时间序列分析——基于R | 第3章 ARMA模型的性质习题代码

时间序列分析——基于R | 第3章 ARMA模型的性质

往期文章

时间序列分析——基于R | 第1章习题代码

时间序列分析——基于R | 第2章时间序列的预处理习题代码

1. 已知某 A R ( 1 ) AR(1) AR(1)模型为： x t = 0.7 x t − 1 + ε t , ε t ∼ W N ( 0 , 1 ) . x_t=0.7x_{t-1}+\varepsilon_t,\varepsilon_t \sim WN(0,1). xt​=0.7xt−1​+εt​,εt​∼WN(0,1).求 E ( x t ) , V a r ( x t ) , ρ 2 E(x_t),Var(x_t),\rho_2 E(xt​),Var(xt​),ρ2​和 ϕ 22 . \phi_{22}. ϕ22​.

   E ( x t ) = ϕ 0 1 − ϕ 1 = 0 1 − 0.7 = 0 E\left(x_t\right)=\frac{\phi_{0}}{1-\phi_{1}}=\frac{0}{1-0.7}=0 E(xt​)=1−ϕ1​ϕ0​​=1−0.70​=0

   V a r ( x t ) = 1 1 − ϕ 1 2 = 1 1 − 0. 7 2 = 1.96 Var(x_t)=\frac{1}{1-\phi_{1}^{2}}=\frac{1}{1-0.7^2}=1.96 Var(xt​)=1−ϕ12​1​=1−0.721​=1.96

   ρ 2 = ϕ 1 2 = 0. 7 2 = 0.49 \rho_2=\phi_1^2=0.7^2=0.49 ρ2​=ϕ12​=0.72=0.49

   ϕ 22 = ∣ 1 ρ 1 ρ 1 ρ 2 ∣ ∣ 1 ρ 1 ρ 1 1 ∣ = 0.49 − 0. 7 2 1 − 0. 7 2 = 0 \phi_{22}=\frac{\left|\begin{array}{cc}1 & \rho_{1} \\\rho_{1} & \rho_{2}\end{array}\right|}{\left|\begin{array}{cc}1 & \rho_{1} \\\rho_{1} & 1\end{array}\right|}=\frac{0.49-0.7^{2}}{1-0.7^{2}}=0 ϕ22​= ​1ρ1​​ρ1​1​ ​ ​1ρ1​​ρ1​ρ2​​ ​​=1−0.720.49−0.72​=0

2. 已知某 AR ⁡ ( 2 ) \operatorname{AR}(2) AR(2) 模型为: x t = ϕ 1 x t − 1 + ϕ 2 x t − 2 + ε t , ε t ∼ W N ( 0 , σ ε 2 ) x_t=\phi_1 x_{t-1}+\phi_2 x_{t-2}+\varepsilon_t, \varepsilon_t \sim W N\left(0, \sigma_{\varepsilon}^2\right) xt​=ϕ1​xt−1​+ϕ2​xt−2​+εt​,εt​∼WN(0,σε2​), 且 ρ 1 = \rho_1= ρ1​= 0.5 , ρ 2 = 0.3 0.5, \rho_2=0.3 0.5,ρ2​=0.3, 求 ϕ 1 , ϕ 2 \phi_1, \phi_2 ϕ1​,ϕ2​ 的值.

   A R ( 2 ) A R(2) AR(2) 模型有:

   { ρ 1 = ϕ 1 1 − ϕ 2 ρ 2 = ϕ 1 ρ 1 + ϕ 2 ⇒ { 0.5 = ϕ 1 1 − ϕ 2 0.3 = 0.5 ϕ 1 + ϕ 2 ⇒ { ϕ 1 = 7 15 , ϕ 2 = 1 15 ϕ 2 = 1 15 \left\{\begin{array} { l } { \rho _ { 1 } = \frac { \phi _ { 1 } } { 1 - \phi _ { 2 } } } \\ { \rho _ { 2 } = \phi _ { 1 } \rho _ { 1 } + \phi _ { 2 } } \end{array} \Rightarrow \left\{\begin{array} { l } { 0 . 5 = \frac { \phi _ { 1 } } { 1 - \phi _ { 2 } } } \\ { 0 . 3 = 0 . 5 \phi _ { 1 } + \phi _ { 2 } } \end{array} \Rightarrow \left\{\begin{array}{l} \phi_1=\frac{7}{15}, \phi_2=\frac{1}{15} \\ \phi_2=\frac{1}{15} \end{array}\right.\right.\right. {ρ1​=1−ϕ2​ϕ1​​ρ2​=ϕ1​ρ1​+ϕ2​​⇒{0.5=1−ϕ2​ϕ1​​0.3=0.5ϕ1​+ϕ2​​⇒{ϕ1​=157​,ϕ2​=151​ϕ2​=151​​

3. 已知某 AR ⁡ ( 2 ) \operatorname{AR}(2) AR(2) 模型为: ( 1 − 0.5 B ) ( 1 − 0.3 B ) x t = ε t , ε t ∼ W N ( 0 , 1 ) (1-0.5 B)(1-0.3 B) x_t=\varepsilon_t, \varepsilon_t \sim W N(0,1) (1−0.5B)(1−0.3B)xt​=εt​,εt​∼WN(0,1), 求 E ( x t ) E\left(x_t\right) E(xt​), Var ⁡ ( x t ) , ρ k , ϕ k k \operatorname{Var}\left(x_t\right), \rho_k, \phi_{k k} Var(xt​),ρk​,ϕkk​, 其中 k = 1 , 2 , 3 k=1,2,3 k=1,2,3.

   (1) ( 1 − 0.5 B ) ( 1 − 0.3 B ) x t = ε t ⇔ x t = 0.8 x t − 1 − 0.15 x t − 2 + ε t (1-0.5 B)(1-0.3 B) x_t=\varepsilon_t \Leftrightarrow x_t=0.8 x_{t-1}-0.15 x_{t-2}+\varepsilon_t (1−0.5B)(1−0.3B)xt​=εt​⇔xt​=0.8xt−1​−0.15xt−2​+εt​

   E ( x t ) = ϕ 0 1 − ϕ 1 − ϕ 2 = 0 E\left(x_t\right)=\frac{\phi_0}{1-\phi_1-\phi_2}=0 E(xt​)=1−ϕ1​−ϕ2​ϕ0​​=0

   (2)

   Var ⁡ ( x t ) = 1 − ϕ 2 ( 1 + ϕ 2 ) ( 1 − ϕ 1 − ϕ 2 ) ( 1 + ϕ 1 − ϕ 2 ) = 1 + 0.15 ( 1 − 0.15 ) ( 1 − 0.8 + 0.15 ) ( 1 + 0.8 + 0.15 ) = 1.98 \begin{aligned} \operatorname{Var}\left(x_t\right) & =\frac{1-\phi_2}{\left(1+\phi_2\right)\left(1-\phi_1-\phi_2\right)\left(1+\phi_1-\phi_2\right)} \\ & =\frac{1+0.15}{(1-0.15)(1-0.8+0.15)(1+0.8+0.15)} \\ & =1.98 \end{aligned} Var(xt​)​=(1+ϕ2​)(1−ϕ1​−ϕ2​)(1+ϕ1​−ϕ2​)1−ϕ2​​=(1−0.15)(1−0.8+0.15)(1+0.8+0.15)1+0.15​=1.98​

   (3)

   ρ 1 = ϕ 1 1 − ϕ 2 = 0.8 1 + 0.15 = 0.70 ρ 2 = ϕ 1 ρ 1 + ϕ 2 = 0.8 × 0.7 − 0.15 = 0.41 ρ 3 = ϕ 1 ρ 2 + ϕ 2 ρ 1 = 0.8 × 0.41 − 0.15 × 0.7 = 0.22 \begin{aligned} & \rho_1=\frac{\phi_1}{1-\phi_2}=\frac{0.8}{1+0.15}=0.70 \\ & \rho_2=\phi_1 \rho_1+\phi_2=0.8 \times 0.7-0.15=0.41 \\ & \rho_3=\phi_1 \rho_2+\phi_2 \rho_1=0.8 \times 0.41-0.15 \times 0.7=0.22 \end{aligned} ​ρ1​=1−ϕ2​ϕ1​​=1+0.150.8​=0.70ρ2​=ϕ1​ρ1​+ϕ2​=0.8×0.7−0.15=0.41ρ3​=ϕ1​ρ2​+ϕ2​ρ1​=0.8×0.41−0.15×0.7=0.22​

   (4)

   ϕ 11 = ρ 1 = 0.7 ϕ 22 = ϕ 2 = − 0.15 ϕ 33 = 0 \begin{aligned} \phi_{11} & =\rho_1=0.7 \\ \phi_{22} & =\phi_2=-0.15 \\ \phi_{33} & =0 \end{aligned} ϕ11​ϕ22​ϕ33​​=ρ1​=0.7=ϕ2​=−0.15=0​

4. 已知 AR ⁡ ( 2 ) \operatorname{AR}(2) AR(2) 序列为 x t = x t − 1 + c x t − 2 + ε t x_t=x_{t-1}+c x_{t-2}+\varepsilon_t xt​=xt−1​+cxt−2​+εt​, 其中 { ε t } \left\{\varepsilon_t\right\} {εt​} 为白噪声序列. 确定 c c c 的取值范围, 以保证 { x t } \left\{x_t\right\} {xt​} 为平稳序列, 并给出该序列 ρ k \rho_k ρk​ 的表达式.

   (1) A R ( 2 ) A R(2) AR(2) 模型的平稳条件是

   { ∣ c ∣