TS Module 6 Stationary autoregressive processes
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Time series autoregressive processes practice problems
** Exercise 6.1: Stationarity of AR(2) process
What are the three conditions for an AR(2) process to be stationary?
Given
ö1, what values of ö2 create a stationary AR(2) process?
Part A:
The three conditions are
ö
1 + ö2 < 1
ö
2 – ö1 < 1
|
ö2| < 1
(See Cryer and Chan page 72, equation 4.3.11)
Part B:
If
ö1 is given, the conditions are
ö
2 < 1 – ö1
ö
2 < 1 + ö1
|
ö2| < 1
Illustration:
If
ö1 = 0.9, then ö2 must be less than 0.1 and more than –1.
*Question 6.2: Characteristic polynomial
An AR(2) process is Yt =
ö1 Yt-1 + ö2 Yt-2 + et .
What is the characteristic polynomial for this time series?
ö
(x) = 1 + ö1(x) + ö2(x2)
ö
(x) = 1 + ö2(x) + ö1(x2)
ö
(x) = 1 – ö1(x) – ö2(x2)
ö
(x) = 1 – ö2(x) + ö1(x2)
ö
(x) = 1 + ö2(x) – ö1(x2)
Answer 6.2: C
(See Cryer and Chan page 71, equation 4.3.9)
The characteristic polynomial reverses the sign of the autoregressive coefficient.
** Exercise 6.3: AR(2) autocorrelations
An AR(2) process with mean zero is Yt =
ö1 Yt-1 + ö2 Yt-2 + åt
What is
ã0, the variance of Yt?
What is
ã1, the covariance of Yt and Yt-1?
What is
ã2, the covariance of Yt and Yt-2?
What is
ñ1, the autocorrelation of lag 1?
What is
ñ2, the autocorrelation of lag 2?
Part A:
Multiply the expression for the AR(2) process by Yt-1 and take expectations to get
E(Yt × Yt-1) =
ö1 E(Yt-1 × Yt-1) + ö2 E(Yt-2 × Yt-1) + E(åt × Yt-1)
The time series observations are uncorrelated with the error terms in subsequent periods, so E(
åt, Yt-1) = 0.
E(Yt, Yt-1) =
ã1, E(Yt-1, Yt-1) = ã0, and E(Yt-2, Yt-1) = ã1, so
ã
1 = ö1 ã0 + ö2 ã1
Divide this equation by
ã0 to get
ñ
1 = ö1 ñ0 + ö2 ñ1
ñ
o is the correlation of a random variable with itself, which is always 1, so we derive that
ñ
o = è1 + è2 ñ1 ñ1 = è1 / (1 – è2)
Part B:
Multiply the expression for the AR(2) process by Yt-2 and take expectations to get
E(Yt × Yt-2) =
ö1 E(Yt-1 × Yt-2) + ö2 E(Yt-2 × Yt-2) + E(åt × Yt-2)
The time series observations are uncorrelated with the error terms in subsequent periods, so E(
åt, Yt-2) = 0.
E(Yt, Yt-2) =
ã2, E(Yt-1, Yt-2) = ã1, and E(Yt-2, Yt-2) = ã0, so
ã
2 = ö1 ã1 + ö2 ã0
Divide by
ã0 to get
ñ
2 = ö1 ñ1 + ö2 ñ0 = ö2 + ö12 / (1 – ö2)
Cryer and Chan write this as
ñ
2 = [ ö2 (1 – ö2) + ö12 ] / (1 – ö2)