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I need to get this resulting price and quantity (housing):

price and quantity functions

It's pretty clear that the denominator of the quantity function is just the price function.

From this utility function:

utility

And this constraint:

constraint

Now this is a monocentric city model function, so these methods follow:

enter image description here

or:

enter image description here

However, I do not see the actual process of doing that. I need to make this calculations for other utility functions, but when I follow the steps, I do not get the desired result.

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  • $\begingroup$ Are you sure the "results" at the beginning are correct? They seem to correspond to the utility function $c^\alpha q^{1-\alpha}$, while your utility function is $c^{1-\alpha}q^\alpha$. $\endgroup$
    – Herr K.
    Jan 26 at 20:48
  • $\begingroup$ I'm not sure until I can calculate the steps. I'm using a programme developed by someone else, so I'm trying to trace back the steps. If that is the case, I will change the "result" to match the correct utility function. $\endgroup$ Jan 26 at 21:36
  • $\begingroup$ @Victor 'I'm using a programme developed by someone else' is that program publicly available? $\endgroup$ Jan 27 at 17:42
  • $\begingroup$ The programme is mathematica, but the script is made by a journal article, where you can ask for it. $\endgroup$ Jan 28 at 10:48

1 Answer 1

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Let $K:=y-(f+fa)-(t+ta)x$, and so $c=K-pq$.

From condition $(\mathrm{A1})$, \begin{align} &&\frac{v_q(\cdot)}{v_c(\cdot)}&=p \\ \quad\Rightarrow\quad && \frac{\alpha[K-pq]^{1-\alpha}q^{\alpha-1}}{(1-\alpha)[K-pq]^{-\alpha}q^\alpha}&=p \\ \quad\Rightarrow\quad && pq&=\alpha K \\ \quad\Rightarrow\quad && q&=\frac{\alpha K}{p}. \tag{*} \end{align} From $(\mathrm{A2})$ and $(*)$, \begin{align} &&[K-pq]^{1-\alpha}q^\alpha &= u \\ \quad\overset{(*)}{\Rightarrow}\quad &&[(1-\alpha)K]^{1-\alpha}\biggl(\frac{\alpha K}{p}\biggr)^\alpha &= u\\ \quad\Rightarrow\quad && (1-\alpha)^{1-\alpha}\alpha^\alpha \frac{K}{u}&=p^\alpha. \end{align} Thus, \begin{align} p(\cdot)&=\left[(1-\alpha)^{1-\alpha}\alpha^\alpha \frac{K}{u}\right]^{1/\alpha}\\ q(\cdot)&=\frac{\alpha K}{\left[(1-\alpha)^{1-\alpha}\alpha^\alpha \frac{K}{u}\right]^{1/\alpha}} \end{align}

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  • $\begingroup$ So in A2 you first substitute in the c from the constraint and then you substitute in all q's with the marshallian demand function. And then just solve for p and q respectively. $\endgroup$ Jan 27 at 8:55
  • $\begingroup$ I've been trying to do this myself with the other utility function. c^a * q^(1-a) because it seems the programme was written with that despite their paper writing the other one, but I don't get the desired result either using that. Instead, I get p = -(a-1)K(u(aK)^(-a))^(1/1-a). Wondering what I'm getting wrong here. $\endgroup$ Jan 27 at 19:17
  • $\begingroup$ @VictorNielsen: If utility is $c^\alpha q^{1-\alpha}$, then the only difference in $p(\cdot)$ is the outer-most exponent. You probably made some algebraic mistakes somewhere, and without seeing the steps, it's hard to tell exactly where you got tripped up. $\endgroup$
    – Herr K.
    Jan 27 at 22:16
  • $\begingroup$ if I switch around the q and c, then I get that the q in A1 is (K-aK)/p. But if the equation's only difference is the exponent, how com your p(⋅) and q(⋅) be so different from the equations in the first picture in my post? $\endgroup$ Jan 28 at 10:52
  • $\begingroup$ @VictorNielsen: $(K-\alpha K)/p=(1-\alpha)K/p$, which is similar to my $\alpha K/p$, except that $\alpha$ switches place with $(1-\alpha)$. And I don't think my results are very different from the coded expression in your post: switch $\alpha$ and $(1-\alpha)$ and plug in for $K$, you'll see they are the same. $\endgroup$
    – Herr K.
    Jan 28 at 14:14

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