By definition the inflation rate is $$\pi=\dfrac{P-P_{-1}}{P_{-1}}\cdot100\%$$ or could be defined in terms of the consumer price index CPI, but in this case I think the former is the one to consider.

From the book Macroeconomia by Blanchard, Amighini, and Giavazzi, the aggregate supply equation is \begin{equation} P=P^e(1+\mu)F(u^-,z^+) \end{equation} From here, and taking a specific $F:$

\begin{align} P&=P^e(1+\mu)e^{-\alpha u+z} \nonumber\\ \ln(P)&=\ln(P^e)+\ln(1+\mu)-\alpha u+z \nonumber\\ \ln(P)-\ln(P_{-1})&=\ln(P^e)-\ln(P_{-1})+\ln(1+\mu)-\alpha u+z \nonumber\\ \ln(P)-\ln(P_{-1})&=\ln(P^e)-\ln(P_{-1})+\mu-\alpha u+z\dots \text{due to $\ln(1+\mu)\approx\mu$ because $\mu$ is close to $0$ } \nonumber \end{align}

Thus \begin{align}\pi=\pi^e+(\mu+z)-\alpha u \end{align}

Why? For example how to pass from $\ln(P)-\ln(P_{-1})$ to $\pi$?

We know that $\ln(P)-\ln(P_{-1})=\ln\dfrac{P}{P_{-1}}\fbox {=?}\dfrac{P}{P_{-1}}-1$

Thank you in advance


1 Answer 1


This is because for small values $x$, $$\ln x_{t+1} - \ln x_{t} \approx \frac{x_{t+1}-x_{t}}{x_{t}}.$$

This holds since the growth rate $g$ can be expressed as follows:

$$g= \frac{x_{t+1}-x_{t}}{x_{t}} \implies x_{t+1} = (1+g)x_{t}$$

taking logs we get that:

$$ \ln x_{t+1} = \ln (1+g)+ \ln x_{t} \implies \\ \ln x_{t+1} -\ln x_{t} = \ln (1+g) $$

Finally, for small values of $g$ we get that $\ln (1+g) \approx g$ thus we get:

$$\ln x_{t+1} -\ln x_{t} \approx g $$

By the same token for small values of inflation (and inflation typically will take values $\pi<0.1$) it is completely reasonable to just define inflation as $\pi = \ln P_t - \ln P_{t-1}$ (although it would be more appropriate to use $\approx$).

  • $\begingroup$ Thanks for your answer. So the expression $\ln x_{t+1} -\ln x_{t} \approx g$ can turn into an equality through 2 ways: 1. By considering $\ln (1+g) = g$ or 2. if $g$ is very small, $\ln x_{t+1} -\ln x_{t} = g?$ $\endgroup$ Commented Apr 5, 2021 at 23:34
  • $\begingroup$ @Verónica no those are not two different ways that is part of the derivation - the reason why $\ln x - \ln x_{-1} \approx g$ works is that $\ln (1+g) \approx g$. those are not two distinct approaches the latter expression is intermediate step in showing why it holds $\endgroup$
    – 1muflon1
    Commented Apr 5, 2021 at 23:45
  • $\begingroup$ oh ok. For a moment it seemed like they were 2 different approaches. $\endgroup$ Commented Apr 5, 2021 at 23:55

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