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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!

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

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(1-h)/sigma*sigma_pref

instead of just

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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

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X+Y=0

will preserve the plus sign.

jpfeifer wrote: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

Code: Select all

(1-h)/sigma*sigma_pref

instead of just

Code: Select all

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

Code: Select all

X+Y=0

will preserve the plus sign.

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.

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.

jpfeifer wrote: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.

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.