Page 1 of 1

Steady State of a Linear Model

PostPosted: Fri Jul 24, 2015 3:48 pm
by kgarb
Hi All,

I see repeatedly that it's optimal to calculate the analytical steady state by hand and then solve it numerically using a non-linear solver, but I also see here and there that linear models (measured in log deviations from the steady state) should be functional taking the initial steady state as being a vector of zeroes.

What could cause a linear model to not run when given a vector of zeroes as initial steady state? Example (using the system of equations from Beidas-Strom and Poghosyan, 2011) code attached.

Thanks!

Re: Steady State of a Linear Model

PostPosted: Fri Jul 24, 2015 5:46 pm
by jpfeifer
Mistakes in the equations.
Code: Select all
(1-caprhho)*betta*bstar=betta*istar+chii*bstar(-1)+chii*X(-1)
   +(cstar_Hx_steady)*chii*pr_H+chii*(delttas-ppi)+((p_XxeB)-betta)
   *X+((p_XxeB)-betta)*pr_H-((p_XxeB)*mX_steady)*M-((p_XxeB)*c_F/X)
   *rer-((p_XxeB)*oX_steady)*prstar_O;

has a division by X. Dynare also complains about
Code: Select all
((1+v_L*phhi_L+sigmma_L*eps_L*(phhi_L+v_L))/(1+sigmma_L*eps_L)*xi_L)*ppi+phhi_L*xi_L*ppi(-1)
   +v_L*ppi(-1)+zetta;

Re: Steady State of a Linear Model

PostPosted: Fri Jul 24, 2015 6:27 pm
by kgarb
Thanks for pointing those out!

Just to clarify though, using a vector of zeroes for the steady state should generally work given a log-linearized model without errors?

Re: Steady State of a Linear Model

PostPosted: Sun Jul 26, 2015 2:07 pm
by jpfeifer
Yes. Linearity is not equal to 0 steady states (there may be a constant term as in the Gali mod-files on my hompage). But given linearity, 0 starting values should always succeed in finding the steady state.