Public infrastructure for production and international trade in a small open economy: a dynamic analysis

Abstract

This paper develops a dynamic trade model with a stock of public infrastructure, which has a property of “unpaid factor of production”. We show that a country with a smaller (larger) labor endowment tends to become an exporter of a good whose productivity is more (less) sensitive to the stock of public infrastructure. We also show that after the opening of trade, the labor-scarce country becomes unambiguously better off but the labor-abundant country may become worse off. Overall, these results contrasts with those obtained in the case of public intermediate goods with a “creation of atmosphere” property.

Appendix

1.1 A.1 Proof of Lemma 1

Making use of (7), the production functions (1) can be rewritten as

$$\begin{aligned} Y_1 = R^{\alpha _1 } \left[ \frac{(1 - \alpha _1 )y}{\theta f'(L_R )} \right] ^{1 - \alpha _1 } , \quad Y_2 = R^{\alpha _2 } \left[ \frac{(1 - \alpha _2 )(1 - y)}{\theta f'(L_R )} \right] ^{1 - \alpha _2 }. \end{aligned}$$

(A.1)

Also, using (7), the full employment condition (3) can be rewritten as

$$\begin{aligned} (1 - \alpha _1 )y + (1 - \alpha _2 )(1 - y) = \theta f'(L_R )(L - L_R ). \end{aligned}$$

(A.2)

Therefore, the temporary equilibrium solutions for $Y_1$, $Y_2$, y, and $L_R$ are derived from the system of Eqs. (A.1), (A.2), and (8). Totally differentiating the system, we have

$$\begin{aligned} \begin{bmatrix} 1&0&- \frac{(1 - \alpha _1 )Y_1}{y}&\frac{(1 - \alpha _1 )Y_1 f''}{f'} \\ 0&1&\frac{(1 - \alpha _2 )Y_2}{1 - y}&\frac{(1 - \alpha _2 )Y_2 f''}{f'} \\ 0&0&\frac{\alpha _2 - \alpha _1}{\theta }&{f' - (L - L_R )f''} \\ - \frac{1 - y}{Y_1}&\frac{1 - y}{Y_2 }&\frac{1}{y}&0 \\ \end{bmatrix}\nonumber \\ \begin{bmatrix} {dY_1 } \\ {dY_2 } \\ {dy} \\ {dL_R } \\ \end{bmatrix} = \begin{bmatrix} \frac{\alpha _1 Y_1 }{R}dR - \frac{(1 - \alpha _1 )Y_1 }{\theta }d\theta \\ \frac{\alpha _2 Y_2 }{R}dR - \frac{(1 - \alpha _2 )Y_2 }{\theta }d\theta \\ \frac{(L - L_R )f'}{\theta }d\theta + f'dL \\ \frac{1 - y}{p}dp \\ \end{bmatrix}. \end{aligned}$$

(A.3)

Of great interest are the effects on the temporary equilibrium solutions for y and $L_R$, which can be obtained by solving (A.3):

$$\begin{aligned} dy= & {} \frac{1}{\Delta }\left\{ \frac{(\alpha _1 - \alpha _2 )[f' - (L - L_R )f'']}{R}dR + \frac{(\alpha _1 - \alpha _2 )f'}{\theta }d\theta \right. \nonumber \\&\left. +\, \frac{f' - (L - L_R )f''}{p}dp + (\alpha _1 - \alpha _2 )f''dL \right\} , \end{aligned}$$

(A.4)

$$\begin{aligned} dL_R= & {} \frac{1}{\Delta }\left\{ \frac{(\alpha _1 - \alpha _2 )^2 }{R\theta }dR + \frac{\alpha _1 (1 - \alpha _2 )(1 - y) + \alpha _2 (1 - \alpha _1 )y}{\theta ^2 y(1 - y)}d\theta \right. \nonumber \\&\left. +\, \frac{\alpha _1 - \alpha _2 }{\theta p}dp + \frac{\alpha _1 (1 - y) + \alpha _2 y}{y(1 - y)}f'dL \right\} , \end{aligned}$$

(A.5)

where

$$\begin{aligned} \Delta\equiv & {} \frac{\alpha _1 (1 - y) + \alpha _2 y}{y(1 - y)}[f' - (L - L_R )f''] - \frac{(\alpha _1 - \alpha _2 )^2 f''}{\theta f'}\\= & {} \frac{\alpha _1 (1 - y) + \alpha _2 y}{y(1 - y)}f' - \frac{\alpha _1 (1 - \alpha _2 )(1 - y) + \alpha _2 (1 - \alpha _1 )y}{y(1 - y)\theta f'}f'' > 0. \end{aligned}$$

It is clear from (A.4) that under Assumption 1, y is increasing in R and $\theta$ but decreasing in L, and that an increase in p unambiguously increases y. It is also verified from (A.5) that $L_R$ is unambiguously increasing in R, $\theta$, and L, and under Assumption 1, an increase in p increases $L_R$. $\square$

1.2 A.2 The linearized dynamic system

Let us denote the right-hand side of (11) and that of (12) by $\varPhi (R,\theta ;L,p)$ and $\varPsi (R,\theta ;L,p)$, respectively. Then, we have

$$\begin{aligned}&\frac{\partial \varPhi }{\partial R} =f'\frac{\partial L^R}{\partial R}-\beta , \end{aligned}$$

(A.6)

$$\begin{aligned}&\frac{\partial \varPhi }{\partial \theta } =f'\frac{\partial L^R}{\partial \theta }>0, \end{aligned}$$

(A.7)

$$\begin{aligned}&\frac{\partial \varPsi }{\partial R} =\frac{1}{R}\left\{ \frac{\alpha _1 y+\alpha _2 (1-y)}{R}-(\alpha _1-\alpha _2)\frac{\partial y}{\partial R}\right\} >0, \end{aligned}$$

(A.8)

$$\begin{aligned}&\frac{\partial \varPsi }{\partial \theta } =\rho +\beta -\frac{1}{R}(\alpha _1-\alpha _2)\frac{\partial y}{\partial \theta }, \end{aligned}$$

(A.9)

where the sign of $\partial \varPsi /\partial R$ comes from

$$\begin{aligned}&\frac{\alpha _1 y+\alpha _2 (1-y)}{R}-(\alpha _1-\alpha _2)\frac{\partial y}{\partial R} \\&\quad =\frac{1}{{\Delta R}}\{ [\alpha _1 y + \alpha _2 (1 - y)]\Delta - (\alpha _1 - \alpha _2 )^2 [f' - (L - L_R )f'']\} \\&\quad> \frac{{f' - (L - L_R )f''}}{{y(1 - y)\Delta R}}\{ [\alpha _1 y + \alpha _2 (1 - y)][\alpha _1 (1 - y) + \alpha _2 y] - (\alpha _1 - \alpha _2 )^2 y(1 - y)\} \\&\quad = \frac{{\alpha _1 \alpha _2 [f' - (L - L_R )f'']}}{{y(1 - y)\Delta R}}>0. \end{aligned}$$

Although the signs of $\partial \varPhi /\partial R$ and $\partial \varPsi /\partial \theta$ are generally ambiguous, we can determine the signs when they are evaluated at the steady state. Eq. (A.6 A.7 A.8) is rewritten as

$$\begin{aligned} \left. \frac{\partial \varPhi }{\partial R}\right| _{\dot{R}=0} =\frac{f(\bar{L}_R)}{\bar{R}} \left\{ \epsilon _f(\bar{L}_R) \frac{\partial L^R}{\partial R}\frac{\bar{R}}{\bar{L}_R} -1 \right\} , \end{aligned}$$

(A.10)

where $\epsilon _f(L_R) \equiv f'(L_R) L_R/f(L_R)<1$ $\forall L_R>0$ because $f(L_R)$ is assumed to be concave. Moreover, from (7) and (A.5), we have

$$\begin{aligned} \frac{\partial L^R }{\partial R}\frac{R}{L_R } - 1= & {} \frac{(\alpha _1 - \alpha _2 )^2 - \Delta \theta L_R }{\Delta \theta L_R } \\= & {} \frac{ (\alpha _1 - \alpha _2 )^2 - \theta L_R \left\{ \frac{\alpha _1 (1 - y) + \alpha _2 y}{y(1 - y)}[f' - (L - L_R )f''] - \frac{(\alpha _1 - \alpha _2 )^2 f''}{\theta f'} \right\} }{\Delta \theta L_R } \\< & {} \frac{(\alpha _1 - \alpha _2 )^2 - L_R \left( \frac{\alpha _1 }{y} + \frac{\alpha _2 }{1 - y} \right) \theta f'}{\Delta \theta L_R } \\= & {} \frac{ (\alpha _1 - \alpha _2 )^2 - \alpha _1 (1 - \alpha _1 )\frac{L - L_1 - L_2 }{L_1 } - \alpha _2 (1 - \alpha _2 )\frac{L - L_1 - L_2 }{L_2 }}{\Delta \theta L_R } \\= & {} \frac{f' \left\{ [\alpha _1 L_2 - \alpha _2 (L - L_1 )]y + [\alpha _2 L_1 - \alpha _1 (L - L_2 )](1 - y) \right\} }{\Delta L_R y(1 - y)} \\= & {} \frac{Lf'}{\Delta L_R y(1 - y)}\left\{ - \alpha _1 (1 - y - \lambda _2 ) + \alpha _2 (\lambda _1 - y) \right\} , \end{aligned}$$

where $\lambda _i \equiv L_i /L$, $i = 1,2$. Notice that $y > \lambda _1$ and $1 - y > \lambda _2$ must hold,¹⁷ and thus, $(\partial L^R /\partial R)(R/L_R )$ must be smaller than 1. Therefore, we have $\partial \varPhi /\partial R < 0$ evaluated at the steady state. Finally, in light of (9) and (A.4), (A.9) is rewritten as

$$\begin{aligned} \left. \frac{\partial \varPsi }{\partial \theta }\right| _{\dot{\theta }=0}= & {} \frac{1}{\bar{R}}\left\{ \frac{\alpha _1 \bar{y} + \alpha _2 (1 - \bar{y})}{\bar{\theta }}- (\alpha _1 - \alpha _2 )\frac{\partial y}{\partial \theta } \right\} \nonumber \\= & {} \frac{[\alpha _1 \bar{y} + \alpha _2 (1 - \bar{y})]\Delta - (\alpha _1 - \alpha _2 )^2 f'}{\Delta \bar{R}\bar{\theta }} \nonumber \\> & {} \frac{f'\{ [\alpha _1 \bar{y} + \alpha _2 (1 - \bar{y})][\alpha _1 \bar{y} + \alpha _2 (1 - \bar{y})] - (\alpha _1 - \alpha _2 )^2 \bar{y}(1 - \bar{y})\} }{\bar{y}(1 - \bar{y})\Delta \bar{R}\bar{\theta }} \nonumber \\= & {} \frac{\alpha _1 \alpha _2 f'}{\bar{y}(1 - \bar{y})\Delta \bar{R}\bar{\theta }} > 0. \end{aligned}$$

(A.11)

1.3 A.3 Comparative statics for the steady-state equilibrium

Totally differentiating the steady-state conditions, we have

$$\begin{aligned} \begin{bmatrix} a_{11}&a_{12} \\ a_{21}&a_{22} \end{bmatrix} \begin{bmatrix} d\bar{R} \\ d\bar{\theta } \end{bmatrix} =- \begin{bmatrix} \varPhi _L \\ \varPsi _L \end{bmatrix} dL- \begin{bmatrix} \varPhi _p \\ \varPsi _p \end{bmatrix} dp, \end{aligned}$$

(A.12)

where $a_{ij}$’s are defined in (18) and

$$\begin{aligned} \varPhi _L\equiv & {} f'\frac{\partial L^R}{\partial L} =\frac{(f')^2 [\alpha _1 (1 - \bar{y}) + \alpha _2 \bar{y}]}{\Delta \bar{y}(1 - \bar{y})}> 0, \\ \varPsi _L\equiv & {} -\frac{\alpha _1-\alpha _2}{R}\frac{\partial y}{\partial L} = - \frac{(\alpha _1 - \alpha _2 )^2 f''}{\bar{R}\Delta }>0,\\ \varPhi _p\equiv & {} f'\frac{\partial L^R}{\partial p} =\frac{(\alpha _1 - \alpha _2 )f'}{\bar{\theta }p\Delta },\\ \varPsi _p\equiv & {} -\frac{\alpha _1-\alpha _2}{R}\frac{\partial y}{\partial p} = - \frac{(\alpha _1 - \alpha _2 )[f' - (L - \bar{L}_R )f'']}{\bar{R}p\Delta }. \end{aligned}$$

Solving (A.12), we have

$$\begin{aligned} d\bar{R}= & {} -\frac{\left( a_{22}f'\frac{\partial L^R}{\partial L} +a_{12}\frac{\alpha _1-\alpha _2}{R}\frac{\partial y}{\partial L} \right) dL +\left( a_{22}f'\frac{\partial L^R}{\partial p} +a_{12}\frac{\alpha _1-\alpha _2}{R}\frac{\partial y}{\partial p} \right) dp}{a_{11}a_{22}-a_{12}a_{21}},\nonumber \\ \end{aligned}$$

(A.13)

$$\begin{aligned} d\bar{\theta }= & {} \frac{\left( a_{21}f'\frac{\partial L^R}{\partial L} +a_{11}\frac{\alpha _1-\alpha _2}{R}\frac{\partial y}{\partial L} \right) dL +\left( a_{21}f'\frac{\partial L^R}{\partial p} +a_{11}\frac{\alpha _1-\alpha _2}{R}\frac{\partial y}{\partial p} \right) dp}{a_{11}a_{22}-a_{12}a_{21}}.\qquad \quad \end{aligned}$$

(A.14)

Substituting the values of $a_{ij}$’s and the derivatives into (A.13), we have the comparative static results for $\bar{R}$, i.e., (19) and (20). The comparative statics for $\bar{\theta }$ can be analogously derived from (A.14).

1.4 A.4 Derivation of (28)

Totally differentiating (27), we have

$$\begin{aligned} \left( \frac{\partial y}{\partial R}\frac{\partial \bar{R}_a}{\partial L} +\frac{\partial y}{\partial \theta }\frac{\partial \bar{\theta }_a}{\partial L} +\frac{\partial y}{\partial L} \right) dL +\frac{\partial y}{\partial p}dp=0. \end{aligned}$$

(A.15)

In light of (A.4), (23), and (26), the coefficient of dL is rewritten as

$$\begin{aligned} \frac{\partial y}{\partial R}\frac{\partial \bar{R}_a}{\partial L} +\frac{\partial y}{\partial \theta }\frac{\partial \bar{\theta }_a}{\partial L} +\frac{\partial y}{\partial L} = \frac{(\alpha _1-\alpha _2)\gamma (1-\gamma )\bar{\theta }_a f'' \beta \bar{R}_a}{\Delta \left( \beta \bar{R}_a+w\frac{\partial L^{Ra}}{\partial \theta }\right) }. \end{aligned}$$

(A.16)

Substituting (A.16) into (A.15) and making use of (A.4), we have (28).