Home market effect, land rent, and welfare

Abstract

This paper develops a general-equilibrium model which features the Home Market Effect and land use for production in the sector of increasing returns to scale. The land rent in the larger region is higher. Meanwhile, the larger region holds a more-than-proportionate share of firms, the so called HME in terms of firm share. These two aspects of spatial inequalities are shown to be equivalent. Moreover, the industrial distribution in the larger region and the land rent differential form bell-shaped patterns in economic integration. We demonstrate that the welfare in the larger region is higher and both regions may benefit from trade liberalization.

Appendices

Appendices

1.1 Appendix 1: Proof of Proposition 1

By dividing Eq. (6) with Eq. (7) and plugging Eqs. (1), (3), (5) and (11) into it, we derive an equation which can be defined as

$$\begin{aligned} \mathcal{F}_1(s_1, s_2, k, \phi )\equiv \mathcal{C}_2 \phi ^2+\mathcal{C}_1 \phi +\mathcal{C}_0=0, , \end{aligned}$$

(15)

where

$$\begin{aligned} \mathcal{C}_2&\equiv (s_1s_2)^{\beta (1-\sigma )-\gamma } \left[ (1-k) (1-\lambda )s_2^{\gamma } (1+r+s_2) -k \lambda s_1^{\gamma } (1+r+s_1) \right] ,\\ \mathcal{C}_1&\equiv (s_1s_2)^{-\gamma } (1+r+s_2+s_1 \lambda -s_2 \lambda )\left[ k s_1^{-2 \beta (\sigma -1)} s_2^{\gamma }-(1-k) s_2^{-2 \beta (\sigma -1)}s_1^{\gamma } \right] ,\\ \mathcal{C}_0&\equiv (s_1s_2)^{\beta (1-\sigma )-\gamma } \left[ (1-k) \lambda (1+r+s_1) s_2^{\gamma } -k(1-\lambda )s_1^{\gamma } (1+r+s_2)\right] , \end{aligned}$$

which is a quadratic function in $\phi$. Remember that r is a linear function of $s_1$ and $s_2$, solved in Eq. (12). Meanwhile, we can solve N from Eqs. (8) and (9):

$$\begin{aligned} N=\frac{S}{[(\sigma -1)\beta +\gamma ]}+\left( \frac{\lambda }{r_1^{1-\gamma }s_1^{\gamma -1}}+\frac{1-\lambda }{r_2^{1-\gamma }s_2^{\gamma -1}} \right) . \end{aligned}$$

Plugging the expression of N together with Eqs. (11) and (12) into Eq. (10) generates another equation defined as

$$\begin{aligned} \mathcal{F}_2(s_1, s_2, k)\equiv s_1^{\gamma }k+s_2^{\gamma }(1-k)-\frac{ r[ (\sigma -1)\beta + \gamma ]}{(1-\gamma )\left[ (1-\lambda )s_2^{1-\gamma }+\lambda s_1^{1-\gamma }\right] }=0. \end{aligned}$$

On ther other hand, by dividing Eq. (8) by Eq. (9), we define

$$\begin{aligned} \mathcal{F}_3(s_1, s_2, k)\equiv \left( \frac{s_1}{s_2}\right) ^{1-\gamma }-\left( \frac{k}{1-k}\right) \Big /\left( \frac{\lambda }{1-\lambda }\right) =0. \end{aligned}$$

Three endogenous variables, $s_1$, $s_2$ and k, are implicitly determined by three equations $\mathcal{F}_1(s_1, s_2, k, \phi )=0$, $\mathcal{F}_2(s_1, s_2, k)=0$ and $\mathcal{F}_3(s_1, s_2, k)=0$. At $\phi =0$ and $\phi =1$, we solve

$$\begin{aligned} s\equiv s_1=s_2=\frac{(1-\alpha ) [\gamma +\beta (\sigma -1)]}{(\sigma -1)(1-\beta )+\alpha [1+\beta (\sigma -1)]}, \quad k=\lambda . \end{aligned}$$

With these solutions, we derive

$$\begin{aligned} \frac{\partial k}{\partial \phi }\biggr |_{\phi =0}=\frac{(2 \lambda -1)(1-\gamma ) \sigma }{\sigma -(1-\alpha ) \gamma -(1-\alpha ) \beta (\sigma -1)}>0 , \end{aligned}$$

(16)

and

$$\begin{aligned} \frac{\partial k}{\partial \phi }\biggr |_{\phi =1}=-\frac{(1-\gamma ) (1-\lambda ) \lambda (2 \lambda -1)}{\gamma +\beta (\sigma -1)}<0, \end{aligned}$$

(17)

where the inequalities come from $\sigma >1$, $1>\gamma >0$, $1>\alpha >0$ and $1>\lambda >1/2.$

For given k, variables $s_1$ and $s_2$ are simultaneously determined by $\mathcal{F}_2(s_1, s_2, k)=0$ and $\mathcal{F}_3(s_1, s_2, k)=0$. Note that the parameter $\phi$ is independent of functions $\mathcal{F}_2(s_1, s_2, k)$ and $\mathcal{F}_3(s_1, s_2, k)$. For simplicity, we do not consider the possibility of multiple solutions. Therefore, we can rewrite equation $\mathcal{F}_1(s_1, s_2, k, \phi )=0$ as $\mathcal{F}_1(s_1(k), s_2(k), k, \phi )=0,$ which is a quadratic function in terms of $\phi$. For any $k^\sharp$, equation $\mathcal{F}_1(s_1(k^\sharp ), s_2(k^\sharp ), k^\sharp ,\phi )=0$ has at most two solutions of $\phi$. In the $\phi$-k plane, the curve $k(\phi )$ crosses any horizontal line $k=k^\sharp$ at most twice. Together with inequalities (16), (17) and $k=\lambda$ at $\phi =0$ and $\phi =1$, we conclude that $k>\lambda$ and $k(\phi )$ evolves in a bell-shaped pattern in terms of $\phi \in (0,1)$. $\square$

1.2 Appendix 2: Proof of Proposition 2

Differentiating $s_1/s_2$ w.r.t. $\phi$ generates

$$\begin{aligned} \frac{\partial (s1/s2)}{\partial \phi }=\frac{\partial (s1/s2)}{\partial k}\frac{\partial k}{\partial \phi }. \end{aligned}$$

On the other hand, by Eq. (13), we have

$$\begin{aligned} \frac{\partial (s1/s2)}{\partial k}=\frac{(s_1/s_2)^{\gamma } (1-\lambda )}{(1-k)^2 (1-\gamma ) \lambda }>0, \end{aligned}$$

where the inequality comes from $\lambda <1$ and $\gamma <1$. It implies that the sign of $\partial (s_1/s_2)/\partial \phi$ is identical to that of $\partial k/\partial \phi$. Because k evolves in a bell-shaped pattern in terms of $\phi$, as shown by Appendix 1. Meanwhile, at $\phi =0$ and $\phi =1$, Appendix 1 indicates $s_1/s_2=1$. Therefore, we have $s_1/s_2>1$ and $s_1/s_2$ evolves in a bell-shaped pattern in $\phi \in (0,1)$.

On the other hand, by equations $\mathcal{F}_1(s_1, s_2, k, \phi )=0$, $\mathcal{F}_2(s_1, s_2, k)=0$ and $\mathcal{F}_3(s_1, s_2, k)=0$, at $\phi =0$, we solve

$$\begin{aligned} \frac{\partial s_1}{\partial \phi }\biggr |_{\phi =0}&=\frac{\sigma (2 \lambda -1) s}{\lambda [\sigma -(1-\alpha ) \gamma -(1-\alpha ) \beta (\sigma -1)] }>0,\\ \frac{\partial s_2}{\partial \phi }\biggr |_{\phi =0}&=-\frac{\sigma (2 \lambda -1) s }{(1-\lambda ) [\sigma -(1-\alpha ) \gamma -(1-\alpha ) (\sigma -1)\beta ] }<0, \end{aligned}$$

whereas at $\phi =1$, we have

$$\begin{aligned} \frac{\partial s_1}{\partial \phi }\biggr |_{\phi =1}&=-\frac{(1-\alpha ) (1-\lambda ) (2 \lambda -1)}{(\sigma -1)(1-\beta )+\alpha [1+\beta (\sigma -1)]}<0,\\ \frac{\partial s_2}{\partial \phi }\biggr |_{\phi =1}&=\frac{(1-\alpha ) \lambda (2 \lambda -1)}{(\sigma -1)(1-\beta )+\alpha [1+\beta (\sigma -1)]}>0, \end{aligned}$$

where the inequalities above come from $\alpha , \gamma , \beta \in (0,1)$, $\lambda \in (1/2,1)$ and $\sigma >1$.

Moreover, for given $s_1$ (resp. $s_2$), variables $s_2$ (resp. $s_1$) and k are simultaneously determined by $\mathcal{F}_2(s_1, s_2, k)=0$ and $\mathcal{F}_3(s_1, s_2, k)=0$. Note that the parameter $\phi$ is independent of functions $\mathcal{F}_2(s_1, s_2, k)$ and $\mathcal{F}_3(s_1, s_2, k)$. Therefore, we can rewrite equation $\mathcal{F}_1(s_1, s_2, k, \phi )=0$ as $\mathcal{F}_1(s_1, s_2(s_1), k(s_1), \phi )=0$ (resp. $\mathcal{F}_1(s_1(s_2), s_2, k(s_2), \phi )=0$) which is a quadratic function in terms of $\phi$. For any $s_1^\sharp$ (resp. $s_2^\sharp$), equation $\mathcal{F}_1(s_1^\sharp , s_2(s_1^\sharp ), k(s_1^\sharp ),\phi )=0$ (resp. $\mathcal{F}_1(s_1(s_2^\sharp ), s_2^\sharp , k(s_2^\sharp ),\phi )=0$) has at most two solutions of $\phi$. In the $\phi$-$s_1$ (resp. $\phi$-$s_2$) plane, the curve $s_1(\phi )$ (resp. $s_2(\phi )$) crosses any horizontal line $s_1=s_1^\sharp$ (resp. $s_2=s_2^\sharp$) at most twice. Together with inequalities $\frac{\partial s_1}{\partial \phi }|_{\phi =0}>0$, $\frac{\partial s_1}{\partial \phi }|_{\phi =1}<0$, $\frac{\partial s_2}{\partial \phi }|_{\phi =0}<0$ and $\frac{\partial s_2}{\partial \phi }|_{\phi =1}>0$, we conclude that $s_1$ evolves in a bell-shaped pattern, while $s_2$ evolves in a U-shaped pattern in $\phi \in (0,1)$. $\square$

1.3 Appendix 3: Proof of Lemma 1

By Eq. (15), we have $\mathcal{C}_2<0$ and $\mathcal{C}_0<0$ where the first inequality comes from $k>\lambda >1/2$ and $s_1>s_2,$ while the second inequality is generated by using Eq. (13) and

$$\begin{aligned} (1-k) \lambda (1+r+s_1) s_2^{\gamma } -k(1-\lambda )s_1^{\gamma } (1+r+s_2)=-\lambda (1-k)s_2^{\gamma -1}(1+r) (s_1-s_2)<0. \end{aligned}$$

We therefore have

$$\begin{aligned} \mathcal{C}_1\equiv k s_1^{-2 \beta (\sigma -1)} s_2^{\gamma }-(1-k) s_2^{-2 \beta (\sigma -1)}s_1^{\gamma }>0, \quad {\text {and}}\quad k s_1^{-2 \beta (\sigma -1)}>(1-k) s_2^{-2 \beta (\sigma -1)}, \end{aligned}$$

where the second inequality comes from $\mathcal{C}_1>0$ and $s_2^{\gamma }<s_1^{\gamma }$. It further derives

$$\begin{aligned} \frac{k}{1-k}>\left( \frac{s_1}{s_2} \right) ^{2\beta (\sigma -1)}>\left( \frac{s_1}{s_2} \right) ^{\beta (\sigma -1)}, \end{aligned}$$

where the second inequality comes from $s_1/s_2>1$. We therefore have $k s_1^{(1-\sigma )\beta }>(1-k)s_2^{(1-\sigma )\beta }.$$\square$

1.4 Appendix 4: Proof of Proposition 4

Lemma 1 implies that the price index in the larger region is lower. Meanwhile, individual income in the larger region is higher because $r_1>r_2$. Therefore, the welfare in the larger region is always higher. On the other hand, with total differentiate $\omega _i$ w.r.t. $\phi$ at $\phi =0$, we obtain

$$\begin{aligned} \frac{\partial \omega _1}{\partial \phi }\biggr |_{\phi =0}&=\mathcal{J}\lambda ^{\left( \frac{1-\alpha }{\sigma -1}\right) }\left\{ \alpha (1-\lambda ) [\gamma +\beta (\sigma -1)]+\lambda [\sigma -\gamma -\beta (\sigma -1)]\right\} \cdot \frac{\partial s_1}{\partial \phi }\biggr |_{\phi =0}>0,\\ \frac{\partial \omega _2}{\partial \phi }\biggr |_{\phi =0}&=-\mathcal{J}(1-\lambda )^{\left( \frac{1-\alpha }{\sigma -1}\right) }\{ (\lambda +\alpha \lambda -1)[\gamma +\beta (\sigma -1)]+\sigma (1-\lambda )\} \cdot \frac{\partial s_2}{\partial \phi }\biggr |_{\phi =0}>0, \end{aligned}$$

where

$$\begin{aligned} \mathcal{J}\equiv \frac{(1-\alpha ) s^{-\beta (1-\alpha )}}{(\sigma -1)(2 \lambda -1) [\gamma +\beta (\sigma -1)] }\left\{ \frac{S (1-\gamma )^{(\gamma -1)}}{[\gamma +\beta (\sigma -1)]^{\gamma }}\right\} ^{\left( \frac{1-\alpha }{\sigma -1}\right) }>0. \end{aligned}$$

In particular, the second inequality comes from the fact that, if $\lambda +\alpha \lambda -1<0$, we have

$$\begin{aligned} (\lambda +\alpha \lambda -1)(\gamma +\beta (\sigma -1))+\sigma (1-\lambda )&=\left[ (1-\lambda )-\gamma (1-\lambda -\alpha \lambda )\right] \\&\quad +(\sigma -1)\left[ (1-\lambda )-\beta (1-\lambda -\alpha \lambda )\right] \\&>\alpha \lambda +(\sigma -1)\alpha \lambda \\&>0.\end{aligned}$$

On the other hand, at $\phi =1$, we have

$$\begin{aligned} \frac{\partial \omega _1}{\partial \phi }\biggr |_{\phi =1}&=\frac{(1-\alpha ) (1-\lambda ) [2 \sigma (1-\lambda )+2\lambda -1]}{[\gamma +\beta (\sigma -1)] (\sigma -1)s^{\beta (1-\alpha )-1}} \\&\quad \times \left\{ \frac{S (1-\gamma )^{(\gamma -1)}}{[\gamma +\beta (\sigma -1)]^{\gamma }}\right\} ^{\left( \frac{1-\alpha }{\sigma -1}\right) }>0,\\ \frac{\partial \omega _2}{\partial \phi }\biggr |_{\phi =1}&=\frac{(1-\alpha )^2 \lambda [1+2 \lambda (\sigma -1)]}{(\sigma -1) [(\sigma -1)(1-\beta )+ \alpha +\alpha \beta (\sigma -1)]s^{\beta (1-\alpha )}} \\&\quad \times \left\{ \frac{S (1-\gamma )^{(\gamma -1)}}{[\gamma +\beta (\sigma -1)]^{\gamma }}\right\} ^{\left( \frac{1-\alpha }{\sigma -1}\right) }>0. \end{aligned}$$

$\square$