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

CC BY-SA 4.0
2024-05-16T17:03:37.937Z
2024-05-16T17:03:37.937Z
tdm
<p>I'm deriving the Antras and Helpman (2004) paper. The model assumes a nested CES utility function
<span class="math-container">$$
U = x_0 + \frac{1}{\mu} \sum_{j=1}^{J} X_j^\mu
$$</span>
where <span class="math-container">$X_j = \left[ \int x_j(i)^\alpha \,di \right]^{1/\alpha}$</span>.</p>
<p>The paper straightforwardly shows that the inverse demand function for each variety i in sector j is
<span class="math-container">$$
p_j(i) = X^{\mu-\alpha}x_j(i)^{\alpha-1}
$$</span></p>
<p>I know this can be obtained by setting up the Lagrangian function and finding the first-order condition with respect to <span class="math-container">$x_j(i)$</span>. But why here we can drop the Lagrangian multiplier <span class="math-container">$\lambda$</span>? What I actually found is
<span class="math-container">$$
p_j(i) = X^{\mu-\alpha}x_j(i)^{\alpha-1} (1/\lambda)
$$</span>
I suspect in this case we can normalise <span class="math-container">$\lambda$</span> 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.</p>

CC BY-SA 4.0
2024-05-15T08:23:41.603Z
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.
Giskard
CC BY-SA 4.0
2024-05-15T11:23:09.053Z
@Giskard Yes, $x_0$ is a numeraire good. Then why do we have $\lambda$ to be one when we have the numeraire good?
Hopeless Economist
CC BY-SA 4.0
2024-05-15T13:09:31.370Z
I don't know, I wrote a suggestion that works if it was not a numeraire.
Giskard
<international-trade><demand><ces-function><lagrangian>
Why can the Lagrangian Multiplier be dropped in the inverse demand function?
Hopeless Economist
# Why can the Lagrangian Multiplier be dropped in the inverse demand function?

(Asked by: username_0 on 2024-05-15)

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.

**Tags:** international-trade, demand, ces-function, lagrangian

> **Comments:**
> **username_1**: 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.  
> **username_0**: @username_1 Yes, $x_0$ is a numeraire good. Then why do we have $\lambda$ to be one when we have the numeraire good?  
> **username_1**: I don't know, I wrote a suggestion that works if it was not a numeraire.  

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## 1 answers

### Answer by username_2 on 2024-05-16. Score: 2 | Accepted answer

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 $\lambda = 1$.


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