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Published byAmber McCann Modified over 9 years ago
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Comparison of the models
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Concentration data
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Its ACF
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Its PACF. AR(2)?
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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
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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
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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
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Increments of the data
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ACF of the increments
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PACF of the increments
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MA(1)? b 1 = 0.701, p-value 0.0023 Portmanteau test 29.14, p-value 0.1879
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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
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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
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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
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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
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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
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Coal Production data
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Its ACF
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Its PACF. AR(2)?
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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
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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
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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
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Profit Margin data
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Its ACF
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Its PACF. AR(1)?
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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
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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
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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
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Parts Availability data
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Its increments
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ACF for the increments. MA(1)?
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PACF for the increments. AR(2)?
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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
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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
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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
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MA(1)? b 1 = 0.7249, p-value 0.0000 Portmanteau test 12.23, p-value 0.9722
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Treasury Bonds Yield data
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Its increments
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ACF for the increments. MA(1)??
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PACF for the increments. AR(1)?
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AR(1)? a 1 = 0.4241, p-value 0.0000 Portmanteau test 29.02, p-value 0.2195
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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
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MA(1)? b 1 = -0.3818, p-value 0.0496 Portmanteau test 36.57, p-value 0.0483
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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
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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
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Crops Prices data
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Take the Logarithm
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Take the difference
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Its ACF
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MA(3) might work?
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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
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Residual ACF for MA(3) model. Looks good, but the estimate for b(3) has a big p-value 0.157
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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
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Residual ACF for MA(2) model, Portmanteau test has p-value 0.0003
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Switch to AR models. Here is PACF of the data.
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AR(8)? Or AR(4)? 8 th value is almost 4 standard deviations
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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
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Residual ACF for AR(4). Portmanteau test is a No (p-value 0.008)
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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
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Residual ACF for AR(8). Portmanteau test is a Yes (p-value 0.215). Finally?
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Since 8 is sort of too many, let’s try mixed models
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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
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Residuals for ARMA(2,1). Portmanteau test is a Yes (p-value 0.161)
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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
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Residual ACF for ARMA(8,1). Note that the first 5 values are practically zeroes, one of the symptoms of over- parametrization
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