TSS module 9 Time series differencing


TSS module 9 Time series differencing

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NEAS
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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. 

 


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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. 


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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?
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