Identification Analysis Error

Dynare help (legacy posts)
1

A suggested in Dynare reference manual I am trying to trigger an identification analysis in a Bayesian estimation environment.

[quote]For example, the placing identification; dynare_sensitivity(identification=1,
morris=2); in the Dynare model file trigger identification analysis[/quote]

WHat I get is this error:

[code]Testing current parameter values
??? Attempted to access x(1,13); index out of bounds because size(x)=[1,12].

Error in ==> hms_est19_dynamic at 42
T159 = y(19)-T15*y(21)-y(25)*params(35)/params(32)-x(it_, 13)*params(30)/params(32);

Error in ==> identification_analysis at 77
[residual, g1 ] = feval([M_.fname,’_dynamic’],yy0, …

Error in ==> dynare_identification at 280
[idehess_point, idemoments_point, idemodel_point, idelre_point, derivatives_info_point, info] =
…

Error in ==> hms_est19 at 624
dynare_identification(options_ident);

Error in ==> dynare at 180
evalin(‘base’,fname) ;[/code]

I am writing the command right before the estimation command:

[code]estimated_params;
…

end;

identification;
dynare_sensitivity(identification=1, morris=2);

estimation(datafile…)

[/code]

Maybe I am just putting the command in the wrong position.

Thank you in advance.

2

Please post or send the mod-file.

3

Thank you, just sent.

4

Identification is not compatible with

Why do you need this?

5

I wrote this code: (varexo_det=g) at the very beginning of my use of dynare. I thought it was the way to set non-stochastic exogenous variables in the model. In particular g is the exogenous government expenditure and is exogenous and costant over time (e.g 0.4) in the model. However I cannot use it as a parameter since it enters in the model as a variable.

Can you suggest me the proper way of setting exogenous, constant variable in a linearized model? If I put the variables between exogenous I end up with a stochastic variable and is not what I need. What I need, in my understanding, is a deterministic exogenous variable this is why I used that command.

Thank you once again

6

I am not sure I understand the point. If it is constant (no time index), it is a parameter, not a variable. Just declare

parameters g,gf; g=0.4; gf=0.4;
and your code should run. Unfortunately, identification does not support estimated correlations unless explicitly specified via e.g. a common shock.

7

Unfortunately it is a variable but exogenous, in the sense that the specified value of public consumption does not come from an optimization since government is not in the model. Households and firms just take the exogenous g value, that could change, but like in deterministic simulation my idea was to give it a certain value (costant) for t that goes from 0 to inifinity.

Since as far I understand this is not possible in a estimation environment, I guess the best choice would be to threat the variable like we normally do with technology stochastic process.

g=rhog*g(-1) + eg or g=rhog*g(-1) +(1-rhog)*g_ss + eg wheire g_ss is the steady state value, in my case equal to 0.40.

Do you think this would make sense? Both option gives me good identification results but the second gives me (strange) different from zero steady states results.

I know I am asking a lot… so feel free to let me alone in this puzzling situation.

best

8

I see. I understood that g is invariant over time. But you are saying it is a deterministic trend.
How about this: If you are not estimating any parameters associated with g (purely exogenous) you can take the identification results for your model with g set as a parameter. It won’t change any results in a meaningful way. Of course for estimation/simulation, it matters that g is an exogenous determinstic variable, but here you won’t need to run identification.

9

Perfect, thank you very much, again.