<p>Simply multiply and divide one <span class="math-container">$\left(a_{F} \boldsymbol{F}\right)^{\frac{(\sigma-1)}{\sigma}}$</span> in the bracket and then take one outside the bracket. And by the way your FOC is incorrect in that <span class="math-container">$\frac{\sigma}{(\sigma-1)}$</span> should be cancelled out.</p>

CC BY-SA 4.0
2021-03-13T15:21:47.987Z
2021-03-13T15:21:47.987Z
Alalalalaki
<p>I am struggling to derive a first order in this model with Cobb-Douglas production function and CES labour aggregator with two types labour (here male and female, but could be equally low and high skill).</p>
<p><strong>Setup</strong>: Cobb-Douglas production function
<span class="math-container">$$ 
Y = A K^{\alpha}L^{(1-\alpha)} \tag{1}\label{1}
$$</span>
where <span class="math-container">$K$</span> is capital and <span class="math-container">$L$</span> is labor, and <span class="math-container">$A$</span> the TFP. There are two types of labor <span class="math-container">$i \in \{F,M\}$</span> that are aggregated with constant elasticity of substitution
<span class="math-container">$$ 
L = \left[(1-\lambda) (a_M M)^{\frac{\sigma-1}{\sigma}} +\lambda(a_F F)^{\frac{\sigma-1}{\sigma}} \right]^{\frac{\sigma}{\sigma-1}} \tag{2}\label{2}
$$</span>
where <span class="math-container">$\sigma$</span> represents elasticity of substitution between <span class="math-container">$M$</span> and <span class="math-container">$F$</span>, <span class="math-container">$a_M$</span> and <span class="math-container">$a_F$</span> are productivity terms, and <span class="math-container">$\lambda$</span> is the share parameter. Combining these two equations gives
<span class="math-container">$$ 
Y = A K^{\alpha} \left[(1-\lambda) (a_M M)^{\frac{\sigma-1}{\sigma}} +\lambda(a_F F)^{\frac{\sigma-1}{\sigma}} \right]^{\frac{(1-\alpha)\sigma}{\sigma-1}} \tag{3}\label{3}
$$</span></p>
<p><strong>Solution</strong>: Setting marginal products of labour equal to wage should give
<span class="math-container">$$ 
W^{F} = (1-\alpha)\lambda a_F A K^{\alpha} (a_FF)^{-\alpha} \times \left[ (1-\lambda) \left( \frac{A_MM}{a_FF}\right)^{\frac{\sigma-1}{\sigma}}+\lambda \right]^{\frac{(1-\alpha)\sigma}{\sigma-1}-1} \tag{4.1}\label{4.1}
$$</span>
<span class="math-container">$$ 
W^{M} = (1-\alpha)(1-\lambda) a_M A K^{\alpha} (a_MM)^{-\alpha} \times \left[ (1-\lambda) \left( \frac{A_FF}{a_MM}\right)^{\frac{\sigma-1}{\sigma}} \right]^{\frac{(1-\alpha)\sigma}{\sigma-1}-1} \tag{4.2}\label{4.2}
$$</span></p>
<p><strong>Steps</strong>: So, I wanted to get the foc of equation <span class="math-container">$ \eqref{3} $</span> to get the marginal product of each type of labor and doing double chain rule for each powered bracket, but my result seems not to simplify to what it should.
<span class="math-container">$$ 
\frac{\partial Y}{\partial F} = (1-\alpha) \lambda a_F  A K^{\alpha} (a_FF)^{\frac{-1}{\sigma}} \times  \left[(1-\lambda) (a_MM)^{\frac{(\sigma-1)}{\sigma}} + \lambda(a_FF)^{\frac{(\sigma-1)}{\sigma}} \right]^{\frac{(1-\alpha)\sigma}{\sigma-1}-1} 
$$</span></p>
<p><strong>Edit</strong>: Edited the FoC for the mistake. And to close this, I am putting here the simplification as suggested in the answer (hopefully without typos):
<span class="math-container">$$ 
 = (1-\alpha) \lambda a_F  A K^{\alpha} (a_FF)^{\frac{-1}{\sigma}} \times  \left[\left((1-\lambda) (a_MM)^{\frac{(\sigma-1)}{\sigma}} + \lambda(a_FF)^{\frac{(\sigma-1)}{\sigma}}\right) \times \frac{(a_FF)^{\frac{(\sigma-1)}{\sigma}}}{(a_FF)^{\frac{(\sigma-1)}{\sigma}}} \right]^{\frac{(1-\alpha)\sigma}{\sigma-1}-1} 
$$</span></p>
<p><span class="math-container">$$ 
= (1-\alpha) \lambda a_F  A K^{\alpha} (a_FF)^{\frac{-1}{\sigma}} \times  \left[(1-\lambda) \left(\frac{A_MM}{a_FF}\right)^{\frac{\sigma-1}{\sigma}}+ \lambda \right]^{\frac{(1-\alpha)\sigma}{\sigma-1}-1} \times (a_FF)^{\frac{1-\alpha \sigma}{\sigma}}
$$</span></p>
<p><span class="math-container">$$ 
 = (1-\alpha)\lambda a_F A K^{\alpha} (a_FF)^{-\alpha} \times \left[ (1-\lambda) \left( \frac{A_MM}{a_FF}\right)^{\frac{\sigma-1}{\sigma}}+\lambda \right]^{\frac{(1-\alpha)\sigma}{\sigma-1}-1} 
$$</span></p>
<p><strong>Sources</strong>:
Cahuc, P., Carcillo, S., &amp; Zylberberg, A. (2014). Labor economics. MIT press. Chapter 3</p>
<p>Acemoglu, D., Autor, D. H., &amp; Lyle, D. (2004). Women, war, and wages: The effect of female labor supply on the wage structure at midcentury. Journal of political Economy, 112(3), 497-551.</p>

<labor-economics><cobb-douglas>
CES First order Condition with two labour types
Papayapap
# CES First order Condition with two labour types

(Asked by: username_0 on 2021-03-13)

I am struggling to derive a first order in this model with Cobb-Douglas production function and CES labour aggregator with two types labour (here male and female, but could be equally low and high skill).

**Setup**: Cobb-Douglas production function
$$
Y = A K^{\alpha}L^{(1-\alpha)} \tag{1}\label{1}
$$
where $K$ is capital and $L$ is labor, and $A$ the TFP. There are two types of labor $i \in \{F,M\}$ that are aggregated with constant elasticity of substitution
$$
L = \left[(1-\lambda) (a\_M M)^{\frac{\sigma-1}{\sigma}} +\lambda(a\_F F)^{\frac{\sigma-1}{\sigma}} \right]^{\frac{\sigma}{\sigma-1}} \tag{2}\label{2}
$$
where $\sigma$ represents elasticity of substitution between $M$ and $F$, $a\_M$ and $a\_F$ are productivity terms, and $\lambda$ is the share parameter. Combining these two equations gives
$$
Y = A K^{\alpha} \left[(1-\lambda) (a\_M M)^{\frac{\sigma-1}{\sigma}} +\lambda(a\_F F)^{\frac{\sigma-1}{\sigma}} \right]^{\frac{(1-\alpha)\sigma}{\sigma-1}} \tag{3}\label{3}
$$

**Solution**: Setting marginal products of labour equal to wage should give
$$
W^{F} = (1-\alpha)\lambda a\_F A K^{\alpha} (a\_FF)^{-\alpha} \times \left[ (1-\lambda) \left( \frac{A\_MM}{a\_FF}\right)^{\frac{\sigma-1}{\sigma}}+\lambda \right]^{\frac{(1-\alpha)\sigma}{\sigma-1}-1} \tag{4.1}\label{4.1}
$$
$$
W^{M} = (1-\alpha)(1-\lambda) a\_M A K^{\alpha} (a\_MM)^{-\alpha} \times \left[ (1-\lambda) \left( \frac{A\_FF}{a\_MM}\right)^{\frac{\sigma-1}{\sigma}} \right]^{\frac{(1-\alpha)\sigma}{\sigma-1}-1} \tag{4.2}\label{4.2}
$$

**Steps**: So, I wanted to get the foc of equation $ \eqref{3} $ to get the marginal product of each type of labor and doing double chain rule for each powered bracket, but my result seems not to simplify to what it should.
$$
\frac{\partial Y}{\partial F} = (1-\alpha) \lambda a\_F A K^{\alpha} (a\_FF)^{\frac{-1}{\sigma}} \times \left[(1-\lambda) (a\_MM)^{\frac{(\sigma-1)}{\sigma}} + \lambda(a\_FF)^{\frac{(\sigma-1)}{\sigma}} \right]^{\frac{(1-\alpha)\sigma}{\sigma-1}-1}
$$

**Edit**: Edited the FoC for the mistake. And to close this, I am putting here the simplification as suggested in the answer (hopefully without typos):
$$
= (1-\alpha) \lambda a\_F A K^{\alpha} (a\_FF)^{\frac{-1}{\sigma}} \times \left[\left((1-\lambda) (a\_MM)^{\frac{(\sigma-1)}{\sigma}} + \lambda(a\_FF)^{\frac{(\sigma-1)}{\sigma}}\right) \times \frac{(a\_FF)^{\frac{(\sigma-1)}{\sigma}}}{(a\_FF)^{\frac{(\sigma-1)}{\sigma}}} \right]^{\frac{(1-\alpha)\sigma}{\sigma-1}-1}
$$

$$
= (1-\alpha) \lambda a\_F A K^{\alpha} (a\_FF)^{\frac{-1}{\sigma}} \times \left[(1-\lambda) \left(\frac{A\_MM}{a\_FF}\right)^{\frac{\sigma-1}{\sigma}}+ \lambda \right]^{\frac{(1-\alpha)\sigma}{\sigma-1}-1} \times (a\_FF)^{\frac{1-\alpha \sigma}{\sigma}}
$$

$$
= (1-\alpha)\lambda a\_F A K^{\alpha} (a\_FF)^{-\alpha} \times \left[ (1-\lambda) \left( \frac{A\_MM}{a\_FF}\right)^{\frac{\sigma-1}{\sigma}}+\lambda \right]^{\frac{(1-\alpha)\sigma}{\sigma-1}-1}
$$

**Sources**:
Cahuc, P., Carcillo, S., & Zylberberg, A. (2014). Labor economics. MIT press. Chapter 3

Acemoglu, D., Autor, D. H., & Lyle, D. (2004). Women, war, and wages: The effect of female labor supply on the wage structure at midcentury. Journal of political Economy, 112(3), 497-551.

**Tags:** labor-economics, cobb-douglas

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## 1 answers

### Answer by username_1 on 2021-03-13. Score: 5 | Accepted answer

Simply multiply and divide one $\left(a\_{F} \boldsymbol{F}\right)^{\frac{(\sigma-1)}{\sigma}}$ in the bracket and then take one outside the bracket. And by the way your FOC is incorrect in that $\frac{\sigma}{(\sigma-1)}$ should be cancelled out.


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