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I am following and trying to fully understand a famous and interesting work of Bentolila and Saint-paul (2003). 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?.

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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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