<p>This is my guess:</p>
<p>We start with the following equation:
<span class="math-container">$$
\pi_t = \frac{(1-\mu)(1-\lambda)}{\mu}\left[\lambda \left(\frac{k_{et}}{s_t}\right)^\rho + (1-\lambda)\right]^{(\sigma-\rho)/\rho}\left(\frac{h_{ut}}{h_{st}}\right)^{(1-\sigma)}\left(\frac{\psi_{st}}{\psi_{ut}}\right)^\sigma
$$</span></p>
<p>Taking logs gives:
<span class="math-container">$$
\ln(\pi_t) = \ln\left(\frac{(1-\mu)(1-\lambda)}{\mu}\right) + \frac{\sigma - \rho}{\rho}\ln\left[\lambda \left(\frac{k_{et}}{s_t}\right)^\rho + (1-\lambda)\right] + (1-\sigma)\ln h_{ut} - (1-\sigma) \ln h_{st} + \sigma\ln \psi_{st} - \sigma \ln \psi_{ut}
$$</span></p>
<p>Take the approximation <span class="math-container">$\ln(1 + x ) \approx x$</span> to simplify the second term to:
<span class="math-container">$$
\frac{\sigma - \rho}{\rho} \left[-\lambda + \lambda \left(\frac{k_{et}}{s_t}\right)^\rho\right]
= -\lambda \frac{\sigma - \rho}{\rho} + \frac{\sigma - \rho}{\rho}\lambda \left(\frac{k_{et}}{s_t}\right)^\rho
$$</span>
This gives:
<span class="math-container">$$
\ln(\pi_t) = \ln\left(\frac{(1-\mu)(1-\lambda)}{\mu}\right) - \lambda \frac{\sigma - \rho}{\rho} + \frac{\sigma - \rho}{\rho}\lambda \left(\frac{k_{et}}{s_t}\right)^\rho\\ + (1-\sigma)\ln h_{ut} - (1-\sigma) \ln h_{st} + \sigma\ln \psi_{st} - \sigma \ln \psi_{ut}
$$</span></p>
<p>Now differatiating with respect to time gives on the left hand side <span class="math-container">$g_{\pi_t}$</span>. The first two terms on the right hand side disappear. For the third term we get:
<span class="math-container">$$
\begin{align*}
&amp;\frac{\sigma - \rho}{\rho} \lambda \rho \left(\frac{k_{et}}{s_t}\right)^{\rho - 1} \frac{\dot k_t s_t - k_t \dot s_t}{(s_t)^2},\\
&amp;= \frac{\sigma - \rho}{\rho} \lambda \left(\frac{k_{et}}{s_t}\right)^\rho \frac{s_t}{k_t} \frac{\dot k_t - k_t g_{s_t}}{s_t},\\
&amp;= \frac{\sigma - \rho}{\rho} \lambda \left(\frac{k_{et}}{s_t}\right)^\rho (g_{k_t} - g_{s_t})
\end{align*}
$$</span>
I think <span class="math-container">$g_{s_t} = g_{h_{st}} + g_{\psi_{st}}$</span>. The other terms give:
<span class="math-container">$$
(1-\sigma) (g_{h_{ut}} - g_{h_{st}}) + \sigma (g_{\psi_{st}} - g_{\psi_{ut}})
$$</span></p>

CC BY-SA 4.0
2021-10-30T11:52:00.730Z
Yep, sorry I'll try to correct.
tdm
CC BY-SA 4.0
2021-10-30T08:40:18.967Z
2021-10-30T11:52:52.600Z
2021-10-30T11:52:52.600Z
tdm
<p>I was reading the following paper:
<a href="https://www.jstor.org/stable/2999442?seq=1#metadata_info_tab_contents" rel="nofollow noreferrer">https://www.jstor.org/stable/2999442?seq=1#metadata_info_tab_contents</a></p>
<p>At one point there are a couple of simple passages that however I did not fully understand.
<a href="https://i.sstatic.net/JX4jN.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/JX4jN.png" alt="enter image description here" /></a></p>
<p>In particular, I can't figure out how a piece of the first term of (3) disappeared in the log linearization. And also how all three terms appear in the differentiation with respect to time (4).</p>
<p>Can anyone help me understand the steps? I offer a virtual beer, points, and much appreciation. Thank you!</p>

<macroeconomics><microeconomics><mathematical-economics><economic-growth>
Help to understand a Log-linearization and subsequent Differentiation with respect to time
Alessandro
# Help to understand a Log-linearization and subsequent Differentiation with respect to time

(Asked by: username_0 on 2021-10-27)

I was reading the following paper:
<https://www.jstor.org/stable/2999442?seq=1#metadata_info_tab_contents>

At one point there are a couple of simple passages that however I did not fully understand.
[![enter image description here](https://i.sstatic.net/JX4jN.png)](https://i.sstatic.net/JX4jN.png)

In particular, I can't figure out how a piece of the first term of (3) disappeared in the log linearization. And also how all three terms appear in the differentiation with respect to time (4).

Can anyone help me understand the steps? I offer a virtual beer, points, and much appreciation. Thank you!

**Tags:** macroeconomics, microeconomics, mathematical-economics, economic-growth

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## 1 answers

### Answer by username_1 on 2021-10-30. Score: 2

This is my guess:

We start with the following equation:
$$
\pi\_t = \frac{(1-\mu)(1-\lambda)}{\mu}\left[\lambda \left(\frac{k\_{et}}{s\_t}\right)^\rho + (1-\lambda)\right]^{(\sigma-\rho)/\rho}\left(\frac{h\_{ut}}{h\_{st}}\right)^{(1-\sigma)}\left(\frac{\psi\_{st}}{\psi\_{ut}}\right)^\sigma
$$

Taking logs gives:
$$
\ln(\pi\_t) = \ln\left(\frac{(1-\mu)(1-\lambda)}{\mu}\right) + \frac{\sigma - \rho}{\rho}\ln\left[\lambda \left(\frac{k\_{et}}{s\_t}\right)^\rho + (1-\lambda)\right] + (1-\sigma)\ln h\_{ut} - (1-\sigma) \ln h\_{st} + \sigma\ln \psi\_{st} - \sigma \ln \psi\_{ut}
$$

Take the approximation $\ln(1 + x ) \approx x$ to simplify the second term to:
$$
\frac{\sigma - \rho}{\rho} \left[-\lambda + \lambda \left(\frac{k\_{et}}{s\_t}\right)^\rho\right]
= -\lambda \frac{\sigma - \rho}{\rho} + \frac{\sigma - \rho}{\rho}\lambda \left(\frac{k\_{et}}{s\_t}\right)^\rho
$$
This gives:
$$
\ln(\pi\_t) = \ln\left(\frac{(1-\mu)(1-\lambda)}{\mu}\right) - \lambda \frac{\sigma - \rho}{\rho} + \frac{\sigma - \rho}{\rho}\lambda \left(\frac{k\_{et}}{s\_t}\right)^\rho\\ + (1-\sigma)\ln h\_{ut} - (1-\sigma) \ln h\_{st} + \sigma\ln \psi\_{st} - \sigma \ln \psi\_{ut}
$$

Now differatiating with respect to time gives on the left hand side $g\_{\pi\_t}$. The first two terms on the right hand side disappear. For the third term we get:
$$
\begin{align\*}
&\frac{\sigma - \rho}{\rho} \lambda \rho \left(\frac{k\_{et}}{s\_t}\right)^{\rho - 1} \frac{\dot k\_t s\_t - k\_t \dot s\_t}{(s\_t)^2},\\
&= \frac{\sigma - \rho}{\rho} \lambda \left(\frac{k\_{et}}{s\_t}\right)^\rho \frac{s\_t}{k\_t} \frac{\dot k\_t - k\_t g\_{s\_t}}{s\_t},\\
&= \frac{\sigma - \rho}{\rho} \lambda \left(\frac{k\_{et}}{s\_t}\right)^\rho (g\_{k\_t} - g\_{s\_t})
\end{align\*}
$$
I think $g\_{s\_t} = g\_{h\_{st}} + g\_{\psi\_{st}}$. The other terms give:
$$
(1-\sigma) (g\_{h\_{ut}} - g\_{h\_{st}}) + \sigma (g\_{\psi\_{st}} - g\_{\psi\_{ut}})
$$

> **Comments:**
> **username_1**: Yep, sorry I'll try to correct.  

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