<p>The aim of log-linearization is to get an expression that is linear in the deviation from steady state <span class="math-container">$x_t$</span>, where <span class="math-container">$x_t:=\log(X_t/X)$</span>, <span class="math-container">$X$</span> is the steady state of <span class="math-container">$X_t$</span> and we have <span class="math-container">$X_t = Xe^{x_t} \approx X(1+x_t)$</span>.</p>
<p>The <strong>general approach of log-linearization is</strong> (1) to take logs of both sides of the equation and then (2) do a Taylor series expansion of these log-functions (one on each side of the equation).</p>
<p>But in many cases, just plugging <span class="math-container">$X(1+x_t)$</span> for <span class="math-container">$X_t$</span> into the expression gives you something that is linear in <span class="math-container">$x_t$</span> right away, but not for exponential expressions such as in your example, as you tried to do. This is easy to see:</p>
<p><span class="math-container">$$
e^{\theta(X_t-X)} \approx e^{\theta((X(1+x_t)-X)} = e^{\theta Xx_t}
$$</span>
is clearly not linear in <span class="math-container">$x_t$</span>, so that approach won't work. However, if <span class="math-container">$\theta X x_t
$</span> is &quot;small&quot;, you can use <span class="math-container">$e^{\theta X x_t}\approx 1 + \theta X x_t$</span> and you are done.</p>
<p>Generally, if one runs into expressions as above, or the result is non-linear otherwise, the general approach needs to be used. This is also what's applied in DSGE models such cases.</p>
<p><strong>General approach:</strong></p>
<p>Take logs, then do a Taylor series expansion. Define <span class="math-container">$f(X_t)=\theta(X_t-X)$</span>.
<span class="math-container">$$
\log\bigl(e^{\theta(X_t-X)}\bigr)=\theta(X_t-X)\approx f(0) + f'(0)(X_t-X)= 0 + \theta(X_t-X) \approx \theta X x_t
$$</span></p>
<p>Here, <span class="math-container">$\theta X x_t$</span> is the log-linearization of <span class="math-container">$e^{\theta(X_t-X)}$</span>. If <span class="math-container">$\theta X x_t$</span> is &quot;small&quot;, you can recover the level version by simply applying the exponential, which gives you
<span class="math-container">$$
e^{\theta(X_t-X)} \approx 1 + \theta X x_t
$$</span>
The general approach gives an expression that is linear in <span class="math-container">$x_t$</span>. Note that you can take a higher order Taylor series, which will also be linear in <span class="math-container">$x_t$</span>, to get a better approximation and capture more of the dynamics, which is sometimes used if the expression is highly non-linear.</p>

CC BY-SA 4.0
2021-07-04T14:41:56.537Z
No I donr think this is completely right. I am sure the expression exp(theta*X) will not exists because we will substitute X(t)=X in steady state which implies exp(theta*(X-X))=exp(0)=1
john smith
CC BY-SA 4.0
2021-07-04T16:33:02.593Z
Indeed I should have mentioned the case where $X_t=X$.  Added this, and a more general solution to my answer.
BrsG
CC BY-SA 4.0
2021-07-04T19:08:03.653Z
What you are doing is linearisation. Approximating exp(theta*X) around steady state X_bar is well documented in basic textbooks. What I am asking is log linearisation of exponential functions. The answer to that starts from what I have explained. I am reaching some complex expressions and thats why I wanted to make sure from people working with advanced economics. Generally people who do DSGE models can confirm this. If there are people in this forum that have encountered log-linearisation of exponential terms while doing DSGE models may please take a look at my problem.
john smith
CC BY-SA 4.0
2021-07-05T08:44:15.447Z
I have edited my answer further to clarify.
BrsG
CC BY-SA 4.0
2021-07-09T16:18:03.543Z
@Sanaya, can you then please accept.
BrsG
CC BY-SA 4.0
2021-07-04T09:28:53.733Z
2021-07-06T16:04:43.177Z
2021-07-06T16:04:43.177Z
BrsG
<p>How do I log-linearize an expression like:</p>
<p><span class="math-container">$e^{(\theta*(x_t-\bar{x}))}$</span></p>
<p>Where <span class="math-container">$\theta$</span> is any constant and <span class="math-container">$\bar{x}$</span> is the steady-state of <span class="math-container">$x$</span></p>
<p>Generally when we log-linearize say <span class="math-container">$x_t$</span>, we write it as</p>
<p><span class="math-container">$x_t = \bar{x} * e^{(log(X_t-log(\bar{x}))}
     = 1 + X_t$</span></p>
<p>where <span class="math-container">$x_t = log(x_t-log(\bar{x}))$</span></p>
<p>But in my current problem, the expression itself is exponential. So is the following right?</p>
<p>exp(theta*(X(t)-X_bar)) = exp(exp(theta*(log(X(t)-log(X_bar))) * exp(exp(-theta*(log(X_bar)-log(X_bar))))</p>
<p>which boils down to exp(exp(theta<em>x(t))) * exp(1) = exp(1 + exp(theta</em>x(t)))
= 1 + 1 + exp(theta*x(t))</p>

<log-linearization>
log linearize exponential expressions
john smith
# log linearize exponential expressions

(Asked by: username_0 on 2021-07-03)

How do I log-linearize an expression like:

$e^{(\theta\*(x\_t-\bar{x}))}$

Where $\theta$ is any constant and $\bar{x}$ is the steady-state of $x$

Generally when we log-linearize say $x\_t$, we write it as

$x\_t = \bar{x} \* e^{(log(X\_t-log(\bar{x}))}
= 1 + X\_t$

where $x\_t = log(x\_t-log(\bar{x}))$

But in my current problem, the expression itself is exponential. So is the following right?

exp(theta\*(X(t)-X\_bar)) = exp(exp(theta\*(log(X(t)-log(X\_bar))) \* exp(exp(-theta\*(log(X\_bar)-log(X\_bar))))

which boils down to exp(exp(theta*x(t))) \* exp(1) = exp(1 + exp(theta*x(t)))
= 1 + 1 + exp(theta\*x(t))

**Tags:** log-linearization


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## 1 answers

### Answer by username_1 on 2021-07-04. Score: 1

The aim of log-linearization is to get an expression that is linear in the deviation from steady state $x\_t$, where $x\_t:=\log(X\_t/X)$, $X$ is the steady state of $X\_t$ and we have $X\_t = Xe^{x\_t} \approx X(1+x\_t)$.

The **general approach of log-linearization is** (1) to take logs of both sides of the equation and then (2) do a Taylor series expansion of these log-functions (one on each side of the equation).

But in many cases, just plugging $X(1+x\_t)$ for $X\_t$ into the expression gives you something that is linear in $x\_t$ right away, but not for exponential expressions such as in your example, as you tried to do. This is easy to see:

$$
e^{\theta(X\_t-X)} \approx e^{\theta((X(1+x\_t)-X)} = e^{\theta Xx\_t}
$$
is clearly not linear in $x\_t$, so that approach won't work. However, if $\theta X x\_t
$ is "small", you can use $e^{\theta X x\_t}\approx 1 + \theta X x\_t$ and you are done.

Generally, if one runs into expressions as above, or the result is non-linear otherwise, the general approach needs to be used. This is also what's applied in DSGE models such cases.

**General approach:**

Take logs, then do a Taylor series expansion. Define $f(X\_t)=\theta(X\_t-X)$.
$$
\log\bigl(e^{\theta(X\_t-X)}\bigr)=\theta(X\_t-X)\approx f(0) + f'(0)(X\_t-X)= 0 + \theta(X\_t-X) \approx \theta X x\_t
$$

Here, $\theta X x\_t$ is the log-linearization of $e^{\theta(X\_t-X)}$. If $\theta X x\_t$ is "small", you can recover the level version by simply applying the exponential, which gives you
$$
e^{\theta(X\_t-X)} \approx 1 + \theta X x\_t
$$
The general approach gives an expression that is linear in $x\_t$. Note that you can take a higher order Taylor series, which will also be linear in $x\_t$, to get a better approximation and capture more of the dynamics, which is sometimes used if the expression is highly non-linear.

> **Comments:**
> **username_0**: No I donr think this is completely right. I am sure the expression exp(theta*X) will not exists because we will substitute X(t)=X in steady state which implies exp(theta*(X-X))=exp(0)=1  
> **username_1**: Indeed I should have mentioned the case where $X_t=X$.  Added this, and a more general solution to my answer.  
> **username_0**: What you are doing is linearisation. Approximating exp(theta*X) around steady state X_bar is well documented in basic textbooks. What I am asking is log linearisation of exponential functions. The answer to that starts from what I have explained. I am reaching some complex expressions and thats why I wanted to make sure from people working with advanced economics. Generally people who do DSGE models can confirm this. If there are people in this forum that have encountered log-linearisation of exponential terms while doing DSGE models may please take a look at my problem.  
> **username_1**: I have edited my answer further to clarify.  
> **username_1**: @Sanaya, can you then please accept.  

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