Farsighted free trade networks

Abstract

The paper examines whether bilateral free trade agreements can lead to global free trade. We reconsider the endogenous tariff model introduced by Goyal and Joshi (2006) who study pairwise stability of free trade networks. We depart from their analysis by adopting the concept of pairwise farsightedly stable networks (Herings et al. 2009, GEB). We show that the complete network (i.e., global free trade) constitutes a pairwise farsightedly stable set. In particular, there is a farsightedly improving path from the empty network (i.e., no free trade agreement in place) to the complete network, which involves link additions only, while farsightedly improving paths from preexisting free trade networks may involve link deletion (i.e., dissolution of some bilateral FTAs). Moreover, we show that pairwise farsightedly stable set of networks is not unique. One implication of our results is that bilateral trade negotiations, if properly channeled, can lead to global free trade, although some bilateral agreements may have to be dissolved first to pave the way towards global free trade.

Appendix

1.1 Proof of Proposition 8

Step 1: Construction. We construct a sequence $P$ from $\mathring{g }$ to $\tilde{g}$ through an iterative procedure. As assumed, there are $n$ symmetric indexed countries. In the first round, starting from country 1, each country sets up a link to the country next to it according to the index. For example, country 1 links to country 2, country 2 links to country 3,..., and country $i$ links to country $i+1$, etc. Finally, country $n$ links to country 1, and a circle is achieved, completing the first round. In the second round, according to the index, each country sets up a link to the country two steps away from it; that is, country 1 links to country 3, country 2 links to country 4,..., and country $n-1$ links to country 1 and country $n$ links to country 2. Follow the same fashion, in the $qth$ round, country $i$ links to country $i+q$, which is $q$ steps away from country $i$. At the end of $qth$ round, a symmetric network is arrived, where each country has $2q+1$ links.

If $n$ is odd, the last round is $\frac{n-1}{2}th$ round. Country $i$ links to country $i+\frac{n-1}{2}$. At the end of the last round, the complete network $\tilde{g}$ is achieved.

If $n$ is even, the second to last round is $\left( \frac{n}{2}-1\right) th$ round. The network at the end of this round is a symmetric network with each country having $n-1$ links. The last round is $\frac{n}{2}th$ round. In this round, countries link diagonally, and the complete network is achieved eventually.

Step 2: Proof of the farsightedly improving path. After constructing the sequence $P$, we proceed to prove $P$ is a farsightedly improving path in the following Lemma.

Lemma 10

The sequence $P$ from $\mathring{g}$ to $\tilde{g}$ is a farsightedly improving path.

Proof

$n$ could be odd or even, then we discuss these two cases respectively:

Case 1 $n$ is odd. Given an arbitrary round $q (1\le q\le \frac{n-1}{2}$), country 1 will link to country $q+1$ who is $q$ steps away from it. Country 2 sets up a link with country $q+2$,..., and so on. It is needed to show that at $q^{th}$ round, two acting countries who are about to link have less welfare than the one in complete network $\tilde{ g}$. The proof takes the following two steps: In the first step, general welfare functions for any two acting countries who are about to link are determined; In the second step, we locate the country with the largest welfare level in all rounds and show that it still has less welfare than $S\left( \tilde{g}\right)$.

In the first step, we derive the welfare formula for all acting countries. From (6), we know welfare is determined by the current graph, i.e., the link structure. So we need to characterize the link structure first. To simplify analysis, we divide countries into two groups. Group 1 consists of the proposing countries, and group 2 consists of the responding countries. For example, at $q^{th}$ round, country 1 links to country $q+1$. We categorize country 1 into group 1 as a proposing country, and country $q+1$ into group 2 as a responding country.

For an arbitrary round $q,$ Tables 2, 3, 4 and 5 list each country’s link structure. In these tables, the first column is country’s index number. The second column indicates the country’s own number of links, $\eta _{i}$. The third column lists the number of linked countries (the first entry) and the number of links for each linked country (the second entry). The fourth column shows the number of non-linked countries (the first entry) and their own number of links (the second entry). Take country 1 as an example. At ${ qth}$ round, she has $2q-1$ links. Meanwhile, her connected countries are of two kinds: one kind has $2q-1$ links and there are $2\left( q-1\right)$ of them, and the other kind has $2q$ links, but there are 0 of them. Her unconnected countries has $2q-1$ links and there are $n-2q+1$ of them.

Table 2 When q is between 1 and n/3, each country’s structure of links in group 1

Table 3 When q is between 1 and n/3, country’s structure of links in group 2

Table 4 When q is between n/3 and n/2, country’s structure of links in group 1

Table 5 When q is between n/3 and n/2, country’s structure of links in group 2

What we do here is to derive welfare of each country in both group 1 and group 2 according to link structure listed in Tables 2, 3, 4 and 5. Our purpose is to guarantee that all acting countries have less welfare than $S\left( \tilde{g}\right)$. Therefore, in the second step, we are going to locate the country with largest welfare among those acting countries, if it is still less than $S\left( \tilde{g}\right)$, then all acting countries have less welfare than $S\left( \tilde{g}\right)$.

In the second step, after a series of comparisons, we find that when $q= \frac{n-1}{2}$, country $\frac{n+1}{2}$ has the largest value among all acting countries.¹⁴ We denote the welfare of this country by $S_{\max }$ and compare it to $S\left( \tilde{g}\right)$

$$\begin{aligned}&S\left( \tilde{g}\right) -S_{\max } \\&= \frac{3\Theta \left( n\right) }{4\left( 2n^{5}+4n^{4}-9n^{3}-19n^{2}+4n+12\right) ^{2}} \left( \alpha -\gamma \right) ^{2} \end{aligned}$$

where $\Theta \left( n\right) =8n^{6}+32n^{5}-58n^{4}-208n^{3}+65n^{2}+266n+21$. When $n\ge 3$, it is easy to see that $8n^{6}\ge 72n^{4}>58n^{4}$ and $32n^{5}\ge 288n^{3}>208n^{3}$. Therefore, $\Theta \left( n\right) >0$, and $S\left( \tilde{g}\right) >S_{\max }$. The acting country with the largest value is still less than $S\left( \tilde{g}\right)$. We conclude that in Case 1 with $n$ being odd number, all acting countries have less welfare than the one in complete network.

Case 2 $n$ is even. Before the last round $\frac{n}{2}$, the process is the same as Case 1 with $n-1$ countries. Therefore, until the second to last round, the country with the largest welfare has $S_{\max }$. Now look at the countries in the $\frac{n}{2}th$ round and compare it to $S_{\max }$. When $q=\frac{n}{2}$, countries link diagonally. All countries’ own number of links are from $n-1$ to $n$, thus any two acting countries are symmetric. They have the same welfare level. In the last round $q=\frac{n }{2}$, when country 1 is about to link with country $\frac{n}{2}+1$, it has the largest welfare, denoted by $S_{1}$. Take the difference between $S_{1}$ and $S_{\max }$,

$$\begin{aligned}&S_{1}-S_{\max } \\&= \frac{3\Lambda \left( n\right) }{2\left( n-2\right) ^{2}\left( n+2\right) ^{2}\left( -2n+2n^{2}-13\right) ^{2}\left( 2n+2n^{2}-3\right) ^{2}} \end{aligned}$$

where $\Lambda \left( n\right) =16n^{8}+80n^{7}-420n^{6}-1168n^{5}+2668n^{4}+5352n^{3}-5369n^{2}-6754n+3774$. To show $\Lambda \left( n\right) >0$, we need to check whether $\Lambda \left( n\right)$ is monotonically increasing,

$$\begin{aligned} \Lambda ^{\prime }\left( n\right)&= 128n^{7}\!+\!560n^{6}\!-\!2520n^{5}\!-\!5840n^{4}\!+\!10\,672 n^{3}\!+\!16\,056n^{2}\!-\!10\,738n\!-\!6754 \\ \Lambda ^{\prime \prime }\left( n\right)&= 896n^{6}+3360n^{5}-12\,600n^{4}-23\,360n^{3}+32\,016 n^{2}+32\,112n-10\,738 \\ \Lambda ^{\prime \prime \prime }\left( n\right)&= 5376n^{5}+16\,800n^{4}-50\,400n^{3}-70\,080n^{2}+ 64\,032n+32\,112\ \\ \Lambda ^{4}\left( n\right)&= 26\,880n^{4}+67\,200n^{3}-151\,200n^{2}-140\,160 n+64\,032 \\ \Lambda ^{5}\left( n\right)&= 107\,520n^{3}+201\,600n^{2}-302\,400n-140\,160\\ \Lambda ^{6}\left( n\right)&= 322\,560n^{2}+403\,200n-302\,400 \end{aligned}$$

It is easy to observe that $\Lambda ^{6}\left( n\right) >0$ when $n\ge 3$, then $\Lambda ^{5}\left( n\right)$ is increasing; The smallest even number which $n$ can take is 4, so we check $\left. \Lambda ^{5}\left( n\right) \right|_{n=4}= 8757\,120>0$, then $\Lambda ^{4}\left( n\right)$ is increasing; Again $\left. \Lambda ^{4}\left( n\right) \right|_{n=4}=8266\,272>0$, so $\Lambda ^{\prime \prime \prime }\left( n\right)$ is increasing; $\left. \Lambda ^{\prime \prime \prime }\left( n\right) \right|_{n=4}=5747\,184>0,$ then $\Lambda ^{\prime \prime }\left( n\right)$ is increasing; $\left. \Lambda ^{\prime }\left( n\right) \right|_{n=4}=3019\,982>0,$ then $\Lambda ^{\prime }\left( n\right)$ is increasing; $\left. \Lambda ^{\prime }\left( n\right) \right|_{n=4}=1205\,590>0$, then $\Lambda \left( n\right)$ is increasing; with $\left. \Lambda \left( n\right) \right|_{n=4}=359\,334>0$, we conclude that $S_{1}-S_{\max }>0$ and $S_{1}$ has the largest value. It remains to show $S_{1}<S\left( \tilde{g}\right)$, then

$$\begin{aligned} S\left( \tilde{g}\right) -S_{1}= \frac{3\left( 4n^{2}+4n-3\right) }{2\left( -2n^{3}-4n^{2}+n+3\right) ^{2}}>0 \end{aligned}$$

Then in case 2 with $n$ being even number, all acting countries has less welfare than $S\left( \tilde{g}\right).$

To summarize, in the sequence $P$ we defined above, the acting countries always have less welfare than the one in the complete network. Hence they have incentive to add a link. The sequence $P$ is a farsightedly improving path from $\mathring{g}$ to $\tilde{g}$. $\square$

Step 3: The third step is to examine the networks which are not on the sequence $P$. For these networks, we want to show they are able to achieve the complete network through a farsightedly improving path. It suffices to show for any network which is not on the sequence $P$, there always exists a country who prefers the welfare in the complete network and would like to cut its links. The following lemma demonstrates the existence of such a country.

Lemma 11

Denote $P$ the set of networks on the farsightedly improving path we constructed above, then $\forall g\in G\backslash P, \exists i$ s.t. $2\le \eta _{i}\left( g\right) \le n$ and $S_{i}\left( g\right) <S\left( \tilde{g}\right)$

Proof

By negation, suppose $\exists g\in G\backslash P$, there doesn’t exist country $i$ such that $2\le \eta _{i}\le n$ and $S_{i}\left( g\right) <S_{i}\left( \tilde{g}\right)$. Then $\forall i,$ if $2\le \eta _{i}\le n$, we must have $S_{i}\left( g\right) \ge S\left( \tilde{g} \right)$, while if $S_{i}\left( g\right) <S\left( \tilde{g}\right) ,$ then $\eta _{i}=1$. Denote the first kind of country type I and the second kind type II. Actually, type I is the member country of some component, and type II is a singleton. So network $g$ must belong to one of the following cases:

• Case 1: $g$ consists of only type I countries, i.e., $2\le \eta _{i}\left( g\right) \le n$ and $S_{i}\left( g\right) \ge S\left( \tilde{g} \right) \forall i\in I$.

  • If all countries are symmetric,

    • If $\eta _{i}=n \forall i\in I$, then $g$ is the complete network. So $g$ is on the sequence $P$, contradicting our assumption.

    • If $2\le \eta _{i}<n$ $\forall i\in N,$we know complete network is the unique efficient one, so $S_{i}\left( g\right) <S\left( \tilde{g}\right)$, $\forall i\in I$, contradicting $S_{i}\left( g\right) \ge S\left( \tilde{ g}\right)$, so this network doesn’t exist.

  • If countries are not symmetric, there exist at least one country such that $S_{i}\left( g\right) >S_{i}\left( \tilde{g}\right)$. Sum over all countries’ welfare, yielding $\sum _{i\in N}S_{i}\left( g\right)>\sum _{i\in N}S\left( \tilde{g}\right)$, contradicting that the complete network is the unique efficient network. This network doesn’t exist either.

• Case 2: $g$ consists of only type II countries, i.e., $S_{i}\left( g\right) <S\left( \tilde{g}\right)$ and $\eta _{i}=1$, $\forall$ $i\in I$. Now $g$ is the empty network. It is on the sequence $P$, contradicting the assumption.

• Case 3: $g$ consists of both type I and type II countries. Now we are ready to prove non-existence of networks of case 3. If we can prove type II (a singleton country) does not have the lowest welfare, or even it has the lowest welfare, type I (a member country of some component) does not have the greater welfare than $S\left( \tilde{g}\right)$, then case 3 does not exist. In the following part, what we do is to compare the welfare difference between a singleton country and a member country within some arbitrary component. To simplify the analysis, we consider the network $g^{\prime }$ with one singleton and one component $c_{n-1}$ with $n-1$ countries. This specification does not affect our result since in an arbitrary network, all other components (except for the singleton and the underlying component) as non-linked parts can be cancelled out when taking the difference. Consider the network $g^{\prime }$, the average value of component $c_{n-1}$ can be represented as follows

  $$\begin{aligned}&\frac{w\left( c_{n-1},g^{\prime }\right) }{n-1} \\&= \frac{\left( \alpha -\gamma \right) ^{2}}{n-1}\sum \limits _{i\in c_{n-1}}\left( \frac{\left( 4n^{2}+20n+13\right) \eta _{i}^{2}+\left( 20-2n-4n^{2}\right) \eta _{i}+\left( n^{2}-4n-2\right) }{2\left( 5\eta _{i}-n+2n\eta _{i}+2\right) ^{2}} \right) \\&+\left( \frac{1}{\left( 2n+5\right) -\left( n-2\right) } \right) ^{2}\left( \alpha -\gamma \right) ^{2} \end{aligned}$$

  At the same time, the welfare of a singleton country can be calculated as

  $$\begin{aligned} S_{1}\left( g^{\prime }\right)&= \frac{\left( 2n+1\right) -\left( n-4\right) }{2\left[ \left( 2n+5\right) -\left( n-2\right) \right] }\left(\alpha -\gamma \right) ^{2} \\&+\sum \limits _{i\in c_{n-1}}\left[ \frac{\left( 2\eta _{i}\left( g^{\prime }\right) -1\right) }{\eta _{i}\left( g^{\prime }\right) \left( 2n+5\right) -\left( n-2\right) }\right] ^{2}\left( \alpha -\gamma \right) ^{2} \end{aligned}$$

  Take the difference between them, yielding

  $$\begin{aligned}&S_{1}\left( g^{\prime }\right) -\frac{w\left( c_{n-1},g^{\prime }\right) }{n-1} \\&= \frac{\left( \alpha -\gamma \right) ^{2}}{n-1}\sum \limits _{i\in c_{n-1}} \frac{\Omega \left( \eta _{i}\right) }{\left( n+7\right) ^{2}\left( 5\eta _{i}\left( g\right) -n+2n\eta _{i}\left( g\right) +2\right) ^{2}} \end{aligned}$$

  where $\Omega \left( \eta _{i}\right) =\left( 6n^{2}+39n-102\right) \eta _{i}^{2}\left( g\right) +\left( 36-144n-18n^{2}\right) \eta _{i}\left( g\right) +\left( 12n^{2}+105n+66\right)$. If $S_{1}\left( g^{\prime }\right) -\frac{w\left( c_{n-1},g^{\prime }\right) }{n-1}>0$, the singleton country doesn’t have the lowest welfare, case 3 does not exist. If $S_{1}\left( g^{\prime }\right) -\frac{w\left( c_{n-1},g^{\prime }\right) }{ n-1}<0$, singleton country may or may not be the country with the lowest welfare. In order to show the non-existence of case 3, we need to prove that a member country in the component has less welfare than complete one. Now we analyze the property of expression $\Phi \left( \eta _{i}\right) =\frac{ \Omega \left( \eta _{i}\right) }{\left( n+7\right) ^{2}\left( 5\eta _{i}\left( g\right) -n+2n\eta _{i}\left( g\right) +2\right) ^{2}}$. Take the first derivative of it with respect to $\eta _{i}$, yielding $\Phi ^{\prime }\left( \eta _{i}\right) =\frac{6\left( \left( 4n-2\right) \eta _{i}-\left( 5n+2\right) \right) }{\left( 5\eta _{i}-n+2n\eta _{i}+2\right) ^{3}} >0$ since $\eta _{i}\ge 2> \frac{\left( 5n+2\right) }{\left( 4n-2\right) }$. Thus $\Phi \left( \eta _{i}\right)$ is an increasing function of $\eta _{i}$.

  • When $\eta _{i}=5$, $\left. \Phi \left( \eta _{i}\right) \right|_{\eta _{i}=5}>0$. Since $\Phi \left( \eta _{i}\right)$ is an increasing function, for any $\eta _{i}\ge 5$, $\Phi \left( \eta _{i}\right) >0$ holds always. In this case, the singleton country does not have the lowest welfare, case 3 does not exist.

  • When $\eta _{i}=4, \left. \Phi \left( \eta _{i}\right) \right|_{\eta _{i}=4}>0$ if $n\ge 5$. So when $n\ge 5,\eta _{i}\ge 4 \forall i\in c_{n-1},$we have $\Phi \left( \eta _{i}\right) >0$, and $S_{1}\left( g^{\prime }\right) -\frac{w\left( c_{n-1},g^{\prime }\right) }{n-1}>0$.

  • When $\eta _{i}=3$, $\left. \Phi \left( \eta _{i}\right) \right|_{\eta _{i}=3}= \frac{\left( 12n^{2}+24n-744\right) }{\left( 5n+17\right) ^{2}\left( n+7\right) ^{2}}>0$ when $n\ge 7$. Therefore, when $n\ge 7$ and $\eta _{i}\ge 3 \forall i\in c_{n-1}$, we have $\Phi \left( \eta _{i}\right) >0$. However, if $n=4,5,6$ and $\eta _{i}\ge 3 \forall i\in c_{n-1},$ we need to examine them case by case.

    • If $n=4$, and $\eta _{i}\ge 3 \forall i\in c_{n-1}$, the only case is the network with one singleton and one circle with 3 countries. It is easy to check $S_{1}\left( g^{\prime }\right) -\frac{w\left( c_{n-1},g^{\prime }\right) }{n-1}<0$. Furthermore, we need to prove the welfare of the member country has less welfare than $S\left( \tilde{g} \right)$. Denote the welfare of member country with 3 links by $S_{c_{3}}\left( g^{\prime }\right)$ , then the difference is

      $$\begin{aligned}&S\left( \tilde{g}\right) -S_{c_{3}}\left( g^{\prime }\right) \\&= \frac{3\left( -368n+1230n^{2}+288n^{3}+13n^{4}-3051\right) }{2\left( n+7\right) ^{2}\left( 5n+17\right) ^{2}\left( n+1\right) ^{2}}\left( \alpha -\gamma \right) ^{2} \end{aligned}$$

      when $n=4$,$\left. S\left( \tilde{g}\right) -S_{c_{3}}\left( g^{\prime }\right) \right|_{n=4}= \frac{110\,751}{8282\,450}>0$. So the welfare of the member country has less welfare than $S\left( \tilde{g} \right)$.

    • If $n=5$, and $\eta _{i}\ge 3 \forall i\in c_{n-1}$, when the component $c_{n-1}$ is symmetric, $\Phi \left( \eta _{i}\right) <0$ and $S_{1}\left( g^{\prime }\right) -\frac{w\left( c_{n-1},g^{\prime }\right) }{ n-1}<0.$ Again we need to prove $S\left( \tilde{g}\right) -S_{c_{3}}\left( g^{\prime }\right) >0$ when $n=5$. Use the same logic, we can get $\left. S\left( \tilde{g}\right) -S_{c_{3}}\left( g^{\prime }\right) \right|_{n=5}=\frac{9}{784}>0$. If the component $c_{n-1}$ is not symmetric, the only asymmetric case with $n=5$ is shown in Fig. 1. In this case, the welfare of country 2 and 5 are $0.463627\left( \alpha -\gamma \right) ^{2}$, the welfare of country 3 and 4 are $0.487478\left( \alpha -\gamma \right) ^{2}$, and $S\left( \tilde{g}\right) =0.486111\left( \alpha -\gamma \right) ^{2}$. it’s easy to see country 2 and 5 prefer complete one.

    • If $n=6$ and $\eta _{i}\ge 3 \forall i\in C_{n-1}$, similarly, if the component $c_{n-1}$ is symmetric, then $S_{1}\left( g^{\prime }\right) - \frac{w\left( c_{n-1},g^{\prime }\right) }{n-1}<0$. Again we have $\left. S\left( \tilde{g}\right) -S_{c_{3}}\left( g^{\prime }\right) \right|_{n=6}>0$. If the component $c_{n-1}$ is not symmetric, there are some cases we need to analyze. All possible networks are illustrated in Figs. 2, 3, 4 and 5. When $n=6,$ there are four kinds of asymmetric networks as shown in Figs. 2, 3, 4 and 5. In network 1, country 3, 5 and 6 have welfare $0.476201\left( \alpha -\gamma \right) ^{2}$, which is less than $S\left( \tilde{g}\right) =0.489\,795\,91\, \left( \alpha -\gamma \right) ^{2}$. In network 2, country 3 and 6 have $0.469176\left( \alpha -\gamma \right) ^{2}$, country 4 and 5 have $0.483194\left( \alpha -\gamma \right) ^{2}$ , less than $S\left( \tilde{g}\right) =0.489\,795\,91\, \left( \alpha -\gamma \right) ^{2}$. In network 3, country 1 has $0.482246\left( \alpha -\gamma \right) ^{2}$, country 2 and 5 have $0.486306\left( \alpha -\gamma \right) ^{2}$, country 3 and 4 have $0.481102\left( \alpha -\gamma \right) ^{2}$ and country 6 has $0.472288\left( \alpha -\gamma \right) ^{2}$. It can be seen that none of them has greater welfare than complete one. In the last network, country 3 and 4 have $0.476169\left( \alpha -\gamma \right) ^{2},$ country 6 has $0.467356\left( \alpha -\gamma \right) ^{2}$. Therefore, country 3,4 and 6 have less welfare and they prefer the complete network.

  • When $\eta _{i}=2$, $\left. \Phi \left( \eta _{i}\right) \right|_{\eta _{i}=2}= -\frac{27n+270}{\left( 3n+12\right) ^{2}\left( n+7\right) ^{2}}<0$. If $c_{n-1}$ is symmetric, then $n=3$, $g^{\prime }$ is the graph with one singleton and a linked pair. Denote the welfare of country with 2 links by $S_{c_{2}}\left( g^{\prime }\right)$. It is easy to see $S\left( \tilde{g}\right) -S_{c_{2}}\left( g^{\prime }\right) = \frac{783}{39\,200}\left( \alpha -\gamma \right) ^{2}>0$. However, if $c_{n-1}\in g^{\prime }$ is not symmetric, then $n>3$. If we can show $S_{c_{2}}\left( g^{\prime }\right) <S_{1}\left( g^{\prime }\right)$, then the singleton country does not have the lowest welfare, case 3 does not exist. Look for $g^{\prime }$ where the country with 2 links can get the maximal value. The welfare of country with 2 links is

    $$\begin{aligned} S_{c_{2}}\left( g^{\prime }\right)&= \frac{2\left( 2n+1\right) -\left( n-4\right) }{2\left[ 2\left( 2n+5\right) -\left( n-2\right) \right] }\left(\alpha -\gamma \right) ^{2} \\&\quad +\left[ \frac{8}{3\left( 2n+5\right) -\left( n-2\right) } \right] ^{2}\left( \alpha -\gamma \right) ^{2} \\&\quad +\left[ \frac{1}{\left( 2n+5\right) -\left( n-2\right) }\right] ^{2}\left( \alpha -\gamma \right) ^{2} \\&\quad +\left[ \frac{\left( 2n-5\right) }{\left( n-2\right) \left( 2n+5\right) -\left( n-2\right) }\right] ^{2}\left( \alpha -\gamma \right)^{2} \\&\quad +\left( n-4\right) \left[ \frac{\left( 2n-7\right) }{\left( n-3\right) \left( 2n+5\right) -\left( n-2\right) }\right] ^{2}\left( \alpha -\gamma \right) ^{2} \end{aligned}$$

    The welfare of the singleton country is

    $$\begin{aligned} S_{1}\left( g^{\prime }\right)&= \frac{\left( 2n+1\right) -\left( n-4\right) }{2\left[ \left( 2n+5\right) -\left( n-2\right) \right] }\left( \alpha -\gamma \right) ^{2} \\&+\left[ \frac{3}{2\left( 2n+5\right) -\left( n-2\right) } \right] ^{2}\left( \alpha -\gamma \right) ^{2} \\&+\left[ \frac{5}{3\left( 2n+5\right) -\left( n-2\right) } \right] ^{2}\left( \alpha -\gamma \right) ^{2} \\&+\left[ \frac{\left( 2n-5\right) }{\left( n-2\right) \left( 2n+5\right) -\left( n-2\right) }\right] ^{2}\left( \alpha -\gamma \right) ^{2} \\&+\left( n-4\right) \left[ \frac{\left( 2n-7\right) }{\left( n-3\right) \left( 2n+5\right) -\left( n-2\right) }\right] ^{2}\left( \alpha -\gamma \right) ^{2} \end{aligned}$$

    Take the difference between them, we have $S_{1}\left( g^{\prime }\right) -S_{c_{2}}\left( g^{\prime }\right)$ $=\frac{3\left( 12n^{4}+209n^{3}+1173n^{2}+2379n+1079\right) }{\left( 5n^{3}+72n^{2}+327n+476\right) ^{2}} >0$. Obviously, the sign of the difference is positive, so it must be the case that singleton country does not have the lowest welfare.

  • To summarize the proof, we prove case 3 with both type I and type II countries does not exist, either type II country does not have the lowest welfare or when it has the lowest welfare, there exists some member country of the component with less welfare than the one in the complete network. Therefore case 3 does not exist.

• So far, we have finished the proof of Lemma 11, for any network which is not on the constructed sequence $P$, there always exists a country with more than 2 links and less welfare than $S\left( \tilde{g}\right)$.

$\square$

Fig. 1

Asymmetric network with 5 countries

Fig. 2

Asymmetric network 1 with 6 countries

Fig. 3

Asymmetric network 2 with 6 countries

Fig. 4

Asymmetric network 3 with 6 countries

Fig. 5

Asymmetric network 4 with 6 countries

The third step is based on Lemma 11, with this lemma, we conclude for any network, there is a farsightedly improving path leading it to the complete one.