Derive and Decompose The Aggregate Elasticity of Substitution in CES Economy

Question

The paper is Oberfield & Raval 2021.

Consumers have standard Dixit-Stiglitz preferences consuming the bundle

$$Y= \left(\sum_{i \in I} D_{i}^{\frac{1}{\varepsilon}} Y_{i}^{\frac{\varepsilon-1}{\varepsilon}}\right)^{\frac{\varepsilon}{\varepsilon-1}}$$

There are monopolistic plants with a common CES between capital and labor inputs, i.e.

$$Y_{i}=\left[\left(A_{i} K_{i}\right)^{\frac{\sigma-1}{\sigma}}+\left(B_{i} L_{i}\right)^{\frac{\sigma-1}{\sigma}}\right]^{\frac{\sigma}{\sigma-1}}$$.

Let $\alpha_{i} \equiv \frac{r K_{i}}{r K_{i}+w L_{i}}$ and $\alpha \equiv \frac{r K}{r K+w L}$ denote the cost shares of capital at plant level and aggregate level, where $K \equiv \sum_{i \in I} K_{i}$ and $L \equiv \sum_{i \in I} L_{i}$.

The plant-level and aggregate elasticity of substitution is

$$\begin{aligned} \sigma-1 &=\frac{\mathrm{d} \ln r K_{i} / w L_{i}}{\mathrm{~d} \ln w / r}=\frac{\mathrm{d} \ln \alpha_{i} /\left(1-\alpha_{i}\right)}{\mathrm{d} \ln w / r}=\frac{1}{\alpha_{i}\left(1-\alpha_{i}\right)} \frac{\mathrm{d} \alpha_{i}}{\mathrm{~d} \ln w / r}, \\ \sigma^{\mathrm{agg}}-1 &=\frac{\mathrm{d} \ln r K / w L}{\mathrm{~d} \ln w / r}=\frac{\mathrm{d} \ln \alpha /(1-\alpha)}{\mathrm{d} \ln w / r}=\frac{1}{\alpha(1-\alpha)} \frac{\mathrm{d} \alpha}{\mathrm{d} \ln w / r} \end{aligned}$$.

The aggregate cost share of capital can be expressed as $$\alpha=\sum_{i \in I} \alpha_{i} \theta_{i}$$ where $\theta_{i} \equiv \frac{r K_{i}+w L_{i}}{r K+w L}$, and by differentiating $$\sigma^{\mathrm{agg}}-1=\frac{1}{\alpha(1-\alpha)} \sum_{i \in I} \alpha_{i}\left(1-\alpha_{i}\right)(\sigma-1) \theta_{i}+\frac{1}{\alpha(1-\alpha)} \sum_{i \in I} \alpha_{i} \theta_{i} \frac{\mathrm{d} \ln \theta_{i}}{\mathrm{~d} \ln w / r}$$.

For the second term in RHS, the authors say "By Shephard's lemma, a plant's cost share of capital $\alpha_{i}$ measures how relative factor prices affect its mariginal cost" and show that $$\frac{\mathrm{d} \ln \theta_{i}}{\mathrm{~d} \ln w / r}=(\varepsilon-1)\left(\alpha_{i}-\alpha\right)$$.

Then the authors show that "the industry elasticity of substitution is a convex combination of the micro ES and elasticity of demand": $$\sigma^{\mathrm{agg}}=(1-\chi) \sigma+\chi \varepsilon$$ where $\chi \equiv \sum_{i \in I} \frac{\left(\alpha_{i}-\alpha\right)^{2}}{\alpha(1-\alpha)} \theta_{i}$.

Question 1: How to illustrate the claim "By Shephard's lemma, a plant's cost share of capital $\alpha_{i}$ measures how relative factor prices affect its mariginal cost" and to derive the $\frac{\mathrm{d} \ln \theta_{i}}{\mathrm{~d} \ln w / r}$?

Question 2: How to derive the $\sigma^{\mathrm{agg}}=(1-\chi) \sigma+\chi \varepsilon$?

Answer 1

I'm the one who asked the original question in another thread, but unfortunately, my narration is very unclear and hence the original one has been closed but I'm glad that there is a new question which more clear than mine.

Here is what I've tried until now.

$$\min rK_{i}+wL_{i}$$ $$\text{subject to}Y_{i}=[(A_{i}K_{i})^{\frac{\sigma-1}{\sigma}}+(B_{i}L_{i})^{\frac{\sigma-1}{\sigma}}]^{\frac{\sigma}{\sigma-1}}$$

$$w=\lambda_{i}[(A_{i}K_{i})^{\frac{\sigma-1}{\sigma}}+(B_{i}L_{i})^{\frac{\sigma-1}{\sigma}}]^{\frac{1}{\sigma-1}}(B_{i}L_{i})^{\frac{-1}{\sigma}}B_{i}$$

$$r=\lambda_{i}[(A_{i}K_{i})^{\frac{\sigma-1}{\sigma}}+(B_{i}L_{i})^{\frac{\sigma-1}{\sigma}}]^{\frac{1}{\sigma-1}}(A_{i}K_{i})^{\frac{-1}{\sigma}}A_{i}$$

get the marginal cost $\lambda_{i}$

$$\lambda_{i}=[(r/A_{i})^{1-\sigma}+(w/B_{i})^{1-\sigma}]^{\frac{1}{1-\sigma}}$$

then

$$\frac{d\log{\lambda_{i}}}{d\log{r/w}}=\frac{(r/A_{i})^{1-\sigma}}{(r/A_{i})^{1-\sigma}+(w/B_{i})^{1-\sigma}}$$

inserting the formulae for $w$ and $r$

$$\frac{d\log{\lambda_{i}}}{d\log{r/w}}=\frac{rK_{i}}{rK_{i}+wL_{i}}=\alpha_{i}$$

Which are the words "a plant's cost share of capital $\alpha_{i}$ measures how relative factor prices affect its marginal cost"

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But I still haven't work it through, if $rK_{i}+wL_{i}$ can be make identical to $P_{i}Y_{i}$，

and if I can get reasonably

$$\frac{d\log{P_{i}/P_{j}}}{d\log{\lambda_{i}/\lambda_{j}}}=1$$

from

$$P_{i}=\frac{-\epsilon}{1-\epsilon}\lambda_{i}$$

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then by inserting consumer's demand function to get

$$\frac{d\log{P_{i}Y_{i}/P_{j}Y_{j}}}{d\log{P_{i}/P_{j}}}=1-\epsilon$$

and then obtains

$$\frac{d\log{\theta_{i}/\theta_{j}}}{d\log{w/r}}$$ $$=\frac{d\log{P_{i}Y_{i}/P_{j}Y_{j}}}{d\log{w/r}}$$ $$=\frac{d\log{P_{i}Y_{i}/P_{j}Y_{j}}}{d\log{P_{i}/P_{j}}}$$ $$\times\frac{d\log{P_{i}/P_{j}}}{d\log{\lambda_{i}/\lambda_{j}}}$$ $$\times -\frac{d\log{\lambda_{i}/\lambda_{j}}}{d\log{r/w}}$$ $$=(\epsilon-1)(\alpha_{i}-\alpha_{j})$$

but I'm afraid my way of thinking is not right.

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update

$$\begin{aligned} \frac{d\log{\theta/\theta_{j}}}{d\log{w/r}}=&\sum_{i}\frac{d\log{\theta_{i}/\theta_{j}}}{d\log{w/r}}\frac{\theta_{i}/\theta_{j}}{(w/r)}\frac{\theta_{j}/\theta}{r/w}\\ =&\sum_{i}\frac{d\log{\theta_{i}/\theta_{j}}}{d\log{w/r}}\frac{\theta_{i}}{\theta}\\ =&\sum_{i}(\epsilon-1)(\alpha_{i}-\alpha_{j})\frac{\theta_{i}}{\theta}\\ =&(\epsilon-1)(\alpha-\alpha_{j})\\ \end{aligned}$$

i.e.

$$\frac{d\log{\theta_{j}/\theta}}{d\log{w/r}}=(\epsilon-1)(\alpha_{j}-\alpha)$$

but $\theta=\sum_{i}\theta_{i}=1$

hence

$$\frac{d\log{\theta_{j}}}{d\log{w/r}}=(\epsilon-1)(\alpha_{j}-\alpha)$$

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since

$$ \begin{aligned} \sigma^{agg}=1&+\frac{1}{\alpha(1-\alpha)}\sum_{i}\alpha_{i}(1-\alpha_{i})(\sigma-1)\theta_{i}\\ &+\frac{1}{\alpha(1-\alpha)}\sum_{i}\alpha_{i}\theta_{i}\frac{d\log{\theta_{i}}}{d(w/r)}\\ =1&+\frac{1}{\alpha(1-\alpha)}\sum_{i}\alpha_{i}(1-\alpha_{i})\sigma\theta_{i}\\ &-\frac{1}{\alpha(1-\alpha)}\sum_{i}\alpha_{i}(1-\alpha_{i})\theta_{i}\\ &+\frac{1}{\alpha(1-\alpha)}\sum_{i}\alpha_{i}\theta_{i}\epsilon(\alpha_{i}-\alpha)\\ &-\frac{1}{\alpha(1-\alpha)}\sum_{i}\alpha_{i}\theta_{i}(\alpha_{i}-\alpha)\\ =&\{1-[\frac{1}{\alpha(1-\alpha)}\sum_{i}\alpha_{i}\theta_{i}(\alpha_{j}-\alpha)+\frac{1}{\alpha(1-\alpha)}\sum_{i}\alpha_{i}\theta_{i}(1-\alpha_{i})]\}\\ &+[\frac{1}{\alpha(1-\alpha)}\sum_{i}\alpha_{i}\theta_{i}(\alpha_{i}-\alpha)]\epsilon\\ &+[\frac{1}{\alpha(1-\alpha)}\sum_{i}\alpha_{i}\theta_{i}(1-\alpha_{i})]\sigma\\ \end{aligned}$$

Define

$$\begin{aligned} \chi:= &\frac{1}{\alpha(1-\alpha)}\sum_{i}\theta_{i}(\alpha_{i}-\alpha)^{2}\\ =&\frac{1}{\alpha(1-\alpha)}\sum_{i}\theta_{i}(\alpha_{i}^{2}+\alpha^{2}-2\alpha\alpha_{i})\\ =&\frac{1}{\alpha(1-\alpha)}\sum_{i}\theta_{i}\alpha_{i}^{2}+\alpha^{2}\frac{1}{\alpha(1-\alpha)}-2\alpha^{2}\frac{1}{\alpha(1-\alpha)}\\ =&\frac{1}{\alpha(1-\alpha)}\sum_{i}\theta_{i}\alpha_{i}^{2}-\alpha^{2}\frac{1}{\alpha(1-\alpha)}\\ =&\frac{1}{\alpha(1-\alpha)}\sum_{i}\theta_{i}\alpha_{i}^{2}-\frac{1}{\alpha(1-\alpha)}\sum_{i}\theta_{i}\alpha\alpha_{i}\\ =&\frac{1}{\alpha(1-\alpha)}\sum_{i}\theta_{i}(\alpha_{i}^{2}-\alpha\alpha_{i})\\ =&\frac{1}{\alpha(1-\alpha)}\sum_{i}\alpha_{i}\theta_{i}(\alpha_{i}-\alpha)\\ \end{aligned}$$

then

$$\sigma^{agg}=(1-\chi)\sigma+\chi\epsilon$$