Time Series Analysis (Best MSE Predictor & Best Linear Predictor)

Time Series Analysis

Best MSE (Mean Square Error) Predictor

对于所有可能的预测函数 $f(X_{n})$，找到一个使 $\mathbb{E}\big[\big(X_{n} - f(X_{n})\big)^{2} \big]$ 最小的 $f$ 的 predictor。这样的 predictor 假设记为 $m(X_{n})$， 称作 best MSE predictor，i.e.，

$$m(X_{n}) = \mathop{\arg\min}\limits_{f} \mathbb{E}\big[ \big( X_{n+h} - f(X_{n}) \big)^{2} \big]$$

我们知道：$\mathop{\arg\min}\limits_{f} \mathbb{E}\big[ \big( X_{n+h} - f(X_{n}) \big)^{2} \big]$ 的解即为：

$$\mathbb{E}\big[ X_{n+h} ~ \big| ~ X_{n} \big]$$

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证明：

基于 $X_{n}$ 求 $\mathbb{E}\big[ \big( X_{n+h} - f(X_{n}) \big)^{2} \big]$ 的最小值，实际上：

$$\mathop{\arg\min}\limits_{f} \mathbb{E}\big[ \big( X_{n+h} - f(X_{n}) \big)^{2} \big] \iff \mathop{\arg\min}\limits_{f} \mathbb{E}\big[ \big( X_{n+h} - f(X_{n}) \big)^{2} ~ \big| ~ X_{n} \big]$$

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• 私以为更严谨的写法是 $\mathop{\text{argmin}}\limits_{f} ~ \mathbb{E}\Big[\Big(X_{n+h} - f\big( X_{n}\big)\Big)^{2} ~ | ~ \mathcal{F}_{n}\Big]$，其中 $\left\{ \mathcal{F}_{t}\right\}_{t\geq 0}$ 为 $\left\{ X_{t} \right\}_{t\geq 0}$ 相关的 natural filtration，but whatever。

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等式右侧之部分：

$$\begin{align*} \mathbb{E}\big[ \big( X_{n+h} - f(X_{n}) \big)^{2} ~ \big| ~ X_{n} \big] & = \mathbb{E}[X_{n+h}^{2} ~ | ~ X_{n}] - 2f(X_{n})\mathbb{E}[X_{n+h} ~ | ~ X_{n}] + f^{2}(X_{n}) \\ \end{align*}$$

其中由于：

$$\begin{align*} Var(X_{n+h} ~ | ~ X_{n}) & = \mathbb{E}\Big[ \big( X_{n+h} - \mathbb{E}\big[ X_{n+h}^{2} ~ | ~ X_{n} \big] \big)^{2} ~ \Big| ~ X_{n} \Big] \\ & = \mathbb{E}\big[ X_{n+h}^{2} ~ \big| ~ X_{n} \big] - 2\mathbb{E}^{2}\big[ X_{n+h}^{2} ~ \big| ~ X_{n} \big] + \mathbb{E}^{2}\big[ X_{n+h}^{2} ~ \big| ~ X_{n} \big] \\ & = \mathbb{E}\big[ X_{n+h}^{2} ~ \big| ~ X_{n} \big] - \mathbb{E}^{2}\big[ X_{n+h}^{2} ~ \big| ~ X_{n} \big] \end{align*}$$

which gives that:

$$\implies Var(X_{n+h} ~ | ~ X_{n}) = \mathbb{E}\big[ X_{n+h}^{2} ~ \big| ~ X_{n} \big] - \mathbb{E}^{2}\big[ X_{n+h} ~ \big| ~ X_{n} \big]$$

因此，

$$\begin{align*} \mathbb{E}\big[ \big( X_{n+h} - f(X_{n}) \big)^{2} ~ \big| ~ X_{n} \big] & = Var(X_{n+h} ~ | ~ X_{n}) + \mathbb{E}^{2}\big[ X_{n+h} ~ \big| ~ X_{n}\big] - 2f(X_{n})\mathbb{E}[X_{n+h} ~ | ~ X_{n}] + f^{2}(X_{n}) \\ & = Var(X_{n+h} ~ | ~ X_{n}) + \Big( \mathbb{E}\big[ X_{n+h} ~ \big| ~ X_{n}\big] - f(X_{n}) \Big)^{2} \end{align*}$$

方差 $Var(X_{n+h} ~ | ~ X_{n})$ 为定值，那么 optimal solution $m(X_{n})$ 显而易见：

$$m(X_{n}) = \mathbb{E}\big[ X_{n+h} ~ \big| ~ X_{n} \big]$$

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此时 $\left\{ X_{t} \right\}$ 为一个 Stationary Gaussian Time Series, i.e.,

$$\begin{pmatrix} X_{n+h}\\ X_{n} \end{pmatrix} \sim N \begin{pmatrix} \begin{pmatrix} \mu \\ \mu \end{pmatrix}, ~ \begin{pmatrix} \gamma(0) & \gamma(h) \\ \gamma(h) & \gamma(0) \end{pmatrix} \end{pmatrix}$$

那么我们有：

$$X_{n+h} ~ | ~ X_{n} \sim N\Big( \mu + \rho(h)\big(X_{n} - \mu\big), ~ \gamma(0)\big(1 - \rho^{2}(h)\big) \Big)$$

其中 $\rho(h)$ 为 $\left\{ X_{t} \right\}$ 的 ACF，因此，

$$\mathbb{E}\big[ X_{n+h} ~ \big| ~ X_{n} \big] = m(X_{n}) = \mu + \rho(h) \big( X_{n} - \mu \big)$$

注意：

若 $\left\{ X_{t} \right\}$ 是一个 Gaussian time series，则一定能计算 best MSE predictor。而若 $\left\{ X_{t} \right\}$ 并非 Gaussian time series，则计算通常十分复杂。

因此，我们通常不找 best MSE predictor，而寻找 best linear predictor。

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Best Linear Predictor (BLP)

在 BLP 假设下，我们寻找一个形如 $f(X_{n}) \propto aX_{n} + b$ 的 predictor。

则目标为：

$$\text{minimize: } ~ S(a,b) = \mathbb{E} \big[ \big( X_{n+h} - aX_{n} -b \big)^{2} \big]$$

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推导：

分别对 $a, b$ 求偏微分：

$$\begin{align*} \frac{\partial}{\partial b} S(a, b) & = \frac{\partial}{\partial b} \mathbb{E} \big[ \big( X_{n+h} - aX_{n} -b \big)^{2} \big] \\ & = -2 \mathbb{E} \big[ X_{n+h} - aX_{n} - b \big] \\ \end{align*}$$

令：

$$\frac{\partial}{\partial b} S(a, b) = 0$$

则：

$$\begin{align*} -2 \cdot & \mathbb{E} \big[ X_{n+h} - aX_{n} - b \big] = 0 \\ \implies & \qquad \mathbb{E}[X_{n+h}] - a\mathbb{E}[X_{n}] - b = 0\\ \implies & \qquad \mu - a\mu - b = 0 \\ \implies & \qquad b^{\star} = (1 - a^{\star}) \mu \end{align*}$$

回代并 take partial derivative on $a$：

$$\begin{align*} \frac{\partial}{\partial a} S(a, b) & = \frac{\partial}{\partial a} \mathbb{E} \big[ \big( X_{n+h} - aX_{n} - (1 - a)\mu \big)^{2} \big] \\ & = \frac{\partial}{\partial a} \mathbb{E} \Big[ \Big( \big(X_{n+h} - \mu \big) - \big( X_{n} - \mu \big) a \Big)^{2} \Big] \\ & = \mathbb{E} \Big[ - \big( X_{n} - \mu \big) \Big( \big(X_{n+h} - \mu \big) - \big( X_{n} - \mu \big) a \Big)\Big] \\ \end{align*}$$

令：

$$\frac{\partial}{\partial a} S(a, b) = 0$$

则：

$$\begin{align*} & \mathbb{E} \Big[ - \big( X_{n} - \mu \big) \Big( \big(X_{n+h} - \mu \big) - \big( X_{n} - \mu \big) a \Big)\Big] = 0 \\ \implies & \qquad \mathbb{E} \Big[\big( X_{n} - \mu \big) \Big( \big(X_{n+h} - \mu \big) - \big( X_{n} - \mu \big) a \Big)\Big] = 0 \\ \implies & \qquad \mathbb{E} \Big[\big( X_{n} - \mu \big) \big(X_{n+h} - \mu \big) - a \big( X_{n} - \mu \big) \big( X_{n} - \mu \big) \Big] = 0 \\ \implies & \qquad \mathbb{E} \Big[\big( X_{n} - \mu \big) \big(X_{n+h} - \mu \big) \Big] = a \cdot \mathbb{E} \Big[\big( X_{n} - \mu \big) \big( X_{n} - \mu \big) \Big] \\ \implies & \qquad \mathbb{E} \Big[\big( X_{n} - \mathbb{E}[X_{n}] \big) \big(X_{n+h} - \mathbb{E}[X_{n+h}] \big) \Big] = a \cdot \mathbb{E} \Big[\big( X_{n} - \mathbb{E}[X_{n}] \big)^{2} \Big] \\ \implies & \qquad \text{Cov}(X_{n}, X_{n+h}) = a \cdot \text{Var}(X_{n}) \\ \implies & \qquad a^{\star} = \frac{\gamma(h)}{\gamma(0)} = \rho(h) \end{align*}$$

综上，time series $\left\{ X_{n} \right\}$ 的 BLP 为：

$$f(X_{n}) = l(X_{n}) = \mu + \rho(h) \big( X_{n} - \mu \big)$$

且 BLP 相关的 MSE 为：

$$\begin{align*} \text{MSE} & = \mathbb{E}\big[ \big( X_{n+h} - l(X_{n}) \big)^{2} \big] \\ & = \mathbb{E} \Big[ \Big( X_{n+h} - \mu - \rho(h) \big( X_{n} - \mu \big) \Big)^{2} \Big] \\ & = \rho(0) \cdot \big( 1 - \rho^{2}(h) \big) \end{align*}$$