3
$\begingroup$

I'm deriving the Antras and Helpman (2004) paper. The model assumes a nested CES utility function $$ U = x_0 + \frac{1}{\mu} \sum_{j=1}^{J} X_j^\mu $$ where $X_j = \left[ \int x_j(i)^\alpha \,di \right]^{1/\alpha}$.

The paper straightforwardly shows that the inverse demand function for each variety i in sector j is $$ p_j(i) = X^{\mu-\alpha}x_j(i)^{\alpha-1} $$

I know this can be obtained by setting up the Lagrangian function and finding the first-order condition with respect to $x_j(i)$. But why here we can drop the Lagrangian multiplier $\lambda$? What I actually found is $$ p_j(i) = X^{\mu-\alpha}x_j(i)^{\alpha-1} (1/\lambda) $$ I suspect in this case we can normalise $\lambda$ to 1 due to some implicit assumption that is not clearly stated in the paper. But I never heard that we could easily drop or normalise the Lagrangian multiplier in any of my previous economics and maths class.

$\endgroup$
3
  • $\begingroup$ Does price/inverse demand have a specified unit, e.g. is $x_0$ a numeraire good? If not, you could normalize the price unit in such a way that $\lambda$ becomes one. $\endgroup$
    – Giskard
    May 15 at 8:23
  • $\begingroup$ @Giskard Yes, $x_0$ is a numeraire good. Then why do we have $\lambda$ to be one when we have the numeraire good? $\endgroup$ May 15 at 11:23
  • $\begingroup$ I don't know, I wrote a suggestion that works if it was not a numeraire. $\endgroup$
    – Giskard
    May 15 at 13:09

1 Answer 1

2
$\begingroup$

The Lagrangian is given by: $$ L = x_0 + \frac{1}{\mu} \sum_{j = 1}^J X_j^\mu - \lambda(x_0 + \sum_j \int_i p_j(i) x_j(i) di - m). $$ The first order condition (for an interior solution) with respect to $x_0$ gives: $$ 1 - \lambda = 0 $$ So $\lambda = 1$.

$\endgroup$

Your Answer

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

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