AR(1) process of shocks


Hi all,
For the AR(1) process of shocks in DSGE model, should we put also the constant term into the AR(1) equation, for example:
lnA = (1 - rhoA) + rhoA* lnA(-1) + stdeA*eA (1)

Or it is fine to just use: lnA = rhoA* lnA(-1) + stdeA*eA (2)

The paper I am trying to replicate using the 1st way (with constant term) but I see almost everyone use the 2nd one (without constant term). So I wonder could I just ignore the constant term to proceed, since it more popular and more convenient?



The two models are different. In the first case the unconditional expectation of \log A_t is equal to one (which is weird because it means also that the deterministic steady state of A is the Euler number e, while in the second equation the unconditional expectation is 0 (ie the deterministic steady state of A is one). If you ignore the constant, you will change the steady state of the model.



Hi Stephane,
Thanks for your reply. I just checked. So the original model is in level form, not log form,
which is:
A = (1 - rhoA) + rhoA* A(-1) + stdeA*eA (1)
So I will have steady-state A = 1, the same as the above 2nd equation.

So, in this case, are two expression equivalent?

Thank you,


Variables A will then have the same deterministic steady state in both equation, still the unconditional expectation and variance of A will be different. Also by construction A is necessarily positive in the second equation but not with the first equation. This is not a problem if you do a first order approximation of the model, but could be an issue if you simulate the model with the extended path approach.



Let me stress the last point by @stepan-a. If your model is solved at first order, the linear level version
A_t = (1 - \rho_A) + \rho_A A_{t-1} + \sigma_A \varepsilon_A
and the log-normal version
\log A_t = \rho_A \log A_{t-1} + \sigma_A \varepsilon_A
are equivalent. At higher order or for nonlinear solution approaches, there will be differences due to Jensen’s Inequality.