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!

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1 Answer 1


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.

  • $\begingroup$ 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! $\endgroup$
    – L_ST
    Commented Oct 17, 2023 at 5:18

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