Comparative Statics - Capital Augmenting Technological Change

Question

I am reading the article by Acemoglu and Restrepo (AEA Papers and Proceedings, 2018) and have encountered some difficulties deriving equation (1). I hope you can help me work through it.

The aggregate output is given by

\begin{equation} Y = F(A_K K, A_L L) \end{equation}

where $K$ denotes capital, $L$ represents labor, and $A_K$ and $A_L$ denote capital-augmenting and labor-augmenting technology, respectively. The production function $F$ is assumed to be continuously differentiable, concave, and to exhibit constant returns to scale.

Let $F_K$ and $F_L$ denote the marginal products of capital and labor, respectively.

Assuming a competitive labor market, the wage rate $W$ must satisfy

\begin{equation} \frac{\partial Y}{\partial L} = W \quad \text{, i.e.,} \quad W = A_L F_L(A_K K, A_L L) \end{equation}

The labor share in national income is denoted by

\begin{equation} s_L = \frac{WL}{Y} \end{equation}

and the capital share is $s_K = 1 - s_L$.

The article then claims that

\begin{equation} \frac{d \ln W}{d \ln A_K} = \frac{s_K}{\epsilon_{KL}} \end{equation}

where $\epsilon_{KL} = - \frac{d \ln (K/L)}{d \ln (F_K/F_L)}$ is the elasticity of substitution between capital and labor.

I attempted to derive the result $\frac{d \ln W}{d \ln A_K} = \frac{s_K}{\epsilon_{KL}}$ as follows:

First, note that $\ln W = \ln A_L + \ln F_L(A_K K, A_L L)$. Therefore,

\begin{equation} \frac{d \ln W}{d \ln A_K} = \frac{\frac{d W}{W}}{\frac{d A_K}{A_K}} = \frac{d W}{d A_K} \cdot \frac{A_K}{W} \end{equation}

Substituting $W = A_L F_L$ into the equation, we have

\begin{equation} \frac{d \ln W}{d \ln A_K} = A_L \cdot \frac{\partial F_L}{\partial (A_K K)} \cdot \frac{\partial (A_K K)}{\partial A_K} \cdot \frac{A_K}{A_L F_L} \end{equation}

This simplifies to:

\begin{equation} \frac{d \ln W}{d \ln A_K} = \frac{A_K K}{F_L} \cdot \frac{\partial F_L}{\partial (A_K K)} \end{equation}

To make further progress, I attempted to multiply and divide by $Y$, resulting in:

\begin{equation} \frac{d \ln W}{d \ln A_K} = \frac{s_K Y}{F_L} \cdot \frac{\partial F_L}{\partial (A_K K)} \end{equation}

However, I am struggling to complete the derivation and arrive at the final equation given in the paper. Could someone help me bridge the gap?

Answer 1

There is a mistake at the end of your working. The capital share is $s_K=\frac{RK}{Y}=\frac{A_K F_K K}{Y}$, not $\frac{A_K K}{Y}$. Fixing this small mistake, your final expression is (with obvious arguments suppressed), $$\frac{d \ln W}{d \ln A_K}=s_K \frac{Y}{F_K F_L} \frac{\partial F_L}{\partial A_K K}= s_K \frac{F F_{KL}}{F_K F_L}$$

For the final step, one wants to show that $\frac{1}{\varepsilon_{KL}}=\frac{F(A_K K, A_L L) F_{LK}(A_K K, A_LL)}{F_K(A_K K, A_L L) F_L(A_K K, A_L L)}$

To this end, since $F$ is homogeneous of degree 1, its partial derivatives are homogeneous of degree 0. So, $$\begin{align*} \ln (F_K/F_L) &=\ln \left(F_K\left(1,\frac{A_L L}{A_K K}\right)\right)-\ln \left(F_L\left(\frac{A_K K}{A_L L},1\right)\right) \\ &=\ln \left(F_K\left(1,\frac{A_L }{A_K}\exp\{-\ln(K/L)\}\right)\right)-\ln \left(F_L\left(\frac{A_K}{A_L}\exp\{\ln(K/L)\},1\right)\right) \end{align*}$$ Whence, $$\frac{d \ln (F_K/F_L)}{d \ln (K/L)}=-\frac{1}{F_K}\frac{A_L L}{A_K K}F_{KL}\left(1,\frac{A_L L}{A_K K}\right)-\frac{1}{F_L}\frac{A_K K}{A_L L}F_{KL}\left(\frac{A_K K}{A_L L},1\right)$$ Since $F$ is homogeneous of degree 1, $F_{KL}$ is homogenous of degree -1, so our expression simplifies to $$\begin{align*}\frac{d \ln (F_K/F_L)}{d \ln (K/L)}&=-\frac{1}{F_K(A_K K,A_L L)}\frac{A_L L}{A_K K}F_{KL}\left(1,\frac{A_L L}{A_K K}\right)-\frac{1}{F_L(A_K K,A_L L)}\frac{A_K K}{A_L L}F_{KL}\left(\frac{A_K K}{A_L L},1\right) \\ &=-\frac{1}{F_K}A_L LF_{KL}\left(A_K K,A_L L\right)-\frac{1}{F_L}A_K KF_{KL}\left(A_K K,A_L L\right) \\ &=- \frac{F_{KL}\left(A_K K,A_L L\right)}{F_K(A_K K,A_L L)F_L(A_K K,A_L L)}\left[A_L L F_L(A_K K,A_L L)+A_KKF_K(A_K K,A_L L)\right] \\ &=-\frac{F_{KL}\left(A_K K,A_L L\right)F(A_K K,A_L L)}{F_K(A_K K,A_L L)F_L(A_K K,A_L L)} \end{align*}$$

Where I have used Euler's theorem in the final line.

Hence, $$\frac{1}{\varepsilon_{KL}}=-\frac{d \ln (F_K/F_L)}{d\ln (K/L)}=\frac{F(A_K K, A_L L) F_{LK}(A_K K, A_LL)}{F_K(A_K K, A_L L) F_L(A_K K, A_L L)}$$ and the result follows.