Myopic or farsighted: bilateral trade agreements among three symmetric countries

Abstract

We examine network formation through bilateral trade agreements (BTA) among three symmetric countries. Each government decides whether to form a link or not via a BTA depending on the differential of the ex-post and ex-ante sum of real wages in the country. Setting the governmental decision in two forms, myopic and farsighted, we analyze the resulting network formation. Firstly, we find that both myopic and farsighted games achieve complete networks. Secondly, networks resulting from myopic games coincide with those resulting from farsighted games.

Appendices

Appendix A Proofs

1.1 A.1 Proof of Lemma 1

Given any network g and any distribution of firms $\lambda ^{g}$, the capital rent at country i, $\pi _{i}\left( \lambda ^{g}\right)$, is expressed by

$$\begin{aligned} \pi _{i}\left( \lambda ^{g}\right) =\frac{\mu }{\sigma }Y\sum _{j}\frac{\phi _{ij}^{g}}{\Delta _{j}^{g}}{,} \end{aligned}$$

where $\Delta _{j}^{g}=\sum _{k}\lambda _{k}^{g}\phi _{kj}^{g}$ for any j. Hence, these equations are combined into

$$\begin{aligned} \left( \begin{array}[c]{c} \pi _{A}\left( \lambda ^{g}\right) \\ \pi _{B}\left( \lambda ^{g}\right) \\ \pi _{C}\left( \lambda ^{g}\right) \end{array} \right) =\frac{\mu }{\sigma }Y\cdot \Phi ^{g}\left( \begin{array}[c]{c} \dfrac{1}{\Delta _{A}^{g}}\\ \dfrac{1}{\Delta _{B}^{g}}\\ \dfrac{1}{\Delta _{C}^{g}} \end{array} \right) {.} \end{aligned}$$

To deform the formula,

$$\begin{aligned} \Phi ^{g}\left( \begin{array}[c]{c} \dfrac{1}{\Delta _{A}^{g}}\\ \dfrac{1}{\Delta _{B}^{g}}\\ \dfrac{1}{\Delta _{C}^{g}} \end{array} \right) =\frac{\sigma }{\mu Y}\left( \begin{array}[c]{c} \pi _{A}\left( \lambda ^{g}\right) \\ \pi _{B}\left( \lambda ^{g}\right) \\ \pi _{C}\left( \lambda ^{g}\right) \end{array} \right) {,} \end{aligned}$$

which means that vector $^{t}\left( \dfrac{1}{\Delta _{A}^{g}},\dfrac{1}{\Delta _{B}^{g}},\dfrac{1}{\Delta _{C}^{g}}\right)$ is a solution x of the linear equation

$$\begin{aligned} \Phi ^{g}x=\frac{\sigma }{\mu Y}\left( \begin{array} [c]{c} \pi _{A}\left( \lambda ^{g}\right) \\ \pi _{B}\left( \lambda ^{g}\right) \\ \pi _{C}\left( \lambda ^{g}\right) \end{array} \right) {.} \end{aligned}$$

(18)

Suppose that $\lambda _{i}^{g}>0$ for each i. From the equilibrium condition, $\pi _{A}\left( \lambda ^{g}\right) =\pi _{B}\left( \lambda ^{g}\right) =\pi _{C}\left( \lambda ^{g}\right) =\xi$ holds for some value of $\xi >0$. Hence, the Eq. (18) is rewritten as

$$\begin{aligned} \Phi ^{g}x=\frac{\sigma \xi }{\mu Y}\left( \begin{array}[c]{c} 1\\ 1\\ 1 \end{array} \right) {.} \end{aligned}$$

Since (18) has a solution $^{t}\left( \dfrac{1}{\Delta _{A}^{g}}, \dfrac{1}{\Delta _{B}^{g}},\dfrac{1}{\Delta _{C}^{g}}\right)$, the following holds:⁸

$$\begin{aligned} {{\text {rank}}} \begin{pmatrix} 1 &{}\quad \phi _{AB}^{g} &{}\quad \phi _{CA}^{g}\\ \phi _{AB}^{g} &{}\quad 1 &{}\quad \phi _{BC}^{g}\\ \phi _{CA}^{g} &{}\quad \phi _{BC}^{g} &{}\quad 1 \end{pmatrix} ={{\text {rank}}} \begin{pmatrix} 1 &{}\quad \phi _{AB}^{g} &{}\quad \phi _{CA}^{g} &{}\quad 1\\ \phi _{AB}^{g} &{}\quad 1 &{}\quad \phi _{BC}^{g} &{}\quad 1\\ \phi _{CA}^{g} &{}\quad \phi _{BC}^{g} &{}\quad 1 &{}\quad 1 \end{pmatrix} {.} \end{aligned}$$

This formula requires $1+2\phi _{AB}^{g}\phi _{BC}^{g}\phi _{CA}^{g}-\left( \phi _{AB}^{g}\right) ^{2}-\left( \phi _{BC}^{g}\right) ^{2}-\left( \phi _{CA}^{g}\right) ^{2}\ne 0$, i.e., $\det \Phi ^{g}\ne 0$. Furthermore, note that $\Delta _{i}^{g}=\sum _{j}\lambda _{j}^{g}\phi _{ji}^{g}>0$ for each i. By Cramer’s rule, we have

$$\begin{aligned} \dfrac{1}{\Delta _{i}^{g}}=\frac{\sigma \xi }{\mu Y}\cdot \frac{\det \Phi _{i}^{g} }{\det \Phi ^{g}}>0 \end{aligned}$$

for each i, so that $\frac{\det \Phi _{i}^{g}}{\det \Phi ^{g}}>0$ must hold for each i. Since $\Delta _{i}^{g}=\sum _{j}\lambda _{j}^{g}\phi _{ji}^{g}$, we have

$$\begin{aligned} \left( \begin{array}[c]{c} \Delta _{A}^{g}\\ \Delta _{B}^{g}\\ \Delta _{C}^{g} \end{array} \right) =\Phi ^{g}\left( \begin{array}[c]{c} \lambda _{A}^{g}\\ \lambda _{B}^{g}\\ \lambda _{C}^{g} \end{array} \right) =\frac{\mu Y}{\sigma \xi }\det \Phi ^{g}\left( \begin{array}[c]{c} \dfrac{1}{\det \Phi _{A}^{g}}\\ \dfrac{1}{\det \Phi _{B}^{g}}\\ \dfrac{1}{\det \Phi _{C}^{g}} \end{array} \right) {.} \end{aligned}$$

(19)

To apply Cramer’s rule again, we have

$$\begin{aligned} \lambda _{i}^{g}=\frac{\mu Y}{\sigma \xi }\det \Phi ^{g}\cdot \frac{\det \hat{\Phi }_{i}^{g}}{\det \Phi ^{g}}=\frac{\mu Y}{\sigma \xi }\det {\hat{\Phi }}_{i}^{g} \end{aligned}$$

for each i, so that $\det {\hat{\Phi }}_{i}^{g}>0$ must hold for each i. Thus, the fact that $\lambda _{i}^{g}>0$ for each i suffices that $\det \Phi ^{g} \ne 0$, $\frac{\det \Phi _{i}^{g}}{\det \Phi ^{g}}>0$ and $\det {\hat{\Phi }}_{i} ^{g}>0$.

Conversely, we shall consider the case that there does not exist an inner solution $\lambda ^{g}$ which satisfies $\pi _{i}\left( \lambda ^{g}\right) =\pi _{j}\left( \lambda ^{g}\right)$ for any i, j. This case can be divided into three cases: the first is that linear equation (19) is not well-defined, i.e. either $\det \Phi _{i}^{g}=0$ or $\frac{\mu Y}{\sigma \xi }\frac{\det \Phi ^{g}}{\det \Phi _{i}^{g}}\le 0$ (while $\Delta _{i}^{g}=\sum _{j}\lambda _{j}^{g}\phi _{ji}^{g}>0$) for some i; the second is that linear equation (19) is well-defined but it does not have any solution, i.e., $\det \Phi ^{g}=0$; and the third is that linear equation (19) is well-defined and has a solution $\lambda$ but $\lambda _{i}\notin \left( 0,1\right)$ for some i, i.e., $\det {\hat{\Phi }}_{i}^{g}\le 0$ for some i. To sum up, if there does not exist a distribution of firms $\lambda$ such that $\lambda _{i}\in \left( 0,1\right)$ for each i, then $\det \Phi ^{g}=0$, $\det \Phi _{i}^{g}=0$, $\frac{\det \Phi ^{g}}{\det \Phi _{i}^{g}}\le 0$ or $\det {\hat{\Phi }}_{i}^{g}\le 0$ for some i. Thus, if $\det \Phi ^{g}\ne 0$, $\frac{\det \Phi ^{g}}{\det \Phi _{i}^{g}}>0$ and $\det {\hat{\Phi }}_{i}^{g}>0$ for each i, there is a distribution of firms $\lambda ^{g}$ such that $\lambda _{i}^{g}\in \left( 0,1\right)$ for each i.

1.2 A.2 Proof of Lemma 2

From the above-mentioned proof of Lemma 1, we have

$$\begin{aligned} \lambda _{i}^{g}=\frac{\mu Y}{\sigma \xi }\det {\hat{\Phi }}_{i}^{g} \end{aligned}$$

and

$$\begin{aligned} \Delta _{i}^{g}=\frac{\mu Y}{\sigma \xi }\cdot \frac{\det \Phi ^{g}}{\det \Phi _{i}^{g}}{.} \end{aligned}$$

Since $\sum _{i}\lambda _{i}^{g}=1$,

$$\begin{aligned} \sum _{i}\frac{\mu Y}{\sigma \xi }\det {\hat{\Phi }}_{i}^{g}=1{.} \end{aligned}$$

To solve the equation for $\xi$,

$$\begin{aligned} \xi =\frac{\mu Y}{\sigma }\sum _{j}\det {\hat{\Phi }}_{j}^{g}{.} \end{aligned}$$

Finally, we show that, for any given network g, we have $\sum _{i}\det {\hat{\Phi }}_{i}^{g}=3$ to obtain $\pi _{i}\left( \lambda ^{g}\right) =\xi =3\frac{\mu }{\sigma }Y$.

For ease of expression, operator $\left| \cdot \right|$ is defined over countries as $\left| A\right| =1$, $\left| B\right| =2$ and $\left| C\right| =3$. Then we have $\det \Phi _{i}^{g}=\sum _{j}\left( -1\right) ^{\left| j\right| +\left| i\right| }\det \Phi _{-ji}^{g}$ and $\det {\hat{\Phi }}_{i}^{g}=\sum _{j}\left( -1\right) ^{\left| j\right| +\left| i\right| }\frac{\det \Phi _{-ji}^{g} }{\det \Phi _{j}^{g}}$ for each i, where $\Phi _{-ji}^{g}$ is a $2\times 2$ submatrix which is obtained by deleting $\left| j\right|$-th row vector and $\left| i\right|$-th column vector from $\Phi ^{g}$. Also, we have $\det \Phi _{-ji}^{g}=\det \Phi _{-ij}^{g}$ since $^{t}\Phi ^{g}=\Phi ^{g}$. Therefore,

$$\begin{aligned} \sum _{i}\det {\hat{\Phi }}_{i}^{g}&=\sum _{i}\sum _{j}\left( -1\right) ^{\left| j\right| +\left| i\right| }\frac{\det \Phi _{-ji}^{g}}{\det \Phi _{j}^{g}}\\&=\sum _{j}\frac{\sum _{i}\left( -1\right) ^{\left| j\right| +\left| i\right| }\det \Phi _{-ji}^{g}}{\det \Phi _{j}^{g}}\\&=\sum _{j}\frac{\sum _{i}\left( -1\right) ^{\left| i\right| +\left| j\right| }\det \Phi _{-ij}^{g}}{\det \Phi _{j}^{g}}\\&=\sum _{j}\frac{\det \Phi _{j}^{g}}{\det \Phi _{j}^{g}}\\&=3{.} \end{aligned}$$

Appendix B The distribution of firms

In this section, we calculate the distribution of firms under each network g. By symmetry, the cases are separated by the network structure, specifically the number of links.

1.1 B.1 The case of the empty network

When $g=\varnothing$,

$$\begin{aligned} \Phi ^{g}=\left( \begin{array}[c]{ccc} 1 &{}\quad \phi &{}\quad \phi \\ \phi &{}\quad 1 &{}\quad \phi \\ \phi &{}\quad \phi &{}\quad 1 \end{array} \right) \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \det \Phi ^{g}&=2\phi ^{3}-3\phi ^{2}+1=\left( 2\phi +1\right) \left( 1-\phi \right) ^{2}>0,\\ \det \Phi _{A}^{g}&=\det \Phi _{B}^{g}=\det \Phi _{C}^{g}=\phi ^{2}-2\phi +1=\left( 1-\phi \right) ^{2}>0,\\ \det {\hat{\Phi }}_{A}^{g}&=\det {\hat{\Phi }}_{B}^{g}=\det {\hat{\Phi }}_{C}^{g}=1. \end{aligned} \end{aligned}$$

From Lemma 2,

$$\begin{aligned} \lambda _{i}^{g}=\frac{1}{3} \end{aligned}$$

and

$$\begin{aligned} \Delta _{i}^{g}=\frac{2\phi ^{3}-3\phi ^{2}+1}{3\left( \phi ^{2}-2\phi +1\right) }=\frac{1}{3}\left( 2\phi +1\right) \end{aligned}$$

for each i.

1.2 B.2 The case of one-link networks

Without loss of generality, assume that $g=\left\{ AB\right\}$. Then,

$$\begin{aligned} \Phi ^{g}=\left( \begin{array}[c]{ccc} 1 &{}\quad \phi \delta &{}\quad \phi \\ \phi \delta &{}\quad 1 &{}\quad \phi \\ \phi &{}\quad \phi &{}\quad 1 \end{array} \right) \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \det \Phi ^{g}&=\left( 1-\delta \phi \right) \left( \left( 1+2\phi \right) \left( 1-\phi \right) +\left( \delta -1\right) \phi \right)>0,\\ \det \Phi _{A}^{g}&=\det \Phi _{B}^{g}=\left( 1-\phi \right) \left( 1-\delta \phi \right)>0,\\ \det \Phi _{C}^{g}&=\left( 1-\delta \phi \right) \left\{ 1-\phi +\phi \left( \delta -1\right) \right\}>0,\\ \det {\hat{\Phi }}_{A}^{g}&=\det {\hat{\Phi }}_{B}^{g}=\frac{\left( 1-\phi \right) ^{2}+\phi \left( \delta -1\right) }{\left( 1-\phi \right) \left\{ 1-\phi +\phi \left( \delta -1\right) \right\} }>0\\ \det {\hat{\Phi }}_{C}^{g}&=\frac{4\phi ^{2}-3\phi +\delta \phi -3\delta \phi ^{2} +1}{\left( 1-\phi \right) \left\{ 1-\phi +\phi \left( \delta -1\right) \right\} }, \end{aligned} \end{aligned}$$

where $\det {\hat{\Phi }}_{C}^{g}$ can be either positive or negative. If $\det {\hat{\Phi }}_{C}^{g}>0$, from Lemma 2

$$\begin{aligned} \begin{aligned} \lambda _{A}^{g}&=\lambda _{B}^{g}=\frac{\phi ^{2}-3\phi +\delta \phi +1}{3\left( 1-\phi \right) \left( \delta \phi -2\phi +1\right) },\\ \lambda _{C}^{g}&=\frac{4\phi ^{2}-3\phi +\delta \phi -3\delta \phi ^{2}+1}{3\left( 1-\phi \right) \left( \delta \phi -2\phi +1\right) }, \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \Delta _{A}^{g}&=\Delta _{B}^{g} =\frac{\left( 1+\delta \phi -2\phi ^{2}\right) }{3\left( 1-\phi \right) },\\ \Delta _{C}^{g}&=\frac{\left( 1+\delta \phi -2\phi ^{2}\right) }{3\left( 1+\delta \phi -2\phi \right) }. \end{aligned} \end{aligned}$$

If $\det {\hat{\Phi }}_{C}^{g}\le 0$, then a distribution of firms $\lambda$ does not satisfy that $\lambda _{i}^{g}\in \left( 0,1\right)$ for each i. Suppose that $\lambda _{A}^{g}<\lambda _{B}^{g}$. Then

$$\begin{aligned} \begin{aligned} \pi _{A}\left( \lambda ^{g}\right)&=\frac{\mu }{\sigma }Y\left\{ \frac{1}{\lambda _{A}^{g}+\delta \phi \lambda _{B}^{g}+\phi \lambda _{C}^{g}} +\frac{\delta \phi }{\delta \phi \lambda _{A}^{g}+\lambda _{B}^{g}+ \phi \lambda _{C}^{g}}+\frac{\phi }{\phi \lambda _{A}^{g}+\phi \lambda _{B}^{g} +\lambda _{C}^{g}}\right\} {,}\\ \pi _{B}\left( \lambda ^{g}\right)&=\frac{\mu }{\sigma }Y\left\{ \frac{\delta \phi }{\lambda _{A}^{g}+\delta \phi \lambda _{B}^{g}+ \phi \lambda _{C}^{g}}+\frac{1}{\delta \phi \lambda _{A}^{g}+\lambda _{B}^{g}+ \phi \lambda _{C}^{g}}+\frac{\phi }{\phi \lambda _{A}^{g}+\phi \lambda _{B}^{g}+ \lambda _{C}^{g}}\right\} {,}\\ \pi _{C}\left( \lambda ^{g}\right)&=\frac{\mu }{\sigma }Y\left\{ \frac{\phi }{\lambda _{A}^{g}+\delta \phi \lambda _{B}^{g}+\phi \lambda _{C}^{g}} +\frac{\phi }{\delta \phi \lambda _{A}^{g}+\lambda _{B}^{g}+ \phi \lambda _{C}^{g}}+\frac{1}{\phi \lambda _{A}^{g}+\phi \lambda _{B}^{g}+\lambda _{C}^{g}}\right\} {.} \end{aligned} \end{aligned}$$

Since $\lambda _{A}^{g}+\delta \phi \lambda _{B}^{g}+\phi \lambda _{C}^{g} <\delta \phi \lambda _{A}^{g}+\lambda _{B}^{g}+\phi \lambda _{C}^{g}$, $\pi _{A}\left( \lambda ^{g}\right) >\pi _{B}\left( \lambda ^{g}\right)$. This inequality implies $\lambda _{B}^{g}=0$, but which contradicts $0\le \lambda _{A}^{g}<\lambda _{B}^{g}$. Similarly, the assumption of $\lambda _{A}^{g}>\lambda _{B}^{g}$ leads to a contradiction. Thus we have $\lambda _{A}^{g}=\lambda _{B}^{g}$.

Since $\lambda _{i}^{g}=0$ for some i, either $\lambda _{A}^{g}=\lambda _{B}^{g}=0$ or $\lambda _{C}^{g}=0$ holds. If $\lambda _{A}^{g}=\lambda _{B} ^{g}=0$, then $\lambda _{C}^{g}=1$ and

$$\begin{aligned} \begin{aligned} \pi _{A}\left( \lambda ^{g}\right)&=\pi _{B}\left( \lambda ^{g}\right) =\frac{\mu }{\sigma }Y\left( \frac{\phi ^{2}+1+\delta \phi }{\phi }\right) {,}\\ \pi _{C}\left( \lambda ^{g}\right)&=3\frac{\mu }{\sigma }Y{.} \end{aligned} \end{aligned}$$

Note that $\frac{\phi ^{2}+1+\delta \phi }{\phi }>3$, we have $\pi _{A}\left( \lambda ^{g}\right) =\pi _{B}\left( \lambda ^{g}\right) >\pi _{C}\left( \lambda ^{g}\right)$, which contradicts $\lambda _{A}=\lambda _{B}<\lambda _{C}$. On the other hand, if $\lambda _{C}^{g}=0$, then $\lambda _{A}^{g} =\lambda _{B}^{g}=1/2$ and

$$\begin{aligned} \begin{aligned} \pi _{A}\left( \lambda ^{g}\right)&=\pi _{B}\left( \lambda ^{g}\right) =3\frac{\mu }{\sigma }Y{,}\\ \pi _{C}\left( \lambda ^{g}\right)&=\frac{\mu }{\sigma }Y\left( \frac{4\phi }{1+\delta \phi }+\frac{1}{\phi }\right) {.} \end{aligned} \end{aligned}$$

Since $\det {\hat{\Phi }}_{C}^{g}\le 0$, i.e., $4\phi ^{2}-3\phi +\delta \phi -3\delta \phi ^{2}+1\le 0$, $\pi _{A}\left( \lambda \right) =\pi _{B}\left( \lambda \right) \ge \pi _{C}\left( \lambda \right)$ holds, which is consistent with the equilibrium condition.

To summarize, when $g=\left\{ ij\right\}$, a distribution of firms $\lambda$ is given by

$$\begin{aligned} \begin{aligned} \lambda _{i}^{g}&=\lambda _{j}^{g}=\frac{\phi ^{2}-3\phi +\delta \phi +1}{3\left( 1-\phi \right) \left( \delta \phi -2\phi +1\right) }{,}\\ \lambda _{k}^{g}&=\frac{4\phi ^{2}-3\phi +\delta \phi -3\delta \phi ^{2}+1}{3\left( 1-\phi \right) \left( \delta \phi -2\phi +1\right) }<\lambda _{A}=\lambda _{B} \end{aligned} \end{aligned}$$

if $1<\delta <\min \left\{ \frac{4\phi ^{2}-3\phi +1}{\phi \left( 3\phi -1\right) },\frac{1}{\phi }\right\}$ and otherwise $\lambda _{i}^{g}=\lambda _{j} ^{g}=\frac{1}{2}$ and $\lambda _{k}^{g}=0$. Here, $\frac{4\phi ^{2}-3\phi +1}{\phi \left( 3\phi -1\right) }$ is denoted by $\overline{\delta _{1}}$ in Fig. 1. $\overline{\delta _{1}}$ equals $\frac{1}{\phi }$ when $\phi =0.5$.

1.3 B.3 The case of star networks

Without loss of generality, assume that $g=\left\{ AB,BC\right\}$. Then,

$$\begin{aligned} \Phi ^{g}=\left( \begin{array}[c]{ccc} 1 &{}\quad \phi \delta &{}\quad \phi \\ \phi \delta &{}\quad 1 &{}\quad \delta \phi \\ \phi &{}\quad \delta \phi &{}\quad 1 \end{array} \right) {.} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \det \Phi ^{g}&=\left( 1-\phi \right) \left( \phi -2\delta ^{2}\phi ^{2}+1\right) ,\\ \det \Phi _{A}^{g}&=\det \Phi _{C}^{g}=\left( 1-\phi \right) \left( 1-\delta \phi \right) >0,\\ \det \Phi _{B}^{g}&=\left( 1-\phi \right) \left( \phi -2\delta \phi +1\right) ,\\ \det {\hat{\Phi }}_{A}^{g}&=\det {\hat{\Phi }}_{C}^{g}=\frac{\delta ^{2}\phi ^{2} -3\delta \phi +\phi +1}{\left( 1-\delta \phi \right) \left( \phi -2\delta \phi +1\right) },\\ \det {\hat{\Phi }}_{B}^{g}&=\frac{4\delta ^{2}\phi ^{2}-3\delta \phi ^{2}-3\delta \phi +\phi +1}{\left( 1-\delta \phi \right) \left( \phi -2\delta \phi +1\right) }, \end{aligned} \end{aligned}$$

where $\det \Phi ^{g}$, $\det \Phi _{B}^{g}$, $\det {\hat{\Phi }}_{A}^{g}$, $\det {\hat{\Phi }}_{A}^{g}$ and $\det {\hat{\Phi }}_{C}^{g}$ can be either positive or negative.

If $1<\delta <\frac{3-\sqrt{5-4\phi }}{2\phi }$, i.e., $\delta ^{2}\phi ^{2}-3\delta \phi +\phi +1>0$, then all the determinant above are positive. Hence, from Lemma 2,

$$\begin{aligned} \begin{aligned} \lambda _{A}^{g}&=\lambda _{C}^{g}=\frac{\delta ^{2}\phi ^{2}-3\delta \phi +\phi +1}{3\left( 1-\delta \phi \right) \left( \phi -2\delta \phi +1\right) },\\ \lambda _{B}^{g}&=\frac{4\delta ^{2}\phi ^{2}-3\delta \phi ^{2}-3\delta \phi +\phi +1}{3\left( 1-\delta \phi \right) \left( \phi -2\delta \phi +1\right) }, \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \Delta _{A}^{g}&=\Delta _{C}^{g}=\frac{-2\delta ^{2}\phi ^{2}+\phi +1}{3\left( 1-\delta \phi \right) },\\ \Delta _{B}^{g}&=\frac{-2\delta ^{2}\phi ^{2}+\phi +1}{3\left( \phi -2\delta \phi +1\right) }. \end{aligned} \end{aligned}$$

If $\frac{3-\sqrt{5-4\phi }}{2\phi }\le \delta <\frac{1}{\phi }$, i.e., $\delta ^{2}\phi ^{2}-3\delta \phi +\phi +1\le 0$, then either $\det {\hat{\Phi }} _{A}^{g}=\det {\hat{\Phi }}_{C}^{g}\le 0$ or $\det \Phi _{B}^{g}\le 0$, so that a distribution of firms $\lambda$ does not satisfy that $\lambda _{i}^{g} \in \left( 0,1\right)$ for each i. Suppose that $\lambda _{A}^{g} <\lambda _{C}^{g}$. Then

$$\begin{aligned} \begin{aligned} \pi _{A}\left( \lambda ^{g}\right)&=\frac{\mu }{\sigma }Y\left\{ \frac{1}{\lambda _{A}^{g}+\delta \phi \lambda _{B}^{g}+\phi \lambda _{C}^{g}} +\frac{\delta \phi }{\delta \phi \lambda _{A}^{g}+\lambda _{B}^{g}+\delta \phi \lambda _{C}^{g}}+\frac{\phi }{\phi \lambda _{A}^{g}+\delta \phi \lambda _{B} ^{g}+\lambda _{C}^{g}}\right\} {,}\\ \pi _{B}\left( \lambda ^{g}\right)&=\frac{\mu }{\sigma }Y\left\{ \frac{\delta \phi }{\lambda _{A}^{g}+\delta \phi \lambda _{B}^{g}+\phi \lambda _{C}^{g} }+\frac{1}{\delta \phi \lambda _{A}^{g}+\lambda _{B}^{g}+\delta \phi \lambda _{C} ^{g}}+\frac{\delta \phi }{\phi \lambda _{A}^{g}+\delta \phi \lambda _{B}^{g} +\lambda _{C}^{g}}\right\} {,}\\ \pi _{C}\left( \lambda ^{g}\right)&=\frac{\mu }{\sigma }Y\left\{ \frac{\phi }{\lambda _{A}^{g}+\delta \phi \lambda _{B}^{g}+ \phi \lambda _{C}^{g}} +\frac{\delta \phi }{\delta \phi \lambda _{A}^{g}+ \lambda _{B}^{g} +\delta \phi \lambda _{C}^{g}}+ \frac{1}{\phi \lambda _{A}^{g}+ \delta \phi \lambda _{B}^{g} +\lambda _{C}^{g}}\right\} {.} \end{aligned} \end{aligned}$$

Since $\lambda _{A}^{g}+\delta \phi \lambda _{B}^{g}+\phi \lambda _{C}^{g} <\phi \lambda _{A}^{g}+\delta \phi \lambda _{B}^{g}+\lambda _{C}^{g}$, $\pi _{A}\left( \lambda ^{g}\right) >\pi _{C}\left( \lambda ^{g}\right)$. This inequality implies $\lambda _{C}^{g}=0$, which contradicts $0\le \lambda _{A}^{g}<\lambda _{C}^{g}$. Similarly, the assumption of $\lambda _{A} ^{g}>\lambda _{C}^{g}$ leads to a contradiction. Thus we have $\lambda _{A}^{g}=\lambda _{C}^{g}$.

Since $\lambda _{i}^{g}=0$ for some i, either $\lambda _{A}^{g}=\lambda _{C}^{g}=0$ or $\lambda _{B}^{g}=0$ holds. If $\lambda _{B}^{g}=0$, $\lambda _{A}^{g}=\lambda _{C}^{g}=1/2$ and

$$\begin{aligned} \begin{aligned} \pi _{A}\left( \lambda ^{g}\right)&=\pi _{B}\left( \lambda ^{g}\right) =3\frac{\mu }{\sigma }Y{,}\\ \pi _{C}\left( \lambda ^{g}\right)&=\frac{\mu }{\sigma }Y\left( \frac{4\delta \phi }{1+\delta \phi }+\frac{1}{\delta \phi }\right) {.} \end{aligned} \end{aligned}$$

Then, $\pi _{A}\left( \lambda \right) =\pi _{B}\left( \lambda \right) \le \pi _{C}\left( \lambda \right)$, which implies $\lambda _{A}^{g}=\lambda _{C}^{g}=0$. This is a contradiction. On the other hand, if $\lambda _{A} ^{g}=\lambda _{C}^{g}=0$, $\lambda _{B}^{g}=1$ and

$$\begin{aligned} \begin{aligned} \pi _{A}\left( \lambda ^{g}\right)&=\pi _{C}\left( \lambda ^{g}\right) =\frac{\mu }{\sigma }Y\left( \frac{\delta ^{2}\phi ^{2}+1+\phi }{\delta \phi }\right) {,}\\ \pi _{B}\left( \lambda ^{g}\right)&=3\frac{\mu }{\sigma }Y{.} \end{aligned} \end{aligned}$$

From the assumption of $\delta ^{2}\phi ^{2}-3\delta \phi +\phi +1>0$, $\pi _{A}\left( \lambda ^{g}\right) =\pi _{B}\left( \lambda ^{g}\right) >\pi _{C}\left( \lambda ^{g}\right)$, which is consistent with the equilibrium condition.

To summarize, when $g=\left\{ ij,jk\right\}$, a distribution of firms $\lambda$ is given by

$$\begin{aligned} \begin{aligned} \lambda _{i}^{g}&=\lambda _{k}^{g}=\frac{\delta ^{2}\phi ^{2}-3\delta \phi +\phi +1}{3\left( 1-\delta \phi \right) \left( \phi -2\delta \phi +1\right) } {,}\\ \lambda _{k}^{g}&=\frac{4\delta ^{2}\phi ^{2}-3\delta \phi ^{2}-3\delta \phi +\phi +1}{3\left( 1-\delta \phi \right) \left( \phi -2\delta \phi +1\right) } <\lambda _{A}=\lambda _{B} \end{aligned} \end{aligned}$$

if $1<\delta <\frac{3-\sqrt{5-4\phi }}{2\phi }$ and otherwise $\lambda _{i} ^{g}=\lambda _{k}^{g}=0$ and $\lambda _{j}^{g}=1$. Here, $\frac{3-\sqrt{5-4\phi }}{2\phi }$ is denoted by $\overline{\delta _{2}}$ in Fig. 1. $\overline{\delta _{2}}$ is smaller than $\frac{1}{\phi }$ whenever $0<\phi <1$.

1.4 B.4 The case of the complete network

When $g=\left\{ AB,BC,CA\right\}$,

$$\begin{aligned} \Phi ^{g}=\left( \begin{array}[c]{ccc} 1 &{}\quad \delta \phi &{}\quad \delta \phi \\ \delta \phi &{}\quad 1 &{}\quad \delta \phi \\ \delta \phi &{}\quad \delta \phi &{}\quad 1 \end{array} \right) \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \det \Phi ^{g}&=2\delta ^{3}\phi ^{3}-3\delta ^{2}\phi ^{2}+1=\left( 2\delta \phi +1\right) \left( 1-\delta \phi \right) ^{2}>0,\\ \det \Phi _{A}^{g}&=\det \Phi _{B}^{g}=\det \Phi _{C}^{g}=\delta ^{2}\phi ^{2} -2\delta \phi +1=\left( 1-\delta \phi \right) ^{2}>0,\\ \det {\hat{\Phi }}_{A}^{g}&=\det {\hat{\Phi }}_{B}^{g}=\det {\hat{\Phi }}_{C}^{g}=1. \end{aligned} \end{aligned}$$

From Lemma 2,

$$\begin{aligned} \lambda _{i}^{g}=\frac{1}{3} \end{aligned}$$

and

$$\begin{aligned} \Delta _{i}^{g}=\frac{2\delta ^{3}\phi ^{3}-3\delta ^{2}\phi ^{2}+1}{3\left( \delta ^{2}\phi ^{2}-2\delta \phi +1\right) }=\frac{1}{3}\left( 2\delta \phi +1\right) \end{aligned}$$

for each i.

Appendix C The outcome networks in the games

1.1 C.1 The proof of Proposition 1

Proof

We check the condition to form a link at each negotiation:

• Negotiation AB$D_{A}^{\phi ,\left\{ AB\right\} }>0$ and $D_{B} ^{\phi ,\left\{ AB\right\} }>0$, so $\delta >1$. Therefore countries A and B always link together.

• Negotiation BC By the result of Negotiation AB, the en route network Negotiation BC faces is $\left\{ AB\right\}$. In any case, $D_{B}^{\left\{ AB\right\} ,\left\{ AB,BC\right\} }>0$ and $D_{C} ^{\left\{ AB\right\} ,\left\{ AB,BC\right\} }>0$ always hold. Therefore countries B and C always link together.

• Negotiation CA By the result of Negotiation AB and Negotiation CA, the en route network Negotiation CA faces is $\left\{ AB,BC\right\}$. In any case, $D_{C}^{\left\{ AB,BC\right\} ,\left\{ AB,BC,CA\right\} }>0$ and $D_{A}^{\left\{ AB,BC\right\} ,\left\{ AB,BC,CA\right\} }>0$ always hold. Therefore countries C and A always link together.

Finally, the outcome network of $\Gamma _{M}\left( \phi ,\delta \right)$ is always complete. $\square$

1.2 C.2 The proof of Proposition 2

Proof

We solve by backward induction. Note that for any distinct i, j and k, $D_{i}^{\phi ,\left\{ ij\right\} }>0$, $D_{i}^{\left\{ jk\right\} ,\left\{ ij,jk\right\} }>0$, $D_{j}^{\left\{ jk\right\} ,\left\{ ij,jk\right\} }>0$ and $D_{i}^{\left\{ ij,jk\right\} ,\left\{ ij,jk,ki\right\} }>0$.

• Negotiation CA For any en route network g, both $D_{C}^{g,g\cup \left\{ CA\right\} }>0$ and $D_{A}^{g,g\cup \left\{ CA\right\} }>0$ always hold and hence countries C and A always link together.

• Negotiation BC Negotiation BC consider the strategy of Negotiation CA. When the en route network Negotiation BC faces is $\phi$, the outcome network is $\left\{ BC,CA\right\}$ if countries B and C conclude a BTA, and $\left\{ CA\right\}$ if not. Since $D_{B}^{\left\{ CA\right\} \left\{ BC,CA\right\} }>0$ and $D_{C}^{\left\{ CA\right\} \left\{ BC,CA\right\} }>0$ in any case, countries B and C decide to link together. When the en route network Negotiation BC faces is $\left\{ AB\right\}$, the outcome network is $\left\{ AB,BC,CA\right\}$ if countries B and C conclude a BTA, and $\left\{ AB,CA\right\}$ if not. Since $D_{B}^{\left\{ AB,CA\right\} \left\{ AB,BC,CA\right\} }>0$ and $D_{C}^{\left\{ AB,CA\right\} \left\{ AB,BC,CA\right\} }>0$ in any case, countries B and C decide to link together.

• Negotiation AB Negotiation AB consider the strategies of Negotiation CA and Negotiation BC. Since the en route network Negotiation AB faces is $\phi$, the outcome network is $\left\{ AB,BC,CA\right\}$ if countries A and B conclude a BTA, and $\left\{ BC,CA\right\}$ if not. Since $D_{A}^{\left\{ BC,CA\right\} \left\{ AB,BC,CA\right\} }>0$ and $D_{B}^{\left\{ BC,CA\right\} \left\{ AB,BC,CA\right\} }>0$ hold in any case, countries A and B decide to link together.

Finally, the outcome network of $\Gamma _{F}\left( \phi ,\delta \right)$ is also always complete. $\square$