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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 $\sigma^{\mathrm{agg}}=(1-\chi) \sigma+\chi \varepsilon$?

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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"


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

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


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.


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

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


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

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

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  • $\begingroup$ I think it would be better to add this to your question. Otherwise people might think that your question was already answered. $\endgroup$
    – tdm
    Jun 4 at 12:34
  • $\begingroup$ Thanks for your trying. Yes, we can derive $\frac{d \log \lambda_{i}}{d \log r / w}=\alpha_{i}$ like this, but what confusing me is how Shephard's lemma is related to this result, which might indicate some easier derivation. And I don't think $r K_{i}+w L_{i}$ can be make identical to $P_{i} Y_{i}$. And why do you solve $d \log \theta_{i} / \theta_{j}$? I think the difficulty in deriving $\frac{\mathrm{d} \ln \theta_{i}}{\mathrm{~d} \ln w / r}$ is the aggregate part. Actually $\frac{\mathrm{~d} \ln r K_{i}+w L_{i}}{\mathrm{~d} \ln w / r}$ can be derived easily following the first result. $\endgroup$ Jun 5 at 11:42
  • $\begingroup$ @Alalalalaki yes. I'm douting these parts of my reasoning as well. Maybe you can post the part of deriving the elasticity of single firm's cost to relative factor price, and we can figure it out together more fluently? The reason I solve i and j 's relative elasticity is that I think (maybe wrongly again 😂) that the aggregate part can be derived from it as a weighted summation. And I'll try it again when I'm using my computer. $\endgroup$
    – Brandon Li
    Jun 5 at 14:18

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