Indeterminacy: source

Hi,
I am solving a var model and try to simulate it. The things are I have 7 variables is inside the unit circle while the other 4 is outside. But I have 8 given initial values. So the steady-state is a source. I did it manually. my questions is:

  1. Even if I do not have unique SS, can I still simulate the var model by plugging in initial values and using backward, forward approach?.
    Thanks.

Sorry, but I cannot follow that description. What exactly is your question? It seems you are having a model with an infinity of solutions.

Hi, thanks for your attention to my question.
Yes, I have 11 variables. For eigenvalues, i get 7 inside, 4 outside. Among 11 variables, 3 of them forward looking, other 8 is backward looking. Besides, i have 8 initial given conditions.
So,the steady state turns out to be source. My question is:
Even if steady state is source, can i simulate and generate data? is simulation only works for those who statisfy Blanchard-Kahn condition or it also works on my case?. ( i do all the process manually on Matlab).
One extra question: These days, i am learning Dynare+Matlab coding. how will include the expectations during simulation process on Dynare?.
Thanks your time and effort.

What you now describe indicates not indeterminacy but instability. In this case, your model does not have a stable solution.

in this case, can i still get simulated data and IRF?. Seems only state solution is 0, once deviated from zero, it will explode. I wrote the dynare code and run it, it says matrix is nearly singular, i checked a lot, but still could not get the simulated data and irf.
Thanks.

Dynare will not be able to give you simulations and IRFs as we only allow unique stable solutions.

but, is it possible get irf and simulation by manually on matlab? or even on matlab, to get simulation and irf, it needs to satisfy Blenchard-Kahn condition?.
Thanks.

I don’t think that you can do this generically. In this case, you are violating a transversality condition for at least one variable. Unless you know which one you are allowing to explode, I would say no.

Really appreciated. Thanks.