# Log-linearisation Euler Consumption Equation

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!

• 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 Jan 21 '17 at 15:22
• Hi, thanks for the reply. See page 18. 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? – BenBernke Jan 21 '17 at 15:29

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!