This might be a dumb question on algebra, but I would be keen on finding out a quick method of factorising (1-B) from polinomials such as this one: (1 - 0.6B - 0.3B^2 - 0.1B^3) The answer in the core reading is: (1 - B) * (1 + 0.4B +0.1B^2) I can't quickly factorise it and I think the core reading contains an error.... Can someone give me a hint how to nicely factorise (1 - B) from polynomials? This is used for determination of ARIMA processes, and I get stuck
this guy has some good videos, i'm pretty sure you can work it out after watching this.. http://www.youtube.com/watch?v=l6_ghhd7kwQ
As an aside you don't need to be able to factorise it in order to rewrite it as an ARIMA. By using the remainder theorem - you simply replace B's by 1's to get (or just sum the coefficients of the original X's): 1 - 0.6 - 0.3 - 0.1 = 0 Since it equals zero, this means that 1 is a root, so 1-B is a factor. We now start rewriting the time series ito differences. We have X(t) - 0.6X(t-1) - 0.3X(t-2) - 0.1X(t-3) So start with [X(t) - X(t-1))] This gives -X(t-1) but we want -0.6X(t-1) so we correct this by adding 0.4X(t-1): So we now have [X(t) - X(t-1)] + 0.4X(t-1) So let's do +0.4 times the next difference: [X(t) - X(t-1)] + 0.4[X(t-1) - X(t-2)] This gives -0.4X(t-2) but we want -0.3X(t-2) so we correct this by adding 0.1X(t-2): So we now have [X(t) - X(t-1)] + 0.4[X(t-1) - X(t-2)] + 0.1X(t-2) So let's do +0.1 times the next difference: [X(t) - X(t-1)] + 0.4[X(t-1) - X(t-2)] + 0.1[X(t-2) - X(t-3)] Hurrah! We're done as we have the correct number of each X. Hope you can follow that as it's far easier to explain face-to-face than in writing!
I passed CT6 last year, but i like to help out wherever i can! I prefer the long division method for doing these problems thats why i took the time to learn it. It seems to be one of those skills that almost never gets used and tends to be 'forgotten' which is a shame because it is so useful.
the method is fast and easy! the guy on youtube has lots of other interesting videos! some of which come handy for CT4 - for example the integration with the integrating factor....
Thank you. I actually thought that in order to determine if a process is ARIMA and what d =? then we would need to factorise out (1-B) once for d=1 etc. Where is the remainder theorem discussed?
I don't think the remainder theorem is in the notes - but it's use (I think) makes questions much quicker than repeated algebraic long division (sorry Devon!). In the case just done - we have differenced once. I'd rewrite it using the difference operator (I'll use D for that here): DX(t) + 0.4DX(t-1) + 0.1DX(t-2) We then look at the coefficients of the difference see if they sum to 0 again: 1 + 0.4 + 0.1 = 1.5 They don't so we can't difference it again. So d=1 (if it is now stationary). If they did sum to 1, we repeat the process described above to get that 2nd difference and so on.
I just wanted to ask if there are any other mathematical tricks/ methods that it is good to know for CT6?
Gosh! That's a broad question! The only one that comes to mind is the integration tricks (when you make it look like the PDF of a gamma) that are used in Ch3 mixture distributions. I'm sure there are more but it's the only one I remember!
John, I have one more question for you. There ia a lot of stuff which you can derive in CT6. Examples include skew(x)....should I memorize the formula because what really maters is to be able to quickly perfom the calcuations, or is it better to derive everything....which jeopardizes finishing the exam on time... The question is really, do examiners give marks for all steps of calculations and derivations or is it better to be quick and make sure to finish on time?
My list of proofs for CT6 can be found on this thread. But most of the time it is just applying the formulae so memorising is important as many students do run out of time in the CT6 exam.
John, thank you for the link. I will most definitely practice all the proofs....just a quick question on the April 2011 paper for CT6. The question on the time series of ARIMA...why in the world the examiners did not expect a derivation of d? The examiners report, states simply ARIMA (2, 0, 0) and no work is shown...where as the core reading always suggest to derive d, and that is what I was trying to do on the exam, wasting a lot of time....
Using the remainder theorem you can quickly see that it can't be differenced and so d=0 IF the time series is stationary.
Thank you. I have another question on proving the existence of R for aggregate claims. Section 3.4 of ruin theory chapter.... I dont see how E[e^RSi] = e ^Rc as a result of E[e^R(Si-c)] = 1 how can I derive it?
E[e^(R(Si-c))] = 1 => E[e^(RSi) e^(-Rc)] = 1 => e^(-Rc) E[e^(RSi)]= 1 (expectation is a linear function) => E[e^(RSi)] = e^(Rc)
John Lee, just a recap on your answer above. I understand that the remainder theorem is much faster than factorising out (1-B), and if B<>1 then (1-B) cannot be factored out. I am just wondering about the case when d > 1 Should we than rewrite the differences of the differences of (difference operator ^2 etc)? in this case ARIMA would have d= 2 or higher....
Yes. Keep going until you can't difference any more. Then check for stationarity. Then classify as an ARMIA(p,d,q)
To check for stationarity of ARIMA would be to: - show that d=0 - or solve the characteristic polynomial of AR or MA portion or BOTH?
