Hi all, Am doing the example on page 676. I can do the question and get the right answer, but it seems i might be missing something obvious - i dont think its easy to see that differencing with remove the root 1 without actually differencing and finding the roots does anyone know how the CMP knows which root will be eliminated? Thanks, Molly
Hi Molly I’m inclined to agree that it may not be immediately obvious! If we consider writing the process in terms of the backwards shift operator, we have: (1 – 11/6 B + B^2 – 1/6 B^3) Xn = en The LHS characteristic polynomial having a root of one is equivalent to us being able to factorise the LHS into the form: (1 – B)(quadratic in B) Xn = en More specifically, here we have: (1 – B)(1 – 5/6 B + 1/6 B^2) Xn = en Now, and this is the important bit, (1 – B) applied to Xn is the same as differencing the process: (1 – B) Xn = Xn – Xn-1 So we can eliminate this root by differencing the series: (1 – B)(1 – 5/6 B + 1/6 B^2) Xn = en <=> (1 – 5/6 B + 1/6 B^2) (1 – B) Xn = en <=> (1 – 5/6 B + 1/6 B^2) Yn = en Where Yn = Xn – Xn -1. The roots for the characteristic polynomial of Yn are therefore the other two roots of the cubic we started with (now a quadratic for Yn). As per the question, these other two roots are 2 and 3. So differencing removes a root of 1 because it is equivalent to taking out a factor of (1 – B). We can't remove any other roots this way, only if we have a root of 1. Hope this helps! Andy
