state space with timevarying parameters

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matmit2
Posts: 4
Joined: Wed Nov 16, 2011 1:32 pm

state space with timevarying parameters

Postby matmit2 » Wed Nov 16, 2011 1:50 pm

I want to specify linear regression with timevarying
parameters, written as
yt = xtβt + et et ∼ N(0, σ2) // measurement equation
βt = βt−1 + vt vt ∼ N(0, Q) // transition equation

where yt = ΔCt and xt = [ 1 ΔYt ΔGt ] and
β't = [ β0t β1t β2t ]

Generally my main problem is how to write an error
Is my solution correct?

@signal dc = sv1 + sv2 * dg + sv3 * dy + [var = exp(c(1))]
@state sv1 = sv1(-1) + [var = exp(c(2))]
@state sv2 = sv2(-1) + [var = exp(c(3))]
@state sv3 = sv3(-1) + [var = exp(c(4))]

Can I possibly write it in that way?
@signal dc = sv1 + sv2 * dg + sv3 * dy + [var = exp(c(1))]
@state sv1 = sv1(-1) + [var = exp(c(2))]
@state sv2 = sv2(-1) + [var = exp(c(2))]
@state sv3 = sv3(-1) + [var = exp(c(2))]

Thank you in advance for your answer.

trubador
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Joined: Thu Nov 20, 2008 12:04 pm

Re: state space with timevarying parameters

Postby trubador » Thu Nov 17, 2011 1:17 am

Your first solution is correct. Second one, however, restricts the variances of time varying coefficents to be equal. If that is not what you want, then you should stick with your first specification.

matmit2
Posts: 4
Joined: Wed Nov 16, 2011 1:32 pm

Re: state space with timevarying parameters

Postby matmit2 » Thu Nov 17, 2011 2:45 pm

Thank you for such quick answer.

I forgot to write that Q in transition equation is covariance matrix. Is there no difference beetwen specification of error in measurement and transition equation?

trubador
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Re: state space with timevarying parameters

Postby trubador » Fri Nov 18, 2011 3:39 am

Yes, the specifications are done in the same manner. However, this specification implies that errors are orthogonal. If you want errors to be correlated, then you have to explicitly specify covariance conditions. For instance:

Code: Select all

@signal dc = sv1 + sv2 * dg + sv3 * dy + [var = exp(c(1))] @state sv1 = sv1(-1) + [var = exp(c(2))] @state sv2 = sv2(-1) + [ename=e3, var = exp(c(3))] @state sv3 = sv3(-1) + [ename=e4, var = exp(c(4))] @evar cov(e3,e4) = c(5)

matmit2
Posts: 4
Joined: Wed Nov 16, 2011 1:32 pm

Re: state space with timevarying parameters

Postby matmit2 » Mon Nov 21, 2011 1:31 pm

I formulated my model in this way:

Code: Select all

@signal dlnc = sv1 + sv2 * dlng + sv3 * dlny + [var = exp(c(1))] @state sv1 = sv1(-1) + [ename=e1, var = exp(c(2))] @state sv2 = sv2(-1) + [ename=e2, var = exp(c(3))] @state sv3 = sv3(-1) + [ename=e3, var = exp(c(4))] @evar cov(e1,e2) = c(5) @evar cov(e1,e3) = c(6) @evar cov(e2,e3) = c(7)
and this is an output:
Coefficient Std. Error z-Statistic Prob.

C(1) -15.26594 3039.961 -0.005022 0.9960
C(2) -3.632407 0.035709 -101.7214 0.0000
C(3) 0.541939 0.169878 3.190163 0.0014
C(4) -1.705834 0.035704 -47.77758 0.0000
C(5) -0.212236 0.084990 -2.497197 0.0125
C(6) -0.069263 0.006863 -10.09257 0.0000
C(7) 0.547998 0.043116 12.70991 0.0000

Final State Root MSE z-Statistic Prob.

SV1 0.002456 0.162696 0.015093 0.9880
SV2 -0.115266 1.461580 -0.078864 0.9371
SV3 -0.044548 0.427692 -0.104159 0.9170

I suppose that means β0t, β1t, β2t (which are state variables SV1, SV2, SV3) are highly insignificant. Am I right?

When I make OLS model β0 and β1 are significant, just like in my theoretical model. I know that is incomparable, but maybe did I something wrong?

Thank you for your help.

trubador
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Joined: Thu Nov 20, 2008 12:04 pm

Re: state space with timevarying parameters

Postby trubador » Wed Nov 23, 2011 8:10 am

Only the final values of state variables (i.e. time varying betas) are insignificant. Nothing inherently wrong with this. Smoothed state estimates are more informative (View/State views/Graph State Series)...

matmit2
Posts: 4
Joined: Wed Nov 16, 2011 1:32 pm

Re: state space with timevarying parameters

Postby matmit2 » Wed Nov 23, 2011 4:53 pm

Eviews shows the message:
Missing value found in state variance matrix.

I tried to set parameters in such way:
param c(1) .0 c(2) .0 c(3) .0 c(4) .0 c(5) .0 c(6) .0 c(7) .0 (I found your tip on forum), but it didn't work.

My another doubt is I have NA for z-Statistics (when I try to check solution for data without logarithms). What does it mean?


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