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I was reading through Jan Brueckner's "Lectures on Urban Economics" (2011), which is a "rigorous, but non-technical" explanation of various topics in urban economics. It was supposed to be light holiday reading, but my goofy existence wondered about how to do the math for a particular section. The consumer model at the start of chapter 2 is based off of his work in a specific chapter from the Handbook of Regional and Urban Economics. You can find the more detailed article here, but a non-paywalled version here may be more useful to refer to. I will distinguish between referring to the simpler "lectures" and the "Muth-Mills article" that I have linked to.


The very simple urban model has everyone commute to a point in the center of the city and everyone has the same preferences. There is a cost of commuting and there are two goods to buy, dwelling size which is rented out, and some all encompassing consumer good.

Mathematically, the problem is:

$$\max_{\{c, q \}} \quad v(c, q)$$ $$\text{s.t.} \quad c + pq = y - tx$$

$c$ is consumption (a numeraire good), $p$ is the rental price per square foot, $q$ is the square feet of dwelling space, $y$ is the identical income everyone gets from working, $t$ is the constant price of driving per mile, and $x$ is the distance in miles from the city center. And of course $v$ is the utility function.

Since all consumers are identical in this model, equilibrium price is characterized by whether you get the same utility no matter where you choose to live. If you live in the suburbs (further from the city center), you pick a lot of dwelling space but less consumption, and your commute is longer. If you live closer to the city you pick a smaller dwelling space and more consumption, with a smaller commute time.

The indifference curve he draws in the lectures implies convex preferences, so we will keep that in mind for characterizing $v$. His main conclusion in the lectures is that $p$ falls as $x$ increases, and $q$ rises and $x$ rises.


Intuitively, this all makes perfect sense to me, but although I can find $\frac{\partial p}{\partial x}$, I cannot find $\frac{\partial q}{\partial x}$, even following along with the Muth-Mills article, there is a total derivative that is giving me trouble.

I found $\frac{\partial p}{\partial x}$ by simply rearranging the budget constraint so that it read

$$p = \frac{y - tx - c}{q}$$

and the derivative is:

$$\frac{\partial p}{\partial x} = - \frac{t}{q}$$

The same cannot be done for $\frac{\partial q}{\partial x}$, as the sign would not be correct, and I thought about why and figured it probably has to do something with the fact that $q$ is chosen conditionally on $p$. It was here that I then turned to the Muth-Mills article.

In it, the maximization substitutes for $c$ by incorporating the budget constraint.

$$\max_q \quad v(y - tx - pq, q)\tag{1}$$

The first order condition sets the marginal rate of substitution equal to the price ratio:

$$\frac{v_2(y - tx - pq, q)}{v_1(y - tx - pq, q)} = p \tag{2}$$

And in equilibrium you have to reach the same utility no matter where you choose to live in the city. (So rather than fix price, we fix utility in this model.)

$$v(y - tx - pq, q) = u \tag{3}$$

The article then asks us to take the total derivative of $(3)$ with respect to $x$ and arrive at:

$$-v_1 \left( t + \frac{\partial p}{\partial x} q + p \frac{\partial q}{\partial x}\right) + v_2 \frac{\partial q}{\partial x} = 0 \tag{4}$$

And since $(2)$ implies $v_2 = pv_1$, the article then plugs this into $(4)$ to find the comparative static for $p$ given a change in $x$:

$$\frac{\partial p}{\partial x} = - \frac{t}{q}\tag{5}$$

Finally, the article reaches some form of the comparative static I am interested in:

"Note that since utility is constant, the increase in $q$ corresponds exactly to the substitution effect of the housing price decrease. Formally, it follows that

$$\frac{\partial q}{\partial x} = \eta\frac{\partial p}{\partial x} \tag{6}$$

where $\eta < 0$ is the slope of the appropriate income-compensated (constant-utility) demand curve."


So my main question really is just how do you fill in the steps to calculate the total derivative of $(3)$? I think I will feel really quite silly when I see the answer, but I cannot seem to work out where the dependencies are in trying to setup the chain rule here.

The closest thing I have gotten to a coherent attempt was to express the total derivative as:

$$\frac{\partial v(y - tx - pq, q)}{\partial x} = \frac{\partial v}{\partial (y - tx - pq)} \cdot \frac{\partial (y - tx - pq)}{\partial x} + \frac{\partial v}{\partial q} \cdot \frac{\partial q}{\partial x}$$

but I am not sure if this is correct or how to proceed (how to further break up the first set of partials into appropriate parts for the chain rule).

Additionally, I am wondering how to arrive at $(6)$, which I figure is some sort of Slutsky's decomposition, but I have not touched this part of the model yet. I will give a bonus for this harder section because typing up a bunch of matrices does not seem like much fun. Hopefully someone out there over the holidays is around to help out, and thanks in advance.


References:

Brueckner, Jan K. "The structure of urban equilibria: A unified treatment of the muth-mills model." In Handbook of Regional and Urban Economics, vol. 2, pp. 821-845. Elsevier, 1987.

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  • $\begingroup$ Is your question: how to go from (3) to (4)? $\endgroup$ – caverac Dec 29 '18 at 9:16
  • $\begingroup$ So travel is a cost, depending on distance, but not a direct source of (dis)utility (which might be assumed to depend on time)? $\endgroup$ – Adam Bailey Dec 29 '18 at 11:42
  • $\begingroup$ @caverac Yes, that is part of my question. $\endgroup$ – Kitsune Cavalry Dec 29 '18 at 15:35
  • $\begingroup$ @AdamBailey Yes there is no disutility from commuting time from loss of spare time for hobbies or anything like that. The model is kept very simple here. $\endgroup$ – Kitsune Cavalry Dec 29 '18 at 15:35
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$\require{cancel}$ Start with the fact that $v = v(c, q)$ and use the triple product rule

$$ \left(\frac{\partial v}{\partial c}\right)\left(\frac{\partial c}{\partial q}\right)\left(\frac{\partial q}{\partial v}\right) = -1\tag{1} $$

Now use

$$ \color{orange}{c = y - tx - pq} ~~\Rightarrow~~~ \frac{\partial c}{\partial q} = -p \tag{2} $$

to conclude

$$ \frac{\displaystyle{\frac{\partial v}{\partial q}}}{\displaystyle{\frac{\partial v}{\partial c}}} = -\frac{\partial c}{\partial p} = +p \tag{3} $$

Now call

\begin{eqnarray} \color{blue}{v_1} &=& \color{blue}{\frac{\partial v}{\partial c}} \\ \color{red}{v_2} &=& \color{red}{\frac{\partial v}{\partial q}} \tag{4} \end{eqnarray}

And you reach to the conclusion

$$ \bbox[5px,border:2px solid blue] { \frac{v_2(c, q)}{v_1(c, q)} = p } \tag{5} $$

now that this is done, we just need to use the chain rule

\begin{eqnarray} \frac{\partial v}{\partial x} &=& \color{blue}{\frac{\partial v}{\partial c}} \frac{\partial c}{\partial x} + \color{red}{\frac{\partial v}{\partial q}} \frac{\partial q}{\partial x} \\&=& \color{blue}{v_1}\frac{\partial}{\partial x}\left(\color{orange}{y - tx - pq} \right) + \color{red}{v_2} \frac{\partial q}{\partial x} \\ &=& v_1\left(\cancelto{0}{\frac{\partial y}{\partial x}} - t\cancelto{1}{\frac{\partial x}{\partial x}} - p\frac{\partial q}{\partial x} - q \frac{\partial p}{\partial x}\right) + v_2 \frac{\partial q}{\partial x} \\ &=& -v_1 \left(t + p \frac{\partial q}{\partial x} + q \frac{\partial p}{\partial x}\right) + v_2 \frac{\partial q}{\partial x} \end{eqnarray}

Since you fix $u$ then $v = u = {\rm const}$, so that the l.h.s of this last expression is 0 and

$$ \bbox[5px,border:2px solid blue] { -v_1 \left(t + p \frac{\partial q}{\partial x} + q \frac{\partial p}{\partial x}\right) + v_2 \frac{\partial q}{\partial x} = 0 } \tag{6} $$

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  • $\begingroup$ My friend, I appreciate the effort, but you have addressed completely the wrong question. Equations 3 and 4 refer to something very different in my question. $\endgroup$ – Kitsune Cavalry Dec 29 '18 at 16:18
  • $\begingroup$ @KitsuneCavalry I made an edit, still the wrong answer? $\endgroup$ – caverac Dec 29 '18 at 16:20
  • $\begingroup$ Okay, I see what is happening. Could you elaborate on the last step, particularly with regards to $\frac{\partial c}{\partial x}$, as it looks like there is another chain rule happening in the final equation. $\endgroup$ – Kitsune Cavalry Dec 29 '18 at 16:30
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    $\begingroup$ @KitsuneCavalry Sure, how about now? $\endgroup$ – caverac Dec 29 '18 at 16:39
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    $\begingroup$ Thank you! This is quite thorough. :^) You have gotten my upvote already. $\endgroup$ – Kitsune Cavalry Dec 29 '18 at 17:03

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