时间序列分析:均方收敛
对于随机过程 {Xn} 和随机变量 X,当 n→∞ 时,若
\begin{equation} E[(X_n – X)^2] \rightarrow 0 \tag{1} \end{equation}
则称 X_n 均方收敛于 X。
前文中将 \mathrm{MA}(1) 过程 X_t = e_t + \theta e_{t-1} 表示为 \mathrm{AR}(\infty) 的形式
\begin{equation} e_t = \theta(B)^{-1} X_t = 1 – \theta X_{t-1} + \theta^2 X_{t-2} – \theta^3 X_{t-3} + \cdots = \sum_{k=0}^\infty (-\theta)^k X_{t-k} \tag{2} \end{equation}
为了使上式有意义,我们希望式 (2) 等号右边的级数均方收敛于 e_t。考虑部分和 A_n = \sum\limits_{k=0}^n (-\theta)^k X_{t-k},我们希望当 n \rightarrow \infty 时,A_n 均方收敛于 e_t。由
\begin{equation} E\bigg[ \bigg(\sum_{k=0}^n (-\theta)^k X_{t-k} – e_t \bigg)^2 \bigg] = E\bigg[ \bigg(\sum_{k=0}^n (-\theta)^k X_{t-k}\bigg)^2 \bigg] – 2E\bigg[ \bigg(\sum_{k=0}^n (-\theta)^k X_{t-k}\bigg) e_t \bigg] + E(e_t^2) \tag{3} \end{equation}
对于式 (3) 中等号右边第一项
\begin{equation} E\bigg[ \bigg(\sum_{k=0}^n (-\theta)^k X_{t-k}\bigg)^2 \bigg] = E\bigg[ \sum_{k=0}^n \theta^{2k} X_{t-k}^2 \bigg] + 2E\bigg[ \sum_{k=0}^{n} \sum_{j=0}^{k-1} (-\theta)^{k} X_{t-k} (-\theta)^{j} X_{t-j} \bigg] \tag{4} \end{equation}
对于式 (4) 中等号右边的第二项,仅在 j = k – 1 时,X_{t-k} = e_{t-k} + \theta e_{t-k-1} 与 X_{t-j} = X_{t-(k-1)} = e_{t-k+1} + \theta e_{t-k} 有相同项 e_{t-k},于是
\begin{align} &2E\bigg[ \sum_{k=0}^{n} \sum_{j=0}^{k-1} (-\theta)^{k} X_{t-k} (-\theta)^{j} X_{t-j} \bigg] \\ &= 2E\bigg[ \sum_{k=0}^{n-1} (-\theta)^{2k+1} X_{t-k} X_{t-k+1} \bigg] \\ &= -2E\bigg[ \sum_{k=0}^{n-1} \theta^{2k+1} X_{t-k} X_{t-k+1} \bigg] \tag{5} \end{align}
于是式 (4) 可以写成
\begin{equation} E\bigg[ \bigg(\sum_{k=0}^n (-\theta)^k X_{t-k}\bigg)^2 \bigg] = E\bigg[ \sum_{k=0}^n (-\theta)^{k} X_{t-k}^2 \bigg] -2E\bigg[ \sum_{k=0}^{n-1} \theta^{2k+1} X_{t-k} X_{t-k+1} \bigg] \tag{6} \end{equation}
对于式 (3) 中等号右边第二项,仅当 k = 0 时,E(X_t e_t) 不为零,于是
\begin{equation} – 2E\bigg[ \bigg(\sum_{k=0}^n (-\theta)^k X_{t-k}\bigg) e_t \bigg] = -2E(X_t e_t) \tag{7} \end{equation}
对于式 (3) 中等号右边第三项,有
\begin{equation} E(e_t^2) = \sigma_e^2 \tag{8} \end{equation}
将式 (6)、(7)、(8) 带入式 (3),并由前文可知对于 \mathrm{MA}(1) 有 \gamma_0 = (1 + \theta^2) \sigma_e^2,\gamma_1 = \theta \sigma_e^2,于是可以得到
\begin{align} &E\bigg[ \bigg(\sum_{k=0}^n (-\theta)^k X_{t-k} – e_t \bigg)^2 \bigg] \\ &= E\bigg[ \sum_{k=0}^n \theta^{2k} X_{t-k}^2 \bigg] – 2E\bigg[ \sum_{k=0}^{n-1} \theta^{2k+1} X_{t-k} X_{t-k+1} \bigg] – 2E[X_t e_t] + \sigma_e^2 \\ &= \sum_{k=0}^n \theta^{2k} e[X_{t-k}^2] – 2\sum_{k=0}^{n-1} \theta^{2k+1} E[X_{t-k} X_{t-k+1}] – 2E(e_t^2 + \theta e_{t-1}e_t) + \sigma_e^2 \\ &= \gamma_0 \sum_{k=0}^n \theta^{2k} + 2\gamma_1 \sum_{k=0}^{n-1} \theta^{2k+1} – 2\sigma_e^2 + \sigma_e^2 \\ &= (1 + \theta^2) \sigma_e^2 \sum_{k=0}^n \theta^{2k} – 2\theta\sigma_e^2 \sum_{k=0}^{n-1} \theta^{2k+1} – \sigma_e^2 \\ &= \sigma_e^2 \theta^{2n + 2} \tag{9} \end{align}
我们希望当 n \rightarrow \infty 时,E\bigg[ \bigg(\sum\limits_{k=0}^n (-\theta)^k X_{t-k} – e_t \bigg)^2 \bigg] \rightarrow 0,即 \sigma_e^2 \theta^{2n + 2} \rightarrow 0,因此要求 |\theta| < 1,此时特征方程 1 + \theta B = 0 的根 |-1 / \beta| > 1。