Near singular matrix ad nauseam...

[codru]

Hi all,
I am hitting the 'near singular matrix' wall almost every time I try to run my models using 'system' object. I am attaching an example with the data and the specification I use (I should be using a var2 but let me just use a var1 as an example). Any suggestions and ideas would be very helpful! i am not sure if it is a bug with Eviews or with my modeling approach.
the system is ran for 1955-2015 and it is:

ws-ws(-1)=-(c(2)*c(100)+c(3)*c(101)+c(4)*c(102))+c(2)*ws(-1)+c(3)*ef(-1)+c(4)*em(-1)
ef-ws(-1)=-(c(12)*c(100)+c(13)*c(101)+c(14)*c(102))+c(12)*ws(-1)+c(13)*ef(-1)+c(14)*em(-1)
em-ws(-1)=-(c(22)*c(100)+c(23)*c(101)+c(24)*c(102))+c(22)*ws(-1)+c(23)*ef(-1)+c(24)*em(-1)

many thanks, codru.

[codru]

any ideas, anyone? Perhaps the Eviews moderators can chime in? As I said this issue has become a mainstay with estimations based on system objects and it greatly reduces the flexibility in using the software. I have been searching for a solution for a while now, with no luck. thank you, c.

[codru]

the mystery continues...a colleague of mine runs a different specification using the file I posted, he sends me back his Eviews file and playing around I run the model above (with a mistake in the 2nd and 3rd equation the dependent variables corrected) and no more near singular matrix error. I copy the beta vector from that file in my original file and voila it works again. am I dealing with instability, or multiple equilibria that Eviews can't handle? any guesses? I have checked the SSCP of the covariance matrix of the regrressors and there is no multicollinearity

[startz]

It is possible that the issue is the starting values in the c vector, although I don't see why that would be true in this case.

Note that these equations can be rewritten so as to be estimated by least squares. Then you will end up with three values for the intercepts which can be used to solve for c(100), c(101), c(102) as three linear equations in three unknowns.

[codru]

many thanks, startz, for your reply. Yes, you are correct that I can solve for the three coefficients from the intercepts and the rest, but how do I get their distribution then? For whatever is worth I am interested in whether they are stat significant as well.

[startz]

Suppose you've estimated the coefficients A and want the coefficients B, where B=LA. Here L is a 3 by 3 known matrix. The variance of B will be Rvar(A)R'.

[codru]

thank you!