# New Keynesian Phillips Curve: log-linearization

I am facing some problems in log-linearizing the following equation:

$Y_{t}\left(\epsilon&space;-1\right)\left[\frac{\epsilon}{\epsilon&space;-1}\frac{W_{t}}{P_{t}}&space;-1&space;\right]&space;-\Psi\pi_{t}\left(1+\pi_{t}\right)+\beta\Psi\pi_{t+1}\left(1+\pi_{t+1}\right)=0$

which is the New Keynesian Phillips Curve (NKPC). The expected result, once log-linearized, is:

$\pi_{t}=\delta&space;mc_{t}+E_{t}{(\pi_{t+1})}$

which is described by equation 46 in this paper.

However, I do not get this result. For now, I have just took the log transformation, but I did not do the Taylor expansion, and I imposed that:

$\pi_{t+1}=ln\left(1+\pi_{t+1}\right)$

$\pi_{t}=ln\left(1+\pi_{t}\right)$

$\frac{W_{t}}{P_{t}}=MC_{t}$

Is this problem due to the way I am log-linearizing (not taking Taylor expansion into consideration)? Or are there other problems?

• It probably has to do with the way you are log linearizing. The way I do it (and it never fails) is: $X_{t} = \bar{X}\left(1+\hat{X}_{t}\right)$ where $X_{t}$ is the variable, $\bar{X}$ its ss value and $\hat{X}_{t}$ the log deviation. Recall that since it is a first order aproximation then it is linear (so don't worry with $\mathbb{E}X_{t+1}$) and that $\hat{X}_{t}\hat{Y}_{t}=0$. Jun 23, 2021 at 22:47
• Thank you for your reply @PedroIgnacioMartinezBruera! Can you just provide an example with my specific case? (I am not asking you to log-linearize everything, but just one term to show me how it works). Besides, do you know where I can find some papers about this topic? Thank you again.
– AVR
Jun 25, 2021 at 9:14

## 1 Answer

[Not a full solution]

Unlike what the name suggests, log-linearization does not involve taking logs of individual terms in an equation. Otherwise, the equation t wouldn't hold. The general approach is to take logs of both sides as a whole, and then do a Taylor expansion of these two log-functions.

However, this doesn't seem to be necessary here. You can just exploit the approximation $$X_t \approx X(1-x_t)$$, where $$X$$ is the steady state value of $$X_t$$, and $$x_t=\log(\frac{X_t}{X})$$ And, as suggested by Pedro in his comment, you then plug in $$X(1+x_t)$$ wherever $$X_t$$ appears in your equation. In many cases, this gives you an equation that is linear in $$x_t$$ (and $$y_t, z_t$$, etc.). As this is what you want, you are done. Only if that doesn't work, you need to use a Taylor expansion.

Simple example: $$\alpha Y_t \frac{W_t}{P_t}\approx \alpha Y \frac{W}{P}(1+y_t)(1+w_t)(1+p_t)^{-1}\approx \alpha Y \frac{W}{P}(1+y_t+w_t-p_t) = \alpha Y \frac{W}{P}(1+y_t+mc_t)$$ In the last step we use $$(1 + y_t)(1+w_t)\approx (1+y_t+w_t)$$, as $$y_t$$ and $$p_t$$ are "small" and the mixed terms $$y_t p_t$$ are therefore even smaller, so are considered negligible. A similar rule applies for division.

Note that in your example a couple more "tricks" are used. The way it is written, $$\pi$$ seems to be the (net) inflation rate (say 0.02). If that's the case, these terms are already linear in $$\pi_t$$, so there is nothing to log-linearize there. Moreover, it appears that the authors are also discarding mixed terms for inflation, because $$\pi_t \pi_t \approx 0$$. So, for example, you get $$\Psi \pi_t(1+\pi_t) \approx \Psi \pi_t$$ Once you have implemented this approach for all terms in the whole equation, you can get the steady state relationship by setting all small caps to zero. You then can remove the steady state from the equation either by division or by multiplication (or both). What's left is a relationship in terms of deviation from steady state. In your example you will still have to find out why $$y_t$$ disappears and what $$\delta$$ stands for.