IRFs, % deviation when model in exp(), order=1

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IRFs, % deviation when model in exp(), order=1

Postby Nice » Fri Jul 11, 2014 7:49 pm

Hi,

If I have expressed my model in logs and not levels, for e.g. in the preamble I declare a variable y and in the model block I express it as exp(y) and I solve the model using an approximation of order 1. Is the interpretation of the y axis of my IRF a percentage deviation from SS?

I also have debt in my model, which I do not express as exp because it can have a negative SS value. So is the IRF interpretation for debt just a deviation from SS?

I referred to https://sites.google.com/site/pfeiferec ... ations.pdf Remark 17 and am still confused.
I have attached my .mod file for reference.

Any help is much appreciated,

Nice
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ROT_Model_5b.mod
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Re: IRFs, % deviation when model in exp(), order=1

Postby jpfeifer » Mon Jul 14, 2014 2:09 pm

All variables in exp() will have the interpretation as percentage deviations from steady state.

Debt, if it is not in exp() will be in absolute deviations from steady state.
------------
Johannes Pfeifer
University of Cologne
https://sites.google.com/site/pfeiferecon/
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Re: IRFs, % deviation when model in exp(), order=1

Postby costa » Mon Jul 14, 2014 11:56 pm

I have a question along the same lines. I estimated a small open economy model , and since the steady state is different from 0, I created new variables to have the deviation from steady state interpretation in the impulse responses. Is this procedure wrong, considering the previous answer ?

Thanks in advance.
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Re: IRFs, % deviation when model in exp(), order=1

Postby jpfeifer » Tue Jul 15, 2014 7:51 am

IRFs are always relative to a baseline value. If the deviation from this baseline is in logs, it has the interpretation of percentage deviations. For IRFs, there is no reason to recenter everything as you do. However, subtracting the mean is not wrong.
------------
Johannes Pfeifer
University of Cologne
https://sites.google.com/site/pfeiferecon/
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