# Comparative Statics in Acemoglu's Article

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: $$$$Y = \left( \int_{N-1}^{N} y(i)^{\frac{\sigma -1}{\sigma}} \, di \right)^{\frac{\sigma }{\sigma -1}}$$$$ where $$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, $$$$y(i) = \eta(i) k(i) \quad\text{with } i \in [N-1, I]$$$$ and $$$$y(i) = \gamma(i) l(i) \quad\text{with } i \in (I, N]$$$$

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

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, $$$$\frac{d \ln Y}{d I} = \frac{d Y}{d I} \frac{1}{Y}$$$$

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, $$$$Y = Z(I)^{\frac{\sigma}{\sigma-1}}$$$$

$$$$\frac{d Y}{d I} = \frac{\sigma}{\sigma -1} Z(I)^{\frac{1}{\sigma -1}} \frac{d Z}{d I}$$$$

$$$$\frac{d Z}{d I} = \frac{1}{\sigma } \left( \int_{N-1}^{I} \eta(i)^{\sigma-1} \, di \right)^{\frac{1- \sigma}{\sigma}} \eta(I)^{\sigma -1} K^{\frac{\sigma-1}{\sigma}} - \frac{1}{\sigma} \left( \int_{I}^{N} \gamma(i)^{\sigma-1} \, di \right)^{\frac{1 - \sigma}{\sigma}} \gamma(I)^{\sigma -1} L^{\frac{\sigma-1}{\sigma}}$$$$

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?

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*}

• Brilliant! I appreciate it! Commented Aug 13 at 13:41
• 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 $\sigma -1$ Commented Aug 13 at 15:08
• @Maximilian Thanks! I've edited it now. Commented Aug 13 at 16:42