<p><br />
Your second step, <span class="math-container">$X^b e^{b \hat{X}_t} \big( \approx X^b (1 + b \hat{X}_t) \big)$</span>, is wrong here.</p>
<p>Specifically, you linearized the first term in RHS, <span class="math-container">$(C_t - b C_{t-1})^{-\sigma}$</span>, as
<span class="math-container">\begin{equation}
(C - b C)^{-\sigma} e^{ -\sigma (\hat{C}_t - b \hat{C}_{t-1})} \approx (C - b C)^{-\sigma} \big(1 -\sigma (\hat{C}_t - b \hat{C}_{t-1})\big)
\end{equation}</span>
Since the value <span class="math-container">$1$</span> here just works as a term which will be canceled out with the steady-state value in both LHS and RHS, let's ignore it. Then what you get here is <span class="math-container">$(C - b C)^{-\sigma} (-\sigma)(\hat{C}_t - b \hat{C}_{t-1})$</span>.</p>
<p>However, a correct one should be
<span class="math-container">\begin{equation}
(C - b C)^{-\sigma} (-\sigma)\Big(\frac{C}{C-bC} \hat{C}_t - \frac{bC}{C-bC} \hat{C}_{t-1}\Big).
\end{equation}</span>
That is, spare the details, you need additional terms which capture weights between <span class="math-container">$C_t$</span> and <span class="math-container">$C_{t-1}$</span>. <br />
What you've done is log-linearization with respect to, say <span class="math-container">$\gamma_t \equiv (C_t - bC_{t-1})$</span>, not with respect to each <span class="math-container">$C_t$</span> and <span class="math-container">$C_{t-1}$</span>.</p>
<p>I have no idea where you get that process, but applying it to some terms like this will give you wrong output.
When you simply apply the ``<span class="math-container">$X^b e^{b \hat{X}_t}$</span> rule'' into something like <span class="math-container">$Y_t = A_t K_t^{\alpha}$</span> or <span class="math-container">$Y_t = C_t + I_t$</span>, it seems to work fine because you don't need to consider the weights.
But if it becomes, e.g. <span class="math-container">$Z_t = (a_x X_t + a_y Y_t)^{b}$</span>, then you should be careful about it.</p>

CC BY-SA 4.0
2023-10-17T05:18:46.210Z
Thanks a lot! I also tried to derive it again with your solution in mind and actually figured out the true cause of the mistake. It is because I did not treat the steady states of $C_{t-1}$, $C_t$, and $C_{t+1}$ as separate variables. When you take the derivative of the whole term $[(1-b)C]^{-\sigma}$ the $(1-b)$ dropped out when I did the Taylor Approximation. Nevertheless, huge thanks again!
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CC BY-SA 4.0
2023-10-16T00:05:17.390Z
2023-10-16T00:05:17.390Z
JiYong Jung
<p>I am trying to derive the log-linearized Euler Equation (EE) of a Model with additive Habit formation. Attached you will find my attempt to derive the EE. I am missing an (1-b) in the denominator. Where is my mistake? I checked it several times and multiple lecture slides or research papers but I could not find a detailed derivation of the EE. Thanks in advance!</p>
<p><a href="https://i.sstatic.net/suaDw.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/suaDw.jpg" alt="enter image description here" /></a></p>

<macroeconomics><new-keynesian-economics><log-linearization>
Log-linearization of the additive habit formation model
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# Log-linearization of the additive habit formation model

(Asked by: username_0 on 2023-10-15)

I am trying to derive the log-linearized Euler Equation (EE) of a Model with additive Habit formation. Attached you will find my attempt to derive the EE. I am missing an (1-b) in the denominator. Where is my mistake? I checked it several times and multiple lecture slides or research papers but I could not find a detailed derivation of the EE. Thanks in advance!

[![enter image description here](https://i.sstatic.net/suaDw.jpg)](https://i.sstatic.net/suaDw.jpg)

**Tags:** macroeconomics, new-keynesian-economics, log-linearization

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## 1 answers

### Answer by username_1 on 2023-10-16. Score: 1 | Accepted answer

Your second step, $X^b e^{b \hat{X}\_t} \big( \approx X^b (1 + b \hat{X}\_t) \big)$, is wrong here.

Specifically, you linearized the first term in RHS, $(C\_t - b C\_{t-1})^{-\sigma}$, as
\begin{equation}
(C - b C)^{-\sigma} e^{ -\sigma (\hat{C}\_t - b \hat{C}\_{t-1})} \approx (C - b C)^{-\sigma} \big(1 -\sigma (\hat{C}\_t - b \hat{C}\_{t-1})\big)
\end{equation}
Since the value $1$ here just works as a term which will be canceled out with the steady-state value in both LHS and RHS, let's ignore it. Then what you get here is $(C - b C)^{-\sigma} (-\sigma)(\hat{C}\_t - b \hat{C}\_{t-1})$.

However, a correct one should be
\begin{equation}
(C - b C)^{-\sigma} (-\sigma)\Big(\frac{C}{C-bC} \hat{C}\_t - \frac{bC}{C-bC} \hat{C}\_{t-1}\Big).
\end{equation}
That is, spare the details, you need additional terms which capture weights between $C\_t$ and $C\_{t-1}$.   
What you've done is log-linearization with respect to, say $\gamma\_t \equiv (C\_t - bC\_{t-1})$, not with respect to each $C\_t$ and $C\_{t-1}$.

I have no idea where you get that process, but applying it to some terms like this will give you wrong output.
When you simply apply the ``$X^b e^{b \hat{X}\_t}$ rule'' into something like $Y\_t = A\_t K\_t^{\alpha}$ or $Y\_t = C\_t + I\_t$, it seems to work fine because you don't need to consider the weights.
But if it becomes, e.g. $Z\_t = (a\_x X\_t + a\_y Y\_t)^{b}$, then you should be careful about it.

> **Comments:**
> **username_0**: Thanks a lot! I also tried to derive it again with your solution in mind and actually figured out the true cause of the mistake. It is because I did not treat the steady states of $C_{t-1}$, $C_t$, and $C_{t+1}$ as separate variables. When you take the derivative of the whole term $[(1-b)C]^{-\sigma}$ the $(1-b)$ dropped out when I did the Taylor Approximation. Nevertheless, huge thanks again!  

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