I am reading 'Cities, Agglomeration and Spatial Equilibrium' by Ed Glaeser.

People live in a monocentric city, where consumers of heterogeneous income $y$ work at the centre of the city. They buy $H(y)$ housing for a price $p(d)$ at distance $d$ from the centre, incurring transport costs $t(y)d$ as they commute to work.

Consumers have a utility function:

$U(C,H)=v(y - t(y)d - p(d)H(y),H(y))$

The budget constraint is:

$C = y - t(y)d - p(d)H(y)$ where C is consumption net of transport and housing.

The tangency condition implies:

$\frac{U_1}{U_2} = p(d)$

where the subscript 1 denotes partial differentiation w.r.t. the first argument etc.

By the spatial equilibrium assumption, $\frac{dU}{dd}=0$: there are no rents to be gained by changing location.

$\frac{dU}{dd} = U_1(-t(y)-p'(d)H(y))=0$

$\therefore p'(d)=\frac{-t(y)}{H(y)}$: the price of housing declines with distance $d$.

If the income elasticity of housing is greater than that of transport (i.e. $H(y)$ increases in $y$ more than $t(y)$ does), the rich (higher $y$) will live in the suburbs. Mathematically, the denominator gets larger than the numerator and the price gradient is less steep. The willingness to pay declines less with distance for the rich.

The book then defines the income associated with each distance $y(d)$ and differentiates again to find the 2nd order condition, $p''(d)>0$.

The book states that this is:


But I believe this is incorrect an should be:


Am I correct?

Also, how can you ignore the fact that $y=y(d)$ when calculating the 1st order condition then introduce it when you differentiate to find the 2nd order condition?

Should you not state $U(y(d)-t(y(d))d-p(d)H(y(d)),H(y(d)))$ and differentiate twice?


You seem to be referring to the section titled Income heterogeneity, in Chapter 2 of the book you mention (pp. 33-40).

$p''(d)=\frac{t(y)H'(y)-H(y)t'(y)}{y'(d)}$ is in this case based on "a simpler version of the model [that] assumes that housing consumption equals $H(y)$ and is a function of income but not price or distance" (p. 35). In other words, when taking the (first- or second order) derivative of $p$ w.r.t $d$, you can treat $H(y)$ as a constant.

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  • $\begingroup$ If you treat $H(y)$ and $t(y)$ as a constant, the first-order condition is $p'(d) = \frac{-t(y)}{H(y)}$ and the second-order condition simply becomes $p''(d) = 0$ $\endgroup$ – StevenRJClarke1985 Mar 9 '18 at 13:41

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