{aligned}$$

(70)

The incentives to form a new link are given by (16) which, when ($g^{\prime }=g+ij$), is rewritten as follows:

$$\begin{aligned} \left[ cov\left( A_{i}^{g^{\prime }},\theta \right) -cov\left( A_{i}^{g},\theta \right) \right] +b\cdot \left[ var\left( a_{i}^{g^{\prime }}\right) -var\left( a_{i}^{g}\right) \right] \end{aligned}$$

(71)

Note that the term inside the first brackets depends only on the covariance of the action of agent $j$ and $\theta$. Therefore using (70) and (18), and noting that passing from $g$ to $g'$ only agents $i$’s and $j$’s coefficients change, we can write (71) as:

$$\begin{aligned} \frac{t^2}{2b}\left[ \frac{n_j\!+\!1}{(p_s\!+\!n_jp_n)} \!-\!\frac{n_j}{(p_s\!+\!(n_j\!-\!1)p_n)} \right] \!+\!\frac{t^2}{4b}\cdot \left[ \frac{n_i\!+\!1}{p_s\!+\!n_i p_n} \!-\!\frac{n_i}{p_s\!+\!(n_i-1) p_n}\right] .\nonumber \\ \end{aligned}$$

(72)

Direct inspection of (72) show that (72) is strictly positive as long as $p_s > p_n$. Therefore every incomplete network is not stable, resulting in the complete network being the unique pairwise stable architecture. $\square$

Appendix 2

Equilibrium coefficients for the star network solve the following system of equations (we have used symmetry where possible):

$$\begin{aligned}&\displaystyle \beta _{ii}^{g^s} = -\frac{1}{2\gamma _{a}}\left( \gamma _{A\theta }k_{1}^{ig^s}+\gamma _{aA}3\beta _{ih}^{g^s}\right)&\end{aligned}$$

(73)

$$\begin{aligned}&\displaystyle \beta _{ih}^{g^s} = -\frac{1}{2\gamma _{a}}\left( \gamma _{A\theta }k_{1}^{ig^s}+\gamma _{aA}\beta _{hh}^{g^s}\right)&\end{aligned}$$

(74)

$$\begin{aligned}&\displaystyle \beta _{hh}^{g^s} = -\frac{1}{2\gamma _{a}}\left( \gamma _{A\theta }k_{1}^{hg^s}+\gamma _{aA}(\beta _{ih}^{g^s}(1+2k_{2}^{hg^s})+2k_{2}^{hg^s}\beta _{hh}^{g^s})\right)&\end{aligned}$$

(75)

$$\begin{aligned}&\displaystyle \beta _{hi}^{g^s} = -\frac{1}{2\gamma _{a}}\left( \gamma _{A\theta }k_{1}^{hg^s}+\gamma _{aA}(\beta _{ii}^{g^s}+\beta _{ih}^{g^s}2k_{2}^{hg^s}+2\beta _{hi}^{g^s}+2k_{2}^{hg^s}\beta _{hh}^{g^s})\right)&\end{aligned}$$

(76)