Tariff diversity and FTA network

Abstract

The number of free trade agreements (FTAs) has increased in the past few decades, and the cross-country variation of the most-favored-nation (MFN) tariffs has decreased. Motivated by these two facts, we study whether the reduced variance in cross-country MFN tariffs can help explain the expansion of FTAs. Using an oligopolistic trade model with the equilibrium concept of pairwise stable networks, we reveal the existence of a set of pairwise stable network equilibria, where the number of FTAs rises as tariffs become increasingly symmetrical. This finding contributes to theoretical FTA literature by showing that an incomplete FTA network can be supported as a pairwise stable network equilibrium.

Appendix A: Characterization of pairwise stability

Proof of Lemma 1

We show that $\beta _{E}(m,\tau )-c(m,\tau )>0$. Note that $\beta _{E}(m,0)-c(m,0)=0$. From direct computation, we find

$$\begin{aligned} \frac{\partial \beta _{E}}{\partial \tau }(m,\tau )-\frac{\partial c}{\partial \tau }(m,\tau )=\frac{1}{(n+1)^{2}}\{a(2n-3)-(8n-10m-7)\tau \}>0 \end{aligned}$$

for a sufficiently large a. Next, we show that $c(m,\tau )>0.$ Again, we observe that $c(m,0)=0.$ In addition,

$$\begin{aligned} \frac{\partial c}{\partial \tau }(m,\tau )=\frac{3a+12n^{2}+4n-7-2m(2n+5))\tau }{(n+1)^{2}}>0 \end{aligned}$$

for a sufficiently large a. $\square$

Proof of Lemma 2

The derivative with respect to m follows from direct computation by treating m as a real number. That is,

$$\begin{aligned} \frac{\partial \beta _{E}}{\partial m}(m,\tau )=-\frac{2n\tau ^{2}}{(n+1)^{2}}<0. \end{aligned}$$

Similarly, we obtain

$$\begin{aligned} \frac{\partial \beta _{E}}{\partial \tau }(m,\tau )=\frac{2n\{a+(n-2m-2)\tau \}}{(n+1)^{2}}>0 \end{aligned}$$

via the assumption $a>(n+1)\tau .$ $\square$

Proof of Lemma 3

From the proof of Lemma 1, we have $\frac{\partial c}{\partial \tau }(m,\tau )>0$ when a is sufficiently large. The second part can be easily expressed by direct computation as follows:

$$\begin{aligned} \frac{\partial }{\partial m}c(m,\tau )=-\frac{(2n+5)\tau ^{2}}{(n+1)^{2}}<0. \end{aligned}$$

$\square$

Proof of Proposition 2

Let a linearly hierarchical network g be given. Note that it is sufficient to consider interior links, $\{23,\,34,\,\ldots ,\,(n-2,n-1)\}.$ In addition, we can restrict our attention to the incentive of the high-tariff country against the low-tariff country (Lemma (4)). First, consider links, $(i,i-1)$ for $i=3,\,4,\,\ldots ,\,n-1.$ Given that ${\underline{\tau }}(2,\tau _{i+1})\le \tau _{i}\text { for }i=2,\,3,\,\ldots ,n-2$, we obtain

$$\begin{aligned} c(2,\tau _{i+1})=\beta _{E}(2,{\underline{\tau }}(2,\tau _{i+1}))&\le \beta _{E}(2,\tau _{i}) \end{aligned}$$

for $i=2,\,3\,,\ldots ,\,n-2.$ This shows that no link in $\{23,\,34,\,\ldots ,\,(n-2,n-1)\}$ is severed, that is, $u_{i}(g)\ge u_{i}(g-(i,i-1))$ for $i=3,\,4,\,\ldots ,\,n-1.$ Second, consider links $(i-1,i+1)$ for $i=3,\,4,\,\ldots ,\,n-2.$ Similarly, because $\tau _{i-1}<{\underline{\tau }}(3,\tau _{i+1})$ for $i=3,\,4,\,\ldots ,\,n-2$, we obtain

$$\begin{aligned} \beta _{E}(3,\tau _{i-1})<\beta _{E}(3,{\underline{\tau }}(3,\tau _{i+1}))=c(3,\tau _{i+1}) \end{aligned}$$

for $i=3,\,4,\,\ldots ,\,n-2.$ This shows that any link between countries i and j,  such as $|i-j|=2$, will not be formed because the net marginal cost of the link is larger than the benefit to the high-tariff country, that is, $u_{i+1}(g)>u_{i+1}(g+(i-1,i+1))$ for $i=3,\,4,\,\ldots ,\,n-2.$ Moreover, increasing the marginal benefit and net marginal cost functions with respect to $\tau$ implies that any link between countries i and j, such as $|i-j|\ge 2$, cannot be formed. Thus, $u_{i}(g)>u_{i}(g+(i-l,i))$ for any i and $l>2$. $\square$

Proof of Proposition 3

Consider network g to be $g_{ij}=P(g)_{ij}P(g)_{ji}$ for all $i,\,j$. Consider two countries i and j. First, assume that $g_{ij}=1$. Then, $P(g)_{ij}=P(g)_{ji}=1$, so $u_{i}(g)\ge u_{i}(g-ij)$ and $u_{j}(g)\ge u_{j}(g-ij)$. On the contrary, assume that $g_{ij}=0$. Then, either $P(g)_{ij}=0$ or $P(g)_{ji}=0$. We may assume that $P(g)_{ij}=0$. We obtain $u_{i}(g+ij)<u_{i}(g)$. Therefore, network g is pairwise stable.

Conversely, suppose that network g is pairwise stable. Note that it is impossible for $u_{i}(g+ij)-u_{i}(g)$ and $u_{j}(g+ij)-u(g)$ to be zero at the same time when a is sufficiently large because

$$\begin{aligned} (u_{i}(g+ij)-u_{i}(g))-(u_{j}(g+ij)-u_{j}(g))=\frac{a(2n+3)(\tau _{j}-\tau _{i})}{(n+1)^{2}}+f(m_i,m_j,\tau _i,\tau _j), \end{aligned}$$

where f is a function that is constant in a. Therefore, for $i,\,j$ such that $i<j$ and $g_{ij}=0$, we may assume that $u_{j}(g)>u_{j}(g+ij)$. We obtain $P(g)_{ji}=0$, implying $P(g)_{ij}P(g)_{ji}=0$. Similarly, for $i,\,j$ such that $g_{ij}=1$, it should be $u_{i}(g)\ge u_{i}(g-ij)$ and $u_{j}(g)\ge u_{j}(g-ij)$. Therefore, we obtain $P(g)_{ij}=P(g)_{ji}=1$, implying $P(g)_{ij}P(g)_{ji}=1$. This completes the proof. $\square$

Proof of Lemma 6

Note that $\beta _{E}(n-1,0)-c(1,0)=0$. From direct computation, we find that

$$\begin{aligned} \frac{\partial \beta _{E}}{\partial \tau }(n-1,\tau )-\frac{\partial c}{\partial \tau }(1,\tau )=\frac{1}{(n+1)^{2}}\{a(2n-3)-(4n^{2}-17)\tau \}, \end{aligned}$$

which is strictly positive if a is sufficiently large. The result follows because $\beta _{E}(m_{j},\tau )-c(m_{i},\tau )\ge \beta _{E}(n-1,\tau )-c(1,\tau )$ for any $m_{i}$ and $m_{j}$. $\square$

Proof of Proposition 4

Consider two countries i and j with $i<j$. Assume that $\tau _{j}/\tau _{i}<2n/3$ to prove the case of (i). We can check that

$$\begin{aligned} \beta _{E}(m_{i},\tau _{i})-c(m_{j},\tau _{j})=\frac{a(4n\tau _{i}-6\tau _{j})}{2(n+1)^{2}}+f(m_{i},m_{j},\tau _{i},\tau _{j}) \end{aligned}$$

for any $m_i$ and $m_j$, where f is a function that is constant in a. Therefore, $\beta _{E}(m_{i},\tau _{i})-c(m_{j},\tau _{j})>0$ for a sufficiently large a, which implies that $P(g)_{ji}=1$ in any network g. Given that $\beta _{E}$ and c are increasing functions with respect to $\tau$, $P(g)_{lk}=1$ for any countries $i\le k<l\le j$. We can apply a similar argument to prove (ii). $\square$

Proof of Proposition 5

The uniqueness is guaranteed by the necessary part of Proposition 4.

(ii) Let g be a complete network. Given that $r<(2n/3)^{1/(n-1)}$, $\tau _{n}/\tau _{1}<2n/3$. Thus, we obtain $P(g)_{ji}=1$ for any $i,\,j$ by Proposition 4. By combining this with Lemma 6, we obtain $g_{ij}=P(g)_{ij}P(g)_{ji}=1$ for all $i,\,j$. Therefore, g is a pairwise stable network due to Proposition 3.

(ii) Let g be a standalone network. Given that $r>2n/3$, we have $P(g)_{i+1,i}=0$ for all $i=1,\,2,\,\dotsc ,\,n-1$, which implies that $P(g)_{ji}=0$ for all $i,\,j$ with $i<j$ by Proposition 4. Combining this with Lemma 6, we have $g_{ij}=P(g)_{ij}P(g)_{ji}=0$ for all $i,\,j$. By Proposition 3, g is pairwise stable.

(iii) Let g be a hierarchical network of size 2m. Given that $\tau _{i+m}/\tau _{i}=r^{m}<2n/3$ for any $i=1,\,2,\,\dotsc ,\,n-m$, we obtain $P(g)_{ji}=1$ for all $i,\,j$ where $i<j\le i+m$. By adopting a similar argument, we can easily verify that $P(g)_{ji}=0$ for all $i,\,j$, where $j>i+m$. Combining this with Lemma 6 provides $g_{ij}=P(g)_{ij}P(g)_{ji}$ for all $i,\,j$. Therefore, g is pairwise stable by Proposition 3. $\square$

Proof of Proposition 6

Given that a is sufficiently large, it suffices to check the incentives of high-tariff countries against those of low-tariff countries. Consider any two countries i and j with $\tau _{i}\le \tau _{j}.$ If the variation of tariffs, $\sigma$, increases to $\sigma '$, the tariff system $\tau$ should be either $\tau _{i}'\ge \tau _{i}$ or $\tau _{i}'\le \tau _{i}.$ In the former case, we can easily verify that $t\le \tau _{i},\tau _{i}',\tau _{j},\tau _{j}'\le 2t$ to ensure $\tau _{1}(t,\sigma ,1)\ge 0$ for any $\sigma \in [0,1].$ We claim that any link between countries $i\le j$ with $\tau _{j}\le 2\tau _{i}$ will be always connected in a pairwise stable network, implying that the number of links included in the former case will not be changed by the increase in $\sigma .$ Here, we write $ij\in g$ if $g_{ij}=1$ and $ij\not \in g$ if $g_{ij}=0$. $\square$

Claim 1

For any $i\le j$ such that $\tau _{j}\le 2\tau _{i}$, $ij\in g( {\tau }).$

Consider i, j such that $\tau _{i}\le \tau _{j}\le 2\tau _{i}$. Note that

$$\begin{aligned} \beta (m_{i}-1,\tau _{i})-c(m_{j}-1,\tau _{j})&\ge \beta (n-1,\tau _{i})-c(1,\tau _{j})\\&\ge \beta (n-1,\tau _{i})-c(1,2\tau _{i}) \end{aligned}$$

for any $m_{i}$ and $m_{j}$. Given that $\beta (n-1,\tau _{i})-c(1,2\tau _{i})>0$ for a sufficiently large a, we obtain $\beta (m_{i}-1,\tau _{i})-c(m_{j}-1,\tau _{j})>0$ as desired.

Next, we claim that the number of links included in the latter case, $\tau _{i}'\le \tau _{i},$ is decreasing (or not increasing) with respect to a variation in tariff levels. Combining this with the first claim will complete the proof.

Claim 2

Consider a link $ij\,(i\le j)$ such that $\tau _{i}'\le \tau _{i}$. If $ij\notin g(\tau ),$ then $ij\notin g(\tau ').$

Note that we are assuming two adjacent pairwise stable networks, $g(T)\,\text{ and } g(T')$. Suppose that $ij\notin g$ but $ij\in g'$. Then $m_{i}'=m_{i}+1$ and $m_{j}'=m_{j}+1$ because the two networks are adjacent. Hence, we obtain

$$\beta (m_{i},\tau _{i}')-c(m_{j},\tau _{j}')\ge 0>\beta (m_{i},\tau _{i})-c(m_{j},\tau _{j}).$$

To obtain a contradiction, we need to show that

$$\beta (m_{i},\tau _{i}')-c(m_{j},\tau _{j}')\le \beta (m_{i},\tau _{i})-c(m_{j},\tau _{j}).$$

Since it is trivial when $\tau _{j}'\ge \tau _{j}$, consider the case of $\tau _{j}>\tau _{j}'.$ Note that $\tau _{i}-\tau _{i}'>\tau _{j}-\tau _{j}'$, $\frac{\partial ^{2}}{\partial m\partial \tau }\beta _{E}(m,\tau )=-\frac{4n\tau }{(n+1)^{2}}<0,$ and $\frac{\partial ^{2}}{\partial m\partial \tau }c(m,\tau )=-\frac{2(2n+5)\tau }{(n+1)^{2}}<0.$ Given that two functions, $\beta _{E}$ and c, are differentiable in $\tau$, the mean value theorem implies that there are two points $\hat{\tau _{i}}\in [\tau _{i}',\,\tau _{i}]$ and $\hat{\tau _{j}}\in [\tau _{j}',\,\tau _{j}]$, such that

$$\beta (m_{i},\tau _{i})-\beta (m_{i},\tau _{i}')=\left[ \frac{\partial }{\partial \tau }\beta _{E}(m_{i},\tau )\right] _{\tau =\hat{\tau _{i}}}(\tau _{i}-\tau _{i}')$$

and

$$c(m_{j},\tau _{j})-c(m_{j},\tau _{j}')=\left[ \frac{\partial }{\partial \tau }c(m_{j},\tau )\right] _{\tau =\hat{\tau _{j}}}(\tau _{j}-\tau _{j}').$$

Thus, we obtain

$$\begin{aligned} \beta (m_{i},\tau _{i})-\beta (m_{i},\tau _{i}')&=\left[ \frac{\partial }{\partial \tau }\beta _{E}(m_{i},\tau )\right] _{\tau =\hat{\tau _{i}}}(\tau _{i}-\tau _{i}')\\&\ge \left[ \frac{\partial }{\partial \tau }\beta _{E}(n-1,\tau )\right] _{\tau =\hat{\tau _{i}}}(\tau _{i}-\tau _{i}')\\&=\frac{2n(a-n\hat{\tau _{i}})}{(n+1)^{2}}(\tau _{i}-\tau _{i}') \end{aligned}$$

and

$$\begin{aligned} c(m_{j},\tau _{j})-c(m_{j},\tau _{j}')&=\left[ \frac{\partial }{\partial \tau }c(m_{j},\tau )\right] _{\tau =\hat{\tau _{j}}}(\tau _{j}-\tau _{j}')\\&\le \left[ \frac{\partial }{\partial \tau }c(1,\tau )\right] _{\tau ={\hat{\tau }}_{j}}(\tau _{j}-\tau _{j}')\\&=\frac{3a+(2n^{2}-17)\hat{\tau _{j}}}{(n+1)^{2}}(\tau _{j}-\tau _{j}'). \end{aligned}$$

Given that

$$\begin{aligned} \frac{2n(a-n\hat{\tau _{i}})}{(n+1)^{2}}(\tau _{i}-\tau _{i}')&-\frac{3a+(2n^{2}-17)\hat{\tau _{j}}}{(n+1)^{2}}(\tau _{j}-\tau _{j}')\\ \ge \frac{(\tau _{j}-\tau _{j}')}{(n+1)^{2}}&\{(2n-3)a-2n^{2}\hat{\tau _{i}}-(2n^{2}-17)\hat{\tau _{j}}\}, \end{aligned}$$

it is positive when a is sufficiently large. Hence,

$$\beta (m_{i},\tau _{i})-\beta (m_{i},\tau _{i}')\ge c(m_{i},\tau _{i})-c(m_{j},\tau _{j}')$$

, which is equivalent to

$$\beta (m_{i},\tau _{i})-c(m_{i},\tau _{i})\ge \beta (m_{i},\tau _{i}')-c(m_{j},\tau _{j}').$$