随机过程 | 鞅论 1.6 独立增量过程

2026-08-13 10:44:14 星期四
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1.6 独立增量过程

Lemma 1.6.3 设 \(\{X_t\}_{t \in \mathbb{R}^+}\) 为 Lévy 过程，定义 \(\varphi_{s,t}(u) = \mathbb{E} e^{iu(X_t - X_s)}\)，则 \(\forall u \in \mathbb{R}, s < t\)，\(\varphi_{s,t}(u) \neq 0\).

证明. 反证法，设存在 \(t_0\)，\(t_0 > s\)，使得 \(\varphi_{s,t_0}(u) = 0\)，令

\[t_1 = \inf \{ t > s \mid \varphi_{s,t}(u) = 0 \}. \]

则

\[0 = \varphi_{s,t_1}(u) = \mathbb{E} e^{iu(X_{t_1} - X_s)} = \mathbb{E} e^{iu(X_{t_1} - X_t)} \mathbb{E} e^{iu(X_t - X_s)} = \varphi_{t,t_1}(u) \underbrace{\varphi_{s,t}(u)}_{\neq 0}, \]

其中 \(s < t < t_1\). \(\therefore \varphi_{t,t_1}(u) = 0\).

而

\[0 = \lim_{t \to t_1} \varphi_{t,t_1}(u) = \varphi_{t_1,t_1}(u) = 1, \]

矛盾.

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Proposition 1.6.4 设 \(\{X_t\}_{t \in \mathbb{R}^+}\) 为 Lévy 过程，令

\[Z_{s,t}(u) = \dfrac{e^{iu(X_t - X_s)}}{\varphi_{s,t}(u)}, \quad s < t, \]

则 \(\{Z_{s,t}(u), \mathcal{F}_t\}_{t > s}\) 为鞅.

证明. \(\forall r > t > s\)，

\[\begin{aligned} \mathbb{E}\left( Z_{s,r}(u) \mid \mathcal{F}_t \right) &= \mathbb{E}\left( \dfrac{1}{\varphi_{s,r}(u)} e^{iu(X_r - X_s)} \mid \mathcal{F}_t \right) \\ &= \dfrac{1}{\varphi_{s,t}(u) \varphi_{t,r}(u)} \mathbb{E}\left( e^{iu(X_r - X_t)} e^{iu(X_t - X_s)} \mid \mathcal{F}_t \right) \\ &= \dfrac{\varphi_{t,r}(u)}{\varphi_{s,t}(u) \varphi_{t,r}(u)} \mathbb{E}\left( e^{iu(X_t - X_s)} \mid \mathcal{F}_t \right) \\ &= Z_{s,t}(u), \quad \text{a.s.} \end{aligned} \]

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Proposition 1.6.5 设 \(\{X_t, \mathcal{F}_t\}_{t \geq 0}\) 为 Lévy 过程，\(\tau\) 为有限停时，\(Y_t \triangleq X_{t+\tau} - X_\tau\)，\(t > 0\).

1. \(\{Y_t\}_{t \geq 0}\) 与 \(\mathcal{F}_\tau\) 独立；
2. \(\{Y_t\}_{t \geq 0}\) 关于 \(\{\mathcal{F}_{t+\tau}\}_{t \in \mathbb{R}^+}\) 是时齐独立增量过程.

证明.

1. 设 \(\varphi_t(u) = \varphi_{0,t}(u)\)，\(Z_t(u) = Z_{0,t}(u)\). 由停时定理，

   \[\mathbb{E}\left( Z_{\tau \wedge n + t} \mid \mathcal{F}_{\tau \wedge n} \right) = Z_{\tau \wedge n}. \tag{1} \]

   根据定义有：

   \[\begin{aligned} Z_{\tau \wedge n + t} &= Z_{\tau \wedge n} \dfrac{e^{iu(X_{\tau \wedge n + t} - X_{\tau \wedge n})}}{\varphi_{\tau \wedge n, \tau \wedge n + t}(u)} \\ &= Z_{\tau \wedge n} \dfrac{e^{iu(X_{\tau \wedge n + t} - X_{\tau \wedge n})}}{\varphi_t(u)} \quad \text{（因为原随机变量族是时齐的）.} \end{aligned} \]

   代回 (1) 的左边，有

   \[\frac{Z_{\tau \wedge n}}{\varphi_t(u)} \mathbb{E}\left( e^{iu(X_{\tau \wedge n + t} - X_{\tau \wedge n})} \mid \mathcal{F}_{\tau \wedge n} \right) = Z_{\tau \wedge n}, \]

   即

   \[\mathbb{E}\left( e^{iu(X_{\tau \wedge n + t} - X_{\tau \wedge n})} \mid \mathcal{F}_{\tau \wedge n} \right) = \varphi_t(u). \tag{2} \]

   为了证明 \(Y_t\) 与 \(\mathcal{F}_\tau\) 独立，我们只需证明：对任意的 \(A \in \mathcal{F}_\tau\)，成立

   \[\mathbb{E}(e^{iuY_t} I_A) = \varphi_t(u) \mathbb{P}(A). \]

   为此，我们取一列递增的集合 \(A_n = A \cap \{ \tau \le n \} \uparrow A\)，可以证明 \(A_n \in \mathcal{F}_{\tau \wedge n}\)，实际上任取 \(t \ge 0\)，我们看 \(A_n \cap \{ \tau \le t \}\) 是否属于 \(\mathcal{F}_t\)：

   \[A_n \cap \{ \tau \le t \} = A \cap \{ \tau \le n \} \cap \{ \tau \le t \} = A \cap \{ \tau \le t \} \in \mathcal{F}_t. \]

   那么借助 (2)，有

   \[\begin{aligned} \varphi_t(u) \mathbb{P}(A_n) &= \mathbb{E}\left( e^{iu(X_{\tau \wedge n + t} - X_{\tau \wedge n})} I_{A_n} \right) \\ &= \mathbb{E}\left( e^{iu(X_{\tau + t} - X_\tau)} I_{A_n} \right) \quad \text{（因为在 $A_n$ 上，$\tau \le n$）} \\ &\xrightarrow{\text{DCT}} \mathbb{E}\left( e^{iu(X_{\tau + t} - X_\tau)} I_A \right) \\ &= \mathbb{E}(e^{iuY_t} I_A). \end{aligned} \]

   左边令 \(n \to \infty\)，由 DCT，得到左边极限是 \(\varphi_t(u) \mathbb{P}(A)\). 于是 \(Y_t\) 与 \(\mathcal{F}_\tau\) 独立.

2. \[Y_{t+s} - Y_s = X_{t+s+\tau} - X_{s+\tau}. \]

   由 (2)，我们可以把有界停时 \(\tau \wedge n\) 换成 \(\tau + s\)：

   \[\mathbb{E}\left( e^{iu(X_{(\tau + s) + t} - X_{\tau + s})} \mathrel{\Big\vert{}} \mathcal{F}_{\tau + s} \right) = \varphi_t(u). \]

   可以发现右边与 \(\mathcal{F}_{\tau + s}\) 无关，而且只与 \(t\) 有关系. 所以是独立增量过程，而且是时齐的！