Hello,
I have obtained my solution to a problem, but I need someone to check whether what I am doing is correct. I need to log-linearize the following equation.
[math]\left(\frac{E_t}{A_t}\right)^{-\epsilon} = (\eta R_t^{-\epsilon} + \xi F_t^{-\epsilon})[/math]
First off, define [imath]z_t \equiv log(Z_t) - log(\bar{Z})[/imath] as the log-deviation of a variable from its steady state.
When log-linearizing sums, we had lecture notes that [imath]Z_t = Z_{1,t} + Z_{2,t}[/imath] becomes [imath]z_t = s_{Z_1}z_{1,t} + s_{Z_2}z_{2,t}[/imath] in log-linearized form, where [imath]s_{Z_1} = \frac{\bar{Z_1}}{\bar{Z_1} + \bar{Z_2}}[/imath] and [imath]s_{Z_2} = \frac{\bar{Z_2}}{\bar{Z_1} + \bar{Z_2}}[/imath]. As a first question, would someone be able to check whether the following derivation of the result above is correct:
Define the growth rate of a variable with respect to its steady state as [imath](1+g^x_t) \equiv \frac{X_t}{\bar{X}}[/imath] and therefore [imath]g^x_t = \frac{X_t - \bar{X}}{\bar{X}}[/imath].
Then we have:
[math]z_t = log(Z_t) - log(\bar{Z}) = log(\frac{Z_t}{\bar{Z}})=log(1+g^z_t)[/math]
When [imath]g^z_t[/imath] is small, then we can use the approximation [math]log(1+g^z_t) \approx g_t^z[/math]. Then we get,
[math]z_t \approx g_t^z = \frac{Z_t - \bar{Z}}{\bar{Z}}[/math].
And since we have that [imath]Z_t = Z_{1,t} + Z_{2,t}[/imath] and [imath]\bar{Z} = \bar{Z_1} + \bar{Z_2}[/imath], we get
[math]z_t \approx \frac{Z_t - \bar{Z}}{\bar{Z}} = \frac{(Z_{1,t} + Z_{2,t}) - (\bar{Z_1} +\bar{Z_2})}{\bar{Z_1} +\bar{Z_2}} = \frac{\bar{Z_1}}{\bar{Z_1} +\bar{Z_2}}\frac{Z_{1,t}-\bar{Z_1}}{\bar{Z_1}} + \frac{\bar{Z_2}}{\bar{Z_1} +\bar{Z_2}}\frac{Z_{2,t}-\bar{Z_2}}{\bar{Z_2}} \approx s_{Z_1}z_{1,t} + s_{Z_2}z_{2,t}[/math]
Now for the other question. Am I allowed to follow the following procedure for the left hand side:
[math]log\left(\left(\frac{E_t}{A_t}\right)^{-\epsilon}\right) = \epsilon log(A_t) - \epsilon log(E_t)[/math]
And then substract the same in steady state to obtain:
[math]\epsilon (log(A_t) - log(\bar{A})) - \epsilon(log(E_t) - log(\bar{E})) = \epsilon a_t - \epsilon e_t[/math]
And equate this to the right hand side, where the procedure for sums from above has been applied:
[math]\frac{\eta \bar{E}^{-\epsilon}}{\eta \bar{E}^{-\epsilon} + \xi \bar{F}^{-\epsilon}} \frac{\eta E_t^{-\epsilon} - \eta \bar{E}^{-\epsilon}}{\eta \bar{E}^{-\epsilon}} + \frac{\xi \bar{F}^{-\epsilon}}{\eta \bar{E}^{-\epsilon} + \xi \bar{F}^{-\epsilon}} \frac{\xi F_t^{-\epsilon} - \xi \bar{F}^{-\epsilon}}{\xi \bar{F}^{-\epsilon}} = s_E\eta e_t^{-\epsilon}+ s_F\xi f_t^{-\epsilon}[/math]
And ultimately write:
[math]\epsilon a_t - \epsilon e_t = s_E\eta e_t^{-\epsilon}+ s_F\xi f_t^{-\epsilon}[/math]
Is this correct?
This is my first time posting, I hope, I chose the correct forum and am happy to further elaborate if something is unclear. I would be very thankful for someone's help in checking this.
I have obtained my solution to a problem, but I need someone to check whether what I am doing is correct. I need to log-linearize the following equation.
[math]\left(\frac{E_t}{A_t}\right)^{-\epsilon} = (\eta R_t^{-\epsilon} + \xi F_t^{-\epsilon})[/math]
First off, define [imath]z_t \equiv log(Z_t) - log(\bar{Z})[/imath] as the log-deviation of a variable from its steady state.
When log-linearizing sums, we had lecture notes that [imath]Z_t = Z_{1,t} + Z_{2,t}[/imath] becomes [imath]z_t = s_{Z_1}z_{1,t} + s_{Z_2}z_{2,t}[/imath] in log-linearized form, where [imath]s_{Z_1} = \frac{\bar{Z_1}}{\bar{Z_1} + \bar{Z_2}}[/imath] and [imath]s_{Z_2} = \frac{\bar{Z_2}}{\bar{Z_1} + \bar{Z_2}}[/imath]. As a first question, would someone be able to check whether the following derivation of the result above is correct:
Define the growth rate of a variable with respect to its steady state as [imath](1+g^x_t) \equiv \frac{X_t}{\bar{X}}[/imath] and therefore [imath]g^x_t = \frac{X_t - \bar{X}}{\bar{X}}[/imath].
Then we have:
[math]z_t = log(Z_t) - log(\bar{Z}) = log(\frac{Z_t}{\bar{Z}})=log(1+g^z_t)[/math]
When [imath]g^z_t[/imath] is small, then we can use the approximation [math]log(1+g^z_t) \approx g_t^z[/math]. Then we get,
[math]z_t \approx g_t^z = \frac{Z_t - \bar{Z}}{\bar{Z}}[/math].
And since we have that [imath]Z_t = Z_{1,t} + Z_{2,t}[/imath] and [imath]\bar{Z} = \bar{Z_1} + \bar{Z_2}[/imath], we get
[math]z_t \approx \frac{Z_t - \bar{Z}}{\bar{Z}} = \frac{(Z_{1,t} + Z_{2,t}) - (\bar{Z_1} +\bar{Z_2})}{\bar{Z_1} +\bar{Z_2}} = \frac{\bar{Z_1}}{\bar{Z_1} +\bar{Z_2}}\frac{Z_{1,t}-\bar{Z_1}}{\bar{Z_1}} + \frac{\bar{Z_2}}{\bar{Z_1} +\bar{Z_2}}\frac{Z_{2,t}-\bar{Z_2}}{\bar{Z_2}} \approx s_{Z_1}z_{1,t} + s_{Z_2}z_{2,t}[/math]
Now for the other question. Am I allowed to follow the following procedure for the left hand side:
[math]log\left(\left(\frac{E_t}{A_t}\right)^{-\epsilon}\right) = \epsilon log(A_t) - \epsilon log(E_t)[/math]
And then substract the same in steady state to obtain:
[math]\epsilon (log(A_t) - log(\bar{A})) - \epsilon(log(E_t) - log(\bar{E})) = \epsilon a_t - \epsilon e_t[/math]
And equate this to the right hand side, where the procedure for sums from above has been applied:
[math]\frac{\eta \bar{E}^{-\epsilon}}{\eta \bar{E}^{-\epsilon} + \xi \bar{F}^{-\epsilon}} \frac{\eta E_t^{-\epsilon} - \eta \bar{E}^{-\epsilon}}{\eta \bar{E}^{-\epsilon}} + \frac{\xi \bar{F}^{-\epsilon}}{\eta \bar{E}^{-\epsilon} + \xi \bar{F}^{-\epsilon}} \frac{\xi F_t^{-\epsilon} - \xi \bar{F}^{-\epsilon}}{\xi \bar{F}^{-\epsilon}} = s_E\eta e_t^{-\epsilon}+ s_F\xi f_t^{-\epsilon}[/math]
And ultimately write:
[math]\epsilon a_t - \epsilon e_t = s_E\eta e_t^{-\epsilon}+ s_F\xi f_t^{-\epsilon}[/math]
Is this correct?
This is my first time posting, I hope, I chose the correct forum and am happy to further elaborate if something is unclear. I would be very thankful for someone's help in checking this.