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Math Help - Expected value and variance of ARMA(1,1) model

  1. #1
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    Expected value and variance of ARMA(1,1) model

    I guess the forum's having latex issues, so I'll try to write the model as best I can.

    y_{t}= 50 + 0.8y_{t-1} + a_{t} - 0.2a_{t-1}

    So, I've been pouring over notes trying to figure E[y_{t}] and VAR[y_{t}] but I just can't find it.

    Can someone help me out please?
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  2. #2
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    can you re-express y(t) in terms of the past errors?

    NB: this may not be the only/easiest method!
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  3. #3
    Moo
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    Hello,

    An ARMA process is stationary, so we have E[y_t]=E[y_{t-1}]=0, don't we ?
    I don't remember for the variance, can't find my notes atm and it's 2am...
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  4. #4
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    i dont think this one is stationary, because of the nonzero constant (50) that is added to each term



    edit the reasoning in this post is wrong, ignore it. theres an agebra mistake too...
    Last edited by SpringFan25; April 29th 2011 at 01:01 PM.
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  5. #5
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    my attempt: not sure if its right (especially the limits on the sums):



    it should (relatively) straightforward to compute the expected value and variance of that using the properties of a_t

    But dont waste time doing that without verifying the expression is correct


    PS: for the variance note that
    Spoiler:





    Edit: Maybe you're supposed to assume the process has an infinitely long history at time t, in which case the sums become infinite and the process might be stationary after all...
    Last edited by SpringFan25; April 29th 2011 at 04:29 AM.
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  6. #6
    Junior Member F.A.P's Avatar
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    Ok, let's have a go at it

    I guess that the a's represent white noise, i.e.

    \{a_{t}\} ~ is ~ WN(0,\sigma^{2})

    By assuming (weak) stationarity, i.e. that the mean and autocovariance function are both independent of t, the expected value can be found by

    E(Y_{t})=E(50+0.8Y_{t-1}+a_{t}-0.2a_{t-1})
    E(Y_{t})=50+0.8E(Y_{t})
    E(Y_{t})=50/(1-0.8)

    The same expected value can be obtained whilst actually checking that the process is stationary by computing its mean and autocovariance function without such prior assumptions (other than that the process has infinite history). For the mean/expectation we can do this by observing that (remember E(a_t) = 0)

    E(Y_{t})=E(50+0.8Y_{t-1}+a_{t}-0.2a_{t-1})
    E(Y_{t})=50+0.8E(50+Y_{t-2}+a_{t-1}-0.2a_{t-2})
    E(Y_{t})=50+0.8(50+0.8E(50+Y_{t-3}+a_{t-2}-0.2a_{t-3})
    ...

    E(Y_{t})=50sum_{k=0}^{n}0.8^{k}+0.8^{n}Y_{t-(n+1)}

    By letting n -> inf, the last term -> 0, and the first term converges to

    E(Y_{t})=50/(1-0.8)


    Now, for the variance: keep in mind that

    Cov(Y_{t},a_{s})=0 ~ for ~ s > t

    but may be nonzero elsewhere.

    Var(Y_{t})=Var(50+0.8Y_{t-1}+a_{t}-0.2a_{t-1})
    Var(Y_{t})=0.8^{2}Var(Y_{t-1})+\sigma^2+0.2^{2}\sigma^2
    \hspace\hspace-2*0.8*0.2*Cov(50+0.8*Y_{t-2}+a_{t-1}-0.2a_{t-2},a_{t-1})
    Var(Y_{t})=0.8^{2}Var(Y_{t-1})+\sigma^2+0.2^{2}\sigma^2-2*0.8*0.2\sigma^2
    Var(Y_{t})=0.8^{2}Var(Y_{t-1})+0.72\sigma^2

    Let n -> inf in the the recursion (first term -> 0)

    Var(Y_t) = (0.8^{2})^{n+1}Var(Y_{t-(n+1)})+0.72\sigma^2sum_{k=0}^{n}(0.8^{2})^{i}

    (or assume stationarity Var(Y_t) = Var(Y_{t-1}), or know something about fixed points(?))

    so that we end up with

    Var(Y_{t})=(0.72\sigma^2)/(1-0.8^2)
    Last edited by F.A.P; May 2nd 2011 at 11:52 PM.
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