1
$\begingroup$

You plan to use the following specification for an empirical study: $$e_i = \alpha_i + \sum_{j=1}^{n} \beta_{ij}p_i + \gamma_iy +\delta_i, i=1,...,n$$ where $e_i$ is the consumer's expenditure on good $i$, $p_i$ is the price of good $i$, $y$ is the income and the $\alpha_i, \beta_{ij}, \gamma_i, \delta_i$ are parameters.

You want the specification to be consistent with standard consumer theory and three friends offer the following opinions:

1) “the equation will work for any values of the parameters;”

(2) “you need to impose the $\beta_{ij}=0,\sum_{i=1}^{n}, \delta_{i}=1$";

(3)“you need to impose restrictions such as $\beta_ij = -\gamma_i\alpha_j, \gamma_i \geq 0,\sum_{i=1}^{n}, \gamma{i}=1,\delta_i=0 $

(a) Explain which of the friends gave sensible advice. (b) Why should you really not listen to the advice of the other friends?

Improve this question
$\endgroup$
3
  • $\begingroup$ As on the LHS of the equality there is $e_i$ indexed by $i$, there should be $\sum_{j=1}^{n} \beta_{ij}p_j$ on the RHS (which depends upon $i$ not $j$). $\endgroup$ – Bertrand Oct 17 '19 at 21:18
  • 2
    $\begingroup$ You could also give some elements of response showing us that you have thought about the question. What are the properties of Marshallian demand functions? $\endgroup$ – Bertrand Oct 17 '19 at 21:24
  • $\begingroup$ I haven't studied the Marshallian demand functions (as this is the 1st week of consumer theory and 3rd week of university) but my thought process was that the expenditure for a good will be price times quantity of that good, so you have to assume that $\alpha$ is the quantity. That means that the first part (alpha times price) is the$e_i$, and the rest has to be zero, so the third friend is right $\endgroup$ – Jacopo Salvatore Oct 18 '19 at 9:41
0
$\begingroup$

You could use the fact that $\sum_{i=1}^{n}e_i=y$ (adding-up property) to see which parameter values are compatible with the above specification and this identity.
Before proceeding, you should replace on the RHS $\sum_{i=1}^{n}\beta_{ij}p_i$ by $\sum_{j=1}^{n}\beta_{ij}p_j$.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Okay thank you, I'm starting to understand it a bit better but I'm still not there. I also have the answer from the back of the textbook but I don't really understand it. If you could give me a few more hints I would really appreciate it. $\endgroup$ – Jacopo Salvatore Oct 19 '19 at 11:22

Not the answer you're looking for? Browse other questions tagged or ask your own question.