Urban wage inequality and economic agglomeration

Abstract

This paper explores the impact of interplay between transport costs and commuting costs on urban wage inequality and economic distribution within a new economic geography model. As in former studies, workers tend at the same time to agglomerate in order to limit transport costs of manufactured good and to disperse in order to alleviate the burden of urban costs. In this paper, we pay special attention to wages and spot light on how the urban wage inequality is determined by interplay between urban costs and transport costs. We also solve analytically the break points and the sustain points and disclose their relationships with transport costs and commuting costs in a general equilibrium model.

Appendices

Appendices

1.1 Appendix A: Proof of Proposition 1

Plugging Eqs. (2) and (7) into (5) and (6), we solve wages as follows:

$$\begin{aligned} w_1&=\frac{2 (1-\lambda )(2+\theta \lambda -\theta )[(\sigma -\lambda \sigma +\lambda ) \phi ^2+\lambda (\sigma -1)]+2 \lambda (2-\theta \lambda ) \sigma \phi }{\lambda (1-\lambda )(2+\theta \lambda -\theta )(2-\theta \lambda ) \left( \sigma ^2-1\right) \phi ^2+{\mathscr {A}}_1},\\ w_2&=\frac{2\lambda (2-\theta \lambda )[(1+\lambda \sigma -\lambda ) \phi ^2+(1-\lambda )(\sigma -1)] +2(1-\lambda )(2+\theta \lambda -\theta ) \sigma \phi }{\lambda (1-\lambda )(2+\theta \lambda -\theta )(2-\theta \lambda ) \left( \sigma ^2-1\right) \phi ^2+{\mathscr {A}}_1}, \end{aligned}$$

where

Treating $w_1/w_2$ as a function of $\phi$, we define $f(\phi )\equiv w_1/w_2$ and have

$$\begin{aligned} f(0)=\frac{2-\theta (1-\lambda )}{2-\theta \lambda }>1 \quad \text {and} \quad f(1)=1, \end{aligned}$$

(A1)

where the inequality comes from $\lambda >1/2.$

Differentiating $f(\phi )$ with respect to $\phi$, we have

$$\begin{aligned} \frac{\partial f}{\partial \phi } \biggr |_{\phi =0}&=\left( \frac{\sigma }{\sigma -1}\right) \left\{ \frac{\left[ 4(1-\theta ) +\theta ^2\left( 1-\lambda +\lambda ^2\right) \right] (2 \lambda -1)}{\lambda (1-\lambda )(2-\theta \lambda )^2}\right\} >0, \end{aligned}$$

(A2)

$$\begin{aligned} \frac{\partial f}{\partial \phi } \biggr |_{\phi =1}&= \left( \frac{2 \lambda -1}{\sigma }\right) \left[ \frac{2 \theta \lambda (1-\lambda )}{ 2-\theta \left( 1-2 \lambda +2\lambda ^2\right) }-\sigma \right] <0, \end{aligned}$$

(A3)

where the first inequality comes from $\lambda >1/2$ and the second inequality comes from $\frac{2 \theta \lambda (1-\lambda )}{ 2-\theta \left( 1-2 \lambda +2\lambda ^2\right) }<1<\sigma .$

On the other hand, the numerators of $w_r$ are quadratic functions of $\phi$. Evidently, for any given parameter ${\mathscr {A}}$, equation $f(\phi )={\mathscr {A}}$ has at most two solutions of $\phi$. In other words, in the $\phi -f(\phi )$ panel, the curve $f(\phi )$ crosses any horizontal line $w_1/w_2={\mathscr {A}}$ at most twice. Given inequalities (A2) and (A3), we know $f(\phi )$ involves in a bell-shaped pattern in terms of $\phi$. Furthermore, we have $w_1/w_2>1$ at $\phi =0$ and $w_1/w_2=1$ at $\phi =1$. Therefore, for $\phi \in (0,1)$, we have $w_1>w_2$.

1.2 Appendix B: Proof of Proposition 2

Differentiating $w_1/w_2$ with respect to $\theta$, we have

$$\begin{aligned}&\frac{\partial f}{\partial \theta }=\frac{ \lambda (1-\lambda )(2\lambda -1) \left( 1-\phi ^2\right) {\mathscr {B}}(\phi )}{\{\lambda (2-\theta \lambda )[(1+\lambda \sigma -\lambda ) \phi ^2+(1-\lambda )(\sigma -1)] +(1-\lambda )(2+\theta \lambda -\theta ) \sigma \phi \}^2}, \end{aligned}$$

where

$$\begin{aligned} {\mathscr {B}}(\phi )\equiv -[\lambda (\sigma -1)+1] [\sigma -\lambda (\sigma -1)] \phi ^2+\lambda (1-\lambda )(\sigma -1)^2. \end{aligned}$$

Evidently, the sign of $\partial (w_1/w_2)/\partial \theta$ depends on ${\mathscr {B}}(\phi )$ which is a quadratic function of $\phi$. Easy to find that ${\mathscr {B}}(0)>0$ and ${\mathscr {B}}(1)=-\sigma <0$, for $\phi \in (0,1)$. There exists one and only one

$$\begin{aligned} \phi ^\sharp \equiv \sqrt{\frac{\lambda (1-\lambda )(\sigma -1)^2}{[(\sigma -1)\lambda +1][\sigma -(\sigma -1)\lambda ]}}, \end{aligned}$$

${\mathscr {B}}(\phi )>0$ if $\phi \in (0,\phi ^\sharp )$ and ${\mathscr {B}}(\phi )<0$ if $\phi \in (\phi ^\sharp ,1)$. Therefore, we have $\partial (w_1/w_2)/\partial \theta >0$ if $\phi \in (0,\phi ^\sharp )$ and $\partial (w_1/w_2)/\partial \theta <0$ if $\phi \in (\phi ^\sharp ,1)$.

1.3 Appendix C: Proof of Proposition 3

Differentiating $V_1/V_2$ with respect to $\lambda$, we have

$$\begin{aligned} \frac{\partial (V_1/V_2)}{\partial \lambda }\biggr |_{\lambda =\frac{1}{2}}=\frac{8 {\mathscr {C}}(\phi )}{(4-\theta ) (\sigma -1) (1+\phi ) (\sigma -1+\phi +\sigma \phi )}, \end{aligned}$$

(C1)

where

$$\begin{aligned} {\mathscr {C}}(\phi )\equiv -2 (3\sigma -\theta \sigma -1) \phi ^2+(\theta +4 \sigma -3 \theta \sigma ) \phi +(2-\theta ) (\sigma -1). \end{aligned}$$

(C2)

Evidently, the sign of Eq. (C1) depends on ${\mathscr {C}}(\phi )$ which is a quadratic function of $\phi$. Easy to find that ${\mathscr {C}}(0)=(2-\theta )(\sigma -1)>0, {\mathscr {C}}'(0)>0$ and ${\mathscr {C}}(1)=-2\theta (\sigma -1)<0$, therefore, for $\phi \in (0,1)$, there exists one and only one

$$\begin{aligned} \phi ^b\equiv \frac{\theta +4 \sigma -3 \theta \sigma +\sqrt{8 (2-\theta ) (\sigma -1) (3\sigma -\theta \sigma -1)+(\theta +4 \sigma -3 \theta \sigma )^2}}{4 (3\sigma -\theta \sigma -1)}. \end{aligned}$$

If $\phi >\phi ^b$, we have $\frac{\partial (V_1/V_2)}{\partial \lambda }\biggr |_{\lambda =\frac{1}{2}}<0$; if $0<\phi <\phi ^b$, we have $\frac{\partial (V_1/V_2)}{\partial \lambda }\biggr |_{\lambda =\frac{1}{2}}>0$.

Furthermore, differentiating $\phi ^b$ with respect to $\theta$ yields

$$\begin{aligned} \frac{\partial \phi ^b}{\partial \theta }&=\frac{(\sigma -1)\left[ -(32-19 \theta ) \sigma ^2+4 (5-3 \theta ) \sigma -4+\theta -(5 \sigma -1){\mathscr {C}}_1 \right] }{4 (1-3\sigma +\theta \sigma )^2 {\mathscr {C}}_1}\\&\le \frac{(\sigma -1)\left[ \left( \frac{-28-12\theta +17 \theta ^2}{32-19 \theta }\right) -(5 \sigma -1){\mathscr {C}}_1 \right] }{4 (1-3\sigma +\theta \sigma )^2 {\mathscr {C}}_1}\\&<0, \end{aligned}$$

where ${\mathscr {C}}_1\equiv \sqrt{8 (2-\theta ) (\sigma -1) (3\sigma -\theta \sigma -1)+(\theta +4 \sigma -3 \theta \sigma )^2}$ and the second inequality comes from $\frac{-28-12\theta +17 \theta ^2}{32-19 \theta }<0.$

1.4 Appendix D: Proof of Proposition 4

For $\sigma >1$ and $1>\phi >0$, solving ${\mathscr {C}}(\phi )<0$ yields

$$\begin{aligned} \theta>\theta ^b\equiv \frac{-2 \phi ^2(3\sigma -1)+4 \sigma \phi +2 (\sigma -1)}{-2 \sigma \phi ^2+(3 \sigma -1)\phi +\sigma -1}>0. \end{aligned}$$

Using properties of quadratic function, we have $\theta ^b<1$ if and only if

$$\begin{aligned} \phi >\frac{\sigma +1+\sqrt{17\sigma ^2 -22\sigma +9}}{4( 2\sigma -1)}\in (0,1). \end{aligned}$$

Differentiating $\theta ^b$ with respect to $\phi$ yields

$$\begin{aligned} \frac{\partial \theta ^b}{\partial \phi }=-\frac{2 (\sigma -1) \left[ (5 \sigma -1) \phi ^2+2(\sigma -1) \phi +\sigma -1\right] }{\left[ 2\sigma \phi ^2+(1-3\sigma )\phi +1-\sigma \right] ^2}<0, \end{aligned}$$

where the inequality comes from $(5 \sigma -1) \phi ^2+2(\sigma -1) \phi +\sigma -1>0.$

1.5 Appendix E: Proof of Proposition 5

Agglomeration is a stable equilibrium if and only if $\frac{V_2}{V_1}\biggr |_{\lambda =1}<1$ that implies

$$\begin{aligned} F(\phi )\equiv \frac{(\sigma -1)}{\sigma }\phi ^{\frac{1}{\sigma -1}}+\frac{2}{ (2-\theta ) \sigma }\phi ^{\frac{\sigma }{\sigma -1}}-1<0. \end{aligned}$$

(E1)

Evidently, $F(\phi )$ is an increasing function of $\phi$, and $F(0)=-1<0$, $F(1)=\frac{\theta }{\sigma (2-\theta )}>0$. Therefore, there exists one and only one $\phi ^s\in (0,1)$ which satisfies $F(\phi ^s)=0$. We have $F(\phi )<0$ if and only if $\phi <\phi ^s$. On the other hand, by using properties of implicit function, we have

$$\begin{aligned} \frac{\partial \phi ^s}{\partial \theta }=\frac{-2 (\sigma -1) \phi ^2}{(2-\theta ) [2 (\sigma -1+\sigma \phi )-\theta (\sigma -1)]}<0. \end{aligned}$$

Therefore, $\phi ^s$ decreases in $\theta$.

1.6 Appendix F: Proof of Proposition 6

Solving $\frac{V_2}{V_1}\biggr |_{\lambda =1}<1$, we obtain

$$\begin{aligned} \theta <\theta ^s\equiv 2\left[ \frac{(\sigma -1) \left( 1-\phi ^{\frac{1}{\sigma -1}}\right) +1- \phi ^{\frac{\sigma }{\sigma -1}}}{\sigma -(\sigma -1) \phi ^{\frac{1}{\sigma -1}}}\right] >0. \end{aligned}$$

To make sure $\theta ^s<1$, it must be hold that

$$\begin{aligned} {\mathscr {F}}(\phi )\equiv 2\phi ^{\frac{\sigma }{\sigma -1}}+(\sigma -1)\phi ^{\frac{1}{\sigma -1}}-\sigma >0. \end{aligned}$$

The above inequality holds if and only if $\phi >\underline{\phi }$ with $\underline{\phi }$ determined by ${\mathscr {F}}(\underline{\phi })=0$. Easy to find that ${\mathscr {F}}(\phi )$ is an increasing function of $\phi$, ${\mathscr {F}}(0)=-\sigma <0$ and ${\mathscr {F}}(1)=1>0$. Therefore, $\underline{\phi }$ is uniquely determined.

On the other hand, differentiating $\theta ^s$ with respect to $\phi$ yields

$$\begin{aligned} \frac{\partial \theta ^s}{\partial \phi }=-\frac{\sigma ^2 \phi ^{\frac{1}{1-\sigma }}-(\sigma -1)^2}{(\sigma -1) \left( 1+\sigma \phi ^{\frac{1}{1-\sigma }}-\sigma \right) ^2}<0, \end{aligned}$$

where the inequality comes from $\phi ^{\frac{1}{1-\sigma } }>1$ and $\sigma >1$. Therefore, $\theta ^s$ decreases in $\phi$. If $\phi <\underline{\phi }$, we know the agglomeration is always stable regardless of commuting costs $\theta$.

1.7 Appendix G: Proof of Proposition 7

We have

$$\begin{aligned} \theta ^s-\theta ^b=\frac{2\sigma \phi ^{\frac{\sigma }{\sigma -1}} {\mathscr {G}}(\phi )}{\left[ \sigma -(\sigma -1) \phi ^{\frac{1}{\sigma -1}}\right] \left[ -2 \sigma \phi ^2+(3 \sigma -1)\phi +\sigma -1\right] }, \end{aligned}$$

where ${\mathscr {G}}(\phi )\equiv (\sigma -1) (1+\phi )\phi ^{\frac{-1}{\sigma -1}}- \left( \sigma -1+\phi +\sigma \phi -2 \phi ^2\right) .$ For $\sigma >1$ and $1>\phi >0$, the denominator is positive, and the sign of $\theta ^s-\theta ^b$ depends on ${\mathscr {G}}(\phi )$. We have ${\mathscr {G}}(0)>0$, ${\mathscr {G}}(1)=0$ and ${\mathscr {G}}'(1)=0$. Furthermore, the twice differential of ${\mathscr {G}}(\phi )$ is

$$\begin{aligned} {\mathscr {G}}''(\phi )=4+\frac{\phi ^{\frac{1}{1-\sigma }-2} (\sigma -\sigma \phi +2 \phi )}{\sigma -1}>0. \end{aligned}$$

Therefore, in the interval of $\phi \in (0,1)$, ${\mathscr {G}}(\phi )$ is convex and ${\mathscr {G}}(\phi )>0$. Then, we have $\theta ^s-\theta ^b>0.$

By construction, $\phi ^s$ is the reciprocal of Eq. (E1), whereas $\phi ^b$ is the reciprocal of Eq. (C2). From Popositions 3 and 5, it follows that $\phi ^s$ and $\phi ^b$ are monotonic with respect to $\theta$. Therefore, it must be that $\phi ^s>\phi ^b$; otherwise, it would have $\theta ^s<\theta ^b$ for some $\phi$, thus contradicting $\theta ^s-\theta ^b>0.$