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# hwk5 - IEOR E4709 Data Analysis for Financial Engineers...

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IEOR E4709 Data Analysis for Financial Engineers Solution 5 1. Problem 2.1 in Tsay’s book: Solution: From the model, 101 = 101 +0 2 100 . Taking conditional expectation at =100 ,wehave the 1-step forecast as 100 (1) = E ( 101 |F (100)) = E ( 101 |F (100)) + E (0 2 100 |F (100)) = 0 + 0 2 100 =0 002 The associated forecast error is 100 (1) = 101 100 (1) = 101 . Therefor, the standard deviation of the forecast error is p Var ( 100 (1)) = std ( 101 )=0 025 For 2-step ahead forecast, we have 102 = 102 2 101 ,andthen 101 (2) = E ( 102 |F (100)) = E ( 102 |F (100)) + E (0 2 101 |F (100)) = 0 The associated forecast error is 100 (2) = 102 100 (2) = 102 2 101 . Therefor, the standard deviation of the forecast error is p ( 100 (2)) = p ( 102 )+0 04 ( 101 )= 1 04 0255 To compute ACF of , use the model to obtain ( ( 2 2 ( 1 )=1 04 2 00065 . Cov (  1 Cov ( 2 1  1 2 2 Cov (0 2 1 1 2 2 = 0 000125 . Cov ( ,for 2 . Therefore, ACF of is 1 192 , for 2 . 2. Problem 2.4 in Tsay’s book Solution: (b) Figures 1 presents the ACF and PACF of Decile 2. We can see that lags 1 and 12 seem to be signi f cant in the acf plot, and lags 1 and 12 seem to be signi f cant in the partial acf plot.

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hwk5 - IEOR E4709 Data Analysis for Financial Engineers...

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