# Reshenia_zadach - 1.2 3.3 4.3 5.4 7.4 9.

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Оглавление Задача 1 ...................................................................................................................................................... 2 Задача 3 ...................................................................................................................................................... 3 Задача 4 ...................................................................................................................................................... 3 Задача 5 ...................................................................................................................................................... 4 Задача 7 ...................................................................................................................................................... 4 Задача 9 ...................................................................................................................................................... 5 Задача 10 .................................................................................................................................................... 7 Задача 11 .................................................................................................................................................. 10 Задача 13 .................................................................................................................................................. 10 Задача 14 .................................................................................................................................................. 11 Задача 15 .................................................................................................................................................. 12 Задача 16 .................................................................................................................................................. 13 Задача 17 .................................................................................................................................................. 14 Задача 18 .................................................................................................................................................. 15

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1 Задача y t = 0.7 y t 1 0.1 y t 2 + ε t 0.35 ε t 1 + 0.025 ε t 2 y t 0.7 y t 1 + 0.1 y t 2 = ε t 0.35 ε t 1 + 0.025 ε t 2 ( 1 0.7 L + 0.1 L 2 ) y t = ( 1 0.35 L + 0.025 L 2 ) ε t 1) MA y t = ( 1 0.35 L + 0.025 L 2 ) ε t Процесс является обратимым, когда все корни z характеристического уравнения B(z)=0 лежат вне единичного круга 1 0.35 z + 0.025 z 2 = 0 D = 0.0225 z 1 = 10 илиz 2 = 4 Процесс является обратимым 2) AR ε t = ( 1 0.7 L + 0.1 L 2 ) y t Процесс стационарен, когда все корни zA(z)=0 лежат вне единичного круга 0.1 z 2 0.7 z + 1 = 0 D = 0.09 z 1 = 5 илиz 2 = 2 Стационарен 3) TS – процесс стационарен относительно тренда (t) У нас DSно порядковой разности нулевой, т.е. S I(0) 5) ARIMA(p,0,q)=ARMA(p,q) У нас ARIMA(2,0,2)= ARMA(2,2) x t = x t 1 0.2 x t 2 + ε t 0.35 ε t 1 + 0.025 ε t 2 x t x t 1 + 0.2 x t 2 = ε t 0.35 ε t 1 + 0.025 ε t 2 ( 1 L + 0.2 L 2 ) x t = ( 1 0.35 L + 0.025 L 2 ) ε t
1) MA x t = ( 1 0.35 L + 0.025 L 2 ) ε t Процесс является обратимым, когда все корни z характеристического уравнения B(z)=0 лежат вне единичного круга 1 0.35 z + 0.025 z 2 = 0 D = 0.0225 z 1 = 10 илиz 2 = 4 Процесс является обратимым 2) AR ε t = ( 1 L + 0.2 L 2 ) x t Процесс стационарен, когда все корни zA(z)=0 лежат вне единичного круга 0.2 z 2 z + 1 = 0 D = 0.2 z 1 = 1.382 или z 2 = 3.618 Стационарен Далее аналогично (1) 3 Задача Для модели ( 1 L ) ( 1 + 0.4 L ) X t = ( 1 0.5 L ) ε t определить параметры p ,d,q. Является липроцесс стационарным? Решение ( 1 L ) ( 1 + 0.4 L ) X t = ( 1 0.5 L ) ε t A ( L ) = ( 1 L ) ( 1 + 0.4 L ) A ( z ) = 0 ( 1 z ) ( 1 + 0.4 z ) = 0 z 1 = 1, z 2 = 1 0.4 →нестационарныйпроцесс ( 1 L + 0.4 L 0.4 L 2 ) X t = ε t 0.5 ε t 1 X t 0.6 X t 1 0.4 X t 2 = ε t 0.5 ε t 1 X t = 0.6 X t 1 + 0.4 X t 2 = ε t 0.5 ε t 1

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X t = 0.6 X t 1 + 0.4 X t 2 + ε t 0.5 ε t 1 X t X t 1 = 0.6 X t 1 X t 1 + 0.4 X t 2 + ε t 0.5 ε t 1 ∆ X t =− 0.4 X t 1 + 0.4 X t 2 + ε t 0.5 ε t 1 X ( ¿¿ t 1 X t 2 )+ ε t 0.5 ε t 1 ∆ X t =− 0.4 ¿ ∆ X t =− 0.4 ∆ X t 1 + ε t 0.5 ε t 1 0.4 < 1 Стационарен Ответ: ARIMA(1, 1, 1) 4 Задача X t = 0,1 X t 1 + E t + 0,2 E t 1 X t =10 E t =0,1 Прогноз на 1 шаг.
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