Consider some time series data $X=\{x_t:t\in[0,\infty)\}$. Define a steady state by \begin{align} x\in X:x_{t+1} - x_t = 0 \end{align}
and the log deviation from steady state with \begin{align} \hat x_t:=\ln x_t - \ln_x. \end{align}
Now we can express $x_t$ as \begin{align} x_t=x\exp(\hat x_t). \end{align}
Consider the the following equation \begin{align} a_t = b_t c_t^\alpha + D\mathbb{E}_t\left[\left(\frac{e_{t+1}}{e_t}\right)\left(\frac{f_t}{g_t}\right)^\beta a_{t+1}\right] \end{align} where $D$ is some constant, which is ought to be log linearized for every time dependent variable around steady state.
First Attempt
Express every variable by its new definition,i.e. \begin{align} a\exp(\hat a_t) = b c^\alpha \exp(\hat b_t + \alpha\hat c_t)+ D\left(\frac{f}{g}\right)^\beta a\mathbb{E}_t\left[\exp(\hat e_{t+1} - \hat e_t + \beta(\hat f_t - \hat g_t) + \hat a_{t+1})\right] \end{align}
which should simplify to \begin{align} \exp(\hat a_t) = \exp(\hat b_t + \alpha\hat c_t)+\mathbb{E}_t\left[\exp(\hat e_{t+1} - \hat e_t + \beta(\hat f_t - \hat g_t) + \hat a_{t+1})\right] \end{align}
- Is this correct?
- What can be done to further improve/simplify the equation at hand?
If someone else would like to answer this using some of my answer (or just their own answer) then that would make my life easier, but if no one has answered it in the next day or so then I will sit down and write something.
– cc7768 Jul 17 '15 at 14:40