考虑MA(1) \[\begin{aligned} X_t = \varepsilon_t + b \varepsilon_{t-1}, \quad t=1,2,\dots \end{aligned}\] 其中\(|b|1\)，抛弃) 用\(\hat\rho_1 = \hat\gamma_1 / \hat\gamma_0\)代替理论值 可得\(b\)的矩估计 \[\begin{aligned} \hat b = \frac{1 - \sqrt{1 - 4\hat\rho_1^2}}{2\hat\rho_1} \end{aligned}\]

如果\(\{\varepsilon_t\}\)是独立同分布的白噪声, 由§16.2定理16.1知道\(\hat\rho_1\)是\(\rho_1\)的强相合估计: \(\lim_{ N\to \infty} \hat \rho_1=\rho_1\) , a.s., 于是 \[ \lim_{N\to \infty} \hat b = \frac{1-\sqrt{1-4 \rho^2_1} }{2 \rho_1} =b, \ a.s., \] 所以\(\hat b\)是\(b\)的强相合估计. 实际上利用§16.2定理16.2还可以证明, 当\(N \to \infty\)时, \(\sqrt{N}( \hat b - b)\) 依分布收敛到正态分布([27]). \[ \text{N}\left(0, \frac{1+b^2+b^4+b^6+b^8}{(1- b^2)^2}\right). \]

考虑可逆MA(\(q\))模型: \[\begin{align} X_t = \varepsilon_t + \sum_{j=1}^q b_j \varepsilon_{t-j}, \quad t \in \mathbb Z, \tag{20.1} \end{align}\] \(\{\varepsilon_t\}\sim\text{WN}(0,\sigma^2)\), 系数满足可逆条件： \[\begin{align} B(z) = 1 + \sum_{j=1}^q b_j z^j \neq 0, \quad |z|\leq 1, \tag{20.2} \end{align}\] 假定\(q\)已知，估计\(\boldsymbol{b} = (b_1,\dots, b_q)^T\)和\(\sigma^2\)。

参数与自协方差函数关系为 \[\begin{align} \gamma_k = \sigma^2(b_0 b_k + b_1 b_{k+1} + \dots + b_{q-k}b_q), \quad 0 \leq k \leq q. \tag{20.3} \end{align}\] (其中\(b_0=1\))

从样本估计出\(\hat\gamma_0, \hat\gamma_1, \dots, \hat\gamma_q\)后， 可以用解非线性方程组的方法求解\(\boldsymbol{b}\)和\(\sigma^2\)， 但不能保证解唯一，也不能保证可逆性条件。 求解方法包括线性迭代方法和Newton-Raphson迭代方法。

还可以根据§12.8计算矩估计。 由MA(\(q\))的\(q\)步截尾性， 可定义 \[\begin{aligned} \tilde \gamma_k = \begin{cases} \hat\gamma_k, & 0 \leq k \leq q,\\ 0, & k > q, \end{cases} \end{aligned}\] 用\(\{\tilde\gamma_k\}\)作为\(\{\gamma_k\}\)的估计代入§12.8的计算公式。 记 \[\begin{align} \begin{array}{llr} A=\left( \begin{array}{llllll} 0 & 1 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 1 & \cdots & 0 & 0 \\ \cdots & \cdots & & \cdots & \cdots & \\ 0 & 0 & 0 & \cdots & 0 & 1 \\ 0 & 0 & 0 & \cdots & 0 & 0 \end{array} \right )_{q\times q}, & C=\left( \begin{array}{c} 1 \\ 0 \\ \vdots \\0 \end{array} \right)_{q \times 1} \\ \\ \Omega_k=\left( \begin{array}{llll} \tilde\gamma_1 & \tilde\gamma_2 & \cdots & \tilde\gamma_k \\ \tilde\gamma_2 & \tilde\gamma_3 & \cdots & \tilde\gamma_{k+1} \\ \cdots & \cdots & \cdots & \cdots \\ \tilde\gamma_q & \tilde\gamma_{q+1} & \cdots & \tilde\gamma_{q+k-1} \end{array} \right), & \boldsymbol{\gamma}_q=\left( \begin{array}{c} \tilde\gamma_1 \\ \tilde\gamma_2 \\ \vdots \\ \tilde\gamma_q \end{array} \right) & \end{array} \tag{20.4} \end{align}\] 矩估计为 \[\begin{align} (\hat b_1, \dots, \hat b_q)^T = \frac{1}{\hat\sigma^2}(\boldsymbol\gamma_q - A \hat\Pi C), \quad \hat\sigma^2 = \hat\gamma_0 - C^T \hat\Pi C \tag{20.5} \end{align}\] 其中 \[ \hat\Pi = \lim_{k\to\infty} \hat\Omega_k \tilde\Gamma_k^{-1} \hat\Omega_k \]

定理20.1 如果模型(20.1)中的\(\{\varepsilon_t\}\)是独立同分布的\(\text{WN}(0,\sigma^2)\), 则几乎必然地当\(N\)充分大后由(20.5)计算的 \(\hat b_1, \hat b_2,\dots,\hat b_q\)满足可逆条件(20.2).

证明 由于当\(N\to \infty\),\(\hat \gamma_k\to \gamma_k\) , a.s., 所以 \[\begin{aligned} \hat f(\lambda) =& \frac{1}{2\pi}\sum_{k=-q}^{q} \hat \gamma_k e^{-ik\lambda} \\ \to& f(\lambda) = \frac{1}{2\pi}\sum_{k=-q}^{q} \gamma_k e^{-ik\lambda}, \ \text{a.s.}, \ \hbox{当}\ N\to \infty \end{aligned}\] 在\([-\pi, \pi]\)上一致成立. 于是当\(N\)充分大后\(\hat f(\lambda)\)恒正. 利用§12.6的引理12.1知道有惟一的\((b_1',b_2',\dots,b_q')\)满足可逆性条件(20.2), 并且使得 \[ \hat f(\lambda) =\frac{\sigma^2_0}{2\pi} |1+\sum_{j=1}^q b_j'e^{-ij\lambda}|^2 \] 是 \[ Y_t = e_t+\sum_{j=1}^q b_j'e_{t-j}, \ \ \{e_t\} \sim \text{WN}(0,\sigma^2_0) \] 的谱密度. 这时, \(\tilde \gamma_k = E(Y_t Y_{t+k})\). 再利用§12.8知道 \[ (\hat b_1, \hat b_2,\dots,\hat b_q) =(b_1',b_2',\cdots,b_q'). \]

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从定理20.1的证明知道这样得到的 \((\hat b_1, \hat b_2,\cdots,\hat b_q)\)满足 \[ 1+\sum_{j=1}^q\hat b_j z^j \neq 0, \ |z|\leq 1 \] 的充分条件是 \[ \sum_{k=-q}^{q} \hat \gamma_k e^{-ik\lambda}>0, \ \lambda \in [-\pi, \pi]. \]