STA3704
Assignment 3
Unique No: 659946
DUE 11 August 2025
, ASSIGNMENT 03
Unique Nr.: 659946
Due date: 11 August 2025
Question 1 [12 Marks]
Given:
• 𝑛 = 100
• 𝑦‾ = 7
• 𝑠𝑦2 = 20
• 𝑟1 = 0.8, 𝑟 2 = 0.6, 𝑟3 = 0.4
• Model: 𝑌𝑡 = 𝜃0 + 𝜙1 𝑌𝑡−1 + 𝜙2 𝑌𝑡−2 + 𝜀𝑡
Yule-Walker equations:
𝛾1
𝑟1 = = 𝜙 1 + 𝜙2 𝑟1
𝛾0
𝑟2 = 𝜙 1 𝑟1 + 𝜙2
Substitute: From first: 0.8 = 𝜙 1 + 0.8𝜙2 From second: 0.6 = 0.8𝜙 1 + 𝜙2
Solve: Multiply first by 0.8: 0.64 = 0.8𝜙 1 + 0.64𝜙2 Subtract from second: 0.6 − 0.64 =
(𝜙2 − 0.64𝜙2 ) ⇒ −0.04 = 0.36𝜙 2 ⇒ 𝜙 2 = −0.1111
Then 𝜙1 = 0.8 − 0.8(−0.1111) = 0.8889
𝜃0 :
𝜃0 𝜃0 𝜃0
𝜇= ⇒ 7= =
1 − 𝜙1 − 𝜙2 1 − 0.8889 + 0.1111 0.2222
⇒ 𝜃0 = 1.5554
𝜎𝜀2 :
Assignment 3
Unique No: 659946
DUE 11 August 2025
, ASSIGNMENT 03
Unique Nr.: 659946
Due date: 11 August 2025
Question 1 [12 Marks]
Given:
• 𝑛 = 100
• 𝑦‾ = 7
• 𝑠𝑦2 = 20
• 𝑟1 = 0.8, 𝑟 2 = 0.6, 𝑟3 = 0.4
• Model: 𝑌𝑡 = 𝜃0 + 𝜙1 𝑌𝑡−1 + 𝜙2 𝑌𝑡−2 + 𝜀𝑡
Yule-Walker equations:
𝛾1
𝑟1 = = 𝜙 1 + 𝜙2 𝑟1
𝛾0
𝑟2 = 𝜙 1 𝑟1 + 𝜙2
Substitute: From first: 0.8 = 𝜙 1 + 0.8𝜙2 From second: 0.6 = 0.8𝜙 1 + 𝜙2
Solve: Multiply first by 0.8: 0.64 = 0.8𝜙 1 + 0.64𝜙2 Subtract from second: 0.6 − 0.64 =
(𝜙2 − 0.64𝜙2 ) ⇒ −0.04 = 0.36𝜙 2 ⇒ 𝜙 2 = −0.1111
Then 𝜙1 = 0.8 − 0.8(−0.1111) = 0.8889
𝜃0 :
𝜃0 𝜃0 𝜃0
𝜇= ⇒ 7= =
1 − 𝜙1 − 𝜙2 1 − 0.8889 + 0.1111 0.2222
⇒ 𝜃0 = 1.5554
𝜎𝜀2 :