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ARMA model

  1. Nov 2, 2007 #1
    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.

  2. jcsd
  3. Nov 2, 2007 #2


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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"?
  4. Nov 2, 2007 #3
    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: Nov 2, 2007
  5. Nov 2, 2007 #4


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    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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