Antidumping as a signaling device under the WTO’s ADA non-disclosure clause

Abstract

Under the WTO Antidumping Agreement (ADA) non-disclosure clause, the investigating AD authority cannot disclose the confidential information it obtains. This paper analyzes how non-disclosure of confidential information leads the government to use the magnitude of the AD duty to signal this information, in casu the costs of the firm under investigation, to the domestic industry. We obtain two main results. First the AD authority sets lower tariffs compared to the full disclosure scenario because it has an incentive to signal that the foreign firm is relatively inefficient. Second, adhering to the non-disclosure clause leads to lower domestic welfare relative to the full disclosure scenario. On the other hand, prices are lower and global welfare is higher under the non-disclosure clause, thus providing an economic rationale for the WTO’s ADA non-disclosure clause.

Appendices

Appendix A: The 3rd period Bayesian–Nash equilibrium outputs

The expected profits of the home and the foreign firm are

$$\begin{aligned} \pi _{H}&= E\left[ \left( \alpha -q_{H}-q_{F}-c\right) q_{H}\mid t\right] ;\\ \pi _{F}&= \left[ \alpha -q_{H}-q_{F}-(c-k)-t\right] q_{F}, \end{aligned}$$

where E is the expectation operator given the beliefs of the home firm about k. The firms choose their outputs to maximize their profits given the beliefs of the home firm about the amount k by which the foreign marginal cost has been reduced due to R&D. The first-order conditions for the Bayesian–Nash equilibrium are

$$\begin{aligned} \partial \pi _{H}/\partial q_{H}&= \alpha -2q_{H}-E(q_{F}\mid t)-c=0; \\ \partial \pi _{F}/\partial q_{F}&= \alpha -q_{H}-2q_{F}-(c-k)-t=0. \end{aligned}$$

(29)

To solve for the equilibrium quantities it is first necessary to solve for $E(q_{F}\mid t)$. This is done by taking the expectation of the first-order condition of the foreign firm and then using the first-order condition of the home firm to obtain

$$\begin{aligned} E(q_{F}\mid t)=\frac{\alpha -c+2(\hat{k}(t)-t)}{3}, \end{aligned}$$

(30)

where $\hat{k}(t)=E(k\mid t)$. Substituting (30) into (29) and solving yields the Bayesian–Nash equilibrium outputs

$$\begin{aligned} q_{H}&= \frac{\alpha +t-c-\hat{k}(t)}{3}; \\ q_{F}&= \frac{\alpha -c+2(k-t)}{3}-\frac{k-\hat{k}(t)}{6}.\end{aligned}$$

$\square$

Appendix B: Regularity conditions of $G^{{\textit{ND}}}(t)$

Belief monotonicity is satisfied whenever $\partial G^{{\textit{ND}}}(t)/\partial \hat{k}(t)$ is strictly positive or negative. From Fig. 2 below we see that $\partial G^{{\textit{ND}}}(t)/\partial \hat{k}(t)<0$; hence, the domestic government would like the home firm to believe that the foreign firm has a low k, i.e. a high cost.

Fig. 2

Obtaining the separating equilibrium

Type monotonicity is satisfied whenever $\partial ^{2}G^{{\textit{ND}}}(t)/\partial k\partial t$ is strictly positive or negative. Using (18), it can be shown that $\partial ^{2}G^{{\textit{ND}}}(t)/\partial t\partial k=1/2>0$. The interpretation of type monotonicity is that the marginal gain for the government from imposing an AD duty, i.e. $\partial G^{{\textit{ND}}}(t)/\partial t$, is increasing in the R&D investment level k. Another way of thinking about type monotonicity is that for a high k, the gain from using a high AD duty becomes so large that it makes it costly for the government to use AD duty to signal the uncompetitiveness of the foreign firm to the home firm. Single crossing is a technical condition that requires $(\partial G^{{\textit{ND}}}(t)/\partial k)/(\partial G^{{\textit{ND}}}(t)/\partial \hat{k}(t))$ to be monotonic in the R&D investment level k. The single crossing condition is satisfied when $(\partial G^{{\textit{ND}}}(t)/\partial k)/(\partial G^{{\textit{ND}}}(t)/\partial \hat{k}(t))$ is monotonic in k. From Eq. (10) it can be shown that:

$$\begin{aligned} \frac{\partial G^{{\textit{ND}}}(t)/\partial t}{\partial G^{{\textit{ND}}}(t)/\partial \hat{k}(t)}=-\frac{4(\alpha -c+k-3t)}{4(\alpha -c)+k-3\hat{k}}. \end{aligned}$$

By taking its derivative with respect to k;

$$\begin{aligned} \frac{\partial \frac{\partial G^{{\textit{ND}}}(t)/\partial t}{\partial G^{{\textit{ND}}}(t)/\partial \hat{k}(t)}}{\partial k}=\frac{12(\alpha -c-\hat{k}+t)}{\left[ 4(\alpha -c)+k-3\hat{k}\right] ^{2}}, \end{aligned}$$

it can be seen that $(\partial G^{{\textit{ND}}}(t)/\partial k)/(\partial G^{{\textit{ND}}}(t)/\partial \hat{k}(t))$ is monotonic in k. The single crossing condition is hence satisfied. $\square$

Appendix C: Finding and explicit solution

The first step in analyzing the differential Eq. (18), as shown in Fig. 2, is to divide the $(k,t^{*})$ space into different regions where the numerator N is positive or negative, and the denominator D is positive or negative. This determines the sign of the differential equation in each region and allows to illustrate the solutions as drawn in Fig. 2.

Note that the $D=0$ locus gives the optimal AD duty under complete information as given by Eq. (22). The numerator is positive (negative) below (above) the $N=0$ locus and the denominator is positive (negative) below (above) the $D=0$ locus. Therefore the sign of the derivative, $d\phi (k)/{\textit{dk}}$ in (22) can be derived from any point in the $(k,t^{*})$ space. The two loci intersect at $k_{0}=2(\alpha -c)$ and $t_{0}=\alpha -c$. This intersection is important because it is used to determine the two linear solutions of the differential Eq. (18), denoted by $L_{1}$ and $L_{2}$, which pass through this intersection.²¹ Noting that $d\phi (k)/{\textit{dk}}=m$, the differential equation can be solved for m, yielding two solutions, one positive $m_{1}=(1+\sqrt{7})/6$, and one negative $m_{2}=(1-\sqrt{7})/6$. These explicit solutions of the two linear equations of the differential equation that pass through the point $\left( k_{0}=2(\alpha -c),\,t_{0}=\alpha -c\right)$ are obtained by positing a linear solution of the form $t-t_{0}=m(k-k_{0})$, where the slope m is set equal to the differential equation $d\phi (k)/{\textit{dk}}$.

First, replace $(k_{0}=2(\alpha -c),\,t_{0}=\alpha -c)$ and rearrange to obtain:

$$\begin{aligned} t=m\left[ k-2(\alpha -c)\right] +\alpha -c. \end{aligned}$$

(31)

Then, set m equal to $d\phi (k)/{\textit{dk}}$ to get

$$\begin{aligned} m&= \frac{d\phi (k)}{{\textit{dk}}}\\ &= \frac{4(\alpha -c)-2k}{4(\alpha -c-3t+k)}; \end{aligned}$$

and solving for t:

$$\begin{aligned} t=\frac{2(\alpha -c+k)m-2(\alpha -c)+k}{6m}. \end{aligned}$$

(32)

Finally, set (31) equal to (32) and solve for m to obtain the two solutions of m:

$$\begin{aligned} m_{1}=\frac{1+\sqrt{7}}{6}\approx 0.61, \end{aligned}$$

and

$$\begin{aligned} m_{2}=\frac{1-\sqrt{7}}{6}\approx - 0.27. \end{aligned}$$

Given the slopes $m_{1}$ and $m_{2}$ the linear equations are then given by $L_{i}\equiv t-t_{0}=m_{i}(k-k_{0})$, $i={1,2}$. The next step in the analysis of the differential equation is to determine the initial condition that selects a particular solution to the differential equation. It will be shown here that the initial condition corresponds to the point where the $N=0$ and $D=0$ loci intersect, i.e. point $(k_{0},t_{0})$. To begin with, note from Eqs. (23) and (24), that the home firm is driven out of the market when $k\geqslant 2(\alpha -c)$. Moreover, observe from Fig. 2 that for all values of $k>2(\alpha -c)$, the numerator is negative ($N<0$) which implies that $\partial G/\partial \hat{k}(t)>0$. Belief monotonicity in this case states that the government wants the home firm to believe that the foreign firm has a high k, but since the home market is no longer active, there is no longer an incentive to signal, and the government imposes the optimal duty as given by Eq. (22). Thus, the only initial condition that can generate a solution is the point where $k=2(\alpha -c)$ and $t=(\alpha -c\big )$. As shown above, starting from the initial condition, there are two possible linear solutions: $L_{1}$ and $L_{2}$. However, the second order condition of the government’s objective function can be used to eliminate the linear solution $L_{2}$. According to Mailath (1987) or Collie and Hviid (1993, 1999a, b), the second order condition can be expressed as:

$$\begin{aligned} \frac{d^{2}G^{{\textit{ND}}}(t)}{{\textit{dt}}^{2}}=-\frac{d\phi ^{-1}}{{\textit{dt}}}\left[ \frac{\partial ^{2}G^{{\textit{ND}}}(t)}{\partial k\partial t}-\frac{\partial ^{2}G^{{\textit{ND}}}(t)}{\partial k\partial \hat{k}(t)}\frac{(\partial G^{{\textit{ND}}}(t)/\partial t)}{(\partial G^{{\textit{ND}}}(t)/\partial \hat{k}(t))}\right] <0. \end{aligned}$$

The first term in the square brackets is positive, $\partial ^{2}G^{{\textit{ND}}}(t)/\partial k\partial t>0$ because of type monotonicity, whilst the second term tends to zero as $k\rightarrow c$, since $\partial G^{{\textit{ND}}}(t)/\partial t\rightarrow 0$ and $\partial G^{{\textit{ND}}}(t)/\partial \hat{k}(t)<0$ because of belief monotonicity. Hence $d\phi ^{-1}/{\textit{dt}}>0$ has to be strictly positive for the second order condition to hold, meaning that the linear solution $L_{2}$ can be ruled out because $D<0$ and $N>0$. The unique separating equilibrium is given by the positively sloped linear solution $L_{1}$ in Fig. 2, where $D>0$ and $N>0$. It is thus given, for all values of $k\leqslant 2(\alpha -c)$, by line $L_{1}$ with equation:²²

$$\begin{aligned} t=\frac{(1+\sqrt{7})k-2(\sqrt{7}-2)(\alpha -c)}{6}. \end{aligned}$$

Appendix D: Proof of Proposition 3

The explicit solution of the optimal duty under incomplete information is:

$$\begin{aligned} t^{*}=\left\{ \begin{array}{ll} \frac{(1+\sqrt{7})k-2(\sqrt{7}-2)(\alpha -c)}{6} & \hbox{iff}\quad k<2(\alpha -c);\\ t^{o}\equiv \frac{\alpha -c+k}{3} & \hbox{iff}\quad k\geqslant 2(\alpha -c). \end{array}\right. \end{aligned}$$

(33)

$\square$

Under the non-disclosure regime, the maximization problem of the foreign firm is given by (25) with $t=t^{*}$. The equilibrium R&D level is given by

$$\begin{aligned} k^{*}=\left\{ \begin{array}{ll} \frac{2\left( 11\sqrt{7}-19\right) (\alpha -c)}{81\gamma +4\left( 5\sqrt{7}-16\right) } & \hbox{if}\quad \gamma >\frac{5-\sqrt{7}}{9};\\ \frac{2(\alpha -c)}{9\gamma -2} & \hbox{if}\quad \underline{\gamma }\leqslant \gamma <\frac{5-\sqrt{7}}{9}, \end{array}\right. \end{aligned}$$

(34)

where $\underline{\gamma }=\frac{2\alpha }{9c}.$ Under antidumping policy and complete information, the maximization problem of the foreign firm is given by (25) with $t=t^{o}$. The equilibrium R&D level is given by

$$\begin{aligned} k^{o}=\left\{ \begin{array}{ll} \frac{8(\alpha -c)}{81\gamma -32} & \hbox{if}\quad \gamma >\frac{4}{9};\\ 2(\alpha -c) &\hbox{if}\quad \frac{1}{3}\leqslant \gamma \leqslant \frac{4}{9};\\ \frac{2(\alpha -c)}{9\gamma -2} &\hbox{if}\quad \underline{\gamma }\leqslant \gamma <\frac{1}{3}, \end{array}\right. \end{aligned}$$

(35)

where $\underline{\gamma }=\frac{2\alpha }{9c}$ so that $k\leqslant c$. We now derive these two results.

1.1 D.i: Proof of Equation 34

Under the non-disclosure regime the maximization program of the foreign firm is given by (25) in which $t=t^{*}$ is given by (19).

$$\begin{aligned} {\mathop {\hbox{max}}\limits _{k}}~ \Pi _{F}=\frac{1}{9}\left[ \alpha -c+2k-\frac{2\left(\sqrt{7}-2\right)(\alpha -c)+\left(1+\sqrt{7}\right)k}{3}\right] ^{2}-\frac{\gamma }{2}k^{2}. \end{aligned}$$

(36)

The FOC of the maximization of (36) with respect to k leads to

$$\begin{aligned} \frac{\partial \Pi _{F}}{\partial k} = \frac{4}{9}\left( 1-\frac{1+\sqrt{7}}{6}\right) \left[ \alpha -c+2k + \frac{2\left(\sqrt{7}-2\right)(\alpha -c)-\left(1+\sqrt{7}\right)k}{3}\right] -\gamma k=0. \end{aligned}$$

Taking the derivative reveals the second-order condition, which is satisfied if and only if $\gamma >4\left( 16-5\sqrt{7}\right) /81$. From the FOC the optimal level of R&D is given by

$$\begin{aligned} k^{*}=\frac{2\left( 11\sqrt{7}-19\right) (\alpha -c)}{81\gamma +4(5\sqrt{7}-16)}. \end{aligned}$$

(37)

Note from Eq. (37) that $k\geqslant 2(\alpha -c)$ when $\gamma \leqslant \frac{5-\sqrt{7}}{9}$. Whenever, $\gamma \leqslant \frac{5-\sqrt{7}}{9}$, the foreign firm monopolizes the market. The computations are therefore the same as point 2 of “Appendix D.ii”. Consequently, putting (37) and (41) (see below) together leads to Eq. (34). $\square$

1.2 D.ii: Proof of Equation 35

Under complete information the maximization program for the foreign firm is given by (25) in which $t=t^{o}$ given by (22)

$$\begin{aligned} {\mathop {\hbox{max}}\limits _{k}}~ \Pi _{F}=\left( \frac{\alpha -c+4k}{9}\right) ^{2}-\frac{\gamma }{2}k^{2}. \end{aligned}$$

(38)

1. The case of duopoly. The FOC of the maximization of (38) with respect to k leads to

$$\begin{aligned} \frac{\partial \Pi _{F}}{\partial k}=\frac{8(\alpha -c)+(32-81\gamma )k}{81}=0. \end{aligned}$$

Taking the derivative reveals the second-order condition, which is satisfied if and only if $\gamma >32/81$. From the FOC the optimal level of R&D is given by

$$\begin{aligned} k^{o}=\frac{8(\alpha -c)}{81\gamma -32}. \end{aligned}$$

(39)

For $c-k^{o}\geqslant 0$ to be satisfied it follows that

$$\begin{aligned} \gamma \geqslant \frac{8\alpha +24c}{81c}. \end{aligned}$$

Finally, note that: $\underline{\gamma }>32/81\Leftrightarrow \alpha >c$, which is a required condition for the firms to be active in the market. So $k^{o}$ is an equilibrium whenever $\gamma \geqslant \underline{\gamma }$. Next, observe from (8) that $q_{H}=0$ if $t^{o}\leqslant k-(\alpha -c)$. By replacing $t^{o}$, given by (25), and rearranging it follows that $q_{H}=0\Leftrightarrow k\geqslant 2(\alpha -c)$. From (39), $k\geqslant 2(\alpha -c)$ whenever $\gamma \leqslant 4/9$. This can only be an equilibrium if

$$\begin{aligned} \frac{4}{9}\geqslant \frac{8\alpha +24c}{81c}\Rightarrow \alpha \leqslant \frac{3}{2}c. \end{aligned}$$

Thus, for $\alpha >3/2c$ monopolization is no longer possible.

2. The case of monopoly. Monopolization is possible when $\gamma <4/9$. The monopoly quantity is given by equation of (9) and the optimal duty by (22). The maximization program is hence given by

$$\begin{aligned} {\mathop {\hbox{max}}\limits _{k}}~ \Pi _{F}=\left( \frac{\alpha -c+k-t^{o}}{2}\right) ^{2}-\frac{\gamma }{2}k^{2}, \end{aligned}$$

(40)

where $t^{o}=(\alpha -c+k)/3$. The FOC of (40) leads to

$$\begin{aligned} \frac{\partial \Pi _{F}}{\partial k}=\frac{2(\alpha -c)+(2-9\gamma )k}{9}=0. \end{aligned}$$

Taking the derivative reveals the second-order condition, which is satisfied if and only if $\gamma >2/9$. From the FOC the optimal level of R&D is given by

$$\begin{aligned} k^{o}=\frac{2(\alpha -c)}{9\gamma -2}. \end{aligned}$$

(41)

For $c-k^{o}\geqslant 0$ to be satisfied it follows that

$$\begin{aligned} \gamma \geqslant \underline{\gamma }\equiv \frac{2\alpha }{9c}. \end{aligned}$$

It can be seen that $k^{o}>2(\alpha -c)\Leftrightarrow \gamma <1/3$. Moreover, note that $\underline{\gamma }<1/3\Leftrightarrow \alpha <3c/2.$ Finally putting (39) and (40) together reveals Eq. (35).

Comparing the top equations of (34) and (35) leads to the following result: a foreign firm with a relatively low (high) R&D investment cost ($\gamma$) ends up investing less (more) in the signaling equilibrium than in the full information case.