jpfeifer wrote:1. It has to do with the way those shocks enter. For example, ev moves Ev. But Ev enters no equation, because
- Code: Select all
log(r/RT)+1=FIP*log(dcpi(+1)/DT)+FIY*log(y/YT)+EV; //18yes
only uses the steady state. Seems you confused yourself by distinguishing three distinct concepts by just a different capitalization.
2. In general, the IRFs are not meaningless, because they are a summary of the model dynamics. If there are problems with them, it signifies problems with the model that may create problems for welfare analysis.
1.Definitely, I am very stupid to make this mistake.
2.The user guide says, "If you linearize your model up to a first order, impulse response functions are simply the algebraic forward iteration of your model's policy or decision rule. If you instead linearize to a second order, impulse response functions will be the result of actual Monte Carlo simulations of future shocks. This is because in second order linear equations, you will have cross terms involving the shocks, so that the effects of the shocks depend on the state of the system when the shocks hit. Thus,
it is impossible to get algebraic average values of all future shocks and their impact. The technique is instead to pull future shocks from their distribution and see how they impact your system, and repeat this procedure a multitude of times in order to draw out an average response. That said,
note that future shocks will not have a significant impact on your results, since they get averaged between each Monte Carlo trial and in the limit should sum to zero, given their mean of zero."
so If I linearize the model to a second order, Can I continue to analysis the IRFS? "that future shocks will not have a significant impact on your results", Does that mean that the results is not so significant as the first order, but it is also meaningful?
Thank you!