Two small time series questions

[Molly]

Hey Guys,

as you may be able to tell by my many posts tonight (Sorry about that !) im really trying to consolidate my time series understanding today.

Just two things i want to clear up

Firstly: when we pull out a factor of (1-B) this is equivalent of differencing once right? so i can say d=1? Equivalently (1-B^2) would be differencing twice etc? so how about if i can pull out (1-B)(1-0.5B) - ive seen solutions say this is differencing once, is it because of the coefficient of the second bracket, so that (1-0.5b) doesnt count as differencing? I hope that makes sense, essentially if i pulled out just (1-0.5B), would i be able to say ive differenced at all?

Secondly: is the error term the same as the white noise e_t term we see in most questions? they seem to be used interchangably in questions, but i could be wrong.

Thanks so much in advance.

 

[Andrew Martin]

Hi Molly

If we have a time series equation written in terms of the backwards shift operator then yes pulling out a factor of (1-B) from the autoregressive terms is equivalent to differencing in that the remaining polynomial (after removing the factor of (1-B)) defines the first differenced series.

For example:

Xt = 0.7Xt-1 + 0.3Xt-2 + et

Writing in terms of B:

(1 - 0.7B - 0.3B^2) Xt = et

Factorising:

(1 - B)(1 + 0.3B) Xt = et

If we let (1-B) Xt = Yt, then we have:

(1 + 0.3B)Yt = et

Or: Yt = -0.3Yt-1 + et

We can also see this by directly differencing the series:

Xt = 0.7Xt-1 + 0.3Xt-2 + et

Xt - Xt-1 = -0.3(Xt-1 - Xt-2) + et

Let Yt = Xt - Xt-1
Yt = -0.3Yt-1 + et as we had above.

Note that pulling out a factor of (1-B) doesn't necessarily mean d = 1. If we are trying to identify the process as an ARIMA process, then we need to keep differencing until we remove all unit roots, this may mean taking out multiple factors of (1-B).

Differencing twice is equivalent to taking out a factor of (1-B)^2 not (1-B^2). The latter would be the lag 2 difference.

Pulling out a factor of (1 - 0.5B) does not relate to a standard lagged difference as it is Xt - 0.5Xt-1.

Yes, without seeing any particular question, I would presume the error terms are referring to the white noise terms.

Hope this helps!

Andy

 

[Molly]

Hi Molly

If we have a time series equation written in terms of the backwards shift operator then yes pulling out a factor of (1-B) from the autoregressive terms is equivalent to differencing in that the remaining polynomial (after removing the factor of (1-B)) defines the first differenced series.

For example:

Xt = 0.7Xt-1 + 0.3Xt-2 + et

Writing in terms of B:

(1 - 0.7B - 0.3B^2) Xt = et

Factorising:

(1 - B)(1 + 0.3B) Xt = et

If we let (1-B) Xt = Yt, then we have:

(1 + 0.3B)Yt = et

Or: Yt = -0.3Yt-1 + et

We can also see this by directly differencing the series:

Xt = 0.7Xt-1 + 0.3Xt-2 + et

Xt - Xt-1 = -0.3(Xt-1 - Xt-2) + et

Let Yt = Xt - Xt-1
Yt = -0.3Yt-1 + et as we had above.

Note that pulling out a factor of (1-B) doesn't necessarily mean d = 1. If we are trying to identify the process as an ARIMA process, then we need to keep differencing until we remove all unit roots, this may mean taking out multiple factors of (1-B).

Differencing twice is equivalent to taking out a factor of (1-B)^2 not (1-B^2). The latter would be the lag 2 difference.

Pulling out a factor of (1 - 0.5B) does not relate to a standard lagged difference as it is Xt - 0.5Xt-1.

Yes, without seeing any particular question, I would presume the error terms are referring to the white noise terms.

Hope this helps!

Andy

That's amazing thank you so so much !