現在回頭再看看 Ito Formula 給我們的 Martingale 判別定理: ============================== Theorem (Martingale PDE condition) 考慮 $t \in [0,T]$,若 $f \in \mathcal{C}^{1,2}(\mathbb{R}^+ \times \mathbb{R})$,且 \[ \frac{{\partial f}}{{\partial t}} + \frac{1}{2}\frac{{{\partial ^2}f}}{{\partial {x^2}}} =0 \]則 $X_t = f(t, B_t)$ 為一個 Local Martingale。 再者,若 ${\frac{{\partial f}}{{\partial x}}\left( {s,{B_s}} \right)}\in \mathcal{H}^2$,亦即 \[ E \left[\int_0^t \left ( {\frac{{\partial f}}{{\partial x}}\left( {s,{B_s}} \right)} \right)^2 ds \right] < \infty \]則 $X_t$ 為一個 Martingale ============================== 現在再看個例子看看 Ito Formula 怎麼幫助我們獲得 Martingale Example 1 令 $B_1(t), B_2(t), B_3 (t),...$ 互為獨立 Standard Brownian Motion。對 $k \in \mathbb{N}$ 定義函數 $g_k$ 與 \[ A_k(t) = \int_0^t g_k(B_1(s), B_2(s), ..., B_k(s))ds \] 現在試求 $A_2$ 使得 \[ B_1(t)^2 B_2(t)^2 - A_2(t) \]為 Martingale。 Hint: 利用 上述 PDE Martingale Condition。 Proof 我們要找 \[ {A_2}(t) = \int_0^t {{g_2}} ({B_1}(s),{B_2}(s))ds \] 使得 $B_1(
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