Inter-industry trade and heterogeneous firms: country size matters

Abstract

This study investigates how industries with different patterns of firm heterogeneity distribute across countries by developing a three-sector general-equilibrium model. We show that the larger country is more specialized in the industry with heterogeneous (homogeneous) firms when trade costs are low (high) and that an increase in the inter-industry difference in firm heterogeneity fosters the larger country’s degree of specialization in the industry with heterogeneous firms. We also disclose the wage inequality and trade patterns across countries and show how they respond to trade liberalization.

Appendices

Appendices

1.1 Appendix A. Proof of Proposition 1

We first consider the interior equilibrium with $n_{1j}>0$ and $n_{2j}>0$. For Eqs. (8), (9), (10), mirror expressions exist for country f, which are, respectively, given as

$$\begin{aligned}&Y_f \alpha \beta =n_{1f}\left( \frac{\tilde{\varphi _f}}{\varphi _f^*} \right) ^{\sigma -1}\sigma f w_f+n_{1h}\left( \frac{\varphi _h^*}{\varphi _{hx}^*} \right) ^k\left( \frac{\tilde{\varphi _{hx}}}{\varphi _{hx}^*} \right) ^{\sigma -1}\sigma f_x w_h, \end{aligned}$$

(A.1)

$$\begin{aligned}&\sigma w_f=\left( \frac{p_{2ff}^{1-\sigma }Y_f}{P_{2f}^{1-\sigma }}+\phi \frac{p_{2ff}^{1-\sigma }Y_h}{P_{2h}^{1-\sigma }}\right) \alpha (1-\beta ), \end{aligned}$$

(A.2)

$$\begin{aligned}&1-\lambda =n_{1f}f+\left( \frac{\varphi _f^*}{\varphi _{fx}^*} \right) ^k n_{1f}f_x+(\varphi _f^*)^k n_{1f}f_e+n_{2f}. \end{aligned}$$

(A.3)

By plugging Eqs. (4)–(6) into Eqs. (8)–(10) and (A.1)–(A.3), the six variables $n_{1h}$, $n_{1f}$, $n_{2h}$, $n_{2f}$, $w_h$ and $w_f$ are endogenously determined by the six equations. Suppose $w_h=w_f$ for $\phi \in (0,1)$, Eqs. (8)–(10) and (A.1)–(A.3) solve

$$\begin{aligned} n_{2h}=-\left( \frac{{{{\mathcal {H}}}}\phi ^{\frac{k-\sigma +1}{\sigma -1}}}{1-{{{\mathcal {H}}}}\phi ^{\frac{k-\sigma +1}{\sigma -1}}}\right) [\lambda -(1-\lambda )\phi ]<0, \end{aligned}$$

which contradicts with $n_{2h}>0$. Therefore, we have $w_h\ne w_f$ for $\phi \in (0,1)$ in the interior equilibrium. On the other hand, at $\phi =0$, Eqs. (8)–(10) and (A.1)–(A.3) uniquely solve

$$\begin{aligned}&w_h=w_f=\frac{L \alpha }{\sigma -\alpha }, \quad n_{1h}=\frac{\beta \lambda (k-\sigma +1)}{f k}, \quad n_{2h}=\lambda (1-\beta ), \end{aligned}$$

(A.4)

$$\begin{aligned}&n_{1f}=\frac{\beta (1-\lambda ) (k-\sigma +1)}{f k}, \quad n_{2f}=(1-\beta )(1-\lambda ). \end{aligned}$$

(A.5)

By total differentiating Eqs. (8)–(10) and (A.1)–(A.3) w.r.t. $\phi$, the derivatives of $w_j$ w.r.t. $\phi$, when $\phi$ is close to zero, are derived as

$$\begin{aligned} \frac{\partial w_h}{\partial \phi }\bigg |_{\phi \rightarrow 0}=\frac{L \alpha \sigma (1-\beta )(2\lambda -1)}{\lambda (\sigma -\alpha )^2}>0, \quad \frac{\partial w_f}{\partial \phi }\bigg |_{\phi \rightarrow 0}=-\frac{L \alpha \sigma (1-\beta )(2\lambda -1)}{(1-\lambda ) (\sigma -\alpha )^2}<0. \end{aligned}$$

(A.6)

Using Eqs. (A.4), (A.6), we derive

$$\begin{aligned} \frac{\partial (w_h/w_f)}{\partial \phi }\bigg |_{\phi \rightarrow 0}=\frac{ \sigma (2 \lambda -1)(1-\beta )}{\lambda (1-\lambda )(\sigma -\alpha )}>0. \end{aligned}$$

Therefore, at $\phi$ close to zero, we have $w_h>w_f$. Together with the result above, because of the continuity, we have $w_h>w_f$ for $\phi \in (0,1)$. Furthermore, at $\phi$ close to zero, we derive

$$\begin{aligned} \frac{\partial (w_h/w_f)}{\partial k}\bigg |_{\phi \rightarrow 0}=\frac{- \beta \sigma (2 \lambda -1) {{{\mathcal {H}}}}\phi ^{\frac{k}{\sigma -1}}}{k \lambda (1-\lambda )(\sigma -\alpha )}<0, \end{aligned}$$

which implies that a smaller k brings to a higher $w_h/w_f$ when $\phi$ is close to zero. Meanwhile, at $\phi =1$, total differentiating Eqs. (8)–(10) and (A.1)–(A.3) w.r.t. $\phi$ yields

$$\begin{aligned} \frac{\partial (w_h/w_f)}{\partial \phi }\bigg |_{\phi =1}=-(2\lambda -1)<0. \end{aligned}$$

It implies that $w_h/w_f$ increases with $\phi$ at $\phi$ close to 0, and decreases with $\phi$ at $\phi$ close to 1.

We then consider the case of corner equilibrium.¹⁸ Denote the threshold value of trade freeness at which $n_{1f}=0$ by $\phi ^\sharp$. For $\phi \in (\phi ^\sharp ,1)$, we have $n_{1f}=0$, $n_{1h}>0$ and $n_{2j}>0$. Suppose $w_h=w_f\equiv w$ when $\phi \in (\phi ^\sharp ,1)$. Plugging $n_{1f}=0$ and Eq. (A.3) into Eqs. (9) and (A.2) solves

$$\begin{aligned} n_{2h}=\frac{(1-\lambda ) [\lambda -(1-\lambda ) \phi ]}{1-\lambda -\lambda \phi } \quad \text {and}\quad w=\frac{L \alpha (1-\beta ) (1-\lambda -\lambda \phi )}{\sigma [1-\lambda -(1-\lambda )\phi ]-\alpha (1-\beta ) (1-\lambda -\lambda \phi )}. \end{aligned}$$

(A.7)

As a result, Eqs. (8), (10), (A.1) together give

$$\begin{aligned} {{{\mathcal {G}}}}\equiv (L+w) \alpha \beta \left[ 1+\frac{\lambda (\sigma -1)\left( 1+ {{{\mathcal {H}}}} \phi ^{\frac{k}{\sigma -1}} \right) }{k-\sigma +1}\right] +(n_{2h}-\lambda ) \left( \frac{k}{k-\sigma +1}\right) \sigma w=0. \end{aligned}$$

By plugging (A.7) into ${{\mathcal {G}}}$, we have

$$\begin{aligned} {{{\mathcal {G}}}}>\frac{\alpha }{(1-\lambda ) (1-\phi )}\left\{ k \phi (2 \lambda -1) -\beta \left[ (1-\lambda )^2 (\sigma -1) (1-\phi )-k (1-\lambda -\lambda \phi )\right] \right\} >0, \end{aligned}$$

where the second inequality comes from the monotonicity of $\beta$. Note that if $(1-\lambda )^2 (\sigma -1) (1-\phi )-k (1-\lambda -\lambda \phi )>0$, at $\beta =1$, we have ${{{\mathcal {G}}}}>\alpha [k-(1-\lambda ) (\sigma -1)]>0.$ It contradicts with ${{{\mathcal {G}}}}=0$, and we therefore have $w_h\ne w_f$ for $\phi \in (\phi ^\sharp ,1)$.

On the other hand, at $\phi =1$, Eqs. (9) and (A.2) together give $w_h=w_f$. By plugging $n_{1f}=0$ and $w_h=w_f$ into Eqs. (8)–(10) and (A.1)–(A.3), we solve $n_{2f}=1-\lambda$ and

$$\begin{aligned}&w_h=w_f=\frac{L \alpha [k+\beta (\sigma -1)(\lambda {{{\mathcal {H}}}}-1+\lambda )]}{ k \sigma -\alpha [k+\beta (\sigma -1)(\lambda {{{\mathcal {H}}}}-1+\lambda ) ]}, \end{aligned}$$

(A.8)

$$\begin{aligned}&n_{2h}=\frac{k (1-\beta )}{k+ \beta (\sigma -1)(\lambda {{{\mathcal {H}}}}-1+\lambda )}-(1-\lambda ), \quad n_{1h}=\frac{\beta \lambda (k-\sigma +1)}{f [k+\beta (H \lambda -1+\lambda ) (\sigma -1)]}. \end{aligned}$$

(A.9)

Total differentiating Eqs. (8)–(10) and (A.2)–(A.3) w.r.t. $w_h$, $w_f$ and $\phi$ at $\phi =1$ and plugging Eqs. (A.8) into it, we derive

$$\begin{aligned} \frac{\partial (w_h/w_f)}{\partial \phi }\bigg |_{\phi =1}=-(2\lambda -1)<0, \end{aligned}$$

which implies that $w_h>w_f$ when $\phi$ is close to 1. Together with the result that $w_h\ne w_f$ for $\phi \in (\phi ^\sharp ,1)$, due to the continuity, we know $w_h>w_f$ when $\phi \in (\phi ^\sharp ,1)$ in the corner equilibrium of $n_{1f}=0$. $\square$

1.2 Appendix B: Proof of Proposition 2

At $\phi =0$, by using Eqs. (A.4) and (A.5), we solve $\eta _{1h}\equiv \frac{n_{1h}}{n_{1h}+n_{1f}}=\lambda$. By total differentiating Eqs. (8)–(10) and (A.1)–(A.3) w.r.t. $\phi$ and plugging Eqs. (A.4), (A.5) into it, using the definition of $n_{1j}$, we derive $\frac{\partial \eta _{1h}}{\partial \phi }\big |_{\phi \rightarrow 0}=-(1-\beta )(2\lambda -1)<0.$ It implies that at $\phi$ close to zero, we have $\eta _{1h}<\lambda$. At $\phi =1$, in interior equilibrium, we solve

$$\begin{aligned}&n_{1h}=\frac{\beta (k-\sigma +1)[\lambda -{{{\mathcal {H}}}}(1-\lambda )]}{f k (1-{{{\mathcal {H}}}}^2)}>0, \quad n_{2h}=\frac{(1-\beta )\lambda (1-{{{\mathcal {H}}}})-\beta (2\lambda -1){{{\mathcal {H}}}}}{1-{{{\mathcal {H}}}}}, \end{aligned}$$

(B.1)

$$\begin{aligned}&n_{1f}=\frac{\beta (k-\sigma +1)(1-\lambda -{{{\mathcal {H}}}}\lambda )}{f k (1-{{{\mathcal {H}}}}^2)}, \quad n_{2f}=\frac{(1-\beta )(1-\lambda )(1-{{{\mathcal {H}}}})+\beta (2\lambda -1){{{\mathcal {H}}}}}{1-{{{\mathcal {H}}}}}>0, \end{aligned}$$

(B.2)

in which the positiveness of $n_{2h}$ and $n_{1f}$ is guaranteed by ${{{\mathcal {H}}}}<\min \{\frac{1-\lambda }{\lambda }, \frac{\lambda -\beta \lambda }{\lambda -\beta (1-\lambda )}\}$. By the definition of $\eta _{1h}$, we have $\eta _{1h}=\lambda +\frac{{{{\mathcal {H}}}}(2\lambda -1)}{1-{{{\mathcal {H}}}}}>\lambda ,$ at $\phi =1$. On the other hand, in the case of corner equilibrium $n_{1f}=0$, for $\phi \in [\phi ^\sharp , 1)$, we have $\eta _{1h}=1>\lambda$ by the definition of $\eta _{1h}$. In both the interior and corner equilibria, because of the continuity of $\eta _{1h}$, there exists a $\phi ^*\in (0,1)$ at which $\eta _{1h}=\lambda$. We have $\eta _{1h}<\lambda$ when $\phi$ is close to 0 and $\eta _{1h}>\lambda$ when $\phi$ is close to 1.

For the Industry 2, at $\phi =0$, we solve $\eta _{2h}\equiv \frac{n_{2h}}{n_{2h}+n_{2f}}=\lambda$ and $\frac{\partial \eta _{2h}}{\partial \phi }\big |_{\phi \rightarrow 0}=\beta (2\lambda -1)>0$. Therefore, at $\phi$ close to zero, we have $\eta _{2h}>\lambda$. At $\phi =1$, in interior equilibrium, $\eta _{2h}$ is solved as

$$\begin{aligned} \eta _{2h}=\frac{(1-{{{\mathcal {H}}}})[\lambda -\beta (1-\lambda )]-\beta (2\lambda -1)}{(1-{{{\mathcal {H}}}})(1-\beta )}<\lambda , \end{aligned}$$

where the inequality is from $\lambda >1/2$. Moreover, in the corner equilibrium, for $\phi \in [\phi ^\sharp , 1)$, we have $n_{1f}=0$ and $n_{2f}=1-\lambda$. Because $n_{1h}>0$, we have $n_{2h}<\lambda$ by Eq. (10), which implies

$$\begin{aligned} \eta _{2h}\equiv \frac{n_{2h}}{n_{2h}+n_{2f}}<\frac{\lambda }{\lambda +(1-\lambda )}=\lambda . \end{aligned}$$

In both the interior and corner equilibria, because of the continuity of $\eta _{2h}$, there exists a $\phi ^\dagger \in (0,1)$ at which $\eta _{2h}=\lambda$. We have $\eta _{2h}>\lambda$ when $\phi$ is close to 0 and $\eta _{2h}<\lambda$ when $\phi$ is close to 1.

Moreover, total differentiating Eqs. (8)–(10) and (A.1)–(A.3) w.r.t. k and plugging Eqs. (A.4), (A.5) into it, by the definition of $n_{vh}$, we derive

$$\begin{aligned} \frac{\partial \eta _{1h}}{\partial k}\bigg |_{\phi \rightarrow 0}=-\frac{{{{\mathcal {H}}}}}{k}(1-\beta )(2\lambda -1)\phi ^{\frac{k}{\sigma -1}}<0\quad \text {and}\quad \frac{\partial \eta _{2h}}{\partial k}\bigg |_{\phi \rightarrow 0}=\frac{{{{\mathcal {H}}}}}{k}\beta (2\lambda -1)\phi ^{\frac{k}{\sigma -1}}>0. \end{aligned}$$

It implies that a smaller k brings to a higher $\eta _{1h}$ and a lower $\eta _{2h}$ when $\phi$ is close to zero. On the other hand, at $\phi =1$, in the interior equilibrium, differentiating $\eta _{vh}$ w.r.t. k yields

$$\begin{aligned} \frac{\partial \eta _{1h}}{\partial k}=-\frac{(2\lambda -1){{{\mathcal {H}}}}\log {\left( \frac{f_x}{f}\right) }}{(\sigma -1)(1-{{{\mathcal {H}}}})^2}<0 \quad \text {and}\quad \frac{\partial \eta _{2h}}{\partial k}=\frac{(2\lambda -1)\beta {{{\mathcal {H}}}}\log {\left( \frac{f_x}{f}\right) }}{(1-\beta )(\sigma -1)(1-{{{\mathcal {H}}}})^2}>0. \end{aligned}$$

In the corner equilibrium of $n_{1f}=0$, we have $\eta _{1h}=1$. By using Eq. (A.9), we derive

$$\begin{aligned} \frac{\partial \eta _{2h}}{\partial k}=\frac{\beta (1-\lambda )}{k^2 (1-\beta )}\left[ {{{\mathcal {H}}}} \lambda (\sigma -1+k \log {(f_x/f)})-(1-\lambda ) (\sigma -1)\right] >0, \end{aligned}$$

where the inequality comes from $k \log {(f_x/f)}>0$ and ${{{\mathcal {H}}}}>\frac{1-\lambda }{\lambda }$ in corner equilibrium. $\square$

1.3 Appendix C: Proof of Proposition 4

At $\phi =0$, we solve $EX_1(0)=EX_2(0)=0$. At $\phi$ close to zero, we derive

$$\begin{aligned} EX_1'(\phi )\big |_{\phi \rightarrow 0}=\frac{L k \alpha \beta \sigma {{{\mathcal {H}}}} (2 \lambda -1)}{(\sigma -\alpha ) (\sigma -1)}\phi ^{\frac{k-\sigma +1}{\sigma -1}}>0, \quad EX_2'(\phi )\big |_{\phi \rightarrow 0}=\frac{L \alpha \sigma (1-\beta ) (2 \lambda -1) }{\sigma -\alpha }>0, \end{aligned}$$

which implies that the larger country is a net exporter of both industrial goods when trade costs are close to the level of autarky.

On the other hand, at $\phi =1$, in the interior equilibrium, by plugging Eqs. (B.1), (B.2) into (11) and (12), we solve

$$\begin{aligned} EX_1(1)=\frac{L \alpha \beta \sigma {{{\mathcal {H}}}}(2\lambda -1)}{(1-{{{\mathcal {H}}}})(\sigma -\alpha )}>0 \quad \text {and}\quad EX_2(1)=-\frac{L\alpha \beta \sigma {{{\mathcal {H}}}}(2\lambda -1)}{(1-{{{\mathcal {H}}}})(\sigma -\alpha )}<0, \end{aligned}$$

which implies that the larger country is a net exporter (importer) of Industry 1 (2) goods when trade costs are close to the level of free trade. Furthermore, total differentiating $EX_1(1)$ and $EX_2(1)$ w.r.t. k derives

$$\begin{aligned} \frac{\partial EX_1(1)}{\partial k}=-\frac{L \alpha \beta \sigma \log (f_x/f){{{\mathcal {H}}}}(2\lambda -1)}{(1-{{{\mathcal {H}}}})^2(\sigma -1)(\sigma -\alpha )}<0, \frac{\partial EX_2(1)}{\partial k}=\frac{L \alpha \beta \sigma \log (f_x/f){{{\mathcal {H}}}} (2\lambda -1)}{(1-{{{\mathcal {H}}}})^2(\sigma -1)(\sigma -\alpha )}>0, \end{aligned}$$

which means that a smaller k increases (decreases) the larger country’s net exports of Industry 1 (2) goods when trade costs are close to the level of free trade. On the other hand, in the corner equilibrium of $n_{1f}=0$, by plugging Eqs. (A.8), (A.9) into (11) and (12), we solve

$$\begin{aligned}&EX_1(1)=\frac{{{{\mathcal {H}}}} k L \alpha \beta \lambda \sigma }{k \sigma -\alpha k-\alpha \beta (\lambda +{{{\mathcal {H}}}} \lambda -1) (\sigma -1)}>0 \quad \text {and} \\&EX_2(1)=-\frac{L \sigma \alpha \beta (1-\lambda ) \left[ k-(1-\lambda -{{{\mathcal {H}}}} \lambda ) (\sigma -1)\right] }{k (\sigma -\alpha )+\alpha \beta (1-\lambda -{{{\mathcal {H}}}} \lambda ) (\sigma -1)}<0, \end{aligned}$$

where the inequalities come from $k>2$ and ${{{\mathcal {H}}}}>\frac{1-\lambda }{\lambda }$. Furthermore, we derive

$$\begin{aligned}&\frac{\partial EX_1(1)}{\partial k}=-\frac{{{{\mathcal {H}}}} L \alpha ^2 \beta ^2\text { }\sigma \lambda (\sigma -1) ({{{\mathcal {H}}}} \lambda -1+\lambda )}{\left[ k (\alpha -\sigma )-\alpha \beta (1-\lambda -{{{\mathcal {H}}}} \lambda ) (\sigma -1)\right] ^2}<0 \quad \text {and}\\&\frac{\partial EX_2(1)}{\partial k}=\frac{L \alpha \beta \sigma (1-\lambda ) (\sigma -1)[\sigma -\alpha (1-\beta )]({{{\mathcal {H}}}} \lambda -1+\lambda ) }{[k (\alpha -\sigma )-\alpha \beta (1-\lambda -{{{\mathcal {H}}}} \lambda ) (\sigma -1)]^2}>0, \end{aligned}$$

where the inequalities come from ${{{\mathcal {H}}}}>\frac{1-\lambda }{\lambda }$ in the corner equilibrium. Therefore, the results are robust in the corner equilibrium of $n_{1f}=0$. In addition, at $\phi$ close to zero, we derive

$$\begin{aligned}&\frac{\partial EX_1(\phi )}{ \partial k}\Big |_{\phi \rightarrow 0}=-\frac{L \alpha \beta \sigma (2\lambda -1){{{\mathcal {H}}}}\phi ^{\frac{k}{\sigma -1}}}{k(\sigma -\alpha )}<0 \quad \text {and}\\&\frac{\partial EX_2(\phi )}{ \partial k}\Big |_{\phi \rightarrow 0}=\frac{L\alpha \beta \sigma (1-\beta )(2\lambda -1)[\alpha \lambda (2\lambda -1)+\sigma -\alpha \lambda ]{{{\mathcal {H}}}}\phi ^{\frac{k+\sigma -1}{\sigma -1}}}{\lambda (1-\lambda )(\sigma -\alpha )^2(k+\sigma -1)}>0, \end{aligned}$$

which imply that a smaller k increases (decreases) the larger country’s net exports of Industry 1 (2) goods when trade costs are close to the level of autarky. $\square$