April 2014 Question 12 ii

[Rieks]

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.

 

[Katherine Young]

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.

 

[Rieks]

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

 

[Sam88]

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

Thanks

 

[John Lee]

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

Thanks

You are correct. there was a typo.