STAT 3830: Time Series and Forecasting Reference answer for Assignment 2 April 27, 2007 1, Using Mathematical induction: (i),Ẑn (1) = α[Zn + (1 − α)Zn−1 + . . .] Ẑn (2) = α[Ẑn (1) + (1 − α)Zn + (1 − α)2 Zn−1 + . . .] = αẐn (1) + (1 − α)[α(Zn + (1 − α)Zn−1 + . . .)] = αẐn (1) + (1 − α)Ẑn (1) = Ẑn (1) (ii), letẐn (i) = Ẑn (1)f ori ≤ k ≤ l we have Ẑn (k + 1) = α[Ẑn (k) + (1 − α)Ẑn (k − 1) + (1 − α)2 Ẑn (k − 2) + . . . + (1 − α)k−1 Ẑn (1) + (1 − α)k Zn + . . .] 1 − (1 − α)k ] + (1 − α)k [αZn + (1 − α)Zn−1 + . . .] = α[Ẑn (1) 1 − (1 − α) = Ẑn (1) − (1 − α)k Ẑn (1) + (1 − α)k Ẑn (1) = Ẑn (1) Consider (i) and (ii), we can conclude from MI that Ẑn (l) = Ẑn (1), for l > 0 2, Ẑ30 (2) = Ẑ30 (1) = 102.5 Ẑ31 (1) = Ẑ30 (1) + α[Z31 − Ẑ30 (1)] = 102.5 + 0.1(105 − 102.5) = 102.75 3, β̂0,24 = 30, β̂1,24 = 2, α = 0.1, w = 0.9 ( [1] [2] [1] 2S24 − S24 = 30 S24 = 12 ⇒ [1] [2] [2] 0.1 [S − S24 ] = 2 S24 ] = −6 0.9 24 1 [1] [2] β̂0,25 = 2S25 − S25 = 31.24 [1] [1] S25 = 0.1Z25 + 0.9S24 = 2.8 + 0.9 × 12 = 13.6 [1] [2] ⇒ β̂1,25 = wa (S25 − S25 ) = 1.96 [2] S = 0.1 × 13.6 + 0.9 × (−6) = −4.04 25 Ẑ25 (1) = 31.24 + 1.96 = 33.2 [1] [2] β̂0,26 = 2S26 − S26 = 32.782 [1] [1] S26 = 0.1Z26 + 0.9S25 = 3.1 + 0.9 × 13.6 = 15.34 [1] [2] ⇒ β̂1,26 = wa (S26 − S26 ) = 1.938 [2] S26 = 0.1 × 15.34 + 0.9 × (−4.04) = −2.102 Ẑ26 (1) = 32.782 + 1.938 = 34.72 [1] [2] β̂0,27 = 2S27 − S27 = 34.9632 [1] [1] S27 = 0.1Z27 + 0.9S26 = 3.6 + 0.9 × 15.34 = 17.406 [2] [1] ⇒ β̂1,27 = wa (S27 − S27 ) = 1.9508 [2] S27 = 0.1 × 17.406 + 0.9 × (−2.102) = −0.1512 Ẑ27 (1) = 34.9632 + 1.9508 = 36.887 4, 1 1 1 0 0 1 0 0 a, f (j + 1) = sin π2 (j + 1) = cos π2 j = f (j) 0 0 1 ⇒ L = 0 0 1 cos π2 (j + 1) − sin π2j 0 −1 0 0 −1 0 1 j+1 b, f (j + 1) = = π sin 2 (j + 1) cos π2 (j + 1) 1 1 = f (j) 0 0 0 0 1 0 0 0 0 −1 0 0 ⇒L= 1 0 2 1 1 0 0 1 j+1 cos π2 j − sin π2j 0 0 1 0 0 0 0 −1 0 0 1 0 1 j+1 π (j + 1) sin 2 c, f (j + 1) = = cos π2 (j + 1) π (j + 1)sin 2 (j + 1) (j + 1)cos π2 (j + 1) 1 j+1 π j cos 2 −sinπ2j (j + 1)cos π2 j −(j + 1)sin π2 j 1 1 0 = f (j) 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 −1 0 0 0 0 1 0 0 −1 0 −1 0 0 0 ⇒L= 0 1 0 1 1 0 0 0 0 1 j+1 (j + 1) sin 2π 2 c, f (j + 1) = = 2π cos 2 (j + 1) sin 4π (j + 1) 2 cos 4π (j + 1) 2 1 1 0 = f (j) 0 0 0 0 0 1 √0 0 23 0 − 21 0 0 0 0 0 0 1 √2 3 2 0 0 0 0 0 0 1 2√ − 3 2 0 0 0 0 √ 1 1 0 ⇒L= 0 3 0 2 1 0 2 3 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 −1 0 0 0 0 1 0 0 −1 0 −1 1 j+1 √ 3 2 sin 2πj + 12 cos 2πj 12 12 √3 cos 2πj − 1 sin 2πj 2 12 √ 2 12 1 4πj 4πj 3 sin + cos 2 12 2 12 √ 4πj 4πj 3 1 cos − sin 2 12 2 12 0 0 1 √0 0 23 0 − 12 0 0 0 0 0 0 1 √2 3 2 0 0 0 0 0 0 1 2√ − 3 2 0 0 0 0 √ 3 2 1 2