TS module 12 AR(2) process and method of moments (practice problem)

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NEAS	NEAS posted 14 Years Ago #10160 #
Supreme Being Group: Administrators Posts: 4.3K, Visits: 1.3K	TS module 12 AR(2) process and method of moments (practice problem) (The attached PDF file has better formatting.) Exercise 12.1: AR(2) process and method of moments The first two sample autocorrelations of an AR(2) process are r₁ = 0.5 and r₂ = 0.4 A. What is the method of moments (Yule-Walker) estimate for ö₁? B. What is the method of moments (Yule-Walker) estimate for ö₂? Solution 12.1: See Cryer and Chan, top of page 150, equation 7.1.2: Part A:* The estimated ö₁ is ( 0.5 × (1 – 0.4) ) / (1 – 0.5²) = 0.4 Part B: The estimated ö₂ is ( 0.4 – 0.5²) ) / (1 – 0.5²) = 0.2 Final exam problems may give ö₁ and ö₂ to derive ñ₁ and ñ₂; they may also give ó²_ε and derive ã₀, ã₁, and ã₂; they may give r₁ and r₂ and derive ö₁ and ö₂; they may also give Var(Y) (ã₀) and derive ó²_ε. The formulas are the same; compare them as you review the modules. ** Exercise 12.2: AR(2) process and method of moments practice problem An AR(2) process with 100 observations has the following observed values: r₁ = 0.8, r₂ = 0.5, = 2, and variance(Y) = 5. A. What is the simple method of moments estimate of ö₁ used by Cryer and Chan? B. What is the simple method of moments estimate of ö₂ used by Cryer and Chan? C. What is the estimate of è₀? D. What is the estimate of ó²_å? Solution 12.2: See Cryer and Chan, top of page 150, equation 7.1.2: Part A: The estimated ö₁ is ( 0.8 × (1 – 0.5) ) / (1 – 0.8²) = 1.111 Part B: The estimated ö₂ is ( 0.5 – 0.8²) ) / (1 – 0.8²) = -0.389 Part C: The estimated mean of the time series is 2. The mean = è₀ / (1 – ö₁ – ö₂), so è₀ = ì × (1 – ö₁ – ö₂) = 2 × (1 – 1.111 – -0.389) = 0.556. Part D: The estimate of ó²_ε is (1 – 1.111 × 0.8 – (-0.389) × 0.5) × 5 = 1.529 Attachments TS module 12 AR(2) process and method of moments practice problem df.pdf (3.1K views, 54.00 KB) 0
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sdeppe80	sdeppe80 posted 14 Years Ago #10243 #
Forum Newbie Group: Forum Members Posts: 2, Visits: 1	Exercise 12.2 is completely wrong, correct? The formulas are correct but the application of the formulas is wrong. You used r₂² in the denominator instead of r₁². Part A: φ₁ should be (0.8 × (1 − 0.5)) / (1 − 0.8²) = 1.11 Part B: φ₂ should be (0.5 – 0.8²) / (1 − 0.8²) = -0.389 Then Part C would be θ₀ = 2 × (1 – 1.11 – (-0.389)) = 0.558 Part D: σ_ε² = (1 – 1.11 × 0.8 – (-0.389) × 0.5) × 5 = 1.5325 Also equation 7.1.18 on page 152 that you refer to is actually equation 7.1.8 on page 151. Thanks! [NEAS: Thank you; typos corrected] 0
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