Page 680 partial autocorrelation function characteristic equation

Discussion in 'CS2' started by Molly, Sep 4, 2022.

  1. Molly

    Molly Ton up Member

    Hey everyone,

    Looking at the example of page 680, its something i havent really seen in the past exam questions yet, but i do understand the idea behind it. Only thing is, what is the rule we apply to the characteristic equation?

    Firstly am wondering, are we finding the characteristic equation of X_t or p_t? And then, im wondering if the rule here isnt the normal z^s=X_(t-s)? because in that instance, by characteristic equation would be 1=5/6z-1/6z^2 but thats not right.

    The notes do mention that the characteristic equation for the difference equation is " unfortunately slightly different to the characteristic equation for the process)", does anyone know the rule??

    Thanks,
    Molly
     
  2. Andrew Martin

    Andrew Martin ActEd Tutor Staff Member

    Hi Molly

    In this example, we're using the fact that we have a difference equation for the rho_k's. We want to solve this difference equation to come up with an explicit formula for the rho_k's.

    To do so, we consider the roots of the characteristic polynomial for this second-order difference equation. This is given on page 4 of the Tables. In particular, we tend to see it in the form:

    a * rho_k = b1 * rho_(k-1) + c1 * rho_(k-2)

    Or, writing it in the form we see in the Tables:

    a * rho_k + b * rho_(k-1) + c * rho_(k-2) = 0

    where b1 = -b and c1 = -c.

    This matches the form of the difference equation on page 4 of the Tables and the relevant polynomial is:

    a * lambda^2 + b * lambda + c.

    In this case we get:

    lam^2 - 5/6 * lam + 1/6 as per the notes.

    As the notes say, unfortunately this is different from the characteristic polynomial of the process terms. The key thing to note here is that we are trying to solve this difference equation, hence we use the version given on page 4 of the Tables.

    Were we checking stationarity for example, then we would use the characteristic polynomial of the process terms. This would be given by:

    1 - 5/6*lam + 1/6 * lam^2

    Interestingly, as an aside (not in the Core Reading or notes), this is actually the reciprocal polynomial (the coefficients are reversed) and there is a relationship between the two. In particular, if r is a root of one, then 1/r must be a root of the other. This means that for a series to be stationary, the roots of the difference equation characteristic polynomial must be all smaller than 1 in magnitude. This makes sense in that, considering the easiest scenario of the roots being real, A * lam1^n + B * lam2^n -> 0 as n -> infinity if the magnitude of the lam's are < 1. Ie the autocorrelations tend to 0 as the lag -> inf, which is what should happen for a stationary series.

    Hope this helps!

    Andy
     
    Last edited: Sep 6, 2022
    Molly likes this.
  3. Molly

    Molly Ton up Member

    Hi Andy,

    Thats amazing that really does help, thank you so so much
     

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