0
$\begingroup$

I am working on a problem 14.2.5 from EMEA by Sydsaeter, Hammond and Strom.

Consider the consumer demand problem: $$ \max_{x,y} U(x,y) = \alpha \ln(x-a) + \beta \ln(y-b) \text{ s.t. } px+qy=m \tag{*} $$ where $\alpha, \beta, a, b, p, q$ and $m$ are positive constants with $\alpha + \beta =1 $, and moreover, with $m>ap+bq$. Show that if $x^*, y^*$ solve problem (*), then expenditure on the two goods is given by the two linear functions: $$ px^*= \alpha m+ pa -\alpha(pa+qb), qy^* = \beta m+ qb - \beta (pa + qb)$$

My attempt:

I set up Lagrangian:

$$ L= \alpha \ln(x-a) + \beta \ln(y-b) - \lambda (px+qy-m) $$

and get FOCs:

$$ \frac{\alpha}{x-1} = p \lambda$$

$$ \frac{\beta}{y-1} = q \lambda$$

$$ px+qy = m$$

I tried to solve the system above and then my plan was to multiply $x^*,y^*$ with $p$ and $q$ but that gives me different consumption function. Book has a hint that I should solve for $px$ and $qy$:

$$px = pa + \frac{\alpha}{\lambda}$$

$$qy = qb + \frac{\beta}{\lambda}$$

Then the book hints I should use budget constraint to eliminate lambda so I plug it there:|

$$ pa + \frac{\alpha}{\lambda} + qb + \frac{\beta}{\lambda} = m $$

$$ \lambda = \frac{\alpha + \beta}{m- pa - qb} $$

When I plug the solution for lambda back to equation for $px$ I just get:

$$ px^* = pa + \frac{\alpha m - \alpha pa - \alpha bq}{\alpha +\beta} $$

And this is again not the consumption expenditure function the textbook says I should get. I don't get why I can't end with the same consumption expenditure as the book. Does anyone know how to solve this?

Improve this question
$\endgroup$
5
  • $\begingroup$ As $\alpha + \beta = 1$, I think these are indeed the same solutions. $\endgroup$
    – tdm
    Commented May 4, 2021 at 18:21
  • $\begingroup$ @tdm I don't get it, can you please explain why? $\endgroup$
    – WilliamT
    Commented May 4, 2021 at 18:21
  • 3
    $\begingroup$ If you put $\alpha + \beta = 1$ in your solution, you get $px = pa + \alpha m - \alpha(pa + bq)$ as in the assignment. There is a typo though, because $1 b$ should be $q b$. $\endgroup$
    – tdm
    Commented May 4, 2021 at 18:25
  • $\begingroup$ @tdm ooooo I see now, I am so dumb... if you want you can post it as an answer I will accept it. Yes, that was just typo it was qb $\endgroup$
    – WilliamT
    Commented May 4, 2021 at 19:00
  • 2
    $\begingroup$ No problem. By the way, the utility function in you example is called a Stone-Geary utility function wiki. It is very similar to a Cobb-Douglas utilities function, with the addition of subsistence levels $a$ and $b$ that provide the minimal levels of both goods that need to be consumed. $\endgroup$
    – tdm
    Commented May 5, 2021 at 5:06

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.