<p>We can easily switch between logged equations and log-deviations. </p>

<p>To see this, note that a given equation always holds, and therefore it also holds in steady state. Hence, you can just subtract the steady state values of both sides of an equation to get log deviations, if you wish.</p>

<p>The reason you want logs, in this case, can be seen in (9). You want <span class="math-container">$i_t - E_t\{\pi_{t+1}\}$</span>, the real interest rate, to deviate around <span class="math-container">$\rho$</span>. This makes more sense economically than deviations around their individual steady states. Mathematically they are equivalent.</p>

CC BY-SA 4.0
2019-08-10T00:55:05.373Z
2019-08-10T00:55:05.373Z
Chris tie
<p><span class="math-container">$$\frac{x_{t+1}}{x_t} \equiv \frac{x_{t}+Δx_{t+1}}{x_t} = 1+\frac{Δx_{t+1}}{x_t}$$</span></p>

<p>Further,</p>

<p><span class="math-container">$$\ln (1+a) \approx a,$$</span></p>

<p>an acceptable approximation for <span class="math-container">$|a| &lt;0.1$</span> and certainly for <span class="math-container">$|a| &lt;0.05$</span>.</p>

CC BY-SA 4.0
2019-08-09T19:52:42.910Z
2019-08-09T20:07:05.843Z
2019-08-09T20:07:05.843Z
Alecos Papadopoulos
<p>In Gali (2015)'s textbook Monetary Policy, Inflation, and the Business Cycle, variables in levels are denoted with capital letters, logged variables with lowercase letters. </p>

<p>However, when a log-linearisation is performed, say of the household optimality condition, why is the result in logs and not in percentage deviations, (lower case letters with a hat, for instance)?</p>

<p>Is there an intermediate step/reason for why we can go from percentage deviations to logs? Thanks!</p>

<p><a href="https://i.sstatic.net/jHKbq.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/jHKbq.png" alt="household optimality condition"></a></p>

CC BY-SA 4.0
2019-08-09T11:47:18.417Z
Are you sure there isn't some steady-state condition mentioned on the steady-state values of $w, c, p, \text{and} \; n$? Referring to equation (6) you should get the percentage deviation (log changes) when log-linearising. I can only understand their result if all these variables are set to be equal to 1.
Ali
<macroeconomics><mathematical-economics><academic-graduate>
Log linearising (Gali textbook)
NuancedPaul
# Log linearising (Gali textbook)

(Asked by: username_0 on 2019-08-09)

In Gali (2015)'s textbook Monetary Policy, Inflation, and the Business Cycle, variables in levels are denoted with capital letters, logged variables with lowercase letters.

However, when a log-linearisation is performed, say of the household optimality condition, why is the result in logs and not in percentage deviations, (lower case letters with a hat, for instance)?

Is there an intermediate step/reason for why we can go from percentage deviations to logs? Thanks!

[![household optimality condition](https://i.sstatic.net/jHKbq.png)](https://i.sstatic.net/jHKbq.png)

**Tags:** macroeconomics, mathematical-economics, academic-graduate

> **Comments:**
> **username_1**: Are you sure there isn't some steady-state condition mentioned on the steady-state values of $w, c, p, \text{and} \; n$? Referring to equation (6) you should get the percentage deviation (log changes) when log-linearising. I can only understand their result if all these variables are set to be equal to 1.  

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## 2 answers

### Answer by username_2 on 2019-08-10. Score: 1 | Accepted answer

We can easily switch between logged equations and log-deviations.

To see this, note that a given equation always holds, and therefore it also holds in steady state. Hence, you can just subtract the steady state values of both sides of an equation to get log deviations, if you wish.

The reason you want logs, in this case, can be seen in (9). You want $i\_t - E\_t\{\pi\_{t+1}\}$, the real interest rate, to deviate around $\rho$. This makes more sense economically than deviations around their individual steady states. Mathematically they are equivalent.


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### Answer by username_3 on 2019-08-09. Score: 2

$$\frac{x\_{t+1}}{x\_t} \equiv \frac{x\_{t}+Δx\_{t+1}}{x\_t} = 1+\frac{Δx\_{t+1}}{x\_t}$$

Further,

$$\ln (1+a) \approx a,$$

an acceptable approximation for $|a| <0.1$ and certainly for $|a| <0.05$.


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