Product Differentiation, Decreasing Costs, and Intra-sectoral Trade

Abstract

This chapter leaves (neo-) classical trade theory in order to be able to address the astonishing phenomenon of enormous international trade among similarly developed industrialized countries. The traditional assumptions of perfect competition in all markets, of trade with standardized homogeneous goods and of constant returns to scale are replaced by monopolistic competition in output markets, product differentiation and decreasing average production costs. This change in the market and cost structure enables us to address intra-industry (intra-sectoral) trade among highly developed countries. In formalizing Linder’s (An essay on trade and transformation. New York: Wiley, 1961) pioneering work on a demand-oriented trade theory in line with the already classic Dixit-Stiglitz (American Economic Review, 67, 297–308, 1977) approach, we present a full-fledged monopolistic equilibrium solution with a 100 % of intra-sectoral trade.

Mathematical Appendix

The Lagrangian of the optimization problem Eq. 11.3 reads as follows:

$$\begin{array}{lllll} \varLambda =\varsigma \ln {{\left[ {\sum\limits_{j=1}^J {{{{\left( {c(j)} \right)}}^{\alpha }}} } \right]}^{{{1 \left/ {\alpha } \right.}}}}+\left( {1-\varsigma } \right)\ln c(0)+\lambda \left[ {e-\sum\limits_{j=1}^J {p(j)c(j)} -c(0)} \right]. \end{array}$$

(11.34)

After differentiating Eq. 11.34 partially with respect to consumption variables and the Lagrange multiplier and setting the derivatives equal to zero, we get:

$$ \frac{{\partial \varLambda }}{{\partial c(j)}}=\frac{{\varsigma c{(j)^{{\alpha -1}}}}}{{\sum\limits_{j=1}^J {c{(j)^{\alpha }}} }}-\lambda p(j)=0,\kern0.75em j=1, \ldots, J, $$

(11.35a)

$$ \frac{{\partial \varLambda }}{{\partial c(0)}}=\frac{{1-\varsigma }}{c(0) }-\lambda =0, $$

(11.35b)

$$ \frac{{\partial \varLambda }}{{\partial \lambda }}=\sum\limits_{j=1}^J {p(j)c(j)+c(0)-e} =0. $$

(11.35c)

We eliminate the Lagrange multiplier by substitution. The resulting equations are:

$$ \frac{{c{(j)^{{\alpha -1}}}}}{{c{{{(j^\prime )}}^{{\alpha -1}}}}}=\frac{p(j) }{{p(j^\prime )}},\kern0.75em j\ne {j}^{\prime}=1, \ldots, J, $$

(11.36a)

$$ \frac{{\varsigma c(0)}}{{\left( {1-\varsigma } \right)\sum\limits_{j=1}^J {c{(j)^{\alpha }}c{(j)^{{1-\alpha }}}} }}=p(j), $$

(11.36b)

$$ \sum\limits_{j=1}^J {p(j)c(j)} +c(0)=e. $$

(11.36c)

Solving Eq. 11.36b for $j=J$ yields:

$$ p(J)c(J)=\frac{{\varsigma c(0)c{(J)^{\alpha }}}}{{\left( {1-\varsigma } \right)\sum\limits_{j=1}^J {c(j){}^{\alpha }} }}. $$

(11.37)

Equation 11.36c can equivalently be written as:

$$ \sum\limits_{j=1}^{J-1 } {p(j)c(j)} +p(J)c(J)+c(0)=e. $$

(11.38)

Equation 11.36a for ${j}^{\prime}=J$ gives:

$$ p(J)c(J)\frac{{c{(j)^{\alpha }}}}{{c{(J)^{\alpha }}}}=p(j)c(j),\kern0.75em j=1, \ldots, J-1. $$

(11.39)

In order to proceed, insert Eq. 11.37 into Eq. 11.39 to obtain:

$$ {{{\left[ {\zeta c(0)c{(j)^{\alpha }}} \right]}} \left/ {{\left[ {(1-\zeta )\sum\limits_{j=1}^J {c{(j)^{\alpha }}} } \right]}} \right.}=p(j)c(j). $$

Insert this result together with Eq. 11.37 into Eq. 11.38 and you will obtain after rearranging and simplifying the solution for $c(0)$ as follows:

$$ c(0)=\left( {1-\varsigma } \right)e. $$

(11.40)

Equation 11.40 together with Eq. 11.36c reveal the following relationship:

$$ \sum\limits_{j=1}^J {p(j)c(j)=\varsigma e} . $$

(11.41)

Equation 11.36a implies:

$$ \frac{c(j) }{{c(j^\prime )}}=\frac{{p{(j)^{{{1 \left/ {{(\alpha -1)}} \right.}}}}}}{{p{{{(j^\prime )}}^{{{1 \left/ {{(\alpha -1)}} \right.}}}}}}\ \Rightarrow\ \frac{p(j)c(j) }{{c(j^\prime )}}=\frac{{p{(j)^{{{\alpha \left/ {{(\alpha -1)}} \right.}}}}}}{{p{{{(j^\prime )}}^{{{1 \left/ {{(\alpha -1)}} \right.}}}}}}. $$

(11.42)

Calculating the sum in Eq. 11.42 over all j gives:

$$ \frac{{\sum\limits_{j=1}^J {p(j)c(j)} }}{{c(j^\prime )}}=\frac{{\sum\limits_{j=1}^J {p{(j)^{{{\alpha \left/ {{(\alpha -1)}} \right.}}}}} }}{{p{{{(j^\prime )}}^{{{1 \left/ {{(\alpha -1)}} \right.}}}}}}. $$

(11.43)

Finally, considering Eqs. 11.41 and 11.5 in Eq. 11.43 yields:

$$ \frac{{\varsigma e}}{{c(j^\prime )}}=\frac{{{q^{{{\alpha \left/ {{(\alpha -1)}} \right.}}}}}}{{p{{{(j^\prime )}}^{{{1 \left/ {{(\alpha -1)}} \right.}}}}}}\;\mathrm{ and}\;c(j)=\varsigma e\frac{{p{(j)^{{{1 \left/ {{(\alpha -1)}} \right.}}}}}}{{{q^{{{\alpha \left/ {{(\alpha -1)}} \right.}}}}}}. $$

(11.44)