How Do Density Ceiling Controls Affect Housing Prices and Urban Boundaries?

Abstract

This article investigates how density ceiling controls in a monocentric city with a stochastic population affect housing prices and the location of the city boundary. We employ a real options model in which each landowner owns one unit parcel of land and chooses the timing and the level of capital intensity for development of the parcel. The landowner incurs up-front development costs that are irreversible and thereafter receives stochastic urban rents. A tighter density ceiling control will extend the urban boundary and decrease housing prices. The urban spatial expansion problem will be intact, but the landowner will delay development when city population is expected to grow less rapidly over time or becomes more volatile.

Appendices

Appendices

Appendix A Proof of Proposition 1

Equation (27) indicates the condition

$$ G\left({x}^{*},{D}_r(t)\right)=\left(\frac{1}{\uprho {\upbeta}_1}-\frac{x^{*}}{\uprho}-\frac{\alpha }{\uprho^2}+\frac{\uptheta {D}_r(t)}{\uprho}\right)\widehat{q}+\upvarepsilon {\widehat{q}}^{\frac{1}{\gamma }}=0, $$

(A1)

while Eq. (32) indicates the condition

$$ H\left({x}^{*},{D}_r(t)\right)=N(t)-\widehat{q}{D}_r(t)=0. $$

(A2)

Totally differentiating Eqs. (A1) and (A2) with respect to $\widehat{q}$ yields

$$ \frac{\partial G}{\partial {x}^{*}}\frac{d{x}^{*}}{d\widehat{q}}+\frac{\partial G}{\partial {D}_r(t)}\frac{d{D}_r(t)}{d\widehat{q}}=\frac{-\partial G}{\partial \widehat{q}}, $$

(A3)

and

$$ \frac{\partial H}{\partial {x}^{*}}\frac{d{x}^{*}}{d\widehat{q}}+\frac{\partial H}{\partial {D}_r(t)}\frac{d{D}_r(t)}{d\widehat{q}}=\frac{-\partial H}{\partial \widehat{q}}, $$

(A4)

where

$$ \frac{\partial G}{\partial {x}^{*}}=-\frac{\widehat{q}}{\uprho}<0, $$

(A5)

$$ \frac{\partial G}{\partial {D}_r(t)}=\frac{\uptheta}{\uprho}\widehat{q}>0, $$

(A6)

$$ \frac{\partial H}{\partial {x}^{*}}=0, $$

(A7)

$$ \frac{\partial H}{\partial {D}_r(t)}=-\widehat{q}<0, $$

(A8)

$$ \frac{\partial G}{\partial \widehat{q}}=\left(\frac{1}{\gamma }-1\right)\upvarepsilon {\widehat{q}}^{\left(\frac{1}{\gamma}\right)-1}>0, $$

(A9)

and

$$ \frac{\partial H}{\partial \widehat{q}}=-{D}_r(t)<0. $$

(A10)

Solving Eqs. (A3) and (A4) simultaneously yields

$$ \begin{array}{l}\frac{d{x}^{*}}{d\widehat{q}}=\frac{\Delta_1}{\Delta}>0,\mathrm{if}\;{\widehat{q}}_0>\widehat{q}>{\left[\frac{\uptheta N(t)\gamma }{\varepsilon \rho \left(1-\gamma \right)}\right]}^{\gamma },\hfill \\ {}=0,\mathrm{if}\;\widehat{q}={\left[\frac{\uptheta N(t)\gamma }{\upvarepsilon \uprho \left(1-\gamma \right)}\right]}^{\gamma },\hfill \\ {}<0,\mathrm{if}\;\widehat{q}<{\left[\frac{\uptheta N(t)\gamma }{\upvarepsilon \uprho \left(1-\gamma \right)}\right]}^{\gamma },\hfill \end{array} $$

(A11)

$$ \frac{d{D}_r(t)}{d\widehat{q}}=\frac{\Delta_2}{\Delta}<0, $$

(A12)

where

$$ \Delta ={\displaystyle \underset{\left(-\right)}{\frac{\partial G}{\partial {x}^{*}}}}{\displaystyle \underset{\left(-\right)}{\frac{\partial H}{\partial {D}_r(t)}}}-{\displaystyle \underset{\left(+\right)}{\frac{\partial G}{\partial {D}_r(t)}}}{\displaystyle \underset{(0)}{\frac{\partial H}{\partial {x}^{*}}}}>0, $$

(A13)

$$ {\Delta}_1=\underset{\left(+\right)}{\frac{\partial G}{\partial {D}_r(t)}}\underset{\left(-\right)}{\frac{\partial H}{\partial \widehat{q}}}-\underset{\left(+\right)}{\frac{\partial G}{\partial \widehat{q}}}\underset{\left(-\right)}{\frac{\partial H}{\partial {D}_r(t)}}=\left[\left(\frac{1}{\gamma }-1\right)\upvarepsilon {\widehat{q}}^{\frac{1}{\gamma }}-\frac{\uptheta N(t)}{\uprho}\right], $$

(A14)

and

$$ {\Delta}_2=\underset{\left(+\right)}{\frac{\partial G}{\partial \widehat{q}}}\underset{(0)}{\frac{\partial H}{\partial {x}^{*}}}-\underset{\left(-\right)}{\frac{\partial G}{\partial {x}^{*}}}\underset{\left(-\right)}{\frac{\partial H}{\partial \widehat{q}}}<0. $$

(A15)

Given that x* = u* − y(t), it follows that the sign of du*/dq, which is the same as that of dx*/dq, is indeterminate.

Q.E.D.

Appendix B Proof of Proposition 2

Note that Eq. (32) indicates that D _r (t) is independent of θ, ε, α _N , and σ _N . Furthermore, recall that

$$ \alpha =\uptheta {\alpha}_N\upeta, $$

(B1)

and

$$ \upsigma =\uptheta {\sigma}_N\upeta . $$

(B2)

Totally differentiating Eqs. (B1) and (B2) with respect to θ, ε, α _N , and σ _N , and then rearranging yields

$$ \frac{\partial \alpha }{\partial \uptheta}=\frac{\Delta_1^{\prime }}{\Delta^{\prime }}>0, $$

(B3)

$$ \frac{\partial \upsigma}{\partial \uptheta}=\frac{\Delta_2^{\prime }}{\Delta^{\prime }}>0, $$

(B4)

$$ \frac{\partial \alpha }{\partial \upvarepsilon}=\frac{\Delta_3^{\prime }}{\Delta^{\prime }}>0, $$

(B5)

$$ \frac{\partial \sigma }{\partial \upvarepsilon}=\frac{\Delta_4^{\prime }}{\Delta^{\prime }}>0, $$

(B6)

$$ \frac{\partial \alpha }{\partial {\alpha}_N}=\frac{\Delta_5^{\prime }}{\Delta^{\prime }}>0, $$

(B7)

$$ \frac{\partial \sigma }{\partial {\alpha}_N}=\frac{\Delta_6^{\prime }}{\Delta^{\prime }}<0, $$

(B8)

$$ \frac{\partial \alpha }{\partial {\upsigma}_N}=\frac{\Delta_7^{\prime }}{\Delta^{\prime }}<0, $$

(B9)

$$ \frac{\partial \upsigma}{\partial {\upsigma}_N}=\frac{\Delta_8^{\prime }}{\Delta^{\prime }}>0, $$

(B10)

$$ {\Delta}^{\prime }=1-\uptheta \underset{\left(+\right)}{\frac{\partial \upeta}{\partial {\upbeta}_1}}\left({\alpha}_N\underset{\left(-\right)}{\frac{\partial {\upbeta}_1}{\partial \alpha }}+{\sigma}_N\underset{\left(-\right)}{\frac{\partial {\upbeta}_1}{\partial \upsigma}}\right)>0, $$

(B11)

$$ {\Delta}_1^{\prime }={\alpha}_N\upeta >0, $$

(B12)

$$ {\Delta}_2^{\prime }={\sigma}_N\upeta >0, $$

(B13)

$$ {\Delta}_3^{\prime }=\uptheta {\alpha}_N{\displaystyle \underset{\left(+\right)}{\frac{\partial \upeta}{\partial \upvarepsilon}}}>0, $$

(B14)

$$ {\Delta}_4^{\prime }=\theta {\sigma}_N{\displaystyle \underset{\left(+\right)}{\frac{\partial \upeta}{\partial \upvarepsilon}}}>0, $$

(B15)

$$ {\Delta}_5^{\prime }=\uptheta \upeta \left(1-\uptheta {\sigma}_N\underset{\left(+\right)}{\frac{\partial \upeta}{\partial {\upbeta}_1}}\underset{\left(-\right)}{\frac{\partial {\upbeta}_1}{\partial \sigma }}\right)>0, $$

(B16)

$$ {\Delta}_6^{\prime }={\uptheta}^2\upeta {\sigma}_N\underset{\left(+\right)}{\frac{\partial \upeta}{\partial {\upbeta}_1}}\underset{\left(-\right)}{\frac{\partial {\upbeta}_1}{\partial \alpha }}<0, $$

(B17)

$$ {\Delta}_7^{\prime }={\uptheta}^2\upeta {\alpha}_N\underset{\left(+\right)}{\frac{\partial \upeta}{\partial {\upbeta}_1}}\underset{\left(-\right)}{\frac{\partial {\upbeta}_1}{\partial \upsigma}}<0, $$

(B18)

$$ {\Delta}_8^{\prime }=\uptheta \upeta \left(1-\uptheta {\alpha}_N\underset{\left(+\right)}{\frac{\partial \upeta}{\partial {\upbeta}_1}}\underset{\left(-\right)}{\frac{\partial {\upbeta}_1}{\partial \alpha }}\right)>0, $$

(B19)

where

$$ \frac{\partial \upeta}{\partial {\upbeta}_1}=\frac{\gamma \eta}{\left(1-\gamma \right){\upbeta}_1}>0, $$

(B20)

$$ \frac{\partial {\upbeta}_1}{\partial \alpha }=\frac{-{\upbeta}_1}{\upsigma^2{\upbeta}_1+\alpha }<0, $$

(B21)

$$ \frac{\partial \upeta}{\partial \upvarepsilon}=\frac{\gamma \eta}{\left(1-\gamma \right)\upvarepsilon}>0, $$

(B22)

and

$$ \frac{\partial {\upbeta}_1}{\partial \upsigma}=\frac{-\sigma {\upbeta}_1^2}{\sigma^2{\upbeta}_1+\alpha }<0. $$

(B23)

We find that β₁β₂ = − 2ρ/σ² and β₁ + β₂ = − 2α/σ² such that $\frac{\alpha }{\rho }-\frac{1}{\upbeta_1}=\frac{1}{\upbeta_2}$. Substituting this relationship into Eq. (30), and then totally differentiating the result with respect to θ, ε α _N , and σ _N yields

$$ \frac{\partial {D}_r}{\partial \uptheta}=\frac{\Delta_3}{\Delta}>0, $$

(B24)

$$ {\Delta}_3=-\underset{\left(+\right)}{\frac{\partial G}{\partial \uptheta}}\underset{\left(-\right)}{\frac{\partial H}{\partial {D}_r}}>0, $$

(B25)

$$ \frac{\partial G}{\partial \uptheta}=\frac{D_r}{\uprho}\widehat{q}-\frac{\widehat{q}}{\rho {\upbeta_1}^2}\left[\underset{\left(-\right)}{\frac{\partial {\upbeta}_1}{\partial \upsigma}}\underset{\left(+\right)}{\frac{\partial \upsigma}{\partial \uptheta}}+\underset{\left(-\right)}{\frac{\partial {\upbeta}_1}{\partial \alpha }}\underset{\left(+\right)}{\frac{\partial \alpha }{\partial \uptheta}}\right]>0, $$

(B26)

$$ \frac{\partial {D}_r(t)}{\partial \varepsilon }=\frac{\Delta_4}{\Delta}>0, $$

(B27)

$$ {\Delta}_4=-\underset{\left(+\right)}{\frac{\partial G}{\partial \upvarepsilon}}\underset{\left(-\right)}{\frac{\partial H}{\partial {D}_r}}>0, $$

(B28)

$$ \frac{\partial G}{\partial \upvarepsilon}={\widehat{q}}^{\frac{1}{\gamma }}-\frac{\widehat{q}}{\uprho {\upbeta_1}^2}\left[\underset{\left(-\right)}{\frac{\partial {\upbeta}_1}{\partial \upsigma}}\underset{\left(+\right)}{\frac{\partial \upsigma}{\partial \upvarepsilon}}+\underset{\left(-\right)}{\frac{\partial {\upbeta}_1}{\partial \alpha }}\underset{\left(+\right)}{\frac{\partial \alpha }{\partial \upvarepsilon}}\right]>0, $$

(B29)

$$ \frac{\partial {D}_r(t)}{\partial {\alpha}_N}=\frac{\Delta_5}{\Delta}<0, $$

(B30)

$$ {\Delta}_5=-\underset{\left(-\right)}{\frac{\partial G}{\partial {\alpha}_N}}\underset{\left(-\right)}{\frac{\partial H}{\partial {D}_r}}<0, $$

(B31)

$$ \frac{\partial G}{\partial {\alpha}_N}=\frac{\widehat{q}}{\rho {\beta_2}^2}\left[{\displaystyle \underset{\left(+\right)}{\frac{\partial {\upbeta}_2}{\partial \upsigma}}}{\displaystyle \underset{\left(-\right)}{\frac{\partial \upsigma}{\partial {\alpha}_N}}}+{\displaystyle \underset{\left(-\right)}{\frac{\partial {\upbeta}_2}{\partial \alpha }}}{\displaystyle \underset{\left(+\right)}{\frac{\partial \alpha }{\partial {\alpha}_N}}}\right]<0, $$

(B32)

$$ \frac{\partial {D}_r(t)}{\partial {\sigma}_N}=\frac{\Delta_6}{\Delta}>0, $$

(B33)

$$ {\Delta}_6=-\underset{\left(+\right)}{\frac{\partial G}{\partial {\upsigma}_N}}\underset{\left(-\right)}{\frac{\partial H}{\partial {D}_r}}>0, $$

(B34)

$$ \frac{\partial G}{\partial {\sigma}_N}=\frac{\widehat{q}}{\rho {\upbeta}_2^2}\left[{\displaystyle \underset{\left(+\right)}{\frac{\partial {\beta}_2}{\partial \upsigma}}}{\displaystyle \underset{\left(+\right)}{\frac{\partial \sigma }{\partial {\upsigma}_N}}}+{\displaystyle \underset{\left(-\right)}{\frac{\partial {\beta}_2}{\partial \alpha }}}{\displaystyle \underset{\left(-\right)}{\frac{\partial \alpha }{\partial {\upsigma}_N}}}\right]>0, $$

(B35)

where

$$ \frac{\partial {\upbeta}_2}{\partial \upsigma}=\frac{-\sigma {\upbeta}_2^2}{\sigma^2{\upbeta}_2+\alpha }>0, $$

(B36)

$$ \frac{\partial {\upbeta}_2}{\partial \alpha }=\frac{-{\upbeta}_2}{\sigma^2{\upbeta}_2+\alpha }<0. $$

(B37)

Q.E.D.

Appendix C Proof of Proposition 3

Differentiating R* in Eq. (29) with respect to $\widehat{q}$ yields

$$ \frac{d{R}^{*}}{d\widehat{q}}=\left(\frac{1}{\gamma }-1\right)\uprho \upvarepsilon {\widehat{q}}^{\frac{1}{\gamma }-2}>0. $$

(C1)

Differentiating V _u (x, D) in Eq. (33) with respect to $\widehat{q}$ yields

$$ \frac{d{V}_u\left(x,D\right)}{d\widehat{q}}={\left(\frac{x}{\uprho}+\frac{\alpha }{\uprho^2}-\frac{\theta D}{\uprho}\right)}^{-\gamma /\left(1-\gamma \right)}>0,\mathrm{for}\;{D}_r>D. $$

(C2)

Differentiating V _a (x, D) in Equation (34) with respect to $\widehat{q}$ yields

$$ \frac{d{V}_a\left(x,D\right)}{d\widehat{q}}=\frac{1}{\uprho {\upbeta}_1}{e}^{\upbeta_1\left(x-{x}^{*}\right)}\left[1-\uprho \upvarepsilon {\upbeta}_1\left(\frac{1}{\gamma }-1\right){\widehat{q}}^{\frac{1}{\gamma }-1}\right]>0, $$

(C3)

given that $\widehat{q}<{\widehat{q}}_0.$

Q.E.D.