Free Trade Networks on Non-tariff Barriers

Abstract

In this paper, we examine the formation of free trade agreements on eliminating non-tariff barriers as a network formation game. A firm faces an import volume quota in a foreign country, which it cannot exceed but can be removed through bilateral free trade agreements. We focus on the stable free trade agreements networks, considering the case where each country determines the quotas of the foreign firms. We show that global free trade may not be achieved in any stable network. Moreover, all countries may be autarkic in a stable network. Finally, we demonstrate that our main results continue to hold regardless of the type of non-tariff barriers, such as import licences, import share quotas, and technical measures for increasing the production costs of foreign firms.

Appendix

Proof of Lemma 1

Goyal and Joshi (2006) show two facts. One is that if η _i (g) ≥ 2, then S _i (g + g _{i j} ) > S _i (g). Two is that if η _i (g) = η _j (g) = 1, then S _i (g + g _{i j} ) > S _i (g). Since we use a stronger stability notion, we need to additionally show several facts. First, we consider g ^c. We show that S _i (g) > S _i (g − g _{i j1}−⋯−g _{j j L} ) for any $\{g_{ij_{1}},{\cdots } ,g_{ij_{L}}\}$ satisfying $g_{ij_{l}}=1$ for all l = 1, ⋯ , L in g ^c. Since S _i (g + g _{i j} ) > S _i (g) if η _i (g) ≥ 2, S _j (g ^c) > S _j (g ^c−g _{j j1}) > S _j (g ^c−g _{j j1}−g _{j j2}) > ⋯ > S _j (g ^c−g _{i j1}−⋯−g _{i j n − 2}). In addition, since

$$\begin{array}{@{}rcl@{}} && S_{i}(g^{c})-S_{i}(g^{c}-g_{ij_{1}}{\cdots} -g_{ij_{n-1}}) = \\ && \frac{1}{2}\left( \frac{nk\left( \alpha -\gamma \right) }{nk+1}\right)^{2}+nk\left( \frac{\alpha -\gamma }{nk+1}\right)^{2}-\frac{1}{2}\left( \frac{k\left( \alpha -\gamma \right) }{k+1}\right)^{2}-k\left( \frac{\alpha -\gamma }{k+1}\right)^{2}>0 \end{array} $$

for all i, the complete network is stable. Next consider g ^ci. Let i be the isolated country in g ^ci. Then, i has no incentive to add any FTA in g ^ci if and only if Eq. 1 holds. This is straightforward from equation (A.8) of Goyal and Joshi (2006). Moreover, since

$$\begin{array}{@{}rcl@{}} && S_{i}(g^{ci})-S_{i}(g^{ci}-\sum\nolimits_{l\neq i}g_{il}) = \frac{1}{2}\left( \frac{\left( n-1\right) k\left( \alpha -\gamma \right) }{ \left( n-1\right) k+1}\right)^{2}\\ && +\left( n-1\right) k\left( \frac{\alpha -\gamma }{\left( n-1\right) k+1}\right)^{2}-\frac{1}{2}\left( \frac{k\left( \alpha -\gamma \right) }{k+1}\right)^{2}-k\left( \frac{\alpha -\gamma }{k+1} \right)^{2}>0 \end{array} $$

for all i ≠ j, no country in the complete component wants to sever any subset of FTAs. □

Proof of Lemma 3

First, we show that for a given g, the maximizer of S _i is either ${q_{i}^{V}}=0$ or $\bar {q}^{V}$. The equilibrium national welfare S _i =

$$\begin{array}{@{}rcl@{}} && \frac{1}{2}\left( Q_{i}\right)^{2}+\left( \alpha -Q_{i}-\gamma \right) \sum\nolimits_{l\in K_{i}}{Q_{i}^{l}}+\sum\nolimits_{j\neq i}\left[ \left( P_{j}-\gamma \right) \sum\nolimits_{l\in K_{i}}{Q_{j}^{l}}\right] \\ &&= \frac{1}{2}\left( \sum\nolimits_{l\in K_{i}}{Q_{i}^{l}}+M_{i}\right)^{2}+\left( \alpha -\sum\nolimits_{l\in K_{i}}{Q_{i}^{l}}-M_{i}-\gamma \right) \sum\nolimits_{l\in K_{i}}{Q_{i}^{l}} \\ &&+\sum\nolimits_{j\neq i}\left[ \left( P_{j}-\gamma \right) \sum\nolimits_{l\in K_{i}}{Q_{j}^{l}}\right] . \end{array} $$

By the first order condition, for all l ∈ K _i ,

$${Q_{i}^{l}}=\frac{\alpha -\gamma -M_{i}}{k+1}\text{ and thus }Q_{i}=\frac{ k\left( \alpha -\gamma \right) +M_{i}}{k+1}. $$

Therefore,

$$\begin{array}{@{}rcl@{}} \frac{\partial S_{i}}{\partial M_{i}} &=&\frac{k\left( \alpha -\gamma \right) +M_{i}}{\left( k+1\right)^{2}}-\frac{2k\left( \alpha -\gamma -M_{i}\right) }{\left( k+1\right)^{2}} \\ &=&\frac{-k\left( \alpha -\gamma \right) +\left( 2k+1\right) M_{i}}{\left( k+1\right)^{2}},\end{array} $$

(4)

$$\frac{\partial^{2}S_{i}}{\partial \left( M_{i}\right)^{2}} =\frac{2k+1}{ \left( k+1\right)^{2}}>0\text{.} $$

(5)

By Eq. 5, for a given g, the maximizer of S _i is either ${q_{i}^{V}}=0$ or $\bar {q}^{V}$.

Second, suppose N _i (g) = {i}. Then,

$$\begin{array}{@{}rcl@{}} S_{i} &=&\frac{1}{2}\left( \frac{k\left( \alpha -\gamma \right) }{k+1} \right)^{2}+k\left( \frac{\alpha -\gamma }{k+1}\right)^{2}+\sum\nolimits_{j\neq i}\left[ \left( P_{j}-\gamma \right) \sum\nolimits_{i\in K_{i}}{Q_{j}^{i}}\right] \text{ if }{q_{i}^{V}}=0, \\ &=&\frac{1}{2}\left( \frac{nk\left( \alpha -\gamma \right) }{nk+1}\right)^{2}+k\left( \frac{\alpha -\gamma }{nk+1}\right)^{2}+\sum\nolimits_{j\neq i} \left[ \left( P_{j}-\gamma \right) \sum\nolimits_{i\in K_{i}}{Q_{j}^{i}}\right] \text{ if }{q_{i}^{V}}=\bar{q}^{V}. \end{array} $$

Therefore, country i chooses ${q_{i}^{V}}=0$ if and only if n ≤ 2k + 3.

Third, suppose η _i (g) ≥ 2. We show that country i chooses ${q_{i}^{V}}=\bar {q}^{V}$ in this case. If ${q_{i}^{V}}=0,$ M _i = (η _i (g)−1)k(α − γ)/(η _i (g)k + 1). By Eq. 4, ∂ S _i /∂ M _i ≥ 0 for all η _i (g) ≥ 2. Thus, if η _i (g) ≥ 2, then country i chooses ${q_{i}^{V}}=\bar {q}^{V}$. (Q.E.D.)

(Proof of Remark 1) Since $q_{1}^{V\ast }(\hat {g}) = q_{2}^{V\ast }(\hat {g}) = \bar {q}^{V}$ and $q_{3}^{V\ast }(\hat {g}) = 0$,

$$\begin{array}{@{}rcl@{}} S_{1}(\hat{g}) &=&S_{2}(\hat{g}) = \frac{1}{2}\left( \frac{3\left( \alpha -\gamma \right) }{4}\right)^{2}+2\left( \frac{\alpha -\gamma }{4}\right)^{2}=\frac{13\left( \alpha -\gamma \right)^{2}}{32}, \\ S_{3}(\hat{g}) &=&\frac{1}{2}\left( \frac{\alpha -\gamma }{2}\right)^{2}+\left( \frac{\alpha -\gamma }{2}\right)^{2}+2\left( \frac{\alpha -\gamma }{4}\right)^{2}=\frac{\left( \alpha -\gamma \right)^{2}}{2}. \end{array} $$

Since $q_{3}^{V\ast }(\hat {g}+g_{13}) = q_{3}^{V\ast }(\hat {g} +g_{23}) = q_{3}^{V\ast }(g^{c}) = \bar {q}^{V},$

$$\begin{array}{@{}rcl@{}} S_{3}(\hat{g}+g_{13}) &=&S_{3}(\hat{g}+g_{23}) = S_{3}(g^{c}) \\ &=&\frac{1}{2}\left( \frac{3\left( \alpha -\gamma \right) }{4}\right)^{2}+3\left( \frac{\alpha -\gamma }{4}\right)^{2}=\frac{15\left( \alpha -\gamma \right)^{2}}{32}. \end{array} $$

We have $S_{3}(\hat {g}+g_{13}) = S_{3}(\hat {g}+g_{23}) = S_{3}(g^{c})<S_{3}(\hat { g})$. Moreover, since $q_{1}^{V\ast }(\hat {g}-g_{12}) = q_{2}^{V\ast }(\hat {g} -g_{12}) = 0,$

$$S_{1}(\hat{g}-g_{12}) = S_{2}(\hat{g}-g_{12}) = \frac{1}{2}\left( \frac{\alpha -\gamma }{2}\right)^{2}+\left( \frac{\alpha -\gamma }{2}\right)^{2}=\frac{ 3\left( \alpha -\gamma \right)^{2}}{8}. $$

We have $S_{1}(\hat {g}) = S_{2}(\hat {g})>S_{1}(\hat {g}-g_{12}) = S_{2}(\hat {g} -g_{12})$. Therefore, $\hat {g}$ is uniquely stable. Furthermore, $S_{3}(\hat { g})>S_{1}(\hat {g}) = S_{2}(\hat {g})$. □

Proof of Proposition 1

At first, we show two facts. First, if two countries i and j that satisfy η _i (g) = η _j (g) = 1, then the FTA between them is not formed if and only if Eq. 3 is satisfied. This is because neither i nor j wants to form an FTA between them if and only if

$$\frac{1}{2}\left( \frac{k\left( \alpha -\gamma \right) }{k+1}\right)^{2}+k\left( \frac{\alpha -\gamma }{k+1}\right)^{2}>\frac{1}{2}\left( \frac{ nk\left( \alpha -\gamma \right) }{nk+1}\right)^{2}+2k\left( \frac{\alpha -\gamma }{nk+1}\right)^{2}, $$

which is equivalent to Eq. 3. Note that if η _i (g) = η _j (g) = 2 and g _{i j} = 1 and Eq. 3 is satisfied, then the FTA between them will be deleted and thus g is not stable. In addition, if Eq. 3 is satisfied, then Eq. 2 is also satisfied.

Second, we show that if there is a connected component including at least two FTAs in a network g, then g is not stable if and only if Eq. 2 is satisfied. We show the necessity of this fact. Suppose that there is a component including at least two FTAs. Let L be the members of the component. Note that at least three members in L. Let i ∈ L be a country such that η _i (g) ≤ η _l (g) for all l ∈ L. Now, suppose i deletes all FTAs that i forms in g and let g′ be the network. If η _i (g) ≥ 3, then η _l (g) ≥ 3 and thus η _l (g′) ≥ 2 for all l ∈ L∖{i}. Moreover, if η _i (g) = 2, then η _j (g′) ≥ 3 for all j such that g _{i j} = 1 because L has at least three countries. Therefore, η _l (g′) ≥ 2 for all l ∈ L∖{i}. By Lemma 3, the firms of country i can freely sells in the countries L∖{i} in g′. Thus, i will sever the FTAs that i forms in g and g is not stable if and only if Eq. 2 is satisfied.

Suppose that Eq. 3 holds. Then, by the facts introduced above, the empty network is uniquely stable. Suppose that Eq. 2 holds and Eq. 3 does not. If there are i and j such that η _i (g) = η _j (g) = 1, then g is not stable. Moreover, if there is a component including at least two FTAs, then g is not stable. Therefore, if n is even, then η _i (g) = 2 for all i in any stable network g. If n is odd, then η _j (g) = 1 and η _i (g) = 2 for all i ≠ j in any stable network g. Finally, if Eq. 2 does not hold, then all networks are stable and all countries adopt the free trade policy. □