“Antitrust Policy” Versus “Industrial Policy”

Abstract

Put in the context of international trade, what is viewed as an industrial policy in the existing literature may be thought of as a type of antitrust policy to seek for a beggar-thy-neighbor effect by permitting (or promoting) its manufacturing sector to take anti-competitive actions. This study demonstrates that in order to retaliate against such an industrial policy, a country may suppress competition in its service sector. For this purpose, we build a simple partial equilibrium version of the Sanyal-Jones model and demonstrate that a state in which a country suppresses competition in its manufacturing sector at the same time that its trading partner country suppresses competition in its service sector can be supported as a Nash equilibrium. In our setting, antitrust policy on the service sector is an effective policy tool only for retaliation. In other words, perfect competition can be maintained throughout the world unless the exporting country adopts an anti-competitive industrial policy, thereby triggering a retaliatory action.

We wish to dedicate this paper to the late Professor Kalyan Sanyal, whose joint work with Ron Jones has heavily influenced our research. We are grateful to an anonymous referee for useful comments. This work has partially been supported by the Keio-Kyoto GCOE program and JSPS Science Grant #23000001.

Appendix

1.1 Proof of Lemma 1

For mathematical simplicity, we introduce the following notations:

$$ \lambda = 1/n,\delta = 1/m,\;{\text{and }}\delta^{ * } = 1/m^{ * } . $$

Perfect competition policies in the three markets can be expressed as $\lambda = 0,\;\delta = 0,$ and $\delta^{ * } = 0.$ Lemma 1 can be rewritten as $\partial \tilde{W}(\lambda ,\delta ,\delta^{ * } )/\partial \delta < 0$ where $W = \tilde{W}(\lambda ,\delta ,\delta^{ * } ).$ From (1) and (2), we obtain

$$ X = \frac{b - q}{a(1 + \delta )}, $$

(A.1)

and

$$ X^{ * } = \frac{b - q}{{a^{ * } (1 + \delta^{ * } )}}. $$

(A.2)

From these and (3), we have

$$ q = - BY + b $$

(A.3)

where

$$ \begin{aligned} B & = \frac{{a(1 + \delta )a^{ * } (1 + \delta^{ * } )}}{{a(1 + \delta ) + a^{ * } (1 + \delta^{ * } )}} \\ & = B(\delta ,\delta^{ * } ). \\ \end{aligned} $$

(A.4)

Then this and (4) yield

$$ Y = \frac{b - h}{B(1 + \lambda ) + g\lambda }. $$

(A.5)

Note that $y = \lambda Y$ and $c^{'} (y) = g\lambda Y + h.$

The total surpluses of the home country are

$$ W = \frac{1}{2}(b - p)X + (p - q)X + Y\left[ {q - (\frac{1}{2}g\lambda Y + h)} \right], $$

where $\frac{1}{2}g\lambda Y + h$ is the average variable cost of a manufacturing firm. A change in $W$ is given by

$$ {\rm{d}}W = (p - q){\rm{d}}X + (Y - X){\rm{d}}q + \left[ {q - c^{'} (\lambda Y)} \right] {\rm{d}}Y, $$

(A.6)

because ${\rm{d}}p = - a{\rm{d}}X$ and $c^{'} (\lambda Y) = g\lambda Y + h.$

Consider first the term $\left[ {q - c^{'} (\lambda Y)} \right]{\rm{d}}Y$ of (A.6) and show that $\left[ {q - c^{'} (\lambda Y)} \right]{\rm{d}}Y \ge 0$ when ${\rm{d}}\delta < 0.$ From (A.4) and (A.5), we have

$$ B_{\delta } = \frac{{aa^{ * 2} (1 + \delta^{ * } )^{2} }}{{\left[ {a(1 + \delta ) + a^{ * } (1 + \delta^{ * } )} \right]^{2} }}. $$

(A.7)

and

$$ Y_{\delta } = - \frac{{\left( {b - h} \right)(1 + \lambda )B_{\delta } }}{{\left[ {(1 + \lambda )B + g\lambda } \right]^{2} }}. $$

(A.8)

Because $B_{\delta } > 0,$ then $Y_{\delta } < 0.$ This implies that ${\rm{d}}Y = Y_{\delta } d\delta > 0$ when ${\rm{d}}\delta < 0.$ We know that $q - c^{'} (\lambda Y) \ge 0$ from (4), so that $\left[ {q - c^{'} (\lambda Y)} \right]dY \ge 0$ when ${\rm{d}}\delta < 0.$

Next we consider the term $(Y - X){\rm{d}}q$ of (A.6) and show that $(Y - X){\rm{d}}q \ge 0$ when ${\rm{d}}\delta < 0.$ Differentiate (A.3) totally and use (A.8) and (A.5) to have

$$ {\rm{d}}q = - B_{\delta } \frac{{\left( {b - h} \right)g\lambda }}{{\left[ {(1 + \lambda )B + g\lambda } \right]^{2} }}{\rm{d}}\delta . $$

(A.9)

This implies that ${\rm{d}}q > 0$ if $\lambda > 0$ and ${\rm{d}}\delta < 0.$ If $\lambda = 0,\,{\rm{d}}q = 0$ because $q$ is fixed at $h.$ Since the home country is an exporting country, $Y - X > 0.$ Thus we have $(Y - X){\rm{d}}q \ge 0$ when ${\rm{d}}\delta < 0.$

Finally consider the term $(p - q){\rm{d}}X$ of (A.6). Differentiate (A.1) totally and use (A.3), (A.4), (A.5), (A.7) and (A.9) to have

$$ \begin{aligned} {\rm{d}}X = & \frac{b - h}{{a(1 + \delta )^{2} \left[ {(1 + \lambda )B + g\lambda } \right]^{2} }}\left\{ {g\lambda \left[ {(1 + \delta )B_{\delta } - B} \right] - B^{2} (1 + \lambda )} \right\}{\rm{d}}\delta \\ = & \frac{b - h}{{a(1 + \delta )^{2} \left[ {(1 + \lambda )B + g\lambda } \right]^{2} }}\left\{ {g\lambda \left[ { - \frac{{a(1 + \delta )a^{ * } (1 + \delta^{ * } )a(1 + \delta )}}{{\left[ {a(1 + \delta ) + a^{ * } (1 + \delta^{ * } )} \right]^{2} }}} \right] - B^{2} (1 + \lambda )} \right\}{\rm{d}}\delta \\ \end{aligned} $$

This shows that ${\rm{d}}X > 0$ when ${\rm{d}}\delta < 0$. Because we consider the case in which ${\rm{d}}\delta < 0,$ we exclude the case in which $\delta = 0,$ that is, $p - q = 0.$ Thus $(p - q){\rm{d}}X > 0$ when ${\rm{d}}\delta < 0.$

Since, in (A.6), the first term is positive and the second and third terms are nonnegative, we can show that ${\rm{d}}W > 0$ when ${\rm{d}}\delta < 0.$ This proves Lemma $1$.

1.2 Proof of Lemma 2

Lemma $2$ can be rewritten as $\partial \tilde{W}^{*} (\lambda ,0,0)/\partial \delta^{ * } > 0$ if $\lambda > 0\;$where $W^{ * } = \tilde{W}^{*} (\lambda ,\delta ,\delta^{ * } )$. The total surpluses of the foreign country are

$$ W^{ * } = \frac{1}{2}(b - p^{ * } )X^{ * } + (p^{ * } - q)X^{ * } . $$

A change in $W^{ * }$ is given by

$$ {\rm{d}}W^{ * } = - X^{ * } {\rm{d}}q. $$

(A.10)

We have used ${\rm{d}}p^{ * } = - X^{ * } {\rm{d}}q,$ and $p^{ * } - q = 0$ because $\delta^{ * } = 0.$ In a similar fashion to (A.9), we have

$$ {\rm{d}}q = - B_{{\delta^{ * } }} \frac{{\left( {b - h} \right)g\lambda }}{{\left[ {(1 + \lambda )B + g\lambda } \right]^{2} }}{\rm{d}}\delta^{ * } , $$

(A.11)

where $B_{{\delta^{ * } }} = \frac{{a^{2} a^{ * } }}{{\left( {a + a^{ * } } \right)^{2} }} > 0.$ From (A.10) and (A.11), we have

$$ {\rm{d}}W^{ * } = X^{ * } B_{{\delta^{ * } }} \frac{{\left( {b - h} \right)g\lambda }}{{\left[ {(1 + \lambda )B + g\lambda } \right]^{2} }}{\rm{d}}\delta^{ * } , $$

which implies that if $\lambda > 0,\,{\rm{d}}W^{ * } > 0$ when ${\rm{d}}\delta^{ * } > 0.$ This proves Lemma 2.

1.3 Proof of Lemma 3

Lemma 3 can be rewritten as $\partial W(0,0,0)/\partial \lambda > 0.\;$A change in $W$ is given by

$$ \begin{aligned} {\rm{d}}W &= \frac{1}{2}\left[ {(b - p){\rm{d}}X -X{\rm{d}}p} \right] + (p - q){\rm{d}}X + X({\rm{d}}p - {\rm{d}}q)\hfill \\ &\quad + \left[ {q -(\frac{1}{2}g\lambda Y + h)} \right]{\rm{d}}Y + Y\left[ {{\rm{d}}q -{\rm{d}}(\frac{1}{2}g\lambda Y + h)} \right] \hfill \\ &= \left[{\frac{1}{2}(b - p) + (p - q)} \right]{\rm{d}}X +\frac{1}{2}X{\rm{d}}p + (Y - X){\rm{d}}q \hfill\\ & \quad + \left[ {q - (\frac{1}{2}g\lambda Y + h)} \right]{\rm{d}}Y -\frac{1}{2}gY(Y{\rm{d}}\lambda + \lambda {\rm{d}}Y) \hfill \\ &=\left[ {\frac{1}{2}(b - p) + (p - q)} \right]{\rm{d}}X +\frac{1}{2}X{\rm{d}}p + (Y - X){\rm{d}}q \hfill \\&\quad + \left[ {q - (g\lambda Y + h)}\right]{\rm{d}}Y - \frac{1}{2}gY^{2} {\rm{d}}\lambda \hfill \\ &=\left[ {\frac{1}{2}(b - p) + (p - q) - \frac{1}{2}aX}\right]{\rm{d}}X + (Y - X){\rm{d}}q + \left[ {q - (g\lambda Y + h)}\right]{\rm{d}}Y - \frac{1}{2}gY^{2} {\rm{d}}\lambda \hfill \\ &=(p - q){\rm{d}}X + (Y - X){\rm{d}}q + \left[ {q - (g\lambda Y + h)}\right]{\rm{d}}Y - \frac{1}{2}gY^{2} {\rm{d}}\lambda \hfill \\ \end{aligned} $$

The coefficient of ${\rm{d}}X$ is zero because $\delta = 0.$ The coefficient of ${\rm{d}}Y$ is also zero because $\lambda = 0.$ Then we have

$$ {\rm{d}}W = (Y - X){\rm{d}}q - \frac{1}{2}gY^{2} {\rm{d}}\lambda . $$

Next we consider the relation between ${\rm{d}}q$ and ${\rm{d}}\lambda .$ Note that in (A.5), $B$ is fixed when $\delta ,\delta^{ * }$ are fixed at zero:

$$ B = \frac{{aa^{ * } }}{{a + a^{ * } }}. $$

(A.12)

From (A.5), we have

$$ Y_{\lambda } = - \frac{(b - h)(B + g)}{{B^{2} }}, $$

when $\lambda = 0.$ From this and $q = - BY + b,$ we have

$$ {\rm{d}}q = \frac{{\left( {b - h} \right)(B + g)}}{B}{\rm{d}}\lambda . $$

Noting that $Y = \frac{b - h}{B}$ with (A.12) and $X = \frac{b - q}{a}$ with $q = h,$ we can rewrite the coefficient of ${\rm{d}}q,Y - X,$ as

$$ Y - X = \frac{b - h}{{a^{ * } }}. $$

Therefore

$$ \begin{aligned} {\rm{d}}W &= \left\{ {(Y - X)\frac{{\left( {b - h} \right)(B + g)}}{B} - \frac{1}{2}g\left( {\frac{b - h}{B}} \right)^{2} } \right\}{\rm{d}}\lambda \\ &= \frac{{\left( {b - h} \right)^{2} }}{B}\left[\frac{B + g}{{a^{ * } }} - \frac{g}{2B}\right]{\rm{d}}\lambda \\ &= \frac{{\left( {b - h} \right)^{2} }}{B}\left[\frac{B}{{a^{ * } }} + \frac{g}{{a^{ * } }} - \frac{g}{2B}\right]{\rm{d}}\lambda \\ &= \frac{{\left( {b - h} \right)^{2} }}{B}\left[\frac{a}{{a + a^{ * } }} + \frac{g}{{a^{ * } }} - \frac{g}{2B}\right]{\rm{d}}\lambda \\ &= \frac{{\left( {b - h} \right)^{2} }}{B}\left[\frac{a}{{a + a^{ * } }} + \frac{g}{{a^{ * } }} - \frac{g}{2}\frac{{a + a^{ * } }}{{aa^{ * } }}\right]{\rm{d}}\lambda \\ &= \frac{{\left( {b - h} \right)^{2} }}{B}\left[\frac{{a^{2} a^{ * } + ga\left( {a + a^{ * } } \right) - \frac{g}{2}\left( {a + a^{ * } } \right)^{2} }}{{aa^{ * } \left( {a + a^{ * } } \right)}}\right]{\rm{d}}\lambda \\ &= \frac{{\left( {b - h} \right)^{2} }}{{Baa^{ * } \left( {a + a^{ * } } \right)}}\left[a^{2} a^{ * } + ga\left( {a + a^{ * } } \right) - \frac{g}{2}\left( {a + a^{ * } } \right)^{2} \right]{\rm{d}}\lambda \\ &= \frac{{\left( {b - h} \right)^{2} }}{{a^{2} a^{ * 2} }}\left[a^{2} a^{ * } + ga\left( {a + a^{ * } } \right) - \frac{g}{2}\left( {a^{2} + 2aa^{ * } + a^{ * 2} } \right)\right]{\rm{d}}\lambda \\ &= \frac{{\left( {b - h} \right)^{2} }}{{a^{2} a^{ * 2} }}\left[a^{2} a^{ * } + \frac{g}{2}a^{2} - \frac{g}{2}a^{ * 2} \right]{\rm{d}}\lambda \\ &= \frac{{\left( {b - h} \right)^{2} }}{{a^{2} a^{ * 2} }}\left[(a^{ * } + \frac{g}{2})a^{2} - \frac{g}{2}a^{ * 2}\right]{\rm{d}}\lambda \\ \end{aligned} $$

Thus $\partial \tilde{W}(0,0,0)/\partial \lambda \begin{array}{l} >\\ =\\ <\end{array}0$ as

$$ (a^{ * } + \frac{g}{2})a^{2} - \frac{g}{2}a^{ * 2}\begin{array}{l} >\\ =\\ <\end{array} 0$$

$$(a^{ * } + \frac{g}{2})a^{2}\begin{array}{l} >\\ =\\ <\end{array} \frac{g}{2}a^{ * 2}$$

$$a\begin{array}{l} >\\ =\\ <\end{array} \sqrt {\frac{\frac{g}{2}}{{a^{ * } + \frac{g}{2}}}} \,a^{ * } $$

Lemma 3 holds if and only if $a > \sqrt {\frac{\frac{g}{2}}{{a^{ * } + \frac{g}{2}}}} {\kern 1pt} a^{ * } .$