Checking Stationarity of ARMA (2,1) Model

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Discussion Overview

The discussion centers on the stationarity of an ARMA (2,1) model represented by the equation xt + 1/6xt-1 – 1/3xt-2 = εt + 0.7εt-1. Participants explore the conditions for stationarity, particularly focusing on the mean, variance, and covariance of the model, as well as the relationship between the AR and MA components.

Discussion Character

  • Exploratory, Technical explanation, Conceptual clarification

Main Points Raised

  • One participant asserts that the mean of the model is zero and questions how to calculate the variance, wondering if the AR and MA components can be combined or must be treated separately.
  • Another participant expresses confusion regarding the assumption that the mean is "obviously zero," indicating a lack of specialization in ARMA models.
  • A later reply clarifies that the time series is a random Gaussian noise, leading to the conclusion that the expectation of each term in the equation is zero, based on the assumption of stationarity.
  • Another participant questions the relationship between the variables Z and x, suggesting that if E[ε{t}^2] = σ^2, one might be able to solve for E[(x{t})^2] under certain assumptions about the equality of expectations across time.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the assumptions regarding the mean and variance of the model. There are differing views on the implications of stationarity and how to approach the calculations involved.

Contextual Notes

Participants express uncertainty about the assumptions underlying their calculations, particularly regarding the independence and distribution of the time series components.

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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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