range of ARMA(2,2) parameters

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range of ARMA(2,2) parameters

Postby Nau2306 on Tue Dec 20, 2011 11:20 am

Hello

Can someone please tell me within which range must the estimated parameters of an ARMA(2,2) lie for it to be stationary and invertible?

Thanks
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Re: range of ARMA(2,2) parameters

Postby startz on Tue Dec 20, 2011 12:21 pm

This isn't a complete answer, but...
EViews computes the inverse roots of the AR and MA polynomials. The requirement is that all the inverse roots be less than one (lie inside the unit circle).

Stationarity depends on combinations of the AR parameters. I think, but may not remember correctly, that the limts are abs(AR(1))<2, abs(ar(2))<1
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Re: range of ARMA(2,2) parameters

Postby Nau2306 on Tue Dec 20, 2011 12:31 pm

Is there a way I can compute the range? From what I have read, the roots of the characteristic polynomial must lie outside the unit disc but with 4 parameters, it is a bit complicated to find the roots of the polynomial.
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Re: range of ARMA(2,2) parameters

Postby startz on Tue Dec 20, 2011 1:19 pm

Nau2306 wrote:Is there a way I can compute the range? From what I have read, the roots of the characteristic polynomial must lie outside the unit disc but with 4 parameters, it is a bit complicated to find the roots of the polynomial.


That's correct. But an ARMA(2,2) has two characteristic polynomials, one AR and one MA, each of order two. So you can solve for the roots using the quadratic formula.
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Re: range of ARMA(2,2) parameters

Postby Nau2306 on Tue Dec 20, 2011 11:27 pm

That means I should estimate the model first, use the estimated parameters to obtain the characteristic polynomial and then solve for it. For stationarity, the roots must be greater than 1. Is that right?
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range of ARMA(2,2) parameters

Postby startz on Wed Dec 21, 2011 11:42 am

No. It means that EViews solves the polynomial for you.

Have you run an ARMA in EViews and looked at the output?
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Re: range of ARMA(2,2) parameters

Postby Nau2306 on Thu Dec 22, 2011 11:43 am

This is the output following an ARMA(2,2) estimation. It seems to me that it gives only the MA roots but not that of the AR. From what I have read, the stationarity of an ARMA model depends solely on the AR part and thus we need only consider the characteristic polynomial of the AR.


Dependent Variable: RETURNS
Method: Least Squares
Date: 12/22/11 Time: 22:41
Sample (adjusted): 1/04/2000 11/30/2011
Included observations: 4349 after adjustments
Convergence achieved after 16 iterations
MA Backcast: 1/02/2000 1/03/2000

Variable Coefficient Std. Error t-Statistic Prob.

C 9.87E-06 3.34E-05 0.295448 0.7677
RETURNS(-1) -0.217256 0.057540 -3.775714 0.0002
RETURNS(-2) 0.748592 0.056903 13.15568 0.0000
MA(1) 0.156197 0.050273 3.107001 0.0019
MA(2) -0.815867 0.049719 -16.40944 0.0000

R-squared 0.010812 Mean dependent var 2.54E-05
Adjusted R-squared 0.009901 S.D. dependent var 0.006483
S.E. of regression 0.006451 Akaike info criterion -7.247997
Sum squared resid 0.180780 Schwarz criterion -7.240665
Log likelihood 15765.77 Hannan-Quinn criter. -7.245409
F-statistic 11.86992 Durbin-Watson stat 2.009067
Prob(F-statistic) 0.000000

Inverted MA Roots .83 -.98
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Re: range of ARMA(2,2) parameters

Postby EViews Gareth on Thu Dec 22, 2011 11:45 am

That's not the output from an ARMA(2,2) estimation. That's the output from an MA(2) estimation with two lagged dependent variables. If you want the output from an ARMA(2,2) estimation, you should include some AR terms in your specification.
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range of ARMA(2,2) parameters

Postby startz on Thu Dec 22, 2011 4:52 pm

Well, it is an ARMA but EViews doesn't know it unless you use the AR command.
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Re: range of ARMA(2,2) parameters

Postby Nau2306 on Fri Dec 23, 2011 11:23 am

Okay that's why i did not obtain the AR roots.
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