Checking Stationarity of ARMA (2,1) Model

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SUMMARY

The discussion centers on determining the stationarity of the ARMA (2,1) model defined by the equation xt + 1/6xt-1 – 1/3xt-2 = εt + 0.7εt-1. Key points include the necessity of checking mean, variance, and covariance to establish stationarity. The mean is confirmed as zero due to the nature of the Gaussian noise, while the variance can be calculated by combining AR and MA components. The expectation of the squared terms is also discussed, emphasizing the relationship between the error terms and the time series.

PREREQUISITES
  • Understanding of ARMA models, specifically ARMA (2,1) structure.
  • Knowledge of statistical concepts such as mean, variance, and covariance.
  • Familiarity with Gaussian noise and its properties in time series analysis.
  • Ability to perform expectation calculations in stochastic processes.
NEXT STEPS
  • Learn how to calculate the variance of ARMA models using the Yule-Walker equations.
  • Study the implications of stationarity in time series analysis.
  • Explore the relationship between AR and MA components in ARMA models.
  • Investigate methods for testing stationarity, such as the Augmented Dickey-Fuller test.
USEFUL FOR

Statisticians, data scientists, and researchers involved in time series analysis, particularly those working with ARMA models and seeking to understand stationarity and its implications.

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Is the following ARMA (2,1) model stationary?

xt + 1/6xt-1 – 1/3xt-2 = εt + 0.7εt-1

Inorder to know if a model is stationary. we check the mean, variance and the covariance and check whether it is dependent on time.

Obviously the mean is zero but my problem is how do i carry out the variance can i combine AR and MA together or do i do it separately?

and another problem is what does E[(xt-1)^2] gives me? I know E [(εt)^2] gives σ^2.

Thx
 
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I am not specialized in ARMA, so I may be missing something when I ask "how is the mean obviously zero"?
 
Well I should have said this earlier.

The time series is a random Gaussian noise so Zt is independent and identically distributed. hence E[Zt] = 0 --> which is the expectation(mean) and E[(Zt)^2] = σ^2 --> expectation variance.

To answer your question the expectation of each term in the equation is 0.
e.g 1/3E[xt-2] = 0
since we assume the process is stationary then E[xt-1] = E[xt] and so E[xt] α E[Zt] = 0
 
Last edited:
How are Z and x related?

If E [ε{t}^2] = σ^2, can't you solve E[(x{t})^2] = E[(-1/6x{t-1} + 1/3x{t-2} + ε{t} + 0.7ε{t-1})^2] if you assume E[(x{t})^2] = E[(x{t-s})^2] ?
 

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