Download presentation

Presentation is loading. Please wait.

Published byAmber McCann Modified about 1 year ago

1
Comparison of the models

2
Concentration data

3
Its ACF

4
Its PACF. AR(2)?

5
AR(1)? m = 17.06, p-value 0.0000 a 1 = 0.5734, p-value 0.0000 Portmanteau test 47.44, p-value 0.003

6
AR(2)? m = 17.06, p-value 0.0000 a 1 = 0.4263, p-value 0.0000 a 2 = 0.2576, p-value 0.0003 Portmanteau test 26.97, p-value 0.2573

7
AR(3)? m = 17.06, p-value 0.0000 a 1 = 0.4058, p-value 0.0000 a 2 = 0.2241, p-value 0.0036 a 3 = 0.0808, p-value 0.2719 Portmanteau test 26.14, p-value 0.2456

8
Increments of the data

9
ACF of the increments

10
PACF of the increments

11
MA(1)? b 1 = 0.701, p-value 0.0023 Portmanteau test 29.14, p-value 0.1879

12
MA(2)? b 1 = 0.6314, p-value 0.0457 b 2 = 0.1209, p-value 0.7005 Portmanteau test 24.77, p-value 0.3625

13
AR(4)? a 1 = -0.5521, p-value 0.0000 a 2 = -0.3161, p-value 0.0001 a 3 = -0.2518, p-value 0.0024 a 4 = -0.1365, p-value 0.0667 Portmanteau test 25.5, p-value 0.0249

14
AR(5)? a 1 = -0.5784, p-value 0.0000 a 2 = -0.3647, p-value 0.0000 a 3 = -0.3147, p-value 0.0002 a 4 = -0.2527, p-value 0.0028 a 5 = -0.2095, p-value 0.0052 Portmanteau test 25.4, p-value 0.1864

15
AR(6)? a 1 = -0.6248, p-value 0.0000 a 2 = -0.4218, p-value 0.0000 a 3 = -0.3855, p-value 0.0000 a 4 = -0.3358, p-value 0.0001 a 5 = -0.3494, p-value 0.0001 a 6 = -0.2386, p-value 0.0015 Portmanteau test 14.81, p-value 0.7346

16
AR(7)? a 1 = -0.6223, p-value 0.0000 a 2 = -0.4182, p-value 0.0000 a 3 = -0.3817, p-value 0.0000 a 4 = -0.331, p-value 0.0003 a 5 = -0.3447, p-value 0.0002 a 6 = -0.2307, p-value 0.0101 a 7 = 0.0127, p-value 0.8716 Portmanteau test 14.83, p-value 0.6739

17
Coal Production data

18
Its ACF

19
Its PACF. AR(2)?

20
AR(1)? m = 3.772, p-value 0.0000 a 1 = 0.7055, p-value 0.0000 Portmanteau test 21.44, p-value 0.6124

21
AR(2)? m = 3.802, p-value 0.0000 a 1 = 0.4896, p-value 0.0000 a 2 = 0.3308, p-value 0.0016 Portmanteau test 12.03, p-value 0.97

22
AR(3)? m = 3.809, p-value 0.0000 a 1 = 0.4593, p-value 0.0000 a 2 = 0.2913, p-value 0.0111 a 3 = 0.0918, p-value 0.4054 Portmanteau test 10.61, p-value 0.9799

23
Profit Margin data

24
Its ACF

25
Its PACF. AR(1)?

26
AR(1)? m = 4.699, p-value 0.0000 a 1 = 0.876, p-value 0.0000 Portmanteau test 22.34, p-value 0.5587 Still, ρ(4) is out of range

27
AR(2)? m = 4.716, p-value 0.0000 a 1 = 1.026, p-value 0.0000 a 2 = -0.173, p-value 0.1311 Portmanteau test 21.98, p-value 0.5214

28
ARMA(1,1)? m = 4.714, p-value 0.0000 a 1 = 0.8281, p-value 0.0026 b 1 = -0.2024, p-value 0.0070 Portmanteau test 19.06, p-value 0.6976

29
Parts Availability data

30
Its increments

31
ACF for the increments. MA(1)?

32
PACF for the increments. AR(2)?

33
AR(1)? a 1 = -0.5605, p-value 0.0000 Portmanteau test 22.4, p-value 0.5552 ρ(2) is way out of range though

34
AR(2)? a 1 = -0.7646, p-value 0.0000 a 2 = -0.4401, p-value 0.0001 Portmanteau test 14.5, p-value 0.9177

35
AR(3)? a 1 = -0.8283, p-value 0.0000 a 2 = -0.5715, p-value 0.0000 a 3 = -0.1966, p-value 0.1046 Portmanteau test 12.24, p-value 0.9522

36
MA(1)? b 1 = 0.7249, p-value 0.0000 Portmanteau test 12.23, p-value 0.9722

37
Treasury Bonds Yield data

38
Its increments

39
ACF for the increments. MA(1)??

40
PACF for the increments. AR(1)?

41
AR(1)? a 1 = 0.4241, p-value 0.0000 Portmanteau test 29.02, p-value 0.2195

42
AR(2)? a 1 = 0.4166, p-value 0.0000 a 2 = 0.0181, p-value 0.8265 Portmanteau test 28.96, p-value 0.1806

43
MA(1)? b 1 = -0.3818, p-value 0.0496 Portmanteau test 36.57, p-value 0.0483

44
MA(2)? b 1 = -0.4097, p-value 0.0487 b 2 = -0.1249, p-value 0.5456 Portmanteau test 30.89, p-value 0.1254

45
ARMA(1,1)? a 1 = 0.4678, p-value 0.0286 b 1 = -0.0532, p-value 0.7571 Portmanteau test 28.94, p-value 0.1823

46
Crops Prices data

47
Take the Logarithm

48
Take the difference

49
Its ACF

50
MA(3) might work?

51
MA(3)? b 1 = 0.03614, p-value 0.8163 b 2 = 0.4499, p-value 0.0012 b 3 = 0.2209, p-value 0.1565 Portmanteau test 27.11, p-value 0.2072

52
Residual ACF for MA(3) model. Looks good, but the estimate for b(3) has a big p-value 0.157

53
MA(2)? b 1 = 0.1188, p-value 0.4004 b 2 = 0.5034, p-value 0.0004 Portmanteau test 54.11, p-value 0.0003

54
Residual ACF for MA(2) model, Portmanteau test has p-value 0.0003

55
Switch to AR models. Here is PACF of the data.

56
AR(8)? Or AR(4)? 8 th value is almost 4 standard deviations

57
AR(4)? a 1 = 0.0394, p-value 0.4487 a 2 = -0.3577, p-value 0.0000 a 3 = -0.1328, p-value 0.0109 a 4 = -0.1394, p-value 0.0082 Portmanteau test 39.62, p-value 0.0083

58
Residual ACF for AR(4). Portmanteau test is a No (p-value 0.008)

59
AR(8)? a 1 = 0.0059, p-value 0.909 a 2 = -0.3927, p-value 0.0000 a 3 = -0.1939, p-value 0.0005 a 4 = -0.2112, p-value 0.0002 a 5 = -0.1203, p-value 0.0323 a 6 = -0.1443, p-value 0.0096 a 7 = 0.0787, p-value 0.1313 a 8 = 0.2046, p-value 0.0001 Portmanteau test 21.27, p-value 0.2145

60
Residual ACF for AR(8). Portmanteau test is a Yes (p-value 0.215). Finally?

61
Since 8 is sort of too many, let’s try mixed models

62
ARMA(2,1)? a 1 = 0.7933, p-value 0.0000 a 2 = -0.3278, p-value 0.0000 b 1 = 0.8421, p-value 0.0000 Portmanteau test 28.44, p-value 0.1614

63
Residuals for ARMA(2,1). Portmanteau test is a Yes (p-value 0.161)

64
Best AIC score = ARMA(8,1) a 1 = 0.6414, p-value 0.0000 a 2 = -0.4063, p-value 0.0007 a 3 = 0.0473, p-value 0.4489 a 4 = -0.0978, p-value 0.2116 a 5 = -0.0073, p-value 0.9147 a 6 = -0.0878, p-value 0.2101 a 7 = -0.0357, p-value 0.6024 a 8 = -0.14470.2046, p-value 0.0225 b 1 = 0.6803, p-value 0.0000 Portmanteau test 12.1, p-value 0.7373

65
Residual ACF for ARMA(8,1). Note that the first 5 values are practically zeroes, one of the symptoms of over- parametrization

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google