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：
\[\mathbb{E}\big[ X_{n+h} ~ \big| ~ X_{n} \big]
\]
\]
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：
\[\mathop{\arg\min}\limits_{f} \mathbb{E}\big[ \big( X_{n+h} - f(X_{n}) \big)^{2} \big] \iff \mathop{\arg\min}\limits_{f} \mathbb{E}\big[ \big( X_{n+h} - f(X_{n}) \big)^{2} ~ \big| ~ X_{n} \big]
\]
\]
- 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。
等式右侧之部分：
\[\begin{align*}
\mathbb{E}\big[ \big( X_{n+h} - f(X_{n}) \big)^{2} ~ \big| ~ X_{n} \big] & = \mathbb{E}[X_{n+h}^{2} ~ | ~ X_{n}] - 2f(X_{n})\mathbb{E}[X_{n+h} ~ | ~ X_{n}] + f^{2}(X_{n}) \\
\end{align*}
\]
\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：
\[\begin{align*}
Var(X_{n+h} ~ | ~ X_{n}) & = \mathbb{E}\Big[ \big( X_{n+h} - \mathbb{E}\big[ X_{n+h}^{2} ~ | ~ X_{n} \big] \big)^{2} ~ \Big| ~ X_{n} \Big] \\
& = \mathbb{E}\big[ X_{n+h}^{2} ~ \big| ~ X_{n} \big] - 2\mathbb{E}^{2}\big[ X_{n+h}^{2} ~ \big| ~ X_{n} \big] + \mathbb{E}^{2}\big[ X_{n+h}^{2} ~ \big| ~ X_{n} \big] \\
& = \mathbb{E}\big[ X_{n+h}^{2} ~ \big| ~ X_{n} \big] - \mathbb{E}^{2}\big[ X_{n+h}^{2} ~ \big| ~ X_{n} \big]
\end{align*}
\]
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，
\[\begin{align*}
\mathbb{E}\big[ \big( X_{n+h} - f(X_{n}) \big)^{2} ~ \big| ~ X_{n} \big] & = Var(X_{n+h} ~ | ~ X_{n}) + \mathbb{E}^{2}\big[ X_{n+h} ~ \big| ~ X_{n}\big] - 2f(X_{n})\mathbb{E}[X_{n+h} ~ | ~ X_{n}] + f^{2}(X_{n}) \\
& = Var(X_{n+h} ~ | ~ X_{n}) + \Big( \mathbb{E}\big[ X_{n+h} ~ \big| ~ X_{n}\big] - f(X_{n}) \Big)^{2}
\end{align*}
\]
\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：
\[m(X_{n}) = \mathbb{E}\big[ X_{n+h} ~ \big| ~ X_{n} \big]
\]
\]
此时 \(\left\{ X_{t} \right\}\) 为一个 Stationary Gaussian Time Series, i.e.,
\[\begin{pmatrix}
X_{n+h}\\
X_{n}
\end{pmatrix} \sim N \begin{pmatrix}
\begin{pmatrix}
\mu \\
\mu
\end{pmatrix}, ~ \begin{pmatrix}
\gamma(0) & \gamma(h) \\
\gamma(h) & \gamma(0)
\end{pmatrix}
\end{pmatrix}
\]
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。
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：
\[\text{minimize: } ~ S(a,b) = \mathbb{E} \big[ \big( X_{n+h} - aX_{n} -b \big)^{2} \big]
\]
\]
推导：
分别对 \(a, b\) 求偏微分：
\[\begin{align*}
\frac{\partial}{\partial b} S(a, b) & = \frac{\partial}{\partial b} \mathbb{E} \big[ \big( X_{n+h} - aX_{n} -b \big)^{2} \big] \\
& = -2 \mathbb{E} \big[ X_{n+h} - aX_{n} - b \big] \\
\end{align*}
\]
\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：
\[\begin{align*}
-2 \cdot & \mathbb{E} \big[ X_{n+h} - aX_{n} - b \big] = 0 \\
\implies & \qquad \mathbb{E}[X_{n+h}] - a\mathbb{E}[X_{n}] - b = 0\\
\implies & \qquad \mu - a\mu - b = 0 \\
\implies & \qquad b^{\star} = (1 - a^{\star}) \mu
\end{align*}
\]
-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 \(a\)：
\[\begin{align*}
\frac{\partial}{\partial a} S(a, b) & = \frac{\partial}{\partial a} \mathbb{E} \big[ \big( X_{n+h} - aX_{n} - (1 - a)\mu \big)^{2} \big] \\
& = \frac{\partial}{\partial a} \mathbb{E} \Big[ \Big( \big(X_{n+h} - \mu \big) - \big( X_{n} - \mu \big) a \Big)^{2} \Big] \\
& = \mathbb{E} \Big[ - \big( X_{n} - \mu \big) \Big( \big(X_{n+h} - \mu \big) - \big( X_{n} - \mu \big) a \Big)\Big] \\
\end{align*}
\]
\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：
\[\begin{align*}
& \mathbb{E} \Big[ - \big( X_{n} - \mu \big) \Big( \big(X_{n+h} - \mu \big) - \big( X_{n} - \mu \big) a \Big)\Big] = 0 \\
\implies & \qquad \mathbb{E} \Big[\big( X_{n} - \mu \big) \Big( \big(X_{n+h} - \mu \big) - \big( X_{n} - \mu \big) a \Big)\Big] = 0 \\
\implies & \qquad \mathbb{E} \Big[\big( X_{n} - \mu \big) \big(X_{n+h} - \mu \big) - a \big( X_{n} - \mu \big) \big( X_{n} - \mu \big) \Big] = 0 \\
\implies & \qquad \mathbb{E} \Big[\big( X_{n} - \mu \big) \big(X_{n+h} - \mu \big) \Big] = a \cdot \mathbb{E} \Big[\big( X_{n} - \mu \big) \big( X_{n} - \mu \big) \Big] \\
\implies & \qquad \mathbb{E} \Big[\big( X_{n} - \mathbb{E}[X_{n}] \big) \big(X_{n+h} - \mathbb{E}[X_{n+h}] \big) \Big] = a \cdot \mathbb{E} \Big[\big( X_{n} - \mathbb{E}[X_{n}] \big)^{2} \Big] \\
\implies & \qquad \text{Cov}(X_{n}, X_{n+h}) = a \cdot \text{Var}(X_{n}) \\
\implies & \qquad a^{\star} = \frac{\gamma(h)}{\gamma(0)} = \rho(h)
\end{align*}
\]
& \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：
\[\begin{align*}
\text{MSE} & = \mathbb{E}\big[ \big( X_{n+h} - l(X_{n}) \big)^{2} \big] \\
& = \mathbb{E} \Big[ \Big( X_{n+h} - \mu - \rho(h) \big( X_{n} - \mu \big) \Big)^{2} \Big] \\
& = \rho(0) \cdot \big( 1 - \rho^{2}(h) \big)
\end{align*}
\]
\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*}
\]
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