Time Series Analysis HW 05 Answer keys - 1 TIME SERIES ANALYSIS Assignment 5 answer keys 3.2(for Problem 3.1(b d as 3.1(b is not causal Lag h

Time Series Analysis HW 05 Answer keys - 1 TIME SERIES...

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Unformatted text preview: 1 TIME SERIES ANALYSIS Assignment 5 answer keys 3.2 (for Problem 3.1(b, d), as 3.1(b) is not causal) Lag h ℎ ℎ 1 -0.994 -0.994 2 0.980 -0.810 3 -0.959 0 4 0.932 0 5 -0.900 0 6 0.866 0 7 -0.830 0 8 0.792 0 9 -0.753 0 10 0.715 0 3.3 (for Problem 3.1(b, d), as 3.1(b) is not causal) =1 = −1.8 = −1.8 = −1.8 = −1.8 3.4 = −1.8 = −1.8 + −0.81 + −0.81 + −0.81 + −0.81 ACF: Multiply both sides of = 2.43 = −2.916 = 3.2805 = −3.54294 − 0.8 = by , ℎ ≥ 0, and take expectations. We have 0 − 0.8 2 = ! 1 − 0.8 1 = 0 2 − 0.8 0 = 0 ⋮ ℎ − 0.8 ℎ − 2 = 0 for ℎ ≥ 3 2 TIME SERIES ANALYSIS 0 = Solve the first 3 equations, we obtain Moreover, with ) 1 = 0, and , ℎ = 0.8 ℎ − 2 for ℎ ≥ 3, we obtain 0.8 ℎ =* It follows that ACF . ℎ =/ PACF: By definition, component of / that is, '( 0 = 1, . 0 . 1 . 1 2 . 0 2 = 0.8 = 6 . 3.8 9 0 0.8 0 25! / 1 =0 0 1 / ℎ = 0 for ℎ > 2. On the other hand, / 2 = '( ) . |ℎ| = 0,2, ⋯ |ℎ| = 1,3, ⋯ |ℎ| = 0,2, ⋯ |ℎ| = 1,3, ⋯ 1 = . 1 = 0, and . 1 1 0 2=3 4 . 2 0 1 3 0 0 4 = 3 4; 0.8 0.8 2 is the last By equation (2.2.11) (textbook, page 54), the stationary solution of the difference equations, =6 + with |6| > 1 & 9 :~<= 0, ! , is given by A It follows that C Now, < = −6 Then, C < = C = 0 and . −6 FGH < @ , < = FGH C = 1+6 Thus, 9< :~<= 0, E( . '( = −>6 D @ ?B ℎ = E( 6 '( = 0 and −6 D @ ℎ −6 | | , D ? @? . . −6 ℎ−1 −6 D ! ℎ + 1 = I6 0 ℎ = 0. ℎ≠0 Note that, the causal representation has a smaller white noise variance than the noncausal representation. 3.9 (a) The ACVF is given by 3 TIME SERIES ANALYSIS K ℎ = FGH +L @ +L @ = ! 1 + L + L 19 + ! L 19 B± : B , +L +L + ! L 1 + ! L L 19 : 9 B± : @ or ! 1+L +L Q ! L O ℎ = K ! LL P ! L O N 0 (b) From R, the sample mean is 28.831 and Lag h X ℎ Lag h X ℎ Lag h X ℎ 0 152670 7 -31165 14 17758 So, LZ = )[ (c) Matching 1 , K [\] K 1 -54327 8 -1088 15 -6200 11 , K = −0.586, LZ 2 -15072 9 15277 16 -9656 2.16 ℎ=0 ℎ = ±1 ℎ = ±11 ℎ = ±12 GRℎSTUVWS 3 14585 10 -12435 17 27981 4 -17178 11 29802 18 -29456 5 6340 12 -50867 19 3693 6 17421 13 13768 20 7569 12 with XK 1 , XK 11 , XK 12 , respectively, we have = That is, the model for 9^ : = ∇∇ ^ = 28.831 + : B± ! L = −54327 I! L L = 29802. ! L = −50867 )[ : is − 0.586 = −0.549, !X = − 0.549 . [\ = 92740. , 9 :~<= 0, 92740 . ...
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