• We are pleased to announce that the winner of our Feedback Prize Draw for the Winter 2024-25 session and winning £150 of gift vouchers is Zhao Liang Tay. Congratulations to Zhao Liang. If you fancy winning £150 worth of gift vouchers (from a major UK store) for the Summer 2025 exam sitting for just a few minutes of your time throughout the session, please see our website at https://www.acted.co.uk/further-info.html?pat=feedback#feedback-prize for more information on how you can make sure your name is included in the draw at the end of the session.
  • Please be advised that the SP1, SP5 and SP7 X1 deadline is the 14th July and not the 17th June as first stated. Please accept out apologies for any confusion caused.

Two small time series questions

Molly

Ton up Member
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.
 
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
 
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 !
 
Back
Top