Dynamic factor model with 2 factors and mixed data

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Bablowski
Posts: 18
Joined: Tue Aug 25, 2015 1:08 am

Dynamic factor model with 2 factors and mixed data

Postby Bablowski » Sat Jan 14, 2017 1:13 am

Hello,

I am beginner in estimation of such a model.
I want to estimate dynamic factor model with 2 unobserved factors where I have 1 quarterly variable "hph" (my object is its forecast) and 6 monthly variables (dtr, ipp, idp, sid, spi, tpc). I am a little confused how to specify the model because in literature there are examples of only 1 factor case.
Quarterly variable is known every third month.

I have the following specification and I would like to ask you to check it whether it is good or how to respecify it to obtain some reasonable results.

@signal dtr = c(1)*f1+c(8)*f2+ [var = exp(c(15))]
@signal ipp = c(2)*f1+c(9)*f2+ [var = exp(c(16))]
@signal idp = c(3)*f1+c(10)*f2+ [var = exp(c(17))]
@signal sid = c(4)*f1+c(11)*f2+[var = exp(c(18))]
@signal spi = c(5)*f1+c(12)*f2+[var = exp(c(19))]
@signal tpc = c(6)*f1+c(13)*f2+[var = exp(c(20))]
@signal hph = c(7)*f1+c(14)*f2 +[var = exp(c(21))]

@state f1 = c(22)*f1(-1) + [var=exp(c(24))]
@state f2 = c(23)*f2(-1) + [var=exp(c(25))]


EViews gives me the following statements:
Failure to improve likelihood (non-zero gradients) after 299 iterations
Coefficient covariance computed using outer product of gradients
WARNING: Singular covariance - coefficients are not unique

All std. errors, z-statistics and p-values are NA

Where is the mistake?

My last question is how to set the starting values for the coefficients?

Thank you for any comments

Krille
Posts: 9
Joined: Tue Dec 13, 2016 11:16 am

Re: Dynamic factor model with 2 factors and mixed data

Postby Krille » Sat Jan 14, 2017 2:04 pm

Looks like an optimization gone numerically wrong.

I have few suggestions that might (or might not) work.

- Setting priors for state values and covariances can help and can be done by:
@mprior v1
@vprior m1

- Initial parameter values can be set (at least for some objects, maybe also the state space object) with
param c(1) 153 c(2) .68 c(3) .15

- Finally, using another form of the variance of disturbances have been proven to work for me. E.g. you could try
[var = c(15)^2] instead of [var = exp(c(15))]


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