【时间序列分析】MA模型公式总结

MA
Time Series Analysis
author：zoxiii

MA

0-模型
- MA(q)
- 中心化MA(q)
- 引入延迟算子B
1-均值
2-方差
3-延迟k自协方差函数
4-延迟k自相关系数
- MA(1)
- MA(2)
- MA(q)
5-延迟k偏自相关系数
6-验证模型可逆性
7-逆函数递推公式

【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)

{xt=μ+εt−θ1εt−1−...−θqεt−qθq≠0E(εt)=0,Var(εt)=σε2,E(εtεs)=0,s≠t\begin{cases} x_t=\mu+\varepsilon_t-\theta_1\varepsilon_{t-1}-...-\theta_q\varepsilon_{t-q} \\ \theta_q\ne 0 \\ E(\varepsilon_t)=0,Var(\varepsilon_t)=\sigma_\varepsilon^2,E(\varepsilon_t\varepsilon_s)=0,s \ne t \end{cases} ⎩⎪⎨⎪⎧xt=μ+εt−θ1εt−1−...−θqεt−qθq=0E(εt)=0,Var(εt)=σε2,E(εtεs)=0,s=t

中心化MA(q)

当μ=0\mu=0μ=0时
xt=εt−θ1εt−1−...−θqεt−qx_t=\varepsilon_t-\theta_1\varepsilon_{t-1}-...-\theta_q\varepsilon_{t-q} xt=εt−θ1εt−1−...−θqεt−q

引入延迟算子B

xt=εt−θ1εt−1−...−θqεt−q=εt−θ1Bεt−...−θqBqεt=Θ(B)εtx_t=\varepsilon_t-\theta_1\varepsilon_{t-1}-...-\theta_q\varepsilon_{t-q} \\ ~=\varepsilon_t-\theta_1B\varepsilon_{t}-...-\theta_qB^q\varepsilon_{t} \\ =\Theta(B)\varepsilon_t~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ xt=εt−θ1εt−1−...−θqεt−q =εt−θ1Bεt−...−θqBqεt=Θ(B)εt

得到q阶移动平均系数多项式：
Θ(B)=1−θ1B−θ2B2−...−θqBq\Theta(B)=1-\theta_1B-\theta_2B^2-...-\theta_qB^q Θ(B)=1−θ1B−θ2B2−...−θqBq

1-均值

E(xt)=μE(x_t)=\mu E(xt)=μ

2-方差

Var(xt)=(1+θ12+...+θq2)σε2Var(x_t)=(1+\theta_1^2+...+\theta_q^2)\sigma_\varepsilon^2 Var(xt)=(1+θ12+...+θq2)σε2

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

γk={(1+θ12+...+θq2)σε2,k=0(−θk+∑i=1q−kθiθk+i)σε2,1≤k≤q0,k>q\gamma_k=\begin{cases} (1+\theta_1^2+...+\theta_q^2)\sigma_\varepsilon^2,~~~~~~~~~k=0 \\ (-\theta_k+\sum_{i=1}^{q-k}{\theta_i\theta_{k+i}})\sigma_\varepsilon^2,~~~1 \le k \le q \\ 0,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~k \gt q \end{cases} γk=⎩⎪⎨⎪⎧(1+θ12+...+θq2)σε2, k=0(−θk+∑i=1q−kθiθk+i)σε2, 1≤k≤q0, k>q

4-延迟k自相关系数

MA(1)

ρk=γkγ0={1,k=0−θ11+θ12,k=10,k≥2\rho_k=\frac{\gamma_k}{\gamma_0}=\begin{cases} 1,~~~~~~~~k=0 \\ \frac{-\theta_1}{1+\theta_1^2},~~~k=1 \\ 0,~~~~~~~~k \ge 2 \end{cases} ρk=γ0γk=⎩⎪⎨⎪⎧1, k=01+θ12−θ1, k=10, k≥2

MA(2)

ρk=γkγ0={1,k=0−θk+θ1θ21+θ12+θ22,k=1−θ21+θ12+θ22,k=20,k≥3\rho_k=\frac{\gamma_k}{\gamma_0}=\begin{cases} 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{cases} ρk=γ0γk=⎩⎪⎪⎪⎪⎨⎪⎪⎪⎪⎧1, k=01+θ12+θ22−θk+θ1θ2, k=11+θ12+θ22−θ2, k=20, k≥3

MA(q)

ρk=γkγ0={1,k=0−θk+∑i=1q−kθiθk+i1+θ12+...+θq2,1≤k≤q0,k>q\rho_k=\frac{\gamma_k}{\gamma_0}=\begin{cases} 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 \gt q \end{cases} ρk=γ0γk=⎩⎪⎪⎨⎪⎪⎧1, k=01+θ12+...+θq2−θk+∑i=1q−kθiθk+i, 1≤k≤q0, k>q

5-延迟k偏自相关系数

MA(q)模型的延迟k偏自相关系数ϕkk\phi_{kk}ϕkk拖尾

ϕkk\phi_{kk} ϕkk

6-验证模型可逆性

已知中心化MA(q)模型为xt=εt−θ1εt−1−...−θqεt−qx_t=\varepsilon_t-\theta_1\varepsilon_{t-1}-...-\theta_q\varepsilon_{t-q}xt=εt−θ1εt−1−...−θqεt−q
∴xt=εt−θ1εt−1−...−θqεt−q=εt−θ1Bεt−...−θqBqεt=Θ(B)εtx_t=\varepsilon_t-\theta_1\varepsilon_{t-1}-...-\theta_q\varepsilon_{t-q}\\ ~=\varepsilon_t-\theta_1B\varepsilon_{t}-...-\theta_qB^q\varepsilon_{t}\\ =\Theta(B)\varepsilon_t~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ xt=εt−θ1εt−1−...−θqεt−q =εt−θ1Bεt−...−θqBqεt=Θ(B)εt
∴得到移动平均系数多项式Θ(B)=1−θ1B−θ2B2−...−θqBq\Theta(B)=1-\theta_1B-\theta_2B^2-...-\theta_qB^qΘ(B)=1−θ1B−θ2B2−...−θqBq

设Θ(B)=0\Theta(B)=0Θ(B)=0的根为λ\lambdaλ，

∴1−θ1λ−θ2λ2−...−θqλq=01-\theta_1\lambda-\theta_2\lambda^2-...-\theta_q\lambda^q=01−θ1λ−θ2λ2−...−θqλq=0

求解得到λ1,λ2,...\lambda_1,\lambda_2,...λ1,λ2,...

当满足∣λ1∣>1且∣λ2∣>1且...|\lambda_1|\gt 1且|\lambda_2|\gt 1且...∣λ1∣>1且∣λ2∣>1且...时，MA(q)模型可逆。

7-逆函数递推公式

逆函数IjI_jIj

如果MA(q)xt=εt−θ1εt−1−...−θqεt−qx_t=\varepsilon_t-\theta_1\varepsilon_{t-1}-...-\theta_q\varepsilon_{t-q}xt=εt−θ1εt−1−...−θqεt−q可逆，有

{Θ(B)εt=xt,[1]εt=I(B)xt,[2]\begin{cases} \Theta(B)\varepsilon_t=x_t,[1] \\ \varepsilon_t=I(B)x_t,[2] \end{cases} {Θ(B)εt=xt,[1]εt=I(B)xt,[2]
其中
Θ(B)=1−θ1B−θ2B2−...−θqBq=1−∑i=1qθiBi\Theta(B)=1-\theta_1B-\theta_2B^2-...-\theta_qB^q=1-\sum_{i=1}^{q}{\theta_iB^i} Θ(B)=1−θ1B−θ2B2−...−θqBq=1−i=1∑qθiBi
I(B)=I1+I1B+I2B2+...=∑i=0∞IiBiI(B)=I_1+I_1B+I_2B^2+...=\sum_{i=0}^{\infty}{I_iB^i} I(B)=I1+I1B+I2B2+...=i=0∑∞IiBi
将[2]式代入[1]式得到Θ(B)I(B)xt=xt\Theta(B)I(B)x_t=x_tΘ(B)I(B)xt=xt,按照待定系数法求得
I0=1Il=∑i=1lθi′Il−iI_0=1 \\ I_l=\sum_{i=1}^{l}{\theta_i'I_{l-i}} I0=1Il=i=1∑lθi′Il−i
其中
l≥1θi′={θi,i≤q0,i>ql \ge 1 \\ \theta_i'=\begin{cases} \theta_i,~~~i \le q \\ 0,~~~~i \gt q \end{cases} l≥1θi′={θi, i≤q0, i>q