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can somebody help me understand how it is possible to log-linearize this equation? $$\frac{1}{1+i_{t}}=\beta E^i_{t}\left[\frac{P_{t}}{P_{t+1}} \cdot \frac{U_{c}\left(C_{t+1}^{i}, \xi_{t+1}\right)}{U_{c}\left(C_{t}^{i} ; \xi_{t}\right)}\right]$$

The author of the paper (Bruce Preston, 2006) is following Woodford for the microfundations, but is assuming heterogeneous expectations. He says he log-linearizes the equation around the steady state and gets $$\hat{C}_{t}^{i}=\hat{E}_{t}^{i} \hat{C}_{t+1}^{i}-\sigma\left(\hat{i}_{t}-\hat{E}_{t}^{i} \hat{\pi}_{t+1}\right)+g_{t}-\hat{E}_{t}^{i} g_{t+1}$$ without further explanations.

For a generic variable $\hat{z}_{t} \equiv \log \left(z_{t} / \bar{z}\right)$ . $\bar{z}$ represents the steady state .

Furthermore he sassumes that in the steady state $\xi_{t}=0$, $Y_{t}=\bar{Y}$ and $\pi_{t}=P_{t} / P_{t-1}=1$. The steady state interest $\bar{i}_{t}=\beta^{-1}-1$$=\log [(1+i) /(1+\bar{\imath})]$ .

I tried to linearize the term $U_c(c_t,\xi_t)$ around the steady state: $$U_c(c_t,\xi_t) \approx log(\frac{c_t}{\bar{c}})* \frac{U_{cc}(\bar{c},\bar{\xi})\bar{c}}{U_{c}(\bar{c},\bar{\xi})} +log(\frac{\xi_t}{\bar{\xi}})* \frac{U_{c\xi}(\bar{c},\bar{\xi})\bar{\xi}}{U_{c}(\bar{c},\bar{\xi})}=$$

$$=\hat{c}* \frac{U_{cc}(\bar{c},\bar{\xi})\bar{c}}{U_{c}(\bar{c},\bar{\xi})} +\hat{\xi}* \frac{U_{c\xi}(\bar{c},\bar{\xi})\bar{\xi}}{U_{c}(\bar{c},\bar{\xi})}=$$

$$=\hat{c}* \frac{U_{cc}(\bar{c},\bar{\xi})\bar{c}}{U_{c}(\bar{c},\bar{\xi})} $$

$\hat{\xi}* \frac{U_{c\xi}(\bar{c},\bar{\xi})\bar{\xi}}{U_{c}(\bar{c},\bar{\xi})}$ is zero because $\bar{\xi}=0$

However substituting back in the original equation and log-linearizing the rest i can't reach the same solution as the author.

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    $\begingroup$ There are many functions in these equations. Also, your question is not clear. Do you want someone to take the logarithm of a formula for you? $\endgroup$
    – Giskard
    Commented Jun 9, 2020 at 10:24
  • $\begingroup$ you are right, i have updated the question with what should be the final result. However there is no explanation on how the derivtion of eq.(3) is done except from saying that it is a log-linearization. $\endgroup$
    – bdvse
    Commented Jun 9, 2020 at 10:41
  • $\begingroup$ @bdvse please do not just post pictures of equations, this site allows you to properly typeset them. Also see the sites policy on good questions and homework questions, you should first try to answer problem yourself economics.meta.stackexchange.com/questions/1465/… $\endgroup$
    – 1muflon1 ♦
    Commented Jun 9, 2020 at 11:10
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    $\begingroup$ yes, sorry. I'm new to this website. I have edited the question $\endgroup$
    – bdvse
    Commented Jun 9, 2020 at 12:04

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