April 2014 Question 12 ii

Discussion in 'CT6' started by Rieks, Sep 20, 2014.

  1. Rieks

    Rieks Member

    Is there perhaps a mistake in this question- two solutions to a1?
    (SACF2= a1* SACF1).Therefore (0.55=a1*0.68), hence a1 = 0.808823529 but on the examiner's report they use (SACF1=a1*SACF0) hence (0.68=a1*0.9) hence a1=.7555555556.
     
  2. Katherine Young

    Katherine Young ActEd Tutor Staff Member

    It's not a mistake. If you're given more than just the first few SACFs, then of course you'll have more than one possible answer. You've used SACF1 and SACF2. But it's always better to go for the lowest lags.

    Therefore you should try to find ACF1 in terms of ACF0. You should be able to work out that:

    \[ \rho_1 = \frac{\gamma_1}{\gamma_0} = a_1 \]

    Then use the method of moments to set this equal to 0.68 so that \[ a_1 = 0.68 \]

    You could have done it the way the examiners did it, but note that they have also used the earlier lags. (It's only an AR(1), so there's no need to calculate anything beyond lag 1.) I doubt you'd score marks by calculating lag 2 as you have.
     
  3. Rieks

    Rieks Member

    Thank you Katherine, I feel it wouldn't be fair not to get the marks, but I'll rather be safe than sorry in the exam:)
     
  4. Sam88

    Sam88 Member

    In this question is there a typo when they write "sample auto-covariance function..."? Should this be the sample auto-correlation function?

    Thanks
     
  5. John Lee

    John Lee ActEd Tutor Staff Member

    You are correct. there was a typo.
     

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