<p>There is a mistake at the end of your working. The capital share is <span class="math-container">$s_K=\frac{RK}{Y}=\frac{A_K F_K K}{Y}$</span>, not <span class="math-container">$\frac{A_K K}{Y}$</span>. Fixing this small mistake, your final expression is (with obvious arguments suppressed), <span class="math-container">$$\frac{d \ln W}{d \ln A_K}=s_K \frac{Y}{F_K F_L} \frac{\partial F_L}{\partial A_K K}= s_K \frac{F F_{KL}}{F_K F_L}$$</span></p>
<p>For the final step, one wants to show that <span class="math-container">$\frac{1}{\varepsilon_{KL}}=\frac{F(A_K K, A_L L) F_{LK}(A_K K, A_LL)}{F_K(A_K K, A_L L) F_L(A_K K, A_L L)}$</span></p>
<p>To this end, since <span class="math-container">$F$</span> is homogeneous of degree 1, its partial derivatives are homogeneous of degree 0. So, <span class="math-container">$$\begin{align*} \ln (F_K/F_L) &amp;=\ln \left(F_K\left(1,\frac{A_L L}{A_K K}\right)\right)-\ln \left(F_L\left(\frac{A_K K}{A_L L},1\right)\right) \\
&amp;=\ln \left(F_K\left(1,\frac{A_L }{A_K}\exp\{-\ln(K/L)\}\right)\right)-\ln \left(F_L\left(\frac{A_K}{A_L}\exp\{\ln(K/L)\},1\right)\right) \end{align*}$$</span>
Whence, <span class="math-container">$$\frac{d \ln (F_K/F_L)}{d \ln (K/L)}=-\frac{1}{F_K}\frac{A_L L}{A_K K}F_{KL}\left(1,\frac{A_L L}{A_K K}\right)-\frac{1}{F_L}\frac{A_K K}{A_L L}F_{KL}\left(\frac{A_K K}{A_L L},1\right)$$</span>
Since <span class="math-container">$F$</span> is homogeneous of degree 1, <span class="math-container">$F_{KL}$</span> is homogenous of degree -1, so our expression simplifies to <span class="math-container">$$\begin{align*}\frac{d \ln (F_K/F_L)}{d \ln (K/L)}&amp;=-\frac{1}{F_K(A_K K,A_L L)}\frac{A_L L}{A_K K}F_{KL}\left(1,\frac{A_L L}{A_K K}\right)-\frac{1}{F_L(A_K K,A_L L)}\frac{A_K K}{A_L L}F_{KL}\left(\frac{A_K K}{A_L L},1\right) \\
&amp;=-\frac{1}{F_K}A_L LF_{KL}\left(A_K K,A_L L\right)-\frac{1}{F_L}A_K KF_{KL}\left(A_K K,A_L L\right) \\
&amp;=- \frac{F_{KL}\left(A_K K,A_L L\right)}{F_K(A_K K,A_L L)F_L(A_K K,A_L L)}\left[A_L L F_L(A_K K,A_L L)+A_KKF_K(A_K K,A_L L)\right] \\
&amp;=-\frac{F_{KL}\left(A_K K,A_L L\right)F(A_K K,A_L L)}{F_K(A_K K,A_L L)F_L(A_K K,A_L L)} \end{align*}$$</span></p>
<p>Where I have used Euler's theorem in the final line.</p>
<p>Hence, <span class="math-container">$$\frac{1}{\varepsilon_{KL}}=-\frac{d \ln (F_K/F_L)}{d\ln (K/L)}=\frac{F(A_K K, A_L L) F_{LK}(A_K K, A_LL)}{F_K(A_K K, A_L L) F_L(A_K K, A_L L)}$$</span> and the result follows.</p>

CC BY-SA 4.0
2024-08-09T09:42:56.853Z
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?
Dimitru
CC BY-SA 4.0
2024-08-09T09:53:55.173Z
@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.
Joseph Basford
CC BY-SA 4.0
2024-08-09T09:58:19.400Z
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
Dimitru
CC BY-SA 4.0
2024-08-09T08:17:36.490Z
2024-08-09T09:47:34.097Z
2024-08-09T09:47:34.097Z
Joseph Basford
<p>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.</p>
<p>The aggregate output is given by</p>
<p><span class="math-container">\begin{equation}
Y = F(A_K K, A_L L)
\end{equation}</span></p>
<p>where <span class="math-container">$K$</span> denotes capital, <span class="math-container">$L$</span> represents labor, and <span class="math-container">$A_K$</span> and <span class="math-container">$A_L$</span> denote capital-augmenting and labor-augmenting technology, respectively. The production function <span class="math-container">$F$</span> is assumed to be continuously differentiable, concave, and to exhibit constant returns to scale.</p>
<p>Let <span class="math-container">$F_K$</span> and <span class="math-container">$F_L$</span> denote the marginal products of capital and labor, respectively.</p>
<p>Assuming a competitive labor market, the wage rate <span class="math-container">$W$</span> must satisfy</p>
<p><span class="math-container">\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}</span></p>
<p>The labor share in national income is denoted by</p>
<p><span class="math-container">\begin{equation}
s_L = \frac{WL}{Y}
\end{equation}</span></p>
<p>and the capital share is <span class="math-container">$s_K = 1 - s_L$</span>.</p>
<p>The article then claims that</p>
<p><span class="math-container">\begin{equation}
\frac{d \ln W}{d \ln A_K} = \frac{s_K}{\epsilon_{KL}}
\end{equation}</span></p>
<p>where <span class="math-container">$\epsilon_{KL} = - \frac{d \ln (K/L)}{d \ln (F_K/F_L)}$</span> is the elasticity of substitution between capital and labor.</p>
<p>I attempted to derive the result <span class="math-container">$\frac{d \ln W}{d \ln A_K} = \frac{s_K}{\epsilon_{KL}}$</span> as follows:</p>
<p>First, note that <span class="math-container">$\ln W = \ln A_L + \ln F_L(A_K K, A_L L)$</span>. Therefore,</p>
<p><span class="math-container">\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}</span></p>
<p>Substituting <span class="math-container">$W = A_L F_L$</span> into the equation, we have</p>
<p><span class="math-container">\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}</span></p>
<p>This simplifies to:</p>
<p><span class="math-container">\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}</span></p>
<p>To make further progress, I attempted to multiply and divide by <span class="math-container">$Y$</span>, resulting in:</p>
<p><span class="math-container">\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}</span></p>
<p>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?</p>

<microeconomics><macroeconomics><mathematical-economics><elasticity><elasticity-of-substitution>
Comparative Statics - Capital Augmenting Technological Change
Dimitru
# Comparative Statics - Capital Augmenting Technological Change

(Asked by: username_0 on 2024-08-08)

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?

**Tags:** microeconomics, macroeconomics, mathematical-economics, elasticity, elasticity-of-substitution


---

## 1 answers

### Answer by username_1 on 2024-08-09. Score: 1 | Accepted answer

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.

> **Comments:**
> **username_0**: 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?  
> **username_1**: @username_0 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.  
> **username_0**: 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  

---

