TS Module 5 moving average MA(2) practice problems
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Time series MA(2) process practice problems
** Exercise 5.1: Variances, autocovariances, and autocorrelations of moving average process of order 2
A moving average process of order 2 is Yt = et – è1 et-1 – è2 et-2, with è1 = 0.7, è2 = 0.5, and óe = 2.
A. What is ã0, the variance of Yt?
B. What is ã1, the autocovariance of Yt and Yt-1?
C. What is ã2, the autocovariance of Yt and Yt-2?
D. What is ñ1, the autocorrelation of Yt and Yt-1?
E. What is ñ2, the autocorrelation of Yt and Yt-2?
Part A: ã0 = (1 + è21 + è22) × ó2e = (1 + 0.49 + 0.25 ) × 22 = 6.960
Yt is the sum of three independent random variables: et, et-1, and et-2, with coefficients of 1, –è1, and –è2.
Each random variable has a variance of ó2ε. The variance of the sum is the sum of the variances of each term times the square of its coefficient.
Part B: ã1 = (–è1 + è1 × è2) × ó2ε = (–0.7 + 0.7 × 0.5) × 22 = -1.400
Jacob: ã1 is the autocovariance of lag 1. Why does è2 affect this autocovariance? If the random variable åt-1 increases one unit, Yt-1 increases one unit and Yt increases 1 × –è1 units. The covariance is in terms of the variances: for each 1 variance change of åt-1, Yt decreases è1 × ó2ε. I understand that è2 affects the two period ahead value of the time series; why does it affect the one period ahead value?
Rachel: Your reasoning shows the effect of åt-1 on Yt. Your conclusion is correct: the covariance of åt-1 and Yt is –è1 × ó2ε.
This exercise asks for the covariance of Yt-1 and Yt.
Yt-1 may change randomly for two reasons: random fluctuations of åt-1 and random fluctuations of åt-2.
A one unit random fluctuation in åt-2 causes a change of –è1 in Yt-1 and –è2 in Yt.
The resulting autocorrelation of Yt-1 and Yt is –è1 × –è2 = è1 × è2.
Jacob: That explanation makes sense, but now I don’t understand the autocovariance for an AR(2) process.
In an AR(2) process, a change in Yt-1 may reflect a random fluctuation in åt-1 or a random fluctuation in åt-2.
A random fluctuation of 1 unit in åt-1 causes a one unit change in Yt-1, which causes a change of ö1 in Yt.
A random fluctuation of one unit in åt-2 causes a change of ö1 in Yt-1 and a change of ö2 in Yt.
By your logic, ñ1 for an AR(2) process should be ö1 + ö1 × ö2, instead of just ö1.
Rachel: For an MA(2) process, è1 considers the random fluctuation in just åt-1. The random fluctuation in åt-2 is picked up by è2. For an AR(2) process, è1 considers all changes in Yt-1, whether they come from random fluctuations in åt-1 or åt-2 or any other error term. The ö1 parameter relates all these changes to Yt.
Jacob: Your explanation makes sense, but now I don’t understand the ñ2 in an AR(2) process.
ö1 is the effect of Yt-1 on Yt.
ö2 is the effect of Yt-1 on Yt+1.
The covariance of lag 2, ã2, is (ö12 + ö2) × ó2ε . This seems to double count. The effect of Yt-1 on Yt+1 includes the two step effect of Yt-1 on Yt and of Yt on Yt+1. Isn’t this two step effect double counted in ã2?
Rachel: You are confusing ö2 with ñ2.
ñ2 is the effect of Yt-1 on Yt+1, including all multi-step effects.
ö2 is the direct effect of Yt-1 on Yt+1, independent of any multi-step effects.
The formula for ñ2 combines all the independent effects of the ö parameters.
Part C: ã2 = (–è2) × ó2 = –0.5 × 22 = -2.000
Part D: ñ1 = (–è1 + è1 × è2) / (1 + è12 + è22) = -0.201
Part E: ñ2 = (–è2) / (1 + è12 + è22) = -0.287
(See Cryer and Chan page 62, equation at bottom of page)
For AR(1), AR(2), MA(1), MA(2), and ARMA(1,1) processes, know how to calculate ã0, ã1, ã2, ñ1, and ñ2 from ö1, ö2, è1, and è2. For an MA(2) process:
ã0 =(1 + è12 + è22) × ó2
ã1 = (–è1 + è1 × è2) × ó2
ã2 = (–è2) × ó2
ñ1 = (–è1 + è1 × è2) / (1 + è12 + è22)
ñ2 = (–è2) / (1 + è12 + è22)
ñk = 0 for k = 3, 4, …
See Cryer and Chan, page 63 (equation 4.2.3)
**Exercise 5.2: Variance and autocorrelations of moving average process
A moving average process of order 2 is Yt = et – 0.6 et-1 – et-2 , with ó2e = 1
A. What is ã0, the variance of Yt?
B. What is ã1, the autocovariance of lag 1?
C. What is ã2, the autocovariance of lag 2?
D. What is ñ1, the autocorrelation of lag 1?
E. What is ñ2, the autocorrelation of lag 2?
Part A: ã0 = (1 + è21 + è22) × ó2ε = (1 + 0.62 + 12) × 1 = 2.36
Intuition: Yt is the sum of three independent random variables, with variances of 1, 0.36, and 1.
See Cryer and Chan, chapter 4, page 62, equation at the bottom of the page.
Part B: The autocovariance of lag 1 is – è1 × (1 – è2) = +0.6 × (1 – 1) = 0.
Part C: The autocovariance of lag 2 is – è2 = –1.
Part D: See Cryer and Chan, chapter 4, page 63, equation 4.2.3.
(–0.6 + 0.6 × 1) / (1 + 0.62 + 12) = 0/2.36
ñ1 is the correlation of et – 0.6 et-1 – et-2 with et-1 – 0.6 et-2 – et-3
The numerator has two non-zero terms: –0.6 e2t-1 and +0.6 e2t-2
The error terms have the same variance, so these two terms cancel out.
Jacob: Does ñ2 = 0 mean that the sample autocorrelation of lag 1 in this time series is zero?
Rachel: Actual time series are finite, so sample autocorrelations have random fluctuations. If the time series has 100 observations, the sample autocorrelation r1 is distributed as a normal random variable with a mean of zero and a standard error of 1/√100 = 0.10. The observed autocorrelation is not zero. But if the time series is infinite, the sample autocorrelation of lag 2 approaches zero. This exercise assumes the moving average parameters are known and asks about the true autocorrelations.
Jacob: How should we think of this? Can one see the correlation intuitively?
Rachel: et is correlated with et-1 two ways:
Directly, with a correlation of –è1.
Indirectly through et-2 and back to et-1, with a correlation of –è2 × –è1.
Moving from et-1 to et-2 is like moving from et-2 to et-1. We deal with these relations in more detail in modules 19 and 20 (seasonal time series), where ñ11 = ñ13.
Part E: See Cryer and Chan, chapter 4, page 63, equation 4.2.3
(–1) / (1 + 0.62 + 12) = –1/2.36 = -0.424
Intuition: ñ2 is the correlation of et – 0.6 et-1 – et-2 with et-2 – 0.6 et-3 – et-4
The numerator has one non-zero term: –1 × e2t-2
The denominator is the variance ã0