Comparative Statics in Acemoglu's Article

Question

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?

Answer 1

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

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