24 时间序列的递推预测

§10.1已经用Y-W方程给出了非决定性平稳序列的预测公式, 这里我们进行更深入的讨论。

  • 非决定性平稳序列可以根据自协方差函数用Levinson递推得到预测系数和预测方差;
  • 对AR(\(p\))模型只要使用自回归系数预测;
  • 对MA模型和ARMA模型则只能递推得到预测系数。

这里研究可以化简MA模型和ARMA模型预测的公式, 以及非平稳时也可用的递推公式。 假设自协方差函数已知, 实际中可以用样本自协方差函数代替。

24.1 一般时间序列的递推预测

24.1.1 递推预测的正交分解

\(\{Y_t\}\)是方差有限的零均值时间序列, 对任何正整数\(n\), 用 \[ L_n=\overline{\text{span}}\{Y_1,Y_2,\dots,Y_n\} \] 表示\(Y_1,\dots,Y_n\)的线性组合的全体. 定义\(\boldsymbol{Y}_n = (Y_1,Y_2,\dots,Y_n)^T\), \[\begin{align} \hat Y_1=0, \ \hat Y_{n}=L(Y_n|\boldsymbol{Y}_{n-1}), \ n=2,\dots. \tag{24.1} \end{align}\] 引入预测误差\(W_n\)及其方差\(\nu_{n-1}\)如下: \[\begin{align} W_n= Y_n-\hat Y_n, \ \ \nu_{n-1}= EW_n^2. \ \ n=1,2,\cdots . \tag{24.2} \end{align}\] 由最佳线性预测的性质 7 知道\(W_n\)\(L_{n-1}\)中的任何随机变量正交, 并且\(W_n \in L_n\). 于是 \(\{W_n\}\)是一个正交序列, 满足 \[ E(W_n W_k)=\nu_{n-1} \delta_{n-k}. \] 这里\(\delta_t\)是Kronecker 函数.

\[ M_n \stackrel{\triangle}{=} \overline{\text{span}}\{W_1,W_2,\dots,W_n\} \] 表示\(W_1,W_2,\cdots,W_n\)的线性组合全体. 则\(M_n \subset L_n\).

\(n\in \mathbb N\), 我们用归纳法证明\(Y_n\in M_n\).

首先\(Y_1=W_1 \in M_1\). 如果对\(k\leq n\)已经证明\(Y_k\in M_k\), 注意\(\hat Y_{n+1}=L(Y_{n+1}| \boldsymbol{Y}_n) \in M_n\), 于是 \[ Y_{n+1}=(Y_{n+1}-\hat Y_{n+1}) + \hat Y_{n+1}= W_{n+1} + \hat Y_{n+1} \in M_{n+1}. \] 这就证明了对\(n\in \mathbb N\), \(Y_n \in M_n\)成立.

于是得到 \[\begin{align} L_n =& \overline{\text{span}}\{Y_1,Y_2,\dots,Y_n\} \\ =& \overline{\text{span}}\{W_1,W_2,\dots,W_n\}=M_n, \\ & n = 1,2,\dots \tag{24.3} \end{align}\]

§22.1性质10和(24.3)式告诉我们用 \(\boldsymbol{W}_n =(W_1,W_2,\dots,W_{n})^T\)\(Y_{n+1}\)进行预测和用 \(\boldsymbol{Y}_n =(Y_1,Y_2,\dots,Y_{n})^T\)\(Y_{n+1}\) 进行预测是等价的. 由于\(\{W_t\}\)是正交序列, 所以用\(\boldsymbol{W}_n\)\(Y_{n+1}\)进行预测有很多的方便. 类似于在正交基上的投影,可以直接计算坐标。

24.1.2 递推预测定理

定理24.1 \(\{Y_t\}\)是零均值时间序列(不要求平稳!). 如果\((Y_1,Y_2,\dots,Y_{m+1})^T\)的协方差矩阵 \[\begin{align} \Big( E(Y_s Y_t)\Big )_{1\leq s,t \leq m+1} \tag{24.4} \end{align}\] 正定, 则最佳线性预测 \[\begin{align} \hat Y_{n+1} \stackrel{\triangle}{=}& L(Y_{n+1}|\boldsymbol Y_n), \ n=1,2,\dots,m \nonumber \\ =& \sum_{j=1}^n \theta_{n,j} W_{n+1-j} = \sum_{k=0}^{n-1} \theta_{n,n-k} W_{k+1} \tag{24.5} \\ = & \theta_{n,1} W_n + \theta_{n,2} W_{n-1} + \dots + \theta_{n,n}W_1 \\ =& \theta_{n,n} W_1 + \theta_{n,n-1} W_2 + \dots + \theta_{n,1} W_n \nonumber \end{align}\] 其中的系数\(\{\theta_{n,j}\}\) 和预测的均方误差\(\nu_n=EW_{n+1}^2\)满足如下的递推公式. \[\begin{align} \begin{cases} & \nu_0=EY_1^2, \\ & \theta_{n,n-k} =\left[ E(Y_{n+1}Y_{k+1}) - \sum_{j=0}^{k-1} \theta_{k,k-j}\theta_{n,n-j}\nu_j \right] / \nu_k, \\ & \qquad\qquad 0\leq k \leq n-1,\\ &\nu_n = EY_{n+1}^2 - \sum_{k=0}^{n-1}\theta_{n,n-k}^2\nu_k, \end{cases} \tag{24.6} \end{align}\] 其中约定\(\sum_{j=0}^{-1}(\cdot)\stackrel{\triangle}{=} 0\).

递推的顺序是 \[ \begin{array}{lllll} \nu_0 \\ \theta_{1,1} & \nu_1 \\ \theta_{2,2} & \theta_{2,1} & \nu_2 \\ \theta_{3,3} & \theta_{3,2} & \theta_{3,1} & \nu_3 \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{array} \] 从协方差可以递推计算系数\(\{\theta_{n,k} \}\)\(\{\nu_n\}\), 并递推计算 \[\begin{aligned} \hat Y_1 =& 0, & W_1 =& Y_1 - \hat Y_1 \\ \hat Y_2 =& \theta_{1,1} W_1, & W_2 =& Y_2 - \hat Y_2 \\ \hat Y_3 =& \theta_{2,2} W_1 + \theta_{2,1} W_2, & W_3 =& Y_3 - \hat Y_3 \\ \hat Y_4 =& \theta_{3,3} W_1 + \theta_{3,2} W_2 + \theta_{3,1} W_3, & W_4 =& Y_4 - \hat Y_4 \\ \cdots & \cdots \cdots & \cdots & \cdots \cdots \end{aligned}\]

证明

从自协方差矩阵(24.4)的正定性知道\(\nu_n=EW_{n+1}^2 >0\). 以下设\(0\leq k\leq n-1\). 在(24.5)两边同乘\(W_{k+1}\)后求数学期望, 由\(\{W_k\}\)的正交性得 \[\begin{align} E(\hat Y_{n+1}W_{k+1}) =& \theta_{n,n-k}\nu_k. \tag{24.7} \end{align}\] 利用\(W_{n+1}=Y_{n+1}-\hat Y_{n+1}\)\(W_{k+1}\)垂直, 得 \[\begin{align} E(Y_{n+1} W_{k+1})= E(\hat Y_{n+1}W_{k+1}) = \theta_{n,n-k} \nu_k \tag{24.8} \end{align}\]

注意到 \[ \hat Y_{k+1} =\sum_{j=1}^{k} \theta_{k,j}W_{k+1-j} =\sum_{j=0}^{k-1} \theta_{k,k-j}W_{j+1}, \] 于是利用(24.8)可得 \[\begin{aligned} \theta_{n,n-k}=& E(Y_{n+1}W_{k+1})/\nu_k \\ =& E\left[ Y_{n+1}(Y_{k+1} - \sum_{j=0}^{k-1} \theta_{k,k-j}W_{j+1}) \right] / \nu_k \\ =& \left[ E(Y_{n+1}Y_{k+1}) - \sum_{j=0}^{k-1} \theta_{k,k-j}E(Y_{n+1} W_{j+1}) \right] / \nu_k \\ =& [E(Y_{n+1}Y_{k+1}) - \sum_{j=0}^{k-1}\theta_{k,k-j}\theta_{n,n-j}\nu_j]/\nu_k. \end{aligned}\]

最后, 利用 \(\nu_n=EW^2_{n+1}= EY_{n+1}^2- E\hat Y_{n+1}^2\)(24.5)得到预测的均方误差公式: \[ \nu_n = EY_{n+1}^2 - \sum_{j=1}^{n}\theta_{n,j}^2\nu_{n-j} = EY_{n+1}^2 - \sum_{j=0}^{n-1}\theta_{n,n-j}^2\nu_j. \]

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24.1.3 多步预报问题

下面考虑用\(\{Y_1,Y_2,\cdots,Y_n\}\)预测\(Y_{n+k+1}\)的问题. 设\((Y_1,Y_2,\dots,Y_{n+k+1})^T\)的自协方差矩阵正定. 仍记\(\hat Y_{n+k+1} = L(Y_{n+k+1}|\boldsymbol{Y}_{n+k})\), 用\(W_{j}\)表示预测误差\(Y_j - L(Y_j|\boldsymbol{Y}_{j-1})\), 则有 \[\begin{align} \hat Y_{n+k+1} = \sum _{j=1}^{n+k}\theta_{n+k,j} W_{n+k+1-j}. \tag{24.9} \end{align}\] 注意, 对\(j \geq 0\), \(W_{n+j+1}\)垂直于\(L_n\), \(W_{n-j} \in L_n\). 根据§22.1中最佳线性预测的性质4、5、8或定理22.2得到 \[\begin{align} &L(Y_{n+k+1}|\boldsymbol{Y}_n) = L(\hat Y_{n+k+1}|\boldsymbol{Y}_n) \nonumber \\ =& L [\sum_{j=1}^{n+k} \theta_{n+k,j} W_{n+k+1-j}\ \Big| \ \boldsymbol{W}_n ] \nonumber \\ =&L [\sum _{j=k+1}^{n+k} \theta_{n+k,j} W_{n+k+1-j} \ \big | \ \boldsymbol W_n ] \nonumber \\ =&\sum _{j=k+1}^{n+k}\theta_{n+k,j} W_{n+k+1-j}. \tag{24.10} \\ =& \sum_{j=0}^{n-1} \theta_{n+k,n+k-j} W_{j+1} \nonumber \end{align}\]

由投影的正交性, 得到预测的均方误差 \[\begin{align} & E[Y_{n+k+1} - L(Y_{n+k+1}|\boldsymbol{Y}_n)]^2 \nonumber \\ =& EY_{n+k+1}^2-E[L(Y_{n+k+1}|\boldsymbol{Y}_n)]^2 \nonumber \\ =& EY_{n+k+1}^2 - \sum _{j=k+1}^{n+k} \theta_{n+k,j}^2 \nu_{n+k-j}, \tag{24.11} \\ =& EY_{n+k+1}^2 - \sum_{j=0}^{n-1} \theta_{n+k, n+k-j}^2 \nu_j^2 \nonumber \end{align}\] 其中的系数\(\theta_{n+k,j}\)和预测的均方误差 \(\nu_{n+k-j}\)可用递推公式(24.6)计算, 只不过因为\(Y_{n+1}, \dots, Y_{n+k}\)未知所以\(W_{n+1}, \dots, W_{n+k}\)不能计算。

24.2 正态时间序列的区间预测

如果\(\{Y_t\}\)是正态时间序列, 则\(\hat Y_{n+1}\)也是最佳预测. \(W_{n+1}=Y_{n+1}-\hat Y_{n+1}\) 作为\(Y_1,Y_2,\cdots,Y_{n+1}\)的线性组合服从正态分布\(N(0,\nu_n)\). 利用 \[ P \left( |Y_{n+1}-\hat Y_{n+1}|/\sqrt{\nu_n} \leq 1.96 \right) = 0.95 \] 可以得到\(Y_{n+1}\)的置信度为0.95的置信区间(预测区间) \[ [\hat Y_{n+1} -1.96\sqrt{\nu_n}, \ \hat Y_{n+1} + 1.96\sqrt{\nu_n}] \]

24.3 平稳序列的递推预测

\(\gamma_k=E(X_{t+k}X_t)\) 是零均值平稳序列\(\{X_t\}\)的自协方差函数, \(\Gamma_n\)\(\{X_t\}\)\(n\)阶自协方差矩阵. 设\(\boldsymbol{X}_n =(X_1,X_2,...,X_n)^T\), \(Z_n=X_n - L(X_n|\boldsymbol{X}_{n-1})\), 可以把定理24.1改述如下.

推论24.1 \(\{X_t\}\)是零均值平稳序列, 对任何\(n \in \mathbb N\), 自协方差矩阵\(\Gamma_n\)正定. 则最佳线性预测 \[\begin{align} \hat X_{n+1} \stackrel{\triangle}{=}& L(X_{n+1}|\boldsymbol{X}_n) \nonumber \\ =& \sum_{j=1}^n \theta_{n,j} Z_{n+1-j}, \tag{24.12} \\ =& \sum_{j=0}^{n-1} \theta_{n,n-j} Z_{j+1} \ n=1,2,\dots \nonumber \end{align}\] 其中的系数\(\{\theta_{n,j}\}\) 和预测的均方误差\(\nu_n=EZ_{n+1}^2\)满足如下的递推公式: \[\begin{align} \begin{cases} & \nu_0=\gamma_0 \\ &\theta_{n,n-k} = [\gamma_{n-k} - \sum_{j=0}^{k-1} \theta_{k,k-j}\theta_{n,n-j}\nu_j]/\nu_k, \\ & \qquad\qquad\qquad \ 0\leq k\leq n-1,\\ & \nu_n = \gamma_0 - \sum_{j=0}^{n-1} \theta_{n,n-j}^2\nu_j, \end{cases} \tag{24.13} \end{align}\] 其中\(\sum_{j=0}^{-1}(\cdot)\stackrel{\triangle}{=} 0\), 递推的顺序是 \[ \begin{array}{lllll} \nu_0 \\ \theta_{1,1} & \nu_1 \\ \theta_{2,2} & \theta_{2,1} & \nu_2 \\ \theta_{3,3} & \theta_{3,2} & \theta_{3,1} & \nu_3 \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{array} \]

由于预测误差\(Z_n=X_n-L(X_{n}|\boldsymbol{X}_{n-1})\)\(\boldsymbol{X}_{n-1}\)正交, 所以是不被\(\boldsymbol X_{n-1}\)的线性组合包含的信息. 基于这个原因, 人们又称\(Z_n\)样本新息. 从§23.2的讨论知道, \[ \nu_n=E[X_1-L(X_1|X_0,X_{-1},\cdots,X_{-n+1})]^2 \to \sigma^2, \ (n\to \infty) \] 这里 \(\sigma^2\)是用全体历史\(X_t,X_{t-1},\dots\)预测\(X_{t+1}\)时的均方误差. \(\sigma^2>0\)表示\(\{X_t\}\)是非决定性的.