0
$\begingroup$

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

$\endgroup$

1 Answer 1

1
$\begingroup$


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.

$\endgroup$
1
  • $\begingroup$ 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

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.