TSS module 9 Time series differencing


TSS module 9 Time series differencing

Author
Message
NEAS
Supreme Being
Supreme Being (5.9K reputation)Supreme Being (5.9K reputation)Supreme Being (5.9K reputation)Supreme Being (5.9K reputation)Supreme Being (5.9K reputation)Supreme Being (5.9K reputation)Supreme Being (5.9K reputation)Supreme Being (5.9K reputation)Supreme Being (5.9K reputation)

Group: Administrators
Posts: 4.5K, Visits: 1.6K

TSS module 9 Time series differencing

 

Cryer and Chan show how differencing converts a non-stationary ARIMA process to a stationary time series.

 

** Question 1.2: Linear over three time points

 

Suppose that Yt = Mt + et     with      Mt = Mt-1 + åt

 

The time series M(t) is linear over three consecutive points, so the least squares estimator of M(t) is ⅓ (Yt-1 + Yt + Yt+1)  

 

Which of the following is stationary?

 


A.      Y(t)

B.       Y(t)

C.       2Y(t)

D.        Y(t)

E.      Y(t) – Y(t-1) + Y(t-2)


 

 

Answer 1.2: C

 


If Mt is linear over three points, the best estimate for Mt is the centered moving average of three Yt values:

 

 

If we remove the trend, the time series is stationary.

 

 

Simplify the right hand side of this equation:

 

= –⅓ (Yt+1 –2 Yt + Yt-1)

 

= –⅓ (Yt+1)

 

= –⅓ 2(Yt+1)

 

See Cryer and Chan, chapter 5, page 91

 


 

**Question 1.3: First difference

 

Suppose Yt = Mt + et and Mt = Mt-1 + åt

 

What is Yt (the first difference of Yt)?

 


 

A.      åt

B.      et

C.      åt – et – et-1

D.     åt + et – et-1

E.      åt + et + et-1

 

Answer 1.3: D

 

(See Cryer and Chan, chapter 5, page 90, Equation 5.1.9)

 

Yt = Yt – Yt-1 = Mt + et – (Mt-1 + et-1) = åt + et – et-1

 

Final exam problems give Yt = k × Mt + et (where k is a scalar) and the variances of ei and åt. You must work out variances, auto-covariances, and autocorrelations.

 


 

**Question 1.4: ARIMA(p,1,q) process

 

An ARIMA(p,1,q) process Yt has first differences Yt that are an ARMA(p,q) process with parameters öi and èj.

 

The ARIMA(p,1,q) process is written as a non-stationary moving average process ARMA(p+1, q).

 

What is the coefficient of Yt-2 in the non-stationary ARMA(p+1, q) process?

 


 

A.      1 – ö2

B.      ö1ö2

C.      ö2ö1

D.     ö2 – 1

E.      1 – ö1ö2

 

Answer 1.4: C

 

See equation 5.2.2 on page 92.

 

Intuition: The ARIMA(p,q) process is

 

Yt – Yt-1 = ö1 (Yt-1 – Yt-2) + ö2 (Yt-2 – Yt-3) + … + öp (Yt-p – Yt-p-1) + åtè1 åt-1è2 åt=2 – … – èq åt-q

 

Rewrite this as

 

Yt = (1 + ö1) Yt-1 + (ö2ö1) Yt-2 + (ö3ö2) Yt-3 + … + (öpöp-1) Yt-p + + åtè1 åt-1è2 åt=2 – … – èq åt-q

 


 

**Question 1.5: ARIMA(p,1,q) process

 

An ARIMA(p,1,q) process Yt has first differences Yt that are an ARMA(p,q) process with parameters öi and èj.

 

The ARIMA(p,1,q) process is written as a non-stationary moving average process ARMA(p+1, q).

 

What is the coefficient of åt-2 in the non-stationary ARMA(p+1, q) process?

 


 

A.      è2

B.      1 – è2

C.      è1è2

D.       è2è1

E.      è2 – 1

 

Answer 1.5: A

 

See equation 5.2.2 on page 92.

 

Intuition: The ARIMA(p,q) process is

 

Yt – Yt-1 = ö1 (Yt-1 – Yt-2) + ö2 (Yt-2 – Yt-3) + … + öp (Yt-p – Yt-p-1) + åtè1 åt-1è2 åt=2 – … – èq åt-q

 

Rewrite this as

 

Yt = (1 + ö1) Yt-1 + (ö2ö1) Yt-2 + (ö3ö2) Yt-3 + … + (öpöp-1) Yt-p + + åtè1 åt-1è2 åt=2 – … – èq åt-q

 

The coefficients of the error terms åt-k remain unchanged; they are the negatives of the moving average parameters. 

 


Attachments
NNact
Forum Newbie
Forum Newbie (2 reputation)Forum Newbie (2 reputation)Forum Newbie (2 reputation)Forum Newbie (2 reputation)Forum Newbie (2 reputation)Forum Newbie (2 reputation)Forum Newbie (2 reputation)Forum Newbie (2 reputation)Forum Newbie (2 reputation)

Group: Forum Members
Posts: 1, Visits: 1
Hi NEAS, In Q 1.3 you note:

"Final exam problems give Yt = k × Mt + et (where k is a scalar) and the variances of ei and åt. You must work out variances, auto-covariances, and autocorrelations."

Can you possibly post an example and work through the solution?  I'd like to make sure I understand what the exam problems will be like on this section.  Thank you. 


minnie53053
Junior Member
Junior Member (13 reputation)Junior Member (13 reputation)Junior Member (13 reputation)Junior Member (13 reputation)Junior Member (13 reputation)Junior Member (13 reputation)Junior Member (13 reputation)Junior Member (13 reputation)Junior Member (13 reputation)

Group: Forum Members
Posts: 11, Visits: 1
NEAS:

Question 1.2: Linear over three time points

Since Yt = Mt + et with Mt = Mt-1 + sigma t, then ,
delta Yt =delta Mt + delta et
=sigma t +et-e(t-1)

So, delta Yt is stationary?
GO
Merge Selected
Merge into selected topic...



Merge into merge target...



Merge into a specific topic ID...





Reading This Topic


Login
Existing Account
Email Address:


Password:


Social Logins

  • Login with twitter
  • Login with twitter
Select a Forum....










































































































































































































































Neas-Seminars

Search