<p>To simplify matters, let's call the right-hand side of your starting equation $X_t$ Then, I start just like you did with</p>

<p>$1 = X_t$</p>

<p>The difference between your solution and Gali's is that you took the Taylor expansion around</p>

<p>$\log(1)=0=log(X_t)$  which implies that also the steady state equals 0, so we can simply subtract it to get log-differences,</p>

<p>whereas Gali used</p>

<p>$1 = \exp(\log(X_t))$ with steady state $1=\exp(\log(X^*))$</p>

<p>Let's define $x_t=\log(X_t)$.</p>

<p>By definition of the Taylor expansion this gives</p>

<p>$T(1) = \exp(x^*)+\exp(x^*)(x_t-x^*)$ </p>

<p>Note that the exponential function is identical to its derivative. We know that $\exp(x^*)=1$ from above, so we can simplify to </p>

<p>$T(1)=1+(x_t-x^*)$ </p>

<p>At this point, you are correct that the $\rho$ cancel out. Instead of defining $\hat{x}_t=x_t-x^*$ but Gali chooses to replace $\rho=\pi +\sigma \gamma-i$, which is the steady state condition, which is why $\rho$ stays on the paper. </p>

<p>Note, however, that the "approximation error" from taking log differences instead of Gali's approach is tiny, i.e. for $\beta=0.99$ it is only $-\log(0.99)\approx 0.01$</p>

<p>Have a look at appendix 2.1 of Gali's book! It is quite complicated, but together with this post, I hope you get it!</p>

CC BY-SA 3.0
2017-01-21T17:27:57.617Z
2017-01-21T17:27:57.617Z
Chris tie
<p>I am trying to log-linearize the Euler consumption equation in Gali's book. He says:</p>

<blockquote>
  <p>A log-linear approximation of $Q_{t}= \beta E_{t}[(\frac{C_{t+1}}{C_{t}})^{-\sigma}(\frac{Z_{t+1}}{Z_{t}})(\frac{P_{t}}{P_{t+1}})]$ (1)</p>
</blockquote>

<p>I spent all of last night trying to learn log-linearization so this might be why my answer isn't correct but here is my attempt:</p>

<p>I re-write $Q_{t}=\frac{1}{1+i_{t}}$ and thus (1) becomes:</p>

<p>$1= \beta E_{t}[(1+i_{t})(\frac{C_{t+1}}{C_{t}})^{-\sigma}(\frac{Z_{t+1}}{Z_{t}})(\frac{P_{t}}{P_{t+1}})]$</p>

<p>Taking logs of both sides (I'm assuming I can ignore the expectation operator, although I'm not really sure why):</p>

<p>$0=ln(\beta) + ln(1+i_{t})- \sigma[ln(c_{t+1}) - ln(c_{t})] + ln(z_{t+1}) - ln(z_{t}) + ln(P_{t}) - ln(P_{t+1})]$ (2)</p>

<p>Approximating around the steady state values:</p>

<p>$0=ln(\beta) + ln(1+i^{*})+(\frac{1}{1+i^{*}})(i_{t}-i^{*})- \sigma[ln(c^{*})+(\frac{1}{c^{*}})(c_{t+1}-c^{*}) - ln(c^{*}) - (\frac{1}{c^{*}})(c_{t}-c^{*})] + ln(z^{*}) + (\frac{1}{z^{*}})(z_{t+1}-z^{*}) - ln(z^{*}) - (\frac{1}{z^{*}})(z_{t}-z^{*})+ ln(P^{*}) + (\frac{1}{P^{*}})(P_{t}-P^{*}) - ln(P^{*}) - (\frac{1}{P^{*}})(P_{t+1}-P^{*})$</p>

<p>That was quite messy but hopefully you see what I am doing, a simple first order approximation around the steady state. Next I used equation (2) to cancel some terms. I was left with:</p>

<p>$(\frac{1}{1+i^{*}})(i_{t}-i^{*})- \sigma(\frac{1}{c^{*}})(c_{t+1}-c^{*}) +  \sigma(\frac{1}{c^{*}})(c_{t}-c^{*}) + (\frac{1}{z^{*}})(z_{t+1}-z^{*}) - (\frac{1}{z^{*}})(z_{t}-z^{*})  + (\frac{1}{P^{*}})(P_{t}-P^{*}) -  (\frac{1}{P^{*}})(P_{t+1}-P^{*}) = 0$ </p>

<p>Re-writing in terms of deviations from steady state:</p>

<p>$\tilde i_{t} - \sigma \tilde c_{t+1} + \sigma \tilde c_{t} + \tilde z_{t+1} - \tilde z_{t} + \tilde p_{t} -  \tilde p_{t+1} = 0$</p>

<p>I can re-arrange this in terms of $\tilde c_{t}$ and put in expectation operators but my answer does not match with Gali's. He says:</p>

<blockquote>
  <p>$c_{t} = E[c_{t+1}] + \frac{1}{\sigma}(i_{t} - E_{t}[\pi_{t+1}] - \rho) + \frac{1}{\sigma}(1-\rho _{z})z_{t} $</p>
</blockquote>

<p>First of all, I don't understand where $\rho$ came from since it is not in (1). Furthermore, am I making a mistake in my linear approximation? Should I not have used (2) to cancel terms?</p>

<p>I just learned log-linearization so my method might be quite naive. I simply "logged" both sides, used first order Taylor approximation around steady state, cancelled terms from the "logging" step and solved for consumption.</p>

<p>Any assistance is greatly appreciated!</p>

CC BY-SA 3.0
2017-01-21T15:22:51.557Z
First, please add the source to your question. I think Gali uses $\rho=-\ln{\beta}$. Second, your answer is really close. If you solve your solution for $\widetilde{c}_t$, you are almost there. Your calculations look ok. Where does $\rho_z$ come from?
Chris tie
CC BY-SA 3.0
2017-01-21T15:29:42.040Z
Hi, thanks for the reply. [See page 18](http://elib.peaceland.edu.ng:8383/greenstone3/sites/localsite/collect/peacelan/index/assoc/HASH0139.dir/doc.pdf). You are right about Gali using $\rho = -ln \beta$, I don't understand why though. But even still, I though the $ln \beta$ term was cancelled when I use equation (2) to cancel terms? Meaning that I set every term in equation (2) equal to zero in the Taylor approximation, hence why $ln \beta$ isn't in my final answer. Should I not have done that?
user11767
<consumption>
Log-linearisation Euler Consumption Equation
user11767
# Log-linearisation Euler Consumption Equation

(Asked by: username_0 on 2017-01-21)

I am trying to log-linearize the Euler consumption equation in Gali's book. He says:

> A log-linear approximation of $Q\_{t}= \beta E\_{t}[(\frac{C\_{t+1}}{C\_{t}})^{-\sigma}(\frac{Z\_{t+1}}{Z\_{t}})(\frac{P\_{t}}{P\_{t+1}})]$ (1)

I spent all of last night trying to learn log-linearization so this might be why my answer isn't correct but here is my attempt:

I re-write $Q\_{t}=\frac{1}{1+i\_{t}}$ and thus (1) becomes:

$1= \beta E\_{t}[(1+i\_{t})(\frac{C\_{t+1}}{C\_{t}})^{-\sigma}(\frac{Z\_{t+1}}{Z\_{t}})(\frac{P\_{t}}{P\_{t+1}})]$

Taking logs of both sides (I'm assuming I can ignore the expectation operator, although I'm not really sure why):

$0=ln(\beta) + ln(1+i\_{t})- \sigma[ln(c\_{t+1}) - ln(c\_{t})] + ln(z\_{t+1}) - ln(z\_{t}) + ln(P\_{t}) - ln(P\_{t+1})]$ (2)

Approximating around the steady state values:

$0=ln(\beta) + ln(1+i^{\*})+(\frac{1}{1+i^{\*}})(i\_{t}-i^{\*})- \sigma[ln(c^{\*})+(\frac{1}{c^{\*}})(c\_{t+1}-c^{\*}) - ln(c^{\*}) - (\frac{1}{c^{\*}})(c\_{t}-c^{\*})] + ln(z^{\*}) + (\frac{1}{z^{\*}})(z\_{t+1}-z^{\*}) - ln(z^{\*}) - (\frac{1}{z^{\*}})(z\_{t}-z^{\*})+ ln(P^{\*}) + (\frac{1}{P^{\*}})(P\_{t}-P^{\*}) - ln(P^{\*}) - (\frac{1}{P^{\*}})(P\_{t+1}-P^{\*})$

That was quite messy but hopefully you see what I am doing, a simple first order approximation around the steady state. Next I used equation (2) to cancel some terms. I was left with:

$(\frac{1}{1+i^{\*}})(i\_{t}-i^{\*})- \sigma(\frac{1}{c^{\*}})(c\_{t+1}-c^{\*}) + \sigma(\frac{1}{c^{\*}})(c\_{t}-c^{\*}) + (\frac{1}{z^{\*}})(z\_{t+1}-z^{\*}) - (\frac{1}{z^{\*}})(z\_{t}-z^{\*}) + (\frac{1}{P^{\*}})(P\_{t}-P^{\*}) - (\frac{1}{P^{\*}})(P\_{t+1}-P^{\*}) = 0$

Re-writing in terms of deviations from steady state:

$\tilde i\_{t} - \sigma \tilde c\_{t+1} + \sigma \tilde c\_{t} + \tilde z\_{t+1} - \tilde z\_{t} + \tilde p\_{t} - \tilde p\_{t+1} = 0$

I can re-arrange this in terms of $\tilde c\_{t}$ and put in expectation operators but my answer does not match with Gali's. He says:

> $c\_{t} = E[c\_{t+1}] + \frac{1}{\sigma}(i\_{t} - E\_{t}[\pi\_{t+1}] - \rho) + \frac{1}{\sigma}(1-\rho \_{z})z\_{t} $

First of all, I don't understand where $\rho$ came from since it is not in (1). Furthermore, am I making a mistake in my linear approximation? Should I not have used (2) to cancel terms?

I just learned log-linearization so my method might be quite naive. I simply "logged" both sides, used first order Taylor approximation around steady state, cancelled terms from the "logging" step and solved for consumption.

Any assistance is greatly appreciated!

**Tags:** consumption

> **Comments:**
> **username_1**: First, please add the source to your question. I think Gali uses $\rho=-\ln{\beta}$. Second, your answer is really close. If you solve your solution for $\widetilde{c}_t$, you are almost there. Your calculations look ok. Where does $\rho_z$ come from?  
> **username_0**: Hi, thanks for the reply. [See page 18](http://elib.peaceland.edu.ng:8383/greenstone3/sites/localsite/collect/peacelan/index/assoc/HASH0139.dir/doc.pdf). You are right about Gali using $\rho = -ln \beta$, I don't understand why though. But even still, I though the $ln \beta$ term was cancelled when I use equation (2) to cancel terms? Meaning that I set every term in equation (2) equal to zero in the Taylor approximation, hence why $ln \beta$ isn't in my final answer. Should I not have done that?  

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## 1 answers

### Answer by username_1 on 2017-01-21. Score: 2 | Accepted answer

To simplify matters, let's call the right-hand side of your starting equation $X\_t$ Then, I start just like you did with

$1 = X\_t$

The difference between your solution and Gali's is that you took the Taylor expansion around

$\log(1)=0=log(X\_t)$ which implies that also the steady state equals 0, so we can simply subtract it to get log-differences,

whereas Gali used

$1 = \exp(\log(X\_t))$ with steady state $1=\exp(\log(X^\*))$

Let's define $x\_t=\log(X\_t)$.

By definition of the Taylor expansion this gives

$T(1) = \exp(x^\*)+\exp(x^\*)(x\_t-x^\*)$

Note that the exponential function is identical to its derivative. We know that $\exp(x^\*)=1$ from above, so we can simplify to

$T(1)=1+(x\_t-x^\*)$

At this point, you are correct that the $\rho$ cancel out. Instead of defining $\hat{x}\_t=x\_t-x^\*$ but Gali chooses to replace $\rho=\pi +\sigma \gamma-i$, which is the steady state condition, which is why $\rho$ stays on the paper.

Note, however, that the "approximation error" from taking log differences instead of Gali's approach is tiny, i.e. for $\beta=0.99$ it is only $-\log(0.99)\approx 0.01$

Have a look at appendix 2.1 of Gali's book! It is quite complicated, but together with this post, I hope you get it!


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