<p>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.</p>
<p>Here is what I've tried until now.</p>
<p><span class="math-container">$$\min rK_{i}+wL_{i}$$</span>
<span class="math-container">$$\text{subject to}Y_{i}=[(A_{i}K_{i})^{\frac{\sigma-1}{\sigma}}+(B_{i}L_{i})^{\frac{\sigma-1}{\sigma}}]^{\frac{\sigma}{\sigma-1}}$$</span></p>
<p><span class="math-container">$$w=\lambda_{i}[(A_{i}K_{i})^{\frac{\sigma-1}{\sigma}}+(B_{i}L_{i})^{\frac{\sigma-1}{\sigma}}]^{\frac{1}{\sigma-1}}(B_{i}L_{i})^{\frac{-1}{\sigma}}B_{i}$$</span></p>
<p><span class="math-container">$$r=\lambda_{i}[(A_{i}K_{i})^{\frac{\sigma-1}{\sigma}}+(B_{i}L_{i})^{\frac{\sigma-1}{\sigma}}]^{\frac{1}{\sigma-1}}(A_{i}K_{i})^{\frac{-1}{\sigma}}A_{i}$$</span></p>
<p>get the marginal cost <span class="math-container">$\lambda_{i}$</span></p>
<p><span class="math-container">$$\lambda_{i}=[(r/A_{i})^{1-\sigma}+(w/B_{i})^{1-\sigma}]^{\frac{1}{1-\sigma}}$$</span></p>
<p>then</p>
<p><span class="math-container">$$\frac{d\log{\lambda_{i}}}{d\log{r/w}}=\frac{(r/A_{i})^{1-\sigma}}{(r/A_{i})^{1-\sigma}+(w/B_{i})^{1-\sigma}}$$</span></p>
<p>inserting the formulae for <span class="math-container">$w$</span> and <span class="math-container">$r$</span></p>
<p><span class="math-container">$$\frac{d\log{\lambda_{i}}}{d\log{r/w}}=\frac{rK_{i}}{rK_{i}+wL_{i}}=\alpha_{i}$$</span></p>
<p>Which are the words &quot;a plant's cost share of capital <span class="math-container">$\alpha_{i}$</span> measures how relative factor prices affect its marginal cost&quot;</p>
<hr />
<p>But I still haven't work it through, if <span class="math-container">$rK_{i}+wL_{i}$</span> can be make identical to <span class="math-container">$P_{i}Y_{i}$</span>，</p>
<p>and if I can get reasonably</p>
<p><span class="math-container">$$\frac{d\log{P_{i}/P_{j}}}{d\log{\lambda_{i}/\lambda_{j}}}=1$$</span></p>
<p>from</p>
<p><span class="math-container">$$P_{i}=\frac{-\epsilon}{1-\epsilon}\lambda_{i}$$</span></p>
<hr />
<p>then by inserting consumer's demand function to get</p>
<p><span class="math-container">$$\frac{d\log{P_{i}Y_{i}/P_{j}Y_{j}}}{d\log{P_{i}/P_{j}}}=1-\epsilon$$</span></p>
<p>and then obtains</p>
<p><span class="math-container">$$\frac{d\log{\theta_{i}/\theta_{j}}}{d\log{w/r}}$$</span>
<span class="math-container">$$=\frac{d\log{P_{i}Y_{i}/P_{j}Y_{j}}}{d\log{w/r}}$$</span>
<span class="math-container">$$=\frac{d\log{P_{i}Y_{i}/P_{j}Y_{j}}}{d\log{P_{i}/P_{j}}}$$</span>
<span class="math-container">$$\times\frac{d\log{P_{i}/P_{j}}}{d\log{\lambda_{i}/\lambda_{j}}}$$</span>
<span class="math-container">$$\times -\frac{d\log{\lambda_{i}/\lambda_{j}}}{d\log{r/w}}$$</span>
<span class="math-container">$$=(\epsilon-1)(\alpha_{i}-\alpha_{j})$$</span></p>
<p>but I'm afraid my way of thinking is not right.</p>
<hr />
<p><strong>update</strong></p>
<p><span class="math-container">$$\begin{aligned}
\frac{d\log{\theta/\theta_{j}}}{d\log{w/r}}=&amp;\sum_{i}\frac{d\log{\theta_{i}/\theta_{j}}}{d\log{w/r}}\frac{\theta_{i}/\theta_{j}}{(w/r)}\frac{\theta_{j}/\theta}{r/w}\\
=&amp;\sum_{i}\frac{d\log{\theta_{i}/\theta_{j}}}{d\log{w/r}}\frac{\theta_{i}}{\theta}\\
=&amp;\sum_{i}(\epsilon-1)(\alpha_{i}-\alpha_{j})\frac{\theta_{i}}{\theta}\\
=&amp;(\epsilon-1)(\alpha-\alpha_{j})\\
\end{aligned}$$</span></p>
<p>i.e.</p>
<p><span class="math-container">$$\frac{d\log{\theta_{j}/\theta}}{d\log{w/r}}=(\epsilon-1)(\alpha_{j}-\alpha)$$</span></p>
<p>but <span class="math-container">$\theta=\sum_{i}\theta_{i}=1$</span></p>
<p>hence</p>
<p><span class="math-container">$$\frac{d\log{\theta_{j}}}{d\log{w/r}}=(\epsilon-1)(\alpha_{j}-\alpha)$$</span></p>
<hr />
<p>since</p>
<p><span class="math-container">$$
\begin{aligned}
\sigma^{agg}=1&amp;+\frac{1}{\alpha(1-\alpha)}\sum_{i}\alpha_{i}(1-\alpha_{i})(\sigma-1)\theta_{i}\\
&amp;+\frac{1}{\alpha(1-\alpha)}\sum_{i}\alpha_{i}\theta_{i}\frac{d\log{\theta_{i}}}{d(w/r)}\\
=1&amp;+\frac{1}{\alpha(1-\alpha)}\sum_{i}\alpha_{i}(1-\alpha_{i})\sigma\theta_{i}\\
&amp;-\frac{1}{\alpha(1-\alpha)}\sum_{i}\alpha_{i}(1-\alpha_{i})\theta_{i}\\
&amp;+\frac{1}{\alpha(1-\alpha)}\sum_{i}\alpha_{i}\theta_{i}\epsilon(\alpha_{i}-\alpha)\\
&amp;-\frac{1}{\alpha(1-\alpha)}\sum_{i}\alpha_{i}\theta_{i}(\alpha_{i}-\alpha)\\
=&amp;\{1-[\frac{1}{\alpha(1-\alpha)}\sum_{i}\alpha_{i}\theta_{i}(\alpha_{j}-\alpha)+\frac{1}{\alpha(1-\alpha)}\sum_{i}\alpha_{i}\theta_{i}(1-\alpha_{i})]\}\\
&amp;+[\frac{1}{\alpha(1-\alpha)}\sum_{i}\alpha_{i}\theta_{i}(\alpha_{i}-\alpha)]\epsilon\\
&amp;+[\frac{1}{\alpha(1-\alpha)}\sum_{i}\alpha_{i}\theta_{i}(1-\alpha_{i})]\sigma\\
\end{aligned}$$</span></p>
<p>Define</p>
<p><span class="math-container">$$\begin{aligned}
\chi:=
&amp;\frac{1}{\alpha(1-\alpha)}\sum_{i}\theta_{i}(\alpha_{i}-\alpha)^{2}\\
=&amp;\frac{1}{\alpha(1-\alpha)}\sum_{i}\theta_{i}(\alpha_{i}^{2}+\alpha^{2}-2\alpha\alpha_{i})\\
=&amp;\frac{1}{\alpha(1-\alpha)}\sum_{i}\theta_{i}\alpha_{i}^{2}+\alpha^{2}\frac{1}{\alpha(1-\alpha)}-2\alpha^{2}\frac{1}{\alpha(1-\alpha)}\\
=&amp;\frac{1}{\alpha(1-\alpha)}\sum_{i}\theta_{i}\alpha_{i}^{2}-\alpha^{2}\frac{1}{\alpha(1-\alpha)}\\
=&amp;\frac{1}{\alpha(1-\alpha)}\sum_{i}\theta_{i}\alpha_{i}^{2}-\frac{1}{\alpha(1-\alpha)}\sum_{i}\theta_{i}\alpha\alpha_{i}\\
=&amp;\frac{1}{\alpha(1-\alpha)}\sum_{i}\theta_{i}(\alpha_{i}^{2}-\alpha\alpha_{i})\\
=&amp;\frac{1}{\alpha(1-\alpha)}\sum_{i}\alpha_{i}\theta_{i}(\alpha_{i}-\alpha)\\
\end{aligned}$$</span></p>
<p>then</p>
<p><span class="math-container">$$\sigma^{agg}=(1-\chi)\sigma+\chi\epsilon$$</span></p>

CC BY-SA 4.0
2021-06-04T12:34:46.753Z
I think it would be better to add this to your question. Otherwise people might think that your question was already answered.
tdm
CC BY-SA 4.0
2021-06-05T11:42:18.673Z
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.
Alalalalaki
CC BY-SA 4.0
2021-06-05T14:18:05.810Z
@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.
Jajajajavis
CC BY-SA 4.0
2021-06-04T02:54:55.133Z
2021-06-07T12:44:19.307Z
2021-06-07T12:44:19.307Z
Jajajajavis
<p>The paper is <a href="https://onlinelibrary.wiley.com/doi/full/10.3982/ECTA12807" rel="nofollow noreferrer">Oberfield &amp; Raval 2021</a>.</p>
<p>Consumers have standard Dixit-Stiglitz preferences consuming the bundle</p>
<p><span class="math-container">$$Y= \left(\sum_{i \in I} D_{i}^{\frac{1}{\varepsilon}} Y_{i}^{\frac{\varepsilon-1}{\varepsilon}}\right)^{\frac{\varepsilon}{\varepsilon-1}}$$</span></p>
<p>There are monopolistic plants with a common CES between capital and labor inputs, i.e.</p>
<p><span class="math-container">$$Y_{i}=\left[\left(A_{i} K_{i}\right)^{\frac{\sigma-1}{\sigma}}+\left(B_{i} L_{i}\right)^{\frac{\sigma-1}{\sigma}}\right]^{\frac{\sigma}{\sigma-1}}$$</span>.</p>
<p>Let <span class="math-container">$\alpha_{i} \equiv \frac{r K_{i}}{r K_{i}+w L_{i}}$</span> and <span class="math-container">$\alpha \equiv \frac{r K}{r K+w L}$</span> denote the cost shares of capital at plant level and aggregate level, where <span class="math-container">$K \equiv \sum_{i \in I} K_{i}$</span> and <span class="math-container">$L \equiv \sum_{i \in I} L_{i}$</span>.</p>
<p>The plant-level and aggregate elasticity of substitution is</p>
<p><span class="math-container">$$\begin{aligned}
\sigma-1 &amp;=\frac{\mathrm{d} \ln r K_{i} / w L_{i}}{\mathrm{~d} \ln w / r}=\frac{\mathrm{d} \ln \alpha_{i} /\left(1-\alpha_{i}\right)}{\mathrm{d} \ln w / r}=\frac{1}{\alpha_{i}\left(1-\alpha_{i}\right)} \frac{\mathrm{d} \alpha_{i}}{\mathrm{~d} \ln w / r}, \\
\sigma^{\mathrm{agg}}-1 &amp;=\frac{\mathrm{d} \ln r K / w L}{\mathrm{~d} \ln w / r}=\frac{\mathrm{d} \ln \alpha /(1-\alpha)}{\mathrm{d} \ln w / r}=\frac{1}{\alpha(1-\alpha)} \frac{\mathrm{d} \alpha}{\mathrm{d} \ln w / r}
\end{aligned}$$</span>.</p>
<p>The aggregate cost share of capital can be expressed as <span class="math-container">$$\alpha=\sum_{i \in I} \alpha_{i} \theta_{i}$$</span> where <span class="math-container">$\theta_{i} \equiv \frac{r K_{i}+w L_{i}}{r K+w L}$</span>, and by differentiating
<span class="math-container">$$\sigma^{\mathrm{agg}}-1=\frac{1}{\alpha(1-\alpha)} \sum_{i \in I} \alpha_{i}\left(1-\alpha_{i}\right)(\sigma-1) \theta_{i}+\frac{1}{\alpha(1-\alpha)} \sum_{i \in I} \alpha_{i} \theta_{i} \frac{\mathrm{d} \ln \theta_{i}}{\mathrm{~d} \ln w / r}$$</span>.</p>
<p>For the second term in RHS, the authors say &quot;By Shephard's lemma, a plant's cost share of capital <span class="math-container">$\alpha_{i}$</span> measures how relative factor prices affect its mariginal cost&quot; and show that <span class="math-container">$$\frac{\mathrm{d} \ln \theta_{i}}{\mathrm{~d} \ln w / r}=(\varepsilon-1)\left(\alpha_{i}-\alpha\right)$$</span>.</p>
<p>Then the authors show that &quot;the industry elasticity of substitution is a convex combination of the micro ES and elasticity of demand&quot;: <span class="math-container">$$\sigma^{\mathrm{agg}}=(1-\chi) \sigma+\chi \varepsilon$$</span> where <span class="math-container">$\chi \equiv \sum_{i \in I} \frac{\left(\alpha_{i}-\alpha\right)^{2}}{\alpha(1-\alpha)} \theta_{i}$</span>.</p>
<p>Question 1: How to illustrate the claim &quot;By Shephard's lemma, a plant's cost share of capital <span class="math-container">$\alpha_{i}$</span> measures how relative factor prices affect its mariginal cost&quot; and to derive the <span class="math-container">$\frac{\mathrm{d} \ln \theta_{i}}{\mathrm{~d} \ln w / r}$</span>?</p>
<p>Question 2: How to derive the <span class="math-container">$\sigma^{\mathrm{agg}}=(1-\chi) \sigma+\chi \varepsilon$</span>?</p>

<macroeconomics><elasticity-of-substitution><aggregate-production><monopolistic-competition>
Derive and Decompose The Aggregate Elasticity of Substitution in CES Economy
Alalalalaki
# Derive and Decompose The Aggregate Elasticity of Substitution in CES Economy

(Asked by: username_0 on 2021-06-03)

The paper is [Oberfield & Raval 2021](https://onlinelibrary.wiley.com/doi/full/10.3982/ECTA12807).

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

**Tags:** macroeconomics, elasticity-of-substitution, aggregate-production, monopolistic-competition


---

## 1 answers

### Answer by username_1 on 2021-06-04. Score: 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"

---

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

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

> **Comments:**
> **username_2**: I think it would be better to add this to your question. Otherwise people might think that your question was already answered.  
> **username_0**: 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.  
> **username_1**: @username_0 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.  

---

