Ask for help: two log-linearization problems

Dynare help (legacy posts)
1

Hello, everyone.

I have two questions about the log-linearization of some conditions in DSGE model. I need some confirmations here.

If anybody can offer help, I’d really appreciate it.

I have posted the questions in the attachement.

Thank you all and Merry Christmas!
Two questions about the log.doc (70.5 KB)

2
  1. The coefficient in front of the preference shock essentially is a rescaling of the standard deviation. As the standard deviation of this shock is estimated, leaving out this factor means that what is estimated is

instead of just

As people are often just interested in the whole value and not the individual components of the variance, often the former is used.
2. Equation 5 is correct. A Taylor approximation of

will preserve the plus sign.

3

[quote=“jpfeifer”]1. The coefficient in front of the preference shock essentially is a rescaling of the standard deviation. As the standard deviation of this shock is estimated, leaving out this factor means that what is estimated is

(1-h)/sigma*sigma_pref

instead of just

sigma_pref

As people are often just interested in the whole value and not the individual components of the variance, often the former is used.
2. Equation 5 is correct. A Taylor approximation of

X+Y=0

will preserve the plus sign.[/quote]

Thank you very much for your reply, jpfeifer.

Thanks to your explaination, the first problem is solved.

however, I still have some questions about the second problem. I have posted my deductions in the attachement. Would you please help to have a look at it? It really bothers me a lot, although appears to be very simple.

Thank you in advance.
the log problem of x+y=0.doc (44.5 KB)

4

My mistake. In a sense, you are correct. The problem comes from doing a log-linearization of a variable with negative steady state. In that case, you should not perform a log-linearization. To see the problem, consider your example of X_t being -4 with the steady state being -3. In this case:
(X_t-X)/X(-4-(-3))/(-3)=1/3

This says that X is 1/3 above its steady state, but -4 is clearly below -3. That is why people redefine the variable to include the minus.

5

[quote=“jpfeifer”]My mistake. In a sense, you are correct. The problem comes from doing a log-linearization of a variable with negative steady state. In that case, you should not perform a log-linearization. To see the problem, consider your example of X_t being -4 with the steady state being -3. In this case:
(X_t-X)/X(-4-(-3))/(-3)=1/3

This says that X is 1/3 above its steady state, but -4 is clearly below -3. That is why people redefine the variable to include the minus.[/quote]

Dear jpfeifer,

Thank you very very much!

You know, this question appears a bit weird since both Y(t)+X(t)=0 and Y(t)-X(t)=0 finally give the exactly same log-linear eqution: x(t)=y(t).

That’s why I could not believe at the first sight when I deducted the result. That’s also why I really need someone to help me confirm it: I kept worring that there might be some mistakes in my deductions.

Thank you again and best wishes for the new coming year.