Time Series Analysis HW 09 Answer keys - TIME SERIES ANALYSIS 1 Assignment 9 answer keys 6.1 We know from Problem 1.10 that 1 Since eliminates

Time Series Analysis HW 09 Answer keys - TIME SERIES...

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Unformatted text preview: TIME SERIES ANALYSIS 1 Assignment 9 answer keys 6.1 We know from Problem 1.10 that 1 − Since , ,⋯, eliminates polynomials in of degree ≤ are independent of , we have 1− This together with 1− 6.4 1− + = + + ⋯+ = 0. implies + + ⋯+ − 1. . = The augmented Dickey–Fuller test yields a p-value of 0.254, so we don’t reject the null hypothesis of AR unit root. Then, assuming AR(1) model is equivalent to assuming ARIMA(0, 1, 0) model and assuming AR(2) model is equivalent to assuming ARIMA(1, 1, 0) model. Results from the two analyses vary. 6.5 |1, , Let determined by 1 ! 2 3 ,⋯, #$ #$ #) = − − − + + − − − − − − From equations (1) and (2), we obtain that is, If = = . + # =0 ⟹ # + # / =0 = 0 or 9 : = 0, then + Therefore, where ,⋯, 4 5 − ⋯− − ⋯− − ⋯− = = 0. |1, #$ #) − − , ,⋯, −⋯− − ⋯− . Then, coefficients , , ∙ 1& = 0 ∙ &=0 . ∙ * + = 0, , = 1, ⋯ , - = 0 or det 6 # = 0. Otherwise, if 9 : are any solution of ; + ⋯+ = 1 = 0, then + ⋯+ + , ,⋯, = / = ∙ 1& = 0 . ∙ * + = 0, , = 1, ⋯ , - On the other hand, as > ? is a causal ARMA process with zero mean, |1, # # +⋯+ , ,⋯, are 7 = 0; + # = 0. 2 TIME SERIES ANALYSIS where So, ,⋯, again are any solution of equations (4) and (5). |1, , ,⋯, = 6.7 (Answers are provided for reference only) |1, ,⋯, = + ⋯+ . (a) (See the following pages for R plots) Take log transformation to stabilize the change in volatility with the level of the series Apply 1 − / to eliminate seasonal component ― Not clear whether 1 − should then be applied but it’s better not to difference more than necessary. So we first try without applying 1 − . Subtract mean since we’ll be fitting a zero-mean model ― Sample ACF and PACF plots suggest MA(12) or AR(13) Fit MA(12) and AR(13) to the mean-corrected differenced log data ― MA(12): AIC = −438.8 ― AR(13): AIC = −429 ― Subset AR(13) with non-zero coefficients at lags 1, 2, 12, & 13: AIC = −439.67 In view of the possibility that we could have applied 1 − ARMA(1, 12) to the data, now try ― Subset ARMA(1, 12) with non-zero coefficients at AR lag 1 and MA lags 1, 3, 4, & 12: AIC = −450.25 ― ARIMA model for the logarithms of the data in ASHORT.TSM is given by 1− / 1 − 0.875 0.050 where > ?~JK 0, 0.001 Randomness tests get passed R output: = − 0.304 0.113 − 0.225 0.102 G − 0.236 0.107 H − 0.671 0.101 / , 3 TIME SERIES ANALYSIS (b) 95% confidence bounds for coefficients: : − 0.875 ± 1.96 0.050 : − 0.304 ± 1.96 0.113 G: H: /: − 0.225 ± 1.96 0.102 − 0.236 ± 1.96 0.107 − 0.671 ± 1.96 0.101 (c) Sample ACF and PACF plots of residuals: 4 TIME SERIES ANALYSIS Ljung-Box test has a p-value of 0.70. (d) Graph of data and forecasts: (e) The 12-step ahead forecast and the corresponding 95% prediction bounds: Step 1 2 3 4 5 6 7 8 9 Forecast 425.70 401.53 467.48 457.81 479.40 557.53 641.79 645.74 538.05 Lower bound 396.67 368.57 423.70 412.73 431.57 501.10 575.79 578.17 480.65 Upper bound 456.86 437.45 515.78 507.82 532.52 620.32 715.35 721.21 602.30 True value 417 391 419 461 472 535 622 606 508 Error ‒8.70 ‒10.53 ‒48.48 3.19 ‒7.40 ‒22.53 ‒19.79 ‒39.74 ‒30.05 5 TIME SERIES ANALYSIS 10 11 12 477.31 414.61 456.73 425.32 368.42 404.62 535.66 466.58 515.56 461 390 432 ‒16.31 ‒24.61 ‒24.73 (f) See table in part (e) for forecast errors. 6.11(a) Based on the discussions in the lecture, a SARMA(0, 1)×(1, 0)12 model is suitable for the series O = 0.8, P = 0.5 (invertible solution), and QR / = 7.2. > ?, where Φ So, for the series > ? we would suggest the following SARIMA(0, 1, 1)×(1, 1, 0)12 model: 1− 1− / 1 − 0.8 / = 1 + 0.5 , > ?~JK 0, 7.2 ...
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  • Spring '17
  • Abolfazl Safikhani
  • Statistics, Time series analysis, Autoregressive integrated moving average

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