Rotemberg vs. Calvo

General DSGE Modeling
1

Rotemberg and Calvo are equivalent under first-order approximation (if the slope is equally calibrated) but does this also hold for the perfect foresight solution?
I have a model where this is true for the stochastic simulation and not for the perfect foresight, so I get different responses between Calvo and Rotemberg for my perfect foresight solution. But the irfs are identical for stochastic simulation order 1.

I am wondering if this right?

2

Yes, as you wrote above, the equivalence result holds at first order. At higher order they differ. That is, e.g., particularly relevant at the zero lower bound.

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