<p>The shock process is already linear. any log linearisation will result in identical expression. The underlying non linear shock process can be something like this:$</p>

<p><span class="math-container">$$ Z_t = Z^{(1-\rho)} Z_{t-1} e^{\epsilon_{t}}$$</span></p>

<p>This when you linearise, you obtain your linear shock processes.</p>

CC BY-SA 4.0
2019-11-11T10:44:15.723Z
2019-11-11T10:44:15.723Z
EndLoop
<p>I am still dealing with DSGE models and with log-linearization of characteristic equations of the model. The one that creates me more doubts is the technology shock, that is usually modeled as an AR(1).</p>

<p>Usually, we are given technology processes like 
\begin{gather}
\ln(z_t)=(1-\rho) \ln(\bar z)+\rho \ln(z_{t-1})+\epsilon_t
\end{gather}
so that it is already linear and we can re-write it in deviations from the steady state just moving $\ln(\bar z)$ on both sides of equation, obtaining
\begin{gather}
\hat z_t=\rho \hat z_{t-1}+\epsilon_t
\end{gather}
where hat variables are log deviations from the steady state. </p>

<p>But when the process is something like
\begin{gather}
\ln(z_t)=(1-\rho) (\bar z)+\rho \ln(z_{t-1})+\epsilon_t
\end{gather}
how should I proceed?
What I cannot understand is how the log linearization should be used, taking into account that $\epsilon_t$ is a white noise process. I mean applying logs and differentiating gives me something I do not understand.</p>

CC BY-SA 3.0
2016-10-10T15:46:20.727Z
I don't think this makes any sense, as $\overline{z}$ is the expected value of the process in the first equation, only. Do you have a source for the second one?
Chris tie
CC BY-SA 3.0
2016-10-10T16:03:15.643Z
@Christie we were actually wondering if it were a typo (having forgotten to put logs in $\bar z$, the problem is that the source is a problem set by my Adv. Macro TA
PostDocing
CC BY-SA 3.0
2016-10-11T08:07:39.423Z
Ok, I have a strong feeling that it is a typo, then
Chris tie
CC BY-SA 3.0
2016-10-11T14:19:14.790Z
As a teacher I find it curious that students first contact messageboards and only if that fails do they maybe contact the TA. Has anyone ever lost points by contacting their TA? (Or is the TA a werewolf or something similarly dangerous?)
Giskard
CC BY-SA 3.0
2016-10-11T15:24:54.017Z
It depends on the TA's availability to answer ;)
PostDocing
<macroeconomics><dsge>
Log linearization of the technology dynamic
PostDocing
# Log linearization of the technology dynamic

(Asked by: username_0 on 2016-10-10)

I am still dealing with DSGE models and with log-linearization of characteristic equations of the model. The one that creates me more doubts is the technology shock, that is usually modeled as an AR(1).

Usually, we are given technology processes like
\begin{gather}
\ln(z\_t)=(1-\rho) \ln(\bar z)+\rho \ln(z\_{t-1})+\epsilon\_t
\end{gather}
so that it is already linear and we can re-write it in deviations from the steady state just moving $\ln(\bar z)$ on both sides of equation, obtaining
\begin{gather}
\hat z\_t=\rho \hat z\_{t-1}+\epsilon\_t
\end{gather}
where hat variables are log deviations from the steady state.

But when the process is something like
\begin{gather}
\ln(z\_t)=(1-\rho) (\bar z)+\rho \ln(z\_{t-1})+\epsilon\_t
\end{gather}
how should I proceed?
What I cannot understand is how the log linearization should be used, taking into account that $\epsilon\_t$ is a white noise process. I mean applying logs and differentiating gives me something I do not understand.

**Tags:** macroeconomics, dsge

> **Comments:**
> **username_1**: I don't think this makes any sense, as $\overline{z}$ is the expected value of the process in the first equation, only. Do you have a source for the second one?  
> **username_0**: @Christie we were actually wondering if it were a typo (having forgotten to put logs in $\bar z$, the problem is that the source is a problem set by my Adv. Macro TA  
> **username_1**: Ok, I have a strong feeling that it is a typo, then  
> **username_2**: As a teacher I find it curious that students first contact messageboards and only if that fails do they maybe contact the TA. Has anyone ever lost points by contacting their TA? (Or is the TA a werewolf or something similarly dangerous?)  
> **username_0**: It depends on the TA's availability to answer ;)  

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## 1 answers

### Answer by username_3 on 2019-11-11. Score: 2

The shock process is already linear. any log linearisation will result in identical expression. The underlying non linear shock process can be something like this:$

$$ Z\_t = Z^{(1-\rho)} Z\_{t-1} e^{\epsilon\_{t}}$$

This when you linearise, you obtain your linear shock processes.


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