# Comparative Statics - Capital Augmenting Technological Change

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

$$$$Y = F(A_K K, A_L L)$$$$

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

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

The labor share in national income is denoted by

$$$$s_L = \frac{WL}{Y}$$$$

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

The article then claims that

$$$$\frac{d \ln W}{d \ln A_K} = \frac{s_K}{\epsilon_{KL}}$$$$

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,

$$$$\frac{d \ln W}{d \ln A_K} = \frac{\frac{d W}{W}}{\frac{d A_K}{A_K}} = \frac{d W}{d A_K} \cdot \frac{A_K}{W}$$$$

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

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

This simplifies to:

$$$$\frac{d \ln W}{d \ln A_K} = \frac{A_K K}{F_L} \cdot \frac{\partial F_L}{\partial (A_K K)}$$$$

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

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

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?

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.

• Thank you for your answer. Could you please clarify some of the notation? In the first line, you define $F_{KL} \equiv \frac{\partial F_L}{\partial (A_K K)}$, but then when you compute $\frac{d \ln \left(\frac{F_K}{F_L}\right)}{d \left(\frac{K}{L}\right)}$, it seems that $F_{KL} \equiv \frac{\partial F_K}{\partial \left(\frac{K}{L}\right)}$. What am I missing? Commented Aug 9 at 9:42
• @Dimitru Those are cross partial derivatives. By $F_{KL}$ I mean the function obtained by differentiating $F$ with respect to $K$ and then with respect to $L$ (or the other way around). The notations $\frac{\partial F_L}{\partial (A_K K)}$ or $\frac{\partial F_K}{\partial (K/L)}$ are confusing since they don't tell us which argument the derivatives are taken with respect to, nor what the arguments are. That is why I explicitly wrote out the arguments of the functions throughout, since these do matter for the derivation. Commented Aug 9 at 9:53
• I got it. There was I typo when you defined the elasticity of substitution (the log before $K/L$ was missing). Let me make all the computations and I'm sure they will work. After that, I will vote the answer. Thanks Commented Aug 9 at 9:58