[隨機分析] Ito-formula 與其應用 (1) - Two variables case

回憶我們在上一篇
[隨機分析] Ito-formula 與其應用 (0) -Simplest Case
所提出的問題先前問題，考慮
$$M_t := \exp(\alpha B_t - \alpha^2 t/2)$$ 我們想要計算此Ito Integral
$$\int_0^t M_s dB_s =?$$ 注意到上式隨機積分中的積分變數 $M_t$ 不只是 $B_t$ 的函數，亦為 $t$的函數(亦即 $M_t = f(t, B_t)$ 為雙變數函數) 故原本的 simplest form of Ito formula  沒辦法直接應用，我們需要進一步修正Ito formula來讓我們可以對付 這種情況。

在修正Ito Formula 之前我們先定義下列函數

Definition: ($f \in \mathcal{C}^{m,n}(\mathbb{R^+} \times \mathbb{R})$)
考慮函數 $(t,x)  \mapsto f(t,x) \in \mathbb{R}$，且其對 $t$ 存在 $m$ 階導數且連續，對 $x$ 存在 $n$ 階導數且連續，則我們稱此函數 $f \in \mathcal{C}^{m,n}(\mathbb{R^+} \times \mathbb{R})$

有了上面的定義，我們可以著手拓展Ito formula到雙變數函數 如下

=================
Theorem (Ito's Formula with Space and Time Variables)
對任意函數 $f \in \mathcal{C}^{1,2}(\mathbb{R}^+ \times \mathbb{R})$，則對應的 Ito's formula 為
$$\small{ f(t,B_t) = f(0,B_0) + \int_0^t \frac{{\partial f}}{{\partial t}}\left( {s,{B_s}} \right)ds +  \int_0^t \frac{{\partial f}}{{\partial x}}\left( {s,{B_s}} \right) dB_s +  \frac{1}{2}\int_0^t \frac{{\partial^2 f}}{{\partial x^2}}\left( {s,{B_s}} \right)ds }$$ =================
Proof: omitted.

有了上面的定理，我們現在可以再回頭瞧瞧原本無法解決的例子：

Example
考慮 $M_t := \exp(\alpha B_t - \alpha^2 t/2)$，試求 $\int_0^t M_s dB_s=?$

Solution
首先定義函數
$$f\left( {t,x} \right): = \exp (\alpha x - {\alpha ^2}t/2)$$ 且
$$\begin{array}{l} \frac{\partial }{{\partial t}}f\left( {t,x} \right) = {e^{\alpha x}}{e^{ - {\alpha ^2}t/2}}\left( {\frac{{ - {\alpha ^2}}}{2}} \right)\\ \frac{\partial }{{\partial x}}f\left( {t,x} \right) = {e^{ - {\alpha ^2}t/2}}{e^{\alpha x}}\alpha \\ \frac{{{\partial ^2}}}{{\partial {x^2}}}f\left( {t,x} \right) = {\alpha ^2}{e^{ - {\alpha ^2}t/2}}{e^{\alpha x}} \end{array}$$ 由 "Ito's Formula with Space and Time Variables"，
$$\small{ f(t,B_t) = f(0,B_0) + \int_0^t \frac{{\partial f}}{{\partial t}}\left( {s,{B_s}} \right)ds +  \int_0^t \frac{{\partial f}}{{\partial x}}\left( {s,{B_s}} \right) dB_s +  \frac{1}{2}\int_0^t \frac{{\partial^2 f}}{{\partial x^2}}\left( {s,{B_s}} \right)ds }$$ 可知
$$\begin{array}{*{20}{l}} {\exp (\alpha {B_t} - {\alpha ^2}t/2) = 1 + \left( {\frac{{ - {\alpha ^2}}}{2}} \right)\int_0^t {{e^{\alpha {B_s}}}{e^{ - {\alpha ^2}s/2}}} ds}\\ {\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; + \alpha \int_0^t {{e^{ - {\alpha ^2}s/2}}{e^{\alpha {B_s}}}} d{B_s} + \frac{{{\alpha ^2}}}{2}\int_0^t {{e^{ - {\alpha ^2}s/2}}{e^{\alpha {B_s}}}} ds}\\ { \Rightarrow \frac{1}{\alpha }\left( {\exp (\alpha {B_t} - {\alpha ^2}t/2) - 1} \right) = \int_0^t {{e^{\alpha {B_s}}}{e^{ - {\alpha ^2}s/2}}} d{B_s}}\\ { \Rightarrow {M_t} = 1 + \alpha \int_0^t {{M_t}} d{B_s}} \end{array}$$
故
$${\int_0^t {{M_t}} d{B_s} = \frac{1}{\alpha }\left( {{M_t} - 1} \right)} \ \ \ \ \square$$

ref:
J. M. Steele, Stochastic Calculus and Financial Applications, Springer