$\mathrm{A}\mathrm{R}\mathrm{M}\mathrm{A}(1,1)$ 模型

模型设定与平稳解

$$X_t=a{X}_{t-1}+{\epsilon }_t+b{\epsilon }_{t-1}$$

特征多项式 $A(z)=1-az$ ，$B(z)=1+bz$ 。

由泰勒级数计算 Wold 系数：
$$\Phi (z)=\frac{B(z)}{A(z)}=\frac{1+bz}{1-az}=(1+bz)\sum _{j=0}^{\infty }{a}^j{z}^j=1+(a+b)\sum _{j=1}^{\infty }{a}^{j-1}{z}^j$$
由 Wold 系数递推公式计算 Wold 系数：
$${\psi }_0=1,{\psi }_1=b+a{\psi }_{1-1}=b+a{\psi }_0=a+b,$$

$${\psi }_j=a{\psi }_{j-1}=\cdots =a^{j-1}{\psi }_1=a^{j-1}(a+b),j=2,3,4,\cdots$$

$\mathrm{A}\mathrm{R}\mathrm{M}\mathrm{A}(1,1)$ 模型的平稳解：
$$X_t={\epsilon }_t+(a+b)\sum _{j=0}^{\infty }{a}^{j-1}{\epsilon }_{t-j}.$$

自协方差函数

由 Wold 系数计算自协方差函数：
$$\begin{array}{rl}{\gamma }_0={\sigma }^2\sum _{j=0}^{\infty }{\psi }_j^2& ={\sigma }^2\left[1+\sum _{j=1}^{\infty }(a+b{)}^2{a}^{2(j-1)}\right]\\ & ={\sigma }^2\left[1+\frac{(a+b{)}^2}{1-a^2}\right]\\ & ={\sigma }^2\frac{1+2ab+b^2}{1-a^2}.\end{array}$$

$$\begin{array}{rl}{\gamma }_1={\sigma }^2\sum _{j=0}^{\infty }{\psi }_j{\psi }_{j+1}& ={\sigma }^2\left[{\psi }_1+\sum _{j=1}^{\infty }a(a+b{)}^2{a}^{2(j-1)}\right]\\ & ={\sigma }^2\left[a+b+a\frac{(a+b{)}^2}{1-a^2}\right]\\ & ={\sigma }^2\frac{(a+b)(1+ab)}{1-a^2}.\end{array}$$

$$\begin{array}{rl}{\gamma }_k& ={\sigma }^2\sum _{j=0}^{\infty }{\psi }_j{\psi }_{j+k}={\sigma }^{2}a\sum _{j=0}^{\infty }{\psi }_j{\psi }_{j+k-1}\\ & =a{\gamma }_{k-1}=a^2{\gamma }_{k-2}=\cdots =a^{k-1}{\gamma }_1=a^{k-1}{\sigma }^2\frac{(a+b)(1+ab)}{1-a^2}.\end{array}$$

用 Yule-Walker 方程计算自协方差函数：

对模型方程两边同乘 $X_{t-k},k=0,1,2,\cdots$ 并取期望，由于 $\mathrm{A}\mathrm{R}\mathrm{M}\mathrm{A}(1,1)$ 序列的合理性：$\mathrm{E}({\epsilon }_t{X}_{t-k})=0,k\ge 1$ ，有
$${\gamma }_0=a{\gamma }_1+\mathrm{E}(X_t{\epsilon }_t)+b\mathrm{E}(X_t{\epsilon }_{t-1}),$$

$${\gamma }_1=a{\gamma }_0+0+b\mathrm{E}(X_{t-1}{\epsilon }_{t-1})$$

$${\gamma }_k=a{\gamma }_{k-1}+0+b\cdot 0,k=2,3,\cdots$$

由 Wold 系数表示知 $\mathrm{E}(X_t{\epsilon }_{t-j})={\sigma }^2{\psi }_j$ ，所以
$${\gamma }_0=a{\gamma }_1+{\sigma }^2[1+b(a+b)],$$

$${\gamma }_1=a{\gamma }_0+{\sigma }^{2}b,$$

$${\gamma }_k=a{\gamma }_{k-1}=a^{k-1}{\gamma }_1,k=2,3,\cdots$$

求解得自协方差函数为
$${\gamma }_k=\{\begin{array}{ll}{\sigma }^2\frac{1+2ab+b^2}{1-a^2},& k=0;\\ & \\ {\sigma }^2\frac{(a+b)(1+ab)}{1-a^2},& k=1;\\ & \\ a^{k-1}{\sigma }^2\frac{(a+b)(1+ab)}{1-a^2},& k=2,3,4,\cdots \end{array}$$

自相关系数和偏相关系数

自相关函数为
$${\rho }_k=\{\begin{array}{ll}1,& k=0;\\ & \\ \frac{(a+b)(1+ab)}{1+2ab+b^2},& k=1;\\ & \\ a{\rho }_{k-1}=a^{k-1}{\rho }_1,& k=2,3,4,\cdots \end{array}$$

用 Yule-Walker 方程计算偏相关系数的前几项
$$a_{1,1}={\rho }_1=\frac{(a+b)(1+ab)}{1+2ab+b^2}.$$
写出 $a_{2,2}$ 满足的 Yule-Walker 方程
$$\left[\begin{array}{cc}{\gamma }_0& {\gamma }_1\\ {\gamma }_1& {\gamma }_0\end{array}\right]\left[\begin{array}{c}{a}_{2,1}\\ a_{2,2}\end{array}\right]=\left[\begin{array}{c}{\gamma }_1\\ {\gamma }_2\end{array}\right]$$

$$\left[\begin{array}{cc}1& {\rho }_1\\ {\rho }_1& 1\end{array}\right]\left[\begin{array}{c}{a}_{2,1}\\ a_{2,2}\end{array}\right]=\left[\begin{array}{c}{\rho }_1\\ {\rho }_2\end{array}\right]=\left[\begin{array}{c}{\rho }_1\\ a{\rho }_1\end{array}\right]$$

由Cramer 法则
$$a_{2,2}=\frac{\left|\begin{array}{cc}1& {\rho }_1\\ {\rho }_1& a{\rho }_1\end{array}\right|}{\left|\begin{array}{cc}1& {\rho }_1\\ {\rho }_1& 1\end{array}\right|}=\frac{{\rho }_1(a-{\rho }_1)}{1-{\rho }_1^2}$$
同理可得
$$a_{3,3}=\frac{\left|\begin{array}{ccc}1& {\rho }_1& {\rho }_1\\ {\rho }_1& 1& {\rho }_2\\ {\rho }_2& {\rho }_1& {\rho }_3\end{array}\right|}{\left|\begin{array}{ccc}1& {\rho }_1& {\rho }_2\\ {\rho }_1& 1& {\rho }_1\\ {\rho }_2& {\rho }_1& 1\end{array}\right|}=\frac{{\rho }_1(a-{\rho }_1{)}^2}{1+2a{\rho }_1^3-{\rho }_1^2(2+a^2)}$$
也可以用 Levinson 递推公式计算偏相关系数的前几项
$$a_{2,2}=\frac{{\gamma }_2-{\gamma }_1{a}_{1,1}}{{\gamma }_0-{\gamma }_1{a}_{1,1}}=\frac{{\rho }_2-{\rho }_1^2}{1-{\rho }_1^2}=\frac{{\rho }_1(a-{\rho }_1)}{1-{\rho }_1^2},$$

$$a_{2,1}=a_{1,1}-a_{2,2}{a}_{1,1}={\rho }_1\left(1-\frac{{\rho }_1(a-{\rho }_1)}{1-{\rho }_1^2}\right)=\frac{{\rho }_1(1-a{\rho }_1)}{1-{\rho }_1^2}$$

$$\begin{array}{rl}{a}_{3,3}& =\frac{{\gamma }_3-{\gamma }_2{a}_{2,1}-{\gamma }_1{a}_{2,2}}{{\gamma }_0-{\gamma }_1{a}_{2,1}-{\gamma }_2{a}_{2,2}}\\ & \\ & =\frac{{\rho }_3-{\rho }_2{a}_{2,1}-{\rho }_1{a}_{2,2}}{1-{\rho }_1{a}_{2,1}-{\rho }_2{a}_{2,2}}\\ & \\ & =\frac{a^2{\rho }_1-\frac{a{\rho }_1^2(1-a{\rho }_1)}{1-{\rho }_1^2}-\frac{{\rho }_1^2(a-{\rho }_1)}{1-{\rho }_1^2}}{1-\frac{{\rho }_1^2(1-a{\rho }_1)}{1-{\rho }_1^2}-\frac{a{\rho }_1^2(a-{\rho }_1)}{1-{\rho }_1^2}}\\ & \\ & =\frac{{\rho }_1(a-{\rho }_1{)}^2}{1+2a{\rho }_1^3-{\rho }_1^2(2+a^2)}\end{array}$$

最后

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