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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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

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

- Nau2306
**Posts:**74**Joined:**Thu Nov 17, 2011 11:51 am

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

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

- startz
- Non-normality and collinearity are NOT problems!
**Posts:**2792**Joined:**Wed Sep 17, 2008 2:25 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.

- Nau2306
**Posts:**74**Joined:**Thu Nov 17, 2011 11:51 am

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.

- startz
- Non-normality and collinearity are NOT problems!
**Posts:**2792**Joined:**Wed Sep 17, 2008 2:25 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?

- Nau2306
**Posts:**74**Joined:**Thu Nov 17, 2011 11:51 am

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

Have you run an ARMA in EViews and looked at the output?

Have you run an ARMA in EViews and looked at the output?

- startz
- Non-normality and collinearity are NOT problems!
**Posts:**2792**Joined:**Wed Sep 17, 2008 2:25 pm

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

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

- Nau2306
**Posts:**74**Joined:**Thu Nov 17, 2011 11:51 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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- EViews Gareth
- Fe ddaethom, fe welon, fe amcangyfrifon
**Posts:**10034**Joined:**Tue Sep 16, 2008 5:38 pm

Well, it is an ARMA but EViews doesn't know it unless you use the AR command.

- startz
- Non-normality and collinearity are NOT problems!
**Posts:**2792**Joined:**Wed Sep 17, 2008 2:25 pm

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

- Nau2306
**Posts:**74**Joined:**Thu Nov 17, 2011 11:51 am

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