<p>One needs to be a bit careful when deriving the factor prices due to the CES aggregation. The factor prices are not simply the integrals of the marginal productivities. In particular, unless <span class="math-container">$\gamma(i)$</span> and <span class="math-container">$\eta(i)$</span> are simultaneously constants, the conditions <span class="math-container">$\frac{\partial y(i)}{\partial k(i)}=R$</span> and <span class="math-container">$\frac{\partial y(i)}{\partial l(i)}=W$</span> cannot be satisfied for all <span class="math-container">$i$</span>.</p>
<p>Aggregate output is a CES aggregation <span class="math-container">$$Y=\left(\int_{N-1}^N y(i)^{\frac{\sigma-1}{\sigma}}di\right)^{\frac{\sigma}{\sigma-1}}$$</span> Given prices <span class="math-container">$p(i)$</span>, we can consider the expenditure minimisation problem of the household to consume aggregate output <span class="math-container">$Y$</span> given price schedule <span class="math-container">$p(i)$</span>. This has associated Lagrangian <span class="math-container">$$\mathcal{L}=\int_{N-1}^N p(i)y(i)di-\lambda\left[\left(\int_{N-1}^Ny(i)^{\frac{\sigma-1}{\sigma}}di\right)^{\frac{\sigma}{\sigma-1}}-Y\right]$$</span> A variational argument gives that <span class="math-container">$$y(i)=\lambda^{\sigma}p(i)^{-\sigma}Y=p(i)^{-\sigma}Y \qquad (\dagger)$$</span> where the lagrange multiplier is 1 since we normalised the aggregate price <span class="math-container">$P=\left[\int_{N-1}^Np(i)^{1-\sigma}di\right]^{\frac{1}{1-\sigma}}=1$</span>.</p>
<p>Then given that tasks are produced competitively, <span class="math-container">$p(i)$</span> is set equal to the minimum per unit cost of production. That is, <span class="math-container">$$p(i)=\begin{cases} \min\Bigg\{\frac{R}{\eta(i)},\frac{W}{\gamma(i)}\Bigg\} &amp;\text{ if } i\leq I \\ \frac{W}{\gamma(i)} &amp; \text{ if } i&gt;I\end{cases}$$</span></p>
<p>Given assumption (A1) of the paper, it is strictly cheaper to use capital in lower indexed tasks, so <span class="math-container">$$p(i)=\begin{cases} \frac{R}{\eta(i)} &amp;\text{ if } i\leq I \\ \frac{W}{\gamma(i)} &amp; \text{ if } i&gt;I\end{cases}$$</span></p>
<p>For <span class="math-container">$i\leq I$</span>, <span class="math-container">$y(i)=\eta(i)k(i)$</span>. Combining <span class="math-container">$(\dagger)$</span> with the given <span class="math-container">$p(i)$</span> for this case gives <span class="math-container">$$\eta(i)k(i)=Y\left(\frac{R}{\eta(i)}\right)^{-\sigma}\iff R^{\sigma}k(i)=Y\eta(i)^{\sigma-1}$$</span> integrating this over <span class="math-container">$i\in [N-1,I]$</span> we get <span class="math-container">$R^{\sigma}K=Y\left(\int_{N-1}^I\eta(i)^{\sigma-1}di\right)$</span>. Therefore, <span class="math-container">$$R=Y^{\frac{1}{\sigma}}K^{-\frac{1}{\sigma}}\left(\int_{N-1}^I\eta(i)^{\sigma-1}di\right)^{\frac{1}{\sigma}}$$</span> An analogous argument over tasks <span class="math-container">$i&gt;I$</span>, mutatis mutandis, shows <span class="math-container">$$W=Y^{\frac{1}{\sigma}}L^{-\frac{1}{\sigma}}\left(\int_{I}^N\gamma(i)^{\sigma-1}di\right)^{\frac{1}{\sigma}}$$</span></p>
<p>Therefore, <span class="math-container">$$\begin{align*} R^{1-\sigma}&amp;=Y^{\frac{1-\sigma}{\sigma}}K^{\frac{\sigma-1}{\sigma}}\left(\int_{N-1}^I\eta(i)^{\sigma-1}di\right)^{\frac{1-\sigma}{\sigma}} \\
W^{1-\sigma}&amp;=Y^{\frac{1-\sigma}{\sigma}}L^{\frac{\sigma-1}{\sigma}}\left(\int_{I}^N\gamma(i)^{\sigma-1}di\right)^{\frac{1-\sigma}{\sigma}} \end{align*}$$</span></p>
<hr />
<p>Now, calculating <span class="math-container">$d\ln Y/dI$</span>, we get <span class="math-container">$$\begin{align*}\frac{d\ln Y}{dI} &amp;= \frac{d}{dI}\left[\frac{\sigma}{\sigma-1}\ln\left(\left(\int_{N-1}^I\eta(i)^{\sigma-1}di\right)^{\frac{1}{\sigma}}K^{\frac{\sigma-1}{\sigma}}+\left(\int_{I}^N\gamma(i)^{\sigma-1}di\right)^{\frac{1}{\sigma}}L^{\frac{\sigma-1}{\sigma}}\right)\right] \\
&amp;=\frac{\sigma}{\sigma-1}\frac{1}{Y^{\frac{\sigma-1}{\sigma}}}\left[\frac{1}{\sigma}\left(\int_{N-1}^I\eta(i)^{\sigma-1}di\right)^{\frac{1}{\sigma}-1}\eta(I)^{\sigma-1}K^{\frac{\sigma-1}{\sigma}}-\frac{1}{\sigma}\left(\int_{I}^N\gamma(i)^{\sigma-1}di\right)^{\frac{1}{\sigma}-1}\gamma(I)^{\sigma-1}L^{\frac{\sigma-1}{\sigma}}\right] \\
&amp;=\frac{1}{\sigma-1}\left[Y^{\frac{1-\sigma}{\sigma}}K^{\frac{\sigma-1}{\sigma}}\left(\int_{N-1}^I\eta(i)^{1-\sigma}di\right)^{\frac{1-\sigma}{\sigma}}\eta(I)^{\sigma-1}-Y^{\frac{1-\sigma}{\sigma}}L^{\frac{\sigma-1}{\sigma}}\left(\int_{I}^N\gamma(i)^{1-\sigma}di\right)^{\frac{1-\sigma}{\sigma}}\gamma(I)^{\sigma-1}\right] \\
&amp;= \frac{1}{\sigma-1}\left[R^{1-\sigma}\eta(I)^{\sigma-1}-W^{1-\sigma}\gamma(I)^{\sigma-1}\right] \\
&amp;= \frac{1}{1-\sigma}\left[\left(\frac{W}{\gamma(I)}\right)^{1-\sigma}-\left(\frac{R}{\eta(I)}\right)^{1-\sigma}\right]
\end{align*}$$</span></p>

CC BY-SA 4.0
2024-08-13T13:41:56.953Z
Brilliant! I appreciate it!
Maximilian
CC BY-SA 4.0
2024-08-13T15:08:51.220Z
Just a little remark, there is a typo. $R^{\sigma} k(i)=Y \eta(i)^{\sigma -1}$, and similarly, for $W$ the exponent of $\gamma(i)$ is $\sigma -1$
Maximilian
CC BY-SA 4.0
2024-08-13T16:42:36.763Z
@Maximilian Thanks! I've edited it now.
Joseph Basford
CC BY-SA 4.0
2024-08-13T09:01:53.487Z
2024-08-13T16:41:51.350Z
2024-08-13T16:41:51.350Z
Joseph Basford
<p>I'm struggling to derive equation 8 from the paper by Acemoglu and Restrepo (AEA Papers and Proceedings, 2018).</p>
<p>Consider an economy where aggregate output is produced in perfect competition according to the following technology:
<span class="math-container">\begin{equation}
Y = \left( \int_{N-1}^{N} y(i)^{\frac{\sigma -1}{\sigma}} \, di \right)^{\frac{\sigma }{\sigma -1}}
\end{equation}</span>
where <span class="math-container">$y(i)$</span> denotes a certain type of task.</p>
<p>Suppose that tasks <span class="math-container">$i \in [N-1, I]$</span> are produced with capital only, and tasks <span class="math-container">$i \in (I, N]$</span> are produced with labor. Thus,
<span class="math-container">\begin{equation}
y(i) = \eta(i) k(i) \quad\text{with } i \in [N-1, I]
\end{equation}</span>
and
<span class="math-container">\begin{equation}
y(i) = \gamma(i) l(i) \quad\text{with } i \in (I, N]
\end{equation}</span></p>
<p>Can you show me how to get the following derivative?
<span class="math-container">\begin{equation}
\frac{d\,\ln Y}{dI} = \frac{1}{1 - \sigma} \left[ \left( \frac{W}{\gamma(I)} \right)^{1-\sigma} - \left( \frac{R}{\eta(I)} \right)^{1-\sigma} \right]
\end{equation}</span></p>
<p>Here's my attempt, and I hope you can point out any mistakes I made.</p>
<p>First, <span class="math-container">$\frac{\partial y(i)}{\partial k(i)}= R$</span>, for <span class="math-container">$i \in [N-1, I]$</span>. Aggregating, <span class="math-container">$\int_{N-1}^{I} \eta(i)\,di = R$</span>, and <span class="math-container">$\int_{N-1}^{I} k(i)\,di = K$</span>. Similarly, <span class="math-container">$\int_{I}^{N} \gamma(i)\,di = W$</span>, and <span class="math-container">$\int_{I}^{N} l(i)\,di = L$</span>.</p>
<p>Then,
<span class="math-container">\begin{equation}
\frac{d \ln Y}{d I} = \frac{d Y}{d I} \frac{1}{Y}
\end{equation}</span></p>
<p>I set <span class="math-container">$Z(I)= \left( \int_{N-1}^{I} \eta(i)^{\sigma-1} \, di \right)^{\frac{1}{\sigma}} K^{\frac{\sigma-1}{\sigma}} 
+ \left( \int_{I}^{N} \gamma(i)^{\sigma-1} \, di \right)^{\frac{1}{\sigma}} L^{\frac{\sigma-1}{\sigma}}$</span>.</p>
<p>Thus,
<span class="math-container">\begin{equation}
Y = Z(I)^{\frac{\sigma}{\sigma-1}}
\end{equation}</span></p>
<p><span class="math-container">\begin{equation}
\frac{d Y}{d I} = \frac{\sigma}{\sigma -1} Z(I)^{\frac{1}{\sigma -1}} \frac{d Z}{d I}
\end{equation}</span></p>
<p><span class="math-container">\begin{equation}
\frac{d Z}{d I} = \frac{1}{\sigma } \left( \int_{N-1}^{I} \eta(i)^{\sigma-1} \, di \right)^{\frac{1- \sigma}{\sigma}} \eta(I)^{\sigma -1} K^{\frac{\sigma-1}{\sigma}} 
- \frac{1}{\sigma} \left( \int_{I}^{N} \gamma(i)^{\sigma-1} \, di \right)^{\frac{1 - \sigma}{\sigma}} \gamma(I)^{\sigma -1} L^{\frac{\sigma-1}{\sigma}}
\end{equation}</span></p>
<p>I want to exploit the FOCs that give me the input factor prices, but it seems not feasible since (for instance) <span class="math-container">$\int_{N-1}^{I} \eta(i)^{\sigma -1} di \ne R^{\sigma-1}$</span>.</p>
<p>Any suggestions?</p>

<microeconomics><macroeconomics><mathematical-economics>
Comparative Statics in Acemoglu's Article
Maximilian
# Comparative Statics in Acemoglu's Article

(Asked by: username_0 on 2024-08-12)

I'm struggling to derive equation 8 from the paper by Acemoglu and Restrepo (AEA Papers and Proceedings, 2018).

Consider an economy where aggregate output is produced in perfect competition according to the following technology:
\begin{equation}
Y = \left( \int\_{N-1}^{N} y(i)^{\frac{\sigma -1}{\sigma}} \, di \right)^{\frac{\sigma }{\sigma -1}}
\end{equation}
where $y(i)$ denotes a certain type of task.

Suppose that tasks $i \in [N-1, I]$ are produced with capital only, and tasks $i \in (I, N]$ are produced with labor. Thus,
\begin{equation}
y(i) = \eta(i) k(i) \quad\text{with } i \in [N-1, I]
\end{equation}
and
\begin{equation}
y(i) = \gamma(i) l(i) \quad\text{with } i \in (I, N]
\end{equation}

Can you show me how to get the following derivative?
\begin{equation}
\frac{d\,\ln Y}{dI} = \frac{1}{1 - \sigma} \left[ \left( \frac{W}{\gamma(I)} \right)^{1-\sigma} - \left( \frac{R}{\eta(I)} \right)^{1-\sigma} \right]
\end{equation}

Here's my attempt, and I hope you can point out any mistakes I made.

First, $\frac{\partial y(i)}{\partial k(i)}= R$, for $i \in [N-1, I]$. Aggregating, $\int\_{N-1}^{I} \eta(i)\,di = R$, and $\int\_{N-1}^{I} k(i)\,di = K$. Similarly, $\int\_{I}^{N} \gamma(i)\,di = W$, and $\int\_{I}^{N} l(i)\,di = L$.

Then,
\begin{equation}
\frac{d \ln Y}{d I} = \frac{d Y}{d I} \frac{1}{Y}
\end{equation}

I set $Z(I)= \left( \int\_{N-1}^{I} \eta(i)^{\sigma-1} \, di \right)^{\frac{1}{\sigma}} K^{\frac{\sigma-1}{\sigma}}
+ \left( \int\_{I}^{N} \gamma(i)^{\sigma-1} \, di \right)^{\frac{1}{\sigma}} L^{\frac{\sigma-1}{\sigma}}$.

Thus,
\begin{equation}
Y = Z(I)^{\frac{\sigma}{\sigma-1}}
\end{equation}

\begin{equation}
\frac{d Y}{d I} = \frac{\sigma}{\sigma -1} Z(I)^{\frac{1}{\sigma -1}} \frac{d Z}{d I}
\end{equation}

\begin{equation}
\frac{d Z}{d I} = \frac{1}{\sigma } \left( \int\_{N-1}^{I} \eta(i)^{\sigma-1} \, di \right)^{\frac{1- \sigma}{\sigma}} \eta(I)^{\sigma -1} K^{\frac{\sigma-1}{\sigma}}
- \frac{1}{\sigma} \left( \int\_{I}^{N} \gamma(i)^{\sigma-1} \, di \right)^{\frac{1 - \sigma}{\sigma}} \gamma(I)^{\sigma -1} L^{\frac{\sigma-1}{\sigma}}
\end{equation}

I want to exploit the FOCs that give me the input factor prices, but it seems not feasible since (for instance) $\int\_{N-1}^{I} \eta(i)^{\sigma -1} di \ne R^{\sigma-1}$.

Any suggestions?

**Tags:** microeconomics, macroeconomics, mathematical-economics

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## 1 answers

### Answer by username_1 on 2024-08-13. Score: 6 | Accepted answer

One needs to be a bit careful when deriving the factor prices due to the CES aggregation. The factor prices are not simply the integrals of the marginal productivities. In particular, unless $\gamma(i)$ and $\eta(i)$ are simultaneously constants, the conditions $\frac{\partial y(i)}{\partial k(i)}=R$ and $\frac{\partial y(i)}{\partial l(i)}=W$ cannot be satisfied for all $i$.

Aggregate output is a CES aggregation $$Y=\left(\int\_{N-1}^N y(i)^{\frac{\sigma-1}{\sigma}}di\right)^{\frac{\sigma}{\sigma-1}}$$ Given prices $p(i)$, we can consider the expenditure minimisation problem of the household to consume aggregate output $Y$ given price schedule $p(i)$. This has associated Lagrangian $$\mathcal{L}=\int\_{N-1}^N p(i)y(i)di-\lambda\left[\left(\int\_{N-1}^Ny(i)^{\frac{\sigma-1}{\sigma}}di\right)^{\frac{\sigma}{\sigma-1}}-Y\right]$$ A variational argument gives that $$y(i)=\lambda^{\sigma}p(i)^{-\sigma}Y=p(i)^{-\sigma}Y \qquad (\dagger)$$ where the lagrange multiplier is 1 since we normalised the aggregate price $P=\left[\int\_{N-1}^Np(i)^{1-\sigma}di\right]^{\frac{1}{1-\sigma}}=1$.

Then given that tasks are produced competitively, $p(i)$ is set equal to the minimum per unit cost of production. That is, $$p(i)=\begin{cases} \min\Bigg\{\frac{R}{\eta(i)},\frac{W}{\gamma(i)}\Bigg\} &\text{ if } i\leq I \\ \frac{W}{\gamma(i)} & \text{ if } i>I\end{cases}$$

Given assumption (A1) of the paper, it is strictly cheaper to use capital in lower indexed tasks, so $$p(i)=\begin{cases} \frac{R}{\eta(i)} &\text{ if } i\leq I \\ \frac{W}{\gamma(i)} & \text{ if } i>I\end{cases}$$

For $i\leq I$, $y(i)=\eta(i)k(i)$. Combining $(\dagger)$ with the given $p(i)$ for this case gives $$\eta(i)k(i)=Y\left(\frac{R}{\eta(i)}\right)^{-\sigma}\iff R^{\sigma}k(i)=Y\eta(i)^{\sigma-1}$$ integrating this over $i\in [N-1,I]$ we get $R^{\sigma}K=Y\left(\int\_{N-1}^I\eta(i)^{\sigma-1}di\right)$. Therefore, $$R=Y^{\frac{1}{\sigma}}K^{-\frac{1}{\sigma}}\left(\int\_{N-1}^I\eta(i)^{\sigma-1}di\right)^{\frac{1}{\sigma}}$$ An analogous argument over tasks $i>I$, mutatis mutandis, shows $$W=Y^{\frac{1}{\sigma}}L^{-\frac{1}{\sigma}}\left(\int\_{I}^N\gamma(i)^{\sigma-1}di\right)^{\frac{1}{\sigma}}$$

Therefore, $$\begin{align\*} R^{1-\sigma}&=Y^{\frac{1-\sigma}{\sigma}}K^{\frac{\sigma-1}{\sigma}}\left(\int\_{N-1}^I\eta(i)^{\sigma-1}di\right)^{\frac{1-\sigma}{\sigma}} \\
W^{1-\sigma}&=Y^{\frac{1-\sigma}{\sigma}}L^{\frac{\sigma-1}{\sigma}}\left(\int\_{I}^N\gamma(i)^{\sigma-1}di\right)^{\frac{1-\sigma}{\sigma}} \end{align\*}$$

---

Now, calculating $d\ln Y/dI$, we get $$\begin{align\*}\frac{d\ln Y}{dI} &= \frac{d}{dI}\left[\frac{\sigma}{\sigma-1}\ln\left(\left(\int\_{N-1}^I\eta(i)^{\sigma-1}di\right)^{\frac{1}{\sigma}}K^{\frac{\sigma-1}{\sigma}}+\left(\int\_{I}^N\gamma(i)^{\sigma-1}di\right)^{\frac{1}{\sigma}}L^{\frac{\sigma-1}{\sigma}}\right)\right] \\
&=\frac{\sigma}{\sigma-1}\frac{1}{Y^{\frac{\sigma-1}{\sigma}}}\left[\frac{1}{\sigma}\left(\int\_{N-1}^I\eta(i)^{\sigma-1}di\right)^{\frac{1}{\sigma}-1}\eta(I)^{\sigma-1}K^{\frac{\sigma-1}{\sigma}}-\frac{1}{\sigma}\left(\int\_{I}^N\gamma(i)^{\sigma-1}di\right)^{\frac{1}{\sigma}-1}\gamma(I)^{\sigma-1}L^{\frac{\sigma-1}{\sigma}}\right] \\
&=\frac{1}{\sigma-1}\left[Y^{\frac{1-\sigma}{\sigma}}K^{\frac{\sigma-1}{\sigma}}\left(\int\_{N-1}^I\eta(i)^{1-\sigma}di\right)^{\frac{1-\sigma}{\sigma}}\eta(I)^{\sigma-1}-Y^{\frac{1-\sigma}{\sigma}}L^{\frac{\sigma-1}{\sigma}}\left(\int\_{I}^N\gamma(i)^{1-\sigma}di\right)^{\frac{1-\sigma}{\sigma}}\gamma(I)^{\sigma-1}\right] \\
&= \frac{1}{\sigma-1}\left[R^{1-\sigma}\eta(I)^{\sigma-1}-W^{1-\sigma}\gamma(I)^{\sigma-1}\right] \\
&= \frac{1}{1-\sigma}\left[\left(\frac{W}{\gamma(I)}\right)^{1-\sigma}-\left(\frac{R}{\eta(I)}\right)^{1-\sigma}\right]
\end{align\*}$$

> **Comments:**
> **username_0**: Brilliant! I appreciate it!  
> **username_0**: Just a little remark, there is a typo. $R^{\sigma} k(i)=Y \eta(i)^{\sigma -1}$, and similarly, for $W$ the exponent of $\gamma(i)$ is $\sigma -1$  
> **username_1**: @Maximilian Thanks! I've edited it now.  

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