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 Y_t = e_t – è₁ e_t-1 – è₂ e_t-2, with è₁ = 0.7, è₂ = 0.5, and ó_e = 2.

A.    What is ã₀, the variance of Y_t?

B.    What is ã₁, the autocovariance of Y_t and Y_t-1?

C.    What is ã₂, the autocovariance of Y_t and Y_t-2?

D.    What is ñ₁, the autocorrelation of Y_t and Y_t-1?

E.    What is ñ₂, the autocorrelation of Y_t and Y_t-2?

Part A: ã₀ = (1 + è²₁ + è²₂) × ó²_e = (1 + 0.49 + 0.25 ) × 2² = 6.960

Y_t is the sum of three independent random variables: e_t, e_t-1, and e_t-2, with coefficients of 1, –è₁, and –è₂.

Each random variable has a variance of ó²_ε. 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) × ó²_ε = (–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 å_t-1 increases one unit, Y_t-1 increases one unit and Y_t increases 1 × –è₁ units. The covariance is in terms of the variances: for each 1 variance change of å_t-1, Y_t decreases è₁ × ó²_ε. 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 å_t-1 on Y_t. Your conclusion is correct: the covariance of å_t-1 and Y_t is –è₁ × ó²_ε.

        This exercise asks for the covariance of Y_t-1 and Y_t.

        Y_t-1 may change randomly for two reasons: random fluctuations of å_t-1 and random fluctuations of å_t-2.

        A one unit random fluctuation in å_t-2 causes a change of –è₁ in Y_t-1 and –è₂ in Y_t.

        The resulting autocorrelation of Y_t-1 and Y_t 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 Y_t-1 may reflect a random fluctuation in å_t-1 or a random fluctuation in å_t-2.

        A random fluctuation of 1 unit in å_t-1 causes a one unit change in Y_t-1, which causes a change of ö₁ in Y_t.

        A random fluctuation of one unit in å_t-2 causes a change of ö₁ in Y_t-1 and a change of ö₂ in Y_t.

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 å_t-1. The random fluctuation in å_t-2 is picked up by è₂. For an AR(2) process, è₁ considers all changes in Y_t-1, whether they come from random fluctuations in å_t-1 or å_t-2 or any other error term. The ö₁ parameter relates all these changes to Y_t.

Jacob: Your explanation makes sense, but now I don’t understand the ñ₂ in an AR(2) process.

        ö₁ is the effect of Y_t-1 on Y_t.

        ö₂ is the effect of Y_t-1 on Y_t+1.

The covariance of lag 2, ã₂, is (ö₁² + ö₂) × ó²_ε . This seems to double count. The effect of Y_t-1 on Y_t+1 includes the two step effect of Y_t-1 on Y_t and of Y_t on Y_t+1. Isn’t this two step effect double counted in ã₂?

Rachel: You are confusing ö₂ with ñ₂.

        ñ₂ is the effect of Y_t-1 on Y_t+1, including all multi-step effects.

        ö₂ is the direct effect of Y_t-1 on Y_t+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)

ñ_k = 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 Y_t = e_t – 0.6 e_t-1 – e_t-2 , with ó²_e = 1

A.    What is ã₀, the variance of Y_t?

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) × ó²_ε = (1 + 0.6² + 1²) × 1 = 2.36

Intuition: Y_t 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.

 or

(–0.6 + 0.6 × 1) / (1 + 0.6² + 1²) = 0/2.36

ñ₁ is the correlation of e_t – 0.6 e_t-1 – e_t-2 with e_t-1 – 0.6 e_t-2 – e_t-3

        The numerator has two non-zero terms: –0.6 e²_t-1 and +0.6 e²_t-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: e_t is correlated with e_t-1 two ways:

        Directly, with a correlation of –è₁.

        Indirectly through e_t-2 and back to e_t-1, with a correlation of –è₂ × –è₁.

Moving from e_t-1 to e_t-2 is like moving from e_t-2 to e_t-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 e_t – 0.6 e_t-1 – e_t-2 with e_t-2 – 0.6 e_t-3 – e_t-4

        The numerator has one non-zero term: –1 × e²_t-2 

        The denominator is the variance ã₀