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Time Series: ARCH model properties

  1. Aug 15, 2011 #1
    Consider an ARCH(1) model:
    Xt = σtZt, where Zt~ i.i.d. N(0,1)
    σt2 = w0 + w1 Xt-12
    Find (i) E(Xt)
    and (ii) the autocovariance function γX(h) for h=0,1,2,3,..., assuming the process is second-order stationary.


    Solution:
    (i) E(Xt) = E[E(Xtt2)] =E[E(σtZtt2)]
    =E[σtE(Ztt2)] = E[σtE(Zt)] = E[σt * 0] = 0
    (here I don't understand why E(Ztt2)=E(Zt).)

    (ii) γX(0)=E(Xt2)
    =E(σt2 Zt2)=E(σt2)E(Zt2)
    =E(w0 + w1 Xt-12) * 1
    = w0 + w1 γX(0)
    Solve for γX(0) => γX(0) = w0/(1-w1)

    γX(h)=E(XtXt+h)
    =E[E(XtXt+h|Zt+h-1,...,Zt+1)] = E[XtE(Xt+h|Zt+h-1,...,Zt+1)]
    (here I don't understand why we can pull the Xt out of the expectation.)
    =E[XtE(σt+hZt+h|Zt+h-1,...,Zt+1)] = E[Xtσt+hE(Zt+h|Zt+h-1,...,Zt+1)]
    (here I don't understand why we can pull the σt+h out of the expectation.)
    =E[Xtσt+hE(Zt+h)] = E[Xtσt+h * 0] = 0 for all h>0.

    I'm cannot follow the reasoning of the three equalities labelled in red above. Can someone explain why they are true?
    Any help would be much appreciated! :)
     
  2. jcsd
  3. Aug 16, 2011 #2

    Stephen Tashi

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    [itex] Z_t [/itex] is independent of [itex] \sigma_t^2 [/itex]. I think this step is an example of the fact that if X and Y are independent random variables then E(X|Y) = E(X).

    The conditional expectation with respect to the sequence of values [itex] Z_{h+t-1}...Z_{t+1} [/itex] amounts to an integration with respect to the joint density of a sequence of independent random variables. These are events that happen after time [itex] t [/itex] and [itex] X_t [/itex] doesn't depend on them. I think this step is an example of the idea that [itex] \int g(x_3) f(x_1,x_2)) dx_1 dx_2 = g( x_3) \int f(x_1,x_2) dx_1 dx_2 [/itex]


    I don't understand that step yet. I wonder if the phrase "assuming the process is second-order stationary" tells us anything important. The definition of the process looks very specific, so it isn't clear to me what that phase adds to it.
     
  4. Aug 17, 2011 #3

    Stephen Tashi

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    I could understand the above if it said:

    [itex] E(X_t E(\sigma_{t+h} X_{t+h}| Z_{t+h-1}...Z_{t+1} ) = E(X_t E(\sigma_{t+h}|Z_{t+h-1}...Z_{t+1}) E(Z_{t+h}|Z_{t+h-1}...Z_{t+1})) [/itex]

    since [itex] Z_{t+h} [/itex] is independent of [itex] \sigma_{t+h} [/itex]. The result will follow because the factor of zero still appears.
     
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