Disinflation in a log-linear model

In a non-linear model with non-linear Taylor rule:

We can simulate a perfect foresight model from an old steady-state (\pi^*_1=0.02) to a new steady-state (\pi^*_2=0.01). Thus a 50% drop in the inflation target.

In a log-linearised model with a linear Taylor rule:

i_t = (i^* - \phi \pi^*) + \phi \pi_t

I can do something like the following?

model;
pi_star = rho*pi_star(-1) + epsilon;

initval;
epsilon = 0;
end;
endval;
epsilon = (1-rho)*log(1 - 0.5);
end;


Thus, in period 1, inflation target, pi_star (\pi^*) decreases by 50% permanently. Sounds correct? Because in log-linearised models, typically there is no intercept in Taylor rule (in DYNARE). Thus, i_t = \phi \pi_t. Introducing ( (i^* - \phi \pi^*)) will cause problems? Or perhaps, disinflation simulations can only be done in a non-linear model? Thanks!

Yes, that should work as epsilon is just the constant of the inflation target process. However, your numerical values for epsilon look strange.

You mean this, right?

epsilon = (1-rho)*log(1 - 0.5);


It should be the following?

epsilon = - (1-rho)*log(1 + 0.5);


No, I am saying that now you are setting pi_star to \log(1-0.5)\approx -0.6931, which is not what you outlined.

Oh yeah, thanks! If I want a drop from π^∗_1=0.02 in an old steady-state to π^∗_2=0.01 in a new steady-state, maybe I don’t even need AR(1) process for π^∗. I can just do

model;
pi_star = epsilon;

initval;
epsilon = 0.02;
end;

Be careful. You need to keep in mind that you linearized your model about a particular steady state already. So epsilon=0 will correspond to that steady state.