MA, AR, ARMA, ARIMA etc

[steph]

I'm a bit confused.... if I am given a time series but the question doesn't specify whether it is moving average, auto regressive, etc.. how can I decide which it is?

 

[John Potter]

AR processes depend on previous values of themselves, eg

Kt =7Kt-1 +3 Kt-2 +Zt is AR(2) because K depends on 2 previous values of itself.

MA processes involve white noise terms from before time t, eg

Kt = Zt + Zt-1 + 0.4Zt-2 is MA(2) because the smallest time involved with a Z term is t-2

ARMA is a mixture of the above, eg

Kt = Kt-1 +0.4Kt-4 +Zt -0.7Zt-3 is ARMA(4,3) because the K depends on a vlaue of itself 4 time periods ago and we have a Z term 3 time periods ago.

Finally ARIMA(p,d,q) is an ARMA(p+d,q) process that can be differenced d times in order that it is stationary. If it is not possible to create a stationary process by differencing then it is not an ARIMA process, eg

Kt = 2Kt-1 +Zt - Zt-1 is an ARMA(1,1) but it is NOT an ARIMA process because it is not stationary and you cannot difference it to create a stationary process. You can see this intuitively, we are multiplying by 2 each time, so it will not settle down.

NB All MA processes are stationary so you only need to concentrate on the AR bit to see if it's stationary. For this, use the characteristic polynomial phi and check that the roots are all greater than one in magnitude.

Try this one...

Kt = Kt-1 + Zt +0.5Zt-2

Classify this process,

Good luck!
John

 

[steph]

Try this one...

Kt = Kt-1 + Zt +0.5Zt-2

Classify this process,

Ok, so the process contains previous values of itself and past white noise terms so I think it is ARMA(1,2) or an ARIMA.

To check if ARIMA I need to check for stationarity. Char eqn is:
phi(L) = 1-L = 0, which gives L = 1. So, not stationary and therefore must be ARMA(1,2)?

Is that right?

 

[MissAussie]

Interesting tips in this thread!

Regarding this question:
Kt = Kt-1 + Zt +0.5Zt-2

I wrote it as (1-B)Kt = Zt +0.5 [/SIZE]Zt-2

Therefore I think it is an ARIMA(0,1,2) process...!?! In the acted notes, we are told that the degree of the term (1-B) is d, while p & q can be read off the highest coefficient on the LHS and RHS respectively.

Steph, I am not sure about the logic: "To check if ARIMA I need to check for stationarity", because I think ARIMA's are not always stationary.

John..... what is the answer please??

 

[Anna Bishop]

Hi everyone

K(t) = K(t-1) + Z(t) +0.5Z(t-2)

This series can be classified in two ways:

1) It is an ARMA(1,2) series (p = 1, q = 2) as the final K term is t-1 giving p = 1 and the final Z term is t-2 giving q = 2)

2) It is an ARIMA(0,1,2) series (p = 0, d = 1, q = 2) as if we difference once (d=1) we get:

upside down triangle K(t) = Z(t) + 0.5Z(t-2).

This differenced series is stationary since it is the sum of white noise, which is stationary. The final K term is t giving p = 0 and the final Z term is t-2 giving q = 2.

Steph is indeed right that she needs to check for stationarity. If the differenced process is not stationary then we cannot say it is ARIMA.

For example:

K(t) - 3K(t-1) + 2K(t-2) = Z(t) + 0.5Z(t-1)

This is an ARMA(2, 1) series.

This series is not stationary as the characteristic equation is 1 - 3y + 2y^2 = 0 to which the solutions are 1 and 0.5, neither of which are greater than 1 in magnitude.

Try differencing it:

upside down triangle K(t) - 2 upside down triangle K(t-1) = Z(t) + 0.5Z(t-1)

This looks as though it should be an ARIMA(1, 1, 1) series but it isn't as it is not stationary. The characteristic equation is 1 - 2y = 0 to which the solution is y=0.5, which is not greater than 1 in magnitude.

So in fact, we cannot classify this series as ARIMA at all!

Anna

 

[gcpgcp]

Hi, Anna,

A) ▽ K(t) - 2 ▽ K(t-1) = Z(t) + 0.5Z(t-1)

B) ▽ K(t) = Z(t) + 0.5Z(t-2)

For A: You used CH EQ for stationarity leading to answer that it is not a stationary process.

For B: You used the fact that it is a sum of white noise so it is stationary.

Questions:

1) Why different approach used ?

2) Can we use sum of white noise appraoch for A ? Which will mean it is a stationary process !!!

3) Can we use CH EQ for B ??? If Yes, what are the steps ?

4) Any thumb rule on which approach to take as a standard.

 

[gcpgcp]

 

[geoff97531]

Hi, Anna,

A) ▽ K(t) - 2 ▽ K(t-1) = Z(t) + 0.5Z(t-1)

B) ▽ K(t) = Z(t) + 0.5Z(t-2)

For A: You used CH EQ for stationarity leading to answer that it is not a stationary process.

For B: You used the fact that it is a sum of white noise so it is stationary.

Questions:

1) Why different approach used ?

2) Can we use sum of white noise appraoch for A ? Which will mean it is a stationary process !!!

3) Can we use CH EQ for B ??? If Yes, what are the steps ?

4) Any thumb rule on which approach to take as a standard.

I'm not great at this stuff and may have missed the point but you only need to check processes with an AR component for stationarity. If the process only has white noise terms in it (that is a moving average process), it IS stationary - an example would be equation B.

For the AR bit of equation A you get

1-2y = 0 which leads to y = 0.5

If that y value is not STRICTLY GREATER than one then the process is not stationary.

 

[howdyrathore]

If we can difference the ARMA time series d times but the differenced term is not stationary, we can only classify it as Non stationary ARMA. I am confused about the values of p and q for non stationary ARMA. Are these p and q of the original ARMA process or the final differenced process obtained?

 

[John Lee]

They'll be the original ARMA process.