<p>You have that in  equilibrium each factor is paid its marginal product, so 
<span class="math-container">$$\tag1\frac{w_i}{p_i}=B_if'(l_i)$$</span> and <span class="math-container">$$\tag2\frac{r_i}{p_i}=f(l_i)-K_if'(l_i)\frac{B_iL_i}{K_i^2}=f(l_i)-l_if'(l_i)$$</span> so deviding (2) by (1) we have:</p>

<p><span class="math-container">$$\tag3\frac{r_i}{w_i}=\frac{f(l_i)-l_if'(l_i)}{B_if'(l_i)}.$$</span></p>

<p>Take the derivative of <span class="math-container">$r_i/w_i$</span> with respect to <span class="math-container">$l_i$</span>:
<span class="math-container">$$\tag4\frac{d(r_i/w_i)}{d(l_i)}=\frac{(f'(l_i)-(f'(l_i)+l_if''(l_i)))B_if'(l_i)-B_if''(l_i)(f(l_i)-l_if'(l_i))}{(B_if'(l_i))^2}=-\frac{f''(l_i)f(l_i)}{B_i(f'(l_i))^2}$$</span></p>

<p>Note that <span class="math-container">$l_i=\frac{B_i}{(K_i/L_i)}$</span> so 
<span class="math-container">$$\tag5\frac{d(l_i)}{d(K_i/L_i)}=-\frac{B_i}{(K_i/L_i)^2}=\frac{l_i}{(K_i/L_i)}.$$</span> Using the chain rule ans substituting (4) and (5) we have:
<span class="math-container">$$\frac{d(r_i/w_i)}{d(K_i/L_i)}=\frac{d(r_i/w_i)}{d(l_i)}\frac{d(l_i)}{d(K_i/L_i)}=\frac{l_if''(l_i)f(l_i)}{(K_i/L_i)B_i(f'(l_i))^2}$$</span> and by the inverse function theorem we get:
<span class="math-container">$$\tag 6\frac{d(K_i/L_i)}{d(r_i/w_i)}=\frac{(K_i/L_i)B_i(f'(l_i))^2}{l_if''(l_i)f(l_i)}$$</span></p>

<p>Finally by multiplying (3) and (6) and dividing  by  <span class="math-container">$(K_i/L_i)$</span> we conclude that </p>

<p><span class="math-container">$$\sigma_i=\frac{d(K_i/L_i)}{d(r_i/w_i)}\frac{r_i/w_i}{K_i/L_i}=\frac{(K_i/L_i)B_i(f'(l_i))^2}{l_if''(l_i)f(l_i)}\cdot\frac{f(l_i)-l_if'(l_i)}{(K_i/L_i)B_if'(l_i)}=\frac{f'(l_i)(f(l_i)-l_if'(l_i))}{l_if''(l_i)f(l_i)}=\frac{f(l_i)}{l_if''(l_i)}\left[1-\frac{l_if'(l_i)}{f(l_i)}\right]$$</span></p>

<p>as desired. <span class="math-container">$Q.E.D.$</span></p>

CC BY-SA 4.0
2020-05-03T19:12:01.210Z
2020-05-03T19:12:01.210Z
Regio
<p>I am following and trying to fully understand a famous and interesting work of <a href="http://piketty.pse.ens.fr/files/BentolilaSaintPaul03.pdf" rel="nofollow noreferrer">Bentolila and Saint-paul (2003)</a>. They try to explain movements of the factor's share in terms of a relationship between the labor share (<span class="math-container">$LS$</span>) and the capital-output ratio (<span class="math-container">$k$</span>). To summarize as much as possible they start from a general production function with labor-augmenting technological progress and constant returns to scale:</p>

<p><span class="math-container">$$Y_{i} = F(K_{i},B_{i}L_{i})=K_{i}f(l_{i}),$$</span></p>

<p>where <span class="math-container">$l_{i}=\frac{B_{i}L_{i}}{K_{i}}.$</span> And show that under the usual microeconomic assumptions of equilibrium (i.e. labor its paid its marginal product), there exists a unique function <span class="math-container">$g(\cdot)$</span> such that:</p>

<p><span class="math-container">$$S_{Li} = g(k_{i}).$$</span></p>

<p>where <span class="math-container">$S_{Li}=\frac{w_iL_i}{p_iY_i}$</span> the labor share in the industries revenues, with <span class="math-container">$w_i$</span> denoting the wage, <span class="math-container">$p_i$</span> the product's price and <span class="math-container">$k_i=\frac{K_i}{Y_i}$</span> is the capital-output ratio. </p>

<p>Then, at page 6 of the paper, they employ the standard definition of elasticity of substitution to production function above, i.e. <span class="math-container">$\sigma_{i}=\frac{d(K_i/L_i)}{d(r/w)}\cdot \frac{r/w}{K_i/L_i}$</span> to obtain the following result:</p>

<p><span class="math-container">$$\sigma_{i}=\frac{f'(l_{i})}{l_{i}f''(l_{i})}\left [1-\frac{l_{i}f'(l_{i})}{f(l_{i})} \right ]$$</span></p>

<p>I am trying to do all the derivations, but I can't find a way to derive this last formula with the same result. Is there someone who could kindly help me and show me the steps?.</p>

<macroeconomics><mathematical-economics><production-function><elasticity-of-substitution>
Derivation of the elasticity of substitution of a general production function with labor-augmenting technological progress
Alessandro
# Derivation of the elasticity of substitution of a general production function with labor-augmenting technological progress

(Asked by: username_0 on 2020-04-27)

I am following and trying to fully understand a famous and interesting work of [Bentolila and Saint-paul (2003)](http://piketty.pse.ens.fr/files/BentolilaSaintPaul03.pdf). They try to explain movements of the factor's share in terms of a relationship between the labor share ($LS$) and the capital-output ratio ($k$). To summarize as much as possible they start from a general production function with labor-augmenting technological progress and constant returns to scale:

$$Y\_{i} = F(K\_{i},B\_{i}L\_{i})=K\_{i}f(l\_{i}),$$

where $l\_{i}=\frac{B\_{i}L\_{i}}{K\_{i}}.$ And show that under the usual microeconomic assumptions of equilibrium (i.e. labor its paid its marginal product), there exists a unique function $g(\cdot)$ such that:

$$S\_{Li} = g(k\_{i}).$$

where $S\_{Li}=\frac{w\_iL\_i}{p\_iY\_i}$ the labor share in the industries revenues, with $w\_i$ denoting the wage, $p\_i$ the product's price and $k\_i=\frac{K\_i}{Y\_i}$ is the capital-output ratio.

Then, at page 6 of the paper, they employ the standard definition of elasticity of substitution to production function above, i.e. $\sigma\_{i}=\frac{d(K\_i/L\_i)}{d(r/w)}\cdot \frac{r/w}{K\_i/L\_i}$ to obtain the following result:

$$\sigma\_{i}=\frac{f'(l\_{i})}{l\_{i}f''(l\_{i})}\left [1-\frac{l\_{i}f'(l\_{i})}{f(l\_{i})} \right ]$$

I am trying to do all the derivations, but I can't find a way to derive this last formula with the same result. Is there someone who could kindly help me and show me the steps?.

**Tags:** macroeconomics, mathematical-economics, production-function, elasticity-of-substitution


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## 1 answers

### Answer by username_1 on 2020-05-03. Score: 2 | Accepted answer

You have that in equilibrium each factor is paid its marginal product, so
$$\tag1\frac{w\_i}{p\_i}=B\_if'(l\_i)$$ and $$\tag2\frac{r\_i}{p\_i}=f(l\_i)-K\_if'(l\_i)\frac{B\_iL\_i}{K\_i^2}=f(l\_i)-l\_if'(l\_i)$$ so deviding (2) by (1) we have:

$$\tag3\frac{r\_i}{w\_i}=\frac{f(l\_i)-l\_if'(l\_i)}{B\_if'(l\_i)}.$$

Take the derivative of $r\_i/w\_i$ with respect to $l\_i$:
$$\tag4\frac{d(r\_i/w\_i)}{d(l\_i)}=\frac{(f'(l\_i)-(f'(l\_i)+l\_if''(l\_i)))B\_if'(l\_i)-B\_if''(l\_i)(f(l\_i)-l\_if'(l\_i))}{(B\_if'(l\_i))^2}=-\frac{f''(l\_i)f(l\_i)}{B\_i(f'(l\_i))^2}$$

Note that $l\_i=\frac{B\_i}{(K\_i/L\_i)}$ so
$$\tag5\frac{d(l\_i)}{d(K\_i/L\_i)}=-\frac{B\_i}{(K\_i/L\_i)^2}=\frac{l\_i}{(K\_i/L\_i)}.$$ Using the chain rule ans substituting (4) and (5) we have:
$$\frac{d(r\_i/w\_i)}{d(K\_i/L\_i)}=\frac{d(r\_i/w\_i)}{d(l\_i)}\frac{d(l\_i)}{d(K\_i/L\_i)}=\frac{l\_if''(l\_i)f(l\_i)}{(K\_i/L\_i)B\_i(f'(l\_i))^2}$$ and by the inverse function theorem we get:
$$\tag 6\frac{d(K\_i/L\_i)}{d(r\_i/w\_i)}=\frac{(K\_i/L\_i)B\_i(f'(l\_i))^2}{l\_if''(l\_i)f(l\_i)}$$

Finally by multiplying (3) and (6) and dividing by $(K\_i/L\_i)$ we conclude that

$$\sigma\_i=\frac{d(K\_i/L\_i)}{d(r\_i/w\_i)}\frac{r\_i/w\_i}{K\_i/L\_i}=\frac{(K\_i/L\_i)B\_i(f'(l\_i))^2}{l\_if''(l\_i)f(l\_i)}\cdot\frac{f(l\_i)-l\_if'(l\_i)}{(K\_i/L\_i)B\_if'(l\_i)}=\frac{f'(l\_i)(f(l\_i)-l\_if'(l\_i))}{l\_if''(l\_i)f(l\_i)}=\frac{f(l\_i)}{l\_if''(l\_i)}\left[1-\frac{l\_if'(l\_i)}{f(l\_i)}\right]$$

as desired. $Q.E.D.$


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