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MA
Time Series Analysis
author：zoxiii

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【参考文献】王燕. 应用时间序列分析-第5版[M]. 中国人民大学出版社, 2019.

0-模型

MA(q)

$$\{\begin{array}{l}{x}_t=\mu +{\epsilon }_t-{\theta }_1{\epsilon }_{t-1}-...-{\theta }_q{\epsilon }_{t-q}\\ {\theta }_q\ne 0\\ E({\epsilon }_t)=0,Var({\epsilon }_t)={\sigma }_{\epsilon }^2,E({\epsilon }_t{\epsilon }_s)=0,s\ne t\end{array}$$

中心化MA(q)

当$\mu =0$时
$$x_t={\epsilon }_t-{\theta }_1{\epsilon }_{t-1}-...-{\theta }_q{\epsilon }_{t-q}$$

引入延迟算子B

$$\begin{gathered} x_t={\epsilon }_t-{\theta }_1{\epsilon }_{t-1}-...-{\theta }_q{\epsilon }_{t-q} \\ ={\epsilon }_t-{\theta }_{1}B{\epsilon }_t-...-{\theta }_q{B}^q{\epsilon }_t \\ =\Theta (B){\epsilon }_t \end{gathered}$$

得到q阶移动平均系数多项式：
$$\Theta (B)=1-{\theta }_{1}B-{\theta }_2{B}^2-...-{\theta }_q{B}^q$$

1-均值

$$E(x_t)=\mu$$

2-方差

$$Var(x_t)=(1+{\theta }_1^2+...+{\theta }_q^2){\sigma }_{\epsilon }^2$$

3-延迟k自协方差函数

• MA(q)自协方差函数只与滞后阶数k有关，且q阶截尾

$${\gamma }_k=\{\begin{array}{l}(1+{\theta }_1^2+...+{\theta }_q^2){\sigma }_{\epsilon }^2,k=0\\ (-{\theta }_k+{\sum }_{i=1}^{q-k}{\theta }_i{\theta }_{k+i}){\sigma }_{\epsilon }^2,1\le k\le q\\ 0,k>q\end{array}$$

4-延迟k自相关系数

MA(1)

$${\rho }_k=\frac{{\gamma }_k}{{\gamma }_0}=\{\begin{array}{l}1,k=0\\ \frac{-{\theta }_1}{1+{\theta }_1^2},k=1\\ 0,k\ge 2\end{array}$$

MA(2)

$${\rho }_k=\frac{{\gamma }_k}{{\gamma }_0}=\{\begin{array}{l}1,k=0\\ \frac{-{\theta }_k+{\theta }_1{\theta }_2}{1+{\theta }_1^2+{\theta }_2^2},k=1\\ \frac{-{\theta }_2}{1+{\theta }_1^2+{\theta }_2^2},k=2\\ 0,k\ge 3\end{array}$$

MA(q)

$${\rho }_k=\frac{{\gamma }_k}{{\gamma }_0}=\{\begin{array}{l}1,k=0\\ \frac{-{\theta }_k+{\sum }_{i=1}^{q-k}{\theta }_i{\theta }_{k+i}}{1+{\theta }_1^2+...+{\theta }_q^2},1\le k\le q\\ 0,k>q\end{array}$$

5-延迟k偏自相关系数

• MA(q)模型的延迟k偏自相关系数${\varphi }_{kk}$拖尾

$${\varphi }_{kk}$$

6-验证模型可逆性

已知中心化MA(q)模型为$x_t={\epsilon }_t-{\theta }_1{\epsilon }_{t-1}-...-{\theta }_q{\epsilon }_{t-q}$
∴$$\begin{gathered} x_t={\epsilon }_t-{\theta }_1{\epsilon }_{t-1}-...-{\theta }_q{\epsilon }_{t-q} \\ ={\epsilon }_t-{\theta }_{1}B{\epsilon }_t-...-{\theta }_q{B}^q{\epsilon }_t \\ =\Theta (B){\epsilon }_t \end{gathered}$$
∴得到移动平均系数多项式$\Theta (B)=1-{\theta }_{1}B-{\theta }_2{B}^2-...-{\theta }_q{B}^q$

设$\Theta (B)=0$的根为$\lambda$，

∴$1-{\theta }_1\lambda -{\theta }_2{\lambda }^2-...-{\theta }_q{\lambda }^q=0$

求解得到${\lambda }_1,{\lambda }_2,...$

当满足$|{\lambda }_1|>1且|{\lambda }_2|>1且...$时，MA(q)模型可逆。

7-逆函数递推公式

逆函数$I_j$

如果MA(q)$x_t={\epsilon }_t-{\theta }_1{\epsilon }_{t-1}-...-{\theta }_q{\epsilon }_{t-q}$可逆，有

$$\{\begin{array}{l}\Theta (B){\epsilon }_t=x_t,[1]\\ {\epsilon }_t=I(B)x_t,[2]\end{array}$$
其中
$$\Theta (B)=1-{\theta }_{1}B-{\theta }_2{B}^2-...-{\theta }_q{B}^q=1-\sum _{i=1}^q{\theta }_i{B}^i$$
$$I(B)=I_1+I_{1}B+I_2{B}^2+...=\sum _{i=0}^{\infty }{I}_i{B}^i$$
将[2]式代入[1]式得到$\Theta (B)I(B)x_t=x_t$,按照待定系数法求得
$$\begin{gathered} I_0=1 \\ {I}_l=\sum _{i=1}^l{\theta }_i^{\prime }{I}_{l-i} \end{gathered}$$
其中
$$\begin{gathered} l\ge 1 \\ {\theta }_i^{\prime }=\{\begin{array}{l}{\theta }_i,i\le q\\ 0,i>q\end{array} \end{gathered}$$