<p>This is because for small values <span class="math-container">$x$</span>, <span class="math-container">$$\ln x_{t+1} - \ln x_{t} \approx \frac{x_{t+1}-x_{t}}{x_{t}}.$$</span></p>
<p>This holds since the growth rate <span class="math-container">$g$</span> can be expressed as follows:</p>
<p><span class="math-container">$$g= \frac{x_{t+1}-x_{t}}{x_{t}} \implies x_{t+1} = (1+g)x_{t}$$</span></p>
<p>taking logs we get that:</p>
<p><span class="math-container">$$ \ln x_{t+1} = \ln (1+g)+ \ln x_{t} \implies \\ \ln x_{t+1} -\ln x_{t} = \ln (1+g) $$</span></p>
<p>Finally, for small values of <span class="math-container">$g$</span> we get that <span class="math-container">$\ln (1+g) \approx g$</span> thus we get:</p>
<p><span class="math-container">$$\ln x_{t+1} -\ln x_{t} \approx g $$</span></p>
<p>By the same token for small values of inflation (and inflation typically will take values <span class="math-container">$\pi&lt;0.1$</span>) it is completely reasonable to just define inflation as <span class="math-container">$\pi = \ln P_t - \ln P_{t-1}$</span> (although it would be more appropriate to use <span class="math-container">$\approx$</span>).</p>

CC BY-SA 4.0
2021-04-05T23:34:03.803Z
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, $\ln x_{t+1} -\ln x_{t} = g?$
Verónica Rmz.
CC BY-SA 4.0
2021-04-05T23:45:17.190Z
@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
1muflon1
CC BY-SA 4.0
2021-04-05T23:55:20.220Z
oh ok. For a moment it seemed like they were 2 different approaches.
Verónica Rmz.
CC BY-SA 4.0
2021-04-05T22:34:25.353Z
2021-04-05T22:34:25.353Z
1muflon1
<p>By definition the  inflation rate is <span class="math-container">$$\pi=\dfrac{P-P_{-1}}{P_{-1}}\cdot100\%$$</span> 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.</p>
<p>From the book Macroeconomia by Blanchard, Amighini, and Giavazzi,
the aggregate supply equation is <span class="math-container">\begin{equation}
P=P^e(1+\mu)F(u^-,z^+)
\end{equation}</span>
From here, and taking a specific <span class="math-container">$F:$</span></p>
<p><span class="math-container">\begin{align}
    P&amp;=P^e(1+\mu)e^{-\alpha u+z} \nonumber\\
    \ln(P)&amp;=\ln(P^e)+\ln(1+\mu)-\alpha u+z  \nonumber\\
    \ln(P)-\ln(P_{-1})&amp;=\ln(P^e)-\ln(P_{-1})+\ln(1+\mu)-\alpha u+z \nonumber\\
     \ln(P)-\ln(P_{-1})&amp;=\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}</span></p>
<p>Thus <span class="math-container">\begin{align}\pi=\pi^e+(\mu+z)-\alpha u \end{align}</span></p>
<p>Why? For example how to pass from  <span class="math-container">$\ln(P)-\ln(P_{-1})$</span> to <span class="math-container">$\pi$</span>?</p>
<p>We know that <span class="math-container">$\ln(P)-\ln(P_{-1})=\ln\dfrac{P}{P_{-1}}\fbox {=?}\dfrac{P}{P_{-1}}-1$</span></p>
<p>Thank you in advance</p>
<hr />

<macroeconomics><mathematical-economics><inflation><price><aggregate-supply>
From the aggregate supply equation to the definition of inflation rate
Verónica Rmz.
# From the aggregate supply equation to the definition of inflation rate

(Asked by: username_0 on 2021-04-05)

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

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**Tags:** macroeconomics, mathematical-economics, inflation, price, aggregate-supply

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## 1 answers

### Answer by username_1 on 2021-04-05. Score: 1 | Accepted answer

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 $\approx$).

> **Comments:**
> **username_0**: 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, $\ln x_{t+1} -\ln x_{t} = g?$  
> **username_1**: @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  
> **username_0**: oh ok. For a moment it seemed like they were 2 different approaches.  

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