This may be a weird question, but I am unfortunately confused by terms. Let us suppose log-linearized New Keynesian model as suggested by Gali here: http://crei.cat/people/gali/pdf_files/monograph/slides-ch3.pdf

My first question is, apparently steady value $Y$ is assumed to do log-linearization of $Y_t$, output, but is this $Y$ a constant value, or the whole steady output path? Equivalently, is $Y$ about how output will evolve if $Y_t$ evolves without stochastic factors and errors according to long-run natural rate?

My second question, related to the first question, is whether $Y_t$ refers to total output or normalized output. That is, if economy has positive output growth rate, will $Y_t$ grow? Or is it normalized output that does not change without stochastic elements?

My third question is, what $y_t$ actually mean. As I understand it, it is just $\log Y_t$. Is this correct?

The fact that consumption euler equation exists seems to support the intuition that $Y$ is steady output path, not constant steady value, as real interest rate is often positive for economy. My confusion all rises from here, and I am not sure if this is correct understanding.


3 Answers 3


The following post explains in a somewhat easier way what exactly is happening when we log linearize a model.


Going through the provided example should make it clear what the single steps are.


The log-linearization is performed in the neighborhood of a steady state with zero inflation, constant output, and constant markups over marginal cost, as stated on slide 11 of the Galí presentation you link. Hence $Y$ is indeed intended to be a constant value, the steady state level of output around which the log-linearization is performed. $Y_t$ is just the level of total output in period $t$, while $y_t=\log Y_t$ is the log value of total output, as you say.

Several additional points that seem to be relevant here:

  • This derivation of the basic New Keynesian model is performed under the assumption that there is no steady-state trend output growth. We can only be sure that the log-linearized equations are approximately correct for situations where any deviation from this zero-growth steady state is sufficiently small. Obviously, since we live in a world with conspicuously positive trend growth, this is potentially a problem - so this is a very valid concern on your part.
  • As it happens, I believe that the equations are very similar when we log-linearize around a steady state with positive trend productivity growth (but maintaining the zero trend inflation assumption). In particular, when stated in terms of the output gap and natural rate of interest as in Galí's equation (10), the intertemporal Euler equation is exactly the same (though note that the steady state natural rate $r^n=\rho+\sigma \psi_{ya}g_a$ is higher, where $g_a$ is the log trend growth rate of productivity). The New Keynesian Phillips curve is a little messier: in the log preferences case $\sigma=1$, there are several nice cancellations and we obtain exactly the same NKPC, but for other $\sigma$ the discount rate on future inflation is no longer $\beta$. This is all much more annoying to deal with, though, which is why Galí avoided it for the simple exposition and stuck with the zero-growth steady state.
  • As mentioned above, neither $Y$, $Y_t$, nor $y_t$ are "normalized output" of any kind. However, the output gap $\tilde{y}_t\equiv y_t-y_t^n$ defined in Galí's equation (7) is effectively normalized log output, subtracting off the log "natural output" $y_t^n$ that we'd expect in a flexible-price world given log productivity $a_t$. In this sense, the model can accommodate fluctuations in productivity; but as stated above, if these fluctuations are too large, then the log-linearization around zero trend growth begins to break down, and if $\sigma\neq 1$ we have to rewrite the NKPC in a different form to account for this.
  • Finally, I'm a little confused by the last paragraph, but you seem to imply that the model may have a positive growth rate, "as real interest rate is often positive for economy". This is a misapprehension: the steady state real interest rate in this model is positive because agents in the model have pure time preference, with a discount rate $\beta<1$. If you look below Galí's equation (10), you see that when there is no productivity change the "natural" real interest rate is $r_t^n = \rho$, where $\rho=-\log \beta$.
  • $\begingroup$ Are you sure $y_t$ is $\log Y_t$? I thought it was usually the percent deviation. $\endgroup$
    – cc7768
    Commented Dec 17, 2014 at 1:32
  • $\begingroup$ Yes, here Galí means $y_t=\log Y_t$, not $y_t=\log Y_t-\log Y$. In this case, the latter would be superfluous because he subtracts by the "natural rate of output" $y_t^n$ anyway to obtain the output gap $\tilde{y}_t$. More generally, I've seen lowercase letters used both ways, sometimes for logs and sometimes for log deviations from steady state (when it's the former, you usually add a hat or something for the latter). Even Galí doesn't use a consistent convention, though when deriving the NK model on page 66 of his text he says "lowercase letters denote the logs of the original variable". $\endgroup$ Commented Dec 17, 2014 at 1:40

Full disclosure: I haven't read through the lecture notes that your provided particularly carefully, but I think I can answer your question.

Edit: Heads up, by not carefully reading the link provided by the question, I missed something.

The standard New Keynesian models (such as the one Gali presented) are modeled without growth. If you write down the model then you can represent it as a difference equation:

$$0 = E_t \left[ F(X_{t+1}, X_t, X_{t-1}, Z_{t}) \right]$$

where $X_t$ contains all relevant variables and $Z_t$ represent the shocks to the economy. The "steady state" typically refers to the state of the world where $X_t$ is constant (think stable solution to a difference/differential equation) and $Z_t = 0$, thus you could write it as the solution to:

$$0 = F(X, X, X, 0)$$

in which case $X$ would be the steady state value (notice not time subscripts --sometimes also done by denoting steady state with overhead bars $\bar{X}$). This is what he is calling $Y$ and it is a constant value.

For the second question, I haven't read carefully, so I can't be 100% sure, but typically when a variable is written as $X_t$ it references the actual value that is taken (aka if you solved the model and simulated it exactly, this is the value it would have).

For the third question, I think a deeper understanding of log-linearization will answer it for you. Log-linearization at its heart is just a Taylor expansion around the steady state. Consider a generic equation $f(X_t, Y_t) = g(Z_t)$. There are 3 basic steps to log-linearization (refreshed my memory here).

  • Take logs
  • First Order Taylor Expansion
  • Algebra

We first take logs,

$$\ln(f(X_t, Y_t)) = \ln(g(Z_t))$$

If we do a First order Taylor expansion around the steady state, then we can write:

$$ \ln(f(X_t, Y_t)) \approx \ln(f(X, Y)) + \frac{f_x(X, Y)}{f(X, Y)} (X_t - X) + \frac{f_y(X, Y)}{f(X, Y)} (Y_t - Y)$$

$$ \ln(g(Z_t)) \approx \ln(g(Z)) + \frac{g_z(Z)}{g(Z)} (Z_t - Z)$$

Thus we can write:

$$\ln(f(X, Y)) + \frac{f_x(X, Y)}{f(X, Y)} (X_t - X) + \frac{f_y(X, Y)}{f(X, Y)} (Y_t - Y) \approx \ln(g(Z)) + \frac{g_z(Z)}{g(Z)} (Z_t - Z)$$

Recall that in the steady state $f(X, Y) = g(Z)$ and I will also multiply by one in several places ($\frac{X}{X}$ etc...), so

$$\frac{X f_x(X, Y)}{f(X, Y)} \frac{(X_t - X)}{X} + \frac{Y f_y(X, Y)}{f(X, Y)} \frac{(Y_t - Y)}{Y} \approx \frac{Z g_z(Z)}{g(Z)} \frac{(Z_t - Z)}{Z}$$

Now define $\hat{x_t} := \frac{(X_t - X)}{X}$, $\hat{y_t} = \frac{(Y_t - Y)}{Y}$, and $\hat{z_t} := \frac{(Z_t - Z)}{Z}$. This is the percentage deviation of $X_t$ from $X$ (and correspondingly for $Y_t$ and $Z_t$). Then you can write the log-linearized equation as:

$$\frac{X f_x(X, Y)}{f(X, Y)} \hat{x_t} + \frac{Y f_y(X, Y)}{f(X, Y)} \hat{y_t} \approx \frac{Z g_z(Z)}{g(Z)} \hat{z_t}$$

Two final things. First, one subtlety that caught me off-guard the first time I was switching between percent deviation and true values and you might want to be aware of; values that aren't normally negative can be negative because it just means that it is that percentage below steady state. Secondly, functional forms usually make these simplify quite nicely as you have probably seen in the log-linearized equations presented.

In this example, Gali is using $y_t := \log Y_t$ as seen in the other answer, so hopefully this provides some intuition for what is happening elsewhere.

Hope this helped.

  • 1
    $\begingroup$ If you look at slide 7, you will see that $y_t$ is just log-output, not percentage deviation. You might want to adjust your answer accordingly to avoid confusion. $\endgroup$ Commented Dec 17, 2014 at 1:43
  • 1
    $\begingroup$ The money demand equation does look like it is just log-output, but then he plugs it directly into the log-linearized Euler equation then is $c_t = \log C_t$? I might just have his notation totally wrong and be stuck in bad habits (especially since I'm not particularly familiar with the NK models). Getting out pen and paper although I highly suspect that @AlecosPapadopoulos and nominally rigid are correct. Be back soon :O $\endgroup$
    – cc7768
    Commented Dec 17, 2014 at 2:00
  • $\begingroup$ I seriously congratulate your approach -"mistrust in authority" sometimes unearths treasures. Waiting on your pen and paper results (still my favorites). $\endgroup$ Commented Dec 17, 2014 at 2:09
  • $\begingroup$ No worries - both conventions are pretty common. I don't think that the money demand equation clinches it in either direction, since it is coherent to interpret the terms in that equation as being either logs or log deviations from steady state. $\endgroup$ Commented Dec 17, 2014 at 2:11
  • $\begingroup$ The one case where Galí is definitely using a lowercase variable as log rather than log deviation from steady state is $i_t$, which as Alecos mentions is defined as $i_t=-\log Q_t$ on slide 7; if this was instead defined as log deviation from steady state $i=\rho$ instead, then we wouldn't have the $\rho$ intercept in the intertemporal Euler equation. But there's still some ambiguity about $y_t$ simply because it doesn't really matter: the equations are true under both interpretations. $\endgroup$ Commented Dec 17, 2014 at 2:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.